Theory on Score-Mismatched Diffusion Models and Zero-Shot Conditional Samplers

## Abstract

Abstract The denoising diffusion model has recently emerged as a powerful generative technique, capable of transforming noise into meaningful data. While theoretical convergence guarantees for diffusion models are well established when the target distribution aligns with the training distribution, practical scenarios often present mismatches. One common case is in zero-shot conditional diffusion sampling, where the target conditional distribution is different from the (unconditional) training distribution. These score-mismatched diffusion models remain largely unexplored from a theoretical perspective. In this paper, we present the first performance guarantee with explicit dimensional dependencies for general score-mismatched diffusion samplers, focusing on target distributions with finite second moments. We show that score mismatches result in an asymptotic distributional bias between the target and sampling distributions, proportional to the accumulated mismatch between the target and training distributions. This result can be directly applied to zero-shot conditional samplers for any conditional model, irrespective of measurement noise. Interestingly, the derived convergence upper bound offers useful guidance for designing a novel bias-optimal zero-shot sampler in linear conditional models that minimizes the asymptotic bias. For such bias-optimal samplers, we further establish convergence guarantees with explicit dependencies on dimension and conditioning, applied to several interesting target distributions, including those with bounded support and Gaussian mixtures. Our findings are supported by numerical studies.

## 1 Introduction

Generative modeling stands as a cornerstone in machine learning, with the goal of producing samples whose distribution emulates that of the training data. Traditional approaches encompass variational autoencoders (VAE) [], generative adversarial networks (GANs) [], normalizing flows [], and others. Recently, diffusion models, especially the denoising diffusion probabilistic models (DDPMs) [, ], have emerged as particularly compelling generative models, gaining widespread acclaim for their stable and cutting-edge performance across various tasks, such as image and video generation [, ].

In ideal situations, the training and target distributions of generative models match each other. However, this often does not hold in practice, where distributional mismatch between the training and target distributions can occur due to various reasons such as possible privacy constraints, need for computational efficiency, and knowledge gap between training and sampling processes. Specifically for diffusion models, such mismatches exhibit between the scores obtained from the training data and the scores of the target distribution from which we want to generate samples. One common scenario that existing studies primarily focus on is conditional diffusion models in image generation tasks (see [, , ] for surveys of diffusion models in computer vision).
Different from unconditional image generation, conditional image samplers aim to generate images that are consistent with the given information, either be a text-prompt (as in text-to-image synthesis) or a sub-image (as in image super-resolution). For example, in image super-resolution, given the input of a low-resolution image, the goal is not to generate some arbitrary high-resolution image but the one whose corresponding low-resolution part matches the given input. Here the diffusion models are well-trained on the unconditional distribution of high-resolution images, whereas the target distribution is the conditional distribution given the low-resolution input. If one uses these well-trained unconditional scores to generate conditional samples, there will be a mismatch at each step of the sampling process.

One class of methods to tackle the conditional sampling problem is to include extra-guided training, where a modified score function is trained with the extra knowledge of the conditioning information [, ]. On the theory side, several recent works [, , ]
provided the performance guarantee for such conditional diffusion samplers, where a score guidance is obtained through extra training based on the conditional information. However, the additional guided training in these samplers requires extra computations and needs to be conducted for every image conditioning, which may not be efficient in practice.

Alternatively, zero-shot conditional image samplers arise as a prevalent approach (e.g., [, , , , , , ]) for training-free conditional generation given well-trained unconditional scores. For each conditioned image, zero-shot samplers require no additional training to modify the scores. Instead, they adjust the scores during sampling by calculating rectified scores based on conditional information to mitigate the mismatch between the oracle conditional scores and the approximated ones. Despite their empirical promise, theoretical guarantee on these zero-shot samplers is largely unexplored.
In [], the authors provided a super-polynomial lower bound for zero-shot sampling as a converse result. In [], the authors proposed and analyzed a plug-and-play conditional sampler. However, their analysis relies on the properties of the Markov transition kernel specific to their plug-and-play model, which does not appear to be applicable to several widely used zero-shot samplers, such as Come-Closer-Diffuse-Faster (CCDF) [] and Denoising Diffusion Null-space Model (DDNM) []. Therefore, there is a need to provide the performance guarantee for those popular zero-shot conditional samplers.

In this paper, we address two key theoretical research gaps in zero-shot score-mismatched diffusion models: (i) We provide performance guarantees for general score-mismatched diffusion models, extending their applicability beyond the primary focus of existing theoretical studies on the special case of conditional image generation. (ii) We analyze zero-shot conditional diffusion models, which are generally applicable to existing zero-shot samplers such as CCDF [] and DDNM [] for which the analysis in [] is not applicable (as we discuss above).

### 1.1 Our Contributions

Technically, the main challenge due to mismatched scores is to analyze the expected tilting factor [] under a mean-perturbed Gaussian, providing an upper bound of the asymptotic orders of all Gaussian non-centralized moments. Our detailed contributions are as follows.

Convergence of General Score-Mismatched DDPM: We provide the first non-asymptotic convergence bound on the KL divergence between the target and generated distributions when there is mismatch between the sampling and target scores in DDPM samplers, for general target distributions having finite second moments. We show that the score mismatch at each diffusion step introduces an asymptotic distributional bias that is proportional to the accumulated mismatch. We also provide the first explicit dimensional dependency when the sixth moment of the target distribution exists. Our result is applicable to general mismatch forms between scores, which greatly extend the focus of the existing theoretical research on conditional score-mismatch diffusion models.

We then apply our results to zero-shot conditional DDPM samplers, as long as the conditioning involves certain deterministic or random transformations of the data. This provides the first theoretical guarantees for several existing zero-shot samplers, such as CCDF [] and DDNM []. Notably, the theory in [] does not apply to these samplers, as their analysis relies on the properties of the Markov transition kernel specific to their plug-and-play model. In contrast, our approach is based on the tilting-factor analysis from [], which is applicable to general score-mismatched DDPM models. Moreover, the theory in [] is limited to cases where the measurement log-likelihood function is differentiable or bounded and does not provide explicit dependencies on the data dimension. In contrast, our results do not require the measurement log-likelihood function to be differentiable or bounded and explicitly characterize the dependencies on the data dimension.

Novel Design of Bias-Optimal Zero-shot Sampler BO-DDNM: Inspired by our convergence analysis of score-mismatched DDPM, we design a novel zero-shot conditional sampler, called the BO-DDNM sampler, which minimizes the asymptotic bias for linear conditional models. Such a sampler coincides with the regular DDNM sampler [] when there is no presence of measurement noise, and achieves faster convergence than both the DDNM and DDNM^+ samplers under measurement noise, as shown by our theory and numerical simulations.

Theory for BO-DDNM with Explicit Parameter Dependencies: We provide the convergence bound for the proposed BO-DDNM sampler with explicit dependencies on the dimension $d$ as well as the conditional information $y$, for various interesting classes of target distributions including those having bounded support and Gaussian mixture.
For the case of Gaussian mixture, we further show that two factors positively affect the asymptotic bias: (1) the averaged distance between $y$ and the mean of each Gaussian component, (2) the variance of the measurement noise, and (3) the corresponding correlation within each Gaussian component.

### 1.2 Related Work

We provide a summary of works addressing unconditional and score-matched diffusion models in [Appendix B](#A2). Below we discuss related works on conditional diffusion models which are closely related to our study here.

Extra-Guided Training: In order to achieve conditional sampling using DDPM models in practice, one method is to introduce conditional guided training, where one either uses an existing classifier (a.k.a., classifier guidance) [] or jointly trains the unconditional and conditional scores (a.k.a., classifier-free guidance) [].
Here a guidance term is obtained to “guide” the diffusion sampling process at each step such that the sampling scores correspond to the true conditional scores.
On the theory side, [] investigates the effect of the guidance strength in classifier-free guidance on Gaussian mixture models. There are other theoretical works that investigate sample complexity bounds for conditional score matching for a variety of target distribution models, including the conditional Ising models [], those supported on a low-dimensional linear subspace [], and Hölder smooth and sub-Gaussian conditional models []. Other than stochastic samplers, a conditional ODE sampler is proposed and studied in [], which also requires extra training of the conditional score function.

Zero-shot Samplers: To achieve conditional DDPM sampling, a popular method is to use zero-shot conditional samplers, with which one generates a conditional sample using approximated scores. These scores are calculated from the unconditional score estimates and the conditional information using simple (usually linear) functions without extra-training [, , , , , , ].
The only theoretical works on the performance upper bounds of zero-shot DDPM conditional samplers are [, ]. In [], a diffusion plug-and-play sampler is proposed which alternates between a diffusion sampling step and a consistency sampling step. The difference of our results from those in [] has been thoroughly discussed in [Section 1.1](#S1.SS1).
From an alternative perspective, [] shows that the sampling complexity with zero-shot samplers can take super-polynomial time for some worst-case distribution (among the set of distributions where smooth scores can be efficiently estimated). In contract, our result shows a consistent fact that there exists a non-vanishing asymptotic distributional bias within polynomial time.

## 2 Problem Setup

In this section, we first provide some background on the typical score-matched DDPMs. Then, we introduce the score-mismatched DDPM samplers and, as a special example, the conditional sampling problem and zero-shot samplers.

### 2.1 Background of Score-Matched DDPMs

The goal of the (typical) score-matched sampling problem is to generate a sample whose distribution is close to the data distribution. To this end, the DDPM algorithm [] is widely used, which consists of a forward process and a reverse process of latent variables. 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.
Let $Q_{0}$ be the data distribution, and let $Q_{t}$ (resp., $Q_{t,t-1}$) be the marginal (resp., joint) latent distribution for all $1\leq t\leq T$.

Forward Process:
In the forward process, white Gaussian noise is gradually added to the data: $x_{t}=\sqrt{1-\beta_{t}}x_{t-1}+\sqrt{\beta_{t}}w_{t},\leavevmode\nobreak\ \forall 1\leq t\leq T$,
where $w_{t}\stackrel{{\scriptstyle i.i.d.}}{{\sim}}\mathcal{N}(0,I_{d})$. Equivalently, this can be expressed as:

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

which means that under $Q$, the Markov chain $X_{0}\to X_{1}\to\dots\to X_{T}$ holds.
Define $\alpha_{t}:=1-\beta_{t}$ and $\bar{\alpha}_{t}:=\prod_{i=1}^{t}\alpha_{i}$ for all $1\leq t\leq T$.
An immediate result by accumulating the steps is that $Q_{t|0}(x_{t}|x_{0})=\mathcal{N}(x_{t};\sqrt{\bar{\alpha}_{t}}x_{0},(1-\bar{\alpha}_{t})I_{d})$,
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 $t$ and is independent of $x_{0}$.
Finally, since each $\bar{w}_{t}$ is Gaussian, each $Q_{t}\leavevmode\nobreak\ (t\geq 1)$ is absolutely continuous w.r.t. the Lebesgue measure. Let the p.d.f. of each $Q_{t}$ be $q_{t}$, and $q_{t,t-1}$, $q_{t|t-1}$, and $q_{t-1|t}$ for $t\geq 1$ can be similarly defined.

Reverse Process: In the reverse process, the latent variable at time $T$ is first drawn from a standard Gaussian distribution: $x_{T}\sim\mathcal{N}(0,I_{d})=:P_{T}$.
Then, each forward step is approximated by a reverse sampling step.
At each time $t=T,T-1,\dots,1$, define the true reverse process as $x_{t-1}=\mu_{t}(x_{t})+\sigma_{t}z_{t}$,
where $z_{t}\sim\mathcal{N}(0,I_{d})$. Here $\sigma_{t}^{2}:=\frac{1-\alpha_{t}}{\alpha_{t}}$. For the typical DDPM sampling process, $\mu_{t}(x_{t})=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})\nabla\log q_{t}(x_{t})\right)$. Equivalently, $P_{t-1|t}=\mathcal{N}(x_{t-1};\mu_{t}(x_{t}),\sigma_{t}^{2}I_{d})$.
Here $\nabla\log q_{t}(x)$ is called the score of $q_{t}$, and $\mu_{t}(x_{t})$ is a function of the score. Let $P_{t}$ be the marginal distributions of $x_{t}$ in the true reverse process, and let $p_{t}$ be its corresponding p.d.f. w.r.t. the Lebesgue measure. Define $p_{t-1|t}$ and $p_{t|t-1}$ in a way similar to the forward process.

In practice, one does not have access to $\nabla\log q_{t}(x_{t})$ and thus $\mu_{t}(x_{t})$. Instead, an estimate of $\nabla\log q_{t}(x_{t})$, denoted as $s_{t}(x_{t})$, is used, which results in an estimated $\widehat{\mu}_{t}(x_{t})$ and the estimated reverse process: $x_{t-1}=\widehat{\mu}_{t}(x_{t})+\sigma_{t}z$.
Let $\widehat{P}_{t}$ be the marginal distributions of $x_{t}$ in the estimated reverse process with the 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 $\widehat{P}_{T}=P_{T}$. Hence, under $P$ and $\widehat{P}$, $X_{T}\to X_{T-1}\to\dots\to X_{0}$ holds.

Performance Metrics: In the case where $Q_{0}$ is absolutely continuous w.r.t. the Lebesgue measure, we are interested in measuring the sampling performance through the KL divergence between $Q_{0}$ and $\widehat{P}_{0}$, defined as

$$ $\textstyle\mathrm{KL}(Q\|P):=\int\log\frac{\mathrm{d}Q}{\mathrm{d}P}\mathrm{d}Q=\mathbb{E}_{X\sim Q}\left[\log\frac{q(X)}{p(X)}\right]\geq 0.$ $$

Indeed, from Pinsker’s inequality, the total-variation (TV) distance
can be upper bounded as $\mathrm{TV}(Q_{0},\widehat{P}_{0})^{2}$ $\leq\frac{1}{2}\mathrm{KL}(Q_{0}\|\widehat{P}_{0})$. When $q_{0}$ does not exist, we use the Wasserstein-2 distance to measure the one-step perturbed performance, which is defined as

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

where $\Pi(Q,P)$ is the set of all joint probability measures on $\mathbb{R}^{d}\times\mathbb{R}^{d}$ with marginal distributions $Q$ and $P$, respectively. Both metrics are widely adopted [, ].

### 2.2 Score-Mismatched DDPMs

Differently from the score-matched sampling problem, the goal of the score-mismatched problem is to sample from a different target distribution from the training distribution with which we estimate the scores. Thus, there will be a mismatch between the target score and the estimated score at each diffusion step.
Let $Q_{t}$ ($t\geq 0$) be the training distributions used for training the score.
Let $\tilde{Q}_{0}$ be the target distribution that one hopes to generate samples from, and let $\tilde{Q}_{t}$ ($t\geq 1$) be its Gaussian-perturbed distributions according to the forward process in ([1](#S2.E1)).
Define the posterior mean under the target distributions as $m_{t}(x_{t}):=\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}[X_{t-1}|x_{t}]$. Note that by Tweedie’s formula [], $m_{t}(x_{t})=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})\nabla\log\tilde{q}_{t}(x_{t})\right)$.
Here we leave the generic definition of $\mu_{t}(x_{t})$ without providing any particular expression. An example of $\mu_{t}(x_{t})$ is given later in ([6](#S2.E6)), yet
our general analysis of score-mismatched DDPM does not require any particular form for $\mu_{t}$.
With these notations, the score mismatch at each step $t\geq 1$ can be defined as

$$ $\textstyle\Delta_{t}(x_{t}):=\frac{\sqrt{\alpha_{t}}}{1-\alpha_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}[X_{t-1}|x_{t}]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[X_{t-1}|x_{t}]\right)=\frac{\sqrt{\alpha_{t}}}{1-\alpha_{t}}(m_{t}(x_{t})-\mu_{t}(x_{t})).$ (2) $$

We use the metrics similar to those defined above to evaluate the performance of the score-mismatched DDPM.

### 2.3 Zero-Shot Conditional DDPMs

One interesting example of score mismatch is the zero-shot conditional sampling problem.
Differently from the unconditional counterpart, the conditional sampling problem aims to obtain a sample that aligns in particular with the provided conditioning. Define $y\in\mathbb{R}^{p}$ to be the conditioned information about $x_{0}$. Specifically, let $y=h(x_{0})$, where $h(\cdot)$ is some arbitrary (deterministic or random) function of only $x_{0}$ (apart from independent noise). Note that general score-mismatched DDPMs can be specialized to zero-shot conditional samplers with the following notations:

$$ $\tilde{Q}_{t}=Q_{t|y},\leavevmode\nobreak\ \leavevmode\nobreak\ m_{t}=m_{t,y},\leavevmode\nobreak\ \leavevmode\nobreak\ \mu_{t}=\mu_{t,y},\leavevmode\nobreak\ \leavevmode\nobreak\ \text{and}\leavevmode\nobreak\ \Delta_{t}=\Delta_{t,y}.$ (3) $$

Linear Conditional Models: In practice, one commonly adopted model is the linear conditional model [, , ], defined as

$$ $y:=Hx_{0}+n,$ (4) $$

where $H\in\mathbb{R}^{p\times d}$ ($p\leq d$) is a deterministic matrix and $n\sim\mathcal{N}(0,\sigma_{y}^{2}I_{p})$ is the measurement noise, which is assumed to be Gaussian and independent of $x_{0}$. For the case where there is no measurement noise, let $\sigma_{y}^{2}=0$ and thus $n=0$ almost surely.
In applications like image super-resolution and inpainting [], $H$ admits a simple form of a 0-1 diagonal matrix, where the 1’s only occur on the diagonal and at those locations corresponding to the provided pixels.
In these scenarios, both $H$ and $y$ are fixed and given.
The linear conditional model is studied in [Section 4](#S4).

