Broadening Target Distributions for Accelerated Diffusion Models via a Novel Analysis Approach

## Abstract

Abstract The denoising diffusion model emerges recently as a powerful generative technique that converts noise into data. Theoretical convergence guarantee has been mainly studied for continuous-time diffusion models, and has been obtained for discrete-time diffusion models only for distributions with bounded support in the literature. In this paper, we establish the convergence guarantee for substantially larger classes of distributions under discrete-time diffusion models and further improve the convergence rate for distributions with bounded support. In particular, we first establish the convergence rates for both smooth and general (possibly non-smooth) distributions having finite second moment. We then specialize our results to a number of interesting classes of distributions with explicit parameter dependencies, including distributions with Lipschitz scores, Gaussian mixture distributions, and distributions with bounded support. We further propose a novel accelerated sampler and show that it improves the convergence rates of the corresponding regular sampler by orders of magnitude with respect to all system parameters. For distributions with bounded support, our result improves the dimensional dependence of the previous convergence rate by orders of magnitude. Our study features a novel analysis technique that constructs tilting factor representation of the convergence error and exploits Tweedie’s formula for handling Taylor expansion power terms.

## 1 Introduction

Generative modeling is a fundamental task in machine learning, aiming to generate samples out of a distribution similar to that of training data. Classical generative models include variational autoencoders (VAE) [], generative adversarial networks (GANs) [], and normalizing flows [], etc. Recently, diffusion models [, ] have arisen as an extremely appealing generative model and have received wide popularity due to their state-of-the-art performance for a variety of tasks, including text/image/video generation and other generative tasks. See [, , ] for surveys of diffusion models in different applications.

The empirical success of diffusion models has also motivated theoretical studies, aiming to characterize the convergence of diffusion model processes. Most theoretical interests have focused on continuous-time diffusion models, i.e., stochastic differential equation (SDE) based formulations [], given their close relationship to Langevin Monte-Carlo samplers and many SDE tools available for use (see [Section 1.2](#S1.SS2)). In particular, given time-averaged $L^{2}$ score estimation error [], polynomial-time convergence guarantees
have been established for wide families of target distributions [, , , , , , ].

However, from the application side, many generative samplers, including those state-of-the-art ones, employ discrete-time denoising diffusion probabilistic models (DDPMs) []. See [, Table 1] and [, Figure 5] for a list of algorithms used in the literature. In contrast to continuous-time diffusion models which are extensively studied, discrete-time diffusion models are largely unexplored theoretically. Existing techniques for continuous-time models are not directly transferable to their discrete-time counterparts.
The only work that analyzes the discrete-time diffusion models is [], which provided a non-asymptotic convergence guarantee for denoising diffusion probabilistic
models (DDPMs), but only for distribution with bounded support set.

Despite the practical importance of discrete-time diffusion models, several fundamental questions remain open: (1) (Unbounded support) The convergence guarantee has not been established for discrete-time diffusion models for challenging distributions with unbounded support, such as the common mixture Gaussian distribution. In particular, the distribution with Lipschitz score function (which can have unbounded support) has been shown to enjoy polynomial time convergence for continuous-time diffusion models, but its convergence has not yet been shown for discrete-time models. It appears that the typical-set based analysis tools in [] does not handle the case with unbounded support. (2) (High dimensional dependence) There has been significant effort on the continuous-time analysis to reduce the dependence of the convergence rate on the variable dimension $d$, but the discrete-time analysis still suffers from the large dependence on $d$ (see [Section 1.2](#S1.SS2)).
This clearly calls for a more refined analysis for discrete-time analysis. (3) (Estimation error) Accelerated sampler has been developed in [], which does not handle estimation error. This seems to be due to their choice of estimator for acceleration, and calls for a new estimation algorithm whose error can be easily analyzed.

In order to comprehensively deal with the aforementioned open challenges, new analysis tools are required, which is the goal of this paper.

### 1.1 Our Contributions

Our main contribution lies in the development of novel analytical techniques to analyze the discrete-time DDPM model and provide non-asymptotic convergence guarantees with a better rate and for a substantially larger class of distributions over the state-of-the-art (see [Table 1](#S1.T1) for a summary of our results).
In particular, our contributions are three-fold.

Regular Sampler: We provide the first non-asymptotic convergence bound on the distance between generated and target distributions for discrete-time DDPM for general distributions having finite second moments. Our bound further specializes to explicit dependence on the system parameters
for various classes of distributions, such as those having Lipschitz scores (with the parameter $L$), Gaussian mixtures (with $N$ components), and those having bounded support, for which the generated distribution converges to the target distribution at the rate of $\mathcal{O}\left(\frac{dL^{2}}{\varepsilon^{2}}\right)$, $\mathcal{O}\left(\frac{d^{2}\min\{d,N\}^{2}}{\varepsilon^{2}}\right)$, and $\mathcal{O}\left(\frac{d^{2}}{\delta^{2}\varepsilon^{2}}\right)$, respectively. In particular, for the case where the distribution has bounded support, we improve the number of convergence steps on the dimension $d$ in [] by a factor of $\mathcal{O}(d^{2})$.

Accelerated Sampler: We propose a new accelerated DDPM sampler by introducing an estimator of the Hessian of log-likelihood at each step, which handles the estimation error and improves the dependence of the convergence rate on both $\epsilon$ and $d$. We show that our estimator achieves a provably faster convergence rate of $\mathcal{O}\left(\frac{dM}{\varepsilon}\right)$, $\mathcal{O}\left(\frac{d^{1.5}\min\{d,N\}^{1.5}}{\varepsilon}\right)$ and $\mathcal{O}\left(\frac{d^{1.5}}{\delta^{1.5}\varepsilon}\right)$ respectively for those having Lipschitz scores and Hessians (with parameter $M$), Gaussian mixtures, and those distributions having bounded support. In particular, the accelerated sampler improves the dependence of the convergence rate on $\varepsilon$ of the regular sampler by a factor of $\mathcal{O}(1/\varepsilon)$ as well as enjoying better dependencies on $d$ and $\delta$. For distributions with bounded support, our accelerated DDPM sampler improves the number of convergence steps in [] by a factor of $\mathcal{O}(d^{1.5})$.

Novel Analysis Technique: We develop a novel technique for analyzing discrete-time diffusion models.
Specifically, our technique features two new elements: (i) characterization of the error incurred at each discrete step of the reverse process using tilting factor; and (ii) analysis of the mean value of tilting factor via Tweedie’s formula to handle power terms in the Taylor expansion.
Such a technique enables us to (a) analyze more general distributions including those with unbounded support; (b) tightly identify the dominant term and obtain the convergence rate with lower dimensional dependence; and (c) handle the estimation error in our proposed accelerated sampler when both score and Hessian estimation errors exist.

**Table 1: Summary of convergence results for discrete-time DDPM samplers in terms of the number of steps needed to achieve $\epsilon$-accuracy, where $d$ is the dimension. (i) The first three rows and the last three rows respectively correspond to the results for regular and accelerated samplers. (ii) For $Q_{0}$ with bounded support, our paper considers $Q_{1}(\delta)$ that satisfies $\mathrm{W}_{2}\left(Q_{0},Q\right)^{2}\lesssim\delta d$, where $\delta$ captures the closeness between $Q_{0}$ and $Q_{1}$, whereas [] does not explicitly capture such a parameter in their result. Nevertheless, our results improve the corresponding results in [] in terms of dependence on $d$.**
| Target distribution $Q_{0}$ | Num of steps | Results |
| --- | --- | --- |
| $\nabla\log q_{t}$ $L$-Lipschitz $\forall t$ | $\mathcal{O}\left(\frac{dL^{2}}{\varepsilon^{2}}\right)$ | [Theorem 2](#Thmtheorem2) (this paper) |
| $Q_{0}$ Gaussian mixture ($N$ components) | $\mathcal{O}\left(\frac{d^{2}\min\{d,N\}^{2}}{\varepsilon^{2}}\right)$ | [Theorem 3](#Thmtheorem3) (this paper) |
| $Q_{0}$ bounded support | $\mathcal{O}\left(\frac{d^{4}}{\varepsilon^{2}}\right)$ | [, Theorem 3] |
|  | $\mathcal{O}\left(\frac{d^{2}}{\delta^{2}\varepsilon^{2}}\right)$ | [Theorem 4](#Thmtheorem4) (this paper) |
| $Q_{0}$ Gaussian mixture (accl.) | $\mathcal{O}\left(\frac{d^{1.5}\min\{d,N\}^{1.5}}{\varepsilon}\right)$ | [Theorem 7](#Thmtheorem7) (this paper) |
| $Q_{0}$ bounded support (accl., no est. err) | $\mathcal{O}\left(\frac{d^{3}}{\varepsilon}\right)$ | [, Theorem 4] |
| $Q_{0}$ bounded support (accl.) | $\mathcal{O}\left(\frac{d^{1.5}}{\delta^{1.5}\varepsilon}\right)$ | [Theorem 8](#Thmtheorem8) (this paper) |

### 1.2 Related Works

Theory on Continuous-time DDPM Samplers: Many works have explored the performance guarantees of generative models from a continuous-time perspective based on a SDE-type analysis.
Specifically, a number of studies perform analyses under the $L^{\infty}$ score estimation error [, ]. Later, under $L^{2}$ score estimation error, [] developed polynomial(^1^11By “polynomial” we mean that the number of steps has polynomial dependence on the score estimation error, along with other parameters.) bounds for distributions that have Lipschitz scores and satisfy log-Sobolev inequality. Soon after, [, ] concurrently developed polynomial bounds for those smooth distributions having Lipschitz scores ($\mathcal{O}(dL^{2}/\varepsilon^{2})$) and those non-smooth distributions having bounded support using early stopping ($\mathcal{O}(d^{3}/\delta^{4}\varepsilon^{2})$). Later, [] improved the number of steps to be $\mathcal{O}(d^{2}\log(1/\delta)^{2}/\varepsilon^{2})$ for those target distributions with finite second moment. This result was recently further improved to linear dimensional dependence []. In [], it is shown that linear dimensional dependence can also be achieved for those target distributions having finite relative Fisher information against a Gaussian distribution. Other than the works above, [] analyzed a different sampling scheme (e.g., predictor-corrector), and [, , ] analyzed sampling errors using a different error measure (the Wasserstein-2 distance).

Theory on Continuous-time Deterministic Samplers:
Beyond stochastic samplers, another line of studies provided theoretical guarantees for the probability flow ordinary differential equation (PF-ODE), which is a deterministic sampler corresponding to the SDE [, ]. [] compared the performance of SDE and PF-ODE and investigated conditions where one might outperform the other. Besides PF-ODE, [, ] provided guarantees for the closely-related flow-matching model, which learns a deterministic coupling between any two distributions. [] proposed a novel ODE for sampling from a conditional distribution. [, ] provided convergence guarantees for the more recent consistency models [].

Theory on Discrete-time DDPM Samplers: Despite flourishing results on the continuous-time DDPM model (i.e., SDE), [] is the only work that addresses the theoretical performance of discrete-time models. In their work, polynomial-time guarantees are shown for both stochastic and deterministic discrete-time samplers under the $L^{2}$ estimation error for target distributions having bounded support. Accelerated stochastic and deterministic variants are also analyzed when there is no score estimation error. Our work substantially broadens the target distributions to include those with unbounded support and improves the results in [] for distributions with bounded support with a completely different analytical technique.

Theory on Score Estimation: In order to achieve an end-to-end analysis, several works developed sample complexity bounds to achieve the $L^{2}$ score estimation error of DDPM for a variety of distributions, including those having bounded support [], Gaussian mixture [], high-dimensional graphical models [] and those supported on a low-dimensional linear subspace []. More recently, [] considered the generalizability of the continuous-time diffusion models.

## 2 Preliminaries of the DDPM Algorithm

In this paper, we are interested in analyzing the theoretical performance of the discrete-time DDPM sampler []. DDPM is trained on a forward (Markov) process, and the samples are generated through a reverse (Markov) process.

### 2.1 DDPM and the Forward Process

Let $x_{0}\in\mathbb{R}^{d}$ be the initial data, and let $x_{t}\in\mathbb{R}^{d},\forall 1\leq t\leq T$ be the latent variables in the DDPM algorithm.
Let $Q_{0}$ be the initial data distribution, and let $Q_{t}$ be the marginal latent distribution at time $t$ in the forward process, for all $1\leq t\leq T$. Denote $Q:=Q_{0,\dots,T}$ as the overall joint distribution. In the DDPM algorithm, white Gaussian noise is gradually added to the data in the forward process: $x_{t}=\sqrt{1-\beta_{t}}x_{t-1}+\sqrt{\beta_{t}}w_{t},\leavevmode\nobreak\ \forall 1\leq t\leq T$,
where $w_{t}\stackrel{{\scriptstyle i.i.d.}}{{\sim}}\mathcal{N}(0,I_{d})$. Equivalently, this can be expressed as a conditional distribution at each time $t$:

$$ $Q_{t|t-1}(x_{t}|x_{t-1})=\mathcal{N}(x_{t};\sqrt{1-\beta_{t}}x_{t-1},\beta_{t}I_{d}),$ (1) $$

which means that under $Q$, $X_{0}\to X_{1}\to\dots\to X_{T}$. Here $\beta_{t}\in(0,1)$ captures the “amount” of noise that is injected at time $t$, and $\beta_{t}$’s are called the noise schedule. Define

$$ $\alpha_{t}:=1-\beta_{t},\quad\bar{\alpha}_{t}:=\prod_{i=1}^{t}\alpha_{i},\quad 1\leq t\leq T.$ $$

An immediate result by accumulating the steps is that

$$ $Q_{t|0}(x_{t}|x_{0})=\mathcal{N}(x_{t};\sqrt{\bar{\alpha}_{t}}x_{0},(1-\bar{\alpha}_{t})I_{d}),$ (2) $$

or, written equivalently, $x_{t}=\sqrt{\bar{\alpha}_{t}}x_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{w}_{t},\leavevmode\nobreak\ \forall 1\leq t\leq T$,
where $\bar{w}_{t}\sim\mathcal{N}(0,I_{d})$ denotes the aggregated noise at time $t$.
Intuitively, for large $T$, since $Q_{T|0}\approx\mathcal{N}(0,I_{d})$ (which is independent of $x_{0}$), it is expected that $Q_{T}\approx\mathcal{N}(0,I_{d})$ when $T$ becomes large, as long as the variance under $Q_{0}$ is finite.
Finally, since the conditional noises are Gaussian, each $Q_{t}(t\geq 1)$ is absolutely continuous w.r.t the Lebesgue measure. Let the corresponding p.d.f. of each $Q_{t}$ be $q_{t}(t\geq 1)$. Similarly define $q_{t,t-1}$, $q_{t|t-1}$, and $q_{t-1|t}$ for $t\geq 1$. In case $Q_{0}$ is also absolutely continuous w.r.t. the Lebesgue measure, let $q_{0}$ be the corresponding p.d.f. of $Q_{0}$.

### 2.2 The Reverse Process

The goal of DDPM is to generate samples approximately from the data distribution $Q_{0}$. We first draw the latent variable at time $T$ from a Gaussian distribution: $x_{T}\sim\mathcal{N}(0,I_{d})=:P_{T}$.
Then, to achieve effective sampling, each forward step is approximated by a reverse sampling step. Define

$$ $\displaystyle\mu_{t}(x_{t}):=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})\nabla\log q_{t}(x_{t})\right),\quad\sigma_{t}^{2}:=\frac{1-\alpha_{t}}{\alpha_{t}}.$ (3) $$

At each time $t=T,T-1,\dots,1$, define the true reverse process as

$$ $x_{t-1}=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})\nabla\log q_{t}(x_{t})\right)+\sqrt{\frac{1-\alpha_{t}}{\alpha_{t}}}z=\mu_{t}(x_{t})+\sigma_{t}z$ $$

where $z\sim\mathcal{N}(0,I_{d})$. Equivalently, this can be written as $P_{t-1|t}=\mathcal{N}(x_{t-1};\mu_{t}(x_{t}),\sigma_{t}^{2}I_{d})$.
Here $\nabla\log q_{t}(x)$ is called the score of $q_{t}$. Let $P_{t}$ be the marginal distributions of $x_{t}$ in the true reverse process, and let $p_{t}$ be the corresponding p.d.f. of $P_{t}$ w.r.t. the Lebesgue measure.

In practice, one does not have access to $\nabla\log q_{t}$. Instead, its estimate, denoted as $s_{t}$, is used, which results in an estimated reverse process:

$$ $x_{t-1}=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})s_{t}(x_{t})\right)+\sqrt{\frac{1-\alpha_{t}}{\alpha_{t}}}z=\widehat{\mu}_{t}(x_{t})+\sigma_{t}z$ $$

in which we define

$$ $\widehat{\mu}_{t}(x_{t}):=x_{t}+(1-\alpha_{t})s_{t}(x_{t}).$ (4) $$

Let $\widehat{P}_{t}$ be the marginal distributions of $x_{t}$ in the estimated reverse process with corresponding p.d.f. $\widehat{p}_{t}$. Note that $\widehat{P}_{t-1|t}=\mathcal{N}(x_{t-1};\widehat{\mu}_{t}(x_{t}),\sigma_{t}^{2}I_{d})$ and $\widehat{P}_{T}=P_{T}$. Define $P$, $\widehat{P}$, $p_{t-1|t}$ and $p_{t|t-1}$ in a way similar to the forward process. Hence, under $P$ and $\widehat{P}$, $X_{T}\to X_{T-1}\to\dots\to X_{0}$.

In case where $Q_{0}$ is absolutely continuous w.r.t. the Lebesgue measure, we are interested in measuring the mismatch between $Q_{0}$ and $\widehat{P}_{0}$ through the total-variation distance, defined as

$$ $\mathrm{TV}(Q_{0},\widehat{P}_{0}):=\sup_{A\subseteq\mathbb{R}_{d}}\left|Q_{0}(A)-\widehat{P}_{0}(A)\right|.$ $$

This metric is commonly used in prior theoretical studies [, ]. In particular, from Pinsker’s inequality, the total-variation (TV) distance is upper bounded as $\mathrm{TV}(Q_{0},\widehat{P}_{0})^{2}$ $\leq\frac{1}{2}\mathrm{KL}(Q_{0}||\widehat{P}_{0})$,
where the KL divergence is defined as

$$ $\mathrm{KL}(Q_{0}||P_{0}):=\int\log\frac{\mathrm{d}Q_{0}}{\mathrm{d}P_{0}}\mathrm{d}Q_{0}=\mathbb{E}_{X\sim Q_{0}}\left[\log\frac{q_{0}(X)}{p_{0}(X)}\right]\geq 0.$ $$

Thus, it is sufficient to control the KL divergence when $q_{0}$ exists.

When $q_{0}$ does not exist (say, when $Q_{0}$ has point masses), one can use the Wasserstein distance to measure the mismatch at $t=0$. The Wasserstein-2 distance is defined as

$$ $\mathrm{W}_{2}(Q_{0},P_{0}):=\left\{\min_{\Gamma\in\Pi(Q_{0},P_{0})}\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\left\lVert x-y\right\rVert^{2}\mathrm{d}\Gamma(x,y)\right\}^{1/2},$ (5) $$

where $\Pi(Q_{0},P_{0})$ is the set of all joint probability measures on $\mathbb{R}^{d}\times\mathbb{R}^{d}$ with marginal distributions $Q_{0}$ and $P_{0}$, respectively.

### 2.3 Other Notations

For any two functions $f(d,\delta,T)$ and $g(d,\delta,T)$, we write $f(d,\delta,T)\lesssim g(d,\delta,T)$ (resp. $f(d,\delta,T)\gtrsim g(d,\delta,T)$) for some universal constant (not depending on $\delta$, $d$ or $T$) $L<\infty$ (resp. $L>0$) if $\limsup_{T\to\infty}|f(d,\delta,T)/$ $g(d,\delta,T)|\leq L$ (resp. $\liminf_{T\to\infty}|f(d,\delta,T)/g(d,\delta,T)|\geq L$). We write $f(d,\delta,T)\asymp g(d,\delta,T)$ when both $f(d,\delta,T)\lesssim g(d,\delta,T)$ and $f(d,\delta,T)\gtrsim g(d,\delta,T)$ hold. Note that the dependence on $\delta$ and $d$ is retained with $\lesssim,\gtrsim,\asymp$. We write $f(d,\delta,T)=O(g(T))$ (resp. $f(d,\delta,T)=\Omega(g(T))$) if $f(d,\delta,T)\lesssim L(d,\delta)g(T)$ (resp. $f(d,\delta,T)\gtrsim L(d,\delta)g(T)$) holds for some $L(d,\delta)$ (possibly depending on $\delta$ and $d$). We write $f(d,\delta,T)=o(g(T))$ if $\limsup_{T\to\infty}|f(d,\delta,T)$ $/g(T)|=0$. We write $f(d,\delta,T)=\tilde{O}(g(T))$ if $f(d,\delta,T)=O(g(T)(\log g(T))^{k})$ for some constant $k$. Note that the big-$O$ notation omits the dependence on $\delta$ and $d$. In the asymptotic when $\varepsilon^{-1}\to\infty$, we write $f(d,\varepsilon^{-1})=\mathcal{O}(g(d,\varepsilon^{-1}))$ if $f(d,\delta,\varepsilon^{-1})\lesssim g(d,\delta,\varepsilon^{-1})(\log g(\varepsilon^{-1}))^{k}$ for some constant $k$. Unless otherwise specified, we write $x^{i}(1\leq i\leq d)$ as the $i$-th element of a vector $x\in\mathbb{R}^{d}$ and $[A]^{ij}$ as the $(i,j)$-th element of a matrix $A$. For a function $f(x):\mathbb{R}^{d}\to\mathbb{R}$, we write $\partial_{i}f(z)$ as a shorthand for $\frac{\partial}{\partial x^{i}}f(x)\Big{|}_{x=z}$,
and similarly for higher moments.
For matrices $A,B$, $\mathrm{Tr}(A)$ is the trace of $A$, and $A\preceq B$ means that $B-A$ is positive semi-definite. For a positive integer $n$, $[n]:=\{1,\dots,n\}$.

## 3 General Q 0 subscript 𝑄 0 Q_{0} : Convergence Bound of Discrete-time DDPMs

In this section, we provide the convergence guarantees for discrete-time DDPM samplers for general $Q_{0}$. We will first establish our main result for smooth $Q_{0}$, and then extend it for more general (possibly nonsmooth) $Q_{0}$. We will also provide a sketch of our proof of the main theorem to describe our key analysis techniques.

### 3.1 Technical Assumptions

The following [Assumptions 1](#Thmassumption1), [2](#Thmassumption2) and [3](#Thmassumption3) are commonly adopted in the literature [, ].

###### Assumption 1 (Finite Second Moment) .

There exists a constant $M_{2}<\infty$ (that does not depend on $d$ and $T$) such that $\mathbb{E}_{X_{0}\sim Q_{0}}[X_{0}X_{0}^{\intercal}]\preceq M_{2}I_{d}$.

###### Assumption 2 (Score Estimation Error) .

The $s_{t}$ estimates satisfy

$$ $\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}\leq\varepsilon^{2},\leavevmode\nobreak\ \text{where}\leavevmode\nobreak\ \varepsilon^{2}=\tilde{O}(T^{-1}).$ $$

###### Assumption 3 (Absolute Continuity) .

$Q_{0}$ is absolutely continuous w.r.t. the Lebesgue measure, and thus $q_{0}$ exists. Also, suppose that $q_{0}$ is analytic.

The following regularity assumption ([Assumption 4](#Thmassumption4)) is needed for our analysis based on Taylor expansion. It is rather soft, and we will show that it is satisfied for Gaussian mixture $Q_{0}$ and for any $Q_{0}$ with bounded support (see [Section 4](#S4)).

###### Assumption 4 (Regular Derivatives) .

For any $k\geq 1$, there exists a function $U_{k}(x)$ such that

$$ $\max_{\bm{a}\in[d]^{k}}\left|\partial^{k}_{\bm{a}}\log q_{t}(x)\right|\leq U_{k}(x),\leavevmode\nobreak\ \forall t\geq 0.$ $$

When $q_{0}$ does not exist, this is required only for $\forall t\geq 1$.
Here $U_{k}(x)$ satisfies:

- $1.$
For any fixed $x$, $U_{k}(x)$ does not depend on $T$. Also, for some constant $C_{k}<\infty$ that does not depend on $T$, $\mathbb{E}_{X_{t}\sim Q_{t}}[U_{k}(X_{t})^{p}]<C_{k}^{p},\leavevmode\nobreak\ \forall p\geq 1$.
- $2.$
In terms of $k$, $U_{k}(x)/k!$ converges absolutely for any finite $x$ and $T$: $\sum_{k=1}^{\infty}\frac{1}{k!}U_{k}(x)<\infty$.

###### Assumption 1 (Finite Second Moment) .

###### Assumption 2 (Score Estimation Error) .

###### Assumption 3 (Absolute Continuity) .

###### Assumption 4 (Regular Derivatives) .

### 3.2 Convergence Bounds

To present the main result, we first define a set of noise schedule as follows.

###### Definition 1 (Noise Schedule) .

For all sufficiently large $T$, the step size $\alpha_{t}$’s satisfy

$$ $1-\alpha_{t}\lesssim\frac{\log T}{T},\leavevmode\nobreak\ \forall 1\leq t\leq T,\quad\text{and}\quad\bar{\alpha}_{T}=\prod_{t=1}^{T}\alpha_{t}=o\left(T^{-1}\right).$ $$

In [Definition 1](#Thmdefinition1), the upper bound on $1-\alpha_{t}$ requires that $\alpha_{t}$ is large enough to control the reverse-step error, while the upper bound on $\bar{\alpha}_{T}$ requires that $\alpha_{t}$ is small enough to control the initialization error.
One example of such $\alpha_{t}$ is $1-\alpha_{t}\equiv\frac{c\log T}{T},\leavevmode\nobreak\ \forall t\geq 1$
for any $c>1$. Then, $\bar{\alpha}_{T}=\left(1-\frac{c\log T}{T}\right)^{T}=\exp\left(T\log\left(1-\frac{c\log T}{T}\right)\right)=O\left(e^{T\frac{-c\log T}{T}}\right)=O\left(T^{-c}\right)$.

We first present the following main theorem of the convergence rate for the discrete-time DDPM.

###### Theorem 1 (Smooth $Q_{0}$ ) .

Under [Assumptions 1](#Thmassumption1), [2](#Thmassumption2), [3](#Thmassumption3) and [4](#Thmassumption4), if the noise schedule satisfies [Definition 1](#Thmdefinition1), the distribution $\widehat{P}_{0}$ generated by the discrete-time DDPM satisfies

$$ $\mathrm{KL}(Q_{0}||\widehat{P}_{0})\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{2}\bigg{(}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\nabla^{2}\log q_{t}(X_{t})\Big{)}\right]\\ +\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\left(\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\right)^{2}\right]\bigg{)}+(\log T)\varepsilon^{2}.$ $$

[Theorem 1](#Thmtheorem1) characterizes the convergence in terms of KL divergence (and thus TV distance) for smooth (possibly unbounded) $Q_{0}$. The bound in [Theorem 1](#Thmtheorem1) will be further instantiated with explicit dependence on system parameters for example distributions $Q_{0}$ in [Section 4](#S4). To further explain the upper bound in [Theorem 1](#Thmtheorem1), the last term arises from the score estimation error, and the first two terms capture the errors accumulated during the reverse steps over $t=T,\dots,1$. In particular, we show that the reverse error can be captured by the tilting factor (see [Section 3.3](#S3.SS3)), and hence the first term in [Theorem 1](#Thmtheorem1) is the dominant term of the mean value gap of the tilting factor between $P_{t-1|t}$ and $Q_{t-1|t}$. The second term is the dominant term in the squared mean value of the tilting factor under $P_{t-1|t}$. More detailed explanation can be found in the proof sketch in [Section 3.3](#S3.SS3).

We note that [Theorem 1](#Thmtheorem1) is the first convergence result for discrete-time DDPMs for general distribution smooth $Q_{0}$ that has finite second moment (along with some mild regularity conditions). The (only) previous analysis on discrete-time DDPM models is given by [], which requires the data distribution to have finite support. There, a typical set is manually crafted and the values of the log-likelihoods in and out of this set are bounded separately. This analysis, however, does not apply to handle target distributions having infinite support. Here we develop a different approach based on titling factor representation and Tweedie’s formula to handle power terms, which
is applicable to a much wider class of target distributions, including those having infinite support. We provide an overview of our analysis in [Section 3.3](#S3.SS3). We will also show in [Section 3.3](#S3.SS3) that when we specialize our result to distributions with bounded support we improve the rate dependence on the dimension $d$ in [].

The following corollary shows a tighter bound (where the second term in [Theorem 1](#Thmtheorem1) can go away) under an additional condition in ([6](#S3.E6)). It is not restrictive as we will show in Section [4](#S4).

###### Corollary 1 (Smooth $Q_{0}$ ) .

Under the same setting of [Theorem 1](#Thmtheorem1), if further that, for any fixed $x_{t}$,

$$ $\left\lVert\nabla\log q_{t-1}(\mu_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})\right\rVert^{2}=O\left(1-\alpha_{t}\right),\leavevmode\nobreak\ \forall t\geq 1$ (6) $$

holds, then the distribution $\widehat{P}_{0}$ generated by the discrete-time DDPM satisfies

$$ $\mathrm{KL}(Q_{0}||\widehat{P}_{0})\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\nabla^{2}\log q_{t}(X_{t})\Big{)}\right]+(\log T)\varepsilon^{2}.$ $$

We next further extend [Theorem 1](#Thmtheorem1) for smooth $Q_{0}$ to general $Q_{0}$ that can be possibly non-smooth and hence the density function $q_{0}$ does not exist. Such distributions occur often in practice; for example, when $Q_{0}$ has a discrete support such as for images, or when $Q_{0}$ is supported on a low-dimensional manifold. For non-smooth $Q_{0}$, its one-step perturbation $Q_{1}$ does have a p.d.f. $q_{1}$, which is further analytic ([Lemma 6](#Thmlemma6)). This enables us to apply [Theorem 1](#Thmtheorem1) on $Q_{1}$ to obtain the following convergence bound. Also, we use the Wasserstein distance to measure the perturbation between $Q_{0}$ and $Q_{1}$, which is widely used in the analysis of continuous-time diffusion models [, , ].

###### Corollary 2 (General (possibly non-smooth) $Q_{0}$ ) .

Under [Assumptions 1](#Thmassumption1), [2](#Thmassumption2) and [4](#Thmassumption4), the distribution $\widehat{P}_{1}$ generated by the discrete-time DDPM satisfies

$$ $\mathrm{KL}(Q_{1}||\widehat{P}_{1})\lesssim\sum_{t=2}^{T}(1-\alpha_{t})^{2}\bigg{(}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\nabla^{2}\log q_{t}(X_{t})\Big{)}\right]\\ +\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\left(\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\right)^{2}\right]\bigg{)}+(\log T)\varepsilon^{2}.$ $$

where $Q_{1}$ is such that $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\lesssim(1-\alpha_{1})d$.

In particular, [Corollary 2](#Thmcorollary2) applies to any general target distribution when the second moment is finite (under regularity conditions).

###### Definition 1 (Noise Schedule) .

###### Theorem 1 (Smooth Q 0 subscript 𝑄 0 Q_{0} ) .

###### Corollary 1 (Smooth Q 0 subscript 𝑄 0 Q_{0} ) .

###### Corollary 2 (General (possibly non-smooth) Q 0 subscript 𝑄 0 Q_{0} ) .

### 3.3 Proof Sketch of Theorem 1

In this subsection, we provide a proof sketch of [Theorem 1](#Thmtheorem1) to describe the idea of our analysis approach. The full proof is provided in [Appendix A](#A1). To begin we decompose the total error as

$$ $\displaystyle\mathrm{KL}(Q_{0}||\widehat{P}_{0})$ $\displaystyle\leq\underbrace{\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p_{T}(X_{T})}\right]}_{\text{initialization error}}+\underbrace{\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]}_{\text{estimation error}}$ $\displaystyle\qquad+\underbrace{\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]}_{\text{reverse-step error}}.$ $$

The initialization error can be bounded easily ([Lemma 3](#Thmlemma3)). Below we focus on the rest two terms in four steps. Our novel development lies in Steps 2-4, which upper-bound the reverse-step error.

Step 1: Bounding estimation error ([Lemma 4](#Thmlemma4)). At each time $t=1,\dots,T$, rather than upper-bounding via typical sets as in [], we directly evaluate the expected value of $\log(p_{t-1|t}(x_{t-1}|x_{t})/\widehat{p}_{t-1|t}(x_{t-1}|x_{t}))$. This is straightforward since $P_{t-1|t}$ and $\widehat{P}_{t-1|t}$ are Gaussian with different means yet the same covariance.

Step 2: Tilting factor expression of log-likelihood ratio ([Lemmas 5](#Thmlemma5), [6](#Thmlemma6) and [21](#A1.E21)). With Bayes’ rule, we show that $q_{t-1|t}$ is an exponentially tilted form of $p_{t-1|t}$ with tilting factor:

$$ $\zeta_{t,t-1}=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}(x_{t-1}-\mu_{t})+\sum_{p=2}^{\infty}T_{p}(\log q_{t-1},x_{t-1},\mu_{t})$ $$

where $T_{p}(f,x,\mu)$ is the $p$-th order Taylor power term of function $f$ around $x=\mu$. This tilting factor thus allows us to express $\log(q_{t-1|t}/p_{t-1|t})$ as follows. For each fixed $x_{t}$,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\log\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[e^{\zeta_{t,t-1}}]$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\log\left(1+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right).$ $$

Suppose that $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ is small. Then, via the approximation $\log(1+z)=z-\frac{z^{2}}{2}+O(z^{3})$ for $z\approx 0$, we obtain

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\\ +\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{2}+O\left(\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{3}\right).$ $$

Steps 3-4 develop explicit expressions for the expected value of $\zeta_{t,t-1}$ under $P_{t-1|t}$ and $Q_{t-1|t}$.

Step 3: Explicit expression for $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ ([Lemma 7](#Thmlemma7)).
Given the expression of $\zeta_{t,t-1}$, under regularity conditions (i.e., [Assumption 4](#Thmassumption4)), this step can be reduced to calculating the expected values of the power terms, which are the Gaussian centralized moments.
Note that $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ is indeed small for large $T$ due to vanishing variance (see ([3](#S2.E3))).

Step 4: Explicit expression for $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[\zeta_{t,t-1}]$ ([Lemmas 10](#Thmlemma10), [9](#Thmlemma9) and [8](#Thmlemma8)).
While the Gaussian centralized moments are easy to calculate, $Q_{t-1|t}$ is not Gaussian in general. As an extension to the original Tweedie’s formula [], we calculate the first four centralized moments (as functions of $x_{t}$ and $\log q_{t}$) and provide the asymptotic order of all higher moments for fixed $x_{t}$’s. These enable us to determine the dominating term of $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$.

Finally, substituting the calculated expected values into the upper bound and noting that the partial derivatives do not affect the rate (by [Assumption 4](#Thmassumption4)) we obtain the desirable result.

## 4 Example Q 0 subscript 𝑄 0 Q_{0} ’s: Convergence Rate with Explicit Parameter Dependence

In this section, we specialize [Theorems 1](#Thmtheorem1) and [2](#Thmcorollary2) to several interesting distribution classes, for which convergence bounds with explicit dependence on system parameters can be derived. The key is to locate the dependence in the dominating terms in the reverse-step error.

### 4.1 Q t subscript 𝑄 𝑡 Q_{t} with Lipschitz Scores

We first investigate the distribution class where all $\nabla\log q_{t}(t\geq 0)$ is norm-Lipschitz. Such a class has been well studied for continuous-time diffusion models [], and the following theorem establishes the first such result for the discrete-time diffusion models.

###### Theorem 2 (Lipschitz Scores) .

Suppose that $\nabla\log q_{t}(x),\leavevmode\nobreak\ \forall t\geq 0$ is norm $L$-Lipschitz, i.e., $\exists L>0$ such that

$$ $\left\lVert\nabla\log q_{t}(x)-\nabla\log q_{t}(y)\right\rVert\leq L\left\lVert x-y\right\rVert,\leavevmode\nobreak\ \forall x,y\in\mathbb{R}^{d},\leavevmode\nobreak\ \forall t\geq 0.$ (7) $$

Then, under the same setting of [Theorem 1](#Thmtheorem1), the distribution $\widehat{P}_{0}$ generated by the discrete-time DDPM satisfies

$$ $\mathrm{KL}(Q_{0}||\widehat{P}_{0})\lesssim\frac{dL^{2}\log^{2}T}{T}+(\log T)\varepsilon^{2}.$ $$

Our result for discrete-time DDPMs exhibits the same dependence on $d$, $L$, and $T$ as those for continuous-time models [, Theorem 1]. However, the discrete-time DDPM process is mathematically different from their continuous-time process, and hence it relies on completely different techniques for upper-bounding the reverse-step error, as we outline in [Section 3.3](#S3.SS3).

###### Theorem 2 (Lipschitz Scores) .

### 4.2 Gaussian Mixture Q 0 subscript 𝑄 0 Q_{0}

Next we investigate the case where $Q_{0}$ is Gaussian mixture. This is a rich class of distributions with strong approximation power [, ]. The following theorem establishes the first convergence result for such a distribution class for discrete-time samplers.

###### Theorem 3 ( $Q_{0}$ Gaussian Mixture) .

Suppose that $Q_{0}$ is Gaussian mixture, whose p.d.f. is given by $q_{0}(x_{0})=\sum_{n=1}^{N}\pi_{n}q_{0,n}(x_{0})$,
where $q_{0,n}$ is the p.d.f. of $\mathcal{N}(\mu_{0,n},\Sigma_{0,n})$ and $\pi_{n}\in[0,1]$ is the mixing coefficient where $\sum_{n=1}^{N}\pi_{n}=1$.
Then, under [Assumption 2](#Thmassumption2), we have

$$ $\mathrm{KL}(Q_{0}||\widehat{P}_{0})\lesssim\frac{d^{2}\min\{d,N\}^{2}\log^{2}T}{T}+(\log T)\varepsilon^{2}.$ $$

If further, $\Sigma_{0,n}\equiv\sigma_{0}^{2}I_{d},\forall n=1,\dots,N$, then the bound becomes independent of $N$ as follows:

$$ $\mathrm{KL}(Q_{0}||\widehat{P}_{0})\lesssim\frac{d^{2}\log^{2}T}{T}+(\log T)\varepsilon^{2}.$ $$

[Theorem 3](#Thmtheorem3) is the first DDPM result directly regarding Gaussian mixture $Q_{0}$, which does not have an existing uniform Lipschitz constant on its score.
In [], it is shown that the sample complexity and the number of gradient descent steps to minimize the DDPM objective when $Q_{0}$ is Gaussian mixture are both polynomial in $d$, $N$, and $\varepsilon^{-1}$. Our result below sheds new light for the end-to-end analysis of Gaussian mixture learning through the discrete-time diffusion process.

The proof of [Theorem 3](#Thmtheorem3) is non-trivial in two ways: (i) In order to show that [Assumption 4](#Thmassumption4) holds for Gaussian mixture distributions, it is generally difficult to evaluate and provide an upper bound for all orders of partial derivatives of the logarithm of a mixture density. To this end, we employ the multivariate Faá di Bruno’s formula [] to develop an explicit bound ([Lemmas 15](#Thmlemma15) and [4](#Thmcorollary4)). (ii) In order to eliminate the second term in [Theorem 1](#Thmtheorem1), our method does non-trivial verification of condition ([6](#S3.E6)) for Gaussian mixture based on explicit forms derived from Gaussian mixture. Note that in general, even with small perturbation between $Q_{t-1}$ and $Q_{t}$, there is no guarantee that their respective score functions are also close almost everywhere.

###### Theorem 3 ( Q 0 subscript 𝑄 0 Q_{0} Gaussian Mixture) .

### 4.3 Q 0 subscript 𝑄 0 Q_{0} with Bounded Support

Next, we specialize [Corollary 2](#Thmcorollary2) to the case with $Q_{0}$ having bounded support (and possibly non-smooth). Such a case has been considered in []. To make our result comparable to that in [], we will employ the noise schedule proposed in []:

$$ $1-\alpha_{1}=\delta,\quad 1-\alpha_{t}=\frac{c\log T}{T}\min\left\{\delta\left(1+\frac{c\log T}{T}\right)^{t},1\right\},\leavevmode\nobreak\ \forall 2\leq t\leq T$ (8) $$

for any constants $c>2$ and $\delta\in(e^{-c},1)$. Note that any noise schedule satisfying the above condition also satisfies that in [Definition 1](#Thmdefinition1), and hence our [Corollary 2](#Thmcorollary2) still holds here. We then specialize [Corollary 2](#Thmcorollary2) to the following result for $Q_{0}$ with bounded support.

###### Theorem 4 ( $Q_{0}$ with Bounded Support) .

Suppose that $\left\lVert X_{0}\right\rVert^{2}\leq R^{2}d$ almost surely under $Q_{0}$, where $R$ is a constant that does not depend on $d$. Under [Assumption 2](#Thmassumption2), with the noise schedule in ([8](#S4.E8)), the distribution $\widehat{P}_{1}$ generated by the discrete-time DDPM satisfies

$$ $\mathrm{KL}(Q_{1}||\widehat{P}_{1})\lesssim\frac{d^{2}\log^{2}T}{\delta^{2}T}+(\log T)\varepsilon^{2},$ $$

where $Q_{1}$ is such that $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\lesssim\delta d$.
Further, if there is non-vanishing probability over the support, i.e., $\forall\gamma>0$ and $\forall x_{0}\in\mathrm{supp}(Q_{0})$,

$$ $\forall B_{\gamma}(x_{0}):=\left\{y\in\mathrm{supp}(Q_{0}):\left\lVert y-x_{0}\right\rVert^{2}\leq\gamma^{2}\right\},\leavevmode\nobreak\ Q_{0}(B_{\gamma}(x_{0}))\geq\epsilon,$ (9) $$

then we can achieve an improved rate as

$$ $\mathrm{KL}(Q_{1}||\widehat{P}_{1})\lesssim\frac{d\log^{2}T}{\delta^{2}T}+(\log T)\varepsilon^{2}.$ $$

The condition ([9](#S4.E9)) can be satisfied easily, for example, when $Q_{0}$ has non-zero probabilities only at discrete values (e.g., images).
This condition is unique to our paper.

Comparison to result in []: [Theorem 4](#Thmtheorem4) indicates that $\mathcal{O}\left(\frac{d^{2}}{\delta^{2}\varepsilon^{2}}\right)$ diffusion steps are needed for discrete-time DDPMs to achieve $\mathcal{O}(\varepsilon^{2})$ error in KL divergence (or $\mathcal{O}(\varepsilon)$ in TV distance). Such a result improves the dependence on $d$ in [] by a factor of $\mathcal{O}(d^{2})$. We further note that we consider $Q_{1}(\delta)$ that satisfies $\mathrm{W}_{2}\left(Q_{0},Q_{1}\right)^{2}\lesssim\delta d$, where $\delta$ captures the closeness between $Q_{0}$ and $Q_{1}$, whereas [] does not explicitly capture such a parameter in their result.

The proof of [Theorem 4](#Thmtheorem4) involves the following novel elements. (i) Verifying [Assumption 4](#Thmassumption4) requires evaluating and providing an upper bound for all orders of partial derivatives of the logarithm of a continuous mixture density. (ii) In order to eliminate the second term in [Theorem 1](#Thmtheorem1), we need to provide and verify an analogous condition to ([6](#S3.E6)) yet for general $Q_{0}$. (iii) We use a change-of-measure argument to find the optimal upper bound on the expected trace of Hessian products.

###### Theorem 4 ( Q 0 subscript 𝑄 0 Q_{0} with Bounded Support) .

## 5 Accelerated DDPM Sampler and Convergence Bounds

To generate samples from the data distribution $Q_{0}$, the idea of DDPM is to design a reverse process in which each reverse sampling step well approximates the corresponding forward step. To achieve this, the conditional mean of the true reverse process is designed to match the posterior mean at each step. In this section, we propose a new accelerated sampler along with a new covariance estimator, in which both the conditional mean and covariance of the reverse process match the corresponding posterior quantities at each step.

### 5.1 A New Accelerated DDPM Sampler

At each time $t=T,T-1,\dots,1$, define the true accelerated reverse process as

$$ $x_{t-1}=\mu_{t}(x_{t})+\Sigma_{t}^{\frac{1}{2}}(x_{t})z$ (10) $$

where $\mu_{t}$ is defined in ([3](#S2.E3)), $z\sim\mathcal{N}(0,I_{d})$, and (cf. [Lemma 8](#Thmlemma8))

$$ $\Sigma_{t}(x_{t}):=\frac{1-\alpha_{t}}{\alpha_{t}}\left(I_{d}+(1-\alpha_{t})\nabla^{2}\log q_{t}(x_{t})\right).$ (11) $$

Equivalently, this can be written as $P^{\prime}_{t-1|t}=\mathcal{N}(x_{t-1};\mu_{t}(x_{t}),\Sigma_{t}(x_{t}))$.
When $(1-\alpha_{t})$ is vanishing for large $T$ (e.g., [Definition 1](#Thmdefinition1)), $\Sigma_{t}(x_{t})\succ 0$ for all large $T$’s, and thus the conditional Gaussian process is well-defined. The above accelerated sampler has a close relationship to Ozaki’s discretization method to approximate a continuous-time stochastic process [].

In practice, one does not know $\nabla^{2}\log q_{t}$ and needs to estimate it. Denote such an estimator by $H_{t}$ (see an estimator $H_{t}$ we propose below based on a Hessian matching method), and suppose that $H_{t}$ is also symmetric. Given $H_{t}$, we can obtain an estimated accelerated reverse process:

$$ $x_{t-1}=\hat{\mu}_{t}(x_{t})+\hat{\Sigma}_{t}^{\frac{1}{2}}(x_{t})z$ $$

where $\hat{\mu}_{t}$ is defined in ([4](#S2.E4)), and

$$ $\hat{\Sigma}_{t}(x_{t}):=\frac{1-\alpha_{t}}{\alpha_{t}}\left(I_{d}+(1-\alpha_{t})H_{t}(x_{t})\right).$ (12) $$

Equivalently, we can write

$$ $\widehat{P}^{\prime}_{t-1|t}=\mathcal{N}(x_{t-1};\hat{\mu}_{t}(x_{t}),\hat{\Sigma}_{t}(x_{t})).$ $$

#### A Hessian matching estimator

One method to obtain $H_{t}(x)$, which estimates $\nabla^{2}\log q_{t}(x)$, is as follows. Note that

$$ $\displaystyle\nabla^{2}\log q_{t}(x)$ $\displaystyle=\frac{\nabla^{2}q_{t}(x)}{q_{t}(x)}-(\nabla\log q_{t}(x))(\nabla\log q_{t}(x))^{\intercal}$ $\displaystyle=\left(\frac{\nabla^{2}q_{t}(x)}{q_{t}(x)}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)-\frac{1}{1-\bar{\alpha}_{t}}I_{d}-(\nabla\log q_{t}(x))(\nabla\log q_{t}(x))^{\intercal}.$ (13) $$

Apart from the original score estimate, we require an additional Hessian estimate:

$$ $\displaystyle v_{t}(x)$ $\displaystyle:=\operatorname*{arg\,min}_{v_{\theta}:\mathbb{R}^{d}\to\mathbb{R}^{d\times d}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert v_{\theta}(X_{t})-\left(\frac{\nabla^{2}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)\right\rVert_{F}^{2}$ $\displaystyle=\operatorname*{arg\,min}_{v_{\theta}:\mathbb{R}^{d}\to\mathbb{R}^{d\times d}}\mathbb{E}_{X_{0}\sim Q_{0},\bar{W}_{t}\sim\mathcal{N}(0,I_{d})}\left\lVert v_{\theta}(\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t})-\frac{1}{1-\bar{\alpha}_{t}}\bar{W}_{t}\bar{W}_{t}^{\intercal}\right\rVert_{F}^{2}.$ $$

Here the second equation is an analogy to score matching, which we call as Hessian matching. The result on Hessian matching is included in the following lemma, the proof of which is provided in [Section I.1](#A9.SS1).