Conditional Forward Process for Linear Models: Write the Moore–Penrose pseudo-inverse of $H$ as $H^{\dagger}$, and note that $H^{\dagger}H$ is an orthogonal projection matrix. With this notation, under ([4](#S2.E4)), we can re-express the forward process in ([1](#S2.E1)) as

$$ $x_{t}=\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)x_{0}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-\sqrt{\bar{\alpha}_{t}}H^{\dagger}n+\sqrt{1-\bar{\alpha}_{t}}\bar{w}_{t}.$ $$

Thus, for fixed $x_{0}$ and $y$, we have that, for all $t\geq 1$,

$$ $Q_{t|0,y}(x_{t}|x_{0},y)=\mathcal{N}(x_{t};\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)x_{0}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y,\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}+(1-\bar{\alpha}_{t})I_{d}).$ (5) $$

Also, since the forward process is a Markov chain, we have that $Q_{t|t-1,y}=Q_{t|t-1}$ for all $t\geq 1$.

Zero-shot Conditional Sampler for Linear Models: We employ the zero-shot conditional sampler for linear conditional models in the following form: $x_{t-1}=\mu_{t,y}(x_{t})+\sigma_{t}z_{t}$, where

$$ $\textstyle\mu_{t,y}(x_{t})=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})g_{t,y}(x_{t})\right),\quad g_{t,y}:=(I_{d}-H^{\dagger}H)\nabla\log q_{t}(x_{t})+f_{t,y}(x_{t}).$ (6) $$

Here $f_{t,y}(x_{t})$ is a simple function of $y$ and $x_{t}$ computable without extra training and such that $(I_{d}-H^{\dagger}H)f_{t,y}(x)\equiv 0$ for all $x\in\mathbb{R}^{d}$.
Intuitively, $f_{t,y}$ characterizes the approximate score in the range space of $H^{\dagger}H$.
Indeed, many zero-shot samplers in the literature have such $f_{t,y}(x_{t})$’s that satisfy ([6](#S2.E6)) (see [Appendix D](#A4)).
Here, with the linear model in ([4](#S2.E4)) and the zero-shot conditional sampler in ([6](#S2.E6)), the score mismatch at each time $t\geq 1$ is equal to

$$ $\Delta_{t,y}(x_{t})=(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}(x_{t})-\nabla\log q_{t}(x_{t}))+(H^{\dagger}H)\nabla\log q_{t|y}(x_{t})-f_{t,y}(x_{t}).$ (7) $$

## 3 DDPM under General Score Mismatch

In this section, we provide convergence guarantees for general score-mismatched DDPM samplers under a general target distribution $\tilde{Q}_{0}$. Throughout this section we keep the definition for score mismatch $\Delta_{t}$ as in ([2](#S2.E2)).

### 3.1 Technical Assumptions

The first [Assumption 1](#Thmassumption1) is commonly adopted in the analyses of unconditional DDPM samplers [, , ].

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

There exists a constant $M_{2}<\infty$ (that does not depend on $d$ and $T$) such that $\mathbb{E}_{X_{0}\sim\tilde{Q}_{0}}\left\lVert X_{0}\right\rVert^{2}\leq dM_{2}$.

The following [Assumption 2](#Thmassumption2) is made for the score estimation error for the general mismatched setting, where we leave generic definitions of $\mu_{t}$ and $\widehat{\mu}_{t}$. For zero-shot conditional samplers in linear models, this assumption is weaker than that for the estimation error for unconditional scores (see ([9](#S4.E9))).

###### Assumption 2 (Posterior Mean Estimation) .

The estimated posterior mean $\widehat{\mu}_{t}$ at $t=1,\dots,T$ satisfy

$$ $\textstyle\frac{1}{T}\sum_{t=1}^{T}\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\widehat{\mu}_{t}(X_{t})-\mu_{t}(X_{t})\right\rVert^{2}\leq\varepsilon^{2},\leavevmode\nobreak\ \text{where}\leavevmode\nobreak\ \varepsilon^{2}=\tilde{O}(T^{-2}).$ $$

The following [Assumption 3](#Thmassumption3) is useful for our tilting-factor based analysis, which guarantees that all (higher-order) Taylor polynomials of $\log\tilde{q}_{t}$ are well controlled in expectation. They will be shown to be satisfied for wide classes of distributions (such as Gaussian mixtures) or with $\alpha_{t}$ in ([8](#S3.E8)).

###### Assumption 3 (Regular Derivatives) .

For all $t\geq 1$ where $\tilde{q}_{t-1}$ exists, $m\geq\frac{1}{2}$, $\ell\geq 1$, and $\bm{a}\in[d]^{p}$ where $\left|\bm{a}\right|=p\geq 1$,

$$ $\displaystyle(1-\alpha_{t})^{m}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left|\partial_{\bm{a}}^{p}\log\tilde{q}_{t}(X_{t})\right|^{\ell}=O\left((1-\alpha_{t})^{m}\right),$ $\displaystyle(1-\alpha_{t})^{m}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left|\partial_{\bm{a}}^{p}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\right|^{\ell}=O\left((1-\alpha_{t})^{m}\right).$ $$

The following [Assumption 4](#Thmassumption4) is used to characterize the amount of mismatch at each time $t\geq 1$. The $\bar{\alpha}_{t}:=\prod_{i=1}^{t}\alpha_{i}$ is necessary for the overall bias to be bounded. In the particular problem of zero-shot sampling under linear conditional models in ([4](#S2.E4)), it will be shown to hold for a variety of target distributions, including those having bounded support and Gaussian mixtures (see [Section 4](#S4)).

###### Assumption 4 (Bounded Mismatch) .

For all $t\geq 1$ where $\tilde{q}_{t-1}$ exists, $\ell\geq 2$ and $m\geq 1$,

$$ $(1-\alpha_{t})^{m}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{\ell}=O(\bar{\alpha}_{t}(1-\alpha_{t})^{m}).$ $$

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

###### Assumption 2 (Posterior Mean Estimation) .

###### Assumption 3 (Regular Derivatives) .

###### Assumption 4 (Bounded Mismatch) .

### 3.2 Convergence Bound

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

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

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

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

An example of $\alpha_{t}$ that satisfies [Definition 1](#Thmdefinition1) is $1-\alpha_{t}\equiv\frac{c\log T}{T},\leavevmode\nobreak\ \forall t\geq 1$ with $c>1$. Then, $\bar{\alpha}_{T}=\left(1-\frac{c\log T}{T}\right)^{T}=\exp\left(T\log\left(1-\frac{c\log T}{T}\right)\right)=O\left(e^{T\frac{-c\log T}{T}}\right)=o\left(T^{-1}\right)$.

The following [Theorem 1](#Thmtheorem1) provides the first convergence result for score-mismatched DDPM samplers for any $\tilde{Q}_{0}$ that has finite second moment (along with some mild regularity conditions).

###### Theorem 1 (DDPM under Score Mismatch) .

Suppose that $\tilde{Q}_{0}$ has a p.d.f. $\tilde{q}_{0}$, and suppose that [Assumptions 1](#Thmassumption1), [3](#Thmassumption3), [4](#Thmassumption4) and [2](#Thmassumption2) are satisfied. Then, with the $\alpha_{t}$ chosen to satisfy [Definition 1](#Thmdefinition1), the distribution $\widehat{P}_{0}$ from the score-mismatched DDPM satisfies

$$ $\mathrm{KL}(\tilde{Q}_{0}\|\widehat{P}_{0})\lesssim\mathcal{W}_{\text{oracle}}+\mathcal{W}_{\text{bias}}+\mathcal{W}_{\text{vanish}},\quad\text{where}$ $$

$$ $\displaystyle\mathcal{W}_{\text{oracle}}$ $\displaystyle\textstyle=\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\nabla^{2}\log\tilde{q}_{t}(X_{t})\Big{)}\right]+(\log T)\varepsilon^{2}$ $\displaystyle\mathcal{W}_{\text{bias}}$ $\displaystyle\textstyle=\sum_{t=1}^{T}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\mathcal{W}_{\text{vanish}}$ $\displaystyle\textstyle=\sum_{t=1}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\bigg{[}(\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t}))^{\intercal}\Delta_{t}(X_{t})\bigg{]}$ $\displaystyle\textstyle-\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\Delta_{t}(X_{t})^{\intercal}\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})\right]$ $\displaystyle\textstyle+\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\bigg{[}3\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{i}$ $\displaystyle\textstyle\qquad+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{j}\bigg{]}+\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}(\log T)\varepsilon.$ $$

When $\tilde{Q}_{0}$ does not have a p.d.f., a similar upper bound is applied to $\mathrm{KL}(\tilde{Q}_{1}\|\widehat{P}_{1})$ such that $\mathrm{W}_{2}(\tilde{Q}_{1},\tilde{Q}_{0})^{2}\lesssim(1-\alpha_{1})d$ (see [Corollary 1](#Thmcorollary1) in [Section E.5](#A5.SS5)).

To explain the three error terms in [Theorem 1](#Thmtheorem1), $\mathcal{W}_{\text{oracle}}$ captures the error assuming that one has access to (an estimate of) $\nabla\log\tilde{q}_{t},\leavevmode\nobreak\ \forall t\geq 1$. This error is independent of the score mismatch $\Delta_{t}$, and it decays as $\tilde{O}(T^{-1})$ under [Assumption 3](#Thmassumption3) [, Theorem 1].
The remaining two error terms $\mathcal{W}_{\text{bias}}$ and $\mathcal{W}_{\text{vanish}}$ arise from the mismatched sampling process. Both terms become zero if $\Delta_{t}\equiv 0$ for all $t\geq 1$, which corresponds to the score-matched case.
Under [Assumptions 3](#Thmassumption3) and [4](#Thmassumption4), $\mathcal{W}_{\text{vanish}}$ decays as $\tilde{O}(T^{-1})$ under an additional mild condition (see [Lemma 5](#Thmlemma5) in [Appendix F](#A6)), and $\mathcal{W}_{\text{bias}}$ asymptotically approaches a constant. Combining all three terms, score mismatch causes an asymptotic distributional bias between $\tilde{Q}_{0}$ and $\widehat{P}_{0}$.

To further understand $\mathcal{W}_{\text{bias}}$, note that $1-\alpha_{t}$ is usually summable under [Assumption 4](#Thmassumption4) (cf. [Lemmas 10](#Thmlemma10) and [7](#Thmlemma7)). Thus, the bias can be further upper-bounded by the maximum step-wise mismatch $\max_{t\geq 1}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}$. In case that $\mu_{t}(x_{t})=\frac{1}{\sqrt{\alpha_{t}}}\left(x_{t}+(1-\alpha_{t})g_{t}(x_{t})\right)$ (as in the zero-shot sampler in ([6](#S2.E6))), define a measure $\tilde{P}_{t}$ such that $g_{t}(x_{t})=\nabla\log\tilde{p}_{t}(x_{t})$.
Then, from ([2](#S2.E2)),

$$ $\textstyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}=\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla\log\frac{\tilde{q}_{t}(X_{t})}{\tilde{p}_{t}(X_{t})}\right\rVert^{2}=:\mathcal{F}(\tilde{Q}_{t}\|\tilde{P}_{t}).$ $$

where $\mathcal{F}(Q\|P)$ denotes the Fisher divergence (or called relative Fisher information) between $Q$ and $P$.
In [Section 4](#S4), this distributional bias $\mathcal{W}_{\text{bias}}$ inspires us to design a novel zero-shot DDPM sampler, the BO-DDNM sampler, that minimizes the asymptotic bias.

Next we provide an upper bound with explicit dimensional dependency, for any $Q_{0}$ that has finite sixth moment such as Gaussian mixture $Q_{0}$’s and those $Q_{0}$’s having bounded support.
To this end, we consider
a special noise schedule first proposed in []:

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

for any constants $(c,\delta)$ such that $c>1$ and $\delta e^{c}>1$.
With the $\alpha_{t}$ in ([8](#S3.E8)), we can
show that the regularity condition [Assumption 3](#Thmassumption3) holds for a quite general set of distributions (see [Lemma 5](#Thmlemma5) in [Appendix F](#A6)).

###### Theorem 2 .

Suppose that $\mathbb{E}_{X_{0}\sim\tilde{Q}_{0}}\left\lVert X_{0}\right\rVert^{6}\lesssim d^{3}$. Further, suppose that $\Delta_{t}$ satisfies that $\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{4}\lesssim\frac{\bar{\alpha}_{t}^{2}}{(1-\bar{\alpha}_{t})^{2r}}d^{2\gamma}$ with some $\gamma,r\geq 1$ for all $t\geq 2$. Then, if the estimation error satisfies [Assumption 2](#Thmassumption2) and if $\Delta_{t}$ satisfies [Assumption 4](#Thmassumption4), with the $\alpha_{t}$ in ([8](#S3.E8)) such that $\delta\ll 1$ and $c\asymp\log(1/\delta)$,
we have

$$ $\textstyle\mathrm{KL}(\tilde{Q}_{1}\|\widehat{P}_{1})\lesssim d^{\gamma}\delta^{-r}\left(1-\frac{2\log(1/\delta)\log T}{T}\right)\\ \textstyle+\max\{d^{(3+\gamma)/2}\delta^{-\frac{r+2}{2}},d^{1+\gamma}\delta^{-(r-1)}\}\frac{(\log T)^{2}}{T}+d^{\gamma/2}\delta^{-r/2}(\log T)\varepsilon.$ $$

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

Note that [Theorem 2](#Thmtheorem2) provides the first performance guarantee with explicit dimensional dependence for general score-mismatched DDPMs. Here the finite sixth moment is a technical condition to guarantee small expected difference of the first-order Taylor polynomial in case of mismatched scores (see [Lemma 5](#Thmlemma5) in [Appendix F](#A6)). Later, [Theorem 2](#Thmtheorem2) will be useful to provide guarantees for zero-shot conditional samplers under linear models ([Theorem 4](#Thmtheorem4)).

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

###### Theorem 1 (DDPM under Score Mismatch) .

###### Theorem 2 .

## 4 Zero-shot Conditional DDPM Samplers

As we discuss before, an important scenario of score-mismatched diffusion models is the zero-shot conditional problem, where certain information $y$ is given. In this section, we apply our general results for score-mismatch DDPMs in [Section 3](#S3) to studying zero-shot conditional DDPM samplers, with the notations in ([3](#S2.E3)) for $y=h(x_{0})$ for some arbitrary (deterministic or random) $h(\cdot)$.

In the following, we are particularly interested in the linear conditional model in ([4](#S2.E4)). We take the same [Assumptions 1](#Thmassumption1), [3](#Thmassumption3) and [4](#Thmassumption4) (albeit with changed notations), and further adopt the following common assumption on the unconditional score estimation [, , ].

###### Assumption 5 (Unconditional Score Estimation Error) .

Let the estimate $s_{t}(x_{t})$ of $\nabla\log q_{t}(x_{t})$
satisfy

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

Note that, with the zero-shot sampler defined in ([6](#S2.E6)), since $\left\lVert I_{d}-H^{\dagger}H\right\rVert=1$, $\forall x\in\mathbb{R}^{d}$,

$$ $\left\lVert\widehat{\mu}_{t,y}-\mu_{t,y}\right\rVert^{2}=\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\left\lVert(I_{d}-H^{\dagger}H)(s_{t}-\nabla\log q_{t})\right\rVert^{2}\leq\frac{(1-\alpha_{t})^{2}}{\alpha_{t}}\left\lVert s_{t}-\nabla\log q_{t}\right\rVert^{2}.$ (9) $$

Therefore, [Assumption 5](#Thmassumption5) directly implies [Assumption 2](#Thmassumption2), and thus [Theorem 1](#Thmtheorem1) (as well as [Corollary 1](#Thmcorollary1)) still holds under [Assumptions 1](#Thmassumption1), [3](#Thmassumption3), [4](#Thmassumption4) and [5](#Thmassumption5).

###### Assumption 5 (Unconditional Score Estimation Error) .

### 4.1 A Novel Bias-Optimal Zero-shot Sampler

Guided by the performance guarantee characterized in [Theorem 1](#Thmtheorem1), we will propose a novel optimized zero-shot condition sampler. With the zero-shot sampler defined in ([6](#S2.E6)), the goal is to choose the $f_{t,y}$ function that minimizes the convergence error for each $y\in\mathbb{R}^{p}$ and $t\geq 1$.

Specifically, it is observed in [Theorem 1](#Thmtheorem1) that the convergence error in terms of the KL-divergence will have an asymptotic distributional bias given by $\mathcal{W}_{\text{bias}}$. As follows, we characterize an optimal $f_{t,y}$ that minimizes $\mathcal{W}_{\text{bias}}$, which thus yields a corresponding bias-optimal zero-shot sampler.

###### Theorem 3 .

Define $\Sigma_{t|0,y}:=\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}+(1-\bar{\alpha}_{t})I_{d}$. For any $Q_{0}$ and $t\geq 1$, we have

$$ $\textstyle\nabla\log q_{t|y}(x_{t})=\Sigma_{t|0,y}^{-1}(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-x_{t})+\frac{\sqrt{\bar{\alpha}_{t}}}{1-\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mathbb{E}_{Q_{0|t,y}}[X_{0}|x_{t},y].$ $$

Also, recall the sampler in ([6](#S2.E6)) and define $f_{t,y}^{*}$ as

$$ $f_{t,y}^{*}(x_{t}):=\Sigma_{t|0,y}^{-1}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx_{t}\right).$ (10) $$

Also recall $\Delta_{t,y}$ from ([7](#S2.E7)). Then, $f_{t,y}^{*}$ satisfies that, for all $t\geq 1$ and fixed $y\in\mathbb{R}^{p}$,

$$ $f_{t,y}^{*}\in\operatorname*{arg\,min}_{f_{t,y}:(I_{d}-H^{\dagger}H)f_{t,y}\equiv 0}\left\lVert\Delta_{t,y}\right\rVert^{2},\quad Q_{t|y}\text{--almost surely}.$ $$

The sampler $f_{t,y}^{*}(x_{t})$ defined in ([10](#S4.E10)) provides a bias-optimal zero-shot conditional DDPM sampler.
In the case with $\sigma_{y}=0$, such an optimal sampler coincides with the regular DDNM sampler in [] (see [Appendix D](#A4)). Thus, we call this sampler as Bias-Optimal (BO) DDNM sampler. With ([10](#S4.E10)), we can also calculate the minimum step-wise mismatch as

$$ $\displaystyle\min_{f_{t,y}:(I_{d}-H^{\dagger}H)f_{t,y}\equiv 0}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{2}$ $\displaystyle\textstyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\nabla\log\frac{q_{t|y}(X_{t})}{q_{t}(X_{t})}\right\rVert^{2}_{(I_{d}-H^{\dagger}H)},$ $$

which is the projected Fisher divergence between $Q_{t|y}$ and $Q_{t}$ on $\mathrm{range}(I_{d}-H^{\dagger}H)$.

Figure: Figure 1: Comparison of BO-DDNM, DDNM and DDNM^+ for Gaussian (left) and Gaussian mixture (right) $Q_{0}$ under measurement noise.
Refer to caption: /html/2410.13746/assets/Figures/gauss_mixture_kl_sigy.png

In the following lemma, we provide a performance bound for BO-DDNM when $Q_{0}$ has bounded support. For comparison, we also provide the theoretical performances of vanilla DDNM.

###### Theorem 4 (BO-DDNM vs. DDNM) .

Suppose that $\left\lVert X_{0}\right\rVert^{2}\leq R^{2}d\leavevmode\nobreak\ \leavevmode\nobreak\ a.s.$ under $Q_{0}$. Suppose that [Assumptions 1](#Thmassumption1) and [5](#Thmassumption5) hold.
Then, with the conditional sampler $f_{t,y}^{*}$ in ([10](#S4.E10)), [Theorem 2](#Thmtheorem2) holds with $\gamma=1$ and $r=2$. Also, with the conditional sampler $f_{t,y}^{N}:=(1-\bar{\alpha}_{t})^{-1}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx_{t}\right)$, if further $\left\lVert H^{\dagger}\right\rVert\lesssim 1$, then [Theorem 2](#Thmtheorem2) holds with $\gamma=1$ and $r=4$.

[Theorem 4](#Thmtheorem4) establishes the first result applicable to DDNM type zero-shot conditional samplers for any linear conditional models on target distributions having bounded support.

Advantage of BO-DDNM over DDNM and DDNM^+: When there is positive measurement noise, [Theorem 4](#Thmtheorem4) indicates that our BO-DDNM sampler that uses $f_{t,y}=f_{t,y}^{*}$ enjoys a smaller asymptotic bias than DDNM that uses $f_{t,y}^{N}$ with the $\alpha_{t}$ in ([8](#S3.E8)) ($\delta^{-2}$ vs. $\delta^{-4}$). Note that the DDNM sampler corresponds to $f_{t,y}=f_{t,y}^{N}$ (see [Appendix D](#A4)). Such an advantage is also demonstrated by our numerical experiment. In [Figure 2](#S4.F2), we numerically compared modified conditional zero-shot sampler (as given in ([10](#S4.E10))) with the DDNM and DDNM^+ sampler for both Gaussian and Gaussian mixture $Q_{0}$’s at different levels of measurement noise. It is observed that
the optimal BO-DDNM sampler achieves a much lower bias than both the DDNM and the DDNM^+ sampler numerically, especially when $\sigma_{y}^{2}$ becomes large.

###### Theorem 3 .

###### Theorem 4 (BO-DDNM vs. DDNM) .

### 4.2 BO-DDNM Sampler for Gaussian Mixture Q 0 subscript 𝑄 0 Q_{0}

In this section, we focus on the convergence dependency on other system parameters of the BO-DDNM sampler, including the chosen $y$. In particular, we restrict our attention to Gaussian mixture $Q_{0}$’s and to a special conditional model, where $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$.
This choice can be seen in many applications, such as image super-resolution and inpainting (after reorganizing the pixels), where $I_{p}$ corresponds to the locations of the given pixels [, ]. We assume positive measurement noise.
We introduce the notation $[\Sigma_{0}]_{ab}$ to denote the variance components that correspond to the space of $a\times b$ where $a,b\in\{y,\bar{y}\}$.

The following lemma provides an upper bound on the asymptotic bias for Gaussian mixture $Q_{0}$.

###### Proposition 1 .

Suppose that $Q_{0}$ is Gaussian mixture with equal variance, 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 $\mathcal{N}(\mu_{0,n},\Sigma_{0})$ and $\pi_{n}\in[0,1]$ is the mixing coefficient with $\sum_{n=1}^{N}\pi_{n}=1$. Suppose that $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$, and adopt $f_{t,y}^{*}$ in ([10](#S4.E10)) and $\alpha_{t}$ in [Definition 1](#Thmdefinition1). Write $\lambda_{1}\geq\dots\geq\lambda_{d}$ and $\tilde{\lambda}_{1}\geq\dots\geq\tilde{\lambda}_{d-p}$ as the largest eigenvalue of $\Sigma_{0}$ and $[\Sigma_{0}]_{\bar{y}\bar{y}}$, respectively.
Then,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}$ $\displaystyle\textstyle\lesssim\bar{\alpha}_{t}d+\bar{\alpha}_{t}^{2}\frac{\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}}{\min\{\tilde{\lambda}_{d-p},1\}^{2}\min\{\lambda_{d},1\}^{2}}\max\left\{d(\lambda_{1}+\sigma_{y}^{2})+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2},d\right\}$ $\displaystyle\textstyle\lesssim\bar{\alpha}_{t}\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right).$ $$

[Proposition 1](#Thmproposition1) indicates that three factors affect (an upper bound on) the asymptotic bias.
(i) The measurement noise variance $\sigma_{y}^{2}$ determined by the system nature has an increasing effect on the bias.
(ii) The averaged distance $\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ between $H^{\dagger}y$ and $H^{\dagger}H\mu_{0,n}$ captures the quadratic dependency in $y$, as illustrated in the left plot of [Figure 2](#S4.F2). (iii) The correlation between $Hx_{0}$ and $(I_{d}-H^{\dagger}H)x_{0}$ of each mixture component contributes positively to the bias, which is contained in the factor $\frac{\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}}{\min\{\lambda_{d},1\}^{2}}$.
To see this, consider $\sigma_{y}^{2}=0$ and a specific Gaussian example with $d=2$, $p=1$, and $\Sigma_{0}=\begin{pmatrix}\sigma_{11}^{2}&\rho\sigma_{11}\sigma_{22}\\
\rho\sigma_{11}\sigma_{22}&\sigma_{22}^{2}\end{pmatrix}$.
As the correlation coefficient $\rho$ increases, $\Sigma_{0}$ becomes closer to be singular, and thus
$\lambda_{d}$ decreases to 0. Also, $\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}=\rho^{2}\sigma_{11}^{2}\sigma_{22}^{2}$ increases quadratically with $\rho$. Hence, this factor $\frac{\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}}{\min\{\lambda_{d},1\}^{2}}$ grows unboundedly as $\rho\to 1$, as does $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}$.
Such dependency on the correlation is illustrated numerically in the right plot of [Figure 2](#S4.F2).

The following theorem characterizes the conditional KL divergence when $Q_{0}$ is mixture Gaussian. In particular, we can show [Assumption 4](#Thmassumption4) holds with any $\alpha_{t}$ that satisfies [Definition 1](#Thmdefinition1) when $Q_{0}$ is Gaussian mixture (see [Lemma 8](#Thmlemma8) in [Section H.5](#A8.SS5)).

###### Theorem 5 .

Suppose the same conditions in [Proposition 1](#Thmproposition1) hold and $\sigma_{y}^{2}>0$. Suppose [Assumption 5](#Thmassumption5) holds. Take $f_{t,y}^{*}$ in ([10](#S4.E10)) and $\alpha_{t}$ that further satisfies $\sum_{t=1}^{T}(1-\alpha_{t})\bar{\alpha}_{t}=1+o(1)$. Then,

$$ $\displaystyle\textstyle\mathrm{KL}(Q_{0|y}\|\widehat{P}_{0|y})\lesssim\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)+$ $\displaystyle\textstyle\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)\frac{(\log T)^{2}}{T}+\sqrt{d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}}(\log T)\varepsilon.$ $$

This is the first convergence result for zero-shot samplers where explicit dependency on the conditioning $y$ is derived for Gaussian mixture targets. Note that the extra condition on $\alpha_{t}$ can be verified for both constant $\alpha_{t}$ ([Lemma 10](#Thmlemma10)) and that in ([8](#S3.E8)) ([Lemma 7](#Thmlemma7)).
Among the three terms in [Theorem 5](#Thmtheorem5), the first term is the asymptotic bias analyzed in [Proposition 1](#Thmproposition1).
Since the last two terms decrease to zero as $T\to\infty$, the asymptotic KL divergence will also approach some non-zero limit of order $d$.

The proof of [Theorem 5](#Thmtheorem5) is non-trivial because from [Theorem 1](#Thmtheorem1) we need to figure out the dependency on $y$ in all first three orders of partial derivatives of a Gaussian mixture density, which is generally hard to express. To this end, we restrict focus to a particular linear model where explicit dependency can be sought.
The result can be extended to the case of $\sigma_{y}^{2}=0$ with the $\alpha_{t}$ in ([8](#S3.E8)) (see [Remark 2](#Thmremark2)).

###### Proposition 1 .

###### Theorem 5 .

## 5 Proof Sketch of Theorem 1

We now provide a proof sketch of Theorem 1 to describe the idea of our analysis approach. The main technical challenge due to mismatched scores is to analyze the expected tilting factor under a mean-perturbed Gaussian, providing an upper bound of the asymptotic orders of all Gaussian non-centralized moments. See the full proof in [Appendix E](#A5).

To begin, with [Lemma 1](#Thmlemma1), we decompose the total error as $\mathrm{KL}(\tilde{Q}_{0}\|\widehat{P}_{0})\leq\mathbb{E}_{X_{T}\sim\tilde{Q}_{T}}\left[\log\frac{\tilde{q}_{T}(X_{T})}{\widehat{p}_{T}(X_{T})}\right]+\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{Q}_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]+\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{Q}_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]$. These three terms correspond respectively to the initialization error, estimation error, and reverse-step error.
The initialization error can be bounded by $\bar{\alpha}_{T}d$ in order using [, Lemma 3] under [Assumption 1](#Thmassumption1). Below we focus on the remaining two terms.

Step 1: Bounding estimation error under mismatch ([Lemma 2](#Thmlemma2)). At each time $t=1,\dots,T$, $\log(p_{t-1|t}(x_{t-1}|x_{t})/\widehat{p}_{t-1|t}(x_{t-1}|x_{t}))$ has an explicit expression since they are conditional Gaussians with the same variance. However, differently from the typical matched case, the mean of $P_{t-1|t}$ (i.e., $\mu_{t}(x_{t})$) is no longer equal to the posterior mean of $\tilde{Q}_{t-1|t}$ (i.e., $m_{t}(x_{t})$). Their difference is contained in $\Delta_{t}(x_{t})$, whose asymptotic order needs to be upper-bounded in light of [Assumption 2](#Thmassumption2).

Step 2: Decomposing reverse-step error under mismatch ([Section E.2](#A5.Ex65)).
First we decompose the tilting factor as $\tilde{\zeta}_{t,t-1}(x_{t},x_{t-1})=\tilde{\zeta}_{\text{mis}}(x_{t},x_{t-1})+\tilde{\zeta}_{\text{van}}(x_{t},x_{t-1})$, where $\tilde{\zeta}_{\text{mis}}:=\sqrt{\alpha_{t}}\Delta_{t}(x_{t})^{\intercal}(x_{t-1}-m_{t}(x_{t}))$ and $\tilde{\zeta}_{\text{van}}:=(\nabla\log\tilde{q}_{t-1}(m_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t}))^{\intercal}(x_{t-1}-m_{t}(x_{t}))+\sum_{p=2}^{\infty}T_{p}(\log\tilde{q}_{t-1},x_{t},m_{t}(x_{t}))$.
Here $\tilde{\zeta}_{\text{mis}}$ captures the factor that contributes to the total bias within $\tilde{\zeta}_{t,t-1}$.
Define the oracle sampling process as $\tilde{P}_{t-1|t}=\mathcal{N}(m_{t,y},\sigma_{t}^{2}I_{d})$. Then, the reverse-step error can be decomposed as $\mathbb{E}_{\tilde{Q}_{t,t-1}}[\tilde{\zeta}_{t,t-1}]-\mathbb{E}_{\tilde{Q}_{t}\times P_{t-1|t}}[\tilde{\zeta}_{t,t-1}]=\mathcal{W}_{\text{oracle, rev-step}}+\mathcal{W}_{\text{bias, rev-step}}+\mathcal{W}_{\text{vanish, rev-step}}$, where $\mathcal{W}_{\text{oracle, rev-step}}:=\mathbb{E}_{\tilde{Q}_{t,t-1}}[\tilde{\zeta}_{t,t-1}]-\mathbb{E}_{\tilde{Q}_{t}\times P_{t-1|t}}[\tilde{\zeta}_{t,t-1}]$, $\mathcal{W}_{\text{bias, rev-step}}:=\mathbb{E}_{\tilde{Q}_{t}\times\tilde{P}_{t-1|t}}[\tilde{\zeta}_{\text{mis}}]-\mathbb{E}_{\tilde{Q}_{t}\times P_{t-1|t}}[\tilde{\zeta}_{\text{mis}}]$, and $\mathcal{W}_{\text{vanish, rev-step}}:=\mathbb{E}_{\tilde{Q}_{t}\times\tilde{P}_{t-1|t}}[\tilde{\zeta}_{\text{van}}]-\mathbb{E}_{\tilde{Q}_{t}\times P_{t-1|t}}[\tilde{\zeta}_{\text{van}}]$.

Step 3: Bounding $\mathcal{W}_{\text{oracle, rev-step}}$ and $\mathcal{W}_{\text{bias, rev-step}}$ ([Equations 16](#A5.E16) and [E.3](#A5.Ex70)).
Under [Assumption 3](#Thmassumption3), the dominant term of $\mathcal{W}_{\text{oracle, rev-step}}$ is given by [, Theorem 1]. Also, the calculation of $\mathcal{W}_{\text{bias, rev-step}}$ is reduced to the difference in conditional mean, which is proportional to $\left\lVert\Delta(x_{t})\right\rVert^{2}$.

Step 4: Bounding $\mathcal{W}_{\text{vanish, rev-step}}$ ([Lemmas 3](#Thmlemma3) and [4](#Thmlemma4)).
To upper-bound $\mathcal{W}_{\text{vanish, rev-step}}$,
with results on the matched case in [], we need only to characterize the mean of $\tilde{\zeta}_{\text{van}}$ under the mismatched posterior $P_{t-1|t}$.
We determine the dominant order in the expected values of all Taylor polynomials, which includes calculating all non-centralized moments. We first calculate the first three non-centralized moments ([Lemma 3](#Thmlemma3)) and then determine the asymptotic order of all higher moments ([Lemma 4](#Thmlemma4)). With these, we can finally locate the terms of dominating order in $\mathcal{W}_{\text{vanish, rev-step}}$.

## 6 Conclusion

In this paper, we have provided convergence guarantees for the general score-mismatched diffusion models, which are specialized to zero-shot conditional samplers. For linear conditional models, we also designed an optimal BO-DDNM sampler that minimizes the asymptotic bias and we showed its dependencies on the system parameters. One future direction is to explore zero-shot samplers that use higher-order derivatives of the log-densities, which potentially achieve better convergence results.

## 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 Full List of 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 a vector (resp. matrix), all norms, if not explicitly specified, are referred to 2-norm (resp. spectral norm). For a vector $x$ and matrix $P$, define $\left\lVert x\right\rVert_{P}:=\sqrt{a^{\intercal}Pa}$.
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 $n$, $[n]:=\{1,\dots,n\}$.

## Appendix B Related Works on Unconditional DDPM Samplers

Given time-averaged $L^{2}$ unconditional score estimation error [], polynomial-time convergence guarantees have been established for wide families of target distributions [, , , , , , ]. For all target distributions with finite second moment, under $L^{2}$ score estimation error, $\mathcal{O}(d\log(1/\delta)^{2}/\varepsilon^{2})$ number of steps are sufficient to achieve $\varepsilon^{2}$ KL divergence between the $\delta$-perturbed target distribution and the generated distribution using the specially designed exponential-decay-then-constant step-sizes [, ].
The analysis usually involves applying the Girsanov change-of-measure framework and the Fokker-Plank equation [, ] to either the original SDE diffusion process or some transformed process [, ], followed by an analysis of the discretization of the continuous-time process.
More recently, similar convergence guarantees have been established using non-SDE-type techniques, such as with typical sets [] and with tilting factor representations [].
Here the new analysis introduced in [] is applicable to a larger set of step-sizes (equivalently, noise schedules) than the ones commonly used in previous analyses [, , ]. In this paper, we employ the same analytical framework as in [].