###### Lemma 1 .

With the forward process in ([1](#S2.E1)), we have

$$ $\operatorname*{arg\,min}_{v_{\theta}:\mathbb{R}^{d}\to\mathbb{R}^{d\times d}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert v_{\theta}(X_{t})-\left(\frac{\nabla^{2}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)\right\rVert_{F}^{2}\\ =\operatorname*{arg\,min}_{v_{\theta}:\mathbb{R}^{d}\to\mathbb{R}^{d\times d}}\mathbb{E}_{X_{0}\sim Q_{0},\bar{W}_{t}\sim\mathcal{N}(0,I_{d})}\left\lVert v_{\theta}(\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t})-\frac{1}{1-\bar{\alpha}_{t}}\bar{W}_{t}\bar{W}_{t}^{\intercal}\right\rVert_{F}^{2}.$ $$

From ([5.1](#S5.Ex22)), an estimate for $\nabla^{2}\log q_{t}(x)$ is given by

$$ $H_{t}(x)=v_{t}(x)-\frac{1}{1-\bar{\alpha}_{t}}I_{d}-s_{t}(x)s_{t}^{\intercal}(x).$ (14) $$

In the literature, several other estimators have been proposed for higher order derivatives of $\log q_{t}(x)$ [, ].
We note that any estimator for $H_{t}$ is applicable to our analysis as long as [Assumption 5](#Thmassumption5) is satisfied.

We note that our accelerated sampler is different from that in [, Section 3.3.2]. Our sampler directly estimates the covariance of the reverse process, and hence has a more intuitive explanation that aligns with the original DDPM [, ], whereas their sampler can be interpreted as jointly estimating the covariance matrix plus noise. Since our sampler directly estimates the covariance, it enables us to handle the estimation errors easily, whereas in [] the estimators are assumed to be perfect which may not be feasible in practice. Further, as we will show below, our sampler enjoys a better convergence rate with improved dependence on $d$ than that in [].

###### Lemma 1 .

### 5.2 Technical Assumptions for Accelerated Sampler

In order to analyze the accelerated sampler, we need a new assumption that extends [Assumption 2](#Thmassumption2) to incorporate the Hessian estimation error.

###### Assumption 5 (Score and Hessian Estimation Error) .

The estimates $s_{t}$’s and $H_{t}$’s satisfy

$$ $\frac{1}{T}\sum_{t=1}^{T}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}\leq\varepsilon^{2},$ $$

such that $\varepsilon^{2}=\tilde{O}(T^{-2})$.
Also,

$$ $\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert H_{t}(X_{t})-\nabla^{2}\log q_{t}(X_{t})\right\rVert^{2}_{F}\leq\varepsilon_{H}^{2},$ $$

such that $\varepsilon_{H}^{2}=\tilde{O}(T^{-1})$.

In the following lemma, we provide sufficient conditions such that $s_{t}$ from regular score matching and $H_{t}$ in ([14](#S5.E14)) satisfy [Assumption 5](#Thmassumption5).

###### Lemma 2 .

Under [Assumption 4](#Thmassumption4), suppose that $v_{t}$ and $s_{t}$ satisfy

$$ $\displaystyle\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert v_{t}(X_{t})-\left(\frac{\nabla^{2}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)\right\rVert_{F}^{2}=\tilde{O}(T^{-1}),$ $\displaystyle\frac{1}{T}\sum_{t=1}^{T}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}=\tilde{O}(T^{-2}).$ $$

Then, $s_{t}$ and $H_{t}$ defined in ([14](#S5.E14)) satisfy [Assumption 5](#Thmassumption5).

###### Assumption 5 (Score and Hessian Estimation Error) .

###### Lemma 2 .

### 5.3 Accelerated Convergence Bounds

We first define a new noise schedule as follows, which will be useful for acceleration.

###### Definition 2 (Noise Schedule for Acceleration) .

The step-size $\alpha_{t}$’s satisfy

$$ $1-\alpha_{t}\lesssim\frac{\log T}{T},\leavevmode\nobreak\ \forall 1\leq t\leq T,\quad\text{while}\quad\bar{\alpha}_{T}=\prod_{t=1}^{T}\alpha_{t}=o\left(T^{-2}\right)$ $$

for all sufficiently large $T$’s.

In particular, $\bar{\alpha}_{T}$ needs to decay at a faster rate compared to that in [Definition 1](#Thmdefinition1). An example of $\alpha_{t}$ that satisfies [Definition 2](#Thmdefinition2) is $1-\alpha_{t}\equiv\frac{c\log T}{T},\leavevmode\nobreak\ \forall t\geq 1$ with $c>2$.

We provide the accelerated convergence results below for smooth $Q_{0}$ and general $Q_{0}$ which may be non-smooth. We will then specialize these results to the convergence rate with explicit parameter dependence for Gaussian mixture distributions and distributions with bounded support.

###### Theorem 5 (Accelerated Sampler for Smooth $Q_{0}$ ) .

Under [Assumptions 1](#Thmassumption1), [4](#Thmassumption4), [5](#Thmassumption5) and [3](#Thmassumption3), with the $\alpha_{t}$’s satisfying [Definition 2](#Thmdefinition2), if further that ([6](#S3.E6)) and

$$ $\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})\right\rVert^{2}=O\left(1-\alpha_{t}\right)$ (15) $$

hold for all $t\geq 1$ and fixed $x_{t}\in\mathbb{R}^{d}$, then we have

$$ $\mathrm{KL}(Q_{0}||\widehat{P}^{\prime}_{0})\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{3}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t}(X_{t}))\partial^{3}_{ijk}\log q_{t}(X_{t})\right]\\ +(\log T)\varepsilon^{2}+\frac{\log^{2}T}{T}\varepsilon_{H}^{2}.$ $$

To explain the above bound, the first term arises from the accumulated reverse-step error over times $t=T,\dots,1$. The second and third terms are due to inaccurate score and Hessian estimation, respectively. Now, under [Assumption 4](#Thmassumption4), the expectation of partial derivatives can be upper-bounded by a constant independent of $T$ (which can still depend on other system parameters such as the dimension $d$, and analyzing this is our further interest when we instantiate this result to specific classes of distributions). This implies that the rate of the reverse-step error is controlled by the coefficient $(1-\alpha_{t})^{3}\lesssim(\log T/T)^{3}$. Thus, after the summation from $t=1,\dots,T$, the first term as well as the entire upper bound is in the order of $\tilde{O}(1/T^{2})$ (note that
$\varepsilon^{2}=\tilde{O}(1/T^{2})$ and $\varepsilon_{H}^{2}=\tilde{O}(1/T)$ by [Assumption 5](#Thmassumption5)).
Thus, in order to achieve $\mathcal{O}(\varepsilon^{2})$ error in KL divergence, the number of (discrete) diffusion steps required is $\mathcal{O}(\varepsilon^{-1})$, which improves the dependence of the convergence rate on $\varepsilon$ of the regular sampler by a factor of $\mathcal{O}(\varepsilon^{-1})$.

###### Corollary 3 (Accelerated Sampler for General $Q_{0}$ ) .

Under [Assumptions 1](#Thmassumption1), [4](#Thmassumption4) and [5](#Thmassumption5),
using the $\alpha_{t}$’s defined in ([8](#S4.E8)) with $c>4$, we have

$$ $\mathrm{KL}(Q_{1}||\widehat{P}^{\prime}_{1})\lesssim\sum_{t=2}^{T}(1-\alpha_{t})^{3}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t}(X_{t}))\partial^{3}_{ijk}\log q_{t}(X_{t})\right]\\ +(\log T)\varepsilon^{2}+\frac{\log^{2}T}{T}\varepsilon_{H}^{2}.$ $$

Here $Q_{1}$ is such that $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\lesssim\delta d$.

In the following, we specialize [Theorems 5](#Thmtheorem5) and [3](#Thmcorollary3) to the interesting cases considered in Section [4](#S4).

###### Theorem 6 (Accelerated Sampler for Lipschitz Score and Hessian) .

Suppose that $\nabla\log q_{t}(x)$ and $\nabla^{2}\log q_{t}(x),\leavevmode\nobreak\ \forall t\geq 0$ are norm $L$-Lipschitz and 2-norm $M$-Lipschitz, respectively. This means that $\exists L,M>0$ such that

$$ $\displaystyle\left\lVert\nabla\log q_{t}(x)-\nabla\log q_{t}(y)\right\rVert$ $\displaystyle\leq L\left\lVert x-y\right\rVert,\leavevmode\nobreak\ \forall x,y\in\mathbb{R}^{d},\leavevmode\nobreak\ \forall t\geq 0,$ $\displaystyle\left\lVert\nabla^{2}\log q_{t}(x)-\nabla^{2}\log q_{t}(y)\right\rVert$ $\displaystyle\leq M\left\lVert x-y\right\rVert,\leavevmode\nobreak\ \forall x,y\in\mathbb{R}^{d},\leavevmode\nobreak\ \forall t\geq 0.$ (16) $$

Then, under [Assumptions 1](#Thmassumption1), [3](#Thmassumption3), [4](#Thmassumption4) and [5](#Thmassumption5), the distribution $\widehat{P}_{0}$ generated by the discrete-time DDPM satisfies

$$ $\mathrm{KL}(Q_{0}||\widehat{P}_{0})\lesssim\frac{d^{2}M^{2}\log^{3}T}{T^{2}}+(\log T)\varepsilon^{2}+\frac{\log^{2}T}{T}\varepsilon_{H}^{2}.$ $$

This is the first result for DDPM samplers to achieve an accelerated convergence result under the additional condition of Lipschitz Hessian, and under score and Hessian estimation errors. Prior analogous results are available in the literature of Langevin Monte-Carlo sampling [].

###### Theorem 7 (Accelerated Sampler for Gaussian Mixture $Q_{0}$ ) .

Suppose that $Q_{0}$ is Gaussian mixture (as defined in [Theorem 3](#Thmtheorem3)). Under [Assumption 5](#Thmassumption5), using the $\alpha_{t}$’s in [Definition 2](#Thmdefinition2), we have

$$ $\mathrm{KL}(Q_{0}||\widehat{P}^{\prime}_{0})\lesssim\frac{d^{3}\min\{d,N\}^{3}\log^{3}T}{T^{2}}+(\log T)\varepsilon^{2}+\frac{\log^{2}T}{T}\varepsilon_{H}^{2}.$ $$

This is the first result for DDPM samplers to achieve an accelerated convergence rate for Gaussian mixture targets under score and Hessian estimation error.

###### Theorem 8 (Accelerated Sampler for $Q_{0}$ having Bounded Support) .

Suppose that $\left\lVert X_{0}\right\rVert^{2}\leq R^{2}d$ almost surely under $Q_{0}$. Under [Assumption 5](#Thmassumption5), using the $\alpha_{t}$’s defined in ([8](#S4.E8)) with $c>4$, we have

$$ $\mathrm{KL}(Q_{1}||\widehat{P}^{\prime}_{1})\lesssim\frac{d^{3}\log^{3}T}{\delta^{3}T^{2}}+(\log T)\varepsilon^{2}+\frac{\log^{2}T}{T}\varepsilon_{H}^{2},$ $$

where $Q_{1}$ is such that $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\lesssim\delta d$.

In other words, when $Q_{0}$ has bounded support, it takes the accelerated algorithm $\mathcal{O}\left(\frac{d^{1.5}}{\delta^{1.5}\varepsilon}\right)$ steps to approximate a perturbed data distribution $Q_{1}$ within $\varepsilon^{2}$ error in KL divergence. Compared to [], our number of convergence steps to achieve $\varepsilon^{2}$-KL divergence has an improved dimensional dependence by a factor of $\mathcal{O}(d^{1.5})$. Our analysis includes both score and Hessian estimation errors, whereas [] does not handle these errors.

###### Definition 2 (Noise Schedule for Acceleration) .

###### Theorem 5 (Accelerated Sampler for Smooth Q 0 subscript 𝑄 0 Q_{0} ) .

###### Corollary 3 (Accelerated Sampler for General Q 0 subscript 𝑄 0 Q_{0} ) .

###### Theorem 6 (Accelerated Sampler for Lipschitz Score and Hessian) .

###### Theorem 7 (Accelerated Sampler for Gaussian Mixture Q 0 subscript 𝑄 0 Q_{0} ) .

###### Theorem 8 (Accelerated Sampler for Q 0 subscript 𝑄 0 Q_{0} having Bounded Support) .

## 6 Conclusion

In this paper, we have developed a novel analysis framework directly over the discrete-time DDPM process, and we have provided convergence guarantees for both smooth $Q_{0}$ and general $Q_{0}$ with early-stopping. One future direction is to further shrink the $\delta$ dependence from polynomial to logarithmic for general $Q_{0}$ under early-stopping. It is also interesting to investigate other families of distributions where explicit parameter dependence can be found.

## Acknowledgements

This work has been supported in part by the U.S. National Science Foundation under the grants: CCF-1900145, NSF AI Institute (AI-EDGE) 2112471, CNS-2312836, CNS-2223452, CNS-2225561, and was sponsored by the Army Research Laboratory under Cooperative Agreement Number W911NF-23-2-0225. The views and conclusions contained in this document are those of the authors and should not be interpreted as representing the official policies, either expressed or implied, of the Army Research Laboratory or the U.S. Government. The U.S. Government is authorized to reproduce and distribute reprints for Government purposes notwithstanding any copyright notation herein.

## Appendix A Proof of Theorem 1

To begin the proof of [Theorem 1](#Thmtheorem1), note that

$$ $\displaystyle\mathrm{KL}(Q||\widehat{P})$ $\displaystyle=\mathbb{E}_{X_{0},\dots,X_{T}\sim Q}\left[\log\frac{q(X_{0},\dots,X_{T})}{\widehat{p}(X_{0},\dots,X_{T})}\right]$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathbb{E}_{X_{0},\dots,X_{T}\sim Q}\left[\log\frac{q_{0}(X_{0})\prod_{t=1}^{T}q_{t|t-1}(X_{t}|X_{t-1})}{\widehat{p}(X_{0},\dots,X_{T})}\right]$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\mathbb{E}_{X_{0},\dots,X_{T}\sim Q}\left[\log\frac{q_{0}(X_{0})\prod_{t=1}^{T}q_{t|t-1}(X_{t}|X_{t-1})}{\widehat{p}_{0}(X_{0})\prod_{t=1}^{T}\widehat{p}_{t|t-1}(X_{t}|X_{t-1})}\right]$ $\displaystyle=\mathbb{E}_{X_{0}\sim Q_{0}}\left[\log\frac{q_{0}(X_{0})}{\widehat{p}_{0}(X_{0})}\right]+\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t|t-1}(X_{t}|X_{t-1})}{\widehat{p}_{t|t-1}(X_{t}|X_{t-1})}\right]$ $\displaystyle=\mathbb{E}_{X_{0}\sim Q_{0}}\left[\log\frac{q_{0}(X_{0})}{\widehat{p}_{0}(X_{0})}\right]+\sum_{t=1}^{T}\mathbb{E}_{X_{t-1}\sim Q_{t-1}}\left[\mathbb{E}_{X_{t}\sim Q_{t|t-1}}\left[\log\frac{q_{t|t-1}(X_{t}|X_{t-1})}{\widehat{p}_{t|t-1}(X_{t}|X_{t-1})}\right]\right]$ $\displaystyle=\mathrm{KL}(Q_{0}||\widehat{P}_{0})+\sum_{t=1}^{T}\mathbb{E}_{X_{t-1}\sim Q_{t-1}}\left[\mathrm{KL}(Q_{t|t-1}(\cdot|X_{t-1})||\widehat{P}_{t|t-1}(\cdot|X_{t-1}))\right].$ $$

Here $(i)$ holds because of the Markov property of the forward process. We explain $(ii)$ below. By the backward Markov property of the reverse process, for any $t\geq 1$, given $X_{t-1}=x_{t-1}$, each of $X_{t-2},\dots,X_{0}$ is independent with $X_{t}$. This implies that

$$ $\widehat{p}_{t|t-1,\dots,0}(x_{t}|x_{t-1},\dots,x_{0})=\widehat{p}_{t|t-1}(x_{t}|x_{t-1}),\quad\forall t\geq 1.$ $$

Thus, $\widehat{p}(x_{0},\dots,x_{T})=\widehat{p}_{0}(x_{0})\prod_{t=1}^{T}\widehat{p}_{t|t-1}(x_{t}|x_{t-1})$. In other words, $X_{0},\dots,X_{t}$ is also forward Markov under $\widehat{P}$.

Following from similar arguments,

$$ $\mathrm{KL}(Q||\widehat{P})=\mathrm{KL}(Q_{T}||\widehat{P}_{T})+\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{KL}(Q_{t-1|t}(\cdot|X_{t})||\widehat{P}_{t-1|t}(\cdot|X_{t}))\right].$ $$

Since KL-divergence is non-negative, an upper bound on $\mathrm{KL}(Q_{0}||\widehat{P}_{0})$ is given by

$$ $\displaystyle\mathrm{KL}(Q_{0}||\widehat{P}_{0})$ $\displaystyle=\mathrm{KL}(Q_{T}||\widehat{P}_{T})+\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{KL}(Q_{t-1|t}(\cdot|X_{t})||\widehat{P}_{t-1|t}(\cdot|X_{t}))\right]$ $\displaystyle\qquad-\sum_{t=1}^{T}\mathbb{E}_{X_{t-1}\sim Q_{t-1}}\left[\mathrm{KL}(Q_{t|t-1}(\cdot|X_{t-1})||\widehat{P}_{t|t-1}(\cdot|X_{t-1}))\right]$ $\displaystyle\leq\mathrm{KL}(Q_{T}||\widehat{P}_{T})+\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{KL}(Q_{t-1|t}(\cdot|X_{t})||\widehat{P}_{t-1|t}(\cdot|X_{t}))\right]$ $\displaystyle=\underbrace{\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p_{T}(X_{T})}\right]}_{\text{Term 1: initialization error}}+\underbrace{\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]}_{\text{Term 2: estimation error}}$ $\displaystyle\qquad+\underbrace{\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]}_{\text{Term 3: reverse-step error}}.$ (17) $$

The last equality holds because $\widehat{p}_{T}=p_{T}$. Next, we bound the above three terms separately in a few steps.

### A.1 Step 0: Bounding term 1 – Initialization Error

###### Lemma 3 .

Suppose $\bar{\alpha}_{T}\searrow 0$ as $T\to\infty$. Then,

$$ $\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p_{T}(X_{T})}\right]\leq\frac{1}{2}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}\bar{\alpha}_{T}+O\left(\bar{\alpha}_{T}^{2}\right),\leavevmode\nobreak\ \text{ as }T\to\infty.$ $$

###### Remark 1 .

Under [Assumption 1](#Thmassumption1), if the noise schedule satisfies [Definition 1](#Thmdefinition1), we have

$$ $\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p_{T}(X_{T})}\right]=o(T^{-1}).$ $$

###### Proof.

See [section C.1](#A3.SS1).
∎

###### Lemma 3 .

###### Remark 1 .

###### Proof.

### A.2 Step 1: Bounding term 2 – Score Estimation Error

###### Lemma 4 .

Under [Assumption 2](#Thmassumption2), with the $\alpha_{t}$’s satisfying [Definition 1](#Thmdefinition1), we have

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim(\log T)\varepsilon^{2}.$ $$

###### Proof.

See [section C.2](#A3.SS2).
∎

###### Lemma 4 .

###### Proof.

### A.3 Step 2: Expressing Log-likelihood Ratio via Tilting Factor

To understand and analyze the reverse-step error, we first investigate the relationship between $q_{t-1|t}$ and $p_{t-1|t}$ for any fixed $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$. In this section write $\mu_{t}(x_{t})=\mu_{t}$ for brevity.
In the following lemma, we show that $q_{t-1|t}$ is an exponentially tilted form of $p_{t-1|t}$, which further enables us to upper-bound the expected log-likelihood ratio using the tilting factor.

###### Lemma 5 .

For any fixed $x_{t}\in\mathbb{R}^{d}$, we have

$$ $q_{t-1|t}(x_{t-1}|x_{t})=\frac{p_{t-1|t}(x_{t-1}|x_{t})e^{\zeta_{t,t-1}(x_{t},x_{t-1})}}{\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[e^{\zeta_{t,t-1}(x_{t},X_{t-1})}]}$ $$

where

$$ $\zeta_{t,t-1}(x_{t},x_{t-1}):=\log q_{t-1}(x_{t-1})-\log q_{t-1}(\mu_{t})-(x_{t-1}-\mu_{t})^{\intercal}(\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})).$ (18) $$

###### Proof.

See [section C.3](#A3.SS3).
∎

In the following we write $\zeta_{t,t-1}=\zeta_{t,t-1}(x_{t},x_{t-1})$ and omit dependencies on $x_{t}$ and $x_{t-1}$ for brevity. For any fixed $x_{t}$, the conditional reverse-step error can be upper-bounded as

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}-\log\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[e^{\zeta_{t,t-1}}]\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\log\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[e^{\zeta_{t,t-1}}]$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\log\left(1+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)$ (19) $$

wherethe last inequality follows because $e^{z}\geq 1+z$ for any $z\in\mathbb{R}$.

Given the definition of $\zeta_{t,t-1}$ in ([18](#A1.E18)), below we analyze $\log q_{t-1}(x)$ around $x=\mu_{t}$ using Taylor expansion. The following lemma guarantees the validity of Taylor expansion for $t\geq 1$.

###### Lemma 6 .

Fix $t\geq 1$. For any $Q_{0}$ (not necessarily having a p.d.f. w.r.t. the Lebesgue measure), given any $k\geq 1$ and any vector of indices $\bm{a}\in[d]^{k}$, $q_{t}$ exists and $|\partial^{k}_{\bm{a}}\log q_{t}(x_{t})|<\infty,\leavevmode\nobreak\ \forall x_{t}\in\mathbb{R}^{d}$ (which possibly depends on $T$). Further, $q_{t}$ and $\log q_{t}$ are both analytic.

###### Proof.

See [section C.4](#A3.SS4).
∎

Recall that $x^{i}(1\leq i\leq d)$ denotes the $i$-th element of a vector $x$. In general, given an analytic function $f(x)$, its Taylor expansion around $x=\mu$ is given by

$$ $\displaystyle f(x)$ $\displaystyle=f(\mu)+\sum_{p=1}^{\infty}T_{p}(f,x,\mu)$ $\displaystyle=f(\mu)+\nabla f(\mu)^{\intercal}(x-\mu)+\frac{1}{2}\sum_{i=1}^{d}\partial^{2}_{ii}f(\mu)(x^{i}-\mu^{i})^{2}+\frac{1}{2}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{2}_{ij}f(\mu)(x^{i}-\mu^{i})(x^{j}-\mu^{j})$ $\displaystyle+\frac{1}{3!}\sum_{i=1}^{d}\partial^{3}_{iii}f(\mu)(x^{i}-\mu^{i})^{3}+\frac{1}{3!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}f(\mu)(x^{i}-\mu^{i})^{2}(x^{j}-\mu^{j})$ $\displaystyle+\frac{1}{3!}\sum_{\begin{subarray}{c}i,j,k=1\\ i,j,k\text{ all differ}\end{subarray}}^{d}\partial^{3}_{ijk}f(\mu)(x^{i}-\mu^{i})(x^{j}-\mu^{j})(x^{k}-\mu^{k})+\frac{1}{4!}\sum_{i=1}^{d}\partial^{4}_{iiii}f(\mu)(x^{i}-\mu^{i})^{4}$ $\displaystyle+\frac{1}{4!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{4}_{iiij}f(\mu)(x^{i}-\mu^{i})^{3}(x^{j}-\mu^{j})+\frac{1}{4!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{4}_{iijj}f(\mu)(x^{i}-\mu^{i})^{2}(x^{j}-\mu^{j})^{2}$ $\displaystyle+\frac{1}{4!}\sum_{\begin{subarray}{c}i,j,k=1\\ i,j,k\text{ all differ}\end{subarray}}^{d}\partial^{4}_{iijk}f(\mu)(x^{i}-\mu^{i})^{2}(x^{j}-\mu^{j})(x^{k}-\mu^{k})$ $\displaystyle+\frac{1}{4!}\sum_{\begin{subarray}{c}i,j,k,l=1\\ i,j,k,l\text{ all differ}\end{subarray}}^{d}\partial^{4}_{ijkl}f(\mu)(x^{i}-\mu^{i})(x^{j}-\mu^{j})(x^{k}-\mu^{k})(x^{l}-\mu^{l})+\sum_{p=5}^{\infty}T_{p}(f,x,\mu)$ $$

where

$$ $\displaystyle T_{p}(f,x,\mu)$ $\displaystyle:=\frac{1}{p!}\sum_{\gamma\in\mathbb{N}^{d}:\sum_{i}\gamma^{i}=p}\partial^{p}_{\bm{a}}f(\mu)\prod_{i=1}^{d}(x^{i}-\mu^{i})^{\gamma^{i}}$ (20) $$

where in $\bm{a}\in[d]^{p}$ the multiplicity of $i(\in[d])$ is $\gamma^{i}$.
Specialize to the case where $f=\log q_{t-1}$, $x=x_{t-1}$, and $\mu=\mu_{t}$. Since $\log q_{t-1}$ is analytic (by [Assumption 3](#Thmassumption3) and [Lemma 6](#Thmlemma6)), Taylor expansion can be applied and is equal to (cf. ([18](#A1.E18)))

$$ $\zeta_{t,t-1}=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}(x_{t-1}-\mu_{t})+\sum_{p=2}^{\infty}T_{p}(\log q_{t-1},x_{t-1},\mu_{t}).$ (21) $$

With all these tools, we are now ready to analyze $\zeta_{t,t-1}$ under $P_{t-1|t}$ and $Q_{t-1|t}$.

###### Lemma 5 .

###### Proof.

###### Lemma 6 .

###### Proof.

### A.4 Step 3: Conditional Expectation of ζ t , t − 1 subscript 𝜁 𝑡 𝑡 1 \zeta_{t,t-1} under P t − 1 | t subscript 𝑃 𝑡 conditional 1 𝑡 P_{t-1|t}

Starting from now, we will need to examine the rate of $\frac{1-\alpha_{t}}{\alpha_{t}}$. A direct implication of [Definition 1](#Thmdefinition1) is that, with some constant $C_{1}$, since $\alpha_{t}\searrow 0$ as $T\to\infty$,

$$ $\frac{(1-\alpha_{t})^{p}}{\alpha_{t}^{q}}\leq\frac{C_{1}^{p}\log^{p}T/T^{p}}{\left(1-C_{1}\log T/T\right)^{q}}\lesssim(1-\alpha_{t})^{p},\quad\forall p,q\geq 1,t\geq 1.$ (22) $$

Also, from the definition of $\mu_{t}$ in ([3](#S2.E3)), if $(1-\alpha_{t})$ satisfies [Definition 1](#Thmdefinition1) and if $\left\lVert\nabla\log q_{t}(x)\right\rVert$ satisfies [Assumption 4](#Thmassumption4), we have $\mu_{t}(x)=x+O(\log T/T)$ for any fixed $x\in\mathbb{R}^{d}$.

The following lemma evaluates $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$. Although we consider smooth $Q_{0}$ in this section, the result is also applicable to non-smooth $Q_{0}$ at $t\geq 2$ since $Q_{t}(t\geq 1)$ is always smooth due to [Lemma 6](#Thmlemma6).

###### Lemma 7 .

Fix $t\geq 1$. Suppose that all partial derivatives of $\log q_{t-1}$ exist.
For any fixed $x_{t}\in\mathbb{R}^{d}$, under [Assumption 4](#Thmassumption4),

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]=\frac{1}{2}\left(\sum_{i=1}^{d}\partial^{2}_{ii}\log q_{t-1}(\mu_{t})\right)\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)$ $\displaystyle+\frac{1}{8}\left(\sum_{i=1}^{d}\partial^{4}_{iiii}\log q_{t-1}(\mu_{t})\right)\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+\frac{1}{4!}\left(\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{4}_{iijj}\log q_{t-1}(\mu_{t})\right)\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+O\left((1-\alpha_{t})^{3}\right).$ $$

###### Remark 2 .

Suppose the noise schedule satisfies [Definition 1](#Thmdefinition1). We have $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]=O(1-\alpha_{t})=\tilde{O}(T^{-1})$, which vanishes as $T\to\infty$.

###### Proof.

See [section C.5](#A3.SS5).
∎

###### Lemma 7 .

###### Remark 2 .

###### Proof.

### A.5 Step 4: Conditional Expectation of ζ t , t − 1 subscript 𝜁 𝑡 𝑡 1 \zeta_{t,t-1} under Q t − 1 | t subscript 𝑄 𝑡 conditional 1 𝑡 Q_{t-1|t}

Although each $Q_{t|t-1}$ is conditionally Gaussian, the posterior $Q_{t-1|t}$ is not Gaussian in general. In the following, we analyze the posterior centralized moments under $Q_{t-1|t}$ using the idea of Tweedie’s formula []. Then, we apply them to analyze $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[\zeta_{t,t-1}]$, again using the Taylor expansion in ([21](#A1.E21)). Again, the result is more generally applicable to non-smooth $Q_{0}$ at $t\geq 2$ due to [Lemma 6](#Thmlemma6).

###### Lemma 8 .

Fix $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$. Define $\tilde{x}_{t}:=\frac{\sqrt{\alpha_{t}}}{1-\alpha_{t}}x_{t}$, and

$$ $\kappa(\tilde{x}_{t}):=\log q_{t}\left(\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\tilde{x}_{t}\right)+\frac{1-\alpha_{t}}{2\alpha_{t}}\left\lVert\tilde{x}_{t}\right\rVert^{2}+\frac{d}{2}\log\left(2\pi(1-\alpha_{t})\right).$ (23) $$

Let $1\leq i,j,k,l\leq d$, which are possibly equal to each other. The first 3 centralized moments under $Q_{t-1|t}$ satisfy

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[X_{t-1}\right]$ $\displaystyle=\nabla\kappa=\mu_{t}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}-\mu_{t})(X_{t-1}-\mu_{t})^{\intercal}\right]$ $\displaystyle=\nabla^{2}\kappa=\frac{1-\alpha_{t}}{\alpha_{t}}I_{d}+\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\nabla^{2}\log q_{t}(x_{t})$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})(X_{t-1}^{k}-\mu_{t}^{k})\right]$ $\displaystyle=\partial^{3}_{ijk}\kappa$ $\displaystyle=\frac{(1-\alpha_{t})^{3}}{\alpha_{t}^{3/2}}\partial^{3}_{ijk}\log q_{t}(x_{t})=O((1-\alpha_{t})^{3}).$ $$

The fourth centralized moment satisfies

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})(X_{t-1}^{k}-\mu_{t}^{k})(X_{t-1}^{l}-\mu_{t}^{l})\right]$ $\displaystyle=(\partial^{2}_{ij}\kappa)(\partial^{2}_{kl}\kappa)+(\partial^{2}_{ik}\kappa)(\partial^{2}_{jl}\kappa)+(\partial^{2}_{il}\kappa)(\partial^{2}_{jk}\kappa)+\partial^{4}_{ijkl}\kappa$ $\displaystyle=\begin{cases}3\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+O((1-\alpha_{t})^{3}),&\text{if}\ i=j=k=l,\\ \left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+O((1-\alpha_{t})^{3}),&\text{if}\ i=k\neq j=l,\\ O((1-\alpha_{t})^{3}),&\text{otherwise}.\end{cases}$ $$

Note that all derivatives above are w.r.t. $\tilde{x}_{t}$.

###### Proof.

See [section C.6](#A3.SS6).
∎

The following lemma provides the correct order (in terms of $(1-\alpha_{t})$) of any posterior centralized moment. In other words, this shows that $Q_{t-1|t}$ has nice (Gaussian) concentration.

###### Lemma 9 .

Fix $t\geq 1$, $x_{t}\in\mathbb{R}^{d}$, and $p\geq 2$. Let $\bm{a}=(a_{1},\dots,a_{p})\in[d]^{p}$ be a vector of indices of length $p$. With the same definitions of $\tilde{x}_{t}$ and $\kappa(\tilde{x}_{t})$ as in Lemma [8](#Thmlemma8), if $p$ is odd,

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{p}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]=O\left((1-\alpha_{t})^{\frac{p+3}{2}}\right),\leavevmode\nobreak\ \forall\bm{a}\in[d]^{p}.$ (24) $$

If $p$ is even,

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{p}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]=O((1-\alpha_{t})^{\frac{p}{2}}),\leavevmode\nobreak\ \forall\bm{a}\in[d]^{p}.$ (25) $$

###### Proof.

See [section C.7](#A3.SS7).
∎

Combining the moment results in [Lemmas 8](#Thmlemma8) and [9](#Thmlemma9), we can derive the following result, again using the Taylor expansion around $\mu_{t}$ (see ([21](#A1.E21))).

###### Lemma 10 .

Fix $t\geq 1$. Suppose that all partial derivatives of $\log q_{t-1}$ exist. For any fixed $x_{t}\in\mathbb{R}^{d}$, under [Assumption 4](#Thmassumption4),

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[\zeta_{t,t-1}]$ $\displaystyle=\frac{1}{2}\sum_{i=1}^{d}\partial^{2}_{ii}\log q_{t-1}(\mu_{t})\left(\frac{1-\alpha_{t}}{\alpha_{t}}+\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\partial^{2}_{ii}\log q_{t}(x_{t})\right)$ $\displaystyle+\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{2}_{ij}\log q_{t-1}(\mu_{t})\partial^{2}_{ij}\log q_{t}(x_{t})+\frac{1}{8}\left(\sum_{i=1}^{d}\partial^{4}_{iiii}\log q_{t-1}(\mu_{t})\right)\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}$ $\displaystyle+\frac{1}{4!}\left(\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{4}_{iijj}\log q_{t-1}(\mu_{t})\right)\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+O\left((1-\alpha_{t})^{3}\right).$ $$

###### Proof.

The proof is similar to that of [Lemma 7](#Thmlemma7) and thus omitted. Note that the exchange of infinite sum and expectation is still guaranteed by [Assumption 4](#Thmassumption4). The only difference is that we substitute a different set of moments according to [Lemmas 8](#Thmlemma8) and [9](#Thmlemma9) into the Taylor expansion in ([21](#A1.E21)).
∎

###### Lemma 8 .

###### Proof.

###### Lemma 9 .

###### Proof.

###### Lemma 10 .

###### Proof.

### A.6 Bounding term 3 – Reverse-step Error

Now, we combine [Lemmas 7](#Thmlemma7) and [10](#Thmlemma10) and continue from ([A.3](#A1.Ex51)) to upper-bound term 3 of the reverse-step error. The Taylor expansion of $\log(1+z)$ around $z=0$ yields that

$$ $\log(1+z)=z-\frac{z^{2}}{2}+O(z^{3}),\leavevmode\nobreak\ \text{when}\leavevmode\nobreak\ z\to 0.$ $$

For each $t\geq 1$, suppose that [Assumption 4](#Thmassumption4) is satisfied. Since $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ is vanishing because $1-\alpha_{t}\searrow 0$ as $T\to\infty$ (see [Remark 2](#Thmremark2)), we can use [Lemmas 7](#Thmlemma7) and [10](#Thmlemma10) and continue ([A.3](#A1.Ex51)) as

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\log\left(1+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{2}$ $\displaystyle\qquad+O\left(\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{3}\right)$ $\displaystyle=\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}^{2}}\left(\mathrm{Tr}\left(\nabla^{2}\log q_{t-1}(\mu_{t})\nabla^{2}\log q_{t}(x_{t})\right)+\mathrm{Tr}\left(\nabla^{2}\log q_{t-1}(\mu_{t})\right)^{2}\right)$ $\displaystyle\qquad+O\left(\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{3}\right)$ $$

where the last line follows from [Lemmas 7](#Thmlemma7) and [10](#Thmlemma10) and the relationship in ([22](#A1.E22)).
The final step is to take the expectation over $X_{t}\sim Q_{t}$. Under [Assumption 4](#Thmassumption4), since all higher order partial derivatives have an upper bound in expectation w.r.t. $Q_{t}$ that does not depend on $T$, the rate of each term does not change after the expectation. In other words, (cf. [Lemma 7](#Thmlemma7))

$$ $\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{p}\right]=O((1-\alpha_{t})^{p}),\quad\forall p\geq 1.$ (26) $$

Therefore, summing over $t\geq 1$, we obtain

$$ $\displaystyle\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]$ $\displaystyle\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\nabla^{2}\log q_{t}(X_{t})\Big{)}+\mathrm{Tr}\left(\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\right)^{2}\right].$ $$

Now, the proof of [Theorem 1](#Thmtheorem1) is complete.

## Appendix B Proof of Corollaries 1 and 2

### B.1 Proof of Corollary 1

The initialization and estimation errors can be bounded in the same way as in [Theorem 1](#Thmtheorem1), and the difference lies in how to bound the reverse-step error.
In order to get rid of the $\mathrm{Tr}(\cdot)^{2}$ term and to achieve a more precise bound, we need to control the expected values of higher (specifically, second and third) moments of $\zeta_{t,t-1}$. From ([21](#A1.E21)), it can be observed that the first-order term of $\zeta_{t,t-1}$ becomes zero after taking expectation under $P_{t-1|t}$ and $Q_{t-1|t}$.
However, this term will be present in the powers of $\zeta_{t,t-1}$ which in turn will affect their expected values. To this end, condition ([6](#S3.E6)) guarantees that the coefficient for the first-order term is small.
In the following lemma, we show that the expected powers of $\zeta_{t,t-1}$ are small when ([6](#S3.E6)) holds.

###### Lemma 11 .

Suppose that [Assumption 4](#Thmassumption4) and ([6](#S3.E6)) hold.
Then, if the noise schedule satisfies [Definition 1](#Thmdefinition1), for all $t\geq 1$ and fixed $x_{t}\in\mathbb{R}^{d}$,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{2}]$ $\displaystyle=O\left((1-\alpha_{t})^{2}\right)=O\left(\frac{\log^{2}T}{T^{2}}\right),\leavevmode\nobreak\ \text{and}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{3}]$ $\displaystyle=O\left((1-\alpha_{t})^{3}\right)=O\left(\frac{\log^{3}T}{T^{3}}\right).$ $$

###### Proof.

See [section C.8](#A3.SS8).
∎

In order to achieve a more precise error control, we also need to expand more power terms for $e^{z}$ than in ([A.3](#A1.Ex51)). Specifically,

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]\\ -\log\left(1+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]\right)$ (27) $$

where we have used the fact that $e^{z}\geq 1+z+\frac{z^{2}}{2}+\frac{z^{3}}{3!}$ for any $z\in\mathbb{R}$.
Then,

$$ $\displaystyle\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]$ $\displaystyle\leq\mathbb{E}_{X_{t}\sim Q_{t}}\Bigg{[}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]$ $\displaystyle\qquad-\log\left(1+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]\right)\Bigg{]}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\Bigg{[}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ $\displaystyle\qquad-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]-\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]$ $\displaystyle\qquad+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]\right)^{2}$ $\displaystyle\qquad+O\left(\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]\right)^{3}\right)\Bigg{]}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathbb{E}_{X_{t}\sim Q_{t}}\Bigg{[}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ $\displaystyle\qquad-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{2}\Bigg{]}+O((1-\alpha_{t})^{3})$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right]+O((1-\alpha_{t})^{3}).$ $$

Here $(i)$ follows from the results in [Lemmas 7](#Thmlemma7) and [11](#Thmlemma11) and [Assumption 4](#Thmassumption4) (see ([26](#A1.E26))), and $(ii)$ follows from Jensen’s inequality.
Thus, again using the results in [Lemmas 7](#Thmlemma7) and [10](#Thmlemma10), we can achieve a faster result:

$$ $\displaystyle\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]$ $\displaystyle\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1}(\mu_{t}(X_{t}))\nabla^{2}\log q_{t}(X_{t})\Big{)}\right].$ $$

###### Lemma 11 .

###### Proof.

### B.2 Proof of Corollary 2

Note that $q_{1}$ always exists and is analytic by [Lemma 6](#Thmlemma6). Therefore, it remains to upper-bound the mismatch between $Q_{0}$ and $Q_{1}$.
In the following lemma we provide such a bound in Wasserstein distance.

###### Lemma 12 .

For any $Q_{0}$,

$$ $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\leq(1-\alpha_{1})(M_{2}+1)d.$ $$

###### Remark 3 .

If $1-\alpha_{1}=\delta$, this implies that

$$ $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\lesssim\delta d.$ $$

###### Proof.

See [section C.9](#A3.SS9).
∎

The proof of this corollary is thus complete.
A consequence of Lemma [12](#Thmlemma12) is that, in order to obtain convergence guarantees for general distributions, one can view $1-\alpha_{1}$ as controlling the mismatch between $Q_{0}$ and $Q_{1}$ (in terms of the Wasserstein distance), and $1-\alpha_{t},\leavevmode\nobreak\ \forall t\geq 2$ as controlling the mismatch between $Q_{1}$ and $\widehat{P}_{1}$ (in terms of the KL-divergence).

###### Lemma 12 .

###### Remark 3 .

###### Proof.

## Appendix C Auxiliary Proofs for Theorems 1 , 1 and 2

In this section, we provide the proofs for those auxiliary lemmas in the proof of [Theorems 1](#Thmtheorem1), [1](#Thmcorollary1) and [2](#Thmcorollary2).

### C.1 Proof of Lemma 3

First, note that

$$ $q_{T}(x_{T})=\mathbb{E}_{X_{0}\sim Q_{0}}[q_{T|0}(x_{T}|X_{0})].$ $$

Also note that the function $f(x)=x\log(x)$ is convex. Thus, by Jensen’s inequality,

$$ $\displaystyle\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log q_{T}(X_{T})\right]$ $\displaystyle=\int\mathbb{E}_{X_{0}\sim Q_{0}}[q_{T|0}(x_{T}|X_{0})]\log\mathbb{E}_{X_{0}\sim Q_{0}}[q_{T|0}(x_{T}|X_{0})]\mathrm{d}x_{T}$ $\displaystyle\leq\int\mathbb{E}_{X_{0}\sim Q_{0}}\left[q_{T|0}(x_{T}|X_{0})\log q_{T|0}(x_{T}|X_{0})\right]\mathrm{d}x_{T}$ $\displaystyle=\mathbb{E}_{X_{0}\sim Q_{0}}\left[\int q_{T|0}(x_{T}|X_{0})\log q_{T|0}(x_{T}|X_{0})\mathrm{d}x_{T}\right].$ $$

Since $Q_{T|0}$ is conditional Gaussian $\mathcal{N}(\sqrt{\bar{\alpha}_{T}}x_{0},(1-\bar{\alpha}_{T})I_{d})$, its negative conditional entropy equals

$$ $\int q_{T|0}(x_{T}|x_{0})\log q_{T|0}(x_{T}|x_{0})\mathrm{d}x_{T}=-\frac{d}{2}-\frac{d}{2}\log(2\pi(1-\bar{\alpha}_{T}))$ $$

for any $x_{0}\in\mathbb{R}^{d}$. On the other hand, since $P_{T}=\mathcal{N}(0,I_{d})$,

$$ $\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log p_{T}(X_{T})\right]=-\frac{d}{2}\log(2\pi)-\frac{1}{2}\mathbb{E}_{X_{T}\sim Q_{T}}\left\lVert X_{T}\right\rVert^{2}$ $$

where

$$ $\displaystyle\mathbb{E}_{X_{T}\sim Q_{T}}\left\lVert X_{T}\right\rVert^{2}$ $\displaystyle=\bar{\alpha}_{T}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}+(1-\bar{\alpha}_{T})\mathbb{E}_{\bar{W}_{T}\sim\mathcal{N}(0,I_{d})}\left\lVert\bar{W}_{T}\right\rVert^{2}$ $\displaystyle=\bar{\alpha}_{T}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}+(1-\bar{\alpha}_{T})d.$ $$

Putting the two together,

$$ $\displaystyle\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p_{T}(X_{T})}\right]=\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log q_{T}(X_{T})\right]-\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log p_{T}(X_{T})\right]$ $\displaystyle\leq-\frac{d}{2}-\frac{d}{2}\log(2\pi(1-\bar{\alpha}_{T}))+\frac{d}{2}\log(2\pi)+\frac{1}{2}\left(\bar{\alpha}_{T}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}+(1-\bar{\alpha}_{T})d\right)$ $\displaystyle=\frac{1}{2}\bar{\alpha}_{T}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}-\frac{d\bar{\alpha}_{T}}{2}-\frac{d}{2}\log(1-\bar{\alpha}_{T}).$ $$

When $T$ is large (and thus when $\bar{\alpha}_{T}$ is small), the Taylor expansion w.r.t. $\bar{\alpha}_{T}$ around 0 yields

$$ $\log(1-\bar{\alpha}_{T})=-\bar{\alpha}_{T}+O\left(\bar{\alpha}_{T}^{2}\right).$ $$

Therefore,

$$ $\displaystyle\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p_{T}(X_{T})}\right]$ $\displaystyle\leq\frac{1}{2}\bar{\alpha}_{T}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}-\frac{d\bar{\alpha}_{T}}{2}-\frac{d}{2}(-\bar{\alpha}_{T})+O\left(\bar{\alpha}_{T}^{2}\right)$ $\displaystyle=\frac{1}{2}\bar{\alpha}_{T}\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert X_{0}\right\rVert^{2}+O\left(\bar{\alpha}_{T}^{2}\right).$ $$

### C.2 Proof of Lemma 4

Note that both $P_{t-1|t}$ and $\widehat{P}_{t-1|t}$ are Gaussian with the same variance. Thus, for each $t=1,\dots,T$,

$$ $\displaystyle\log\frac{p_{t-1|t}(x_{t-1}|x_{t})}{\widehat{p}_{t-1|t}(x_{t-1}|x_{t})}$ $\displaystyle=\frac{\alpha_{t}}{2(1-\alpha_{t})}\left(\left\lVert x_{t-1}-\widehat{\mu}_{t}(x_{t})\right\rVert^{2}-\left\lVert x_{t-1}-\mu_{t}(x_{t})\right\rVert^{2}\right)$ $\displaystyle=\frac{\alpha_{t}}{(1-\alpha_{t})}(x_{t-1}-\mu_{t}(x_{t}))^{\intercal}(\mu_{t}(x_{t})-\widehat{\mu}_{t}(x_{t}))+\frac{\alpha_{t}}{2(1-\alpha_{t})}\left\lVert\mu_{t}(x_{t})-\widehat{\mu}_{t}(x_{t})\right\rVert^{2}.$ $$

For the first term above,

$$ $\displaystyle\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[(X_{t-1}-\mu_{t}(X_{t}))^{\intercal}(\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t}))\right]$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\left[(\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[X_{t-1}|X_{t}]-\mu_{t}(X_{t}))^{\intercal}(\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t}))\right]$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\left[(\mu_{t}(X_{t})-\mu_{t}(X_{t}))^{\intercal}(\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t}))\right]=0$ $$

where the last line follows from Lemma [8](#Thmlemma8). Thus,

$$ $\displaystyle\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim Q_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]$ $\displaystyle=\sum_{t=1}^{T}\frac{\alpha_{t}}{2(1-\alpha_{t})}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}$ $\displaystyle=\sum_{t=1}^{T}\frac{1-\alpha_{t}}{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\lesssim}}\frac{\log T}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}(\log T)\varepsilon^{2}$ $$

where $(i)$ follows because $1-\alpha_{t}\lesssim\log T/T$, and $(ii)$ holds by [Assumption 2](#Thmassumption2). The proof is now complete.