Some other works analyzed sampling errors using a different measure (the Wasserstein-2 distance) [, , ]. Beyond stochastic samplers, another line of studies provided theoretical guarantees for the deterministic sampler corresponding to DDPM [, , ].
Besides, [, , , ] provided guarantees for the closely-related flow-matching model, which learns a deterministic coupling between any two distributions.
Also, [, ] provided convergence guarantees for the closely-related consistency models []. Finally, in order to achieve an end-to-end analysis, several works also developed sample complexity bounds to achieve the $L^{2}$ score estimation error for a variety of distributions [, , , , , , , ].

## Appendix C Detail of Numerical Simulations

In [Figure 2](#S4.F2), we compared the performances of our optimal BO-DDNM sampler (with the $f_{t,y}^{*}$ in ([10](#S4.E10))) against the DDNM and DDNM^+ samplers [] at different levels of $\sigma_{y}^{2}$. For Gaussian, we use $\mu_{0}=0$, $d=4$, $p=2$, and $y=\begin{pmatrix}0.5&0.5\end{pmatrix}$. We first randomly generate a positive definite matrix $\Sigma_{0}$ and uniformly sample $\rho\in[0.4,0.7)$, and then this correlation coefficient is enforced for any $[\Sigma_{0}]^{ij}$ where $i\in[p]$ and $j\in\{p+1,\dots,d\}$. We use the noise schedule in ([8](#S3.E8)) with $c=3$ and $\delta=0.0001$ for Gaussian $Q_{0}$. For Gaussian mixture, we use $N=2$, $d=2$, $p=1$, and $y=1$. We set $\pi_{n}=\begin{pmatrix}0.4&0.6\end{pmatrix}$, $\text{diag}(\Sigma_{0})=\begin{pmatrix}0.1&1\end{pmatrix}$, and $\rho=0.6$. We further uniformly sample $\{\mu_{0,n}\}_{n=1}^{N}$ in the space $[-1,1)\times[-1,1)$. We use the noise schedule in ([8](#S3.E8)) with $c=4$ and $\delta=0.02$ for Gaussian mixture $Q_{0}$. We use 150000 samples to estimate the divergence when $Q_{0}$ is Gaussian mixture.

In [Figure 2](#S4.F2), we numerically verify the exact bias in KL divergence as a function of $y$ and $\rho$ for Gaussian $Q_{0}$. Here $Q_{0}=\mathcal{N}(0,\Sigma_{0})$, $d=4$ and $p=2$. Suppose that $\sigma_{y}^{2}=0$. We assume that each element of $y$ has equal values. The correlation coefficient $\rho$ is enforced for any pair of $x^{i}$ and $x^{j}$ where $i\in[p]$ and $j\in\{p+1,\dots,d\}$. We first randomly generate a positive definite matrix $\Sigma_{0}$ and then enforce the correlation condition for any $x^{i}$ and $x^{j}$ where $i\in[p]$ and $j\in\{p+1,\dots,d\}$. We use a sufficiently large number of steps $T=20000$. The conditional sampler is set as $f_{t,y}=f_{t,y}^{*}$ given in ([10](#S4.E10)). The noise schedule in ([8](#S3.E8)) with $c=3$ and $\delta=0.0001$ is used.

## Appendix D Derivation of Score Bias for Existing Zero-shot DDPM Samplers

In this section we show some examples of zero-shot conditional samplers proposed in the literature and in particular how they are related to the formulation of interest in ([6](#S2.E6)). We recall the notations $H$, $y$, and $\sigma_{t}$ from [Section 2.3](#S2.SS3). Also denote

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

which is the mean of the unconditional reverse-step at time $t\geq 1$.

### D.1 Come-Closer-Diffuse-Faster (CCDF)

We first examine the Come-Closer-Diffuse-Faster (CCDF) algorithm []. The CCDF algorithm using DDPM samplers gives that

$$ $\displaystyle x_{t-1}^{\prime}$ $\displaystyle=\mu_{t}+\sigma_{t}z_{t,1},$ $\displaystyle x_{t-1}$ $\displaystyle=(I-H^{\dagger}H)x_{t-1}^{\prime}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y+\sqrt{1-\bar{\alpha}_{t}}z_{t,2}$ $$

where $z_{t,1},z_{t,2}\stackrel{{\scriptstyle i.i.d.}}{{\sim}}\mathcal{N}(0,I_{d})$ are standard Gaussian random variables. Thus, the conditional mean of the update is

$$ $\displaystyle\mu_{t,y}$ $\displaystyle=(I-H^{\dagger}H)\mu_{t}+\sqrt{\bar{\alpha}_{t-1}}H^{\dagger}y$ $\displaystyle=\frac{1}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(I-H^{\dagger}H)\nabla\log q_{t}(x_{t})+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-\frac{1}{\sqrt{\alpha_{t}}}H^{\dagger}Hx_{t}$ $$

in which

$$ $f_{t,y}(x_{t})=\frac{1}{1-\alpha_{t}}\left(\sqrt{\alpha_{t}}\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx_{t}\right).$ $$

### D.2 DDNM and DDNM +

Next, we examine the DDNM algorithm and its modified version DDNM^+ []. We first note that the unconditional DDPM satisfies that (cf. [, Equations (7) and (11)]),

$$ $\displaystyle\mu_{t}$ $\displaystyle:=\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}x_{0|t}+\frac{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}x_{t},$ $\displaystyle x_{0|t}$ $\displaystyle:=\frac{1}{\sqrt{\bar{\alpha}_{t}}}x_{t}+\frac{1-\bar{\alpha}_{t}}{\sqrt{\bar{\alpha}_{t}}}\nabla\log q_{t}(x_{t})=\mathbb{E}_{X_{0}\sim Q_{0|t}(\cdot|x_{t})}[X_{0}|x_{t}].$ (11) $$

Combining these two lines, we have $\mu_{t}=\frac{1}{\sqrt{\alpha_{t}}}(x_{t}+(1-\alpha_{t})\nabla\log q_{t}(x_{t}))$.
In DDNM, $x_{0|t}$ is projected along the direction of the given $y$, which yields

$$ $x_{0|t,y}:=H^{\dagger}y+(I_{d}-H^{\dagger}H)x_{0|t},$ $$

and the corresponding conditional mean of the update becomes

$$ $\mu_{t,y}=\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}x_{0|t,y}+\frac{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}x_{t}.$ $$

Thus,

$$ $\displaystyle\mu_{t,y}$ $\displaystyle=\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}\left(H^{\dagger}y+(I_{d}-H^{\dagger}H)x_{0|t}\right)+\frac{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}x_{t}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}(I_{d}-H^{\dagger}H)\mu_{t}+H^{\dagger}\left(\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}y+\frac{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}Hx_{t}\right)$ $\displaystyle=\frac{1}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(I_{d}-H^{\dagger}H)\nabla\log q_{t}(x_{t})$ $\displaystyle\qquad+\left(\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}H^{\dagger}y+\frac{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}H^{\dagger}Hx_{t}-\frac{1}{\sqrt{\alpha_{t}}}H^{\dagger}Hx_{t}\right)$ $$

where $(i)$ follows from ([D.2](#A4.Ex32)).
Thus, to express this conditional mean in the form of ([6](#S2.E6)),

$$ $\displaystyle f_{t,y}(x_{t})$ $\displaystyle=\frac{\sqrt{\bar{\alpha}_{t}}}{1-\bar{\alpha}_{t}}H^{\dagger}y+\frac{1}{1-\alpha_{t}}\left(\frac{\alpha_{t}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}-1\right)H^{\dagger}Hx_{t}$ $\displaystyle=\frac{1}{1-\bar{\alpha}_{t}}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx_{t}\right).$ $$

Here note that $f_{t,y}(x_{t})$ is supported on $\mathrm{range}(H^{\dagger}H)$. Also note that for DDNM, $f_{t,y}=f_{t,y}^{*}$, which is the BO-DDNM sampler defined in ([10](#S4.E10)), when there is no measurement noise (i.e., $\sigma_{y}^{2}=0$).

Next we investigate its modified version, DDNM^+, in particular when $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$. To relate the notations of [, Section 3.3 and Appendix I] with ours, note that $\Sigma=A=H$, $U=I_{p}$, $V=I_{d}$, $s_{1},\dots,s_{p}=1$, $s_{p+1},\dots,s_{d}=0$, and $a=\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}$. If $\sigma_{t}\geq a\sigma_{y}$, we have

$$ $\Sigma_{t}=I_{d},\quad\Phi_{t}=\begin{pmatrix}(\sigma_{t}^{2}-a^{2}\sigma_{y}^{2})I_{p}&0\\ 0&\sigma_{t}^{2}I_{d-p}\end{pmatrix}.$ $$

Otherwise, if $\sigma_{t}<a\sigma_{y}$, we have

$$ $\Sigma_{t}=\begin{pmatrix}\frac{\sigma_{t}}{a\sigma_{y}}I_{p}&0\\ 0&I_{d-p}\end{pmatrix},\quad\Phi_{t}=\begin{pmatrix}0&0\\ 0&\sigma_{t}^{2}I_{d-p}\end{pmatrix}.$ $$

Observe that the only difference is on the space that supports $H^{\dagger}H$.

From [, Equations (17) and (18)], we can write

$$ $\hat{x}_{0|t,y}:=(I_{d}-H^{\dagger}H)x_{0|t}+\underbrace{\Sigma_{t}H^{\dagger}y+(I_{d}-\Sigma_{t})H^{\dagger}Hx_{0|t}}_{\text{supported on }\mathrm{range}(H^{\dagger}H)},$ $$

Thus, with similar arguments above,

$$ $\displaystyle\mu_{t,y}$ $\displaystyle=\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}\hat{x}_{0|t,y}+\frac{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t-1})}{1-\bar{\alpha}_{t}}x_{t}$ $\displaystyle=\frac{1}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(I_{d}-H^{\dagger}H)\nabla\log q_{t}(x_{t})$ $\displaystyle+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}(1-\bar{\alpha}_{t})}\underbrace{\left(\sqrt{\bar{\alpha}_{t}}(\Sigma_{t}H^{\dagger}y+(I_{d}-\Sigma_{t})H^{\dagger}Hx_{0|t})-H^{\dagger}Hx_{t}\right)}_{\text{supported on }\mathrm{range}(H^{\dagger}H)}$ $$

where $f_{t,y}$ is again supported on $\mathrm{range}(H^{\dagger}H)$.

### D.3 Samplers Using Higher-Order Derivatives

Before we end this section, we note that the formulation in ([6](#S2.E6)) only uses (estimates of) first-order derivatives of (unconditional) log-p.d.f.s (a.k.a. unconditional score functions). This might not correspond to the optimal zero-shot sampler, and in practice there have been methods that use both first- and second-order derivatives (namely, in $\partial x_{0|t}(x_{t})/\partial x_{t}$) to achieve better zero-shot sampling performance [, ]. Nevertheless, the second-order derivatives might be hard to obtain, which require extra machine time and memory in the calculation. We leave investigations to use second-order derivatives in zero-shot conditional samplers as future work.

## Appendix E Proof of Theorem 1 and Corollary 1

Overall, the structure of the proof of [Theorem 1](#Thmtheorem1) is similar to that for [, Theorem 1]. To start, we note that with similar arguments in [, Equation 13],
an upper bound on $\mathrm{KL}(\tilde{Q}_{0}\|\widehat{P}_{0})$ is given by

$$ $\displaystyle\mathrm{KL}(\tilde{Q}_{0}\|\widehat{P}_{0})$ $\displaystyle=\mathrm{KL}(\tilde{Q}_{T}\|\widehat{P}_{T})+\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\mathrm{KL}(\tilde{Q}_{t-1|t}(\cdot|X_{t})\|\widehat{P}_{t-1|t}(\cdot|X_{t}))\right]$ $\displaystyle\qquad-\sum_{t=1}^{T}\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1}}\left[\mathrm{KL}(\tilde{Q}_{t|t-1}(\cdot|X_{t-1})\|\widehat{P}_{t|t-1}(\cdot|X_{t-1}))\right]$ $\displaystyle\leq\mathrm{KL}(\tilde{Q}_{T}\|\widehat{P}_{T})+\sum_{t=1}^{T}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\mathrm{KL}(\tilde{Q}_{t-1|t}(\cdot|X_{t})\|\widehat{P}_{t-1|t}(\cdot|X_{t}))\right]$ $\displaystyle=\underbrace{\mathbb{E}_{X_{T}\sim\tilde{Q}_{T}}\left[\log\frac{\tilde{q}_{T}(X_{T})}{\widehat{p}_{T}(X_{T})}\right]}_{\text{Term 1: initialization error}}+\underbrace{\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{Q}_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]}_{\text{Term 2: estimation error}}$ $\displaystyle\qquad+\underbrace{\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{Q}_{t,t-1}}\left[\log\frac{\tilde{q}_{t-1|t}(X_{t-1}|X_{t})}{p_{t-1|t}(X_{t-1}|X_{t})}\right]}_{\text{Term 3: reverse-step error}}.$ (12) $$

The last equality holds because $\widehat{p}_{T}=p_{T}$.
Now, we provide an upper bound for the reverse-step error that is ready for further analysis.
In the following lemma, we show that the mismatched $\tilde{q}_{t-1|t}$ is an exponentially tilted form of $p_{t-1|t}$.

###### Lemma 1 .

Fixed $t\geq 1$. For any fixed $x_{t}\in\mathbb{R}^{d}$, as long as $\tilde{q}_{t-1}$ exists, we have

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

where

$$ $\displaystyle\tilde{\zeta}_{t,t-1}(x_{t},x_{t-1})$ $\displaystyle:=\sqrt{\alpha_{t}}\Delta_{t}(x_{t})^{\intercal}(x_{t-1}-m_{t}(x_{t}))+(\nabla\log\tilde{q}_{t-1}(m_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t}))^{\intercal}(x_{t-1}-m_{t}(x_{t}))$ $\displaystyle+\sum_{p=2}^{\infty}T_{p}(\log\tilde{q}_{t-1},x_{t-1},m_{t}(x_{t})).$ $$

Here we define the $p$-th order term in the Taylor expansion of $f(x)$ around $\mu$ as

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

where $\bm{a}\in[d]^{p}$ are the indices of differentiation in which the multiplicity of $i\in[d]$ is $\gamma^{i}$.

###### Proof.

See [Section G.1](#A7.SS1).
∎

We abbreviate $\tilde{\zeta}_{t,t-1}=\tilde{\zeta}_{t,t-1}(x_{t},x_{t-1})$. Given the expression of $\tilde{\zeta}_{t,t-1}$, the conditional reverse-step error can be upper-bounded for any fixed $x_{t}$ as

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}\left[\log\frac{\tilde{q}_{t-1|t}(X_{t-1}|x_{t})}{p_{t-1|t}(X_{t-1}|x_{t})}\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}\left[\tilde{\zeta}_{t,t-1}-\log\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[e^{\tilde{\zeta}_{t,t-1}}]\right]$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\leq}}\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}\left[\tilde{\zeta}_{t,t-1}\right]+\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[-\log e^{\tilde{\zeta}_{t,t-1}}\right]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}[\tilde{\zeta}_{t,t-1}]-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\tilde{\zeta}_{t,t-1}]$ (13) $$

where in $(i)$ we use Jensen’s inequality and note that $-\log(\cdot)$ is convex. Thus, from ([E](#A5.Ex47)), we have an upper bound as

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

Here, using [, Lemma 3], the initialization error can be upper-bounded as, when $T\to\infty$,

$$ $\mathbb{E}_{X_{T}\sim\tilde{Q}_{T}}\left[\log\frac{\tilde{q}_{T}(X_{T})}{\widehat{p}_{T}(X_{T})}\right]\leq\frac{1}{2}\mathbb{E}_{X_{0}\sim\tilde{Q}_{0}}\left\lVert X_{0}\right\rVert^{2}\bar{\alpha}_{T}+O\left(\bar{\alpha}_{T}^{2}\right).$ $$

This implies that, under [Assumption 1](#Thmassumption1) and if $c>1$,

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

Also, under [Assumption 3](#Thmassumption3), the higher-order Taylor polynomials enjoy exponential rate of decay in expectation, which is contained in powers of $(1-\alpha_{t})$. Thus, we are allowed to exchange the limit (of Taylor expansion) and the expectation operators (cf. [, Lemma 11]).

Now, we upper-bound the estimation error and reverse-step error under score mismatch separately.

###### Lemma 1 .

###### Proof.

### E.1 Step 1: Bounding estimation error under mismatch

The following lemma provides an upper bound for the estimation error under score mismatch.

###### Lemma 2 .

Under [Assumptions 2](#Thmassumption2) and [4](#Thmassumption4), with the $\alpha_{t}$ satisfying [Definition 1](#Thmdefinition1), we have

$$ $\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{Q}_{t,t-1}}\left[\log\frac{p_{t-1|t}(X_{t-1}|X_{t})}{\widehat{p}_{t-1|t}(X_{t-1}|X_{t})}\right]\lesssim\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}(\log T)\varepsilon+(\log T)\varepsilon^{2}.$ $$

###### Proof.

See [Section G.2](#A7.SS2).
∎

###### Lemma 2 .

###### Proof.

### E.2 Step 2: Decomposing reverse-step error under mismatch

Now, we decompose $\tilde{\zeta}_{t,t-1}(x_{t},x_{t-1})=\tilde{\zeta}_{\text{mis}}+\tilde{\zeta}_{\text{van}}$ where

$$ $\displaystyle\tilde{\zeta}_{\text{mis}}$ $\displaystyle:=\sqrt{\alpha_{t}}\Delta_{t}(x_{t})^{\intercal}(x_{t-1}-m_{t}(x_{t})),$ $\displaystyle\tilde{\zeta}_{\text{van}}$ $\displaystyle:=(\nabla\log\tilde{q}_{t-1}(m_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t}))^{\intercal}(x_{t-1}-m_{t}(x_{t}))+\sum_{p=2}^{\infty}T_{p}(\log\tilde{q}_{t-1},x_{t-1},m_{t}(x_{t})).$ (14) $$

Here $\tilde{\zeta}_{\text{van}}$ is the same tilting factor without score bias (cf. []).
Also, define an auxiliary conditional probability $\tilde{P}_{t-1|t}$ such that

$$ $\tilde{P}_{t-1|t}:=\mathcal{N}\left(m_{t}(x_{t}),\frac{1-\alpha_{t}}{\alpha_{t}}I_{d}\right),$ $$

which corresponds to the oracle reverse process that knows the true scores of the perturbed target distributions. Thus, we can decompose the expected value of ([E](#A5.Ex57)) in the following way:

$$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)[\tilde{\zeta}_{t,t-1}]$ $\displaystyle=\underbrace{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}\right)[\tilde{\zeta}_{t,t-1}]}_{\mathcal{W}_{\text{oracle, rev-step}}}$ $\displaystyle\quad+\underbrace{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)[\tilde{\zeta}_{\text{mis}}]}_{\mathcal{W}_{\text{bias, rev-step}}}$ $\displaystyle\quad+\underbrace{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)[\tilde{\zeta}_{\text{van}}]}_{\mathcal{W}_{\text{vanish, rev-step}}}.$ (15) $$

### E.3 Step 3: Bounding 𝒲 oracle, rev-step subscript 𝒲 oracle, rev-step \mathcal{W}_{\text{oracle, rev-step}} and 𝒲 bias, rev-step subscript 𝒲 bias, rev-step \mathcal{W}_{\text{bias, rev-step}}

Among the terms above, [, Theorem 1] shows that, under [Assumption 3](#Thmassumption3) and using the $\alpha_{t}$ in [Definition 1](#Thmdefinition1),

$$ $\mathcal{W}_{\text{oracle, rev-step}}\lesssim\sum_{t=1}^{T}(1-\alpha_{t})^{2}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\nabla^{2}\log\tilde{q}_{t}(X_{t})\Big{)}\right].$ (16) $$

Also, for $\mathcal{W}_{\text{bias, rev-step}}$, since direct calculation yields

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}[\tilde{\zeta}_{\text{mis}}(x_{t},X_{t-1})]$ $\displaystyle=\mathbb{E}_{X_{t-1}\sim\tilde{Q}_{t-1|t}}[\tilde{\zeta}_{\text{mis}}(x_{t},X_{t-1})]=0,$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\tilde{\zeta}_{\text{mis}}(x_{t},X_{t-1})]$ $\displaystyle=-(1-\alpha_{t})\left\lVert\Delta_{t}(x_{t})\right\rVert^{2},$ $$

we have

$$ $\displaystyle\mathcal{W}_{\text{bias, rev-step}}$ $\displaystyle=\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)[\tilde{\zeta}_{\text{mis}}(X_{t},X_{t-1})]$ $\displaystyle=(1-\alpha_{t})\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}.$ (17) $$

### E.4 Step 4: Bounding 𝒲 vanish, rev-step subscript 𝒲 vanish, rev-step \mathcal{W}_{\text{vanish, rev-step}}

Next, for $\mathcal{W}_{\text{vanish, rev-step}}$, we first note that $\tilde{P}_{t-1|t}$ is conditional Gaussian. Thus, under [Assumption 3](#Thmassumption3), we are able to exchange the limit (from Taylor series) and the expectation due to Gaussian-like moments (cf. [, Lemma 11]), which gives us

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}[\tilde{\zeta}_{\text{van}}(x_{t},X_{t-1})]$ $\displaystyle\quad=\frac{1-\alpha_{t}}{2\alpha_{t}}\mathrm{Tr}(\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(x_{t})))+\sum_{p=4}^{\infty}\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}\left[T_{p}(\log\tilde{q}_{t-1},X_{t-1},m_{t}(x_{t}))\right].$ $$

Here the expected value at $p=3$ is zero because all odd-order centralized moments of Gaussian vanish.

Now it remains to characterize the expectation of $\tilde{\zeta}_{\text{van}}(x_{t},x_{t-1})$ under $P_{t-1|t}$. To this end, we introduce the following notation.

###### Definition 2 (Big-O in $\mathcal{L}^{p}$ space) .

For a random variable $Z_{T}$, we say that $Z_{T}(x)=O_{\mathcal{L}^{p}(Q)}(1)$ if $\left(\mathbb{E}_{X\sim Q}\left|Z_{T}(X)\right|^{p}\right)^{1/p}=O(1)$ for all $p\geq 1$ as $T\to\infty$. Define $\tilde{O}_{\mathcal{L}^{p}(Q)}$ likewise.

One property is that if $Z_{T}(x)=O_{\mathcal{L}^{p}(Q)}(1)$ then $\mathbb{E}_{X\sim Q}\left|Z_{T}(X)\right|=O(1)$.
Another property is that if $Z_{1}=O_{\mathcal{L}^{p}(Q)}(a_{T})$ and $Z_{2}=O_{\mathcal{L}^{p}(Q)}(b_{T})$ for all $p\geq 1$, applying Cauchy-Schwartz inequality we get, for all $p\geq 1$,

$$ $\left(\mathbb{E}_{X\sim Q}\left|Z_{1}Z_{2}\right|^{p}\right)^{1/p}\leq\left(\mathbb{E}_{X\sim Q}Z_{1}^{2p}\mathbb{E}_{X\sim Q}Z_{2}^{2p}\right)^{1/(2p)}=O(a_{T}b_{T}),$ $$

which implies that $O_{\mathcal{L}^{p}(Q)}(a_{T})O_{\mathcal{L}^{p}(Q)}(b_{T})=O_{\mathcal{L}^{p}(Q)}(a_{T}b_{T})$. Now, with this notation, the first lines of [Assumption 3](#Thmassumption3) can be equivalently written as

$$ $\displaystyle(1-\alpha_{t})^{m}\left|\partial_{\bm{a}}^{k}\log q_{t}(X_{t})\right|$ $\displaystyle=O_{\mathcal{L}^{p}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{m}\right),\leavevmode\nobreak\ \forall p\geq 1,$ $\displaystyle(1-\alpha_{t})^{m}\left|\partial_{\bm{a}}^{k}\log q_{t-1}(m_{t}(X_{t}))\right|$ $\displaystyle=O_{\mathcal{L}^{p}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{m}\right),\leavevmode\nobreak\ \forall p\geq 1.$ $$

Also, [Assumption 4](#Thmassumption4) can be equivalently written as

$$ $(1-\alpha_{t})^{m}\left\lVert\Delta_{t,y}(X_{t})\right\rVert=O_{\mathcal{L}^{p}(\tilde{Q}_{t})}(\bar{\alpha}_{t}(1-\alpha_{t})^{m}),\leavevmode\nobreak\ \forall p\geq 1.$ $$

With these notations, the following lemma characterizes the expectation of $\tilde{\zeta}_{\text{van}}(x_{t},x_{t-1})$ under $P_{t-1|t}$, which involves non-centralized Gaussian moments.

###### Lemma 3 .

As long as $\tilde{q}_{t-1}$ is defined, with the definition of $\tilde{\zeta}_{\text{van}}$ in ([E.2](#A5.Ex63)), under [Assumptions 3](#Thmassumption3) and [4](#Thmassumption4), we have $\forall\ell\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\tilde{\zeta}_{\text{van}}(x_{t},X_{t-1})]$ $\displaystyle=-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(\nabla\log\tilde{q}_{t-1}(m_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t}))^{\intercal}\Delta_{t}(x_{t})$ $\displaystyle\quad+\frac{1-\alpha_{t}}{2\alpha_{t}}\mathrm{Tr}(\nabla^{2}\log\tilde{q}_{t-1}(m_{t}))+\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\Delta_{t}(x_{t})^{\intercal}\nabla^{2}\log\tilde{q}_{t-1}(m_{t})\Delta_{t}(x_{t})$ $\displaystyle\quad-\frac{1}{3!}\left(\frac{(1-\alpha_{t})^{2}}{{\alpha_{t}}^{3/2}}\right)\left(3\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t})\Delta_{t}^{i}+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t})\Delta_{t}^{j}\right)$ $\displaystyle\quad+\sum_{p=4}^{\infty}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{p}(\log\tilde{q}_{t-1},X_{t-1},m_{t}(x_{t}))\right]+O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{3}\right).$ $$

###### Proof.

See [Section G.3](#A7.SS3).
∎

The following lemma provides the rate of decay of the difference in expectation of all Taylor polynomials with order $p\geq 4$.

###### Lemma 4 .

As long as $\tilde{q}_{t-1}$ is defined, under [Assumptions 3](#Thmassumption3) and [4](#Thmassumption4), we have, $\forall p\geq 4,\ell\geq 1$,

$$ $\left(\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)[T_{p}(\log\tilde{q}_{t-1},X_{t-1},m_{t}(X_{t}))]=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{3}\right).$ $$

###### Proof.

See [Section G.4](#A7.SS4).
∎

Thus, with the help of [Lemmas 3](#Thmlemma3) and [4](#Thmlemma4), we can identify the dominating terms in $\mathcal{W}_{\text{vanish, rev-step}}$ when $1-\alpha_{t}$ is small. The dominating term is

$$ $\displaystyle\mathcal{W}_{\text{vanish, rev-step}}$ $\displaystyle=\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left(\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)[\tilde{\zeta}_{\text{van}}(X_{t},X_{t-1})]$ $\displaystyle=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[(\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t}))^{\intercal}\Delta_{t}(X_{t})\right]$ $\displaystyle\quad-\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\Delta_{t}(X_{t})^{\intercal}\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})\right]$ $\displaystyle\quad+\frac{1}{3!}\left(\frac{(1-\alpha_{t})^{2}}{{\alpha_{t}}^{3/2}}\right)\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\bigg{[}3\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{i}$ $\displaystyle\qquad\quad+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{j}\bigg{]}$ $\displaystyle\quad+O((1-\alpha_{t})^{3}).$ (18) $$

Therefore, with the decomposition in ([E.2](#A5.Ex65)) in mind, an upper bound on the reverse-step error is achieved by summing up ([16](#A5.E16)), ([E.3](#A5.Ex70)) and ([E.4](#A5.Ex83)). The proof of [Theorem 1](#Thmtheorem1) is now complete.

###### Definition 2 (Big-O in ℒ p superscript ℒ 𝑝 \mathcal{L}^{p} space) .

###### Lemma 3 .

###### Proof.

###### Lemma 4 .

###### Proof.

### E.5 Corollary 1 and its proof

Below we state and prove a corollary of [Theorem 1](#Thmtheorem1) when $\tilde{q}_{0}$ does not exist. By [, Lemma 6], $\tilde{q}_{1}$ always exists, which provides us with the following convergence result with early-stopping.

###### Corollary 1 .

Suppose that [Assumptions 1](#Thmassumption1), [3](#Thmassumption3), [4](#Thmassumption4) and [2](#Thmassumption2) are satisfied. Then, with the $\alpha_{t}$ according to [Definition 1](#Thmdefinition1), the distribution $\widehat{P}_{1}$ from the discrete-time DDPM under score bias satisfies

$$ $\displaystyle\mathrm{KL}(\tilde{Q}_{1}\|\widehat{P}_{1})$ $\displaystyle\lesssim\sum_{t=2}^{T}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}$ $\displaystyle+\sum_{t=2}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\bigg{[}(\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t}))^{\intercal}\Delta_{t}(X_{t})\bigg{]}$ $\displaystyle+\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\left(\nabla^{2}\log\tilde{q}_{t}(X_{t})-\Delta_{t}(X_{t})\Delta_{t}(X_{t})^{\intercal}\right)\Big{)}\right]$ $\displaystyle+\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[3\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{i}+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{j}\right]$ $\displaystyle+\max_{t\geq 2}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}(\log T)\varepsilon+(\log T)\varepsilon^{2},$ $$

where $\mathrm{W}_{2}(\tilde{Q}_{1},\tilde{Q}_{0})^{2}\lesssim(1-\alpha_{1})d$.

###### Proof.

The result directly follows with the same arguments as in the proof of [Theorem 1](#Thmtheorem1). The only difference is the guarantee under the Wasserstein distance, which can be obtained using [, Lemma 12].
∎

###### Corollary 1 .

###### Proof.

## Appendix F Proof of Theorem 2

We first recall some of the properties of the noise schedule in ([8](#S3.E8)). By [Lemma 6](#Thmlemma6), the noise schedule in ([8](#S3.E8)) satisfies that $\frac{1-\alpha_{t}}{(1-\bar{\alpha}_{t-1})^{p}}\lesssim\frac{\log T\log(1/\delta)}{\delta^{p-1}T}$ for all $p\geq 1$ while $\bar{\alpha}_{T}=o(T^{-1})$, and thus such $\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1) when $t\geq 2$.
Further, with the $\alpha_{t}$ in ([8](#S3.E8)), [, Lemmas 15 and 17] show that for
any $Q_{0}$ with finite variance under early-stopping, $\forall p,\ell\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left|\partial_{\bm{a}}^{p}\log\tilde{q}_{t}(X_{t})\right|^{\ell}$ $\displaystyle=O\left(\frac{1}{(1-\bar{\alpha}_{t})^{p\ell/2}}\right),$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left|\partial_{\bm{a}}^{p}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\right|^{\ell}$ $\displaystyle=O\left(\frac{1}{(1-\bar{\alpha}_{t-1})^{p\ell/2}}\right).$ $$

Thus, using [Lemma 6](#Thmlemma6), [Assumption 3](#Thmassumption3) is satisfied (since $\delta$ is constant). In the following, we further verify the last relationship in [Assumption 3](#Thmassumption3) holds.

Therefore, since [Assumptions 1](#Thmassumption1), [2](#Thmassumption2) and [4](#Thmassumption4) have been satisfied, we can invoke [Corollary 1](#Thmcorollary1) and get $\mathrm{KL}(\tilde{Q}_{1}\|\widehat{P}_{1})\lesssim\mathcal{W}_{\text{oracle}}+\mathcal{W}_{\text{bias}}+\mathcal{W}_{\text{vanish}}$.
Now, we investigate the dimensional dependence for each term of the upper bound in [Corollary 1](#Thmcorollary1).

To start, from [, Theorem 3], for any $\tilde{Q}_{0}$ having finite variance, with the $\alpha_{t}$ in ([8](#S3.E8)), we have

$$ $\mathcal{W}_{\text{oracle}}\lesssim\frac{d^{2}\log^{2}(1/\delta)(\log T)^{2}}{T}+(\log T)\varepsilon^{2}.$ $$

Also, since by assumption $\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}\lesssim\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t})^{r}}d^{\gamma}$, with the $\alpha_{t}$ in ([8](#S3.E8)), we have from [Lemma 7](#Thmlemma7) that when $\delta\ll 1$,

$$ $\mathcal{W}_{\text{bias}}\lesssim\frac{d^{\gamma}}{\delta^{r}}\left(1-\frac{2\log(1/\delta)\log T}{T}\right).$ $$

Now we investigate each term in $\mathcal{W}_{\text{vanish}}$. The following lemma is useful to determine the rate of difference of the first-order Taylor polynomials.

###### Lemma 5 .

When $\mathbb{E}_{X_{0}\sim\tilde{Q}_{0}}\left\lVert X_{0}\right\rVert^{6}\lesssim d^{3}$, with the $\alpha_{t}$ in ([8](#S3.E8)), we have

$$ $(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2}}\lesssim\frac{d^{3/2}(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{3}}.$ $$

As a result, [Assumption 3](#Thmassumption3) holds.