### C.3 Proof of Lemma 5

We remind readers that throughout this proof $x_{t}$ is fixed (roughly speaking, like a constant). We want to know how $q_{t-1|t}(x_{t-1}|x_{t})$ changes with $x_{t-1}$ given fixed $x_{t}$.
Before we begin the derivation, we first provide some useful equations:

$$ $\displaystyle p_{t-1|t}(x_{t-1}|x_{t})\propto\exp\left(-\frac{\left\lVert x_{t-1}-\mu_{t}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}}\right),$ (28) $\displaystyle\frac{\left\lVert x_{t}-\sqrt{\alpha_{t}}x_{t-1}\right\rVert^{2}}{2(1-\alpha_{t})}=\frac{\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}},$ (29) $$

where $\propto$ means proportional to the change of $x_{t-1}$ (for fixed $x_{t}$), and

$$ $\displaystyle\left\lVert x_{t-1}-\mu_{t}\right\rVert^{2}-\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}=$ $\displaystyle\left\lVert x_{t-1}-\frac{x_{t}}{\sqrt{\alpha_{t}}}+\frac{x_{t}}{\sqrt{\alpha_{t}}}-\mu_{t}\right\rVert^{2}-\left\lVert x_{t-1}-\frac{x_{t}}{\sqrt{\alpha_{t}}}\right\rVert^{2}$ $\displaystyle=$ $\displaystyle 2(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}(x_{t}/\sqrt{\alpha_{t}}-\mu_{t})+\left\lVert\frac{x_{t}}{\sqrt{\alpha_{t}}}-\mu_{t}\right\rVert^{2},$ (30) $\displaystyle\frac{(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}(x_{t}/\sqrt{\alpha_{t}}-\mu_{t})}{(1-\alpha_{t})/\alpha_{t}}=$ $\displaystyle-(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})\ \text{ (by \lx@cref{creftype~refnum}{eq:def_mu_sigma_t})}$ $\displaystyle=$ $\displaystyle-\sqrt{\alpha_{t}}x_{t-1}^{\intercal}\nabla\log q_{t}(x_{t})+x_{t}^{\intercal}\nabla\log q_{t}(x_{t}).$ (31) $$

Now we are ready to investigate $q_{t-1|t}(x_{t-1}|x_{t})$.
By Bayes’ rule, we have

$$ $\displaystyle q_{t-1|t}(x_{t-1}|x_{t})$ $\displaystyle=\frac{q_{t|t-1}(x_{t}|x_{t-1})q_{t-1}(x_{t-1})}{q_{t}(x_{t})}$ $\displaystyle\propto q_{t-1}(x_{t-1})q_{t|t-1}(x_{t}|x_{t-1})$ (again, $\propto$ means proportional to the change of $x_{t-1}$ ; here $q_{t}(x_{t})$ is omitted since $x_{t}$ is fixed) $\displaystyle\propto q_{t-1}(x_{t-1})\exp\left(-\frac{\left\lVert x_{t}-\sqrt{\alpha_{t}}x_{t-1}\right\rVert^{2}}{2(1-\alpha_{t})}\right)$ $\displaystyle\propto q_{t-1}(x_{t-1})p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{\left\lVert x_{t-1}-\mu_{t}\right\rVert^{2}-\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}}\right)\ \text{ (by \lx@cref{creftypeplural~refnum}{eq.temp_112901} and~\lx@cref{refnum}{eq.temp_112902})}$ $\displaystyle\propto q_{t-1}(x_{t-1})p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}(x_{t}/\sqrt{\alpha_{t}}-\mu_{t})}{(1-\alpha_{t})/\alpha_{t}}\right)\ \text{ (by \lx@cref{creftype~refnum}{eq.temp_112903})}$ $\displaystyle\propto p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\log q_{t-1}(x_{t-1})-\sqrt{\alpha_{t}}x_{t-1}^{\intercal}\nabla\log q_{t}(x_{t})\right)\ \text{ (by \lx@cref{creftype~refnum}{eq.temp_112904})}$ $\displaystyle\propto p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\zeta_{t,t-1}(x_{t},x_{t-1})\right),$ $$

where $\zeta_{t,t-1}(x_{t},x_{t-1})=\log q_{t-1}(x_{t-1})-\sqrt{\alpha_{t}}x_{t-1}^{\intercal}\nabla\log q_{t}(x_{t})+\mathrm{fn}(x_{t})$ and $\mathrm{fn}(x_{t})$ can be some arbitrary function of $x_{t}$.
In other words, $q_{t-1|t}$ is an exponentially tilted form of $p_{t-1|t}$ with tilting factor $\zeta_{t,t-1}(x_{t},x_{t-1})$. Now, choose $\mathrm{fn}(x_{t})=-\log q_{t-1}(\mu_{t})+\mu_{t}^{\intercal}\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})$ and we have

$$ $\zeta_{t,t-1}(x_{t},x_{t-1}):=\log q_{t-1}(x_{t-1})-\log q_{t-1}(\mu_{t})-(x_{t-1}-\mu_{t})^{\intercal}(\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})).$ $$

Thus, for fixed $x_{t}$,

$$ $q_{t-1|t}(x_{t-1}|x_{t})=\frac{p_{t-1|t}(x_{t-1}|x_{t})e^{\zeta_{t,t-1}(x_{t},x_{t-1})}}{\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[e^{\zeta_{t,t-1}(x_{t},X_{t-1})}]}.$ $$

### C.4 Proof of Lemma 6

Recall [Equation 2](#S2.E2).
Let $\tilde{Q}_{0}$ denote the distribution of $\sqrt{\bar{\alpha}_{t}}x_{0}$, and let $g(z)$ denote the p.d.f. (w.r.t. the Lebesgue measure) of the distribution of $\sqrt{1-\bar{\alpha}_{t}}\bar{w}_{t}$. Note that $g$ is a scaled version of the unit Gaussian p.d.f., and $\int_{z\in\mathbb{R}^{d}}g(z)\mathrm{d}z=1<\infty$.
Now, for any event $A\subseteq\mathcal{B}(\lambda)$,

$$ $Q_{t}(A)=\int_{x\in A}\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}g(x-\tilde{x}_{0})\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})\mathrm{d}x=\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}\left(\int_{x\in A}g(x-\tilde{x}_{0})\mathrm{d}x\right)\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})$ $$

by Fubini’s theorem. If $A$ has Lebesgue measure 0, by continuity of $g(x)$ we get $\int_{x\in A}g(x-\tilde{x}_{0})\mathrm{d}x=0$, and thus $Q_{t}(A)=0$. This shows that $Q_{t}$ is absolutely continuous w.r.t. the Lebesgue measure, and its p.d.f. exists, denoted as $q_{t}$.

Now, since any order of derivative of the Gaussian p.d.f. is bounded away from infinity and $\tilde{Q}_{0}$ is a probability measure, we can invoke the dominated convergence theorem here to change the order of derivative and integral as

$$ $\partial^{k}_{\bm{a}}q_{t}(x)=\partial^{k}_{\bm{a}}\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}g(x-\tilde{x}_{0})\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})=\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}\partial^{k}_{\bm{a}}g(x-\tilde{x}_{0})\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0}).$ (32) $$

Thus, for any $k\geq 1$ and any vector of indices $\bm{a}\in[d]^{k}$, we have

$$ $\left|\partial^{k}_{\bm{a}}q_{t}(x)\right|\leq\sup_{x\in\mathbb{R}^{d}}\left|\partial^{k}_{\bm{a}}g(x)\right|\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})=\sup_{x\in\mathbb{R}^{d}}\left|\partial^{k}_{\bm{a}}g(x)\right|<\infty.$ $$

This also implies that the Taylor term $\left|T_{k}(q_{t},x,\mu)\right|<\infty$ for any $x$ and $\mu$, and

$$ $\displaystyle q_{t}(x)$ $\displaystyle=\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}g(x-\tilde{x}_{0})\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})\stackrel{{\scriptstyle(i)}}{{=}}\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}\lim_{p\to\infty}\sum_{k=0}^{p}T_{k}(g(x-\tilde{x}_{0}),x,\mu)\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\lim_{p\to\infty}\int_{\tilde{x}_{0}\in\mathbb{R}^{d}}\sum_{k=0}^{p}T_{k}(g(x-\tilde{x}_{0}),x,\mu)\mathrm{d}\tilde{Q}_{0}(\tilde{x}_{0})$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}\lim_{p\to\infty}\sum_{k=0}^{p}T_{k}(q_{t},x,\mu)$ $$

where $(i)$ follows because (scaled) Gaussian density is analytic, $(ii)$ follows from dominated convergence theorem and the fact that $g$ is a Gaussian density and has an upper bound independent of $\tilde{x}_{0}$, and $(iii)$ follows from ([32](#A3.E32)). This shows that $q_{t}$ is analytic.

Finally, since $\partial^{k}_{\bm{a}}\log q_{t}$ is a smooth function of $q_{t},\partial^{1}q_{t},\dots,\partial^{k}q_{t}$, we have $\partial^{k}_{\bm{a}}\log q_{t}(x_{t})<\infty$ (possibly depending on $T$) for all $k\geq 1$ and fixed (finite) $x_{t}\in\mathbb{R}^{d}$. Also, $\log q_{t}$ is analytic because $\log(\cdot)$ is analytic and $q_{t}(x_{t})>0,\leavevmode\nobreak\ \forall x_{t}\in\mathbb{R}^{d}$.

### C.5 Proof of Lemma 7

Recall the Taylor expansion of $\zeta_{t,t-1}$ from ([21](#A1.E21)) and that $P_{t-1|t}=\mathcal{N}(x_{t-1};\mu_{t},\sigma_{t}^{2}I_{d})$. In other words, under $P_{t-1|t}$, given $x_{t}$, $X_{t-1}^{i}$ is conditionally independent of $X_{t-1}^{j}$ for any $1\leq i,j\leq d$ with $i\neq j$. Thus, using the result of (conditionally independent) Gaussian moments, we obtain

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[(X_{t-1}^{i}-\mu_{t}^{i})^{p}]$ $\displaystyle=0,\quad\text{$\forall p\geq 1$ odd}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[(X_{t-1}^{i}-\mu_{t}^{i})^{2}]$ $\displaystyle=\frac{1-\alpha_{t}}{\alpha_{t}}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[(X_{t-1}^{i}-\mu_{t}^{i})^{4}]$ $\displaystyle=3\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[(X_{t-1}^{i}-\mu_{t}^{i})^{2}(X_{t-1}^{j}-\mu_{t}^{j})^{2}]$ $\displaystyle=\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\prod_{\begin{subarray}{c}i\in\bm{a}\\ \text{s.t. }\bm{a}\in[d]^{p}\end{subarray}}(X_{t-1}^{i}-\mu_{t}^{i})\right]$ $\displaystyle\asymp\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{p/2},\quad\text{$\forall p\geq 2$ even}.$ (33) $$

Therefore, following from ([21](#A1.E21)), for any fixed $x_{t}$,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\frac{1}{2}\sum_{i=1}^{d}\partial^{2}_{ii}\log q_{t-1}(\mu_{t})\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}(X_{t-1}^{i}-\mu_{t}^{i})^{2}$ $\displaystyle\qquad+\frac{1}{4!}\sum_{i=1}^{d}\partial^{4}_{iiii}\log q_{t-1}(\mu_{t})\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}(X_{t-1}^{i}-\mu_{t}^{i})^{4}$ $\displaystyle\qquad+\frac{1}{4!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{4}_{iijj}\log q_{t-1}(\mu_{t})\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}(X_{t-1}^{i}-\mu_{t}^{i})^{2}(X_{t-1}^{j}-\mu_{t}^{j})^{2}$ $\displaystyle\qquad+\sum_{p=6}^{\infty}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[T_{p}(\log q_{t-1},X_{t-1},\mu_{t})]$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\frac{1}{2}\sum_{i=1}^{d}\partial^{2}_{ii}\log q_{t-1}(\mu_{t})\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)+\frac{1}{8}\sum_{i=1}^{d}\partial^{4}_{iiii}\log q_{t-1}(\mu_{t})\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}$ $\displaystyle\qquad+\frac{1}{4!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{4}_{iijj}\log q_{t-1}(\mu_{t})\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+O\left(\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}\right).$ $$

Here $(ii)$ follows from [Assumption 4](#Thmassumption4) and because $\left|\partial^{k}_{\bm{a}}\log q_{t-1}(x)\right|(\forall\bm{a}:\left|\bm{a}\right|=k)$ admits an upper bound $U_{k}(x)$ which is independent of $T$ for all $k\geq 1$ and $x\in\mathbb{R}^{d}$.
We now explain step $(i)$ below. Since $\log q_{t-1}$ is analytic, all its partial derivatives exist. It remains to show that the infinite sum and the expectation can be exchanged. This follows because

$$ $\displaystyle\lim_{k\to\infty}\left|\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\sum_{p=2}^{k}T_{p}(\log q_{t-1},X_{t-1},\mu_{t})\right]\right|$ $\displaystyle\leq\lim_{k\to\infty}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left|\zeta_{t,t-1}-\sum_{p=2}^{k}T_{p}(\log q_{t-1},X_{t-1},\mu_{t})\right|$ $\displaystyle\leq\lim_{k\to\infty}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\sum_{p=k+1}^{\infty}\left|T_{p}(\log q_{t-1},X_{t-1},\mu_{t})\right|\right]$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\lim_{k\to\infty}\sum_{p=k+1}^{\infty}\left|T_{p}(\log q_{t-1},X_{t-1},\mu_{t})\right|\right]\stackrel{{\scriptstyle(iv)}}{{=}}0.$ $$

Here both $(iii)$ and $(iv)$ are because of the following. Under [Assumption 4](#Thmassumption4), for any $x_{t-1}$, $\mu_{t}$, and $T$, the infinite sum is convergent:

$$ $\sum_{p=1}^{\infty}\left|T_{p}(\log q_{t-1},x_{t-1},\mu_{t})\right|\leq\sum_{p=1}^{\infty}\frac{d^{p}}{p!}U_{p}(\mu_{t})\left|x_{t-1}-\mu_{t}\right|^{p}<\infty.$ $$

Thus, $(iv)$ directly follows. This also implies that for any $\epsilon>0$ there exists some large enough $k_{\epsilon}$ such that

$$ $\sum_{p=1}^{\infty}\left|T_{p}(\log q_{t-1},x_{t-1},\mu_{t})\right|\leq\sum_{p=1}^{k_{\epsilon}}\left|T_{p}(\log q_{t-1},x_{t-1},\mu_{t})\right|+\epsilon.$ $$

This shows that $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\sum_{p=1}^{\infty}\left|T_{p}(\log q_{t-1},x_{t-1},\mu_{t})\right|\right]<\infty$ since all centralized moments exist under Gaussian distribution. Thus, $(iii)$ immediately follows from the dominated convergence theorem, and $(i)$ is established.
The proof is complete since $\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{p}=O\left((1-\alpha_{t})^{p}\right),\leavevmode\nobreak\ \forall p\geq 1$ by ([22](#A1.E22)).

### C.6 Proof of Lemma 8

Note that $q_{t|t-1}(x_{t}|x_{t-1})=\frac{1}{(2\pi(1-\alpha_{t}))^{d/2}}\exp\left(-\frac{\left\lVert x_{t}-\sqrt{\alpha_{t}}x_{t-1}\right\rVert^{2}}{2(1-\alpha_{t})}\right)$. Following from the idea of Tweedie [], we have

$$ $\displaystyle q_{t-1|t}(x_{t-1}|x_{t})$ $\displaystyle=\frac{q_{t-1}(x_{t-1})}{q_{t}(x_{t})}q_{t|t-1}(x_{t}|x_{t-1})$ $\displaystyle=\frac{q_{t-1}(x_{t-1})}{q_{t}(x_{t})}q_{t|t-1}(x_{t}|0)\exp\left(\frac{\sqrt{\alpha_{t}}}{1-\alpha_{t}}x_{t}^{\intercal}x_{t-1}-\frac{\alpha_{t}}{2(1-\alpha_{t})}\left\lVert x_{t-1}\right\rVert^{2}\right)$ $\displaystyle=\left(q_{t-1}(x_{t-1})e^{-\frac{\alpha_{t}}{2(1-\alpha_{t})}\left\lVert x_{t-1}\right\rVert^{2}}\right)\exp\left(\frac{\sqrt{\alpha_{t}}}{1-\alpha_{t}}x_{t}^{\intercal}x_{t-1}-\log q_{t}(x_{t})+\log q_{t|t-1}(x_{t}|0)\right)$ $\displaystyle=:f(x_{t-1})\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)$ (34) $$

where we have used the definitions of $\tilde{x}_{t}$ and $\kappa(\tilde{x}_{t})$ in ([23](#A1.E23)). This shows that $x_{t-1}$ is a conditional exponential family given $\tilde{x}_{t}$.
Thus, the first moment can be found as (cf. Prop. 11.1 in [])

$$ $\displaystyle 0$ $\displaystyle=\nabla_{\tilde{x}_{t}}\int q_{t-1|t}(x_{t-1}|x_{t})\mathrm{d}x_{t-1}=\nabla_{\tilde{x}_{t}}\int f(x_{t-1})\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)\mathrm{d}x_{t-1}$ $\displaystyle=\int f(x_{t-1})\nabla_{\tilde{x}_{t}}\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)\mathrm{d}x_{t-1}$ $\displaystyle=\int f(x_{t-1})\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)\left(x_{t-1}-\nabla_{\tilde{x}_{t}}\kappa(\tilde{x}_{t})\right)\mathrm{d}x_{t-1}$ $\displaystyle=\int f(x_{t-1})\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)x_{t-1}\mathrm{d}x_{t-1}-\nabla_{\tilde{x}_{t}}\kappa(\tilde{x}_{t})$ $$

which implies that

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[X_{t-1}\right]=\nabla\kappa.$ (35) $$

For the second moment,

$$ $\displaystyle 0$ $\displaystyle=\partial^{2}_{ij}\int q_{t-1|t}(x_{t-1}|x_{t})\mathrm{d}x_{t-1}$ $\displaystyle=\int f(x_{t-1})\frac{\partial}{\partial\tilde{x}_{t}^{j}}\Big{(}\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)\left(x_{t-1}^{i}-\partial_{i}\kappa(\tilde{x}_{t})\right)\Big{)}\mathrm{d}x_{t-1}$ $\displaystyle=\int f(x_{t-1})\exp\left(x_{t-1}^{\intercal}\tilde{x}_{t}-\kappa(\tilde{x}_{t})\right)\left((x_{t-1}^{i}-\partial_{i}\kappa(\tilde{x}_{t}))(x_{t-1}^{j}-\partial_{j}\kappa(\tilde{x}_{t}))-\partial^{2}_{ij}\kappa(\tilde{x}_{t})\right)\mathrm{d}x_{t-1}$ $$

which yields

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}-\mu_{t})(X_{t-1}-\mu_{t})^{\intercal}\right]=\nabla^{2}\kappa=\frac{1-\alpha_{t}}{\alpha_{t}}I_{d}+\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\nabla^{2}\log q_{t}(x_{t}).$ (36) $$

Below, we write $x=x_{t-1}$ and $\kappa=\kappa(\tilde{x}_{t})$ for brevity. We remind readers that all derivatives are w.r.t. $\tilde{x}_{t}$ instead of $x=x_{t-1}$. For the third moment,

$$ $0=\partial^{3}_{ijk}\int q_{t-1|t}\mathrm{d}x=:\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{3}(x,\tilde{x}_{t})\mathrm{d}x$ $$

where

$$ $\displaystyle D_{3}(x,\tilde{x}_{t})$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{k}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)\left((x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)-\partial^{2}_{ij}\kappa\right)\Big{)}$ $\displaystyle=(x^{k}-\partial_{k}\kappa)\left((x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)-\partial^{2}_{ij}\kappa\right)$ $\displaystyle\qquad+(-\partial^{2}_{ik}\kappa)(x^{j}-\partial_{j}\kappa)+(-\partial^{2}_{jk}\kappa)(x^{i}-\partial_{i}\kappa)-\partial^{3}_{ijk}\kappa.$ (37) $$

Now, for any function $\mathrm{fn}(\tilde{x}_{t})$ and $1\leq i\leq d$,

$$ $\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)\mathrm{fn}(\tilde{x}_{t})(x^{i}-\partial_{i}\kappa)\mathrm{d}x=0$ $$

by the first moment result ([35](#A3.E35)). Thus, we get

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})(X_{t-1}^{k}-\mu_{t}^{k})\right]=\partial^{3}_{ijk}\kappa.$ $$

For the fourth moment, we have

$$ $0=\partial^{4}_{ijkl}\int q_{t-1|t}\mathrm{d}x=:\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{4}(x,\tilde{x}_{t})\mathrm{d}x$ $$

where

$$ $\displaystyle D_{4}(x,\tilde{x}_{t})$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{l}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)((x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)$ $\displaystyle\qquad-\partial^{2}_{ij}\kappa(x^{k}-\partial_{k}\kappa)-\partial^{2}_{ik}\kappa(x^{j}-\partial_{j}\kappa)-\partial^{2}_{jk}\kappa(x^{i}-\partial_{i}\kappa)-\partial^{3}_{ijk}\kappa)\Big{)}$ $\displaystyle=(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)(x^{l}-\partial_{l}\kappa)+\partial_{l}\Big{(}(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)\Big{)}$ $\displaystyle\qquad-\partial^{2}_{ij}\kappa(x^{k}-\partial_{k}\kappa)(x^{l}-\partial_{l}\kappa)-\partial^{3}_{ijl}\kappa(x^{k}-\partial_{k}\kappa)+\partial^{2}_{ij}\kappa\partial^{2}_{kl}\kappa$ $\displaystyle\qquad-\partial^{2}_{ik}\kappa(x^{j}-\partial_{j}\kappa)(x^{l}-\partial_{l}\kappa)-\partial^{3}_{ikl}\kappa(x^{j}-\partial_{j}\kappa)+\partial^{2}_{ik}\kappa\partial^{2}_{jl}\kappa$ $\displaystyle\qquad-\partial^{2}_{jk}\kappa(x^{i}-\partial_{i}\kappa)(x^{l}-\partial_{l}\kappa)-\partial^{3}_{jkl}\kappa(x^{i}-\partial_{i}\kappa)+\partial^{2}_{jk}\kappa\partial^{2}_{il}\kappa$ $\displaystyle\qquad-\partial^{3}_{ijk}\kappa(x^{l}-\partial_{l}\kappa)-\partial^{4}_{ijkl}\kappa$ (38) $$

and

$$ $\displaystyle\partial_{l}\Big{(}(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)\Big{)}$ $\displaystyle=-\partial^{2}_{il}\kappa(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)-\partial^{2}_{jl}\kappa(x^{i}-\partial_{i}\kappa)(x^{k}-\partial_{k}\kappa)-\partial^{2}_{kl}\kappa(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa).$ $$

Using the first and second moment results in ([35](#A3.E35)) and ([36](#A3.E36)), we get

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})(X_{t-1}^{k}-\mu_{t}^{k})(X_{t-1}^{l}-\mu_{t}^{l})\right]=\\ (\partial^{2}_{ij}\kappa)(\partial^{2}_{kl}\kappa)+(\partial^{2}_{ik}\kappa)(\partial^{2}_{jl}\kappa)+(\partial^{2}_{il}\kappa)(\partial^{2}_{jk}\kappa)+\partial^{4}_{ijkl}\kappa.$ $$

And the fourth moment result follows directly by applying ([36](#A3.E36)) to each of the terms.

### C.7 Proof of Lemma 9

We first introduce some notations used in the proof. We write $x=x_{t-1}$ and $\kappa=\kappa(\tilde{x}_{t})$. Given a set of indices $A$, define its bipartition as

$$ $\mathrm{bipart}(A):=\{(B,C):A=B\sqcup C\}$ $$

where $B$ and $C$ are both sets of indices (and therefore the order of indices within each of $B$ and $C$ does not matter). Here $\sqcup$ refers to the disjoint union of the two sets (which is only defined when the two sets are disjoint). Next, given a set $B$, define $\mathrm{allpart_{\geq 2}}(B)$ as a set containing all partitions of $B$ such that there are at least 2 elements in each part of the partition. As an example, $\mathrm{allpart_{\geq 2}(\{1,2,3,4\})}=\{\{\{1,2\},\{3,4\}\},\{\{1,3\},\{2,4\}\},\{\{1,4\},\{2,3\}\}$, and $\{\{1\},\{2,3,4\}\}\notin\mathrm{allpart_{\geq 2}(\{1,2,3,4\})}$ despite the fact that it is a valid partition.
For each partition $b\in\mathrm{allpart_{\geq 2}}(B)$, define

$$ $\bm{\partial}_{b}\kappa:=\prod_{\xi\in b}\partial^{\left|\xi\right|}_{a_{\xi}}\kappa.$ $$

Here note that $\xi$ is also a set, and $\bm{\partial}_{b}\kappa$ is well defined since the order of indices to take partial derivative with does not matter.
Define

$$ $\displaystyle D_{0}(x,\tilde{x}_{t})$ $\displaystyle:=1$ $\displaystyle D_{p}(x,\tilde{x}_{t})$ $\displaystyle:=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{a_{p}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{p-1}(x,\tilde{x}_{t})\Big{)}$ $$

for all $p\geq 1$.
We again remind readers that all derivatives are w.r.t. $\tilde{x}_{t}$ instead of $x=x_{t-1}$.

By working out the derivative, a direct implication of the definition of $D_{p}$ is a recursive relationship:

$$ $D_{p}(x,\tilde{x}_{t})=(x^{a_{p}}-\partial_{a_{p}}\kappa)D_{p-1}(x,\tilde{x}_{t})+\partial_{a_{p}}D_{p-1}(x,\tilde{x}_{t}).$ $$

Also, if we unroll the recursion of $D_{p}$, we get

$$ $\displaystyle D_{p}(x,\tilde{x}_{t})$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{a_{p}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{p-1}(x,\tilde{x}_{t})\Big{)}$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{a_{p}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{a_{p-1}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{p-2}(x,\tilde{x}_{t})\Big{)}\Big{)}$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial^{2}_{a_{p},a_{p-1}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{p-2}(x,\tilde{x}_{t})\Big{)}$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial^{p}_{a_{p},\dots,a_{1}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)\Big{)}$ $$

and thus

$$ $\displaystyle 0=\partial^{p}_{a_{1},\dots,a_{p}}\int q_{t-1|t}\mathrm{d}x$ $\displaystyle=\int f(x)\partial^{p}_{a_{1},\dots,a_{p}}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)\Big{)}\mathrm{d}x$ $\displaystyle=\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{p}(x,\tilde{x}_{t})\mathrm{d}x$ (39) $$

where we recall the definition of $f(x)$ back in ([C.6](#A3.Ex155)).

In the following, we present the entire proof into two parts. In part 1, we inductively show that each $D_{p}(x,\tilde{x}_{t})$ satisfies a particular polynomial form. In part 2, we inductively show that this polynomial form results in the desired rates.

Part 1 of the proof of Lemma [9](#Thmlemma9):
The first step toward proving the desired results is to obtain the form of $D_{p}$ for all $p\geq 2$. Now, we aim to show inductively that

$$ $D_{p}(x,\tilde{x}_{t})=\prod_{i=1}^{p}(x^{a_{i}}-\partial_{a_{i}}\kappa)-\sum_{(B,C)\in\mathrm{bipart}([p])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{p}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)$ (40) $$

where $d_{p}(b,C)$ is a constant from combinatorics, which is possibly 0 and which only depends on $p$. From Lemma [8](#Thmlemma8), the bases cases have been established that (cf. ([C.6](#A3.Ex166)) and ([C.6](#A3.Ex171)))

$$ $\displaystyle D_{2}(x,\tilde{x}_{t})=(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)-\partial^{2}_{ij}\kappa$ $\displaystyle D_{3}(x,\tilde{x}_{t})=(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)$ $\displaystyle\quad-\partial^{2}_{ij}\kappa(x^{k}-\partial_{k}\kappa)-\partial^{2}_{ik}\kappa(x^{j}-\partial_{j}\kappa)-\partial^{2}_{jk}\kappa(x^{i}-\partial_{i}\kappa)-\partial^{3}_{ijk}\kappa$ $\displaystyle D_{4}(x,\tilde{x}_{t})=(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)(x^{l}-\partial_{l}\kappa)$ $\displaystyle\quad-\partial^{2}_{ij}\kappa(x^{k}-\partial_{k}\kappa)(x^{l}-\partial_{l}\kappa)-\partial^{2}_{ik}\kappa(x^{j}-\partial_{j}\kappa)(x^{l}-\partial_{l}\kappa)-\partial^{2}_{jk}\kappa(x^{i}-\partial_{i}\kappa)(x^{l}-\partial_{l}\kappa)$ $\displaystyle\quad+\partial_{l}((x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa))-\partial^{3}_{ijk}\kappa(x^{l}-\partial_{l}\kappa)-\partial^{3}_{ijl}(x^{k}-\partial_{k}\kappa)$ $\displaystyle\quad-\partial^{3}_{ikl}(x^{j}-\partial_{j}\kappa)-\partial^{3}_{jkl}(x^{i}-\partial_{i}\kappa)+\partial^{2}_{ij}\kappa\partial^{2}_{kl}\kappa+\partial^{2}_{ik}\kappa\partial^{2}_{jl}\kappa+\partial^{2}_{jk}\kappa\partial^{2}_{il}\kappa-\partial^{4}_{ijkl}\kappa.$ $$

In particular, each term of $D_{p}$ ($p=2,3,4$) is in the form of either $\prod_{i=1}^{p}(x^{a_{i}}-\partial_{a_{i}}\kappa)$ or $(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)$, where $\left|\xi\right|\geq 2,\leavevmode\nobreak\ \forall\xi\in b$, and $(\sqcup_{\xi\in b}\xi)\sqcup C=[p]$. Therefore, $D_{2},D_{3},D_{4}$ all satisfy the hypothesis ([40](#A3.E40)).

Turning to the inductive step, we suppose that $D_{k}$ satisfies ([40](#A3.E40)), i.e.,

$$ $D_{k}(x,\tilde{x}_{t})=\prod_{i=1}^{k}(x^{a_{i}}-\partial_{a_{i}}\kappa)-\sum_{(B,C)\in\mathrm{bipart}([k])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{k}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa).$ $$

Then, using the recursive relationship, we have

$$ $\displaystyle D_{k+1}(x,\tilde{x}_{t})$ $\displaystyle=(x^{a_{k+1}}-\partial_{a_{k+1}}\kappa)D_{k}(x,\tilde{x}_{t})+\partial_{a_{k+1}}D_{k}(x,\tilde{x}_{t})$ $\displaystyle=\underbrace{\prod_{i=1}^{k+1}(x^{a_{i}}-\partial_{a_{i}}\kappa)}_{T_{1}}-\underbrace{\sum_{(B,C)\in\mathrm{bipart}([k])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{k}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)(x^{a_{k+1}}-\partial_{a_{k+1}}\kappa)}_{T_{2}}$ $\displaystyle\quad-\underbrace{\partial_{a_{k+1}}\left(-\prod_{i=1}^{k}(x^{a_{i}}-\partial_{a_{i}}\kappa)\right)}_{T_{3}}-\underbrace{\sum_{(B,C)\in\mathrm{bipart}([k])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{k}(b,C)(\bm{\partial}_{b}\kappa)\left(\partial_{a_{k+1}}\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)\right)}_{T_{4}}$ $\displaystyle\quad-\underbrace{\sum_{(B,C)\in\mathrm{bipart}([k])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{k}(b,C)\left(\partial_{a_{k+1}}(\bm{\partial}_{b}\kappa)\right)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)}_{T_{5}}$ $\displaystyle=T_{1}-T_{2}-T_{3}-T_{4}-T_{5}$ $$

where we define each term as $T_{1},\dots,T_{5}$.
Now we discuss these terms separately:

- 1.
$T_{1}$ (and only $T_{1}$) is in the form $\prod_{i=1}^{k+1}(x^{a_{i}}-\partial_{a_{i}}\kappa)$.
- 2.
$T_{2}$ is a summation of individual terms: $(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)(x^{a_{k+1}}-\partial_{a_{k+1}}\kappa)$. Here $b\in\mathrm{allpart}_{\geq 2}(B)$ and $(B,C)\in\mathrm{bipart}([k])$. Thus, by definition of $\mathrm{bipart}$ and $\mathrm{allpart}_{\geq 2}$, for each $\xi\in b$, $\left|\xi\right|\geq 2$ and $(\sqcup_{\xi\in b}\xi)\sqcup C=[k]$. Therefore, $k+1\notin B\sqcup C$ and
$(\sqcup_{\xi\in b}\xi)\sqcup C\sqcup\{k+1\}=[k]\sqcup\{k+1\}=[k+1].$
This implies that each individual term of $T_{2}$ is in the form of $(\bm{\partial}_{b}\kappa)\prod_{c\in C_{2}}(x^{c}-\partial_{c}\kappa)$ where $b\in\mathrm{allpart}_{\geq 2}(B_{2})$, such that $B_{2}:=B$ and $C_{2}:=C\sqcup\{k+1\}$. Here $C_{2}$ is well defined because $k+1\notin C$. Since $(B_{2},C_{2})\in\mathrm{bipart}([k+1])$,
$T_{2}=\sum_{(B,C)\in\mathrm{bipart}([k+1])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{2}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)$
for some constant $d_{2}(b,C)$.
- 3.
$T_{3}$ is the derivative of product, which is a summation of individual terms: $(\partial^{2}_{a_{j},a_{k+1}}\kappa)\prod_{\begin{subarray}{c}i=1\\
i\neq j\end{subarray}}^{d}(x^{a_{i}}-\partial_{a_{i}}\kappa),\leavevmode\nobreak\ j=1,\dots,k$. Therefore, for each $j=1,\dots,k$, each term is of the form $(\bm{\partial}_{b}\kappa)\prod_{c\in C_{3}}(x^{a_{c}}-\partial_{a_{c}}\kappa)$ where $b\in\mathrm{allpart}_{\geq 2}(B_{3})$, such that $B_{3}:=\{j,k+1\}$ and $C_{3}:=[k]\setminus\{j\}$. Since $(B_{3},C_{3})\in\mathrm{bipart}([k+1])$,
$T_{3}=\sum_{(B,C)\in\mathrm{bipart}([k+1])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{3}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)$
for some constant $d_{3}(b,C)$.
- 4.
$T_{4}$ is a summation of individual terms: $(\bm{\partial}_{b}\kappa)\left(\partial_{a_{k+1}}\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)\right)$ where $b\in\mathrm{allpart}_{\geq 2}(B)$ and $(B,C)\in\mathrm{bipart}([k])$. Now,
$\displaystyle(\bm{\partial}_{b}\kappa)\left(\partial_{a_{k+1}}\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)\right)$
$\displaystyle=-(\bm{\partial}_{b}\kappa)(\partial^{2}_{a_{j},a_{k+1}}\kappa)\prod_{\begin{subarray}{c}i\in C\\
i\neq c\end{subarray}}(x^{a_{i}}-\partial_{a_{i}}\kappa)$
$\displaystyle=-(\bm{\partial}_{b_{4}}\kappa)\prod_{i\in C_{4}}(x^{a_{i}}-\partial_{a_{i}}\kappa)$
where $b_{4}:=b\sqcup\{k+1,c\}$ and $C_{4}:=C\setminus\{c\}$. Here $b_{4}$ is well defined because $k+1,c\notin b$. Define $B_{4}:=[k+1]\setminus C_{4}$, and we have $b_{4}\in\mathrm{allpart}_{\geq 2}(B_{4})$. Since $(B_{4},C_{4})$ is a valid partition of $[k+1]$, we have
$T_{4}=\sum_{(B,C)\in\mathrm{bipart}([k+1])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{4}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)$
for some constant $d_{4}(b,C)$.
- 5.
$T_{5}$ is a summation of individual terms: $\left(\partial_{a_{k+1}}(\bm{\partial}_{b}\kappa)\right)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa)$, where $b\in\mathrm{allpart}_{\geq 2}(B)$ and $(B,C)\in\mathrm{bipart}([k])$. From definition of $\bm{\partial}_{b}\kappa$,
$\partial_{a_{k+1}}(\bm{\partial}_{b}\kappa)=\partial_{a_{k+1}}\left(\prod_{\xi\in b}\partial^{\left|\xi\right|}_{a_{\xi}}\kappa\right)=\sum_{\xi\in b}\left(\partial^{\left|\xi\right|+1}_{a_{\xi},a_{k+1}}\kappa\right)\prod_{\begin{subarray}{c}\zeta\in b\\
\zeta\neq\xi\end{subarray}}\partial^{\left|\zeta\right|}_{a_{\zeta}}\kappa=\sum_{\xi\in b}\bm{\partial}_{b_{\xi}}\kappa$
where, for each $\xi\in b$, we have defined a new partition $b_{\xi}$ such that $k+1$ is added to the $\xi$ in the partition $b$. Formally, define $b_{\xi}:=b\setminus\xi\sqcup\{\xi\sqcup\{k+1\}\}$, which is well defined because $\xi\notin(b\setminus\xi)$ and $k+1\notin B$. Define $B_{5}:=B\sqcup\{k+1\}$ and $C_{5}:=C$, and note that $(B_{5},C_{5})$ is a valid partition of $[k+1]$. Since $\left|\zeta\right|\geq 2,\leavevmode\nobreak\ \forall\zeta\in b$, we have $\left|\zeta^{\prime}\right|\geq 2,\leavevmode\nobreak\ \forall\zeta^{\prime}\in b_{\xi}$. Since $b\in\mathrm{allpart}_{\geq 2}(B)$, we have $b_{\xi}\in\mathrm{allpart}_{\geq 2}(B_{5})$ for all $\xi\in b$. Therefore, for any fixed $C(=C_{5})$
$\displaystyle\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{k}(b,C)\left(\partial_{a_{k+1}}(\bm{\partial}_{b}\kappa)\right)$
$\displaystyle=\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}\sum_{\xi\in b}d_{k}(b,C)\bm{\partial}_{b_{\xi}}\kappa$
$\displaystyle=\sum_{b_{5}\in\mathrm{allpart}_{\geq 2}(B_{5})}d_{5}(b_{5},C)\bm{\partial}_{b_{5}}\kappa$
for some constant $d_{5}(b_{5},C)$, and thus
$T_{5}=\sum_{(B,C)\in\mathrm{bipart}([k+1])}\sum_{b\in\mathrm{allpart}_{\geq 2}(B)}d_{5}(b,C)(\bm{\partial}_{b}\kappa)\prod_{c\in C}(x^{a_{c}}-\partial_{a_{c}}\kappa).$

Finally, letting

$$ $d_{k+1}(b,C):=\sum_{j=2}^{5}d_{j}(b,C)$ $$

for each $b\in\mathrm{allpart}_{\geq 2}(B)$ and $C$ such that $(B,C)\in\mathrm{bipart}([k+1])$, we have shown that if $D_{k}(x,\tilde{x}_{t})$ is in the form of ([40](#A3.E40)), $D_{k+1}(x,\tilde{x}_{t})$ is also in this form. Thus, claim ([40](#A3.E40)) is valid for all $p\geq 2$.

Part 2 of the proof of Lemma [9](#Thmlemma9):
First, we remind readers of the definition of $\kappa(\tilde{x}_{t})$ in ([23](#A1.E23)). Note that $\nabla\kappa=\mu_{t}$ from direct differentiation. From ([C.7](#A3.Ex188)) and ([40](#A3.E40)), for fixed $x_{t}$, we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{p}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]$ $\displaystyle=O\left(\sup_{\begin{subarray}{c}(B,C)\in\mathrm{bipart}([p])\\ b\in\mathrm{allpart}_{\geq 2}(B)\end{subarray}}\bm{\partial}_{b}\kappa(\tilde{x}_{t})\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{c\in C}(X_{t-1}^{a_{c}}-\mu_{t}^{a_{c}})\right]\right)$ $\displaystyle=O\left(\sup_{(B,C)\in\mathrm{bipart}([p])}\left(\sup_{b\in\mathrm{allpart}_{\geq 2}(B)}\bm{\partial}_{b}\kappa(\tilde{x}_{t})\right)\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{c\in C}(X_{t-1}^{a_{c}}-\mu_{t}^{a_{c}})\right]\right).$ (41) $$

We first consider the term $\sup_{b\in\mathrm{allpart}_{\geq 2}(B)}\bm{\partial}_{b}\kappa(\tilde{x}_{t})$. Given a partition $b\in\mathrm{allpart}_{\geq 2}(B)$, direct differentiation yields

$$ $\displaystyle\partial^{\left|\xi\right|}_{a_{\xi}}\kappa$ $\displaystyle=\frac{1-\alpha_{t}}{\alpha_{t}}+\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\partial^{2}_{a_{\xi}}\log q_{t}(x_{t})=O(1-\alpha_{t}),\quad\text{if }\left|\xi\right|=2\text{ and }\xi_{1}=\xi_{2}$ $\displaystyle\partial^{\left|\xi\right|}_{a_{\xi}}\kappa$ $\displaystyle=\frac{(1-\alpha_{t})^{\left|\xi\right|}}{\alpha_{t}^{\left|\xi\right|/2}}\partial^{\left|\xi\right|}_{\xi}\log q_{t}(x_{t})=O((1-\alpha_{t})^{\left|\xi\right|}),\quad\text{for all other }\xi.$ $$

Since by definition $\bm{\partial}_{b}\kappa=\prod_{\xi\in b}\partial^{\left|\xi\right|}_{a_{\xi}}\kappa$ and $\sqcup_{\xi\in b}\xi=B$, the slowest rate of $\bm{\partial}_{b}\kappa$ (as a function of $B$) is determined by the partition $b$ containing the most number of equal pairs. The slowest rate is

$$ $\sup_{b\in\mathrm{allpart}_{\geq 2}(B)}\bm{\partial}_{b}\kappa(\tilde{x}_{t})=\begin{cases}O\left((1-\alpha_{t})^{(\left|B\right|-1)/2}(1-\alpha_{t})^{3}\right)=O\left((1-\alpha_{t})^{(\left|B\right|+5)/2}\right)&\text{if }\left|B\right|\text{is odd}\\ O\left((1-\alpha_{t})^{\left|B\right|/2}\right)&\text{if }\left|B\right|\text{is even}\end{cases}$ $$

To proceed, we will again use induction to find the overall rate. From Lemma [8](#Thmlemma8), base cases have been established that

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{2}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]$ $\displaystyle=O\left(1-\alpha_{t}\right),\leavevmode\nobreak\ \forall a\in[d]^{2}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{3}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]$ $\displaystyle=O\left((1-\alpha_{t})^{3}\right),\leavevmode\nobreak\ \forall a\in[d]^{3}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{4}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]$ $\displaystyle=O\left((1-\alpha_{t})^{2}\right),\leavevmode\nobreak\ \forall a\in[d]^{4}.$ $$

These rates satisfy ([24](#A1.E24)) and ([25](#A1.E25)) when $p=2,3,4$. Now we turn to the inductive step. Suppose $k\geq 4$ is even. For purpose of induction, suppose ([24](#A1.E24)) and ([25](#A1.E25)) hold for all $p=2,\dots,k$. Then, following ([C.7](#A3.Ex214)), for $p=k+1$ (odd number), we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{k+1}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]$ $\displaystyle=O\left(\sup_{\begin{subarray}{c}(B,C)\in\mathrm{bipart}([k+1])\\ \left|B\right|\text{ odd},\leavevmode\nobreak\ \left|C\right|\text{ even}\end{subarray}}(1-\alpha_{t})^{(\left|B\right|+5)/2}(1-\alpha_{t})^{\left|C\right|/2}+\sup_{\begin{subarray}{c}(B,C)\in\mathrm{bipart}([k+1])\\ \left|B\right|\text{ even},\leavevmode\nobreak\ \left|C\right|\text{ odd}\end{subarray}}(1-\alpha_{t})^{\left|B\right|/2}(1-\alpha_{t})^{(\left|C\right|+3)/2}\right)$ $\displaystyle=O\left((1-\alpha_{t})^{(k+1)/2+5/2}+(1-\alpha_{t})^{(k+1)/2+3/2}\right)$ $\displaystyle=O\left((1-\alpha_{t})^{(k+1)/2+3/2}\right).$ $$

Then, for $p=k+2$ (even number), we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i=1}^{k+2}(X_{t-1}^{a_{i}}-\mu_{t}^{a_{i}})\right]$ $\displaystyle=O\left(\sup_{\begin{subarray}{c}(B,C)\in\mathrm{bipart}([k+2])\\ \left|B\right|\text{ odd},\leavevmode\nobreak\ \left|C\right|\text{ odd}\end{subarray}}(1-\alpha_{t})^{(\left|B\right|+5)/2}(1-\alpha_{t})^{(\left|C\right|+3)/2}+\sup_{\begin{subarray}{c}(B,C)\in\mathrm{bipart}([k+1])\\ \left|B\right|\text{ even},\leavevmode\nobreak\ \left|C\right|\text{even}\end{subarray}}(1-\alpha_{t})^{\left|B\right|/2}(1-\alpha_{t})^{\left|C\right|/2}\right)$ $\displaystyle=O\left((1-\alpha_{t})^{(k+2)/2+4}+(1-\alpha_{t})^{(k+2)/2}\right)$ $\displaystyle=O\left((1-\alpha_{t})^{(k+2)/2}\right).$ $$

These show the validity of the claims ([24](#A1.E24)) and ([25](#A1.E25)). The proof is now complete.

### C.8 Proof of Lemma 11

Fix $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$. We write $g(x_{t})=\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})$ for brevity, and from ([6](#S3.E6)) we have $\left\lVert g(x_{t})\right\rVert^{2}=O(1-\alpha_{t})$ for any fixed $x_{t}$. The exchange of the infinite sum and the expectation is again guaranteed by [Assumption 4](#Thmassumption4). From ([21](#A1.E21)) and ([C.5](#A3.Ex139)), it can be verified that the expected second moment is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{2}]=\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\left(T_{1}+T_{2}+T_{3}\right)^{2}\right]+O(\sigma_{t}^{6})$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{1}^{2}\right]+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{1}T_{3}\right]+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{2}^{2}\right]+O(\sigma_{t}^{6})$ $\displaystyle=\sigma_{t}^{2}\left\lVert g(x_{t})\right\rVert^{2}+3\sigma_{t}^{4}\sum_{i=1}^{d}\left(\frac{1}{4}\left(\partial^{2}_{ii}\log q_{t-1}(\mu_{t})\right)^{2}+\frac{1}{3}g^{i}(x_{t})\partial^{3}_{iii}\log q_{t-1}(\mu_{t})\right)$ $\displaystyle\qquad+\sigma_{t}^{4}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\bigg{(}\frac{1}{4}\left(\partial^{2}_{ii}\log q_{t-1}(\mu_{t})\right)\left(\partial^{2}_{jj}\log q_{t-1}(\mu_{t})\right)+\frac{1}{2}\left(\partial^{2}_{ij}\log q_{t-1}(\mu_{t})\right)^{2}$ $\displaystyle\qquad+g^{i}(x_{t})\partial^{3}_{ijj}\log q_{t-1}(\mu_{t})\bigg{)}+O(\sigma_{t}^{6}).$ (42) $$

Since $\sigma_{t}^{2}=(1-\alpha_{t})/\alpha_{t}$ and since all partial derivatives are well controlled according to [Assumption 4](#Thmassumption4), we immediately get $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{2}]=O\left((1-\alpha_{t})^{2}\right)$ for the expected second moment.

The expected third moment is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{3}]=\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\left(T_{1}+T_{2}+T_{3}\right)^{3}\right]+O(\sigma_{t}^{6})$ $\displaystyle=3\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{1}^{2}T_{2}\right]+O(\sigma_{t}^{6})$ $\displaystyle=3\sigma_{t}^{4}\sum_{i=1}^{d}\frac{3}{2}(g^{i}(x_{t}))^{2}(\partial^{2}_{ii}\log q_{t-1}(\mu_{t}))+3\sigma_{t}^{4}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}(g^{i}(x_{t}))(g^{j}(x_{t}))(\partial^{2}_{ij}\log q_{t-1}(\mu_{t}))$ $\displaystyle\qquad+\sigma_{t}^{4}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\frac{3}{2}(g^{i}(x_{t}))^{2}(\partial^{2}_{jj}\log q_{t-1}(\mu_{t}))+O(\sigma_{t}^{6}).$ (43) $$

Now, by Cauchy-Schwartz inequality we get

$$ $\displaystyle\left|\sum_{i=1}^{d}(g^{i}(x_{t}))^{2}(\partial^{2}_{ii}\log q_{t-1}(\mu_{t}))\right|\leq\left\lVert g(x_{t})\right\rVert_{4}^{2}\left\lVert\mathrm{diag}(\nabla^{2}\log q_{t-1}(\mu_{t}))\right\rVert$ $\displaystyle\qquad\leq\left\lVert g(x_{t})\right\rVert^{2}\left\lVert\mathrm{diag}(\nabla^{2}\log q_{t-1}(\mu_{t}))\right\rVert=O(1-\alpha_{t})$ $\displaystyle\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}(g^{i}(x_{t}))(g^{j}(x_{t}))(\partial^{2}_{ij}\log q_{t-1}(\mu_{t}))$ $\displaystyle\qquad=\sum_{i,j=1}^{d}(g^{i}(x_{t}))(g^{j}(x_{t}))(\partial^{2}_{ij}\log q_{t-1}(\mu_{t}))-\sum_{i=1}^{d}(g^{i}(x_{t}))^{2}(\partial^{2}_{ii}\log q_{t-1}(\mu_{t}))$ $\displaystyle\qquad\leq\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})\right\rVert\left\lVert g(x_{t})\right\rVert^{2}+O(1-\alpha_{t})=O(1-\alpha_{t})$ $\displaystyle\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}(g^{i}(x_{t}))^{2}(\partial^{2}_{jj}\log q_{t-1}(\mu_{t}))=\sum_{i,j=1}^{d}(g^{i}(x_{t}))^{2}(\partial^{2}_{jj}\log q_{t-1}(\mu_{t}))-\sum_{i=1}^{d}(g^{i}(x_{t}))^{2}(\partial^{2}_{ii}\log q_{t-1}(\mu_{t}))$ $\displaystyle\qquad\leq\left\lVert g(x_{t})\right\rVert^{2}\left|\mathrm{Tr}(\nabla^{2}\log q_{t-1}(\mu_{t}))\right|+O(1-\alpha_{t})=O(1-\alpha_{t})$ $$

where $\mathrm{diag}(A)$ is a vector of the diagonal elements of matrix $A$.
Under [Assumption 4](#Thmassumption4), we obtain $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{3}]=O\left((1-\alpha_{t})^{3}\right)$ for the expected third moment for $t\geq 1$.

### C.9 Proof of Lemma 12

From ([1](#S2.E1)), the forward process at the first step is

$$ $x_{1}=\sqrt{\alpha_{1}}x_{0}+\sqrt{1-\alpha_{1}}w_{1}$ $$

where $w_{1}\sim\mathcal{N}(0,I_{d})$ is independent of $Q_{0}$. Thus,

$$ $\displaystyle\mathbb{E}_{X_{1}\sim Q_{1},X_{0}\sim Q_{0}}\left\lVert X_{1}-X_{0}\right\rVert^{2}$ $\displaystyle=\mathbb{E}_{W_{1}\sim\mathcal{N}(0,I_{d}),X_{0}\sim Q_{0}}\left\lVert\sqrt{1-\alpha_{1}}W_{1}+(\sqrt{\alpha_{t}}-1)X_{0}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathbb{E}_{W_{1}\sim\mathcal{N}(0,I_{d})}\left\lVert\sqrt{1-\alpha_{1}}W_{1}\right\rVert^{2}+\mathbb{E}_{X_{0}\sim Q_{0}}\left\lVert(\sqrt{\alpha_{t}}-1)X_{0}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}(1-\alpha_{1})d+(\sqrt{\alpha_{1}}-1)^{2}M_{2}d$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\leq}}(1-\alpha_{1})(M_{2}+1)d$ $$

where $(i)$ follows from independence, $(ii)$ follows from [Assumption 1](#Thmassumption1), and $(iii)$ follows because $(\sqrt{z}-1)^{2}\leq 1-z$ for all $z\in[0,1]$. The proof is complete since $\mathrm{W}_{2}(Q_{0},Q_{1})^{2}\leq\mathbb{E}_{X_{1}\sim Q_{1},X_{0}\sim Q_{0}}\left\lVert X_{1}-X_{0}\right\rVert^{2}$ by the definition of Wasserstein-2 distance.

## Appendix D Proof of Theorems 2 and 3

In this section, we instantiate [Theorem 1](#Thmtheorem1) (along with its corollaries) to provide upper bounds that have explicit parameter dependence for a number of interesting distribution classes.
In order to obtain an upper bound that explicitly depends on system parameters, we need only to provide an explicit bound on the reverse-step error, which is the main topic that we address in the following subsections.

### D.1 Proof of Theorem 2

We first establish a useful result under the distribution class in ([7](#S4.E7)).
Consider $y=x+\xi u$ where $u\in\mathbb{R}^{d}$ satisfies $\left\lVert u\right\rVert^{2}=1$ and $\xi\in\mathbb{R}$ is some small constant. With some $x^{*}=\gamma x+(1-\gamma)y$ where $\gamma\in(0,1)$, we have

$$ $\nabla\log q_{t}(y)-\nabla\log q_{t}(x)=\nabla^{2}\log q_{t}(x^{*})(y-x)=\xi\nabla^{2}\log q_{t}(x^{*})u$ $$

$$ $\implies\left\lVert\xi\nabla^{2}\log q_{t}(x^{*})u\right\rVert=\left\lVert\nabla\log q_{t}(y)-\nabla\log q_{t}(x)\right\rVert\leq L\left\lVert y-x\right\rVert=L\xi\left\lVert u\right\rVert.$ $$

Thus, taking the limit of $\xi$ to 0, by continuity of $\nabla^{2}\log q_{t}(x)$, we get $\left\lVert\nabla\log q_{t}(x)u\right\rVert\leq L\left\lVert u\right\rVert$. Since $u$ is an arbitrary unit vector, this implies that all absolute eigenvalues of $\nabla^{2}\log q_{t}(x)$ are upper-bounded by $L$.
Since $\nabla^{2}\log q_{t}(x)$ is real symmetric, this means that all the singular values of $\nabla^{2}\log q_{t}(x)$ are no more than $L$. Using this fact, we can verify condition ([6](#S3.E6)), which is contained in the following lemma.

###### Lemma 13 .

If $\nabla\log q_{t}$ is norm-Lipschitz for all $t\geq 0$ (as in ([7](#S4.E7))), we have, for any $x_{t}\in\mathbb{R}^{d}$,

$$ $\left\lVert\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})\right\rVert^{2}=O\left(1-\alpha_{t}\right),\leavevmode\nobreak\ \forall t\geq 1.$ $$

As a result, condition ([6](#S3.E6)) holds.

###### Proof.

See [section F.1](#A6.SS1).
∎

Now, to derive an explicit upper bound on the reverse-step error, we apply Von Neumann’s trace inequality and obtain

$$ $\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1}(\mu_{t})\nabla^{2}\log q_{t}(x_{t})\Big{)}\leq dL^{2},\quad\forall t\geq 1.$ $$

Therefore, following from [Corollary 1](#Thmcorollary1), we obtain

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{dL^{2}\log^{2}T}{T}.$ $$

###### Lemma 13 .

###### Proof.

### D.2 Proof of Theorem 3

We first introduce some relevant notations. Given that $Q_{0}$ is Gaussian mixture, the p.d.f. of $q_{t}$ at each time $t\geq 1$ can be calculated as

$$ $\displaystyle q_{t}(x)$ $\displaystyle=\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\sum_{n=1}^{N}\pi_{n}q_{0,n}(x_{0})\mathrm{d}x_{0}$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})q_{0,n}(x_{0})\mathrm{d}x_{0}=:\sum_{n=1}^{N}\pi_{n}q_{t,n}(x).$ $$

Since the convolution of two Gaussian density is still Gaussian, we have that $q_{t,n}$ is the p.d.f. of $\mathcal{N}(\mu_{t,n},\Sigma_{t,n})$, where $\mu_{t,n}:=\sqrt{\bar{\alpha}_{t}}\mu_{0,n}$ and $\Sigma_{t,n}:=\bar{\alpha}_{t}\Sigma_{0,n}+(1-\bar{\alpha}_{t})I_{d}$. Note that $\Sigma_{t,n}$ has full rank.
We check condition ([6](#S3.E6)) in the following lemma.

###### Lemma 14 .

When $Q_{0}$ is Gaussian mixture (see [Theorem 3](#Thmtheorem3)), we have, for any $x_{t}\in\mathbb{R}^{d}$,

$$ $\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})=O\left(1-\alpha_{t}\right),\leavevmode\nobreak\ \forall t\geq 1.$ $$

As a result, ([6](#S3.E6)) holds.

###### Proof.

See [section F.2](#A6.SS2).
∎

Next, we verify [Assumption 4](#Thmassumption4) for Gaussian mixture $Q_{0}$. The following lemma provides an upper bound on any order of partial derivative of a Gaussian mixture density, as long as each mixture component is well controlled. With this we further verify [Assumption 4](#Thmassumption4) for Gaussian mixture in [Corollary 4](#Thmcorollary4).

###### Lemma 15 .

Let $g(x|z)$ be the conditional Gaussian p.d.f. of $\mathcal{N}(\mu_{z},\Sigma_{z})$. Define $q(x):=\int g(x|z)\mathrm{d}\Pi(z)$, where $\Pi(z)$ is a mixing distribution (and denote $\mathcal{Z}$ its support).
Suppose $b:=\sup_{z\in\mathcal{Z}}\left\lVert\mu_{z}\right\rVert<\infty$, and
suppose the following conditions on $\Sigma_{z}$ hold for all $z\in\mathcal{Z}$:

- 1.
There exist $u,U\in\mathbb{R}$ such that $u\leq\det(\Sigma_{z})\leq U$;
- 2.
There exists $V\in\mathbb{R}$ such that $\left\lVert\Sigma_{z}^{-1}\right\rVert\leq V$;
- 3.
There exists $w\in\mathbb{R}$ such that $\sup_{z\in\mathcal{Z},i,j\in[d]^{2}}\left|[\Sigma_{z}^{-\frac{1}{2}}]^{ij}\right|\leq w$.

Then,

$$ $\left|\partial_{\bm{a}}^{k}\log q(x)\right|\leq\min\left\{C^{k}B_{k}\frac{d^{2k}\max\left\{w,1\right\}^{k}}{u^{k/2}}U^{k}e^{k\frac{V}{2}(\left\lVert x\right\rVert^{2}+b^{2})},B_{k}\frac{d^{2k}\max\left\{w,1\right\}^{k}}{u^{k/2}}\left|\mathrm{poly}_{k}(x)\right|\right\},$ $$

where $B_{k}$ is the Bell number, $C$ is some constant, and $\mathrm{poly}_{k}(x)$ is some $k$-th order polynomial in $x$.

###### Proof.

See [section F.3](#A6.SS3).
∎

###### Corollary 4 .

When $Q_{0}$ is Gaussian mixture (see [Theorem 3](#Thmtheorem3)), [Assumption 4](#Thmassumption4) is satisfied.

###### Proof.

See [section F.4](#A6.SS4).
∎

Now, we can calculate

$$ $\displaystyle\nabla^{2}\log q_{t}(x)$ $\displaystyle=\frac{1}{q_{t}^{2}(x)}\Bigg{(}q_{t}(x)\left(\sum_{n}\pi_{n}q_{t,n}(x)\left(\Sigma_{t,n}^{-1}(x-\mu_{t,n})(x-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}-\Sigma_{t,n}^{-1}\right)\right)$ $\displaystyle\qquad-\left(\sum_{n}\pi_{n}q_{t,n}(x)\Sigma_{t,n}^{-1}(x-\mu_{t,n})\right)\left(\sum_{n}\pi_{n}q_{t,n}(x)\Sigma_{t,n}^{-1}(x-\mu_{t,n})\right)^{\intercal}\Bigg{)}.$ (44) $$

Putting it differently, we have

$$ $\displaystyle\partial^{2}_{ij}\log q_{t}(x)$ $\displaystyle=\frac{1}{q_{t}^{2}(x)}\Bigg{(}q_{t}(x)\left(\sum_{n}\pi_{n}q_{t,n}(x)\left([\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{i}[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{j}-[\Sigma_{t,n}^{-1}]^{ij}\right)\right)$ $\displaystyle\qquad-\left(\sum_{n}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{i}\right)\left(\sum_{n}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{j}\right)\Bigg{)}.$ $$

We can take $\max_{n}$ within each summation here and get

$$ $\displaystyle\left|\partial^{2}_{ij}\log q_{t}(x)\right|$ $\displaystyle\leq\max_{n}\left([\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{i}[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{j}\right)-\min_{n}[\Sigma_{t,n}^{-1}]^{ij}$ $\displaystyle\qquad+\max_{n}\left|[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{i}\right|\max_{n}\left|[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{j}\right|$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\leq}}2\max_{n}\left|[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{i}\right|\max_{n}\left|[\Sigma_{t,n}^{-1}(x-\mu_{t,n})]^{j}\right|+\max_{n}\left|[\Sigma_{t,n}^{-1}]^{ij}\right|$ $\displaystyle\leq 2\left(\max_{n}\left|z_{t,n}^{i}(x)\right|\max_{n}\left|z_{t,n}^{j}(x)\right|+\max_{n}\left|[\Sigma_{t,n}^{-1}]^{ij}\right|\right),$ $$

where for brevity we write $z_{t,n}(x):=\Sigma_{t,n}^{-1}(x-\mu_{t,n})$.
Here $(i)$ follows because $\max_{n}(a_{n}b_{n})\leq\max_{n}|a_{n}|\max_{n}|b_{n}|$. Below we write $\xi_{t}(x,i):=\max_{n}\left|z_{t,n}^{i}(x)\right|$ and $\bar{\Sigma}$ to be a matrix such that $\bar{\Sigma}^{ij}:=\max_{n}\left|[\Sigma_{t,n}^{-1}]^{ij}\right|$ for brevity. For later purposes we want to show that $\xi_{t-1}(\mu_{t},i)-\xi_{t}(x_{t},i)=O(1-\alpha_{t})$.
Note that for each $n$ and fixed $x_{t}$ (writing $\mu_{t}(x_{t})=\mu_{t}$),

$$ $\displaystyle z_{t-1,n}(\mu_{t})-z_{t,n}(x_{t})$ $\displaystyle=\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})-\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})$ $\displaystyle=(\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1})(\mu_{t}-\mu_{t-1,n})-\Sigma_{t,n}^{-1}((x_{t}-\mu_{t,n})-(\mu_{t}-\mu_{t-1,n})).$ (45) $$

Here, since $\Sigma_{t-1,n}$ is real symmetric, we can write the eigen-decomposition as $\Sigma_{t-1,n}=UDU^{\intercal}$, where $U$ is an orthonormal matrix (having unit 2-norm) and $D$ is a diagonal matrix (with all diagonal elements positive). In the same notation, $\Sigma_{t-1,n}^{-1}=UD^{-1}U^{\intercal}$, and $\Sigma_{t,n}^{-1}=(\alpha_{t}\Sigma_{t-1,n}+(1-\alpha_{t})I_{d})^{-1}=U(\alpha_{t}D+(1-\alpha_{t})I_{d})^{-1}U^{\intercal}$. Since

$$ $\displaystyle\left|[D^{-1}]^{ii}-[(\alpha_{t}D+(1-\alpha_{t})I_{d})^{-1}]^{ii}\right|$ $\displaystyle=\left|\frac{1}{D^{ii}}-\frac{1}{\alpha_{t}D^{ii}+(1-\alpha_{t})}\right|$ $\displaystyle\leq\frac{(1-\alpha_{t})(\left|D^{ii}\right|+1)}{\alpha_{t}(D^{ii})^{2}+(1-\alpha_{t})D^{ii}}$ $\displaystyle=O(1-\alpha_{t}),$ $$

the following holds:

$$ $\left\lVert\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1}\right\rVert=O(1-\alpha_{t}).$ $$

Denote $[A]^{i*}$ as the $i$-th row of a matrix $A$. Thus, following from ([D.2](#A4.Ex266)), for any $i\in[d]$,

$$ $\displaystyle\left\lVert[\Sigma_{t-1,n}^{-1}]^{i*}-[\Sigma_{t,n}^{-1}]^{i*}\right\rVert\stackrel{{\scriptstyle(ii)}}{{\leq}}\left\lVert\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1}\right\rVert$ $\displaystyle=O(1-\alpha_{t}),$ $\displaystyle\left|\mu_{t}^{i}-x_{t}^{i}\right|=\left|\frac{1-\sqrt{\alpha_{t}}}{\sqrt{\alpha_{t}}}x_{t}^{i}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\partial_{i}\log q_{t}(x_{t})\right|$ $\displaystyle=O(1-\alpha_{t}),$ $\displaystyle\left|\mu_{t,n}^{i}-\mu_{t-1,n}^{i}\right|=\left|(1-\sqrt{\alpha_{t}})\mu_{t-1,n}^{i}\right|$ $\displaystyle=O(1-\alpha_{t}),$ $$

where $(ii)$ follows from the definition of matrix 2-norm and from the fact that $[\Sigma_{t,n}^{-1}]^{i*}=\Sigma_{t,n}^{-1}\bm{1}_{i}$ ($\bm{1}_{i}$ is the unit vector where the $i$-th element is 1, and recall that $\Sigma_{t,n}^{-1}$ is symmetric). This implies that $\left|z_{t-1,n}^{i}(\mu_{t})-z_{t,n}^{i}(x_{t})\right|=O(1-\alpha_{t}),\forall i$. Thus,

$$ $\displaystyle\xi_{t-1}(\mu_{t},i)-\xi_{t}(x_{t},i)$ $\displaystyle=\max_{n}\left|z_{t-1,n}^{i}(\mu_{t})\right|-\max_{n}\left|z_{t,n}^{i}(x_{t})\right|$ $\displaystyle\leq\max_{n}\left|z_{t-1,n}^{i}(\mu_{t})-z_{t,n}^{i}(x_{t})\right|=O(1-\alpha_{t}),$ (46) $$

where the last inequality follows because $\max_{n}|a_{n}|+\max_{n}|b_{n}|\geq\max_{n}(|a_{n}|+|b_{n}|)\geq\max_{n}|a_{n}+b_{n}|$.

Now, following from [Theorem 1](#Thmtheorem1), we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j=1}^{d}\partial^{2}_{ij}\log q_{t-1}(\mu_{t}(X_{t}))\partial^{2}_{ij}\log q_{t}(X_{t})\right]$ $\displaystyle\leq 4\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j=1}^{d}\left(\xi_{t-1}(\mu_{t}(X_{t}),i)\xi_{t-1}(\mu_{t}(X_{t}),j)+\bar{\Sigma}^{ij}\right)\left(\xi_{t}(X_{t},i)\xi_{t}(X_{t},j)+\bar{\Sigma}^{ij}\right)\right]$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\asymp}}4\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j=1}^{d}\left(\xi_{t}(X_{t},i)\xi_{t}(X_{t},j)+\bar{\Sigma}^{ij}\right)^{2}\right]$ $\displaystyle\leq 8\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j=1}^{d}\xi_{t}(X_{t},i)^{2}\xi_{t}(X_{t},j)^{2}\right]+8\sum_{i,j=1}^{d}(\bar{\Sigma}^{ij})^{2}$ $\displaystyle=8\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left(\sum_{i=1}^{d}\xi_{t}(X_{t},i)^{2}\right)^{2}\right]+8\sum_{i,j=1}^{d}(\bar{\Sigma}^{ij})^{2},$ (47) $$

where $(iii)$ follows from the asymptotic result in ([D.2](#A4.Ex275)).

In the following, we provide two different ways to upper-bound the two terms in ([D.2](#A4.Ex276)). The first approach can be applied when $N$ is not too large compared to $d$. For the first term,

$$ $\displaystyle\sum_{i=1}^{d}\xi(x,i)^{2}$ $\displaystyle=\sum_{i=1}^{d}\max_{n}([\Sigma_{t,n}^{-1}]^{i*}(x-\mu_{t,n}))^{2}\leq\sum_{i=1}^{d}\sum_{n=1}^{N}([\Sigma_{t,n}^{-1}]^{i*}(x-\mu_{t,n}))^{2}$ $\displaystyle=\sum_{n=1}^{N}\left\lVert\Sigma_{t,n}^{-1}(x-\mu_{t,n})\right\rVert^{2}\leq N\max_{n}\left\lVert\Sigma_{t,n}^{-1}\right\rVert^{2}\max_{n}\left\lVert x-\mu_{t,n}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{\lesssim}}N\max_{n}\left\lVert x-\mu_{t,n}\right\rVert^{2},$ $$

where $(iv)$ follows because of the following.
Since $\Sigma_{t,n}$ is a (full-rank) covariance matrix, all its eigenvalues are positive. Let $\lambda_{n,\text{min}}>0$ be the smallest eigenvalue of $\Sigma_{0,n}$, and thus

$$ $\max_{n}\left\lVert\Sigma_{t,n}^{-1}\right\rVert_{2}\leq\frac{1}{\bar{\alpha}_{t}\min_{n}\lambda_{n,\text{min}}+(1-\bar{\alpha}_{t})}\leq\frac{1}{\min\{1,\min_{n}\lambda_{n,\text{min}}\}}<\infty.$ (48) $$

In particular, this bound does not depend on $d$ or $T$.
Also, for the second term,

$$ $\sum_{i,j=1}^{d}(\bar{\Sigma}^{ij})^{2}=\sum_{i,j=1}^{d}\max_{n}([\Sigma_{t,n}^{-1}]^{ij})^{2}\leq\sum_{i,j=1}^{d}\sum_{n=1}^{N}([\Sigma_{t,n}^{-1}]^{ij})^{2}=\sum_{n=1}^{N}\left\lVert\Sigma_{t,n}^{-1}\right\rVert_{F}^{2}\lesssim Nd,$ $$

where the last inequality follows from ([48](#A4.E48)) and the fact that for any matrix full-rank $A$, $\left\lVert A\right\rVert_{F}\leq\sqrt{d}\left\lVert A\right\rVert_{2}$.

When $N$ is large, we can instead bound the first term as

$$ $\displaystyle\sum_{i=1}^{d}\xi(x,i)^{2}=\sum_{i=1}^{d}\max_{n}([\Sigma_{t,n}^{-1}]^{i*}(x-\mu_{t,n}))^{2}$ $\displaystyle\stackrel{{\scriptstyle(v)}}{{\leq}}\sum_{i=1}^{d}\max_{n}\left(\left\lVert[\Sigma_{t,n}^{-1}]^{i*}\right\rVert^{2}\left\lVert x-\mu_{t,n}\right\rVert^{2}\right)\leq\sum_{i=1}^{d}\max_{n}\left\lVert[\Sigma_{t,n}^{-1}]^{i*}\right\rVert^{2}\max_{n}\left\lVert x-\mu_{t,n}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(vi)}}{{\leq}}\sum_{i=1}^{d}\max_{n}\left\lVert\Sigma_{t,n}^{-1}\right\rVert^{2}\max_{n}\left\lVert x-\mu_{t,n}\right\rVert^{2}\stackrel{{\scriptstyle(vii)}}{{\lesssim}}d\max_{n}\left\lVert x-\mu_{t,n}\right\rVert^{2}.$ $$

Here $(v)$ follows from Cauchy-Schwartz inequality, $(vi)$ holds for the same reason for $(ii)$,
and $(vii)$ follows again from ([48](#A4.E48)). Also, for the second term, we can obtain an alternative upper bound as follows.
Write the eigen-decomposition as $\Sigma_{0,n}=Q_{n}\mathrm{diag}(\lambda_{n,1},\dots,\lambda_{n,d})Q_{n}^{\intercal}$, where $Q_{n}$ here is an orthonormal matrix (that does not depend on $T$). Then,

$$ $\displaystyle\Sigma_{t,n}^{-1}$ $\displaystyle=Q_{n}(\bar{\alpha}_{t}\mathrm{diag}(\lambda_{n,1},\dots,\lambda_{n,d})+(1-\bar{\alpha}_{t})I_{d})^{-1}Q_{n}^{\intercal}$ $\displaystyle=Q_{n}\mathrm{diag}((\bar{\alpha}_{t}\lambda_{n,1}+(1-\bar{\alpha}_{t}))^{-1},\dots,(\bar{\alpha}_{t}\lambda_{n,d}+(1-\bar{\alpha}_{t}))^{-1})Q_{n}^{\intercal},$ $$

and thus

$$ $\displaystyle\max_{n\in[N]}\left|[\Sigma_{t,n}^{-1}]^{ij}\right|$ $\displaystyle=\max_{n\in[N]}\left|\sum_{k=1}^{d}(\bar{\alpha}_{t}\lambda_{n,k}+(1-\bar{\alpha}_{t}))^{-1}Q_{n}^{ik}Q_{n}^{kj}\right|$ $\displaystyle\leq(\min\{1,\min_{n}\lambda_{n,\text{min}}\})^{-1}\max_{n\in[N],i,j\in[d]}\left|(Q_{n}^{i*})^{\intercal}(Q_{n}^{j*})\right|$ $\displaystyle\leq(\min\{1,\min_{n}\lambda_{n,\text{min}}\})^{-1}\max_{n\in[N],i\in[d]}\left\lVert Q_{n}^{i*}\right\rVert^{2}$ $\displaystyle=(\min\{1,\min_{n}\lambda_{n,\text{min}}\})^{-1},$ $$

where the last line follows because $Q_{n}$ is orthonormal for all $n\in[N]$. Note that this is a uniform bound that does not depend on $N$, $T$ or $d$, which further implies that

$$ $\sum_{i,j=1}^{d}(\bar{\Sigma}^{ij})^{2}\lesssim d^{2}.$ $$

Combining the two cases, we get

$$ $\displaystyle\sum_{i=1}^{d}\xi(x,i)^{2}$ $\displaystyle\lesssim\min\{d,N\}\max_{n}\left\lVert x-\mu_{t,n}\right\rVert^{2},$ (49) $\displaystyle\sum_{i,j=1}^{d}(\bar{\Sigma}^{ij})^{2}$ $\displaystyle\lesssim d\min\{d,N\}.$ (50) $$

Now, following from ([D.2](#A4.Ex276)), we obtain

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left(\sum_{i=1}^{d}\xi_{t}(X_{t},i)^{2}\right)^{2}\right]$ $\displaystyle\lesssim\min\{d,N\}^{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\max_{n}\left\lVert X_{t}-\mu_{t,n}\right\rVert^{4}\right]$ $\displaystyle\leq\min\{d,N\}^{2}\left(\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left\lVert X_{t}\right\rVert^{4}\right]+\max_{n}\left\lVert\mu_{t,n}\right\rVert^{4}\right),$ $$

in which

$$ $\max_{n}\left\lVert\mu_{t,n}\right\rVert^{4}\leq\max_{n}\left\lVert\mu_{0,n}\right\rVert^{4}\lesssim d^{2}$ $$

because $\mu_{0,n}<\infty$ is a fixed vector that does not depend on $T$.
For the first term, the expected fourth power of the norm can be upper-bounded as

$$ $\displaystyle\mathbb{E}\left[\left(\left\lVert\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t}\right\rVert^{2}\right)^{2}\right]$ $\displaystyle\leq\mathbb{E}\left[\left(\left\lVert X_{0}\right\rVert^{2}+2\left\lVert X_{0}\right\rVert\left\lVert\bar{W}_{t}\right\rVert+\left\lVert\bar{W}_{t}\right\rVert^{2}\right)^{2}\right]$ $\displaystyle\leq 4\left(\mathbb{E}\left\lVert X_{0}\right\rVert^{4}+\mathbb{E}\left\lVert\bar{W}_{t}\right\rVert^{4}\right).$ $$

Recall that $Q_{0}$ is a Gaussian mixture. Hence,

$$ $\int\left\lVert x_{0}\right\rVert^{4}q_{0}(x_{0})\mathrm{d}x_{0}=\sum_{n=1}^{N}\pi_{n}\int\left\lVert x_{0}\right\rVert^{4}q_{0,n}(x_{0})\mathrm{d}x_{0}\asymp d^{2},$ $$

and

$$ $\mathbb{E}\left\lVert\bar{W}_{t}\right\rVert^{4}=\mathbb{E}\left[\left(\sum_{i=1}^{d}(\bar{W}_{t}^{i})^{2}\right)^{2}\right]=\sum_{i=1}^{d}\mathbb{E}(\bar{W}_{t}^{i})^{4}+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\mathbb{E}(\bar{W}_{t}^{i})^{2}(\bar{W}_{t}^{j})^{2}\asymp d^{2}.$ $$

Therefore, we obtain an upper bound on the reverse-step error as:

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d^{2}\min\{d,N\}^{2}\log^{2}T}{T}.$ $$

Next we consider a special case: where all Gaussian mixtures have the same covariance matrix, and further the covariance is a scaled version of the identity, i.e., $\Sigma_{0,n}\equiv\sigma_{0}^{2}I_{d},\forall n=1,\dots,N$. In this case, $\Sigma_{t,n}\equiv(\bar{\alpha}_{t}\sigma_{0}^{2}+(1-\bar{\alpha}_{t}))I_{d}=:\sigma_{t}^{2}I_{d},\forall n$, and ([D.2](#A4.Ex258)) can be simplified to

$$ $\displaystyle\nabla^{2}\log q_{t}(x)$ $\displaystyle=\frac{1}{q_{t}(x)^{2}}\Bigg{(}q_{t}(x)\left(\sum_{n}\pi_{n}q_{t,n}(x)\left(\sigma_{t}^{-4}(x-\mu_{t,n})(x-\mu_{t,n})^{\intercal}-\sigma_{t}^{-2}I_{d}\right)\right)$ $\displaystyle\qquad-\sigma_{t}^{-4}\left(\sum_{n}\pi_{n}q_{t,n}(x)(x-\mu_{t,n})\right)\left(\sum_{n}\pi_{n}q_{t,n}(x)(x-\mu_{t,n})\right)^{\intercal}\Bigg{)}$ $\displaystyle=\frac{1}{\sigma_{t}^{4}q_{t}(x)^{2}}\Bigg{(}q_{t}(x)\left(\sum_{n}\pi_{n}q_{t,n}(x)\mu_{t,n}\mu_{t,n}^{\intercal}\right)$ $\displaystyle\qquad-\left(\sum_{n}\pi_{n}q_{t,n}(x)\mu_{t,n}\right)\left(\sum_{n}\pi_{n}q_{t,n}(x)\mu_{t,n}\right)^{\intercal}\Bigg{)}-\sigma_{t}^{-2}I_{d}$ $\displaystyle=\frac{1}{\sigma_{t}^{4}}\sum_{n}\tilde{\pi}_{n}(x)(\mu_{t,n}-\tilde{\mu}_{t}(x))(\mu_{t,n}-\tilde{\mu}_{t}(x))^{\intercal}-\sigma_{t}^{-2}I_{d}$ $$

in which we have defined

$$ $\tilde{\pi}_{n}(x):=\frac{\pi_{n}q_{t,n}(x)}{q_{t}(x)},\quad\tilde{\mu}_{t}(x):=\sum_{n=1}^{N}\tilde{\pi}_{n}(x)\mu_{t,n}.$ $$

Note that here $\tilde{\pi}_{n}(x)$ is a p.m.f. in $n$ for any fixed $x$ since $\sum_{n=1}^{N}\tilde{\pi}_{n}(x)=1$. Also, note that $\sigma_{t}^{-4}\leq\max\{\sigma_{0}^{-4},1\}$ and $\mu_{t,n}=\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\leq\mu_{0,n}$. Therefore, using a similar technique in ([E.4](#A5.Ex340)), we can obtain

$$ $\displaystyle\sum_{i,j=1}^{d}\partial^{2}_{ij}\log q_{t-1}(\mu_{t})\partial^{2}_{ij}\log q_{t}(x_{t})$ $\displaystyle\leq\max\{\sigma_{0}^{-4},1\}\left(\max_{a,b\in[n]}\left\lVert\mu_{0,a}-\mu_{0,b}\right\rVert^{4}+2d\max_{a,b\in[n]}\left\lVert\mu_{0,a}-\mu_{0,b}\right\rVert^{2}+d^{2}\right)$ $\displaystyle\lesssim d^{2}.$ $$

Therefore, when all Gaussian mixtures have the same independent covariance, we can upper-bound the reverse-step error as

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d^{2}\log^{2}T}{T}.$ $$

###### Lemma 14 .

###### Proof.

###### Lemma 15 .

###### Proof.

###### Corollary 4 .

###### Proof.

## Appendix E Proof of Theorem 4

Throughout this section we adopt the noise schedule $\alpha_{t}$’s defined in ([8](#S4.E8)). We first investigate some nice properties of the noise schedule in ([8](#S4.E8)). Below, [Lemma 16](#Thmlemma16) (along with its proof) is similar to [, Eq (39)]. We include a full proof here for the sake of completeness.

###### Lemma 16 .

The $\alpha_{t}$’s defined in ([8](#S4.E8)) (with $c>2$) satisfy

$$ $\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{p}}\lesssim\frac{\log T}{\delta^{p-1}T}\leavevmode\nobreak\ \leavevmode\nobreak\ \text{while}\leavevmode\nobreak\ \leavevmode\nobreak\ \bar{\alpha}_{T}=o(T^{-1}),\quad\forall 2\leq t\leq T,\leavevmode\nobreak\ p\geq 1.$ $$

###### Remark 4 .

When $p=2$, the $\alpha_{t}$’s in ([8](#S4.E8)) (with $c>2$) satisfy the later condition ([51](#A5.E51)).

###### Proof.

See [section F.5](#A6.SS5).
∎

Similar to ([22](#A1.E22)), another property of the $\alpha_{t}$’s in ([8](#S4.E8)) is that

$$ $\frac{(1-\alpha_{t})^{p}}{\alpha_{t}^{q}(1-\bar{\alpha}_{t-1})^{2p}}\lesssim\frac{(1-\alpha_{t})^{p}}{(1-\bar{\alpha}_{t-1})^{2p}},\quad\forall p,q\geq 1,t\geq 2.$ $$

Additionally,

$$ $\frac{1-\bar{\alpha}_{t}}{1-\bar{\alpha}_{t-1}}-1=\frac{\bar{\alpha}_{t-1}(1-\alpha_{t})}{1-\bar{\alpha}_{t-1}}\leq\frac{1-\alpha_{t}}{1-\bar{\alpha}_{t-1}}\lesssim\frac{\log T}{T},\quad\forall 2\leq t\leq T.$ $$

Since the parameter dependence is clear in the bound for the initialization and estimation errors ([Lemmas 3](#Thmlemma3) and [4](#Thmlemma4)), it remains to provide a bound on the reverse-step error that depends explicitly on the system parameters, which is the main topic below. In the following, we first provide a more refined bound than [Corollary 2](#Thmcorollary2) when $Q_{0}$ has bounded support.

###### Lemma 16 .

###### Remark 4 .

###### Proof.

### E.1 Verifying Assumption 4

When $Q_{0}$ is smooth, we can verify [Assumption 4](#Thmassumption4) when $Q_{0}$ is a Gaussian mixture. It turns out, then, that [Assumption 4](#Thmassumption4) is still satisfied for bounded $Q_{0}$ since $q_{t}$ is always a Gaussian mixture where the mixing distribution is $Q_{0}$. It remains to show that each mixture component has well-controlled mean and variance, which is indeed the case if a non-vanishing $1-\alpha_{1}$ is chosen (i.e., with early-stopping). This is shown in the following lemma.

###### Lemma 17 .

If $1-\alpha_{1}=\delta$, and suppose that $Q_{0}$ has bounded support, then [Assumptions 4](#Thmassumption4) and [4](#Thmassumption4) is satisfied.

###### Proof.

See [section F.6](#A6.SS6).
∎

###### Lemma 17 .

###### Proof.

### E.2 Bounding Moments of ζ t , t − 1 subscript 𝜁 𝑡 𝑡 1 \zeta_{t,t-1} under P t − 1 | t subscript 𝑃 𝑡 conditional 1 𝑡 P_{t-1|t}

When $Q_{0}$ is smooth, we need ([6](#S3.E6)) to show that the expected second and third powers of $\zeta_{t,t-1}$ under $P_{t-1|t}$ is small, and as a result we can get rid of the term $\mathrm{Tr}(\cdot)^{2}$ (see [Corollary 1](#Thmcorollary1)). For general $Q_{0}$ (possibly non-smooth), we can show that a similar condition as ([6](#S3.E6)) holds, which further allows us to show that the expected second and third powers of $\zeta_{t,t-1}$ under $P_{t-1|t}$ are still small.

###### Lemma 18 .

Fix $t\geq 2$ and $x_{t}\in\mathbb{R}^{d}$. Suppose that [Assumption 4](#Thmassumption4) holds and that the $\alpha_{t}$’s satisfy

$$ $\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}=O\left(\frac{\log T}{T}\right),\quad\forall t\geq 2.$ (51) $$

Then, if $Q_{0}$ satisfies [Assumption 1](#Thmassumption1) (not necessarily having a p.d.f.), we have

$$ $\nabla\log q_{t-1}(\mu_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{3}}\right).$ $$

###### Proof.

See [section F.7](#A6.SS7).
∎

###### Corollary 5 .

Under the same condition as [Lemma 18](#Thmlemma18),
if the $\alpha_{t}$’s satisfy ([51](#A5.E51)), then

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{2}]$ $\displaystyle=O\left(\frac{(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{4}}\right)=O\left(\frac{\log^{2}T}{T^{2}}\right),\leavevmode\nobreak\ \text{and}$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}^{3}]$ $\displaystyle=O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right)=O\left(\frac{\log^{3}T}{T^{3}}\right).$ $$

###### Proof.

We will use the result in [Lemma 18](#Thmlemma18). Write $g(x_{t})=\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})$. The proof is similar to [Lemma 11](#Thmlemma11). The only difference is that the rate of $g(x_{t})$ is slightly different. We still use the expected second and third moment expressions from ([C.8](#A3.Ex230)) and ([C.8](#A3.Ex234)), and we note that $\sigma_{t}^{2}=(1-\alpha_{t})/\alpha_{t}$ and $(1-\alpha_{t})/(1-\bar{\alpha}_{t-1})^{2}=O(\log T/T)$ (by ([51](#A5.E51))). Since [Assumption 4](#Thmassumption4) is satisfied (by [Lemma 17](#Thmlemma17)), the partial derivatives do not affect the rate, and we immediately obtain the desired result.
∎

###### Lemma 18 .

###### Proof.

###### Corollary 5 .

###### Proof.

### E.3 More Refined Bound for Q 0 subscript 𝑄 0 Q_{0} with Bounded Support

Now we combine the ingredients together to show a more refined bound than that in [Corollary 2](#Thmcorollary2). The technique to bound $\mathrm{KL}(Q_{1}||\widehat{P}_{1})$ is similar to that in [Corollary 1](#Thmcorollary1).
In particular, we provide a refined bound for the reverse-step error. As in [Corollary 1](#Thmcorollary1), we start from ([27](#A2.E27)).

If we adopt the $\alpha_{t}$’s in ([8](#S4.E8)), then [Assumption 4](#Thmassumption4) is satisfied when $Q_{0}$ has bounded support by [Lemma 17](#Thmlemma17).
As a result, all the results for conditional expectations are applicable ([Lemmas 7](#Thmlemma7), [8](#Thmlemma8) and [9](#Thmlemma9)), and higher moments under $P_{t-1|t}$ are also guaranteed by [Corollary 5](#Thmcorollary5).
Therefore, we can continue from ([27](#A2.E27)) and proceed as follows. For each $t\geq 2$ and fixed $x_{t}\in\mathbb{R}^{d}$, we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]-\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]$ $\displaystyle\qquad+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]\right)^{2}$ $\displaystyle\qquad+O\left(\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{3}_{t,t-1}]\right)^{3}\right)$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{2}_{t,t-1}]+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]\right)^{2}$ $\displaystyle\qquad+O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right)$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta_{t,t-1}]+O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right)$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathrm{Tr}\left(\nabla^{2}\log q_{t-1}(\mu_{t})\nabla^{2}\log q_{t}(x_{t})\right)+O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right).$ (52) $$

Here $(i)$ follows from [Corollaries 5](#Thmcorollary5) and [7](#Thmlemma7), $(ii)$ follows from Jensen’s inequality, and $(iii)$ follows from [Lemmas 7](#Thmlemma7), [8](#Thmlemma8) and [9](#Thmlemma9). Next we take the expectation over $X_{t}\sim Q_{t}$, which does not affect the rate of each term because of [Assumption 4](#Thmassumption4) (shown to be satisfied here, by [Lemma 17](#Thmlemma17)). Taking a summation over $t=2,\dots,T$, we have thus obtained a refined bound for $Q_{0}$ with bounded support.

### E.4 Explicit Parameter Dependence

Finally, we provide the explicit parameter dependence when $Q_{0}$ has bounded support. Again, we begin by investigating $\nabla^{2}\log q_{t}(t\geq 1)$.
We derive the Hessian of $\log q_{t}(x)$ as

$$ $\displaystyle\nabla^{2}\log q_{t}(x)=\frac{\partial}{\partial x}\left(\frac{\int_{x_{0}\in\mathbb{R}^{d}}\nabla q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})}{\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})}\right)$ $\displaystyle=\frac{q_{t}(x)\int_{x_{0}\in\mathbb{R}^{d}}\nabla^{2}q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})-\left(\int_{x_{0}\in\mathbb{R}^{d}}\nabla q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})\right)\left(\int_{x_{0}\in\mathbb{R}^{d}}\nabla q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}}{q_{t}^{2}(x)}$ $\displaystyle=\frac{1}{(1-\bar{\alpha}_{t})^{2}q_{t}^{2}(x)}\bigg{(}q_{t}(x)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\left((x-\sqrt{\bar{\alpha}_{t}}x_{0})(x-\sqrt{\bar{\alpha}_{t}}x_{0})^{\intercal}-(1-\bar{\alpha}_{t})I_{d}\right)\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\quad-\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(x-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(x-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}\bigg{)}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}-\frac{1}{1-\bar{\alpha}_{t}}I_{d}+\frac{1}{(1-\bar{\alpha}_{t})^{2}q_{t}^{2}(x)}\bigg{(}q_{t}(x)\Big{(}q_{t}(x)xx^{\intercal}-\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\sqrt{\bar{\alpha}_{t}}x_{0}\mathrm{d}Q_{0}(x_{0})\right)x^{\intercal}$ $\displaystyle\quad-x\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\sqrt{\bar{\alpha}_{t}}x_{0}\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}+\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\left((\sqrt{\bar{\alpha}_{t}}x_{0})(\sqrt{\bar{\alpha}_{t}}x_{0})^{\intercal}\right)\mathrm{d}Q_{0}(x_{0})\Big{)}$ $\displaystyle\quad-\left(q_{t}(x)x-\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)\left(q_{t}(x)x-\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}\bigg{)}$ $\displaystyle=-\frac{1}{1-\bar{\alpha}_{t}}I_{d}+\frac{1}{(1-\bar{\alpha}_{t})^{2}q_{t}^{2}(x)}\bigg{(}q_{t}(x)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\left((\sqrt{\bar{\alpha}_{t}}x_{0})(\sqrt{\bar{\alpha}_{t}}x_{0})^{\intercal}\right)\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\quad-\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}\bigg{)}$ (53) $$

where $(i)$ follows because for any function $\mathrm{fn}(x)$ of $x$,

$$ $\frac{1}{q_{t}(x)}\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\mathrm{fn}(x)\mathrm{d}Q_{0}(x_{0})=\frac{1}{q_{t}(x)}\left(\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})\right)\mathrm{fn}(x)=\mathrm{fn}(x).$ $$

For ease of presentation, define

$$ $h_{t}(x_{0}|x):=\frac{q_{t|0}(x|x_{0})}{q_{t}(x)}=\frac{q_{t|0}(x|x_{0})}{\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})}.$ $$

Here $h_{t}$ is the likelihood ratio between the conditional and the marginal at time $t$. Note that $\int_{x_{0}\in\mathbb{R}^{d}}h_{t}(x_{0}|x)\mathrm{d}Q_{0}(x_{0})=1$. Therefore, given any $x$, there exists a probability measure $\tilde{Q}_{0|t}(x_{0}|x)$ (absolutely continuous w.r.t. $Q_{0}$) such that $\frac{\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)}{\mathrm{d}Q_{0}(x_{0})}=h_{t}(x_{0}|x)$ for all $x_{0}$. Since $h_{t}(x_{0}|x)>0$, $\tilde{Q}_{0|t}(x_{0}|x)$ has the same support as $Q_{0}$ for all $x$. Also define $m_{t}(x):=\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t}(\cdot|x)}[X_{0}]$. With these notations, we can use change-of-measure to get

$$ $\displaystyle\frac{1}{1-\bar{\alpha}_{t}}I_{d}+\nabla^{2}\log q_{t}(x)$ $\displaystyle=\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t})^{2}}\bigg{(}\int_{x_{0}\in\mathbb{R}^{d}}h_{t}(x_{0}|x)x_{0}x_{0}^{\intercal}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad-\left(\int_{x_{0}\in\mathbb{R}^{d}}h_{t}(x_{0}|x)x_{0}\mathrm{d}Q_{0}(x_{0})\right)\left(\int_{x_{0}\in\mathbb{R}^{d}}h_{t}(x_{0}|x)x_{0}\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}\bigg{)}$ $\displaystyle=\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t})^{2}}\int_{x_{0}\in\mathbb{R}^{d}}(x_{0}-m_{t}(x))(x_{0}-m_{t}(x))^{\intercal}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x).$ (54) $$

In particular, for each $i,j\in[d]$ and fixed $x$, we have

$$ $\displaystyle\left|\partial^{2}_{ij}\log q_{t}(x)\right|$ $\displaystyle\leq\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t})^{2}}\int_{x_{0}\in\mathbb{R}^{d}}\left|x_{0}^{i}-m_{t}^{i}(x)\right|\left|x_{0}^{j}-m_{t}^{j}(x)\right|\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)+\frac{1}{1-\bar{\alpha}_{t}}.$ $$

Now consider $Q_{0}$ with bounded support. In this case $\tilde{Q}_{0|t}(x_{0}|x)$ has the same bounded support as $Q_{0}$ for all $x$. Note that $m_{t}(x)$ is also contained in this support. We have

$$ $\displaystyle\sum_{i,j=1}^{d}\partial^{2}_{ij}\log q_{t-1}(\mu_{t})\partial^{2}_{ij}\log q_{t}(x_{t})$ $\displaystyle\leq\frac{\bar{\alpha}_{t}\bar{\alpha}_{t-1}}{(1-\bar{\alpha}_{t})^{2}(1-\bar{\alpha}_{t-1})^{2}}\times\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\sum_{i,j=1}^{d}|x_{a}^{i}-m_{t-1}^{i}(\mu_{t})||x_{a}^{j}-m_{t-1}^{j}(\mu_{t})|$ $\displaystyle\qquad|x_{b}^{i}-m_{t}^{i}(x_{t})||x_{b}^{j}-m_{t}^{j}(x_{t})|\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t-1}}{(1-\bar{\alpha}_{t})(1-\bar{\alpha}_{t-1})^{2}}\times\int_{x_{a}\in\mathbb{R}^{d}}\sum_{i,j=1}^{d}|x_{a}^{i}-m_{t-1}^{i}(\mu_{t})||x_{a}^{j}-m_{t-1}^{j}(\mu_{t})|\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})^{2}}\times\int_{x_{b}\in\mathbb{R}^{d}}\sum_{i,j=1}^{d}|x_{b}^{i}-m_{t}^{i}(x_{t})||x_{b}^{j}-m_{t}^{j}(x_{t})|\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\qquad+\frac{d^{2}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}$ $\displaystyle=\frac{\bar{\alpha}_{t}\bar{\alpha}_{t-1}}{(1-\bar{\alpha}_{t})^{2}(1-\bar{\alpha}_{t-1})^{2}}\times$ $\displaystyle\quad\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\left(\sum_{i=1}^{d}(x_{a}^{i}-m_{t-1}^{i}(\mu_{t}))(x_{b}^{i}-m_{t}^{i}(x_{t}))\right)^{2}\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t-1}}{(1-\bar{\alpha}_{t})(1-\bar{\alpha}_{t-1})^{2}}\times\int_{x_{a}\in\mathbb{R}^{d}}\left(\sum_{i=1}^{d}|x_{a}^{i}-m_{t-1}^{i}(\mu_{t})|\right)^{2}\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})^{2}}\times\int_{x_{b}\in\mathbb{R}^{d}}\left(\sum_{i=1}^{d}|x_{b}^{i}-m_{t}^{i}(x_{t})|\right)^{2}\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\qquad+\frac{d^{2}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\frac{\bar{\alpha}_{t}\bar{\alpha}_{t-1}}{(1-\bar{\alpha}_{t})^{2}(1-\bar{\alpha}_{t-1})^{2}}\times$ $\displaystyle\quad\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\left\lVert x_{a}-m_{t-1}(\mu_{t})\right\rVert^{2}\left\lVert x_{b}-m_{t}(x_{t})\right\rVert^{2}\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t-1}}{(1-\bar{\alpha}_{t})(1-\bar{\alpha}_{t-1})^{2}}\times\int_{x_{a}\in\mathbb{R}^{d}}d\left\lVert x_{a}-m_{t-1}(\mu_{t})\right\rVert^{2}\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})^{2}}\times\int_{x_{b}\in\mathbb{R}^{d}}d\left\lVert x_{b}-m_{t}(x_{t})\right\rVert^{2}\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\qquad+\frac{d^{2}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\leq}}(1-\bar{\alpha}_{t-1})^{-4}\times$ $\displaystyle\qquad\bigg{(}\max_{x_{a},x_{b}:\left\lVert x_{a}\right\rVert^{2},\left\lVert x_{b}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}-x_{b}\right\rVert^{4}+2d\max_{x_{a},x_{b}:\left\lVert x_{a}\right\rVert^{2},\left\lVert x_{b}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}-x_{b}\right\rVert^{2}+d^{2}\bigg{)}$ (55) $$

where $(ii)$ follows from Cauchy–Schwartz inequality and $(iii)$ follows from the assumption that $Q_{0}$ has bounded support and because $\bar{\alpha}_{t}<\bar{\alpha}_{t-1}<1$. Since

$$ $\max_{x_{a},x_{b}:\left\lVert x_{a}\right\rVert^{2},\left\lVert x_{b}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}-x_{b}\right\rVert\leq 2\max_{x_{a}:\left\lVert x_{a}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}\right\rVert\lesssim R\sqrt{d},$ $$

we can continue from ([E.4](#A5.Ex340)) and upper-bound each summation term in [section E.3](#A5.Ex319) as

$$ $\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j=1}^{d}\partial^{2}_{ij}\log q_{t-1}(\mu_{t})\partial^{2}_{ij}\log q_{t}(x_{t})\right]\lesssim\frac{(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{4}}d^{2}R^{4}.$ $$

Therefore, for the reverse-step error, we finally obtain

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d^{2}R^{4}\log^{2}T}{\delta^{2}T}.$ $$

We remind readers that $R$ does not depend on $d$.