###### Proof.

See [Section G.5](#A7.SS5).
∎

In other words, combining [Lemma 5](#Thmlemma5) and [Lemma 6](#Thmlemma6), we have

$$ $(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2}}=\tilde{O}\left(\frac{1}{T^{2}}\right).$ (19) $$

Now, by Cauchy-Schwartz inequality and [Lemma 5](#Thmlemma5),

$$ $\displaystyle\sum_{t=2}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\bigg{[}(\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t}))^{\intercal}\Delta_{t}(X_{t})\bigg{]}$ $\displaystyle\leq\sum_{t=2}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\times$ $\displaystyle\qquad\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2}}$ $\displaystyle\lesssim\frac{d^{\gamma/2}}{(1-\bar{\alpha}_{t})^{r/2}}\times\frac{d^{3/2}(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{3}}$ $\displaystyle\lesssim\sum_{t=2}^{T}\frac{d^{\frac{3+\gamma}{2}}\log(1/\delta)^{2}(\log T)^{2}}{\delta^{1+r/2}T^{2}}$ $\displaystyle\leq\frac{d^{\frac{3+\gamma}{2}}\log(1/\delta)^{2}(\log T)^{2}}{\delta^{1+r/2}T}.$ $$

To proceed for higher orders of Taylor polynomials, we first note that from [, Section G.2], the second and third derivatives of $\log\tilde{q}_{t}$ are

$$ $\displaystyle\nabla^{2}\log\tilde{q}_{t}(x)$ $\displaystyle=-\frac{1}{1-\bar{\alpha}_{t}}I_{d}+\frac{1}{(1-\bar{\alpha}_{t})^{2}}\bigg{(}\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t}(\cdot|x)}\left[(x-\sqrt{\bar{\alpha}_{t}}X_{0})(x-\sqrt{\bar{\alpha}_{t}}X_{0})^{\intercal}\right]$ $\displaystyle\quad-\left(\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t}(\cdot|x)}\left[x-\sqrt{\bar{\alpha}_{t}}X_{0}\right]\right)\left(\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t}(\cdot|x)}\left[x-\sqrt{\bar{\alpha}_{t}}X_{0}\right]\right)^{\intercal}\bigg{)}$ $\displaystyle\partial^{3}_{ijk}\log\tilde{q}_{t}(x)$ $\displaystyle=-\int z^{i}z^{j}z^{k}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)$ $\displaystyle\qquad+\sum_{\begin{subarray}{c}a_{1}=i,j,k\\ a_{2}<a_{3},\leavevmode\nobreak\ a_{2},a_{3}\neq a_{1}\end{subarray}}\int z^{a_{1}}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)\int z^{a_{2}}z^{a_{3}}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)$ $\displaystyle\qquad-2\int z^{i}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)\int z^{j}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)\int z^{k}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x)$ $$

where $z:=\frac{x-\sqrt{\bar{\alpha}_{t}}x_{0}}{1-\bar{\alpha}_{t}}$.
Thus, in order to provide an upper bound on the expected norm of the second-order derivative of $\log\tilde{q}_{t-1}(m_{t})$, we can calculate

$$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t-1}(\cdot|m_{t})}\left[(m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0})(m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0})^{\intercal}\right]\right\rVert_{F}^{2}$ $\displaystyle\leq\mathbb{E}_{X_{t}\sim\tilde{Q}_{t},X_{0}\sim\tilde{Q}_{0|t-1}(\cdot|m_{t})}\left\lVert(m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0})(m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0})^{\intercal}\right\rVert_{F}^{2}$ $\displaystyle=\mathbb{E}_{X_{t}\sim\tilde{Q}_{t},X_{0}\sim\tilde{Q}_{0|t-1}(\cdot|m_{t})}\left\lVert m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0}\right\rVert^{4}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\lesssim}}d^{2}(1-\bar{\alpha}_{t-1})^{2},$ $$

and

$$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\left(\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t-1}}\left[m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0}\right]\right)\left(\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t-1}}\left[m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0}\right]\right)^{\intercal}\right\rVert_{F}^{2}$ $\displaystyle=\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mathbb{E}_{X_{0}\sim\tilde{Q}_{0|t-1}(\cdot|m_{t})}\left[m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0}\right]\right\rVert^{4}$ $\displaystyle\leq\mathbb{E}_{X_{t}\sim\tilde{Q}_{t},X_{0}\sim\tilde{Q}_{0|t-1}(\cdot|m_{t})}\left\lVert m_{t}-\sqrt{\bar{\alpha}_{t-1}}X_{0}\right\rVert^{4}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}d^{2}(1-\bar{\alpha}_{t-1})^{2},$ $$

where both $(i)$ and $(ii)$ follow from [, Lemma 16].
Thus,

$$ $\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\right\rVert_{F}^{2}\lesssim\frac{1}{(1-\bar{\alpha}_{t-1})^{2}}d^{2}.$ $$

For third-order derivatives, we can similarly use [, Lemma 16] and get (cf. [, Section G.2])

$$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\sum_{i=1}^{d}(\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t}(X_{t})))^{2}\right]$ $\displaystyle\lesssim\frac{1}{(1-\bar{\alpha}_{t-1})^{3}}\sum_{i=1}^{d}\mathbb{E}(Z^{i})^{6}\lesssim\frac{d}{(1-\bar{\alpha}_{t-1})^{3}},$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\sum_{i,j=1}^{d}(\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(X_{t})))^{2}\right]$ $\displaystyle\lesssim\frac{1}{(1-\bar{\alpha}_{t-1})^{3}}\sum_{i,j=1}^{d}\left(\mathbb{E}(Z^{i})^{6}\right)^{2/3}\left(\mathbb{E}(Z^{j})^{6}\right)^{1/3}$ $\displaystyle\lesssim\frac{d^{2}}{(1-\bar{\alpha}_{t-1})^{3}}.$ $$

Here we denote $Z\sim\mathcal{N}(0,I_{d})$, and note that $\mathbb{E}(Z^{i})^{6},\mathbb{E}(Z^{j})^{6}\lesssim 1$.

Therefore, by Cauchy-Schwartz inequality,

$$ $\displaystyle\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\Delta_{t}(X_{t})^{\intercal}\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})\right]$ $\displaystyle\leq\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{4}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\right\rVert^{2}}$ $\displaystyle\lesssim\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\frac{d^{\gamma}}{(1-\bar{\alpha}_{t})^{r}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla^{2}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\right\rVert_{F}^{2}}$ $\displaystyle\lesssim\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\frac{d^{\gamma}}{(1-\bar{\alpha}_{t})^{r}}\sqrt{\frac{d^{2}}{(1-\bar{\alpha}_{t-1})^{2}}}$ $\displaystyle\lesssim\frac{d^{1+\gamma}\log(1/\delta)^{2}(\log T)^{2}}{\delta^{r-1}T},$ $$

and

$$ $\displaystyle\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[3\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{i}\right]$ $\displaystyle\leq\sum_{t=2}^{T}\frac{3(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\sum_{i=1}^{d}(\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t}(X_{t})))^{2}}$ $\displaystyle\lesssim\sum_{t=2}^{T}(1-\alpha_{t})^{2}\frac{d^{\gamma/2}}{(1-\bar{\alpha}_{t})^{r/2}}\sqrt{\frac{d}{(1-\bar{\alpha}_{t-1})^{3}}}$ $\displaystyle\lesssim\frac{d^{\frac{1+\gamma}{2}}\log(1/\delta)^{2}(\log T)^{2}}{\delta^{\frac{r-1}{2}}T},$ $$

and, with $M$ being a matrix such that $M^{ij}(x):=\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(x))$,

$$ $\displaystyle\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(X_{t}))\Delta_{t}(X_{t})^{j}\right]$ $\displaystyle\leq\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert M(X_{t})\Delta_{t}(X_{t})\right\rVert_{1}$ $\displaystyle\leq\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\sqrt{d}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert M(X_{t})\right\rVert^{2}}$ $\displaystyle\leq\sum_{t=2}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\sqrt{d}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\sum_{i,j=1}^{d}(\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t}(X_{t})))^{2}\right]}$ $\displaystyle\lesssim\sum_{t=2}^{T}(1-\alpha_{t})^{2}\frac{d^{(1+\gamma)/2}}{(1-\bar{\alpha}_{t})^{r/2}}\sqrt{\frac{d^{2}}{(1-\bar{\alpha}_{t-1})^{3}}}$ $\displaystyle\lesssim\frac{d^{\frac{3+\gamma}{2}}\log(1/\delta)^{2}(\log T)^{2}}{\delta^{\frac{r-1}{2}}T},$ $$

and

$$ $\max_{t\geq 2}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}(\log T)\varepsilon\lesssim\frac{d^{\gamma/2}}{\delta^{r/2}}(\log T)\varepsilon.$ $$

Therefore, combining all the above, we get

$$ $\displaystyle\mathrm{KL}(\tilde{Q}_{1}\|\widehat{P}_{1})$ $\displaystyle\lesssim d^{\gamma}\delta^{-r}\left(1-\frac{2\log(1/\delta)\log T}{T}\right)$ $\displaystyle\qquad+\max\{d^{(3+\gamma)/2}\delta^{-\frac{r+2}{2}},d^{1+\gamma}\delta^{-(r-1)}\}\frac{(\log T)^{2}}{T}$ $\displaystyle\qquad+d^{\gamma/2}\delta^{-r/2}(\log T)\varepsilon.$ $$

###### Lemma 5 .

###### Proof.

## Appendix G Auxiliary Lemmas and Proofs in Section 3

### G.1 Proof of Lemma 1

We remind readers that throughout this proof $x_{t}$ is fixed. For brevity write $m_{t}=m_{t}(x_{t})$, $\mu_{t}=\mu_{t}(x_{t})$, and $\Delta_{t}(x_{t})=\Delta_{t}$. Recall that $m_{t}=x_{t}/\sqrt{\alpha_{t}}+(1-\alpha_{t})/\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t})$.
By Bayes’ rule, we have

$$ $\displaystyle\tilde{q}_{t-1|t}(x_{t-1}|x_{t})$ $\displaystyle=\frac{\tilde{q}_{t|t-1}(x_{t}|x_{t-1})\tilde{q}_{t-1}(x_{t-1})}{\tilde{q}_{t}(x_{t})}$ $\displaystyle\propto\tilde{q}_{t-1}(x_{t-1})\tilde{q}_{t|t-1}(x_{t}|x_{t-1})$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\propto}}\tilde{q}_{t-1}(x_{t-1})\exp\left(-\frac{\left\lVert x_{t}-\sqrt{\alpha_{t}}x_{t-1}\right\rVert^{2}}{2(1-\alpha_{t})}\right)$ $\displaystyle\propto\tilde{q}_{t-1}(x_{t-1})p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{\left\lVert x_{t-1}-\mu_{t}\right\rVert^{2}-\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}}\right)$ $\displaystyle=\tilde{q}_{t-1}(x_{t-1})p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}+x_{t}/\sqrt{\alpha_{t}}-\mu_{t}\right\rVert^{2}-\left\lVert x_{t-1}-x_{t}/\sqrt{\alpha_{t}}\right\rVert^{2}}{2(1-\alpha_{t})/\alpha_{t}}\right)$ $\displaystyle\propto\tilde{q}_{t-1}(x_{t-1})p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\frac{(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}(x_{t}/\sqrt{\alpha_{t}}-\mu_{t})}{(1-\alpha_{t})/\alpha_{t}}\right)$ $$

where $(i)$ follows because the forward process is Markov and $\tilde{q}_{t|t-1}=q_{t|t-1}$.
Here, the exponent is equal to

$$ $\displaystyle\frac{(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}(x_{t}/\sqrt{\alpha_{t}}-\mu_{t})}{(1-\alpha_{t})/\alpha_{t}}$ $\displaystyle=\frac{(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}(m_{t}-\mu_{t})}{(1-\alpha_{t})/\alpha_{t}}-\frac{(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}((1-\alpha_{t})/\sqrt{\alpha_{t}})\nabla\log\tilde{q}_{t}(x_{t})}{(1-\alpha_{t})/\alpha_{t}}$ $\displaystyle=\sqrt{\alpha_{t}}\Delta_{t}^{\intercal}(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})-\sqrt{\alpha_{t}}(x_{t-1}-x_{t}/\sqrt{\alpha_{t}})^{\intercal}\nabla\log\tilde{q}_{t}(x_{t}).$ $$

Thus,

$$ $\tilde{q}_{t-1|t}(x_{t-1}|x_{t})\propto p_{t-1|t}(x_{t-1}|x_{t})\exp\left(\tilde{\zeta}_{t,t-1}(x_{t},x_{t-1})\right)$ $$

where

$$ $\tilde{\zeta}_{t,t-1}(x_{t},x_{t-1})=\sqrt{\alpha_{t}}\Delta_{t}^{\intercal}(x_{t-1}-m_{t})+\log\tilde{q}_{t-1}(x_{t-1})-\sqrt{\alpha_{t}}(x_{t-1}-m_{t})^{\intercal}\nabla\log\tilde{q}_{t}(x_{t}).$ $$

Finally, since all partial derivatives of $\tilde{q}_{t-1}$ exists for any $t\geq 2$ (See [, Lemma 6]), the Taylor expansion of $\log\tilde{q}_{t-1}$ around $m_{t}$ gives the desirable result.

### G.2 Proof of Lemma 2

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,

$$ $\displaystyle\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{Q}_{t,t-1}}\left[(X_{t-1}-\mu_{t}(X_{t}))^{\intercal}(\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t}))\right]$ $\displaystyle=\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[(m_{t}(X_{t})-\mu_{t}(X_{t}))^{\intercal}(\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t}))\right]$ $\displaystyle=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left[\Delta_{t}(X_{t})^{\intercal}(\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t}))\right]$ $\displaystyle\leq\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}}.$ $$

Here we recall the definition of $\Delta_{t}$ from ([2](#S2.E2)) where $m_{t}(x)-\mu_{t}(x)=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}(x)$.
Thus,

$$ $\displaystyle\sum_{t=1}^{T}\mathbb{E}_{X_{t},X_{t-1}\sim\tilde{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\lesssim\sum_{t=1}^{T}\left(\sqrt{\alpha_{t}}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}}+\frac{\alpha_{t}}{1-\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}\right)$ $\displaystyle=\sum_{t=1}^{T}(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\times\sqrt{\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}}$ $\displaystyle\qquad+\sum_{t=1}^{T}(1-\alpha_{t})\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{\lesssim}}\frac{\log T}{T}\sum_{t=1}^{T}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\times\sqrt{\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}}$ $\displaystyle\qquad+\frac{\log T}{T}\sum_{t=1}^{T}\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\leq\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}\frac{\log T}{T}\sum_{t=1}^{T}\sqrt{\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}}$ $\displaystyle\qquad+\frac{\log T}{T}\sum_{t=1}^{T}\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}(\log T)\sqrt{\frac{1}{T}\sum_{t=1}^{T}\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}}$ $\displaystyle\qquad+\frac{\log T}{T}\sum_{t=1}^{T}\frac{\alpha_{t}}{(1-\alpha_{t})^{2}}\mathbb{E}_{X_{t}\sim Q_{t}}\left\lVert\mu_{t}(X_{t})-\widehat{\mu}_{t}(X_{t})\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\lesssim}}\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\Delta_{t}(X_{t})\right\rVert^{2}}(\log T)\varepsilon+(\log T)\varepsilon^{2}$ $$

where $(i)$ follows from
[Definition 1](#Thmdefinition1), $(ii)$ follows from the fact that for any non-negative sequence $a_{t}$, $\frac{1}{T}\sum_{t=1}^{T}\sqrt{a_{t}}\leq\sqrt{\frac{1}{T}\sum_{t=1}^{T}a_{t}}$ by Jensen’s inequality, and $(iii)$ follows from [Assumption 2](#Thmassumption2). The proof is complete.

### G.3 Proof of Lemma 3

Recall that $P_{t-1|t}=\mathcal{N}(\mu_{t},\frac{1-\alpha_{t}}{\alpha_{t}}I_{d})$, and thus $\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[X_{t-1}-m_{t}(x_{t})]=-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}(x_{t})$.
Note that we can change the limit and the expectation under [Assumption 3](#Thmassumption3).
Now, we can calculate that

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}[\tilde{\zeta}_{\text{van}}(x_{t},X_{t-1})]$ $\displaystyle=-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(\nabla\log\tilde{q}_{t-1}(m_{t}(x_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t}))^{\intercal}\Delta_{t}(x_{t})$ $\displaystyle+\sum_{p=2}^{\infty}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{p}(\log\tilde{q}_{t-1},X_{t-1},m_{t}(x_{t}))\right].$ $$

Below we write $m_{t}=m_{t}(x_{t})$. Since $T_{2}$ is in quadratic form, the expected value under $P_{t-1|t}$ for this term is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{2}(\log\tilde{q}_{t-1},X_{t-1},m_{t})\right]$ $\displaystyle\quad=\frac{1-\alpha_{t}}{2\alpha_{t}}\mathrm{Tr}(\nabla^{2}\log\tilde{q}_{t-1}(m_{t}))+\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\Delta_{t}(x_{t})^{\intercal}\nabla^{2}\log\tilde{q}_{t-1}(m_{t})\Delta_{t}(x_{t}).$ $$

Recall the formula for Gaussian non-centralized third moment. If $Z\sim\mathcal{N}(\mu,\sigma^{2})$, then $\mathbb{E}[Z^{2}]=\mu^{2}+\sigma^{2}$ and $\mathbb{E}[Z^{3}]=\mu^{3}+3\mu\sigma^{2}$. Thus, the expected value under $P_{t-1|t}$ for $T_{3}$ is

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[T_{3}(\log\tilde{q}_{t-1},X_{t-1},m_{t})\right]$ $\displaystyle=\frac{1}{3!}\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t})\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}(X_{t-1}^{i}-m_{t}^{i})^{3}$ $\displaystyle\quad+\frac{1}{3!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t})\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}(X_{t-1}^{i}-m_{t}^{i})^{2}(X_{t-1}^{j}-m_{t}^{j})$ $\displaystyle\quad+\frac{1}{3!}\sum_{\begin{subarray}{c}i,j,k=1\\ i,j,k\text{ all differ}\end{subarray}}^{d}\partial^{3}_{ijk}\log\tilde{q}_{t-1}(m_{t})\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}(X_{t-1}^{i}-m_{t}^{i})(X_{t-1}^{j}-m_{t}^{j})(X_{t-1}^{k}-m_{t}^{k})$ $\displaystyle=\frac{1}{3!}\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t})\left(\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{i}\right)^{3}+3\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{i}\right)\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)\right)$ $\displaystyle\quad+\frac{1}{3!}\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t})\left(\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{i}\right)^{2}+\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)\right)\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{j}\right)$ $\displaystyle\quad+\frac{1}{3!}\sum_{\begin{subarray}{c}i,j,k=1\\ i,j,k\text{ all differ}\end{subarray}}^{d}\partial^{3}_{ijk}\log\tilde{q}_{t-1}(m_{t})\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{i}\right)\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{j}\right)\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{k}\right)$ $\displaystyle=\frac{1}{3!}\left(-\frac{(1-\alpha_{t})^{2}}{{\alpha_{t}}^{3/2}}\right)\left(3\sum_{i=1}^{d}\partial^{3}_{iii}\log\tilde{q}_{t-1}(m_{t})\Delta_{t}^{i}+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log\tilde{q}_{t-1}(m_{t})\Delta_{t}^{j}\right)$ $\displaystyle\quad+O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{3}\right),\leavevmode\nobreak\ \forall\ell\geq 1.$ $$

Here the last line follows because $(1-\alpha_{t})^{3}\left|\partial^{3}_{ijk}\log\tilde{q}_{t-1}(m_{t})\right|=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{3}\right)$ under [Assumption 3](#Thmassumption3) and $(1-\alpha_{t})^{3}\left\lVert\Delta_{t}\right\rVert=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}((1-\alpha_{t})^{3})$ under [Assumption 4](#Thmassumption4), both for all $\ell\geq 1$. The proof is now complete.