In the following, we show that the result can be further improved under the additional assumption in ([9](#S4.E9)). The goal is to provide an upper bound on $h_{t}(x_{0}|x)$ for all $x$.
From ([9](#S4.E9)),

$$ $\displaystyle\int_{y\in\mathbb{R}^{d}}q_{t|0}(x|y)\mathrm{d}Q_{0}(y)$ $\displaystyle\geq\int_{y\in B_{\gamma}(x_{0})}q_{t|0}(x|y)\mathrm{d}Q_{0}(y)$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{\geq}}\epsilon\inf_{y\in B_{\gamma}(x_{0})}q_{t|0}(x|y)$ $$

where $(iv)$ holds by ([9](#S4.E9)). Taking the limit $\gamma\to 0$, by continuity of $q_{t|0}(x|x_{0})$ in $x_{0}$, we obtain that

$$ $\int_{y\in\mathbb{R}^{d}}q_{t|0}(x|y)\mathrm{d}Q_{0}(y)\geq\epsilon\cdot q_{t|0}(x|x_{0}),$ $$

and thus $h_{t}(x_{0}|x)\leq\epsilon^{-1}$ for all $x\in\mathbb{R}^{d}$ and $x_{0}\in\mathrm{supp}(Q_{0})$. Thus, following from ([E.4](#A5.Ex336)), we have

$$ $\displaystyle-\frac{1}{1-\bar{\alpha}_{t}}I_{d}\preceq\nabla^{2}\log q_{t}(x)$ $\displaystyle\preceq\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t})^{2}}\int_{x_{0}\in\mathbb{R}^{d}}x_{0}x_{0}^{\intercal}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)$ $\displaystyle\preceq\epsilon^{-1}\frac{1}{(1-\bar{\alpha}_{t})^{2}}\int_{x_{0}\in\mathbb{R}^{d}}x_{0}x_{0}^{\intercal}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\preceq\frac{1}{(1-\bar{\alpha}_{t})^{2}}M_{2}I_{d},$ $$

where the last line follows from [Assumption 1](#Thmassumption1). This implies that

$$ $\left\lVert\nabla^{2}\log q_{t}(x_{t})\right\rVert_{2}\leq\max\left\{\frac{1}{1-\bar{\alpha}_{t}},\frac{1}{(1-\bar{\alpha}_{t})^{2}}M_{2}\right\}\leq\frac{1}{(1-\bar{\alpha}_{t})^{2}}(M_{2}+1).$ $$

Hence, using Von Neumann’s trace inequality, we obtain

$$ $\displaystyle\mathrm{Tr}\left(\nabla^{2}\log q_{t}(x_{t})\nabla^{2}\log q_{t-1}(\mu_{t})\right)$ $\displaystyle\leq d\left\lVert\nabla^{2}\log q_{t}(x_{t})\right\rVert_{2}\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})\right\rVert_{2}$ $\displaystyle\leq\frac{1}{(1-\bar{\alpha}_{t})^{2}(1-\bar{\alpha}_{t-1})^{2}}d(M_{2}+1)^{2}.$ $$

Note that here $M_{2}$ does not depend on $d$. Therefore, we obtain an improved bound on the reverse-step error under condition ([9](#S4.E9)) as follows:

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d\log^{2}T}{\delta^{2}T}.$ $$

## Appendix F Auxiliary Proofs for Theorems 2 , 3 and 4

In this section, we provide the proofs for the lemmas in the proofs for [Theorems 2](#Thmtheorem2), [3](#Thmtheorem3) and [4](#Thmtheorem4).

### F.1 Proof of Lemma 13

Following from [, Lemma 20], we have

$$ $\nabla\log q_{t}(x)=\frac{1}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x)}[\nabla\log q_{t-1}(X_{t-1})].$ $$

Hence,

$$ $\displaystyle\left\lVert\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})\right\rVert^{2}$ $\displaystyle=\left\lVert\nabla\log q_{t-1}(\mu_{t})-\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}[\nabla\log q_{t-1}(X_{t-1})]\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\leq}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\nabla\log q_{t-1}(\mu_{t})-\nabla\log q_{t-1}(X_{t-1})\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}L^{2}\cdot\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\mu_{t}-X_{t-1}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}L^{2}\cdot\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert X_{t-1}-\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}[X_{t-1}]\right\rVert^{2}$ $$

where $(i)$ follows from Jensen’s inequality, $(ii)$ follows from [Equation 7](#S4.E7), and $(iii)$ follows from [Lemma 8](#Thmlemma8).
Finally, following from [Lemma 8](#Thmlemma8), we obtain

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert X_{t-1}-\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}[X_{t-1}]\right\rVert^{2}$ $\displaystyle=\mathrm{Tr}\left(\mathrm{Cov}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}(X_{t-1})\right)$ $\displaystyle=\frac{1-\alpha_{t}}{\alpha_{t}}d+\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\mathrm{Tr}\left(\nabla^{2}\log q_{t}(x_{t})\right).$ $$

Here $\left|\mathrm{Tr}\left(\nabla^{2}\log q_{t}(x_{t})\right)\right|<dL$ since all absolute values of the eigenvalues of $\nabla^{2}\log q_{t}(x)$ are upper-bounded by $L$. The proof is now complete.

### F.2 Proof of Lemma 14

Fix $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$. We have that

$$ $\displaystyle q_{t}(x)$ $\displaystyle=\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\sum_{n=1}^{N}\pi_{n}q_{0,n}(x_{0})\mathrm{d}x_{0}$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})q_{0,n}(x_{0})\mathrm{d}x_{0}=:\sum_{n=1}^{N}\pi_{n}q_{t,n}(x).$ $$

Here $q_{t,n}$ is the p.d.f. of $\mathcal{N}(\mu_{t,n},\Sigma_{t,n})$ where $\mu_{t,n}:=\sqrt{\bar{\alpha}_{t}}\mu_{0,n}$ and $\Sigma_{t,n}:=\bar{\alpha}_{t}\Sigma_{0,n}+(1-\bar{\alpha}_{t})I_{d}$. Thus, we have

$$ $\displaystyle\nabla\log q_{t}(x)$ $\displaystyle=\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\Sigma_{t,n}^{-1}(x-\mu_{t,n})}{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)}$ $\displaystyle=\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\Sigma_{t,n}^{-1}}{q_{t}(x)}x-\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\Sigma_{t,n}^{-1}\mu_{t,n}}{q_{t}(x)}.$ $$

This implies that

$$ $\displaystyle\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})$ $\displaystyle=\underbrace{\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}}{q_{t-1}(\mu_{t})}\mu_{t}-\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}}{q_{t}(x_{t})}\sqrt{\alpha_{t}}x_{t}}_{\text{term 1}}$ $\displaystyle\qquad-\underbrace{\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}\mu_{t-1,n}}{q_{t-1}(\mu_{t})}-\sqrt{\alpha_{t}}\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}\mu_{t,n}}{q_{t}(x_{t})}\right)}_{\text{term 2}}.$ (56) $$

The first term in ([F.2](#A6.Ex383)) is equal to

$$ term 1 $\displaystyle=\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}-\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t,m}(x_{t})}\right)\mu_{t}$ $\displaystyle\qquad+\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}}{q_{t}(x_{t})}\left(\mu_{t}-\sqrt{\alpha_{t}}x_{t}\right)$ $\displaystyle=\frac{\sum_{m,n=1}^{N}\pi_{m}\pi_{n}\left(q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})\Sigma_{t-1,n}^{-1}-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}\right)}{q_{t-1}(\mu_{t})q_{t}(x_{t})}\mu_{t}$ $\displaystyle\qquad+\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}}{q_{t}(x_{t})}\left(\mu_{t}-\sqrt{\alpha_{t}}x_{t}\right)$ $$

where

$$ $\displaystyle\mu_{t}-\sqrt{\alpha_{t}}x_{t}$ $\displaystyle=\frac{x_{t}}{\sqrt{\alpha_{t}}}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t}(x_{t})-\sqrt{\alpha_{t}}x_{t}$ $\displaystyle=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(x_{t}+\nabla\log q_{t}(x_{t}))=O(1-\alpha_{t}).$ $$

The second term in ([F.2](#A6.Ex383)) is equal to

$$ term 2 $\displaystyle=\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}\mu_{t-1,n}}{q_{t-1}(\mu_{t})}-\alpha_{t}\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}\mu_{t-1,n}}{q_{t}(x_{t})}\right)$ $\displaystyle=(1-\alpha_{t})\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}\mu_{t-1,n}}{q_{t-1}(\mu_{t})}\right)$ $\displaystyle\qquad+\alpha_{t}\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}\mu_{t-1,n}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}-\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}\mu_{t-1,n}}{\sum_{m=1}^{N}\pi_{m}q_{t,m}(x_{t})}\right)$ $\displaystyle=(1-\alpha_{t})\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}\mu_{t-1,n}}{q_{t-1}(\mu_{t})}\right)$ $\displaystyle\qquad+\alpha_{t}\frac{\sum_{m,n=1}^{N}\pi_{m}\pi_{n}\left(q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})\Sigma_{t-1,n}^{-1}-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}\right)\mu_{t-1,n}}{q_{t-1}(\mu_{t})q_{t}(x_{t})}.$ $$

Observe that there is a common factor in both terms above, namely, $q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})\Sigma_{t-1,n}^{-1}-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}$. The proof would be complete if this common factor is of order $(1-\alpha_{t})$, which we now investigate. Note that

$$ $\displaystyle\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1}$ $\displaystyle=\Sigma_{t-1,n}^{-1}-\left(\alpha_{t}\Sigma_{t-1,n}+(1-\alpha_{t})I_{d}\right)^{-1}$ $\displaystyle=\Sigma_{t-1,n}^{-1}-\left((\alpha_{t}\Sigma_{t-1,n})(I_{d}+\frac{1-\alpha_{t}}{\alpha_{t}}\Sigma_{t-1,n}^{-1})\right)^{-1}$ $\displaystyle=\Sigma_{t-1,n}^{-1}-\frac{1}{\alpha_{t}}(I_{d}+\frac{1-\alpha_{t}}{\alpha_{t}}\Sigma_{t-1,n}^{-1})^{-1}\Sigma_{t-1,n}^{-1}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\Sigma_{t-1,n}^{-1}-\frac{1}{\alpha_{t}}\left(I_{d}-\frac{1-\alpha_{t}}{\alpha_{t}}\Sigma_{t-1,n}^{-1}+O((1-\alpha_{t})^{2})\right)\Sigma_{t-1,n}^{-1}$ $\displaystyle=\frac{1-\alpha_{t}}{\alpha_{t}}\Sigma_{t-1,n}^{-1}+\frac{1-\alpha_{t}}{\alpha_{t}^{2}}\Sigma_{t-1,n}^{-2}+O((1-\alpha_{t})^{2})=O(1-\alpha_{t})$ $$

where $(i)$ follows from Taylor expansion and note that $1-\alpha_{t}=O(\log T/T)$ and $\Sigma_{t-1,n}^{-1}=O(1)$. Also,

$$ $\displaystyle q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})$ $\displaystyle=q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\left(\frac{q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})}{q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})}-1\right)$ $\displaystyle=:q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})(e^{\Delta}-1)$ $$

where

$$ $\displaystyle\Delta$ $\displaystyle=-\frac{1}{2}(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})+\frac{1}{2}(x_{t}-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})$ $\displaystyle\qquad-\frac{1}{2}(x_{t}-\mu_{t,m})^{\intercal}\Sigma_{t,m}^{-1}(x_{t}-\mu_{t,m})+\frac{1}{2}(\mu_{t}-\mu_{t-1,m})^{\intercal}\Sigma_{t-1,m}^{-1}(\mu_{t}-\mu_{t-1,m}).$ $$

Here, for any $i\in[d]$, we have

$$ $\displaystyle\left|(\mu^{i}_{t}-\mu^{i}_{t-1,n})-(x^{i}_{t}-\mu^{i}_{t,n})\right|$ $\displaystyle=\left|\frac{1-\sqrt{\alpha_{t}}}{\sqrt{\alpha_{t}}}x^{i}_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}[\nabla\log q_{t}(x_{t})]^{i}-(1-\sqrt{\alpha_{t}})\mu^{i}_{t-1,n}\right|$ $\displaystyle\leq\frac{1-\sqrt{\alpha_{t}}}{\sqrt{\alpha_{t}}}\left|x^{i}_{t}\right|+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\left|[\nabla\log q_{t}(x_{t})]^{i}\right|+(1-\sqrt{\alpha_{t}})\left|\mu^{i}_{t-1,n}\right|$ $\displaystyle\leq\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\left|x^{i}_{t}\right|+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\left|[\nabla\log q_{t}(x_{t})]^{i}\right|+(1-\alpha_{t})\left|\mu^{i}_{t-1,n}\right|$ $\displaystyle=O(1-\alpha_{t}).$ $$

Thus, $(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})=O(1-\alpha_{t})$ and

$$ $\displaystyle\left\lVert(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})\right\rVert$ $\displaystyle=\Big{\|}(\mu_{t}-\mu_{t-1,n})^{\intercal}(\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1})(\mu_{t}-\mu_{t-1,n})$ $\displaystyle\qquad+(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t,n}^{-1}\big{(}(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})\big{)}$ $\displaystyle\qquad+\big{(}(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})\big{)}^{\intercal}\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})\Big{\|}$ $\displaystyle\leq\Big{\|}(\mu_{t}-\mu_{t-1,n})^{\intercal}(\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1})(\mu_{t}-\mu_{t-1,n})\Big{\|}$ $\displaystyle\qquad+\Big{\|}(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t,n}^{-1}\Big{\|}\Big{\|}(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})\Big{\|}$ $\displaystyle\qquad+\Big{\|}(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})\Big{\|}\Big{\|}\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})\Big{\|}$ $\displaystyle=O(1-\alpha_{t}),$ $$

which further implies that $\Delta=O(1-\alpha_{t})$. Then,

$$ $q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})=O(e^{\Delta}-1)=O(\Delta)=O(1-\alpha_{t}).$ $$

Hence, the common factor in both terms of ([F.2](#A6.Ex383)) satisfies

$$ $\displaystyle q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})\Sigma_{t-1,n}^{-1}-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}$ $\displaystyle=q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})(\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1})+(q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t}))\Sigma_{t,n}^{-1}$ $\displaystyle=O(1-\alpha_{t}).$ $$

The proof is now complete.

### F.3 Proof of Lemma 15

Fix $k\geq 1$ and $\bm{a}\in[d]^{k}$. Recall that $u\leq\det(\Sigma_{z})\leq U$, $\left\lVert\Sigma_{z}^{-1}\right\rVert\leq V$, and $\sup_{z\in\mathcal{Z},i,j\in[d]^{2}}\left|[\Sigma_{z}^{-\frac{1}{2}}]^{ij}\right|\leq w$ for all $z\in\mathcal{Z}$.
Also write $\phi(y)$ as the p.d.f. of the unit Gaussian.
We are interested in upper-bounding the absolute partial derivatives of $\log q(x)$ with a function of $x$ where

$$ $q(x)=\int g(x|z)\mathrm{d}\Pi(z),$ $$

where, using the change-of-variable formula,

$$ $g(x|z)=\frac{1}{\det(\Sigma_{z})^{\frac{1}{2}}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right).$ (57) $$

We first identify an upper bound on the absolute partial derivatives of $q(x)$.
Now,

$$ $\displaystyle\partial^{k}_{\bm{a}}q(x)$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\int\partial^{k}_{\bm{a}}g(x|z)\mathrm{d}\Pi(z)$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\frac{1}{\inf_{z\in\mathcal{Z}}\det(\Sigma_{z})^{\frac{1}{2}}}\int\partial^{k}_{\bm{a}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\mathrm{d}\Pi(z)$ $$

where $(i)$ follows from the dominated convergence theorem (see ([32](#A3.E32))), and $(ii)$ follows from ([57](#A6.E57)).
To obtain an upper bound on the $k$-th derivative of Gaussian density, we invoke the multivariate version of the Faá di Bruno’s formula [, Theorem 2.1]. Since $y=\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})$ is linear in $x$, only the first-order partial derivative is non-zero and is equal to an entry in $\Sigma_{z}^{-\frac{1}{2}}$. Thus, we have

$$ $\displaystyle\left|\partial^{k}_{\bm{a}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\right|$ $\displaystyle=\left|\sum_{\bm{a}^{\prime}\in[d]^{k}}\phi^{(k)}_{\bm{a}^{\prime}}(y)\prod_{s=1}^{k}\frac{\partial}{\partial x_{a_{s}}}[\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})]^{a^{\prime}_{s}}\right|$ $\displaystyle\leq\left|\sum_{\bm{a}^{\prime}\in[d]^{k}}\phi^{(k)}_{\bm{a}^{\prime}}\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\right|\max\left\{w,1\right\}^{k},\leavevmode\nobreak\ \forall\bm{a}:\left|\bm{a}\right|=k.$ $$

Here we define $\phi^{(k)}_{\bm{a}}(y):=\partial^{k}_{\bm{a}}\phi(y)$. Since $\phi(y)$ is a Gaussian density which is infinitely differentiable and decays exponentially at the tail, its $k$-th order derivative satisfies $\phi^{(k)}_{\bm{a}}(y)=\mathrm{poly}_{k}(y)\phi(y)$ where $\mathrm{poly}_{k}(y)$ is a $k$-th order polynomial function in $y_{1},\dots,y_{d}$ (and thus in $x_{1},\dots,x_{d}$ by linearity). Also note that, for any $\bm{a}\in[d]^{k}$,

$$ $\lim_{\left\lVert y\right\rVert\to\infty}\left|\phi^{(k)}_{\bm{a}}(y)\right|=\lim_{\left\lVert y\right\rVert\to\infty}\left|\mathrm{poly}_{k}(y)\phi(y)\right|=0.$ $$

By the continuity of $\phi^{(k)}_{\bm{a}}(y)$, there exists $\bar{y}_{\bm{a}}$ such that $\left|\phi^{(k)}_{\bm{a}}(y)\right|\leq\left|\phi^{(k)}_{\bm{a}}(\bar{y}_{\bm{a}})\right|\leq\mathrm{poly}_{k}(\bar{y}_{\bm{a}})$ for all $y\in\mathbb{R}^{d}$.
Now, for all $x\in\mathbb{R}^{d}$,

$$ $\displaystyle\left|\partial^{k}_{\bm{a}}q(x)\right|$ $\displaystyle\leq\int\det(\Sigma_{z})^{-\frac{1}{2}}\left|\partial^{k}_{\bm{a}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\right|\mathrm{d}\Pi(z)$ $\displaystyle\leq\max\left\{w,1\right\}^{k}\int\det(\Sigma_{z})^{-\frac{1}{2}}\left(\sum_{\bm{a}\in[d]^{k}}\left|\mathrm{poly}_{k}\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\right|\right)\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\mathrm{d}\Pi(z)$ (58) $\displaystyle\leq\frac{d^{k}\max\left\{w,1\right\}^{k}}{\sqrt{u}}\left|\mathrm{poly}_{k}\left(\bar{y}_{\bm{a}}\right)\right|\phi\left(\bar{y}_{\bm{a}}\right).$ (59) $$

We have thus obtained a constant upper bound on all partial derivatives of $q(x)$ of order $k$.

Next, we convert the partial derivative bound into that for $\log q(x)$. We again invoke Faá di Bruno’s formula []. Note that

$$ $\partial^{k}_{\bm{a}}\log q(x)=q(x)^{-k}\sum_{\bm{b_{1}},\dots,\bm{b_{k}}}\prod_{j=1}^{k}\partial^{\left|\bm{b_{j}}\right|}_{\bm{b_{j}}}q(x)=:\sum_{\bm{b_{1}},\dots,\bm{b_{k}}}r_{\bm{b_{1}},\dots,\bm{b_{k}}}(x)$ (60) $$

in which we define each summation term as $r$.
Here $\{\bm{b_{1}},\dots,\bm{b_{k}}\}$ is some (possibly empty) partition of $\bm{a}$, i.e., $\sum_{j}\bm{b_{j}}=\bm{a}$ and $\sum_{j}\left|\bm{b_{j}}\right|=k$ (thus, at most $k$ partitions). We order this partition such that $k\geq\left|\bm{b_{1}}\right|\geq\dots\geq\left|\bm{b_{k}}\right|\geq 0$.
Note that the total number of partition can be upper-bounded by $d^{k}\sum_{l=1}^{k}B_{k,l}(1,\dots,1)=d^{k}B_{k}$, where $B_{k,l}(\cdot)$ and $B_{k}$ are the Bell polynomials and the Bell number, respectively.

We first showcase a simple yet useful upper bound.
From ([59](#A6.E59)), we get,

$$ $\displaystyle\left|\prod_{j=1}^{k}\partial^{\left|\bm{b_{j}}\right|}_{\bm{b_{j}}}q(x)\right|$ $\displaystyle\leq\prod_{j=1}^{k}\left|\partial^{\left|\bm{b_{j}}\right|}_{\bm{b_{j}}}q(x)\right|$ $\displaystyle\leq\frac{(d\max\{w,1\})^{\sum_{j}\left|\bm{b_{j}}\right|}}{\min\{u,1\}^{k/2}}\max\{\max_{y}\phi(y),1\}^{k}\prod_{j=1}^{k}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}(\bar{y}_{\bm{b_{j}}})\right|$ $\displaystyle\leq\frac{(d\max\{w,1\})^{\sum_{j}\left|\bm{b_{j}}\right|}}{\min\{u,1\}^{k/2}}\max\{\max_{y}\phi(y),1\}^{k}\max_{j}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}(\bar{y}_{\bm{b_{j}}})\right|^{k}$ $\displaystyle\leq C_{\bm{b_{1}},\dots,\bm{b_{j}}}^{k}\frac{d^{k}\max\left\{w,1\right\}^{k}}{\min\{u,1\}^{k/2}}$ $$

where, as noted above, $\bar{y}_{\bm{b_{j}}}$ does not depend on $x$.
Here $C_{\bm{b_{1}},\dots,\bm{b_{j}}}$ is some constant which depends only on the partition $\{\bm{b_{1}},\dots,\bm{b_{j}}\}$ and is independent of $x$.
On the other hand, we can also obtain a simple lower bound on $q(x)$. Observe that $q(x)$ is continuous and always positive. Recall that $b=\sup_{z\in\mathcal{Z}}\left\lVert\mu_{z}\right\rVert$. Thus,

$$ $\displaystyle q(x)$ $\displaystyle=\int_{\mathcal{Z}}g(x|z)\mathrm{d}\Pi(z)$ $\displaystyle\geq\frac{1}{(2\pi)^{d/2}\sup_{z\in\mathcal{Z}}\det(\Sigma_{z})^{\frac{1}{2}}}\int_{\mathcal{Z}}\exp\left(-\frac{1}{2}\sup_{z\in\mathcal{Z}}(x-\mu_{z})^{\intercal}\Sigma_{z}^{-1}(x-\mu_{z})\right)\mathrm{d}\Pi(z)$ $\displaystyle\geq\frac{1}{(2\pi)^{d/2}\sup_{z\in\mathcal{Z}}\det(\Sigma_{z})^{\frac{1}{2}}}\int_{\mathcal{Z}}\exp\left(-\frac{1}{2}\sup_{z\in\mathcal{Z}}\left\lVert\Sigma_{z}^{-1}\right\rVert(\left\lVert x\right\rVert^{2}+\left\lVert\mu_{z}\right\rVert^{2})\right)\mathrm{d}\Pi(z)$ $\displaystyle\geq\frac{1}{(2\pi)^{d/2}\sup_{z\in\mathcal{Z}}\det(\Sigma_{z})^{\frac{1}{2}}}\int_{\mathcal{Z}}\exp\left(-\frac{1}{2}\sup_{z\in\mathcal{Z}}\left\lVert\Sigma_{z}^{-1}\right\rVert(\left\lVert x\right\rVert^{2}+b^{2})\right)\mathrm{d}\Pi(z)$ $\displaystyle\geq\frac{1}{(2\pi)^{d/2}U}\exp\left(-\frac{V}{2}(\left\lVert x\right\rVert^{2}+b^{2})\right).$ $$

Therefore, if we set $C:=\max_{\bm{b_{1}},\dots,\bm{b_{j}}}C_{\bm{b_{1}},\dots,\bm{b_{j}}}$, we obtain

$$ $\left|\partial^{k}_{\bm{a}}\log q(x)\right|\leq C^{k}B_{k}\frac{d^{2k}\max\left\{w,1\right\}^{k}}{\min\{u,1\}^{k/2}}U^{k}e^{k\frac{V}{2}(\left\lVert x\right\rVert^{2}+b^{2})}.$ (61) $$

The upper bound above, though it depends only on parameters $u,U,V,w$, has an exponential dependence on $x$, which is not desirable. We next derive a more refined bound in $x$. For brevity of analysis, we re-express $r$ (defined in ([60](#A6.E60))) to avoid empty partitions:

$$ $r_{\bm{b_{1}},\dots,\bm{b_{k}}}(x)=q(x)^{-p}\prod_{j=1}^{p}\partial^{\left|\bm{b_{j}}\right|}_{\bm{b_{j}}}q(x),\leavevmode\nobreak\ \text{s.t.}\leavevmode\nobreak\ \left|\bm{b_{p+1}}\right|=\dots=\left|\bm{b_{k}}\right|=0.$ $$

Now, by the boundedness of $\left\lVert\mu_{z}\right\rVert$ and $\left\lVert\Sigma_{z}^{-1/2}\right\rVert$ on $\mathcal{Z}$, for each $x$, there exist (bounded) $\bar{\Sigma}_{\bm{b_{j}}}$ and $\bar{\mu}_{\bm{b_{j}}}$ such that, $\forall z\in\mathcal{Z}$,

$$ $\sum_{\bm{b}\in[d]^{\left|\bm{b_{j}}\right|}}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\right|\leq\sum_{\bm{b}\in[d]^{\left|\bm{b_{j}}\right|}}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}\left(\bar{\Sigma}_{\bm{b_{j}}}^{-\frac{1}{2}}(x-\bar{\mu}_{\bm{b_{j}}})\right)\right|<\infty.$ $$

Then, following from ([58](#A6.E58)), we obtain

$$ $\displaystyle\left|r_{\bm{b_{1}},\dots,\bm{b_{k}}}(x)\right|=q(x)^{-p}\left|\prod_{j=1}^{p}\partial^{\left|\bm{b_{j}}\right|}_{\bm{b_{j}}}q(x)\right|\leq q(x)^{-p}\prod_{j=1}^{p}\left|\partial^{\left|\bm{b_{j}}\right|}_{\bm{b_{j}}}q(x)\right|$ $\displaystyle\leq\frac{(d\max\{w,1\})^{\sum_{j=1}^{p}\left|\bm{b_{j}}\right|}}{u^{p/2}}\times$ $\displaystyle\qquad\prod_{j=1}^{p}\frac{\int\det(\Sigma_{z})^{-\frac{1}{2}}\sum_{\bm{c_{j}}\in[d]^{\left|\bm{b_{j}}\right|}}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\right|\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\mathrm{d}\Pi(z)}{\int\det(\Sigma_{z})^{-\frac{1}{2}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\mathrm{d}\Pi(z)}$ $\displaystyle\leq\frac{(d\max\{w,1\})^{k}}{\min\{1,u\}^{k/2}}\times$ $\displaystyle\qquad\prod_{j=1}^{p}\frac{\sum_{\bm{c_{j}}\in[d]^{\left|\bm{b_{j}}\right|}}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}\left(\bar{\Sigma}_{\bm{b_{j}}}^{-\frac{1}{2}}(x-\bar{\mu}_{\bm{b_{j}}})\right)\right|\left(\int\det(\Sigma_{z})^{-\frac{1}{2}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\mathrm{d}\Pi(z)\right)}{\int\det(\Sigma_{z})^{-\frac{1}{2}}\phi\left(\Sigma_{z}^{-\frac{1}{2}}(x-\mu_{z})\right)\mathrm{d}\Pi(z)}$ $\displaystyle=\frac{(d\max\{w,1\})^{k}}{\min\{1,u\}^{k/2}}\prod_{j=1}^{p}\sum_{\bm{c_{j}}\in[d]^{\left|\bm{b_{j}}\right|}}\left|\mathrm{poly}_{\left|\bm{b_{j}}\right|}\left(\bar{\Sigma}_{\bm{b_{j}}}^{-\frac{1}{2}}(x-\bar{\mu}_{\bm{b_{j}}})\right)\right|$ $$

Note that for each $j$, the number of terms in the summation above is upper-bounded by $d^{\left|\bm{b_{j}}\right|}$. Thus, expanding the product of summations would result in no more than $\prod_{j=1}^{p}d^{\left|\bm{b_{j}}\right|}=d^{k}$ terms. Also, since $\left|\mathrm{poly}_{k_{1}}(y)\right|\cdot\left|\mathrm{poly}_{k_{2}}(y)\right|=\left|\mathrm{poly}_{k_{1}+k_{2}}(y)\right|$, and since any $\bar{\Sigma}_{\bm{b_{j}}}^{-\frac{1}{2}}(x-\bar{\mu}_{\bm{b_{j}}})$ is linear in $x$ and independent in $z$, each product term is a $k$-th order polynomial in $x$. Therefore, we obtain

$$ $\left|r_{\bm{b_{1}},\dots,\bm{b_{k}}}(x)\right|\leq\frac{d^{2k}\max\{w,1\}^{k}}{\min\{1,u\}^{k/2}}\max_{\bm{c_{j}}\in[d]^{\left|\bm{b_{j}}\right|},\forall j=1,\dots,p}\left|\mathrm{poly}_{k}\left(x\right)\right|$ $$

and thus

$$ $\left|\partial^{k}_{\bm{a}}\log q(x)\right|\leq B_{k}\frac{d^{2k}\max\{w,1\}^{k}}{\min\{1,u\}^{k/2}}\max_{\bm{b_{1}},\dots,\bm{b_{k}}}\max_{\bm{c_{j}}\in[d]^{\left|\bm{b_{j}}\right|},\forall j=1,\dots,p}\left|\mathrm{poly}_{k}\left(x\right)\right|.$ (62) $$

We have thus identified an upper bound on $\left|\partial^{k}_{\bm{a}}\log q(x)\right|$ which is polynomial in $x$.
The proof is now complete by combining ([61](#A6.E61)) and ([62](#A6.E62)).

### F.4 Proof of Corollary 4

We first identify $u,U,V,w$ for $\Sigma_{t,n}$ such that they are independent of $T$ and $k$ for all $t\geq 1$. Fix $t\geq 1$. We use the fact that $\Sigma_{t,n}=\bar{\alpha}_{t}\Sigma_{0,n}+(1-\bar{\alpha}_{t})I_{d}$. If we let $\lambda_{n,1}\geq\dots\geq\lambda_{n,d}>0$ as the eigenvalues of $\Sigma_{0,n}$ (which do not depend on $T$), the eigenvalues of $\Sigma_{t,n}$ are $\{\bar{\alpha}_{t}\lambda_{n,i}+(1-\bar{\alpha}_{t})\}_{i=1}^{d}$. Therefore, for any $n=1,\dots,N$ and $t\geq 1$,

$$ $(u:=)\prod_{i=1}^{d}\min\{\min_{n}\lambda_{n,i},1\}\leq\det(\Sigma_{t,n})\leq\prod_{i=1}^{d}\max\{\max_{n}\lambda_{n,i},1\}(=:U).$ $$

Also, following from ([48](#A4.E48)), we have $V:=\frac{1}{\min\{1,\min_{n}\lambda_{n,d}\}}$.
Next, write the eigen-decomposition as $\Sigma_{0,n}=Q_{n}\mathrm{diag}(\lambda_{n,1},\dots,\lambda_{n,d})Q_{n}^{\intercal}$, where $Q_{n}$ here is an orthonormal matrix (that does not depend on $T$). Then, for any $t\geq 1$,

$$ $\displaystyle\Sigma_{t,n}^{-\frac{1}{2}}$ $\displaystyle=Q_{n}(\bar{\alpha}_{t}\mathrm{diag}(\lambda_{n,1},\dots,\lambda_{n,d})+(1-\bar{\alpha}_{t})I_{d})^{-\frac{1}{2}}Q_{n}^{\intercal}$ $\displaystyle=Q_{n}\mathrm{diag}((\bar{\alpha}_{t}\lambda_{n,1}+(1-\bar{\alpha}_{t}))^{-\frac{1}{2}},\dots,(\bar{\alpha}_{t}\lambda_{n,d}+(1-\bar{\alpha}_{t}))^{-\frac{1}{2}})Q_{n}^{\intercal}$ $$

and thus, for all $t\geq 1$,

$$ $\displaystyle[\Sigma_{t,n}^{-\frac{1}{2}}]^{ij}$ $\displaystyle=\sum_{k=1}^{d}(\bar{\alpha}_{t}\lambda_{n,k}+(1-\bar{\alpha}_{t}))^{-\frac{1}{2}}Q_{n}^{ik}Q_{n}^{kj}$ $\displaystyle\leq(\min\{1,\min_{n}\lambda_{n,d}\})^{-\frac{1}{2}}\max_{n\in[N],i,j\in[d]}\left|\sum_{k=1}^{d}Q_{n}^{ik}Q_{n}^{kj}\right|=:w.$ $$

Since the identified $u,U,V,w$ are all independent of $T$ and $k$, by [Lemma 15](#Thmlemma15) we have obtained an upper bound on $\left|\partial^{k}_{\bm{a}}\log q(x)\right|$ for any fixed $x$ which is independent of $T$. Also, in terms of $k$, we can use the exponential generating function for $B_{k}$ and show that, for any $r\in\mathbb{R}$,

$$ $\sum_{p=1}^{\infty}\frac{B_{p}}{p!}r^{p}=e^{e^{r}}-1<\infty.$ $$

This shows that the infinite sum of $U_{k}(x)$ converges after scaled by $k!$.
Now,
note that under Gaussian $g(x_{t}|z)$ the expected value of any $k$-th order polynomial term in $x_{t}$ is finite and is equal to the $k$-th order non-centralized moment. Using the results on Gaussian non-centralized moment,
we have from [Lemma 15](#Thmlemma15), for all $p\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}[U_{k}(X_{t})^{p}]$ $\displaystyle\leq C_{k}^{p}\int\int\left|\mathrm{poly}_{pk}(x_{t})\right|g(x_{t}|z)\mathrm{d}x_{t}\mathrm{d}\Pi(z)$ $\displaystyle=C_{k}^{p}\int\left|\mathrm{poly}_{pk}(\mu_{z},\Sigma_{z})\right|\mathrm{d}\Pi(z)$ $\displaystyle\leq(C_{k}^{\prime})^{p}\max\left\{1,\max_{z\in\mathcal{Z}}\max_{i\in[d]}\left|\mu_{z}^{i}\right|\max_{i,j\in[d]}\left|[\Sigma_{z}]^{ij}\right|\right\}^{kp}<\infty$ $$

where $C_{k}$ and $C^{\prime}_{k}$ are some constants that depend only on $k$. This is finite because each element of $\mu_{z}$ and $\Sigma_{z}$ has a finite upper bound that is independent of $T$. Combining all the above, we have shown [Assumption 4](#Thmassumption4).

### F.5 Proof of Lemma 16

The proof is similar to that of [, Eq (39)]. First, note that if $T$ is large,

$$ $\delta\left(1+\frac{c\log T}{T}\right)^{\frac{T}{\log T}}\asymp\delta e^{c}>1.$ $$

Thus, for $t\geq\frac{T}{2}(\geq\frac{T}{\log T})$, we have

$$ $1-\alpha_{t}=\frac{c\log T}{T}\min\left\{\delta\left(1+\frac{c\log T}{T}\right)^{t},1\right\}=\frac{c\log T}{T}.$ $$

As a result,

$$ $\bar{\alpha}_{T}\leq\prod_{t=\left\lfloor T/2\right\rfloor}^{T}\alpha_{t}=\left(1-\frac{c\log T}{T}\right)^{\left\lceil T/2\right\rceil}\asymp\exp\left(\left\lceil\frac{T}{2}\right\rceil\left(-\frac{c\log T}{T}\right)\right)=T^{-c/2}.$ (63) $$

This shows that $\alpha_{t}$ satisfies $\bar{\alpha}_{T}=o\left(T^{-1}\right)$ if $c>2$.

Define $\tau$ such that

$$ $\delta\left(1+\frac{c\log T}{T}\right)^{\tau}\leq 1<\delta\left(1+\frac{c\log T}{T}\right)^{\tau+1}.$ (64) $$

Here $\tau$ is unique since $1-\alpha_{t}$ is non-decreasing. In other words, $\tau$ is the last time that $1-\alpha_{t}$ is exponentially growing.
Assume that $T$ is large enough such that $\tau\geq 2$. Below, we show that

$$ $1-\bar{\alpha}_{t-1}\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t},\quad\forall 2\leq t\leq\tau.$ (65) $$

If $t=2$,

$$ $1-\bar{\alpha}_{t-1}=1-\bar{\alpha}_{1}=1-\alpha_{1}=\delta\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right).$ $$

Here the last inequality holds when $T$ is sufficiently large. For $t>2$, suppose for purpose of contradiction that there exists $2<t_{0}\leq\tau$ such that

$$ $1-\bar{\alpha}_{t_{0}-1}<\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}}\leavevmode\nobreak\ \text{while}\leavevmode\nobreak\ 1-\bar{\alpha}_{t-1}\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t},\leavevmode\nobreak\ \forall 2\leq t\leq t_{0}-1.$ $$

In words, $t_{0}$ is defined as the first time that ([65](#A6.E65)) is violated. To arrive at a contradiction, we first write

$$ $\displaystyle 1-\bar{\alpha}_{t_{0}-1}$ $\displaystyle=(1-\bar{\alpha}_{t_{0}-2})\left(1+\frac{\bar{\alpha}_{t_{0}-2}(1-\alpha_{t_{0}-1})}{1-\bar{\alpha}_{t_{0}-2}}\right)$ $\displaystyle\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}-1}\left(1+\frac{\bar{\alpha}_{t_{0}-2}(1-\alpha_{t_{0}-1})}{1-\bar{\alpha}_{t_{0}-2}}\right).$ $$

Here the inequality holds because $t_{0}$ is the first time that ([65](#A6.E65)) is violated, and thus ([65](#A6.E65)) stills holds for $t=t_{0}-1$. Also,

$$ $1-\bar{\alpha}_{t_{0}-2}\leq 1-\bar{\alpha}_{t_{0}-1}\stackrel{{\scriptstyle(i)}}{{<}}\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}}\stackrel{{\scriptstyle(ii)}}{{\leq}}\frac{1}{2}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}-1}\stackrel{{\scriptstyle(iii)}}{{\leq}}\frac{1}{2}$ $$

where $(i)$ holds because ([65](#A6.E65)) is violated at $t=t_{0}$, $(ii)$ holds when $T$ is sufficiently large, and $(iii)$ holds because $t_{0}-1\leq\tau$ and by the definition of $\tau$ in ([64](#A6.E64)). Thus,

$$ $\frac{\bar{\alpha}_{t_{0}-2}(1-\alpha_{t_{0}-1})}{1-\bar{\alpha}_{t_{0}-2}}\geq\frac{\frac{1}{2}\frac{c\log T}{T}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}-1}}{\frac{1}{2}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}-1}}=\frac{c\log T}{T},$ $$

and thus

$$ $1-\bar{\alpha}_{t_{0}-1}\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}-1}\left(1+\frac{\bar{\alpha}_{t_{0}-2}(1-\alpha_{t_{0}-1})}{1-\bar{\alpha}_{t_{0}-2}}\right)\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t_{0}}.$ $$

We have reached a contradiction. Therefore, we have shown that ([65](#A6.E65)) holds.

Now, ([65](#A6.E65)) implies that

$$ $1-\bar{\alpha}_{t-1}\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t}\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t/p},\quad\forall 2\leq t\leq\tau.$ $$

There are two cases:

- •
If $2\leq t\leq\tau$, then
$\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{p}}\leq\frac{\frac{c\log T}{T}\delta\left(1+\frac{c\log T}{T}\right)^{t}}{\frac{1}{3^{p}}\delta^{p}\left(1+\frac{c\log T}{T}\right)^{t}}=\frac{3^{p}c\log T}{\delta^{p-1}T}.$
- •
If $t>\tau$, then
$\displaystyle\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{p}}\leq\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{\tau-1})^{p}}\leq\frac{\frac{c\log T}{T}}{\frac{1}{3^{p}}\delta^{p}\left(1+\frac{c\log T}{T}\right)^{\tau}}$
$\displaystyle=\frac{\frac{c\log T}{T}\left(1+\frac{c\log T}{T}\right)}{3^{-p}\delta^{p-1}\left(1+\frac{c\log T}{T}\right)^{\tau+1}}$
$\displaystyle<\frac{3^{p}c\log T}{\delta^{p-1}T}\left(1+\frac{c\log T}{T}\right).$

In both cases, if $T$ is large enough,

$$ $\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{p}}\leq\frac{4^{p}c\log T}{\delta^{p-1}T}\lesssim\frac{\log T}{\delta^{p-1}T},\quad\forall 2\leq t\leq T$ $$

because $p$ and $c$ are constants (that do not depend on $T$, $d$, and $\delta$).
The proof is now complete.

### F.6 Proof of Lemma 17

We use the same notation as in [Lemma 6](#Thmlemma6): $\tilde{Q}_{0}$ denotes the distribution of $\sqrt{\bar{\alpha}_{t}}x_{0}$, $g(x)$ denotes the p.d.f. (w.r.t. the Lebesgue measure) of the distribution of $\sqrt{1-\bar{\alpha}_{t}}\bar{w}_{t}$, and

$$ $q_{t}(x)=\int g(x-x_{0})\mathrm{d}\tilde{Q}_{0}(x_{0}).$ $$

Now, we first need to identify $u,U,V,w$ from $\Sigma_{t,x_{0}}\equiv(1-\bar{\alpha}_{t})I_{d},\leavevmode\nobreak\ \forall x_{0}\in\mathrm{supp}(\tilde{Q}_{0}),\leavevmode\nobreak\ \forall t\geq 1$.
Note that $1-\bar{\alpha}_{t}\geq 1-\alpha_{1}=\delta$. Therefore, for all $t\geq 1$,

$$ $\displaystyle u:=\delta^{d}\leq\det(\Sigma_{t,x_{0}})\leq 1=:U$ $\displaystyle\left\lVert\Sigma_{t,x_{0}}^{-1}\right\rVert\leq\frac{1}{\delta}=:V$ $\displaystyle\left|[\Sigma_{t,x_{0}}^{-\frac{1}{2}}]^{ij}\right|\leq\frac{1}{\sqrt{\delta}}=:w.$ $$

Since the identified $u,U,V,w$ are all independent of $T$ and $k$, by [Lemma 15](#Thmlemma15) we have obtained an upper bound on $\left|\partial^{k}_{\bm{a}}\log q_{t}(x)\right|$ for all $k\geq 1$ and $t\geq 1$, a bound that is independent of $T$ and polynomial in $x$. With the same line of argument in [Corollary 4](#Thmcorollary4), the infinite sum of $U_{k}(x)$ converges if scaled by $k!$.

Finally, since $[\Sigma_{t,x_{0}}]^{ij}\leq 1-\bar{\alpha}_{t}\leq 1$, we have from [Lemma 15](#Thmlemma15), for all $p\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}[U_{k}(X_{t})^{p}]$ $\displaystyle\leq C_{k}^{p}\int\int\left|\mathrm{poly}_{pk}(x_{t})\right|g(x_{t}-x_{0})\mathrm{d}x_{t}\mathrm{d}\tilde{Q}_{0}(x_{0})$ $\displaystyle\leq C_{k}^{p}\int\left|\mathrm{poly}_{pk}(x_{0})\right|\mathrm{d}\tilde{Q}_{0}(x_{0})$ $\displaystyle\leq C_{k}^{p}\max\left\{1,\sup_{x_{0}\in\mathrm{supp}(\tilde{Q}_{0}),i\in[d]}\left|x_{0}^{i}\right|\right\}^{pk}<\infty,$ $$

where $C_{k}$ is some constants that only depends on $k$. Since the support of $\tilde{Q}_{0}$ is finite, we have shown [Assumption 4](#Thmassumption4).

### F.7 Proof of Lemma 18

Fix $t\geq 2$ and $x_{t}\in\mathbb{R}^{d}$.
Recall that $q_{t|0}(x|x_{0})$ is the p.d.f. of $\mathcal{N}(\sqrt{\bar{\alpha}_{t}}x_{0},(1-\bar{\alpha}_{t})I_{d})$. The gradient of $\log q_{t}(x)$ is

$$ $\nabla\log q_{t}(x)=\frac{\int_{x_{0}\in\mathbb{R}^{d}}\nabla q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})}{\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})}=\frac{-\frac{1}{1-\bar{\alpha}_{t}}\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})(x-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})}{\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})\mathrm{d}Q_{0}(x_{0})}.$ (66) $$

Thus,

$$ $\displaystyle\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})$ $\displaystyle=\frac{-\frac{1}{1-\bar{\alpha}_{t-1}}\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)(\mu_{t}-\sqrt{\bar{\alpha}_{t}}x)\mathrm{d}Q_{0}(x)}{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)\mathrm{d}Q_{0}(x)}-\frac{-\frac{\sqrt{\alpha_{t}}}{1-\bar{\alpha}_{t}}\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)(x_{t}-\sqrt{\bar{\alpha}_{t}}y)\mathrm{d}Q_{0}(y)}{\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)\mathrm{d}Q_{0}(y)}$ $\displaystyle=-\frac{1}{1-\bar{\alpha}_{t}}\Bigg{(}\left(\frac{1-\bar{\alpha}_{t}}{1-\bar{\alpha}_{t-1}}-1\right)\frac{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)}{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)\mathrm{d}Q_{0}(x)}$ $\displaystyle\qquad+\frac{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)}{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)\mathrm{d}Q_{0}(x)}-\sqrt{\alpha_{t}}\frac{\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)(x_{t}-\sqrt{\bar{\alpha}_{t}}y)\mathrm{d}Q_{0}(y)}{\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)\mathrm{d}Q_{0}(y)}\Bigg{)}$ $\displaystyle=\frac{1}{1-\bar{\alpha}_{t}}\Big{(}\left(1-\bar{\alpha}_{t}-(1-\bar{\alpha}_{t-1})\right)\nabla\log q_{t-1}(\mu_{t})\Big{)}$ $\displaystyle\qquad-\frac{1}{1-\bar{\alpha}_{t}}\Bigg{(}\frac{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)}{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)\mathrm{d}Q_{0}(x)}-\sqrt{\alpha_{t}}\frac{\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)(x_{t}-\sqrt{\bar{\alpha}_{t}}y)\mathrm{d}Q_{0}(y)}{\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)\mathrm{d}Q_{0}(y)}\Bigg{)}$ $\displaystyle\leq\frac{1-\alpha_{t}}{1-\bar{\alpha}_{t}}\left|\nabla\log q_{t-1}(\mu_{t})\right|+\frac{1}{1-\bar{\alpha}_{t}}\underbrace{\left|\mu_{t}-\sqrt{\alpha_{t}}x_{t}\right|}_{\text{term 1}}$ $\displaystyle\qquad+\frac{1}{1-\bar{\alpha}_{t}}\underbrace{\left|\frac{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)(\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)}{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)\mathrm{d}Q_{0}(x)}-\frac{\alpha_{t}\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)(\sqrt{\bar{\alpha}_{t-1}}y)\mathrm{d}Q_{0}(y)}{\int_{y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|y)\mathrm{d}Q_{0}(y)}\right|}_{\text{term 2}}.$ (67) $$

Term 1 in ([F.7](#A6.Ex478)) is

$$ $\displaystyle\mu_{t}-\sqrt{\alpha_{t}}x_{t}$ $\displaystyle=\frac{x_{t}}{\sqrt{\alpha_{t}}}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t}(x_{t})-\sqrt{\alpha_{t}}x_{t}$ $\displaystyle=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(x_{t}+\nabla\log q_{t}(x_{t})).$ $$

To investigate term 2, we provide some short-hand notation. Define

$$ $\displaystyle h_{t-1}(x)$ $\displaystyle:=\frac{q_{t-1|0}(\mu_{t}|x)}{\int_{x\in\mathbb{R}^{d}}q_{t-1|0}(\mu_{t}|x)\mathrm{d}Q_{0}(x)}$ $\displaystyle h_{t}(x)$ $\displaystyle:=\frac{q_{t|0}(x_{t}|x)}{\int_{x\in\mathbb{R}^{d}}q_{t|0}(x_{t}|x)\mathrm{d}Q_{0}(x)}.$ $$

Then, term 2 (without absolute value) in ([F.7](#A6.Ex478)) becomes

$$ $\displaystyle\int_{x\in\mathbb{R}^{d}}h_{t-1}(x)(\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)-\alpha_{t}\int_{x\in\mathbb{R}^{d}}h_{t}(x)(\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)$ $\displaystyle=(1-\alpha_{t})\int_{x\in\mathbb{R}^{d}}h_{t-1}(x)(\sqrt{\bar{\alpha}_{t-1}}x)\mathrm{d}Q_{0}(x)+\alpha_{t}\sqrt{\bar{\alpha}_{t-1}}\int_{x\in\mathbb{R}^{d}}\left(h_{t-1}(x)-h_{t}(x)\right)x\mathrm{d}Q_{0}(x).$ (68) $$

Here we obtained a coefficient $(1-\alpha_{t})$ in the first term of ([F.7](#A6.Ex489)). The reason why this term is small will be discussed below. For the second term in ([F.7](#A6.Ex489)),

$$ $\displaystyle\int_{x\in\mathbb{R}^{d}}(h_{t-1}(x)-h_{t}(x))x\mathrm{d}Q_{0}(x)$ $\displaystyle=\frac{1}{q_{t-1}(\mu_{t})q_{t}(x_{t})}\int_{x,y\in\mathbb{R}^{d}}\left(q_{t-1|0}(\mu_{t}|x)q_{t|0}(x_{t}|y)-q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)\right)x\mathrm{d}Q_{0}(y)\mathrm{d}Q_{0}(x)$ $$

where

$$ $\displaystyle q_{t-1|0}(\mu_{t}|x)q_{t|0}(x_{t}|y)-q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)$ $\displaystyle=q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)\left(\frac{q_{t-1|0}(\mu_{t}|x)q_{t|0}(x_{t}|y)}{q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)}-1\right)$ $\displaystyle=q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)\times$ $\displaystyle\quad\left(\exp\left(-\frac{\left\lVert\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x\right\rVert^{2}}{2(1-\bar{\alpha}_{t-1})}-\frac{\left\lVert x_{t}-\sqrt{\bar{\alpha}_{t}}y\right\rVert^{2}}{2(1-\bar{\alpha}_{t})}+\frac{\left\lVert\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}y\right\rVert^{2}}{2(1-\bar{\alpha}_{t-1})}+\frac{\left\lVert x_{t}-\sqrt{\bar{\alpha}_{t}}x\right\rVert^{2}}{2(1-\bar{\alpha}_{t})}\right)-1\right)$ $\displaystyle=:q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)(e^{\Delta}-1)$ $$

in which we defined the exponent as $\Delta$. Now,

$$ $\displaystyle\Delta$ $\displaystyle=-\frac{\left\lVert\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x\right\rVert^{2}}{2(1-\bar{\alpha}_{t-1})}-\frac{\left\lVert x_{t}-\sqrt{\bar{\alpha}_{t}}y\right\rVert^{2}}{2(1-\bar{\alpha}_{t})}+\frac{\left\lVert\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}y\right\rVert^{2}}{2(1-\bar{\alpha}_{t-1})}+\frac{\left\lVert x_{t}-\sqrt{\bar{\alpha}_{t}}x\right\rVert^{2}}{2(1-\bar{\alpha}_{t})}$ $\displaystyle=\frac{\sqrt{\bar{\alpha}_{t-1}}(x-y)^{\intercal}\mu_{t}+\bar{\alpha}_{t-1}\left\lVert y\right\rVert^{2}-\bar{\alpha}_{t-1}\left\lVert x\right\rVert^{2}}{2(1-\bar{\alpha}_{t-1})}-\frac{\sqrt{\bar{\alpha}_{t}}(x-y)^{\intercal}x_{t}+\bar{\alpha}_{t}\left\lVert y\right\rVert^{2}-\bar{\alpha}_{t}\left\lVert x\right\rVert^{2}}{2(1-\bar{\alpha}_{t})}$ $\displaystyle=\frac{1}{2}\left(\frac{\sqrt{\bar{\alpha}_{t}}(1-\alpha_{t})}{\alpha_{t}(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}x_{t}+\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}\nabla\log q_{t}(x_{t})\right)^{\intercal}(x-y)$ $\displaystyle\qquad+\frac{\bar{\alpha}_{t-1}(1-\alpha_{t})}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}(\left\lVert y\right\rVert^{2}-\left\lVert x\right\rVert^{2}).$ $$

Following from assumption ([51](#A5.E51)),

$$ $\displaystyle\frac{\sqrt{\bar{\alpha}_{t}}(1-\alpha_{t})}{\alpha_{t}(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}$ $\displaystyle=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}\right)=O\left(\frac{\log T}{T}\right),$ $\displaystyle\frac{1-\alpha_{t}}{1-\bar{\alpha}_{t-1}}$ $\displaystyle=O\left(\frac{\log T}{T}\right),$ $\displaystyle\frac{\bar{\alpha}_{t-1}(1-\alpha_{t})}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}$ $\displaystyle=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}\right)=O\left(\frac{\log T}{T}\right).$ $$

For fixed $x_{t}$, under [Assumption 4](#Thmassumption4), this implies that $\Delta=O(\log T/T)$, and thus

$$ $e^{\Delta}-1=\Delta+O(\Delta^{2})=\Delta+O\left(\frac{\log^{2}T}{T^{2}}\right).$ $$

Now, in order to upper-bound the integrals that appear in the two terms in ([F.7](#A6.Ex489)), we note that for any $p>0$,

$$ $\int_{x\in\mathbb{R}^{d}}q_{t|0}(x_{t}|x)\left\lVert x\right\rVert^{p}\mathrm{d}Q_{0}(x)=\int_{x\in\mathbb{R}^{d}}\exp\left(-\frac{\left\lVert x-x_{t}/\sqrt{\bar{\alpha}_{t}}\right\rVert^{2}}{2(1-\bar{\alpha}_{t})/\bar{\alpha}_{t}}\right)\left\lVert x\right\rVert^{p}\mathrm{d}Q_{0}(x).$ $$

Write $u(x):=\exp\left(-\frac{\left\lVert x-x_{t}/\sqrt{\bar{\alpha}_{t}}\right\rVert^{2}}{2(1-\bar{\alpha}_{t})/\bar{\alpha}_{t}}\right)\left\lVert x\right\rVert^{p}$. Note that $u(x)>0$ and that $u(x)\searrow 0$ as $\left\lVert x\right\rVert\to\infty$. Define a subset $B:=\left\{x:u(x)\leq\left\lVert x\right\rVert^{2}\right\}$. Since $B$ is bounded away from infinity, there exists a compact set $\overline{B}$ such that $B\subseteq\overline{B}$. By continuity of $u(x)$, $\sup_{x\in\overline{B}}u(x)<\infty$. Thus, since the second moment of $Q_{0}$ is bounded by [Assumption 1](#Thmassumption1), we have

$$ $\displaystyle\int_{x\in\mathbb{R}^{d}}u(x)\mathrm{d}Q_{0}(x)$ $\displaystyle=\int_{x\in\overline{B}}u(x)\mathrm{d}Q_{0}(x)+\int_{x\notin\overline{B}}u(x)\mathrm{d}Q_{0}(x)$ $\displaystyle\leq\sup_{x\in\overline{B}}u(x)+\int_{x\in\mathbb{R}^{d}}\left\lVert x\right\rVert^{2}\mathrm{d}Q_{0}(x)<\infty.$ $$

This guarantees that

$$ $\displaystyle\int_{x,y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)\left\lVert x-y\right\rVert\left\lVert x\right\rVert\mathrm{d}Q_{0}(x)\mathrm{d}Q_{0}(y)$ $\displaystyle<\infty,$ $\displaystyle\int_{x,y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)\left\lVert y\right\rVert^{2}\left\lVert x\right\rVert\mathrm{d}Q_{0}(x)\mathrm{d}Q_{0}(y)$ $\displaystyle<\infty,$ $\displaystyle\int_{x,y\in\mathbb{R}^{d}}q_{t|0}(x_{t}|x)q_{t-1|0}(\mu_{t}|y)\left\lVert x\right\rVert^{3}\mathrm{d}Q_{0}(x)\mathrm{d}Q_{0}(y)$ $\displaystyle<\infty.$ $$

Therefore, combining the results above, we conclude that the second term in ([F.7](#A6.Ex489)) is in the order
$O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}\right)$. Hence, plugging into ([F.7](#A6.Ex478)), we obtain

$$ $\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{3}}\right).$ $$

## Appendix G Proof of Theorem 5

Using the same technique in ([A](#A1.Ex42)) to decompose $\mathrm{KL}(Q||\widehat{P})$, we also need to deal with the three error terms in ([A](#A1.Ex42)) for the accelerated reverse process. With the $\alpha_{t}$’s defined in [Definition 2](#Thmdefinition2), the initialization error satisfies

$$ $\mathbb{E}_{X_{T}\sim Q_{T}}\left[\log\frac{q_{T}(X_{T})}{p^{\prime}_{T}(X_{T})}\right]=o(T^{-2}).$ $$

In the sequel, we focus on improving the reverse-step error and the estimation error in a few steps.

### G.1 Step 1: Improving Estimation Error

We first bound the estimation error, which includes the errors that come from the score and the Hessian estimation. Throughout this subsection we perform analysis under [Assumption 4](#Thmassumption4), where, for fixed $x_{t}$ and in expectation of $X_{t}\sim Q_{t}$, all partial derivatives have upper bounds that do not depend on $T$. In particular, this assumption guarantees that $(1-\alpha_{t})^{p}\partial^{k}_{\bm{a}}\log q_{t}(x)=O((1-\alpha_{t})^{p}),\leavevmode\nobreak\ \forall k,p\geq 1$ for any fixed $x\in\mathbb{R}^{d}$.

To start, note that both $P^{\prime}_{t-1|t}$ and $\widehat{P}^{\prime}_{t-1|t}$ are Gaussian (yet having different mean and covariance). Thus, for each $t=1,\dots,T$,

$$ $\displaystyle\log\frac{p_{t-1|t}(x_{t-1}|x_{t})}{\widehat{p}_{t-1|t}(x_{t-1}|x_{t})}$ $\displaystyle=\log\left(\det(\Sigma_{t})^{-\frac{1}{2}}\right)-\log\left(\det(\hat{\Sigma}_{t})^{-\frac{1}{2}}\right)$ $\displaystyle\qquad-\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\Sigma_{t}^{-1}(x_{t-1}-\mu_{t})+\frac{1}{2}(x_{t-1}-\hat{\mu}_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(x_{t-1}-\hat{\mu}_{t})$ $\displaystyle=\frac{1}{2}\left(\log(\det(\hat{\Sigma}_{t}))-\log(\det(\Sigma_{t}))\right)+\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}(\hat{\Sigma}_{t}^{-1}-\Sigma_{t}^{-1})(x_{t-1}-\mu_{t})$ $\displaystyle\qquad+\frac{1}{2}(x_{t-1}-\hat{\mu}_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(x_{t-1}-\hat{\mu}_{t})-\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(x_{t-1}-\mu_{t})$ $\displaystyle=\frac{1}{2}\left(\log(\det(\hat{\Sigma}_{t}))-\log(\det(\Sigma_{t}))\right)+\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}(\hat{\Sigma}_{t}^{-1}-\Sigma_{t}^{-1})(x_{t-1}-\mu_{t})$ $\displaystyle\qquad+\frac{1}{2}(\mu_{t}-\hat{\mu}_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(x_{t-1}-\mu_{t})+\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(\mu_{t}-\hat{\mu}_{t})+\frac{1}{2}(\mu_{t}-\hat{\mu}_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(\mu_{t}-\hat{\mu}_{t}).$ (69) $$

There are five terms in ([G.1](#A7.Ex513)). We first consider the third and the fourth term, for which we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(\mu_{t}-\hat{\mu}_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(X_{t-1}-\mu_{t})\right]=(\mu_{t}-\hat{\mu}_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[X_{t-1}-\mu_{t}\right]=0,$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}-\mu_{t})^{\intercal}\hat{\Sigma}_{t}^{-1}(\mu_{t}-\hat{\mu}_{t})\right]=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[X_{t-1}-\mu_{t}\right]^{\intercal}\hat{\Sigma}_{t}^{-1}(\mu_{t}-\hat{\mu}_{t})=0.$ $$

Now consider the expectation of the last term in ([G.1](#A7.Ex513)).
From the definition of $\hat{\Sigma}_{t}$ in ([12](#S5.E12)), for small $1-\alpha_{t}$ we have $\hat{\Sigma}_{t}\succ 0$, and we can define $\hat{B}_{t}:=I_{d}-(I_{d}+(1-\alpha_{t})H_{t})^{-1}$, and thus $\hat{\Sigma}_{t}^{-1}=\frac{1}{\sigma_{t}^{2}}(I_{d}-\hat{B}_{t})$. From Taylor expansion, we have $\hat{B}_{t}=(1-\alpha_{t})H_{t}+O((1-\alpha_{t})^{2})$, since from [Assumption 4](#Thmassumption4), for any fixed $x_{t}$, $H_{t}(x_{t})$ has an upper bound independent of $T$.
Thus, for each $t\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left[(\mu_{t}(X_{t})-\hat{\mu}_{t}(X_{t}))^{\intercal}\hat{\Sigma}_{t}^{-1}(X_{t})(\mu_{t}(X_{t})-\hat{\mu}_{t}(X_{t}))\right]$ $\displaystyle=(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t}}\left[(s_{t}(X_{t})-\nabla\log q_{t}(X_{t}))^{\intercal}(I_{d}-\hat{B}_{t}(X_{t}))(s_{t}(X_{t})-\nabla\log q_{t}(X_{t}))\right]$ $\displaystyle=(1-\alpha_{t})\Big{(}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\qquad-\mathbb{E}_{X_{t}\sim Q_{t}}\left[(s_{t}(X_{t})-\nabla\log q_{t}(X_{t}))^{\intercal}\hat{B}_{t}(X_{t})(s_{t}(X_{t})-\nabla\log q_{t}(X_{t}))\right]\Big{)}$ $$

and

$$ $\displaystyle\left|\mathbb{E}_{X_{t}\sim Q_{t}}\left[(s_{t}(X_{t})-\nabla\log q_{t}(X_{t}))^{\intercal}\hat{B}_{t}(X_{t})(s_{t}(X_{t})-\nabla\log q_{t}(X_{t}))\right]\right|$ $\displaystyle\leq\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}\left((1-\alpha_{t})\left\lVert H_{t}(X_{t})\right\rVert+O((1-\alpha_{t})^{2})\right)\right]$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}\left\lVert H_{t}(X_{t})\right\rVert\right]+O((1-\alpha_{t})^{2})$ $\displaystyle\leq(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert H_{t}(X_{t})\right\rVert^{2}}+O((1-\alpha_{t})^{2})$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}+O((1-\alpha_{t})^{2}),$ $$

where $(i)$ and $(ii)$ both follow from [Assumption 4](#Thmassumption4).
Therefore, the expectation of the last term in ([G.1](#A7.Ex513)) can be bounded as

$$ $\displaystyle\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left[(\mu_{t}(X_{t})-\hat{\mu}_{t}(X_{t}))^{\intercal}\hat{\Sigma}_{t}^{-1}(X_{t})(\mu_{t}(X_{t})-\hat{\mu}_{t}(X_{t}))\right]$ $\displaystyle\lesssim\sum_{t=1}^{T}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\qquad+\sum_{t=1}^{T}(1-\alpha_{t})^{2}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}+O((1-\alpha_{t})^{3})$ $\displaystyle\lesssim\frac{\log T}{T}\sum_{t=1}^{T}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}$ $\displaystyle\leq(\log T)\varepsilon^{2},$ (70) $$

where the last line follows by [Assumption 5](#Thmassumption5), and note that $\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}\leq\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}$ following from Jensen’s inequality.

Next we turn to the first two terms in ([G.1](#A7.Ex513)). The first term is given by

$$ $\log(\det(\hat{\Sigma}_{t}))-\log(\det(\Sigma_{t}))=\log(\det(I_{d}+(1-\alpha_{t})H_{t}))-\log(\det(I_{d}+(1-\alpha_{t})\nabla^{2}\log q_{t}(x_{t}))).$ $$

When $(1-\alpha_{t})$ is small, we can use Taylor expansion for the functions $\det(\cdot)$ and $\log(\cdot)$ to get

$$ $\displaystyle\log(\det(I_{d}+(1-\alpha_{t})H_{t}))$ $\displaystyle=\log\left(1+(1-\alpha_{t})\mathrm{Tr}(H_{t})+\frac{(1-\alpha_{t})^{2}}{2}(\mathrm{Tr}(H_{t})^{2}-\mathrm{Tr}(H_{t}^{2}))+O((1-\alpha_{t})^{3})\right)$ $\displaystyle=(1-\alpha_{t})\mathrm{Tr}(H_{t})+\frac{(1-\alpha_{t})^{2}}{2}(\mathrm{Tr}(H_{t})^{2}-\mathrm{Tr}(H_{t}^{2}))-\frac{(1-\alpha_{t})^{2}}{2}\mathrm{Tr}(H_{t})^{2}+O((1-\alpha_{t})^{3})$ $\displaystyle=(1-\alpha_{t})\mathrm{Tr}(H_{t})-\frac{(1-\alpha_{t})^{2}}{2}\mathrm{Tr}(H_{t}^{2})+O((1-\alpha_{t})^{3}).$ $$

Similar expression can be obtained for $\log(\det(I_{d}+(1-\alpha_{t})\nabla^{2}\log q_{t}(x_{t})))$. Thus, the first term in ([G.1](#A7.Ex513)) is equal to

$$ $\displaystyle\log(\det(\hat{\Sigma}_{t}))-\log(\det(\Sigma_{t}))$ $\displaystyle=(1-\alpha_{t})\left(\mathrm{Tr}(H_{t})-\mathrm{Tr}(\nabla^{2}\log q_{t}(x_{t}))\right)-\frac{(1-\alpha_{t})^{2}}{2}\left[\mathrm{Tr}(H_{t}^{2})-\mathrm{Tr}((\nabla^{2}\log q_{t}(x_{t}))^{2})\right]$ $\displaystyle\qquad+O((1-\alpha_{t})^{3}).$ $$

For the second term in ([G.1](#A7.Ex513)), we first take expectation over $x_{t-1}$ and get

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}-\mu_{t})^{\intercal}(\hat{\Sigma}_{t}^{-1}-\Sigma_{t}^{-1})(X_{t-1}-\mu_{t})\right]=\mathrm{Tr}\left((\hat{\Sigma}_{t}^{-1}-\Sigma_{t}^{-1})\Sigma_{t}\right).$ $$

To proceed, note that

$$ $(I_{d}+(1-\alpha_{t})H_{t})^{-1}\stackrel{{\scriptstyle(iii)}}{{=}}I_{d}-(1-\alpha_{t})H_{t}+(1-\alpha_{t})^{2}H_{t}^{2}+O((1-\alpha_{t})^{3}).$ $$

To see $(iii)$, we write $S_{t}$ as the true inverse of $I_{d}+(1-\alpha_{t})H_{t}$. Its existence is guaranteed if $(1-\alpha_{t})$ is small. Since

$$ $(I_{d}+(1-\alpha_{t})H_{t})(I_{d}-(1-\alpha_{t})H_{t}+(1-\alpha_{t})^{2}H_{t}^{2})=I_{d}+O((1-\alpha_{t})^{3}),$ $$

we have

$$ $(I_{d}+(1-\alpha_{t})H_{t})(I_{d}-(1-\alpha_{t})H_{t}+(1-\alpha_{t})^{2}H_{t}^{2}-S_{t})=O((1-\alpha_{t})^{3})$ $$

which implies that $S_{t}=I_{d}-(1-\alpha_{t})H_{t}+(1-\alpha_{t})^{2}H_{t}^{2}+O((1-\alpha_{t})^{3})$. This shows the validity of $(iii)$. Therefore,

$$ $\displaystyle\mathrm{Tr}\left((\hat{\Sigma}_{t}^{-1}-\Sigma_{t}^{-1})\Sigma_{t}\right)=\mathrm{Tr}(\hat{\Sigma}_{t}^{-1}\Sigma_{t}-I_{d})$ $\displaystyle=\mathrm{Tr}\left(\left[I_{d}-(1-\alpha_{t})H_{t}+(1-\alpha_{t})^{2}H_{t}^{2}+O((1-\alpha_{t})^{3})\right]\left[I_{d}+(1-\alpha_{t})\nabla^{2}\log q_{t}(x_{t})\right]-I_{d}\right)$ $\displaystyle=(1-\alpha_{t})\left[\mathrm{Tr}(\nabla^{2}\log q_{t}(x_{t}))-\mathrm{Tr}(H_{t})\right]$ $\displaystyle\qquad+(1-\alpha_{t})^{2}\left[\mathrm{Tr}(H_{t}^{2})-\mathrm{Tr}(H_{t}\nabla^{2}\log q_{t}(x_{t}))\right]+O((1-\alpha_{t})^{3}).$ $$

Adding this to the first term of ([G.1](#A7.Ex513)) and taking expectation over $X_{t}\sim Q_{t}$ (noting [Assumption 4](#Thmassumption4) here), we get

$$ $\displaystyle\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\Big{[}\left(\log(\det(\hat{\Sigma}_{t}(X_{t})))-\log(\det(\Sigma_{t}(X_{t})))\right)$ $\displaystyle\qquad+(X_{t-1}-\mu_{t}(X_{t}))^{\intercal}(\hat{\Sigma}_{t}^{-1}(X_{t})-\Sigma_{t}^{-1}(X_{t}))(X_{t-1}-\mu_{t}(X_{t}))\Big{]}$ $\displaystyle=\frac{(1-\alpha_{t})^{2}}{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\mathrm{Tr}(H_{t}(X_{t})^{2})-2\mathrm{Tr}(H_{t}(X_{t})\nabla^{2}\log q_{t}(X_{t}))+\mathrm{Tr}((\nabla^{2}\log q_{t}(X_{t}))^{2})\right]$ $\displaystyle\qquad+O((1-\alpha_{t})^{3})$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{=}}\frac{(1-\alpha_{t})^{2}}{2}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert H_{t}(X_{t})-\nabla^{2}\log q_{t}(X_{t})\right\rVert_{F}^{2}+O((1-\alpha_{t})^{3}),$ $$

where $(iv)$ follows because for two symmetric matrices $A$ and $B$,

$$ $\displaystyle\mathrm{Tr}(A^{2})-2\mathrm{Tr}(AB)+\mathrm{Tr}(B^{2})=\mathrm{Tr}(A^{2})-\mathrm{Tr}(AB)-\mathrm{Tr}(BA)+\mathrm{Tr}(B^{2})$ $\displaystyle=\mathrm{Tr}((A-B)(A-B))=\mathrm{Tr}((A-B)^{\intercal}(A-B))=\left\lVert A-B\right\rVert_{F}^{2}.$ $$

Thus, following from [Assumption 5](#Thmassumption5),

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\Big{[}\left(\log(\det(\hat{\Sigma}_{t}(X_{t})))-\log(\det(\Sigma_{t}(X_{t})))\right)\\ +(X_{t-1}-\mu_{t}(X_{t}))^{\intercal}(\hat{\Sigma}_{t}^{-1}(X_{t})-\Sigma_{t}^{-1}(X_{t}))(X_{t-1}-\mu_{t}(X_{t}))\Big{]}\lesssim\frac{\log^{2}T}{T}\varepsilon_{H}^{2}.$ (71) $$

Here $\varepsilon_{H}$ is the Hessian estimation error. Combining ([G.1](#A7.Ex530)) and ([71](#A7.E71)) yields the desired result for the accelerated estimation error, which is in the order $\tilde{O}(1/T^{2})$.

### G.2 Step 2: Re-expressing Log-likelihood Ratio in Accelerated Process

Next we focus on the reverse-step error for the accelerated process. Recall that $Q_{0}$ is smooth.
We introduce the following notations for analysis. Let

$$ $A_{t}(x_{t}):=(1-\alpha_{t})\nabla^{2}\log q_{t}(x_{t}),\quad B_{t}(x_{t}):=I_{d}-(I_{d}+A_{t})^{-1},$ (72) $$

which imply that

$$ $\Sigma_{t}(x_{t})=\sigma_{t}^{2}(I_{d}+A_{t}(x_{t})),\quad\Sigma_{t}^{-1}(x_{t})=\sigma_{t}^{-2}(I_{d}-B_{t}(x_{t})).$ $$

When $(1-\alpha_{t})$ is small and $\left\lVert\nabla^{2}\log q_{t}(x_{t})\right\rVert$ satisfies [Assumption 4](#Thmassumption4), we can perform Taylor expansion on $B_{t}$ around $A_{t}$ (for fixed $x_{t}$) and obtain

$$ $B_{t}=A_{t}+O((1-\alpha_{t})^{2}).$ (73) $$

Fix $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$. Write $\mu_{t}(x_{t})=\mu_{t}$, $A_{t}(x_{t})=A_{t}$, and $B_{t}(x_{t})=B_{t}$ for brevity.

###### Lemma 19 .

For any fixed $x_{t}\in\mathbb{R}^{d}$, we have

$$ $q_{t-1|t}(x_{t-1}|x_{t})=\frac{p^{\prime}_{t-1|t}(x_{t-1}|x_{t})e^{\zeta^{\prime}_{t,t-1}(x_{t},x_{t-1})}}{\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[e^{\zeta^{\prime}_{t,t-1}(x_{t},X_{t-1})}]},$ $$

where

$$ $\displaystyle\zeta^{\prime}_{t,t-1}(x_{t},x_{t-1})$ $\displaystyle:=\zeta_{t,t-1}(x_{t},x_{t-1})-\frac{1}{2\sigma_{t}^{2}}(x_{t-1}-\mu_{t})^{\intercal}B_{t}(x_{t-1}-\mu_{t})$ $\displaystyle=\log q_{t-1}(x_{t-1})-\log q_{t-1}(\mu_{t})-(x_{t-1}-\mu_{t})^{\intercal}(\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))$ $\displaystyle\qquad-\frac{1}{2\sigma_{t}^{2}}(x_{t-1}-\mu_{t})^{\intercal}B_{t}(x_{t-1}-\mu_{t}),$ (74) $$

and $\zeta_{t,t-1}(x_{t},x_{t-1})$ is as defined in ([18](#A1.E18)).

###### Proof.

See [section I.3](#A9.SS3).
∎

Compared with the tilting factor for the regular process (in ([18](#A1.E18))), an additional term that is related to $\Sigma_{t}$ (and thus $B_{t}$) is introduced in $\zeta^{\prime}_{t,t-1}$. Through this extra term, we can further control the second-order term in the Taylor expansion of $\log q_{t-1}$ around $\mu_{t}$.

To continue, similar to ([27](#A2.E27)), we obtain

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|x_{t})}{p^{\prime}_{t-1|t}(X_{t-1}|x_{t})}\right]\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]\\ -\log\left(1+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]\right).$ (75) $$

Note that $\log q_{t-1}(t\geq 1)$ is still analytic by [Lemma 6](#Thmlemma6) and [Assumption 3](#Thmassumption3). Thus, the Taylor expansion of $\zeta^{\prime}_{t,t-1}(x_{t},x_{t-1})$ around $x_{t-1}=\mu_{t}$ is (cf. ([21](#A1.E21)))

$$ $\displaystyle\zeta^{\prime}_{t,t-1}$ $\displaystyle=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}(x_{t-1}-\mu_{t})$ $\displaystyle\qquad+\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\left(\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}B_{t}\right)(x_{t-1}-\mu_{t})$ $\displaystyle\qquad+\sum_{p=3}^{\infty}T_{p}(\log q_{t-1},x_{t-1},\mu_{t})$ (76) $$

where $T_{p}(f,x,\mu)$ is defined in ([20](#A1.E20)). Now we move on to investigate the expected values of $\zeta^{\prime}_{t,t-1}$ under $P^{\prime}_{t-1|t}$ and $Q_{t-1|t}$. In order to differentiate from the $T_{2}$ in [Appendix A](#A1), we write

$$ $T^{\prime}_{2}:=\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\left(\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}B_{t}\right)(x_{t-1}-\mu_{t}).$ $$

###### Lemma 19 .

###### Proof.

### G.3 Step 3: Conditional Expectation of ζ t , t − 1 ′ subscript superscript 𝜁 ′ 𝑡 𝑡 1 \zeta^{\prime}_{t,t-1} under P t − 1 | t ′ subscript superscript 𝑃 ′ 𝑡 conditional 1 𝑡 P^{\prime}_{t-1|t} and Q t − 1 | t subscript 𝑄 𝑡 conditional 1 𝑡 Q_{t-1|t}

Below we calculate the centralized moments under $P^{\prime}_{t-1|t}$. Calculation in the accelerated process becomes more involved since the $P^{\prime}_{t-1|t}$ is no longer independent. Fortunately, we can employ Isserlis’s Theorem for our help, which constitutes the main idea in the lemma below. Note that the results in this subsection hold for both smooth ($\forall t\geq 1$) and general ($\forall t\geq 2$) $Q_{0}$’s.

###### Lemma 20 .

Fix $t\geq 1$. For brevity write $Z_{i}=X_{t-1}^{i}-\mu_{t}^{i},\leavevmode\nobreak\ \forall i\in[d]$, $A=A_{t}(x_{t})$, and $\mathbb{E}[\cdot]$ as a shorthand for $\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\cdot]$. Recall the accelerated reverse process in ([10](#S5.E10)). Under [Assumption 4](#Thmassumption4), we have $A_{t}=O(1-\alpha_{t})$, and the following results hold:

$$ $\displaystyle\mathbb{E}\left[\prod_{i\in\bm{a}}Z_{i}\right]$ $\displaystyle=0,\quad\forall\bm{a}:\left|\bm{a}\right|\text{ odd},$ $\displaystyle\mathbb{E}\left[\prod_{i\in\bm{a}}Z_{i}\right]$ $\displaystyle=O\left((1-\alpha_{t})^{\frac{\left|\bm{a}\right|}{2}}\right),\quad\forall\bm{a}:\left|\bm{a}\right|\text{ even}.$ $$

Specifically, for $i,j,k,l\in[d]$ all differ, the fourth moment is

$$ $\displaystyle\mathbb{E}[Z_{i}^{4}]$ $\displaystyle=3\sigma_{t}^{4}(1+A^{ii})^{2}$ $\displaystyle\mathbb{E}[Z_{i}^{3}Z_{j}]$ $\displaystyle=3\sigma_{t}^{4}A^{ij}(1+A^{ii})$ $\displaystyle\mathbb{E}[Z_{i}^{2}Z_{j}^{2}]$ $\displaystyle=\sigma_{t}^{4}(1+A^{ii})(1+A^{jj})+O((1-\alpha_{t})^{4})$ $\displaystyle\mathbb{E}[Z_{i}^{2}Z_{j}Z_{k}]$ $\displaystyle=\sigma_{t}^{4}(1+A^{ii})A^{jk}+O((1-\alpha_{t})^{4})$ $\displaystyle\mathbb{E}[Z_{i}Z_{j}Z_{k}Z_{l}]$ $\displaystyle=O((1-\alpha_{t})^{4}).$ $$

For $i,j,k\in[d]$ all differ, the sixth moment is

$$ $\displaystyle\mathbb{E}[Z_{i}^{6}]$ $\displaystyle=15\sigma_{t}^{6}(1+A^{ii})^{3}$ $\displaystyle\mathbb{E}[Z_{i}^{4}Z_{j}^{2}]$ $\displaystyle=3\sigma_{t}^{6}(1+A^{ii})^{2}(1+A^{jj})+O((1-\alpha_{t})^{4})$ $\displaystyle\mathbb{E}[Z_{i}^{2}Z_{j}^{2}Z_{k}^{2}]$ $\displaystyle=\sigma_{t}^{6}(1+A^{ii})(1+A^{jj})(1+A^{kk})+O((1-\alpha_{t})^{4}),$ $$

and $\mathbb{E}\left[\prod_{i\in\bm{a}:\left|\bm{a}\right|=6}Z_{i}\right]=O((1-\alpha_{t})^{4})$ otherwise.

###### Proof.

See [section I.4](#A9.SS4).
∎

Next we turn to calculate the fifth and sixth centralized moment under $Q_{t-1|t}$, again drawing the idea of Tweedie’s formula []. This is a direct extension to [Lemma 8](#Thmlemma8).

###### Lemma 21 .

Fix $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$. Under [Assumption 4](#Thmassumption4), we have $A_{t}=O(1-\alpha_{t})$. With the same definitions of $\tilde{x}_{t}$ and $\kappa(\tilde{x}_{t})$ as in [Lemma 8](#Thmlemma8), the fifth centralized moment is

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})(X_{t-1}^{k}-\mu_{t}^{k})(X_{t-1}^{l}-\mu_{t}^{l})(X_{t-1}^{m}-\mu_{t}^{m})\right]\\ =\sum_{\xi\in\binom{\{i,j,k,l,m\}}{2}}(\partial^{2}_{\xi}\kappa)(\partial^{3}_{\{i,j,k,l,m\}\setminus\xi}\kappa)+\partial^{5}_{ijklm}\kappa=O((1-\alpha_{t})^{4})$ $$

where, given a set $A$, we define

$$ $\binom{A}{2}:=\Big{\{}\{a_{1},a_{2}\}:a_{1},a_{2}\in A,\leavevmode\nobreak\ a_{1}\neq a_{2}\Big{\}}.$ $$

Let $P_{n}^{k}$ be the set that contains all distinct size-$k$ partitions of $[n]$. Define

$$ $\mathrm{part}_{2}(A):=\{((a_{i},a_{j}):\{i,j\}\in p):p\in P_{\left|A\right|}^{2}\}.$ $$

The sixth centralized moment is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})(X_{t-1}^{k}-\mu_{t}^{k})(X_{t-1}^{l}-\mu_{t}^{l})(X_{t-1}^{m}-\mu_{t}^{m})(X_{t-1}^{n}-\mu_{t}^{n})\right]$ $\displaystyle=\sum_{(\xi_{1},\xi_{2},\xi_{3})\in\mathrm{part}_{2}(\{i,j,k,l,m,n\})}(\partial^{2}_{\xi_{1}}\kappa)(\partial^{2}_{\xi_{2}}\kappa)(\partial^{2}_{\xi_{3}}\kappa)+O((1-\alpha_{t})^{4})$ $\displaystyle=\begin{cases}15\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}+O((1-\alpha_{t})^{4}),&\text{if}\ i=j=k=l=m=n\\ 3\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}+O((1-\alpha_{t})^{4}),&\text{if}\ i=k=m=n\neq j=l\\ \left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}+O((1-\alpha_{t})^{4}),&\text{if}\ i=l,j=m,k=n\leavevmode\nobreak\ \text{while $i,j,k$ all differ}\\ O((1-\alpha_{t})^{4}),&\text{otherwise}\end{cases}$ $$

Again note that all derivatives above are w.r.t. $\tilde{x}_{t}$.

###### Proof.

See [section I.5](#A9.SS5).
∎

In the following lemma, we use the results in the previous two lemmas to control the difference $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[\zeta^{\prime}_{t,t-1}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]$, which will be shown to be the term that dominates the rate in the reverse-step error.

###### Lemma 22 .

Suppose that all derivatives of $\log q_{t-1}$ exist and that [Assumption 4](#Thmassumption4) holds. Then, for any fixed $x_{t}\in\mathbb{R}^{d}$,

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[\zeta^{\prime}_{t,t-1}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]\\ =\frac{(1-\alpha_{t})^{3}}{3!\alpha_{t}^{3/2}}\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})+O((1-\alpha_{t})^{4}).$ $$

###### Proof.

See [section I.6](#A9.SS6).
∎

###### Lemma 20 .

###### Proof.

###### Lemma 21 .

###### Proof.

###### Lemma 22 .

###### Proof.

### G.4 Step 4: Conditional Moments of ζ t , t − 1 ′ subscript superscript 𝜁 ′ 𝑡 𝑡 1 \zeta^{\prime}_{t,t-1} under P t − 1 | t ′ subscript superscript 𝑃 ′ 𝑡 conditional 1 𝑡 P^{\prime}_{t-1|t}

In [Corollary 1](#Thmcorollary1), we improve the convergence result in [Theorem 1](#Thmtheorem1) by introducing ([6](#S3.E6)), which guarantees that the first order term in $\zeta_{t,t-1}$ is small for any fixed $x_{t}$. Now, in order to obtain a better convergence rate for the accelerated sampler, we use the condition ([15](#S5.E15)) to show that the second order term in ([G.2](#A7.Ex561)) is also small.

###### Lemma 23 .

Suppose that both ([6](#S3.E6)) and ([15](#S5.E15)) hold. Under [Assumption 4](#Thmassumption4), with the noise schedule according to [Definition 2](#Thmdefinition2), we have, for any $t\geq 1$ and fixed $x_{t}$,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]$ $\displaystyle=O\left((1-\alpha_{t})^{1.5}\right),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]$ $\displaystyle=O\left((1-\alpha_{t})^{2}\right),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]$ $\displaystyle=O\left((1-\alpha_{t})^{3.5}\right).$ $$

###### Proof.

See [section I.7](#A9.SS7).
∎

###### Lemma 23 .

###### Proof.

### G.5 Improving Reverse-step Error

Below, we combine [Lemmas 22](#Thmlemma22) and [23](#Thmlemma23) to upper-bound the reverse-step error in the accelerated process.
Recall that $Q_{0}$ is smooth ([Assumption 3](#Thmassumption3)).
Following from ([75](#A7.E75)), for each $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$, we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p^{\prime}_{t-1|t}(X_{t-1}|X_{t})}\right]$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]$ $\displaystyle\qquad-\log\left(1+\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]\right)$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]-\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]$ $\displaystyle\qquad+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]\right)^{2}$ $\displaystyle\qquad+O\left(\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]\right)^{3}\right)$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]\right)^{2}$ $\displaystyle\qquad+O((1-\alpha_{t})^{3.5})$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+O((1-\alpha_{t})^{3.5})$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}\frac{(1-\alpha_{t})^{3}}{3!\alpha_{t}^{3/2}}\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})+O((1-\alpha_{t})^{3.5}).$ $$

Here $(i)$ follows from [Lemma 23](#Thmlemma23), $(ii)$ follows from Jensen’s inequality, and $(iii)$ follows from [Lemma 22](#Thmlemma22). Also, recall that under [Assumption 4](#Thmassumption4) we can take the outer expectation over $X_{t}\sim Q_{t}$ while maintaining the rate of each term.
Therefore, we can sum up over $t\geq 1$ and obtain

$$ $\displaystyle\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]$ $\displaystyle\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{3}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})\right].$ (77) $$

This completes the proof of [Theorem 5](#Thmtheorem5).

## Appendix H Proof of Corollaries 3 , 6 , 7 and 8

### H.1 Proof of Corollary 3

We now extend [Theorem 5](#Thmtheorem5) to general $Q_{0}$. We will use the $\alpha_{t}$’s in [Equation 8](#S4.E8), yet with $c>4$. From ([63](#A6.E63)), we can show that this choice guarantees that

$$ $\bar{\alpha}_{T}=o(T^{-2}).$ $$

Since all $q_{t}(t\geq 1)$ exist (by [Lemma 6](#Thmlemma6)), if [Assumption 4](#Thmassumption4) holds, the result in [Lemma 22](#Thmlemma22) still applies. It remains to show that the expected powers of $\zeta^{\prime}_{t,t-1}$ is small under $P^{\prime}_{t-1|t}$. When $Q_{0}$ is smooth, this is guaranteed by ([6](#S3.E6)) and ([15](#S5.E15)). While an analogy to ([6](#S3.E6)) is shown in [Lemma 18](#Thmlemma18), there is still a need to show a counterpart to ([15](#S5.E15)), which we pursue below.

###### Lemma 24 .

Fix $t\geq 2$.
Suppose that [Assumption 4](#Thmassumption4) holds and that the $\alpha_{t}$’s satisfy ([51](#A5.E51)).
For any $Q_{0}$ that satisfies [Assumption 1](#Thmassumption1) (not necessarily having a p.d.f.) and any fixed $x_{t}\in\mathbb{R}^{d}$,

$$ $\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{4}}\right).$ $$

###### Proof.

See [section I.8](#A9.SS8).
∎

Using [Lemmas 18](#Thmlemma18) and [24](#Thmlemma24), we can show that the expected second and third powers of $\zeta^{\prime}_{t,t-1}$ are small under $P^{\prime}_{t-1|t}$, which is contained in the following lemma.

###### Lemma 25 .

Under [Assumption 4](#Thmassumption4), suppose that both

- 1.
$\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})=O((1-\alpha_{t})/(1-\bar{\alpha}_{t-1})^{3})$, and
- 2.
$\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})=O((1-\alpha_{t})/(1-\bar{\alpha}_{t-1})^{4})$

hold for any fixed $x_{t}$. Then, if the $\alpha_{t}$’s satisfy ([51](#A5.E51)), for $t\geq 2$, we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]$ $\displaystyle=O\left((1-\alpha_{t})^{2}/(1-\bar{\alpha}_{t-1})^{4}\right)=O\left(\frac{\log^{2}T}{T^{2}}\right),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]$ $\displaystyle=O\left((1-\alpha_{t})^{3}/(1-\bar{\alpha}_{t-1})^{6}\right)=O\left(\frac{\log^{3}T}{T^{3}}\right),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]$ $\displaystyle=O\left((1-\alpha_{t})^{5}/(1-\bar{\alpha}_{t-1})^{10}\right)=O\left(\frac{\log^{5}T}{T^{5}}\right).$ $$

###### Proof.

See [section I.9](#A9.SS9).
∎

We now continue from ([75](#A7.E75)) to provide an upper bound on the reverse-step error for general $Q_{0}$. Note that the $\alpha_{t}$’s in ([8](#S4.E8)) satisfy $\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}=O(\log T/T)$ by [Lemma 16](#Thmlemma16). Thus, for all $t\geq 2$,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p^{\prime}_{t-1|t}(X_{t-1}|X_{t})}\right]$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]-\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]$ $\displaystyle\qquad+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]\right)^{2}$ $\displaystyle\qquad+O\left(\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{3!}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]\right)^{3}\right)$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]-\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]+\frac{1}{2}\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]\right)^{2}$ $\displaystyle\qquad+O\left(\frac{(1-\alpha_{t})^{4}}{(1-\bar{\alpha}_{t-1})^{8}}\right)$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\zeta^{\prime}_{t,t-1}\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]+O\left(\frac{(1-\alpha_{t})^{4}}{(1-\bar{\alpha}_{t-1})^{8}}\right)$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}\frac{(1-\alpha_{t})^{3}}{3!\alpha_{t}^{3/2}}\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})+O\left(\frac{(1-\alpha_{t})^{4}}{(1-\bar{\alpha}_{t-1})^{8}}\right).$ $$

Here $(i)$ follows from [Lemma 25](#Thmlemma25), $(ii)$ follows from Jensen’s inequality, and $(iii)$ follows from [Lemma 22](#Thmlemma22). Since taking the expectation with $X_{t}\sim Q_{t}$ does not change the rate of each term ([Assumption 4](#Thmassumption4)), the proof is now complete.

###### Lemma 24 .

###### Proof.

###### Lemma 25 .

###### Proof.

### H.2 Proof of Theorem 6

First we check conditions ([6](#S3.E6)) and ([15](#S5.E15)). In [Lemma 13](#Thmlemma13), it is shown that ([6](#S3.E6)) is satisfied under Lipschitz gradient. It remains to show ([15](#S5.E15)) under the additional condition on Lipschitz Hessian.

###### Lemma 26 .

Suppose that $\nabla\log q_{t}$ and $\nabla^{2}\log q_{t}$ are, respectively, norm-Lipschitz and 2-norm-Lipschitz for all $t\geq 0$ (as in ([6](#S5.Ex30))). Then, for any fixed $x_{t}\in\mathbb{R}^{d}$, we have

$$ $\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})\right\rVert^{2}=O\left(1-\alpha_{t}\right),\leavevmode\nobreak\ \forall t\geq 1.$ $$

As a result, condition ([15](#S5.E15)) holds.