### G.4 Proof of Lemma 4

Fix $x_{t}\in\mathbb{R}^{d}$. For brevity write $m_{t}=m_{t}(x_{t})$, $\mu_{t}=\mu_{t}(x_{t})$, and $\Delta_{t}=\Delta_{t}(x_{t})$. Recall that

$$ $T_{p}(\log\tilde{q}_{t-1},x_{t-1},m_{t})=\frac{1}{p!}\sum_{\gamma\in\mathbb{N}^{d}:\sum_{i}\gamma^{i}=p}\partial^{p}_{\bm{a}}\log\tilde{q}_{t-1}(m_{t})\prod_{i=1}^{d}(x_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}$ $$

where $\bm{a}\in[d]^{p}$ are the indices of differentiation in which the multiplicity of $i$ is $\gamma^{i}$.
First, for the expectation under $\tilde{P}_{t-1|t}$ (i.e., Gaussian centralized moments),

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}\left[\prod_{i=1}^{d}(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]$ $\displaystyle=\prod_{i=1}^{d}\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}\left[(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]$ $\displaystyle=\prod_{i=1}^{d}\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{\gamma^{i}/2}(\gamma^{i}-1)!!\mathds{1}\{\gamma^{i}\text{ is even}\}$ $\displaystyle=\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{p/2}\prod_{i=1}^{d}(\gamma^{i}-1)!!\mathds{1}\{\gamma^{i}\text{ is even}\},$ $$

where we use the convention that $(-1)!!=1$.
Next, for the expectation under $P_{t-1|t}$ (i.e., Gaussian non-centralized moments),

$$ $\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\prod_{i=1}^{d}(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]$ $\displaystyle=\prod_{\begin{subarray}{c}i=1\end{subarray}}^{d}\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[(X_{t-1}^{i}-\mu_{t}^{i}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{i})^{\gamma^{i}}\right]$ $\displaystyle=\prod_{\begin{subarray}{c}i=1\end{subarray}}^{d}\sum_{\begin{subarray}{c}\ell=0\\ \ell\text{ even}\end{subarray}}^{\gamma^{i}}\binom{\gamma^{i}}{\ell}\left(-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Delta_{t}^{i}\right)^{\gamma^{i}-\ell}\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{\ell/2}(\ell-1)!!$ $$

To investigate their difference, we divide into the following few cases. Note that under [Assumption 4](#Thmassumption4), $(1-\alpha_{t})^{m}\left\lVert\Delta_{t}(x_{t})\right\rVert=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}((1-\alpha_{t})^{m})$ for any $m\geq 1/2$ and $\ell\geq 1$.

- 1.
Case 1: $p$ is even and all elements of $\gamma^{i}$ are even. Then,
$\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\prod_{i=1}^{d}(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]$
$\displaystyle=\prod_{\begin{subarray}{c}i=1\end{subarray}}^{d}\left(\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{\gamma^{i}/2}(\gamma^{i}-1)!!+O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{\gamma^{i}/2+1}\right)\right)$
$\displaystyle=\left(\frac{1-\alpha_{t}}{\alpha_{t}}\right)^{p/2}\prod_{i=1}^{d}(\gamma^{i}-1)!!+O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{p/2+1}\right)$
- 2.
Case 2: $p$ is even and $\exists i^{*}$ such that $\gamma^{i^{*}}$ is odd. Since $\sum_{i}\gamma^{i}=p$, there exists $j^{*}$ such that $\gamma^{j^{*}}$ is also odd. Then,
$\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[(X_{t-1}^{i^{*}}-m_{t}^{i^{*}})^{\gamma^{i^{*}}}\right]$
$\displaystyle=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left(\left(1-\alpha_{t}\right)^{(\gamma^{i^{*}}+1)/2}\right),$
$\displaystyle\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[(X_{t-1}^{j^{*}}-m_{t}^{j^{*}})^{\gamma^{j^{*}}}\right]$
$\displaystyle=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left(\left(1-\alpha_{t}\right)^{(\gamma^{j^{*}}+1)/2}\right),$
which implies that
$\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\prod_{i=1}^{d}(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left(\left(1-\alpha_{t}\right)^{p/2+1}\right)$
- 3.
Case 3: $p$ is odd and $\exists i^{*}$ such that $\gamma^{i^{*}}$ is odd. Then,
$\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[(X_{t-1}^{i^{*}}-m_{t}^{i^{*}})^{\gamma^{i^{*}}}\right]=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left(\left(1-\alpha_{t}\right)^{(\gamma^{i^{*}}+1)/2}\right),$
which implies that
$\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\left[\prod_{i=1}^{d}(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]=O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left(\left(1-\alpha_{t}\right)^{(p+1)/2}\right)$

Combining these cases, we get

$$ $\left(\mathbb{E}_{X_{t-1}\sim\tilde{P}_{t-1|t}}-\mathbb{E}_{X_{t-1}\sim P_{t-1|t}}\right)\left[\prod_{i=1}^{d}(X_{t-1}^{i}-m_{t}^{i})^{\gamma^{i}}\right]\\ =\begin{cases}O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{\frac{p}{2}+1}\right),&\forall p\geq 4\leavevmode\nobreak\ \text{even}\\ O_{\mathcal{L}^{\ell}(\tilde{Q}_{t})}\left((1-\alpha_{t})^{\frac{p+1}{2}}\right),&\forall p\geq 4\leavevmode\nobreak\ \text{odd}\end{cases}$ $$

The proof is complete by noting that the rate does not change when we take the expectation over $\tilde{Q}_{t}$ under [Assumptions 3](#Thmassumption3) and [4](#Thmassumption4).

### G.5 Proof of Lemma 5

Note that $\tilde{q}_{t|0}(x|x_{0})=q_{t|0}(x|x_{0})$ is the p.d.f. of $\mathcal{N}(\sqrt{\bar{\alpha}_{t}}x_{0},(1-\bar{\alpha}_{t})I_{d})$.
Thus, the gradient of $\log\tilde{q}_{t}(x)$ equals

$$ $\nabla\log\tilde{q}_{t}(x)=\frac{\int_{x_{0}\in\mathbb{R}^{d}}\nabla\tilde{q}_{t|0}(x|x_{0})\mathrm{d}\tilde{Q}_{0}(x_{0})}{\tilde{q}_{t}(x)}=-\frac{1}{1-\bar{\alpha}_{t}}\int_{x_{0}\in\mathbb{R}^{d}}(x-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x).$ (20) $$

Thus,

$$ $\displaystyle\nabla\log\tilde{q}_{t-1}(m_{t})-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(x_{t})$ $\displaystyle=-\frac{1}{1-\bar{\alpha}_{t-1}}\int_{x_{0}\in\mathbb{R}^{d}}(m_{t}-\sqrt{\bar{\alpha}_{t-1}}x_{0})\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})+\frac{\sqrt{\alpha_{t}}}{1-\bar{\alpha}_{t}}\int_{x_{0}\in\mathbb{R}^{d}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})$ $\displaystyle=-\frac{1}{1-\bar{\alpha}_{t}}\Bigg{(}\left(\frac{1-\bar{\alpha}_{t}}{1-\bar{\alpha}_{t-1}}-1\right)\int_{x_{0}\in\mathbb{R}^{d}}(m_{t}-\sqrt{\bar{\alpha}_{t-1}}x_{0})\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})$ $\displaystyle\qquad+\int_{x_{0}\in\mathbb{R}^{d}}(m_{t}-\sqrt{\bar{\alpha}_{t-1}}x_{0})\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})-\sqrt{\alpha_{t}}\int_{x_{0}\in\mathbb{R}^{d}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})\Bigg{)}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\frac{1}{1-\bar{\alpha}_{t}}\Big{(}\left(1-\bar{\alpha}_{t}-(1-\bar{\alpha}_{t-1})\right)\nabla\log\tilde{q}_{t-1}(m_{t})\Big{)}$ $\displaystyle\qquad-\frac{1}{1-\bar{\alpha}_{t}}\Bigg{(}\int_{x_{0}\in\mathbb{R}^{d}}(m_{t}-\sqrt{\bar{\alpha}_{t-1}}x_{0})\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})-\sqrt{\alpha_{t}}\int_{x_{0}\in\mathbb{R}^{d}}(x_{t}-\sqrt{\bar{\alpha}_{t}}x_{0})\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})\Bigg{)}$ $\displaystyle=\underbrace{\frac{\bar{\alpha}_{t-1}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}\nabla\log\tilde{q}_{t-1}(m_{t})}_{\text{term 1}}-\underbrace{\frac{1}{1-\bar{\alpha}_{t}}(m_{t}-\sqrt{\alpha_{t}}x_{t})}_{\text{term 2}}+\underbrace{\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})}_{\text{term 3}}$ $\displaystyle\qquad+\underbrace{\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\left(\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})-\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})\right)}_{\text{term 4}}$ (21) $$

where $(i)$ follows from Tweedie’s formula. Among the four terms in ([G.5](#A7.Ex199)), the first term satisfies that

$$ $\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\frac{\bar{\alpha}_{t-1}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}\nabla\log\tilde{q}_{t-1}(m_{t})\right\rVert^{2}\lesssim\frac{d(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t})^{2}(1-\bar{\alpha}_{t-1})}$ $$

by [, Lemma 17]. In the second term in ([G.5](#A7.Ex199)), by Tweedie’s formula,

$$ $\displaystyle m_{t}-\sqrt{\alpha_{t}}x_{t}$ $\displaystyle=\frac{x_{t}}{\sqrt{\alpha_{t}}}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log\tilde{q}_{t}(x_{t})-\sqrt{\alpha_{t}}x_{t}$ $\displaystyle=\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}(x_{t}+\nabla\log\tilde{q}_{t}(x_{t})).$ $$

Thus, by [, Lemma 15] and [Assumption 1](#Thmassumption1), the second term satisfies that

$$ $\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\frac{1}{1-\bar{\alpha}_{t}}(m_{t}-\sqrt{\alpha_{t}}x_{t})\right\rVert^{2}\lesssim\frac{d(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t})^{3}}.$ $$

The third term in ([G.5](#A7.Ex199)) satisfies that

$$ $\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\frac{\sqrt{\bar{\alpha}_{t-1}}(1-\alpha_{t})}{1-\bar{\alpha}_{t}}\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})\right\rVert^{2}\lesssim\frac{d(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t})^{2}}$ $$

by Jensen’s inequality and [Assumption 1](#Thmassumption1).

To deal with the last term in ([G.5](#A7.Ex199)), note that

$$ $\displaystyle\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})$ $\displaystyle=\frac{\tilde{q}_{t-1|0}(m_{t}|x_{0})}{\tilde{q}_{t-1}(m_{t})}\mathrm{d}\tilde{Q}_{0}(x_{0})=\frac{\tilde{q}_{t-1|0}(m_{t}|x_{0})}{\int_{y\in\mathbb{R}^{d}}\tilde{q}_{t-1|0}(m_{t}|y)\mathrm{d}\tilde{Q}_{0}(y)}\mathrm{d}\tilde{Q}_{0}(x_{0}),$ $\displaystyle\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})$ $\displaystyle=\frac{\tilde{q}_{t|0}(x_{t}|x_{0})}{\tilde{q}_{t}(x_{t})}\mathrm{d}\tilde{Q}_{0}(x_{0})=\frac{\tilde{q}_{t|0}(x_{t}|x_{0})}{\int_{y\in\mathbb{R}^{d}}\tilde{q}_{t|0}(x_{t}|y)\mathrm{d}\tilde{Q}_{0}(y)}\mathrm{d}\tilde{Q}_{0}(x_{0}).$ $$

Thus, the last term in ([G.5](#A7.Ex199)) is equal to

$$ $\displaystyle\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\left(\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})-\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})\right)$ $\displaystyle=\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\cdot\frac{1}{\tilde{q}_{t-1}(m_{t})\tilde{q}_{t}(x_{t})}\left(\int_{x,y\in\mathbb{R}^{d}}x(\tilde{q}_{t-1|0}(m_{t}|x)\tilde{q}_{t|0}(x_{t}|y)-\tilde{q}_{t|0}(x_{t}|x)\tilde{q}_{t-1|0}(m_{t}|y))\mathrm{d}\tilde{Q}_{0}(x)\mathrm{d}\tilde{Q}_{0}(y)\right)$ $$

where

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

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

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

Now, with the $\alpha_{t}$ defined in ([8](#S3.E8)), following from [Lemma 6](#Thmlemma6),

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

Thus, for fixed $x,y,x_{t}$, $\Delta\to 0$ as $T\to\infty$, and thus when $T$ becomes large,

$$ $e^{\Delta}-1=\Delta+O(\Delta^{2})\lesssim\left|\Delta\right|,\quad\forall x_{t}\in\mathbb{R}^{d}.$ $$

Also, since $\tilde{q}_{t|0}(x_{t}|x)$ and $\tilde{q}_{t-1|0}(m_{t}|y)$ decay exponentially in terms of $x$ and $y$ (for any fixed $x_{t}$), we have

$$ $\displaystyle\int\tilde{q}_{t|0}(x_{t}|x)\mathrm{poly}(x)\mathrm{d}\tilde{Q}_{0}(x)$ $\displaystyle<\infty,$ $\displaystyle\int\tilde{q}_{t-1|0}(m_{t}|y)\mathrm{poly}(y)\mathrm{d}\tilde{Q}_{0}(y)$ $\displaystyle<\infty.$ $$

Thus, the limit and the integral can be exchanged due to Dominated Convergence Theorem. Thus, the fourth term in ([G.5](#A7.Ex199)) gives us

$$ $\displaystyle\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\left(\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t-1}(x_{0}|m_{t})-\int_{x_{0}\in\mathbb{R}^{d}}x_{0}\mathrm{d}\tilde{Q}_{0|t}(x_{0}|x_{t})\right)$ $\displaystyle\lesssim\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\cdot\frac{1}{\tilde{q}_{t-1}(m_{t})\tilde{q}_{t}(x_{t})}\left(\int_{x,y\in\mathbb{R}^{d}}x\tilde{q}_{t|0}(x_{t}|x)\tilde{q}_{t-1|0}(m_{t}|y)\left|\Delta\right|\mathrm{d}\tilde{Q}_{0}(x)\mathrm{d}\tilde{Q}_{0}(y)\right)$ $\displaystyle=\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\left(\int_{x,y\in\mathbb{R}^{d}}(x\cdot\left|\Delta\right|)\mathrm{d}\tilde{Q}_{0|t}(x|x_{t})\mathrm{d}\tilde{Q}_{0|t-1}(y|m_{t})\right)$ $$

and, from definition of $\Delta$ and using Cauchy-Schwartz and Jensen’s inequality, we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t}}\int_{x,y\in\mathbb{R}^{d}}(x\cdot\left|\Delta\right|)\mathrm{d}\tilde{Q}_{0|t}(x|X_{t})\mathrm{d}\tilde{Q}_{0|t-1}(y|m_{t}(X_{t}))\right\rVert^{2}$ $\displaystyle\lesssim\frac{(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{2}(1-\bar{\alpha}_{t})^{4}}\cdot$ $\displaystyle\qquad\mathbb{E}_{\begin{subarray}{c}X_{t}\sim\tilde{Q}_{t}\\ X\sim\tilde{Q}_{0|t}(\cdot|X_{t})\\ Y\sim\tilde{Q}_{0|t-1}(\cdot|m_{t}(X_{t}))\end{subarray}}\bigg{[}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{2}\bigg{(}(\left\lVert X_{t}\right\rVert^{2}+(1-\bar{\alpha}_{t})^{2}\left\lVert\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2})$ $\displaystyle\qquad(\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{2}+\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{2})+(\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{4}+\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{4})\bigg{)}\bigg{]}$ $\displaystyle=\frac{(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{2}(1-\bar{\alpha}_{t})^{4}}\cdot$ $\displaystyle\qquad\mathbb{E}_{\begin{subarray}{c}X_{t}\sim\tilde{Q}_{t}\\ X\sim\tilde{Q}_{0|t}(\cdot|X_{t})\\ Y\sim\tilde{Q}_{0|t-1}(\cdot|m_{t}(X_{t}))\end{subarray}}\bigg{[}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{4}(\left\lVert X_{t}\right\rVert^{2}+(1-\bar{\alpha}_{t})^{2}\left\lVert\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2})$ $\displaystyle\qquad+\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{2}\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{2}(\left\lVert X_{t}\right\rVert^{2}+(1-\bar{\alpha}_{t})^{2}\left\lVert\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2})$ $\displaystyle\qquad+\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{6}+\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{2}\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{4}\bigg{]}$ $\displaystyle\leq\frac{(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{2}(1-\bar{\alpha}_{t})^{4}}\cdot$ $\displaystyle\qquad\left(\mathbb{E}_{X\sim\tilde{Q}_{0}}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{6}\right)^{2/3}\left(\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}(\left\lVert X_{t}\right\rVert^{6}+(1-\bar{\alpha}_{t})^{6}\left\lVert\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{6})\right)^{1/3}$ $\displaystyle\qquad+\left(\mathbb{E}_{X\sim\tilde{Q}_{0}}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{6}\right)^{1/3}\left(\mathbb{E}_{\begin{subarray}{c}X_{t}\sim\tilde{Q}_{t}\\ Y\sim\tilde{Q}_{0|t-1}(\cdot|m_{t}(X_{t}))\end{subarray}}\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{6}\right)^{1/3}$ $\displaystyle\qquad\quad\left(\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}(\left\lVert X_{t}\right\rVert^{6}+(1-\bar{\alpha}_{t})^{6}\left\lVert\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{6})\right)^{1/3}$ $\displaystyle\qquad+\mathbb{E}_{X\sim\tilde{Q}_{0}}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{6}+\left(\mathbb{E}_{X\sim\tilde{Q}_{0}}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{6}\right)^{1/3}\left(\mathbb{E}_{\begin{subarray}{c}X_{t}\sim\tilde{Q}_{t}\\ Y\sim\tilde{Q}_{0|t-1}(\cdot|m_{t}(X_{t}))\end{subarray}}\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{6}\right)^{2/3}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}\frac{d^{3}(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{2}(1-\bar{\alpha}_{t})^{4}},$ $$