###### Proof.

See [section I.10](#A9.SS10).
∎

In order to continue from [Theorem 5](#Thmtheorem5) (in particular, ([G.5](#A7.Ex592))), we need to introduce some useful notations for the distribution class in ([6](#S5.Ex30)). For a matrix $A$, define its vectorization as $\mathrm{vec}(A):=[A^{11},\dots,A^{1d},\dots,A^{d1},\dots,A^{dd}]^{\intercal}\in\mathbb{R}^{d^{2}}$. Define $K_{t}\in\mathbb{R}^{d^{2}\times d}$ to be the matrix that reorganizes the third-order partial derivative tensor, i.e.,

$$ $[K_{t}(x)]^{mk}:=\partial^{3}_{ijk}\log q_{t}(x),\leavevmode\nobreak\ \text{s.t.}\leavevmode\nobreak\ m=(i-1)d+j,\leavevmode\nobreak\ \forall i,j,k\in[d].$ $$

With these notations, consider $y=x+\xi u$ where $u\in\mathbb{R}^{d}$ satisfies $\left\lVert u\right\rVert^{2}=1$ and $\xi\in\mathbb{R}$ is some small constant. Then,

$$ $\mathrm{vec}(\nabla^{2}\log q_{t}(y))-\mathrm{vec}(\nabla^{2}\log q_{t}(x))=K_{t}(x^{*})(y-x)=\xi K_{t}(x^{*})u.$ $$

Here $x^{*}=\gamma x+(1-\gamma)y$ for some $\gamma\in(0,1)$. Also, we have

$$ $\displaystyle\left\lVert\mathrm{vec}(\nabla^{2}\log q_{t}(y))-\mathrm{vec}(\nabla^{2}\log q_{t}(x))\right\rVert$ $\displaystyle=\left\lVert\nabla^{2}\log q_{t}(y)-\nabla^{2}\log q_{t}(x)\right\rVert_{F}$ $\displaystyle\leq\sqrt{d}\left\lVert\nabla^{2}\log q_{t}(y)-\nabla^{2}\log q_{t}(x)\right\rVert\leq\sqrt{d}M\left\lVert y-x\right\rVert$ $$

where the last inequality comes from ([6](#S5.Ex30)). Thus, noting that $y=x+\xi u$ and that $\left\lVert u\right\rVert^{2}=1$, we take the limit of $\xi$ to 0 and get

$$ $\left\lVert K_{t}(x)\right\rVert\leq\sqrt{d}M,\quad\forall x\in\mathbb{R}^{d},\leavevmode\nobreak\ \forall t\geq 0.$ $$

We now derive an explicit upper bound on the reverse-step error. Using Cauchy-Schwartz inequality, for any $t\geq 1$ and $x_{t}\in\mathbb{R}^{d}$, we have

$$ $\displaystyle\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})$ $\displaystyle\leq\sqrt{\sum_{i,j,k=1}^{d}(\partial^{3}_{ijk}\log q_{t-1}(\mu_{t}))^{2}}\sqrt{\sum_{i,j,k=1}^{d}(\partial^{3}_{ijk}\log q_{t}(x_{t}))^{2}}$ $\displaystyle=\left\lVert K_{t-1}(\mu_{t})\right\rVert_{F}\times\left\lVert K_{t}(x_{t})\right\rVert_{F}$ $\displaystyle\leq(\sqrt{d}\left\lVert K_{t-1}(\mu_{t})\right\rVert)\times(\sqrt{d}\left\lVert K_{t}(x_{t})\right\rVert)$ $\displaystyle\leq d^{2}M^{2}.$ $$

Therefore, following from [Theorem 5](#Thmtheorem5), we obtain

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d^{2}M^{2}\log^{3}T}{T^{2}}.$ $$

###### Lemma 26 .

###### Proof.

### H.3 Proof of Theorem 7

We remind readers that [Assumption 4](#Thmassumption4) is satisfied from [Corollary 4](#Thmcorollary4).
This proof is similar to that of [Theorem 3](#Thmtheorem3), and we remind readers of the notations $q_{t,n}$, $\mu_{t,n}$, and $\Sigma_{t,n}$ in [Theorem 3](#Thmtheorem3). Also, condition ([6](#S3.E6)) has been checked in [Lemma 14](#Thmlemma14). In order to invoke [Theorem 5](#Thmtheorem5), we need to check the additional condition ([15](#S5.E15)), as shown in the following lemma.

###### Lemma 27 .

When $Q_{0}$ is Gaussian mixture (see [Theorem 3](#Thmtheorem3)), then, for all fixed $x_{t}$,

$$ $\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})=O\left(1-\alpha_{t}\right).$ $$

As a result, ([15](#S5.E15)) holds.

###### Proof.

See [section I.11](#A9.SS11).
∎

Now we continue from [Theorem 5](#Thmtheorem5) to work for an explicit dependence on $d$. Again write $z_{t,n}(x):=\Sigma_{t,n}^{-1}(x-\mu_{t,n})$. Note that $\partial_{k}z_{t,n}^{i}=[\Sigma_{t,n}^{-1}]^{ik}$, and that $\partial_{k}q_{t,n}(x)=q_{t,n}(x)(-z_{t,n}^{k}(x))$.

From ([D.2](#A4.Ex258)), the second partial derivative of its log-p.d.f. is

$$ $\displaystyle\partial^{2}_{ij}\log q_{t}(x)$ $\displaystyle=\frac{1}{q_{t}^{2}(x)}\Bigg{(}q_{t}(x)\underbrace{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left(z_{t,n}^{i}(x)z_{t,n}^{j}(x)-[\Sigma_{t,n}^{-1}]^{ij}\right)}_{\text{N1}}$ $\displaystyle\qquad-\underbrace{\left(\sum_{n}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\right)\left(\sum_{n}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)\right)}_{\text{N2}}\Bigg{)}.$ $$

To calculate the third partial derivative of its log-p.d.f., we need first to calculate the partial derivative of N1 and N2. The derivative for N1 is given by

$$ $\displaystyle\partial_{k}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left(z_{t,n}^{i}(x)z_{t,n}^{j}(x)-[\Sigma_{t,n}^{-1}]^{ij}\right)$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)(-z_{t,n}^{k}(x))\left(z_{t,n}^{i}(x)z_{t,n}^{j}(x)-[\Sigma_{t,n}^{-1}]^{ij}\right)$ $\displaystyle\qquad+\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ik}z_{t,n}^{j}(x)+\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{jk}z_{t,n}^{i}(x),$ $$

and the derivative for term N2 is given by

$$ $\displaystyle\partial_{k}\left(\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)\right)$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left((-z_{t,n}^{k}(x))z_{t,n}^{i}(x)+[\Sigma_{t,n}^{-1}]^{ik}\right)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)$ $\displaystyle\qquad+\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left((-z_{t,n}^{k}(x))z_{t,n}^{j}(x)+[\Sigma_{t,n}^{-1}]^{jk}\right).$ $$

Combining these, the derivative for the numerator is

$$ $\displaystyle\partial_{k}(q_{t}(x)\text{N1}-\text{N2})=\partial_{k}(q_{t}(x))\text{N1}+q_{t}(x)\partial_{k}(\text{N1})-\partial_{k}(\text{N2})$ $\displaystyle=-\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left(z_{t,n}^{i}(x)z_{t,n}^{j}(x)-[\Sigma_{t,n}^{-1}]^{ij}\right)$ $\displaystyle\qquad+q_{t}(x)\Bigg{(}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)(-z_{t,n}^{k}(x))\left(z_{t,n}^{i}(x)z_{t,n}^{j}(x)-[\Sigma_{t,n}^{-1}]^{ij}\right)$ $\displaystyle\qquad+\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ik}z_{t,n}^{j}(x)+\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{jk}z_{t,n}^{i}(x)\Bigg{)}$ $\displaystyle\qquad-\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left((-z_{t,n}^{k}(x))z_{t,n}^{i}(x)+[\Sigma_{t,n}^{-1}]^{ik}\right)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)$ $\displaystyle\qquad-\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)\left((-z_{t,n}^{k}(x))z_{t,n}^{j}(x)+[\Sigma_{t,n}^{-1}]^{jk}\right)$ $\displaystyle=-q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)z_{t,n}^{j}(x)z_{t,n}^{k}(x)-\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)z_{t,n}^{j}(x)$ $\displaystyle\qquad+\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)z_{t,n}^{k}(x)+\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)z_{t,n}^{k}(x)$ $\displaystyle\qquad+q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ij}z_{t,n}^{k}(x)+\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ij}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)$ $\displaystyle\qquad+q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ik}z_{t,n}^{j}(x)-\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ik}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)$ $\displaystyle\qquad+q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{jk}z_{t,n}^{i}(x)-\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{jk}.$ $$

Since

$$ $\displaystyle\partial^{3}_{ijk}\log q_{t}(x)$ $\displaystyle=\partial_{k}\left(\frac{q_{t}(x)\text{N1}-\text{N2}}{q_{t}^{2}(x)}\right)$ $\displaystyle=\frac{1}{q_{t}^{3}(x)}\left(\partial_{k}(q_{t}(x)\text{N1}-\text{N2})q_{t}(x)+2(q_{t}(x)\text{N1}-\text{N2})\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)\right),$ $$

we get

$$ $\displaystyle q_{t}^{3}(x)\partial^{3}_{ijk}\log q_{t}(x)$ $\displaystyle=-q^{2}_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)z_{t,n}^{j}(x)z_{t,n}^{k}(x)+q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)z_{t,n}^{j}(x)$ $\displaystyle\qquad+q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)z_{t,n}^{k}(x)$ $\displaystyle\qquad+q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)z_{t,n}^{k}(x)$ $\displaystyle\qquad-2\left(\sum_{n}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\right)\left(\sum_{n}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)\right)\left(\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)\right)$ $\displaystyle\qquad+q_{t}^{2}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ij}z_{t,n}^{k}(x)-q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ij}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{k}(x)$ $\displaystyle\qquad+q_{t}^{2}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ik}z_{t,n}^{j}(x)-q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{ik}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{j}(x)$ $\displaystyle\qquad+q_{t}^{2}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{jk}z_{t,n}^{i}(x)-q_{t}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)z_{t,n}^{i}(x)\sum_{n=1}^{N}\pi_{n}q_{t,n}(x)[\Sigma_{t,n}^{-1}]^{jk}.$ $$

Below write $\xi_{t}(x,i):=\max_{n}\left|z_{t,n}^{i}(x)\right|$ and $\bar{\Sigma}$ to be a matrix such that $\bar{\Sigma}^{ij}:=\max_{n}\left|[\Sigma_{t,n}^{-1}]^{ij}\right|$. Also write $h_{t,n}(x)=\pi_{n}q_{t,n}(x)/q_{t}(x)$. Note that for any $x$, $\sum_{n=1}^{N}h_{t,n}(x)=1$.
Therefore, we take $\max_{n}$ within each summation above and get

$$ $\left|\partial^{3}_{ijk}\log q_{t}(x)\right|\leq 6\xi_{t}(x,i)\xi_{t}(x,j)\xi_{t}(x,k)+2\bar{\Sigma}^{ij}\xi_{t}(x,k)+2\bar{\Sigma}^{ik}\xi_{t}(x,j)+2\bar{\Sigma}^{jk}\xi_{t}(x,i).$ $$

Following from [Theorem 5](#Thmtheorem5), we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t}(X_{t}))\partial^{3}_{ijk}\log q_{t}(X_{t})\right]$ $\displaystyle\leq\mathbb{E}_{X_{t}\sim Q_{t}}\Bigg{[}\sum_{i,j,k=1}^{d}\Big{(}6\xi(\mu_{t}(X_{t}),i)\xi(\mu_{t}(X_{t}),j)\xi(\mu_{t}(X_{t}),k)+2\bar{\Sigma}^{ij}\xi(\mu_{t}(X_{t}),k)+2\bar{\Sigma}^{ik}\xi(\mu_{t}(X_{t}),j)$ $\displaystyle\qquad+2\bar{\Sigma}^{jk}\xi(\mu_{t}(X_{t}),i)\Big{)}\left(6\xi(X_{t},i)\xi(X_{t},j)\xi(X_{t},k)+2\bar{\Sigma}^{ij}\xi(X_{t},k)+2\bar{\Sigma}^{ik}\xi(X_{t},j)+2\bar{\Sigma}^{jk}\xi(X_{t},i)\right)\Bigg{]}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\asymp}}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\left(\xi(X_{t},i)\xi(X_{t},j)\xi(X_{t},k)+\bar{\Sigma}^{ij}\xi(X_{t},k)+\bar{\Sigma}^{ik}\xi(X_{t},j)+\bar{\Sigma}^{jk}\xi(X_{t},i)\right)^{2}\right]$ $\displaystyle\leq 2\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\xi(X_{t},i)^{2}\xi(X_{t},j)^{2}\xi(X_{t},k)^{2}+(\bar{\Sigma}^{ij})^{2}\xi(X_{t},k)^{2}+(\bar{\Sigma}^{ik})^{2}\xi(X_{t},j)^{2}+(\bar{\Sigma}^{jk})^{2}\xi(X_{t},i)^{2}\right]$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}\min\{d,N\}^{3}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left\lVert X_{t}\right\rVert^{6}+\max_{n}\left\lVert\mu_{t,n}\right\rVert^{6}\right]+(d\min\{d,N\})(d\min\{d,N\})$ $$

where $(i)$ follows from ([D.2](#A4.Ex275)), and $(ii)$ follows from ([49](#A4.E49)) and ([50](#A4.E50)).
Here

$$ $\max_{n}\left\lVert\mu_{t,n}\right\rVert^{6}\leq\max_{n}\left\lVert\mu_{0,n}\right\rVert^{6}\lesssim d^{3}$ $$

since $\mu_{0,n}<\infty$ is a fixed vector, and the expected sixth power of the norm can be bounded as

$$ $\mathbb{E}\left\lVert X_{t}\right\rVert^{6}=\mathbb{E}\left[\left(\left\lVert\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t}\right\rVert^{2}\right)^{3}\right]\lesssim\left(\mathbb{E}\left\lVert X_{0}\right\rVert^{6}+\mathbb{E}\left\lVert\bar{W}_{t}\right\rVert^{6}\right).$ $$

Since $Q_{0}$ is a Gaussian mixture,

$$ $\int\left\lVert x_{0}\right\rVert^{6}q_{0}(x_{0})\mathrm{d}x_{0}=\sum_{n=1}^{N}\pi_{n}\int\left\lVert x_{0}\right\rVert^{6}q_{0,n}(x_{0})\mathrm{d}x_{0}\asymp d^{3},$ $$

and also $\mathbb{E}\left\lVert\bar{W}_{t}\right\rVert^{6}\asymp d^{3}$.
Therefore, following from [Theorem 5](#Thmtheorem5), we finally obtain a bound on the reverse-step error with explicit system parameters:

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p^{\prime}_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d^{3}\min\{d,N\}^{3}\log^{3}T}{T^{2}}.$ $$

###### Lemma 27 .

###### Proof.

### H.4 Proof of Theorem 8

We remind readers that if we use the $\alpha_{t}$’s in ([8](#S4.E8)), [Assumption 4](#Thmassumption4) is satisfied when $Q_{0}$ has bounded support (by [Lemma 17](#Thmlemma17)).
We again begin by investigating $\partial^{3}_{ijk}\log q_{t}(x)(t\geq 1)$ for general $Q_{0}$. Write $z^{i}:=\frac{x^{i}-\sqrt{\bar{\alpha}_{t}}x^{i}_{0}}{1-\bar{\alpha}_{t}}$. Note that $\partial_{k}q_{t|0}(x|x_{0})=q_{t|0}(x|x_{0})(-z^{k})$. Following from ([E.4](#A5.Ex326)), the second partial derivative of $\log q_{t}(x)$ is given by

$$ $\partial^{2}_{ij}\log q_{t}(x)=\frac{1}{q_{t}^{2}(x)}\bigg{(}q_{t}(x)\underbrace{\int q_{t|0}(x|x_{0})z^{i}z^{j}\mathrm{d}Q_{0}(x_{0})}_{\text{N1}}\\ -\underbrace{\int q_{t|0}(x|x_{0})z^{i}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{j}\mathrm{d}Q_{0}(x_{0})}_{\text{N2}}\bigg{)}-\frac{1}{1-\bar{\alpha}_{t}}I_{d}.$ $$

Note that the last term is a constant. The derivative for term N1 is given by

$$ $\displaystyle\partial_{k}\int q_{t|0}(x|x_{0})z^{i}z^{j}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle=\int q_{t|0}(x|x_{0})(-z^{k})z^{i}z^{j}+\mathds{1}(k=i)q_{t|0}(x|x_{0})(1-\bar{\alpha}_{t})^{-1}z^{j}$ $\displaystyle\qquad+\mathds{1}(k=j)q_{t|0}(x|x_{0})(1-\bar{\alpha}_{t})^{-1}z^{i}\mathrm{d}Q_{0}(x_{0}),$ $$

and the derivative for term N2 is given by

$$ $\displaystyle\partial_{k}\left(\int q_{t|0}(x|x_{0})z^{i}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{j}\mathrm{d}Q_{0}(x_{0})\right)$ $\displaystyle=\int q_{t|0}(x|x_{0})\left((-z^{k})z^{i}+\mathds{1}(k=i)(1-\bar{\alpha}_{t})^{-1}\right)\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{j}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad+\int q_{t|0}(x|x_{0})z^{i}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})\left((-z^{k})z^{j}+\mathds{1}(k=j)(1-\bar{\alpha}_{t})^{-1}\right)\mathrm{d}Q_{0}(x_{0})$ $\displaystyle=\left(\int q_{t|0}(x|x_{0})(-z^{k})z^{i}\mathrm{d}Q_{0}(x_{0})+\mathds{1}(k=i)(1-\bar{\alpha}_{t})^{-1}q_{t}(x)\right)\int q_{t|0}(x|x_{0})z^{j}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad+\int q_{t|0}(x|x_{0})z^{i}\mathrm{d}Q_{0}(x_{0})\left(\int q_{t|0}(x|x_{0})(-z^{k})z^{j}\mathrm{d}Q_{0}(x_{0})+\mathds{1}(k=j)(1-\bar{\alpha}_{t})^{-1}q_{t}(x)\right).$ $$

Combining these, the derivative for the numerator is given by

$$ $\displaystyle\partial_{k}(q_{t}(x)\text{N1}-\text{N2})=\partial_{k}(q_{t}(x))\text{N1}+q_{t}(x)\partial_{k}(\text{N1})-\partial_{k}(\text{N2})$ $\displaystyle=-q_{t}(x)\int q_{t|0}(x|x_{0})z^{i}z^{j}z^{k}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad-\int q_{t|0}(x|x_{0})z^{k}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{i}z^{j}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad+\int q_{t|0}(x|x_{0})z^{j}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{i}z^{k}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad+\int q_{t|0}(x|x_{0})z^{i}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{j}z^{k}\mathrm{d}Q_{0}(x_{0}).$ $$

Thus,

$$ $\displaystyle\partial^{3}_{ijk}\log q_{t}(x)=\partial_{k}\frac{q_{t}(x)\text{N1}-\text{N2}}{q_{t}^{2}(x)}$ $\displaystyle=\frac{1}{q_{t}^{3}(x)}\left(\partial_{k}(q_{t}(x)\text{N1}-\text{N2})q_{t}(x)+2(q_{t}(x)\text{N1}-\text{N2})\int q_{t|0}(x|x_{0})z^{k}\mathrm{d}Q_{0}(x_{0})\right)$ $\displaystyle=\frac{1}{q_{t}^{3}(x)}\bigg{(}-q_{t}^{2}(x)\int q_{t|0}(x|x_{0})z^{i}z^{j}z^{k}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad+q_{t}(x)\sum_{\begin{subarray}{c}a_{1}=i,j,k\\ a_{2}<a_{3},\leavevmode\nobreak\ a_{2},a_{3}\neq a_{1}\end{subarray}}\int q_{t|0}(x|x_{0})z^{a_{1}}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{a_{2}}z^{a_{3}}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad-2\int q_{t|0}(x|x_{0})z^{i}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{j}\mathrm{d}Q_{0}(x_{0})\int q_{t|0}(x|x_{0})z^{k}\mathrm{d}Q_{0}(x_{0})\bigg{)}.$ $$

Now, for ease of presentation, define $h_{t}(x_{0}|x):=\frac{q_{t|0}(x|x_{0})}{q_{t}(x)}$. Then,

$$ $\displaystyle\partial^{3}_{ijk}\log q_{t}(x)$ $\displaystyle=-\int h_{t}(x_{0}|x)z^{i}z^{j}z^{k}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad+\sum_{\begin{subarray}{c}a_{1}=i,j,k\\ a_{2}<a_{3},\leavevmode\nobreak\ a_{2},a_{3}\neq a_{1}\end{subarray}}\int h_{t}(x_{0}|x)z^{a_{1}}\mathrm{d}Q_{0}(x_{0})\int h_{t}(x_{0}|x)z^{a_{2}}z^{a_{3}}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\qquad-2\int h_{t}(x_{0}|x)z^{i}\mathrm{d}Q_{0}(x_{0})\int h_{t}(x_{0}|x)z^{j}\mathrm{d}Q_{0}(x_{0})\int h_{t}(x_{0}|x)z^{k}\mathrm{d}Q_{0}(x_{0}).$ $$

Plugging the definition of $z^{i},z^{j},z^{k}$, we obtain

$$ $\displaystyle\partial^{3}_{ijk}\log q_{t}(x)$ $\displaystyle=\frac{\bar{\alpha}_{t}^{3/2}}{(1-\bar{\alpha}_{t})^{3}}\int h_{t}(x_{0}|x)\left(x_{0}^{i}-\int h_{t}(x_{0}|x)x_{0}^{i}\mathrm{d}Q_{0}(x_{0})\right)$ $\displaystyle\qquad\left(x_{0}^{j}-\int h_{t}(x_{0}|x)x_{0}^{j}\mathrm{d}Q_{0}(x_{0})\right)\left(x_{0}^{k}-\int h_{t}(x_{0}|x)x_{0}^{k}\mathrm{d}Q_{0}(x_{0})\right)\mathrm{d}Q_{0}(x_{0})$ $$

Note that $\int h_{t}(x_{0}|x)\mathrm{d}Q_{0}(x_{0})=1$. Therefore, given any $x$, we can similarly define a probability measure $\tilde{Q}_{0|t}(x_{0}|x)$ (absolutely continuous w.r.t. $Q_{0}$) such that $\frac{\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)}{\mathrm{d}Q_{0}(x_{0})}=h_{t}(x_{0}|x)$ for all $x_{0}$. Similarly write $m_{t}(x):=\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t}(\cdot|x)}[X_{0}]$. With these notations, we can write

$$ $\partial^{3}_{ijk}\log q_{t}(x)=\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t}(\cdot|x)}\left[(X_{0}^{i}-m_{t}^{i}(x))(X_{0}^{j}-m_{t}^{j}(x))(X_{0}^{k}-m_{t}^{k}(x))\right].$ $$

Now consider $Q_{0}$ with bounded support. In this case $\tilde{Q}_{0|t}$ has the same (bounded) support as $Q_{0}$ for all $x$. Note that $m_{t}(x)$ is also contained in this support. Then,

$$ $\displaystyle\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})$ $\displaystyle=\frac{\bar{\alpha}_{t}^{3/2}\bar{\alpha}_{t-1}^{3/2}}{(1-\bar{\alpha}_{t})^{3}(1-\bar{\alpha}_{t-1})^{3}}\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\sum_{i,j,k=1}^{d}(x_{a}^{i}-m_{t-1}^{i}(\mu_{t}))(x_{b}^{i}-m_{t}^{i}(x_{t}))$ $\displaystyle\qquad(x_{a}^{j}-m_{t-1}^{j}(\mu_{t}))(x_{b}^{j}-m_{t}^{j}(x_{t}))(x_{a}^{k}-m_{t-1}^{k}(\mu_{t}))(x_{b}^{k}-m_{t}^{k}(x_{t}))\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle=\frac{\bar{\alpha}_{t}^{3/2}\bar{\alpha}_{t-1}^{3/2}}{(1-\bar{\alpha}_{t})^{3}(1-\bar{\alpha}_{t-1})^{3}}\times$ $\displaystyle\quad\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\left(\sum_{i=1}^{d}(x_{a}^{i}-m_{t-1}^{i}(\mu_{t}))(x_{b}^{i}-m_{t}^{i}(x_{t}))\right)^{3}\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\leq}}\frac{\bar{\alpha}_{t}^{3/2}\bar{\alpha}_{t-1}^{3/2}}{(1-\bar{\alpha}_{t})^{3}(1-\bar{\alpha}_{t-1})^{3}}\times$ $\displaystyle\quad\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\left(\left\lVert x_{a}-m_{t-1}(\mu_{t})\right\rVert^{2}\left\lVert x_{b}-m_{t}(x_{t})\right\rVert^{2}\right)^{\frac{3}{2}}\mathrm{d}\tilde{Q}_{0|t-1}(x_{a}|\mu_{t})\mathrm{d}\tilde{Q}_{0|t}(x_{b}|x_{t})$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}(1-\bar{\alpha}_{t})^{-3}(1-\bar{\alpha}_{t-1})^{-3}\max_{x_{a},x_{b}:\left\lVert x_{a}\right\rVert^{2},\left\lVert x_{b}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}-x_{b}\right\rVert^{6}$ $$

where $(i)$ follows from Cauchy–Schwartz inequality, and $(ii)$ follows from the assumption of bounded support. Since

$$ $\max_{x_{a},x_{b}:\left\lVert x_{a}\right\rVert^{2},\left\lVert x_{b}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}-x_{b}\right\rVert\leq 2\max_{x_{a}:\left\lVert x_{a}\right\rVert^{2}\leq R^{2}d}\left\lVert x_{a}\right\rVert\lesssim R\sqrt{d},$ $$

the reverse-step error satisfies

$$ $(1-\alpha_{t})^{3}\mathbb{E}_{X_{t}\sim Q_{t}}\left[\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t})\right]\lesssim\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{3}(1-\bar{\alpha}_{t})^{3}}d^{3}R^{6}.$ $$

Thus, the reverse-step error in [Corollary 3](#Thmcorollary3) has an upper bound with explicit parameter dependence:

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t-1},X_{t}\sim Q_{t-1,t}}\left[\log\frac{q_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\frac{d^{3}R^{6}\log^{3}T}{\delta^{3}T^{2}}.$ $$

Again note that here $R$ does not depend on $d$.

## Appendix I Auxiliary Proofs for Theorems 5 , 3 , 6 , 7 and 8

### I.1 Proof of Lemma 1

The idea is similar to score matching. Define $v^{\prime}_{\theta}(x):=v_{\theta}(x)-\frac{1}{1-\bar{\alpha}_{t}}I_{d}$. For each $i,j\in[d]$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left(v_{\theta}^{ij}(X_{t})-\left(\frac{\partial^{2}_{ij}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{\mathds{1}\left\{i=j\right\}}{1-\bar{\alpha}_{t}}\right)\right)^{2}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\left([v^{\prime}_{\theta}(X_{t})]^{ij}-\frac{\partial^{2}_{ij}q_{t}(X_{t})}{q_{t}(X_{t})}\right)^{2}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\left([v^{\prime}_{\theta}(X_{t})]^{ij}\right)^{2}-2\mathbb{E}_{X_{t}\sim Q_{t}}\left[[v^{\prime}_{\theta}(X_{t})]^{ij}\frac{\partial^{2}_{ij}q_{t}(X_{t})}{q_{t}(X_{t})}\right]+\mathrm{const}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\left([v^{\prime}_{\theta}(X_{t})]^{ij}\right)^{2}-2\int[v^{\prime}_{\theta}(x_{t})]^{ij}\partial^{2}_{ij}q_{t}(x_{t})\mathrm{d}x_{t}+\mathrm{const}$ $$

where $\mathrm{const}$ denotes terms that are independent of $\theta$, and

$$ $\displaystyle\int[v^{\prime}_{\theta}(x_{t})]^{ij}\partial^{2}_{ij}q_{t}(x_{t})\mathrm{d}x_{t}$ $\displaystyle=\int[v^{\prime}_{\theta}(x_{t})]^{ij}\int\partial^{2}_{ij}q_{t|0}(x_{t}|x_{0})\mathrm{d}Q_{0}(x_{0})\mathrm{d}x_{t}$ $\displaystyle=\int\int q_{t|0}(x_{t}|x_{0})[v^{\prime}_{\theta}(x_{t})]^{ij}\frac{\partial^{2}_{ij}q_{t|0}(x_{t}|x_{0})}{q_{t|0}(x_{t}|x_{0})}\mathrm{d}Q_{0}(x_{0})\mathrm{d}x_{t}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\int\int q_{t|0}(x_{t}|x_{0})[v^{\prime}_{\theta}(x_{t})]^{ij}\left(\partial^{2}_{ij}\log q_{t|0}(x_{t}|x_{0})+\partial_{i}\log q_{t|0}(x_{t}|x_{0})\partial_{j}\log q_{t|0}(x_{t}|x_{0})\right)\mathrm{d}Q_{0}(x_{0})\mathrm{d}x_{t}$ $\displaystyle=\int\int q_{t|0}(x_{t}|x_{0})[v^{\prime}_{\theta}(x_{t})]^{ij}\left(-\frac{\mathds{1}\left\{i=j\right\}}{1-\bar{\alpha}_{t}}+\frac{x_{t}^{i}-\sqrt{\bar{\alpha}_{t}}x_{0}^{i}}{1-\bar{\alpha}_{t}}\cdot\frac{x_{t}^{j}-\sqrt{\bar{\alpha}_{t}}x_{0}^{j}}{1-\bar{\alpha}_{t}}\right)\mathrm{d}Q_{0}(x_{0})\mathrm{d}x_{t}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\mathbb{E}_{\begin{subarray}{c}X_{0}\sim Q_{0},\bar{W}_{t}\sim\mathcal{N}(0,I_{d})\\ X_{t}=\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t}\end{subarray}}\left[[v^{\prime}_{\theta}(X_{t})]^{ij}\left(-\frac{\mathds{1}\left\{i=j\right\}}{1-\bar{\alpha}_{t}}+\frac{1}{1-\bar{\alpha}_{t}}\bar{W}_{t}^{i}\bar{W}_{t}^{j}\right)\right]$ $$

where $(i)$ follows because for any function $f(x)$ we have $\partial^{2}_{ij}\log f(x)=\frac{\partial^{2}_{ij}f(x)}{f(x)}-(\partial_{i}\log f(x))(\partial_{j}\log f(x))$, and $(ii)$ follows because $x_{t}=\sqrt{\bar{\alpha}_{t}}x_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{w}_{t}$.
Therefore,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left(v_{\theta}^{ij}(X_{t})-\left(\frac{\partial^{2}_{ij}q_{t}(X_{t})}{q_{t}(X_{t})}-\frac{\mathds{1}\left\{i=j\right\}}{1-\bar{\alpha}_{t}}\right)\right)^{2}$ $\displaystyle=\mathbb{E}_{\begin{subarray}{c}X_{0}\sim Q_{0},\bar{W}_{t}\sim\mathcal{N}(0,I_{d})\\ X_{t}=\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t}\end{subarray}}\left([v^{\prime}_{\theta}(X_{t})]^{ij}-\left(-\frac{\mathds{1}\left\{i=j\right\}}{1-\bar{\alpha}_{t}}+\frac{1}{1-\bar{\alpha}_{t}}\bar{W}_{t}^{i}\bar{W}_{t}^{j}\right)\right)^{2}+\mathrm{const}$ $\displaystyle=\mathbb{E}_{\begin{subarray}{c}X_{0}\sim Q_{0},\bar{W}_{t}\sim\mathcal{N}(0,I_{d})\\ X_{t}=\sqrt{\bar{\alpha}_{t}}X_{0}+\sqrt{1-\bar{\alpha}_{t}}\bar{W}_{t}\end{subarray}}\left([v_{\theta}(X_{t})]^{ij}-\frac{1}{1-\bar{\alpha}_{t}}\bar{W}_{t}^{i}\bar{W}_{t}^{j}\right)^{2}+\mathrm{const}$ $$

and the result follows immediately after we sum up over $i,j\in[d]$.

### I.2 Proof of Lemma 2

Recall that

$$ $H_{t}(x)=v_{t}(x)-\frac{1}{1-\bar{\alpha}_{t}}I_{d}-s_{t}(x)s_{t}^{\intercal}(x).$ $$

Denote

$$ $\displaystyle\varepsilon_{H,1}^{2}$ $\displaystyle:=\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert v_{t}(X_{t})-\left(\frac{\nabla^{2}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)\right\rVert_{F}^{2}=\tilde{O}(T^{-1}),$ $\displaystyle\varepsilon_{H,2}^{2}$ $\displaystyle:=\frac{1}{T}\sum_{t=1}^{T}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}=\tilde{O}(T^{-2}).$ $$

The goal is to show that $H_{t}$ is close to $\nabla^{2}\log q_{t}$ (i.e., the second relationship in [Assumption 5](#Thmassumption5)). Given that $\nabla^{2}\log q_{t}(x)=\frac{\nabla^{2}q_{t}(x)}{q_{t}(x)}-(\nabla\log q_{t}(x))(\nabla\log q_{t}(x))^{\intercal}$, the key is to control the error incurred by $s_{t}(x)s_{t}(x)^{\intercal}$, which is

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\sum_{i,j=1}^{d}\left(s_{t}^{i}(X_{t})s_{t}^{j}(X_{t})-[\nabla\log q_{t}(X_{t})]^{i}[\nabla\log q_{t}(X_{t})]^{j}\right)^{2}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t}}\sum_{i,j=1}^{d}\left((s_{t}^{i}(X_{t})-[\nabla\log q_{t}(X_{t})]^{i})s_{t}^{j}(X_{t})+[\nabla\log q_{t}(X_{t})]^{i}(s_{t}^{j}(X_{t})-[\nabla\log q_{t}(X_{t})]^{j})\right)^{2}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\leq}}2\mathbb{E}_{X_{t}\sim Q_{t}}\sum_{i,j=1}^{d}(s_{t}^{i}(X_{t})-[\nabla\log q_{t}(X_{t})]^{i})^{2}(s_{t}^{j}(X_{t}))^{2}+([\nabla\log q_{t}(X_{t})]^{i})^{2}(s_{t}^{j}(X_{t})-[\nabla\log q_{t}(X_{t})]^{j})^{2}$ $\displaystyle=2\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{2}(\left\lVert\nabla\log q_{t}(X_{t})\right\rVert^{2}+\left\lVert s_{t}(X_{t})\right\rVert^{2})\right]$ $$

where $(i)$ follows because $(a+b)^{2}=a^{2}+b^{2}+2ab\leq 2a^{2}+2b^{2}$. To continue, we use the Cauchy-Schwartz inequality and obtain

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})s_{t}^{\intercal}(X_{t})-(\nabla\log q_{t}(X_{t}))(\nabla\log q_{t}(X_{t}))^{\intercal}\right\rVert_{F}^{2}$ $\displaystyle\qquad\leq 2\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}\sqrt{2\mathbb{E}_{X_{t}\sim Q_{t}}\left[\left\lVert\nabla\log q_{t}(X_{t})\right\rVert^{4}+\left\lVert s_{t}(X_{t})\right\rVert^{4}\right]}.$ $$

Here the second term is

$$ $\displaystyle\mathbb{E}[\left\lVert s_{t}(X_{t})\right\rVert^{4}]$ $\displaystyle\leq 8\mathbb{E}[\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}]+8\mathbb{E}[\left\lVert\nabla\log q_{t}(X_{t})\right\rVert^{4}]$ $\displaystyle\lesssim\mathbb{E}[\left\lVert\nabla\log q_{t}(X_{t})\right\rVert^{4}]<\infty$ $$

which has an upper bound that is independent of $T$ by [Assumption 4](#Thmassumption4).
Therefore,

$$ $\displaystyle\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert H_{t}(X_{t})-\nabla^{2}\log q_{t}(X_{t})\right\rVert^{2}_{F}$ $\displaystyle\leq\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert v_{\theta}(X_{t})-\left(\frac{\nabla^{2}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)\right\rVert^{2}_{F}$ $\displaystyle\qquad+\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})s_{t}^{\intercal}(X_{t})-(\nabla\log q_{t}(X_{t}))(\nabla\log q_{t}(X_{t}))^{\intercal}\right\rVert^{2}_{F}$ $\displaystyle\lesssim\frac{1}{T}\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert v_{\theta}(X_{t})-\left(\frac{\nabla^{2}q_{t}(X_{t})}{q_{t}(X_{t})}+\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right)\right\rVert^{2}_{F}$ $\displaystyle\qquad+\frac{1}{T}\sum_{t=1}^{T}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert s_{t}(X_{t})-\nabla\log q_{t}(X_{t})\right\rVert^{4}}$ $\displaystyle=\varepsilon_{H,1}^{2}+\varepsilon_{H,2}^{2}=\tilde{O}(T^{-1}).$ $$

The proof is now complete.

### I.3 Proof of Lemma 19

By Bayes’ rule, for any $x_{t-1}$ given fixed $x_{t}$, we have

$$ $\displaystyle q_{t-1|t}(x_{t-1}|x_{t})$ $\displaystyle\propto q_{t-1}(x_{t-1})\exp\left(-\frac{\left\lVert x_{t}-\sqrt{\alpha_{t}}x_{t-1}\right\rVert^{2}}{2(1-\alpha_{t})}\right)$ $\displaystyle\propto q_{t-1}(x_{t-1})p^{\prime}_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\Sigma_{t}^{-1}(x_{t-1}-\mu_{t})-\frac{\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}}\right)$ $\displaystyle=q_{t-1}(x_{t-1})p^{\prime}_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{1}{2\sigma_{t}^{2}}(x_{t-1}-\mu_{t})^{\intercal}(I_{d}-B_{t})(x_{t-1}-\mu_{t})-\frac{\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}}\right)$ (by Equation 72 ) $\displaystyle\propto p^{\prime}_{t-1|t}(x_{t-1}|x_{t})\exp\left(\zeta_{t,t-1}(x_{t},x_{t-1})-\frac{1}{2\sigma_{t}^{2}}(x_{t-1}-\mu_{t})^{\intercal}B_{t}(x_{t-1}-\mu_{t})\right),$ $$

Now, with the definition of $\zeta^{\prime}_{t,t-1}(x_{t},x_{t-1})$ in [Lemma 19](#A7.Ex559), we have

$$ $q_{t-1|t}(x_{t-1}|x_{t})=\frac{p^{\prime}_{t-1|t}(x_{t-1}|x_{t})e^{\zeta^{\prime}_{t,t-1}(x_{t},x_{t-1})}}{\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[e^{\zeta^{\prime}_{t,t-1}(x_{t},X_{t-1})}]}.$ $$

### I.4 Proof of Lemma 20

The result follows directly from Isserlis’s Theorem, which says that

$$ $\mathbb{E}\left[\prod_{i=1}^{n}Z_{i}\right]=\sum_{p\in P_{n}^{2}}\prod_{\{i,j\}\in p}\mathbb{E}[Z_{i}Z_{j}]=\sum_{p\in P_{n}^{2}}\prod_{\{i,j\}\in p}\mathrm{Cov}(Z_{i},Z_{j})$ $$

since each $Z_{i}$ is centered. Here $P_{n}^{2}$ is the set that contains all distinct size-2 partitions of $[n]$. For example, $P_{4}^{2}=\{(\{1,2\},\{3,4\}),(\{1,3\},\{2,4\}),(\{1,4\},\{2,3\})\}$. Thus, since $A_{t}=O(1-\alpha_{t})$ under [Assumption 4](#Thmassumption4),

$$ $\displaystyle\mathbb{E}\left[\prod_{i=1}^{n}Z_{i}\right]$ $\displaystyle=0,\leavevmode\nobreak\ \text{if $n$ is odd}$ $\displaystyle\mathbb{E}\left[\prod_{i=1}^{n}Z_{i}\right]$ $\displaystyle=O\left(\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{\frac{n}{2}}\right)=O\left((1-\alpha_{t})^{\frac{n}{2}}\right),\leavevmode\nobreak\ \text{if $n$ is even}.$ $$

More specifically, following from Isserlis’s Theorem, the fourth moment is

$$ $\mathbb{E}[Z_{i}Z_{j}Z_{k}Z_{l}]=\mathrm{Cov}(Z_{i},Z_{j})\mathrm{Cov}(Z_{k},Z_{l})\\ +\mathrm{Cov}(Z_{i},Z_{k})\mathrm{Cov}(Z_{j},Z_{l})+\mathrm{Cov}(Z_{i},Z_{l})\mathrm{Cov}(Z_{j},Z_{k}),\leavevmode\nobreak\ \forall i,j,k,l\in[d].$ $$

Here $\mathrm{Cov}(Z_{i},Z_{j})=\frac{1-\alpha_{t}}{\alpha_{t}}(\mathds{1}\left\{i=j\right\}+(1-\alpha_{t})A^{ij})$. The fourth moment result follows immediately by plugging into the formula. Turning to the sixth moment, we note that we are interested only in the coefficients for the terms that grow at a rate $O((1-\alpha_{t})^{3})$. Since the sixth moment consists of sum of product terms in which three covariances are multiplied (giving us a rate at least $O((1-\alpha_{t})^{3})$), at least one product term in the sum must take covariance values only on the diagonal of the matrix. Therefore, only $\mathbb{E}[Z_{i}^{6}]$, $\mathbb{E}[Z_{i}^{4}Z_{j}^{2}]$, and $\mathbb{E}[Z_{i}^{2}Z_{j}^{2}Z_{k}^{2}]$ with $i,j,k$ all differ satisfy this requirement, and we immediately get the desired result from Isserlis’s Theorem.

### I.5 Proof of Lemma 21

The proof is similar to that of [Lemma 8](#Thmlemma8). The idea is to use the inductive relationship (provided in the proof of [Lemmas 8](#Thmlemma8) and [9](#Thmlemma9)):

$$ $\displaystyle D_{5}(x,\tilde{x}_{t})$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{m}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{4}(x,\tilde{x}_{t})\Big{)}$ $\displaystyle=(x^{m}-\partial_{m}\kappa)D_{4}(x,\tilde{x}_{t})+\partial_{m}D_{4}(x,\tilde{x}_{t})$ $\displaystyle D_{6}(x,\tilde{x}_{t})$ $\displaystyle=\exp\left(-x^{\intercal}\tilde{x}_{t}+\kappa\right)\partial_{n}\Big{(}\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)D_{5}(x,\tilde{x}_{t})\Big{)}$ $\displaystyle=(x^{n}-\partial_{n}\kappa)D_{5}(x,\tilde{x}_{t})+\partial_{n}D_{5}(x,\tilde{x}_{t}).$ $$

Let $P_{n}^{k}$ be the set that contains all distinct size-$k$ partitions of $[n]$. We use the definitions:

$$ $\displaystyle\binom{A}{k}$ $\displaystyle:=\Big{\{}\{a_{1},\dots,a_{k}\}:a_{1},\dots,a_{k}\in A,\leavevmode\nobreak\ a_{1},\dots,a_{k}\leavevmode\nobreak\ \text{all differ}\Big{\}},\leavevmode\nobreak\ k\leq\left|A\right|$ $\displaystyle\mathrm{part}_{k}(A)$ $\displaystyle:=\{((a_{i},a_{j}):\{i,j\}\in p):p\in P_{\left|A\right|}^{k}\}.$ $$

Recall the formula for $D_{4}$ in ([C.6](#A3.Ex171)), which can be abbreviated as (here $\left|\bm{a}\right|=4$):

$$ $\displaystyle D_{4}(x,\tilde{x}_{t})$ $\displaystyle=\prod_{i\in\bm{a}}(x^{i}-\partial_{i}\kappa)-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\prod_{i\in\bm{a}\setminus\bm{b}}(x^{i}-\partial_{i}\kappa)+\sum_{(\bm{b},\bm{c})\in\mathrm{part}_{2}(\bm{a})}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa$ $\displaystyle\qquad-\sum_{i\in\bm{a}}\partial^{3}_{\bm{a}\setminus\{i\}}\kappa(x^{i}-\partial_{i}\kappa)-\partial^{4}_{\bm{a}}\kappa.$ $$

Also recall the definition of $f(x)$ in [Lemma 8](#Thmlemma8) and that $\int f(x)e^{x^{\intercal}\tilde{x}_{t}-\kappa}D_{p}(x,\tilde{x}_{t})\mathrm{d}x=0$, through which we can find the expected $p$-th moments of $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i\in\bm{a}}(X^{i}_{t-1}-\mu^{i}_{t})\right]$. For reference, the first four moments are

$$ $\displaystyle\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)(x^{i}-\partial_{i}\kappa)\mathrm{d}x=0$ $\displaystyle\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)\mathrm{d}x=\partial^{2}_{ij}\kappa=O(1-\alpha_{t})$ $\displaystyle\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)\mathrm{d}x=\partial^{3}_{ijk}\kappa=O((1-\alpha_{t})^{3})$ $\displaystyle\int f(x)\exp\left(x^{\intercal}\tilde{x}_{t}-\kappa\right)(x^{i}-\partial_{i}\kappa)(x^{j}-\partial_{j}\kappa)(x^{k}-\partial_{k}\kappa)(x^{l}-\partial_{l}\kappa)\mathrm{d}x$ $\displaystyle\qquad=(\partial^{2}_{ij}\kappa)(\partial^{2}_{kl}\kappa)+(\partial^{2}_{ik}\kappa)(\partial^{2}_{jl}\kappa)+(\partial^{2}_{il}\kappa)(\partial^{2}_{jk}\kappa)+\partial^{4}_{ijkl}\kappa=O((1-\alpha_{t})^{2})$ $$

where we note that $\partial^{k}_{\bm{a}}\kappa=O((1-\alpha_{t})^{k})$ for all $k\geq 3$.

We can calculate $D_{5}$ as (with $\left|\bm{a}\right|=5$):

$$ $\displaystyle D_{5}(x,\tilde{x}_{t})$ $\displaystyle=(x^{a_{5}}-\partial_{a_{5}}\kappa)D_{4}(x,\tilde{x}_{t})+\partial_{a_{5}}D_{4}(x,\tilde{x}_{t})$ $\displaystyle=\prod_{i\in\bm{a}}(x^{i}-\partial_{i}\kappa)-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\prod_{i\in\bm{a}\setminus\bm{b}}(x^{i}-\partial_{i}\kappa)-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa\prod_{i\in\bm{b}}(x^{i}-\partial_{i}\kappa)$ $\displaystyle\qquad+\sum_{\begin{subarray}{c}i\in\bm{a}\\ (\bm{b},\bm{c})\in\mathrm{part}_{2}(\bm{a}\setminus\{i\})\end{subarray}}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa(x^{i}-\partial_{i}\kappa)$ $\displaystyle\qquad-\sum_{i\in\bm{a}}\partial^{4}_{\bm{a}\setminus\{i\}}\kappa(x^{i}-\partial_{i}\kappa)+\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa-\partial^{5}_{\bm{a}}\kappa.$ $$

Therefore,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i\in\bm{a}:\left|\bm{a}\right|=5}(X_{t-1}^{i}-\mu_{t}^{i})\right]$ $\displaystyle=\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa+\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa\partial^{2}_{\bm{b}}\kappa-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa+\partial^{5}_{\bm{a}}\kappa$ $\displaystyle=\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa+\partial^{5}_{\bm{a}}\kappa=O((1-\alpha_{t})^{4}).$ $$

Now we turn to calculate $D_{6}$ (and let $\left|\bm{a}\right|=6$):

$$ $\displaystyle D_{6}(x,\tilde{x}_{t})=(x^{a_{6}}-\partial_{a_{6}}\kappa)D_{5}(x,\tilde{x}_{t})+\partial_{a_{6}}D_{5}(x,\tilde{x}_{t})$ $\displaystyle=\prod_{i\in\bm{a}}(x^{i}-\partial_{i}\kappa)-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\prod_{i\in\bm{a}\setminus\bm{b}}(x^{i}-\partial_{i}\kappa)-\sum_{\bm{b}\in\binom{\bm{a}}{3}}\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa\prod_{i\in\bm{b}}(x^{i}-\partial_{i}\kappa)$ $\displaystyle\qquad-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{4}_{\bm{a}\setminus\bm{b}}\kappa\prod_{i\in\bm{b}}(x^{i}-\partial_{i}\kappa)+\sum_{\begin{subarray}{c}\bm{b}\in\binom{\bm{a}}{2}\\ (\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a}\setminus\bm{b})\end{subarray}}\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa\prod_{i\in\bm{b}}(x^{i}-\partial_{i}\kappa)+\sum_{i\in\bm{a}}\mathrm{fn}(\kappa)(x^{i}-\partial_{i}\kappa)$ $\displaystyle\qquad-\sum_{(\bm{b},\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a})}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa+\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{4}_{\bm{a}\setminus\bm{b}}\kappa+\sum_{(\bm{b},\bm{c})\in\mathrm{part}_{3}(\bm{a})}\partial^{3}_{\bm{b}}\kappa\partial^{3}_{\bm{c}}\kappa-\partial^{6}_{\bm{a}}\kappa.$ $$

Here $\mathrm{fn}(\kappa)$ is a function of $\kappa$ which does not depend on $x$. Note that $\mathrm{fn}$ does not affect the expected value because $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[X_{t-1}-\mu_{t}]=0$. Therefore, we have

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[\prod_{i\in\bm{a}:\left|\bm{a}\right|=6}(X_{t-1}^{i}-\mu_{t}^{i})\right]$ $\displaystyle=\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\left(\sum_{(\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a}\setminus\bm{b})}\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa+\partial^{4}_{\bm{a}\setminus\bm{b}}\kappa\right)+\sum_{\bm{b}\in\binom{\bm{a}}{3}}\partial^{3}_{\bm{a}\setminus\bm{b}}\kappa\partial^{3}_{\bm{b}}\kappa$ $\displaystyle\qquad+\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{4}_{\bm{a}\setminus\bm{b}}\kappa\partial^{2}_{\bm{b}}\kappa-\sum_{\begin{subarray}{c}\bm{b}\in\binom{\bm{a}}{2}\\ (\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a}\setminus\bm{b})\end{subarray}}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa$ $\displaystyle\qquad+\sum_{(\bm{b},\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a})}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa-\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{4}_{\bm{a}\setminus\bm{b}}\kappa-\sum_{(\bm{b},\bm{c})\in\mathrm{part}_{3}(\bm{a})}\partial^{3}_{\bm{b}}\kappa\partial^{3}_{\bm{c}}\kappa+\partial^{6}_{\bm{a}}\kappa$ $\displaystyle=\sum_{\bm{b}\in\binom{\bm{a}}{2}}\partial^{2}_{\bm{b}}\kappa\partial^{4}_{\bm{a}\setminus\bm{b}}\kappa+\sum_{(\bm{b},\bm{c})\in\mathrm{part}_{3}(\bm{a})}\partial^{3}_{\bm{b}}\kappa\partial^{3}_{\bm{c}}\kappa+\sum_{(\bm{b},\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a})}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa+\partial^{6}_{\bm{a}}\kappa$ $\displaystyle=\sum_{(\bm{b},\bm{c},\bm{e})\in\mathrm{part}_{2}(\bm{a})}\partial^{2}_{\bm{b}}\kappa\partial^{2}_{\bm{c}}\kappa\partial^{2}_{\bm{e}}\kappa+O((1-\alpha_{t})^{5}).$ $$

The proof is now complete.

### I.6 Proof of Lemma 22

Under [Assumption 4](#Thmassumption4), we can use a similar analysis as in the proof of [Lemma 7](#Thmlemma7) to show the absolute convergence of the Taylor series (for any fixed $x_{t-1}$ and $\mu_{t}$), and thus we are still able to exchange the infinite sum and the expectation under both $P^{\prime}_{t-1|t}$ and $Q_{t-1|t}$. Thus, we obtain

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]$ $\displaystyle=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[X_{t-1}-\mu_{t}]$ $\displaystyle\qquad+\frac{1}{2}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(X_{t-1}-\mu_{t})^{\intercal}\left(\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}B_{t}\right)(X_{t-1}-\mu_{t})\right]$ $\displaystyle\qquad+\sum_{p=3}^{\infty}\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[T_{p}(\log q_{t-1},X_{t-1},\mu_{t})]$ $$

and similar under $Q_{t-1|t}$ due to its Gaussian-like concentration of centralized moments.

We investigate each term in the Taylor expansion ([G.2](#A7.Ex561)) separately. First note that

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{1}\right]=0=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}\right]$ $$

as previously. Also, for $T^{\prime}_{2}$, note that for any random variable $Z$ (regardless of its distribution) with $\mathbb{E}Z=0$ and $\mathrm{Cov}(Z)=\Sigma$, the mean of the quadratic form (with fixed matrix $\Xi$) is

$$ $\mathbb{E}[Z^{\intercal}\Xi Z]=\mathbb{E}[\mathrm{Tr}\left(Z^{\intercal}\Xi Z\right)]=\mathrm{Tr}\left(\Xi\Sigma\right).$ $$

This implies that

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[T^{\prime}_{2}]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(X_{t-1}-\mu_{t})^{\intercal}\left(\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}B_{t}\right)(X_{t-1}-\mu_{t})\right]$ $\displaystyle=\mathrm{Tr}\left(\left(\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}B_{t}\right)\Sigma_{t}\right)$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[(X_{t-1}-\mu_{t})^{\intercal}\left(\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}B_{t}\right)(X_{t-1}-\mu_{t})\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[T^{\prime}_{2}].$ $$

Using [Lemmas 8](#Thmlemma8) and [20](#Thmlemma20), the rate for $T_{3}$ is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{3}(\log q_{t-1},X_{t-1},\mu_{t})\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{3}(\log q_{t-1},X_{t-1},\mu_{t})\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{3}(\log q_{t-1},X_{t-1},\mu_{t})\right]$ $\displaystyle=\frac{(1-\alpha_{t})^{3}}{3!\alpha_{t}^{3/2}}\sum_{i,j,k=1}^{d}\partial^{3}_{ijk}\log q_{t-1}(\mu_{t})\partial^{3}_{ijk}\log q_{t}(x_{t}).$ $$

Using [Lemmas 9](#Thmlemma9) and [20](#Thmlemma20), and when the partial derivatives satisfy [Assumption 4](#Thmassumption4), the rate for $T_{5}$, $T_{7}$, and $T_{p}(p\geq 8)$ can also be determined:

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{5}(\log q_{t-1},X_{t-1},\mu_{t})\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{5}(\log q_{t-1},X_{t-1},\mu_{t})\right]$ $\displaystyle\quad=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{5}(\log q_{t-1},X_{t-1},\mu_{t})\right]=O((1-\alpha_{t})^{4}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{7}(\log q_{t-1},X_{t-1},\mu_{t})\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{7}(\log q_{t-1},X_{t-1},\mu_{t})\right]$ $\displaystyle\quad=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{7}(\log q_{t-1},X_{t-1},\mu_{t})\right]=O((1-\alpha_{t})^{5}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{p}(\log q_{t-1},X_{t-1},\mu_{t})\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{p}(\log q_{t-1},X_{t-1},\mu_{t})\right]$ $\displaystyle\quad=O((1-\alpha_{t})^{4}),\leavevmode\nobreak\ \forall p\geq 8.$ $$

The remaining orders are $T_{4}$ and $T_{6}$. The following proof will draw from the results in [Lemmas 8](#Thmlemma8), [21](#Thmlemma21) and [20](#Thmlemma20). Write $Z_{i}=X_{t-1}^{i}-\mu_{t}^{i}$ and $A^{ij}=[A_{t}]^{ij}$ for $i,j\in[d]$. For $T_{4}$, let $i,j,k,l\in[d]$ all differ, and the difference (in expectation) of each term of $T_{4}$ is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{4}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{4}]$ $\displaystyle\qquad=3\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+6\frac{(1-\alpha_{t})^{3}}{\alpha_{t}^{2}}\partial^{2}_{ii}\log q_{t}(x_{t})-3\sigma_{t}^{4}(1+A^{ii})^{2}+O((1-\alpha_{t})^{4}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{3}Z_{j}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{3}Z_{j}]$ $\displaystyle\qquad=3\frac{(1-\alpha_{t})^{3}}{\alpha_{t}^{2}}\partial^{2}_{ij}\log q_{t}(x_{t})-3\sigma_{t}^{4}A^{ij}(1+A^{ii})+O((1-\alpha_{t})^{4}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{2}Z_{j}^{2}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{2}Z_{j}^{2}]$ $\displaystyle\qquad=\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{2}+\frac{(1-\alpha_{t})^{3}}{\alpha_{t}^{2}}\left(\partial^{2}_{ii}\log q_{t}(x_{t})+\partial^{2}_{jj}\log q_{t}(x_{t})\right)-\sigma_{t}^{4}(1+A^{ii})(1+A^{jj})+O((1-\alpha_{t})^{4})$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{2}Z_{j}Z_{k}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{2}Z_{j}Z_{k}]$ $\displaystyle\qquad=\frac{(1-\alpha_{t})^{3}}{\alpha_{t}^{2}}\partial^{2}_{jk}\log q_{t}(x_{t})-\sigma_{t}^{4}(1+A^{ii})A^{jk}+O((1-\alpha_{t})^{4}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}Z_{j}Z_{k}Z_{l}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}Z_{j}Z_{k}Z_{l}]=O((1-\alpha_{t})^{4}).$ $$

Recall from ([72](#A7.E72)) that $A_{t}=(1-\alpha_{t})\nabla^{2}\log q_{t}(x_{t})=O(1-\alpha_{t})$ under [Assumption 4](#Thmassumption4). Hence, many low-order terms above are cancelled, and we get

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{4}(\log q_{t-1},X_{t-1},\mu_{t})\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{4}(\log q_{t-1},X_{t-1},\mu_{t})\right]=O((1-\alpha_{t})^{4}).$ $$

Now we turn to $T_{6}$. Let $i,j,k\in[d]$ all differ, and the difference (in expectation) of each lowest-order term of $T_{6}$ is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{6}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{6}]$ $\displaystyle\qquad=15\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}-15\sigma_{t}^{6}(1+A^{ii})^{3}+O((1-\alpha_{t})^{4}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{4}Z_{j}^{2}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{4}Z_{j}^{2}]$ $\displaystyle\qquad=3\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}-3\sigma_{t}^{6}(1+A^{ii})^{2}(1+A^{jj})+O((1-\alpha_{t})^{4}),$ $\displaystyle\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}[Z_{i}^{2}Z_{j}^{2}Z_{k}^{2}]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[Z_{i}^{2}Z_{j}^{2}Z_{k}^{2}]$ $\displaystyle\qquad=\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{3}-\sigma_{t}^{6}(1+A^{ii})(1+A^{jj})(1+A^{kk})+O((1-\alpha_{t})^{4}).$ $$

Also, by [Lemmas 20](#Thmlemma20) and [21](#Thmlemma21), the rest of the terms already satisfy $O((1-\alpha_{t})^{4})$. Hence,

$$ $\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}}\left[T_{6}(\log q_{t-1},X_{t-1},\mu_{t})\right]-\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{6}(\log q_{t-1},X_{t-1},\mu_{t})\right]=O((1-\alpha_{t})^{4}).$ $$

Therefore, the lowest order term above is $T_{3}$, whose order is $O((1-\alpha_{t})^{3})$. The proof is now complete.

### I.7 Proof of Lemma 23

In this proof we write $g(x_{t})=\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})$ and $\Xi(x_{t})=\nabla^{2}\log q_{t-1}(\mu_{t})-\sigma_{t}^{-2}B_{t}(x_{t})$ for brevity. Following from ([15](#S5.E15)) and ([73](#A7.E73)), we obtain

$$ $\left\lVert\Xi(x_{t})\right\rVert^{2}\leq 2\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\frac{1}{\sigma_{t}^{2}}A_{t}(x_{t})\right\rVert^{2}+2\left\lVert\frac{1}{\sigma_{t}^{2}}(A_{t}(x_{t})-B_{t}(x_{t}))\right\rVert^{2}=O\left(1-\alpha_{t}\right).$ $$

Also, note that $\left\lVert g(x_{t})\right\rVert_{\infty}\leq\left\lVert g(x_{t})\right\rVert=O\left((1-\alpha_{t})^{0.5}\right)$, and $\left\lVert\Xi(x_{t})\right\rVert_{\text{max}}\leq\left\lVert\Xi(x_{t})\right\rVert=O\left((1-\alpha_{t})^{0.5}\right)$.

Recall the Taylor expansion of $\zeta^{\prime}_{t,t-1}$ from ([G.2](#A7.Ex561)), which is

$$ $\zeta^{\prime}_{t,t-1}=T_{1}+T^{\prime}_{2}+\sum_{p=3}^{\infty}T_{p}(\log q_{t-1},x_{t-1},\mu_{t}),$ $$

where

$$ $\displaystyle T_{1}$ $\displaystyle=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}(x_{t-1}-\mu_{t})$ $\displaystyle T^{\prime}_{2}$ $\displaystyle=\frac{1}{2}(x_{t-1}-\mu_{t})^{\intercal}\Xi(x_{t})(x_{t-1}-\mu_{t})$ $$

and in the following we write $T_{p}$ as a shorthand for $T_{p}(\log q_{t-1},x_{t-1},\mu_{t})$. Using the moment results in [Lemma 20](#Thmlemma20), it can be verified that the expected first moment for fixed $x_{t}$ is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]$ $\displaystyle=\frac{1}{2}\mathrm{Tr}(\Xi(x_{t})\Sigma_{t}(x_{t}))+O(\sigma_{t}^{4})$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\leq}}\frac{1}{2}\left\lVert\Xi(x_{t})\right\rVert\mathrm{Tr}(\Sigma_{t}(x_{t}))+O(\sigma_{t}^{4})$ $\displaystyle=O((1-\alpha_{t})^{1.5}).$ $$

Here $(i)$ directly follows from Von Neumann’s trace inequality, and note that $\Sigma_{t}(x_{t})$ is symmetric positive definite (whose eigenvalues equal singular values) and that $\left\lVert\Xi(x_{t})\right\rVert$ is the largest singular value of $\Xi(x_{t})$.
The final rate follows from the condition that $\Sigma_{t}(x_{t}),\left\lVert\Xi(x_{t})\right\rVert^{2}=O(1-\alpha_{t})$.

Recall the expansion for powers of $\zeta_{t,t-1}$ in ([C.8](#A3.Ex230)) and ([C.8](#A3.Ex234)). The expansion for $\zeta^{\prime}_{t,t-1}$ is similar except that $T_{2}$ is replaced by $T^{\prime}_{2}$. Under [Assumption 4](#Thmassumption4), it is still valid to exchange the infinite sum and the expectation. Combining with [Lemma 20](#Thmlemma20), we have the expected second moment as

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[\left(T_{1}+T^{\prime}_{2}+\sum_{p=3}^{\infty}T_{p}\right)^{2}\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[\left(T_{1}+T^{\prime}_{2}+T_{3}\right)^{2}\right]+O(\sigma_{t}^{6})$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}+2T_{1}T_{3}+(T^{\prime}_{2})^{2}\right]+O(\sigma_{t}^{6})$ (78) $\displaystyle=O\left((1-\alpha_{t})^{2}\right)+O\left((1-\alpha_{t})^{2.5}\right)+O\left((1-\alpha_{t})^{3}\right)+O(\sigma_{t}^{6})$ $\displaystyle=O\left((1-\alpha_{t})^{2}\right).$ $$

Here the orders can be determined in the following way. Take $\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}\right]$ as an example. The total order of this term is the sum of (i) that comes from the vanishing coefficient $g^{i}(x_{t})g^{j}(x_{t})(\forall i,j\in[d])$, which (after summation) yields $O\left(1-\alpha_{t}\right)$ by ([6](#S3.E6)), and (ii) that comes from the expectation $\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}$ $[(X_{t-1}^{i}-\mu_{t}^{i})(X_{t-1}^{j}-\mu_{t}^{j})](\forall i,j\in[d])$, which is $O(1-\alpha_{t})$ for each $i,j\in[d]$ by [Lemma 20](#Thmlemma20).

Under [Assumption 4](#Thmassumption4), again using [Lemma 20](#Thmlemma20), we get the expected third moment for any fixed $x_{t}$ as the following. Note that all moments with odd orders are zero.

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1}+T^{\prime}_{2}+T_{3}+T_{4}+T_{5}+T_{6})^{3}\right]+O(\sigma_{t}^{10})$ $\displaystyle=O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}T^{\prime}_{2}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}T_{4}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}T_{6}\right]\right)$ $\displaystyle\quad+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}T^{\prime}_{2}T_{3}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}T^{\prime}_{2}T_{5}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}T_{3}T_{4}\right]\right)$ $\displaystyle\quad+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T^{\prime}_{2})^{3}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T^{\prime}_{2})^{2}T_{4}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T^{\prime}_{2}(T_{3})^{2}\right]\right)$ $\displaystyle\quad+O(\sigma_{t}^{10})$ (79) $\displaystyle=O\left((1-\alpha_{t})^{3.5}\right)+O\left((1-\alpha_{t})^{4}\right)+O\left((1-\alpha_{t})^{5}\right)+O\left((1-\alpha_{t})^{4}\right)+O\left((1-\alpha_{t})^{5}\right)$ $\displaystyle\quad+O\left((1-\alpha_{t})^{4.5}\right)+O\left((1-\alpha_{t})^{4.5}\right)+O\left((1-\alpha_{t})^{5}\right)+O\left((1-\alpha_{t})^{4.5}\right)+O(\sigma_{t}^{10})$ $\displaystyle=O\left((1-\alpha_{t})^{3.5}\right).$ $$

The proof is now complete.

### I.8 Proof of Lemma 24

For $t\geq 2$, note that $\log q_{t-1}$ and $\log q_{t}$ exist by [Lemma 6](#Thmlemma6).
We remind readers of the definitions of $h_{t-1}(x)$ and $h_{t}(x)$ from in the proof of [Lemma 18](#Thmlemma18).
Also from there, we have shown that, for any fixed $x\in\mathbb{R}^{d}$,

$$ $\displaystyle\int_{x\in\mathbb{R}^{d}}(h_{t-1}(x)-h_{t}(x))\mathrm{poly}(x)\mathrm{d}Q_{0}(x)=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}\right),$ (80) $\displaystyle\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{3}}\right)$ (81) $$

where $\mathrm{poly}(x)$ is a polynomial in $x$ of arbitrary order.
We will build on these results in this proof.

From ([E.4](#A5.Ex326)), the Hessian of $\log q_{t}(x)$ is given by

$$ $\displaystyle\nabla^{2}\log q_{t}(x)$ $\displaystyle=\frac{1}{(1-\bar{\alpha}_{t})^{2}}\int_{x_{0}\in\mathbb{R}^{d}}\frac{q_{t|0}(x|x_{0})}{q_{t}(x)}(x-\sqrt{\bar{\alpha}_{t}}x_{0})(x-\sqrt{\bar{\alpha}_{t}}x_{0})^{\intercal}\mathrm{d}Q_{0}(x_{0})-\frac{1}{1-\bar{\alpha}_{t}}I_{d}$ $\displaystyle\quad-\frac{1}{(1-\bar{\alpha}_{t})^{2}}\left(\int_{x_{0}\in\mathbb{R}^{d}}\frac{q_{t|0}(x|x_{0})}{q_{t}(x)}(x-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)\left(\int_{x_{0}\in\mathbb{R}^{d}}\frac{q_{t|0}(x|x_{0})}{q_{t}(x)}(x-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}Q_{0}(x_{0})\right)^{\intercal}$ $\displaystyle=\underbrace{\frac{1}{(1-\bar{\alpha}_{t})^{2}}\int_{x_{0}\in\mathbb{R}^{d}}\frac{q_{t|0}(x|x_{0})}{q_{t}(x)}(x-\sqrt{\bar{\alpha}_{t}}x_{0})(x-\sqrt{\bar{\alpha}_{t}}x_{0})^{\intercal}\mathrm{d}Q_{0}(x_{0})}_{\text{term 1}}$ $\displaystyle\quad-\underbrace{\frac{1}{1-\bar{\alpha}_{t}}I_{d}}_{\text{term 2}}-\underbrace{(\nabla\log q_{t}(x))(\nabla\log q_{t}(x))^{\intercal}}_{\text{term 3}}.$ $$

The difference corresponding to term 3 is

$$ $\displaystyle(\nabla\log q_{t-1}(\mu_{t}))(\nabla\log q_{t-1}(\mu_{t}))^{\intercal}-\alpha_{t}(\nabla\log q_{t}(x_{t}))(\nabla\log q_{t}(x_{t}))^{\intercal}$ $\displaystyle=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))(\nabla\log q_{t-1}(\mu_{t}))^{\intercal}$ $\displaystyle\qquad+(\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}$ $\displaystyle=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{3}}\right).$ $$

Here the rate follows from ([81](#A9.E81)) under [Assumption 4](#Thmassumption4).

For term 2, the corresponding difference is

$$ $\left(\frac{1}{1-\bar{\alpha}_{t-1}}-\alpha_{t}\frac{1}{1-\bar{\alpha}_{t}}\right)I_{d}=\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})(1-\bar{\alpha}_{t})}I_{d}=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}\right).$ $$

For term 1, the corresponding difference is

$$ $\displaystyle\frac{1}{(1-\bar{\alpha}_{t-1})^{2}}\int_{x\in\mathbb{R}^{d}}h_{t-1}(x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)^{\intercal}\mathrm{d}Q_{0}(x)$ $\displaystyle\qquad-\frac{\alpha_{t}}{(1-\bar{\alpha}_{t})^{2}}\int_{x\in\mathbb{R}^{d}}h_{t}(x)(x_{t}-\sqrt{\bar{\alpha}_{t}}x)(x_{t}-\sqrt{\bar{\alpha}_{t}}x)^{\intercal}\mathrm{d}Q_{0}(x)$ $\displaystyle=\left[\frac{1}{(1-\bar{\alpha}_{t-1})^{2}}-\frac{1}{(1-\bar{\alpha}_{t})^{2}}\right]\int_{x\in\mathbb{R}^{d}}h_{t-1}(x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)^{\intercal}\mathrm{d}Q_{0}(x)$ $\displaystyle\qquad+\frac{1}{(1-\bar{\alpha}_{t})^{2}}\int_{x\in\mathbb{R}^{d}}\left[h_{t-1}(x)-h_{t}(x)\right](\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)^{\intercal}\mathrm{d}Q_{0}(x)$ $\displaystyle\qquad+\frac{1}{(1-\bar{\alpha}_{t})^{2}}\int_{x\in\mathbb{R}^{d}}h_{t}(x)\left[(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)-\sqrt{\alpha_{t}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x)\right](\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)^{\intercal}\mathrm{d}Q_{0}(x)$ $\displaystyle\qquad+\frac{1}{(1-\bar{\alpha}_{t})^{2}}\int_{x\in\mathbb{R}^{d}}h_{t}(x)\sqrt{\alpha_{t}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x)\left[(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)-\sqrt{\alpha_{t}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x)\right]^{\intercal}\mathrm{d}Q_{0}(x).$ $$

Here

$$ $\displaystyle\frac{1}{(1-\bar{\alpha}_{t-1})^{2}}-\frac{1}{(1-\bar{\alpha}_{t})^{2}}$ $\displaystyle=\frac{1}{(1-\bar{\alpha}_{t})^{2}}\left(\frac{1}{1-\bar{\alpha}_{t-1}}+\frac{1}{1-\bar{\alpha}_{t}}\right)\left(\frac{1}{1-\bar{\alpha}_{t-1}}-\frac{1}{1-\bar{\alpha}_{t}}\right)$ $\displaystyle=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t})^{3}(1-\bar{\alpha}_{t-1})}\right),$ $\displaystyle(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)-\sqrt{\alpha_{t}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x)$ $\displaystyle=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(x_{t}+\nabla\log q_{t}(x_{t}))-(1-\alpha_{t})\sqrt{\bar{\alpha}_{t-1}}x$ $\displaystyle=O(1-\alpha_{t}).$ $$

Also, by ([80](#A9.E80)), since the second moment under $Q_{0}$ is finite, we have

$$ $\int_{x\in\mathbb{R}^{d}}\left[h_{t-1}(x)-h_{t}(x)\right](\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)(\mu_{t}-\sqrt{\bar{\alpha}_{t-1}}x)^{\intercal}\mathrm{d}Q_{0}(x)=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{2}}\right).$ $$

Therefore, we have

$$ $\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{4}}\right).$ $$

### I.9 Proof of Lemma 25

The proof is similar to that for [Lemma 25](#Thmlemma25). The sole difference is in the rate of $g(x_{t})(=\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))$ and $\Xi(x_{t})(=\nabla^{2}\log q_{t-1}(\mu_{t})-\sigma_{t}^{-2}B_{t}(x_{t}))$. Note that $\Xi(x_{t})=O((1-\alpha_{t})/(1-\bar{\alpha}_{t-1})^{4})$
by ([73](#A7.E73)).

Since $\Sigma_{t}(x_{t})=O(1-\alpha_{t})$ and $\Xi(x_{t})=O\left(\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{4}}\right)$, for the first moment we get

$$ $\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[\zeta^{\prime}_{t,t-1}]=\frac{1}{2}\mathrm{Tr}(\Xi(x_{t})\Sigma_{t}(x_{t}))+O(\sigma_{t}^{4})=O\left(\frac{(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{4}}\right).$ $$

For the second moment, we can continue from ([I.7](#A9.Ex808)) and get

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{2}]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}+2T_{1}T_{3}+(T^{\prime}_{2})^{2}\right]+O(\sigma_{t}^{6})$ $\displaystyle=O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right)+O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right)+O\left(\frac{(1-\alpha_{t})^{4}}{(1-\bar{\alpha}_{t-1})^{8}}\right)+O(\sigma_{t}^{6})$ $\displaystyle=O\left(\frac{(1-\alpha_{t})^{3}}{(1-\bar{\alpha}_{t-1})^{6}}\right).$ $$

For the third moment, we can continue from ([I.7](#A9.Ex812)) and get

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}[(\zeta^{\prime}_{t,t-1})^{3}]$ $\displaystyle=O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}T^{\prime}_{2}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}T_{4}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T_{1})^{2}T_{6}\right]\right)$ $\displaystyle\quad+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}T^{\prime}_{2}T_{3}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}T^{\prime}_{2}T_{5}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T_{1}T_{3}T_{4}\right]\right)$ $\displaystyle\quad+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T^{\prime}_{2})^{3}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[(T^{\prime}_{2})^{2}T_{4}\right]\right)+O\left(\mathbb{E}_{X_{t-1}\sim P^{\prime}_{t-1|t}}\left[T^{\prime}_{2}(T_{3})^{2}\right]\right)$ $\displaystyle\quad+O(\sigma_{t}^{10})$ $\displaystyle=O\left(\frac{(1-\alpha_{t})^{5}}{(1-\bar{\alpha}_{t-1})^{10}}\right)+O\left(\frac{(1-\alpha_{t})^{5}}{(1-\bar{\alpha}_{t-1})^{6}}\right)+O\left(\frac{(1-\alpha_{t})^{6}}{(1-\bar{\alpha}_{t-1})^{6}}\right)+O\left(\frac{(1-\alpha_{t})^{5}}{(1-\bar{\alpha}_{t-1})^{7}}\right)$ $\displaystyle\quad+O\left(\frac{(1-\alpha_{t})^{6}}{(1-\bar{\alpha}_{t-1})^{7}}\right)+O\left(\frac{(1-\alpha_{t})^{5}}{(1-\bar{\alpha}_{t-1})^{3}}\right)+O\left(\frac{(1-\alpha_{t})^{6}}{(1-\bar{\alpha}_{t-1})^{12}}\right)+O\left(\frac{(1-\alpha_{t})^{6}}{(1-\bar{\alpha}_{t-1})^{8}}\right)$ $\displaystyle\quad+O\left(\frac{(1-\alpha_{t})^{5}}{(1-\bar{\alpha}_{t-1})^{4}}\right)+O(\sigma_{t}^{10}).$ $\displaystyle=O\left(\frac{(1-\alpha_{t})^{5}}{(1-\bar{\alpha}_{t-1})^{10}}\right).$ $$

The proof is now complete.

### I.10 Proof of Lemma 26

Using a similar procedure as in [, Lemma 22], we can obtain

$$ $\nabla^{2}\log q_{t}(x)=\\ \frac{1}{\alpha_{t}}\left(\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x)}[\nabla^{2}\log q_{t-1}(X_{t-1})]+\mathrm{Cov}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x)}(\nabla\log q_{t-1}(X_{t-1}))\right).$ $$

Therefore,

$$ $\displaystyle\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})$ $\displaystyle=\underbrace{\nabla^{2}\log q_{t-1}(\mu_{t})-\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}[\nabla^{2}\log q_{t-1}(X_{t-1})]}_{\text{term 1}}$ $\displaystyle\qquad-\underbrace{\mathrm{Cov}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}(\nabla\log q_{t-1}(X_{t-1}))}_{\text{term 2}}.$ $$

For term 1, by Jensen’s inequality, we have

$$ $\displaystyle\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}[\nabla^{2}\log q_{t-1}(X_{t-1})]\right\rVert^{2}$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\nabla^{2}\log q_{t-1}(X_{t-1})\right\rVert^{2}$ $\displaystyle\leq M^{2}\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\mu_{t}-X_{t-1}\right\rVert^{2}$ $\displaystyle=O(1-\alpha_{t}),$ $$

where the rate follows from [Lemma 8](#Thmlemma8).
For term 2, by Jensen’s inequality, we have

$$ $\displaystyle\left\lVert\mathrm{Cov}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}(\nabla\log q_{t-1}(X_{t-1}))\right\rVert$ $\displaystyle\leq\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\nabla\log q_{t-1}(X_{t-1})-\mathbb{E}_{\tilde{X}_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\nabla\log q_{t-1}(\tilde{X}_{t-1})\right\rVert^{2}$ $\displaystyle\leq\mathbb{E}_{X_{t-1},\tilde{X}_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\nabla\log q_{t-1}(X_{t-1})-\nabla\log q_{t-1}(\tilde{X}_{t-1})\right\rVert^{2}$ $\displaystyle\leq L^{2}\cdot\mathbb{E}_{X_{t-1},\tilde{X}_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert X_{t-1}-\tilde{X}_{t-1}\right\rVert^{2}$ $\displaystyle\leq 2L^{2}\cdot\left(\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert X_{t-1}-\mu_{t}\right\rVert^{2}+\mathbb{E}_{\tilde{X}_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}\left\lVert\tilde{X}_{t-1}-\mu_{t}\right\rVert^{2}\right)$ $\displaystyle=O(1-\alpha_{t}).$ $$

Here the rate again follows from [Lemma 8](#Thmlemma8).
Combining the two terms, we obtain

$$ $\displaystyle\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\alpha_{t}\nabla^{2}\log q_{t}(x_{t})\right\rVert^{2}$ $\displaystyle\leq 2\left\lVert\nabla^{2}\log q_{t-1}(\mu_{t})-\mathbb{E}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}[\nabla^{2}\log q_{t-1}(X_{t-1})]\right\rVert^{2}$ $\displaystyle\qquad+2\left\lVert\mathrm{Cov}_{X_{t-1}\sim Q_{t-1|t}(\cdot|x_{t})}(\nabla\log q_{t-1}(X_{t-1}))\right\rVert^{2}$ $\displaystyle=O(1-\alpha_{t})+O((1-\alpha_{t})^{2})=O(1-\alpha_{t}).$ $$

The proof is now complete.

### I.11 Proof of Lemma 27

From [Corollary 4](#Thmcorollary4), [Assumption 4](#Thmassumption4) is satisfied. We remind readers of the notations $\pi_{n}$, $q_{t,n}$, $\mu_{t,n}$, and $\Sigma_{t,n}$ in the proof of [Lemma 14](#Thmlemma14). There, we also proved that, for all $m,n\in[N]$,

$$ $\displaystyle\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1}=O(1-\alpha_{t}),$ (82) $\displaystyle q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})=O(1-\alpha_{t}),$ (83) $\displaystyle(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})=O(1-\alpha_{t}),$ (84) $\displaystyle q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})\Sigma_{t-1,n}^{-1}-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}=O(1-\alpha_{t}),$ (85) $\displaystyle\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t})=O(1-\alpha_{t}).$ (86) $$

From ([D.2](#A4.Ex258)), we have

$$ $\displaystyle\nabla^{2}\log q_{t}(x)$ $\displaystyle=\left(\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n}(x)}{q_{t}(x)}\Sigma_{t,n}^{-1}(x-\mu_{t,n})(x-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}\right)-\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n}(x)}{q_{t}(x)}\Sigma_{t,n}^{-1}$ $\displaystyle\qquad-\left(\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n}(x)}{q_{t}(x)}\Sigma_{t,n}^{-1}(x-\mu_{t,n})\right)\left(\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n}(x)}{q_{t}(x)}\Sigma_{t,n}^{-1}(x-\mu_{t,n})\right)^{\intercal}$ $\displaystyle=\underbrace{\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n}(x)}{q_{t}(x)}\Sigma_{t,n}^{-1}(x-\mu_{t,n})(x-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}}_{\text{term 1}}-\underbrace{\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n}(x)}{q_{t}(x)}\Sigma_{t,n}^{-1}}_{\text{term 2}}-\underbrace{(\nabla\log q_{t}(x))(\nabla\log q_{t}(x))^{\intercal}}_{\text{term 3}}.$ $$

We consider the difference associated with each term separately.
For term 3, the difference is

$$ $\displaystyle(\nabla\log q_{t-1}(\mu_{t}))(\nabla\log q_{t-1}(\mu_{t}))^{\intercal}-\alpha_{t}(\nabla\log q_{t}(x_{t}))(\nabla\log q_{t}(x_{t}))^{\intercal}$ $\displaystyle=(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))(\nabla\log q_{t-1}(\mu_{t}))^{\intercal}$ $\displaystyle\qquad+(\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))(\nabla\log q_{t-1}(\mu_{t})-\sqrt{\alpha_{t}}\nabla\log q_{t}(x_{t}))^{\intercal}$ $\displaystyle=O(1-\alpha_{t}).$ $$

Here the rate follows from ([86](#A9.E86)).
For term 2, the difference is

$$ $\displaystyle\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}-\alpha_{t}\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t,m}(x_{t})}$ $\displaystyle=(1-\alpha_{t})\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}+\alpha_{t}\left(\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}-\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t,m}(x_{t})}\right)$ $\displaystyle=(1-\alpha_{t})\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}$ $\displaystyle\qquad+\alpha_{t}\frac{\sum_{m,n=1}^{N}\pi_{m}\pi_{n}(q_{t-1,n}(\mu_{t})q_{t,m}(x_{t})\Sigma_{t-1,n}^{-1}-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1})}{q_{t-1}(\mu_{t})q_{t}(x_{t})}$ $\displaystyle=O(1-\alpha_{t}).$ $$

Here the rate in the last line follows from ([85](#A9.E85)).
For term 1, the difference is

$$ $\displaystyle\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}$ $\displaystyle\qquad-\alpha_{t}\frac{\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})(x_{t}-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t,m}(x_{t})}$ $\displaystyle=(1-\alpha_{t})\frac{\sum_{n=1}^{N}\pi_{n}q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}}{\sum_{m=1}^{N}\pi_{m}q_{t-1,m}(\mu_{t})}$ $\displaystyle\qquad+\alpha_{t}\frac{\sum_{m,n=1}^{N}\pi_{m}\pi_{n}D_{m,n}}{q_{t-1}(\mu_{t})q_{t}(x_{t})}$ $$

where

$$ $\displaystyle D_{m,n}$ $\displaystyle:=q_{t,m}(x_{t})q_{t-1,n}(\mu_{t})\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}$ $\displaystyle\quad-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})(x_{t}-\mu_{t,n})^{\intercal}\Sigma_{t,n}^{-1}$ $\displaystyle=\left[q_{t,m}(x_{t})q_{t-1,n}(\mu_{t})-q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\right]\Sigma_{t-1,n}^{-1}(\mu_{t}-\mu_{t-1,n})(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}$ $\displaystyle\quad+q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\left[\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1}\right](\mu_{t}-\mu_{t-1,n})(\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}$ $\displaystyle\quad+q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}\left[(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})\right](\mu_{t}-\mu_{t-1,n})^{\intercal}\Sigma_{t-1,n}^{-1}$ $\displaystyle\quad+q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})\left[(\mu_{t}-\mu_{t-1,n})-(x_{t}-\mu_{t,n})\right]^{\intercal}\Sigma_{t-1,n}^{-1}$ $\displaystyle\quad+q_{t-1,m}(\mu_{t})q_{t,n}(x_{t})\Sigma_{t,n}^{-1}(x_{t}-\mu_{t,n})(x_{t}-\mu_{t,n})^{\intercal}\left[\Sigma_{t-1,n}^{-1}-\Sigma_{t,n}^{-1}\right]$ $\displaystyle=O(1-\alpha_{t}).$ $$

Since the differences in all three terms are well controlled using ([82](#A9.E82))–([86](#A9.E86)), the proof is complete.

## Peer Reviews

confidence: 3
contribution: 3
initial_rating: 6
presentation: 3
questions: The proof is quite lengthy, and I did not have the opportunity to examine all the details. It would be beneficial if the author could elaborate on how the accelerated convergence is achieved using the proposed novel proof technique (specifically, in Step 2 regarding the tilting factor). Additionally, restating Tweedie's formula and clarifying its extension in this context would help enhance the clarity of the proof for readers.
soundness: 3
strengths: The paper introduces a novel accelerated stochastic DDPM sampler and demonstrates its applicability to various distribution classes while relying on weaker smoothness conditions and the bounded second moment condition. Additionally, it establishes the first acceleration guarantees for Gaussian mixture models.
summary: This paper introduces a novel accelerated stochastic DDPM sampler that leverages the tilting factor representation and Tweedie's formula. This approach is designed to accommodate a broader class of target distributions, requiring weaker conditions than previous methods.
weaknesses: Assumption 4 plays a crucial role in the construction of the tilting factor, requiring that the score function has derivatives of all orders and that the $p$-th moment is bounded $p \ge 1$. In my view, this condition is not as "rather soft" as suggested in the submission. The requirement for all derivatives to exist means that the score function is not merely smooth but infinitely differentiable, a level of regularity that is often absent in practical score functions, particularly for complex or non-smooth distributions.
Furthermore, the requirement that the $p$-th moment of the derivatives is bounded indicates that the derivatives exist and do not grow excessively. This can be quite a restrictive requirement, especially for distributions with heavy tails or irregularities. While this assumption is essential for maintaining theoretical rigor, it may limit the applicability of the models to a narrower range of problems.

---

confidence: 4
contribution: 2
initial_rating: 5
presentation: 2
questions: I have several questions regarding Theorem 1 and Corollary 1, which form the central results of the paper. 

In Theorem 1, constants are encapsulated within the notation $\lesssim$. I assume these constants depend on the distribution $Q_0$. It would be valuable to clarify which specific characteristics of $Q_0$ influence these constants. In particular, given that the theorem assumes the density $q_0$ of $Q_0$ is analytic, it would be helpful to know whether the constants depend on the derivatives of $q_0$.

If these constants are independent of $q_0$ and its derivatives, this would imply that the claim of Theorem 1 might not require the assumption that $q_0$ is infinitely smooth. Conversely, if the constants do depend on $q_0$ or its derivatives, then it raises the possibility that Theorem 1's claim may not be applicable in proving Corollary 1.
soundness: 2
strengths: 1. The paper deals with an important and timely problem.
2. The idea of considering the second-order derivatives to accelerate the convergence is interesting and the results presented in the paper show that it leads to some improvements.
3. The authors did a great job in discussing the prior work and in comparing their results to previous ones.
summary: The paper proposes a new approach to diffusion-based generative modeling that leverages second-order derivatives of the log-density function to achieve accelerated convergence. The framework assumes that the underlying distribution has a finite second-order moment and an infinitely smooth density with respect to the Lebesgue measure. For cases where the distribution lacks infinite smoothness, the algorithm is shown to converge to a convolution of the original distribution with a Gaussian distribution.

The main results provide an upper bound on the KL-divergence between the learned and target distributions. This bound depends on the number of iterations, the error in estimating the gradient and Hessian of the log-density, and a remainder term involving higher-order derivatives of the smoothed density function. The paper also examines Gaussian mixtures and distributions with Lipschitz-continuous log-densities in detail, explicitly characterizing the dependence of the remainder term on the dimension in these cases.
weaknesses: ### **Main concern**
My primary concern with this paper is the quality of the writing, which requires significant polishing to reach a publishable standard.

The main contributions of this work are clearly mathematical and reside in the theorems presented. The proofs are highly technical, involving lengthy and intricate computations. However, the current presentation makes the proofs challenging to follow and verify. This poses a problem, as reviewers often lack the time to check such detailed proofs thoroughly. If an error were to be found in a proof, it would substantially diminish the paper’s value.

I strongly recommend that the authors dedicate additional time and effort to revising the manuscript for clarity, making the technical sections easier to parse and verify.

### **Specific comments/typos**
- Lines 176-177: Avoid writing expressions such as "for all $1 \leqslant t$" or "$\forall 1 \leqslant t$", as these read ambiguously as "for all 1 that are less than $t$," which is clearly not the intended meaning. Instead, use "for all integers $t \in [0, T]$" or "for all $t \in \\{1, \ldots, T\\}$" to convey the intended range of values.
- line 215: I find it worrisome that the total variation is defined as the supremum over **all** subsets $A$ of $\mathbb R^d$. This should be replaced by all the measurable subsets of $\mathbb R^d$. 
- Lines 213–218: I do not see any reason for using the notation $\hat P'_0$ in this paragraph. It would be preferable to maintain consistency with the next paragraph by defining the KL-divergence between two measures $Q$ and $P$ instead.
- Line 220: Instead of 'when $q_0$ exists', it would be more accurate to write "when $Q_0$ is absolutely continuous with respect to $\hat P_0'$." The existence of $q_0$ alone does not ensure that the KL-divergence is finite.
- Line 243: For the Ozaki discretization, it would be more suitable to cite the papers
  1.  T. Ozaki. A bridge between nonlinear time series models and nonlinear stochastic dynamical systems: a local linearization approach. Statistica Sinica, 2(1):113–135, 1992.
  2. O. Stramer and R. L. Tweedie. Langevin-type models. II. Self-targeting candidates for MCMC algorithms. Methodol. Comput. Appl. Probab., 1(3):307–328, 1999b
- Line 244: "estimate" -> "estimates"
- Eq. (6): There is no guarantee that this matrix is positive semi-definite and, therefore, can be used as a covariance matrix. I agree that for most $x_t$ this will be the case for large $T$, but it is necessary here to have a definition that works for **every** $x_t$. 
- Line 251: "denote as Hessian matching" -> "refer to as Hessian matching".
- Line 262: It is better to remove the second equality here and to keep it in the claim of Lemma 1 only.
- Line 269 (and in several places in the proof): instead of writing the marginal distributions of $X_0$ and $\bar W_t$, it is necessary to indicate the joint distribution of these random variables. It suffices to write $(X_0,\bar W_t)\sim Q_0\otimes \mathcal N(0,I_d)$. 
- line 280: "satisfy" -> "satisfies"
- lines 277-289: It is generally not recommended to refer to material, such as an assumption, before it has been introduced. In this case, Assumption 3 is mentioned before it is properly stated. Similarly, within the text of Assumption 3, Assumption 4 is referenced prior to its statement. I strongly recommend that the authors adjust the flow of the text to avoid such inversions.
- Lemma 2: In the statement of this lemma and in other places, the meaning of the Landau notation $O(\cdot)$ is unclear. It should be specified that this notation is understood in the asymptotic limit $T \to \infty$. Additionally, in Eq. (10), there should be no term depending on $t$ on the right-hand side. Either $\alpha_t$ should be replaced with $\alpha_T$, or, preferably, the left-hand side should be divided by $(1 - \alpha_t)^2$ within the maximum. **My recommendation would be to avoid these notation and to prefer more conventional writing** ``there exist constants $C_1$ and $C_2$ such that ... for all $T\in\mathbb N$''. 
- Line 281: $\alpha_t$ is not defined in (1). I guess you wanted to write Definition 1 instead of (1).
- Lines 287–288: Since $H_t$ first appears on line 287, it would be clearer to move the indication "defined by (8)" forward to accompany its first mention.
- Line 287: It is unclear why the factor $(1 - \alpha_t)$ is retained on the left-hand side while $O(1 - \alpha_t)$ is used on the right-hand side. Rather than writing $(1 - \alpha_t) g_t = O(1 - \alpha_t)$, wouldn’t it be clearer to simply write $g_t = O(1)$? Additionally, should this relation hold uniformly for $\ell \geqslant 1$?
- Line 300: What is the precise meaning of *analytic* in this context? I was unable to find a definition in the text, either in the main body or in the appendix.
- Assumption 4: Same remark as above, I guess it is possible to suppress the factors $(1-\alpha_t)^{p\ell}/2$ on both sides of the equalities.
- Assumption 4: What is $p$ in this assumption? Should the boundedness of $p$-th order derivatives hold for every $p$? Should the upper bound be independent of $p$?
- Definition 1: This should be stated as a condition rather than a definition.
- Line 342: I did not find the definition of $\widehat P_0'$. 
- Line 376: "titling" -> "tilting"
- Line 389: "rest two terms" -> "remaining two terms".
 
- I guess in the claim of Theorem 1 it is also necessary to assume that $q_0(x)>0$ for all $x$. Indeed, if this condition is not satisfied, the term $\log q_0(\mu_1(X_1))$ appearing in the right-hand side of the equation on line 344 is meaningless.

---

confidence: 4
contribution: 2
initial_rating: 5
presentation: 3
questions: - A recent work of [Gatmiry-Kelner-Lee](https://arxiv.org/abs/2404.18869) also established bounds on higher-order scores of Gaussian mixtures, though specifically in the context of identity-covariance components. How do their bounds compare to the ones proved in this submission?
- To get intuition for the assumption for Theorem 4: Is there an example of a second-order smooth data distribution for which the scores along the trajectory are not all first-order smooth?
- Perhaps it is worth including in the related work some discussion on the use of higher-order scores for diffusion models in practice like GENIE?
soundness: 4
strengths: The result in the finite variance setting (Theorem 3) improves upon prior work in a few directions. Firstly, among works that achieve scaling $O(1/\epsilon)$ or better, theirs achieves the best dimension dependence. Secondly, unlike the most relevant preceding work of Li et al. '24c, the present work can tolerate score error. Finally, compared to those prior works, the present work applies in a more general setting where the distribution only has finite variance rather than bounded support.

The Gaussian mixture result is interesting as it requires proving new bounds on the higher-order derivatives of the score function for Gaussian mixtures, which are of independent interest.
summary: This paper continues the study of proving convergence bounds for diffusion models under the assumption of accurate score estimation, with a focus on improved dependence on the target error. The "standard" rate achieved by DDPMs is an iteration complexity of $O(d/\epsilon^2)$ for sampling any distribution with finite variance to $\epsilon$ error in TV (after it has been perturbed with a small amount of noise), due to Benton et al. '23 and Conforti et al. '23. A number of recent works have tried to improve the $\epsilon$ dependence to $1/\epsilon$, but only in more specialized settings (e.g. assuming smoothness or bounded support). One of the main results of this work is to show such a result under only the assumption of finite variance. They achieve an iteration complexity of $O(d^{3/2}/\epsilon)$ using a diffusion-based sampler that has approximate access to both the score function *and its Jacobian*. In the smooth setting, they obtain samplers with similar scaling under one of two assumptions: 1) second-order of the smoothness, and 2) Gaussian mixture.
weaknesses: One weakness is that the sampler requires approximate access to *higher-order* score functions (namely, the Jacobian of the score). While there are examples of diffusion generative models in practice that use higher-order scores essentially by applying automatic differentiation to the first-order score (e.g. GENIE), this is not standard and introduces significant computational overhead. Additionally, these models in practice are not actually trained according to some kind of explicit Hessian matching loss, so it is unclear why one should expect the Jacobian of the score estimates to be close to the Jacobian of the scores. That said, the prior work of Li et al. on acceleration, which appeared in a previous ICLR, also works with second-order access, so this weakness is somewhat justifiable.

Also, not a weakness per se, but note that the result for second-order smoothness (Theorem 4) is technically incomparable to the other previous works getting $1/\epsilon$ scaling because while it does not assume smoothness along the trajectory, it needs to assume that the *Hessian* of the log-density is Lipschitz. And as mentioned above, this result requires second-order scores whereas the probability flow ODE paper of Chen et al. only uses first-order scores.

Another note: while the authors are careful to clarify the chronology of the various related works, it is worth pointing out that in the last month or so (and before the ICLR deadline), there have been important developments in this literature that supersede some of the main results of this submission. For instance, [Li & Yan](https://arxiv.org/pdf/2409.18959) showed that even for the standard SDE-based sampler (without second-order score), one can achieve an accelerated rate of $O(d/\epsilon)$, in fact under only a *first* moment bound.

Meta-review: This paper considers the problem of establishing convergence bounds for diffusion models.
Authors assume that score can be estimated accurately, and provide guarantees in terms of error tolerance. Authors improve the existing rates of diffusion models under finite variance condition, with better  dimension and tolerance dependence. This is achieved by assuming access to Jacobian.


This paper was reviewed by three expert reviewers the following Scores/Confidence: 6/3, 6/3, 6/4. I think the paper is studying an interesting topic and the results are relevant to ICLR community. The following concerns were brought up by the reviewers:

- Several typos and ambiguous statements were pointed by the reviewers. These should be carefully addressed. I strongly encourage the authors to revise their manuscript, and improve the text based on the feedback.
  
- Assumed access to Jacobian may not be readily avaialble in practice. This should be clarified/discussed in detail in the paper.


Authors should carefully go over reviewers' suggestions and address any remaining concerns in their final revision. Based on the reviewers' suggestion, as well as my own assessment of the paper, I recommend including this paper to the ICLR 2025 program.

Decision: Accept (Poster)