where $(ii)$ follows because, following [, Lemmas 15–17] and by the lemma assumption that $\mathbb{E}_{X_{0}\sim\tilde{Q}_{0}}\left\lVert X_{0}\right\rVert^{6}\lesssim d^{3}$, we have

$$ $\displaystyle\mathbb{E}_{X\sim\tilde{Q}_{0}}\left\lVert\sqrt{\bar{\alpha}_{t}}X\right\rVert^{6}\lesssim d^{3},$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert X_{t}\right\rVert^{6}\leq\mathbb{E}_{X_{0}\sim\tilde{Q}_{0}}\left\lVert\sqrt{\bar{\alpha}_{t}}X_{0}\right\rVert^{6}+(1-\bar{\alpha}_{t})^{3}\mathbb{E}_{\bar{W}\sim\mathcal{N}(0,I_{d})}\left\lVert\bar{W}\right\rVert^{6}\lesssim d^{3},$ $\displaystyle\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{6}\lesssim\frac{d^{3}}{(1-\bar{\alpha}_{t})^{3}},$ $\displaystyle\mathbb{E}_{\begin{subarray}{c}X_{t}\sim\tilde{Q}_{t}\\ Y\sim\tilde{Q}_{0|t-1}(\cdot|m_{t}(X_{t}))\end{subarray}}\left\lVert\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{6}$ $\displaystyle\quad\leq\mathbb{E}_{\begin{subarray}{c}X_{t}\sim\tilde{Q}_{t}\\ Y\sim\tilde{Q}_{0|t-1}(\cdot|m_{t}(X_{t}))\end{subarray}}\left\lVert m_{t}-\sqrt{\bar{\alpha}_{t-1}}Y\right\rVert^{6}+\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert m_{t}\right\rVert^{6}\lesssim d^{3}.$ $$

Hence, combining the rates of all parts, we obtain that

$$ $(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim\tilde{Q}_{t}}\left\lVert\nabla\log\tilde{q}_{t-1}(m_{t}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log\tilde{q}_{t}(X_{t})\right\rVert^{2}}\lesssim\frac{d^{3/2}(1-\alpha_{t})^{2}}{(1-\bar{\alpha}_{t-1})^{3}}.$ $$

### G.6 Lemma 6 and its proof

###### Lemma 6 .

The $\alpha_{t}$ defined in ([8](#S3.E8)) (with $c>1$) satisfy

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

###### Proof.

The proof is similar to that of [, Eq (39)]. We first prove the second relationship. 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, with any fixed $r\in(0,1)$ such that $t\geq rT\leavevmode\nobreak\ (\geq\frac{T}{\log T})$, 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,

$$ $\bar{\alpha}_{T}\leq\prod_{t=\left\lfloor rT\right\rfloor}^{T}\alpha_{t}=\left(1-\frac{c\log T}{T}\right)^{\left\lceil(1-r)T\right\rceil}\asymp\exp\left(\left\lceil(1-r)T\right\rceil\left(-\frac{c\log T}{T}\right)\right)=O(T^{-(1-r)c}).$ (22) $$

Given any $c>1$, we can always find some $r$ such that $(1-r)c>1$ (say, $r=(c-1)/2$ if $c\in(1,2)$ and $r=1/4$ if $c\geq 2$). This shows that $\alpha_{t}$ satisfies $\bar{\alpha}_{T}=o\left(T^{-1}\right)$ if $c>1$.

Now, for the first relatinoship, define $\tau$ such that

$$ $\delta\left(1+\frac{c\log T}{T}\right)^{\tau}\leq 1<\delta\left(1+\frac{c\log T}{T}\right)^{\tau+1}.$ (23) $$

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

$$ $1-\bar{\alpha}_{t-1}\geq\frac{1}{3}\delta\left(1+\frac{c\log T}{T}\right)^{t},\quad\forall 2\leq t\leq\tau.$ (24) $$

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 ([24](#A7.E24)) 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 ([24](#A7.E24)) is violated, and thus ([24](#A7.E24)) 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 ([24](#A7.E24)) 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 ([23](#A7.E23)). 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 ([24](#A7.E24)) holds.

Now, ([24](#A7.E24)) 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, noting that $c\gtrsim\log(1/\delta)$, we have

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

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

###### Lemma 6 .

###### Proof.

### G.7 Lemma 7 and its proof

###### Lemma 7 .

With the $\alpha_{t}$ defined in ([8](#S3.E8)), given any $p>0$, if $\delta p<1$,

$$ $\sum_{t=2}^{T}(1-\alpha_{t})\bar{\alpha}_{t}^{p}\leq\left(\frac{1}{p}(1-\delta)^{p}e^{-p\delta\log(1/\delta)}+(1-\delta)^{p}\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}\right)\left(1+O\left(\frac{\log T}{T}\right)\right).$ $$

Further, when $\delta\ll 1$,

$$ $\sum_{t=2}^{T}(1-\alpha_{t})\bar{\alpha}_{t}^{p}\leq\frac{1}{p}-\left(1+\frac{p+1}{2p}\right)\frac{c\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right).$ $$

###### Proof.

Define the sum as $s_{T}$. Recall that

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

We first note a relationship that for fixed $\delta\neq 0$ and $p>0$. As $z\to\infty$,

$$ $(1-\delta z^{-1})^{pz}=e^{pz\log(1-\delta z^{-1})}=e^{pz(-\delta z^{-1}+\delta^{2}z^{-2}/2+O(z^{-3}))}=e^{-\delta p}(1+\delta^{2}pz^{-1}/2)+O(z^{-2}).$ (25) $$

We also use the fact from binomial series that

$$ $(1-z^{-1})^{p}=1-pz^{-1}+\frac{p(p-1)}{2}z^{-2}+O(z^{-3}).$ (26) $$

Define $t^{*}:=\sup\left\{t\in[1,T]:\delta\left(1+\frac{c\log T}{T}\right)^{t}\leq 1\right\}$. Thus, $\alpha_{t}\equiv 1-\frac{c\log T}{T}$ for all $t>t^{*}$. Note that when $T$ becomes large, $t^{*}=\Theta\left(\frac{T}{\log T}\right)$.
To further understand the big-$\Theta$ term, note that using ([25](#A7.E25)),

$$ $\displaystyle\delta\left(1+\frac{c\log T}{T}\right)^{\frac{T\log(1/\delta)}{c\log T}-\log(1/\delta)}$ $\displaystyle=\left(1+\frac{\log(1/\delta)c\log T}{2T}+\tilde{O}\left(\frac{1}{T^{2}}\right)\right)\left(1-\frac{\log(1/\delta)c\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right)\right)$ $\displaystyle=1-\frac{\log(1/\delta)c\log T}{2T}+\tilde{O}\left(\frac{1}{T^{2}}\right)$ $\displaystyle<1\leavevmode\nobreak\ \leavevmode\nobreak\ \text{as $T\to\infty$}.$ $$

This implies that

$$ $t^{*}\geq\log(1/\delta)\left(\frac{T}{c\log T}-1\right).$ (27) $$

To start, we suppose $T>t^{*}$ is large enough and decompose the sum as

$$ $\displaystyle s_{T}$ $\displaystyle=\sum_{t=2}^{t^{*}}(1-\alpha_{t})\bar{\alpha}_{t}^{p}+\sum_{t=t^{*}+1}^{T}(1-\alpha_{t})\bar{\alpha}_{t}^{p}$ $\displaystyle=\frac{c\log T}{T}\delta(1-\delta)^{p}\sum_{t=2}^{t^{*}}\left(1+\frac{c\log T}{T}\right)^{t}\prod_{i=2}^{t}\left(1-\delta\frac{c\log T}{T}\left(1+\frac{c\log T}{T}\right)^{i}\right)^{p}$ $\displaystyle\qquad+\bar{\alpha}_{t^{*}}^{p}\frac{c\log T}{T}\sum_{t=t^{*}+1}^{T}\left(1-\frac{c\log T}{T}\right)^{p(t-t^{*})}.$ (28) $$

Now we first focus on the second term in ([G.7](#A7.Ex275)).

$$ $\displaystyle\frac{c\log T}{T}\sum_{t=t^{*}+1}^{T}\left(1-\frac{c\log T}{T}\right)^{p(t-t^{*})}$ $\displaystyle=\left(1-\frac{c\log T}{T}\right)^{p}\frac{c\log T}{T}\cdot\frac{1-\left(1-\frac{c\log T}{T}\right)^{p(T-t^{*})}}{1-\left(1-\frac{c\log T}{T}\right)^{p}}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\left(1-\frac{c\log T}{T}\right)^{p}\frac{c\log T}{T}\cdot\frac{1-\left(1-\frac{c\log T}{T}\right)^{p(T-t^{*})}}{1-\left(1-\frac{pc\log T}{T}+p(p-1)\frac{c^{2}(\log T)^{2}}{2T^{2}}+\tilde{O}\left(\frac{1}{T^{3}}\right)\right)}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\frac{1}{p}\left(1-\frac{pc\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right)\right)\left(1+\frac{(p-1)c\log T}{2T}+\tilde{O}\left(\frac{1}{T^{2}}\right)-O\left(\frac{1}{T^{pc/2}}\right)\right)$ $\displaystyle=\frac{1}{p}\left(1-\frac{(p+1)c\log T}{2T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right)$ $$

where $(i)$ follows from ([26](#A7.E26)), and $(ii)$ is because $t^{*}=\Theta(T/\log T)$ and thus $T-t^{*}>T/2$ for large $T$.
Also, for all $t=2,\dots,t^{*}$,

$$ $\displaystyle\bar{\alpha}_{t}^{p}$ $\displaystyle=(1-\delta)^{p}\prod_{i=2}^{t}\left(1-\delta\frac{c\log T}{T}\left(1+\frac{c\log T}{T}\right)^{i}\right)^{p}$ $\displaystyle\leq(1-\delta)^{p}\left(1-\delta\frac{c\log T}{T}\right)^{p(t-1)}$ $$

Thus, we have

$$ $\displaystyle\bar{\alpha}_{t^{*}}^{p}$ $\displaystyle\leq(1-\delta)^{p}\left(1-\delta\frac{c\log T}{T}\right)^{p(t^{*}-1)}$ $\displaystyle=(1-\delta)^{p}\left(1+\delta p\frac{c\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right)\right)\left(1-\delta\frac{c\log T}{T}\right)^{pt^{*}}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\leq}}(1-\delta)^{p}\left(1+\delta p(1+\log(1/\delta))\frac{c\log T}{T}\right)\times$ $\displaystyle\qquad\quad e^{-p\delta\log(1/\delta)}\left(1+\delta^{2}p\log(1/\delta)\frac{c\log T}{2T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right)$ $\displaystyle=(1-\delta)^{p}e^{-p\delta\log(1/\delta)}\left(1+\delta p\left(1+\log(1/\delta)+(\delta/2)\log(1/\delta)\right)\frac{c\log T}{T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right).$ $$

Here $(iii)$ follows because using ([27](#A7.E27)) and ([25](#A7.E25)), we have that

$$ $\displaystyle\left(1-\delta\frac{c\log T}{T}\right)^{pt^{*}}\leq\left(1-\delta\frac{c\log T}{T}\right)^{p\log(1/\delta)\left(\frac{T}{c\log T}-1\right)}$ $\displaystyle=\left(1+\delta p\log(1/\delta)\frac{c\log T}{T}\right)e^{-p\delta\log(1/\delta)}\left(1+\delta^{2}p\log(1/\delta)\frac{c\log T}{2T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right).$ (29) $$

Thus, the second term in ([G.7](#A7.Ex275)) satisfies that

$$ $\displaystyle\sum_{t=t^{*}+1}^{T}(1-\alpha_{t})\bar{\alpha}_{t}^{p}$ $\displaystyle\leq\frac{1}{p}(1-\delta)^{p}e^{-p\delta\log(1/\delta)}\left(1-\left(\frac{p+1}{2}-\delta p\left(1+\log(1/\delta)+(\delta/2)\log(1/\delta)\right)\right)\frac{c\log T}{T}\right)$ $\displaystyle\qquad+\tilde{O}\left(\frac{1}{T^{2}}\right).$ (30) $$

Now we turn to the first term in ([G.7](#A7.Ex275)), in which the summation can be upper-bounded as

$$ $\displaystyle\sum_{t=2}^{t^{*}}\left(1+\frac{c\log T}{T}\right)^{t}\prod_{i=2}^{t}\left(1-\delta\frac{c\log T}{T}\left(1+\frac{c\log T}{T}\right)^{i}\right)^{p}$ $\displaystyle\leq\sum_{t=2}^{t^{*}}\left(1+\frac{c\log T}{T}\right)^{t}\left(1-\delta\frac{c\log T}{T}\right)^{p(t-1)}=:\left(1+\frac{c\log T}{T}\right)\sum_{t=1}^{t^{*}-1}q^{t}$ $$

where

$$ $\displaystyle q$ $\displaystyle:=\left(1+\frac{c\log T}{T}\right)\left(1-\delta\frac{c\log T}{T}\right)^{p}$ $\displaystyle=\left(1+\frac{c\log T}{T}\right)\left(1-\delta p\frac{c\log T}{T}+\delta^{2}p(p-1)\frac{c^{2}(\log T)^{2}}{2T^{2}}+\tilde{O}\left(\frac{1}{T^{3}}\right)\right)$ $\displaystyle=1+(1-\delta p)\frac{c\log T}{T}+\left(\frac{\delta^{2}p(p-1)}{2}-\delta p\right)\frac{c^{2}(\log T)^{2}}{T^{2}}+\tilde{O}\left(\frac{1}{T^{3}}\right).$ $$

Note that by assumption $\delta p<1$.
Also, by definition of $t^{*}$ and ([G.7](#A7.Ex289)),

$$ $\displaystyle\delta q^{t^{*}}\leq\left(1-\delta\frac{c\log T}{T}\right)^{pt^{*}}$ $\displaystyle\leq e^{-p\delta\log(1/\delta)}\left(1+\delta p\log(1/\delta)(1+\delta/2)\frac{c\log T}{T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right).$ $$

Thus, we have

$$ $\displaystyle\delta\frac{c\log T}{T}\sum_{t=1}^{t^{*}-1}q^{t}=\frac{c\log T}{T}\times\frac{\delta q^{t^{*}}-\delta q}{q-1}$ $\displaystyle\leq\frac{c\log T}{T}\times\frac{e^{-p\delta\log(1/\delta)}\left(1+\delta p\log(1/\delta)(1+\delta/2)\frac{c\log T}{T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right)-\delta q}{q-1}$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{=}}\frac{c\log T}{T}\times\frac{e^{-p\delta\log(1/\delta)}\left(1+\delta p\log(1/\delta)(1+\delta/2)\frac{c\log T}{T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right)-1-(1-\delta p)\frac{c\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right)}{(1-\delta p)\frac{c\log T}{T}+\left(\frac{\delta^{2}p(p-1)}{2}-\delta p\right)\frac{c^{2}(\log T)^{2}}{T^{2}}+\tilde{O}\left(\frac{1}{T^{3}}\right)}$ $\displaystyle=\left(\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}+\left(\frac{\delta p\log(1/\delta)(1+\delta/2)}{1-\delta p}-1\right)\frac{c\log T}{T}\right)\times$ $\displaystyle\qquad\left(1+\frac{\delta p-\frac{\delta^{2}p(p-1)}{2}}{1-\delta p}\cdot\frac{c\log T}{T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right)$ $\displaystyle=\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}+\left(\frac{\delta p\log(1/\delta)(1+\delta/2)}{1-\delta p}-1+\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}\cdot\frac{\delta p(1-\delta(p-1)/2)}{1-\delta p}\right)\frac{c\log T}{T}$ $\displaystyle\qquad+\tilde{O}\left(\frac{1}{T^{2}}\right).$ $$

where $(iv)$ follows from ([26](#A7.E26)).
Therefore,

$$ $\displaystyle\sum_{t=2}^{t^{*}}(1-\alpha_{t})\bar{\alpha}_{t}^{p}=(1-\delta)^{p}\left(1+\frac{c\log T}{T}\right)\left(\delta\frac{c\log T}{T}\sum_{t=1}^{t^{*}-1}q^{t}\right)$ $\displaystyle\leq(1-\delta)^{p}\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}$ $\displaystyle\qquad+(1-\delta)^{p}\bigg{(}\frac{\delta p\log(1/\delta)(1+\delta/2)}{1-\delta p}-1$ $\displaystyle\qquad\quad+\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}\left(1+\frac{\delta p(1-\delta(p-1)/2)}{1-\delta p}\right)\bigg{)}\frac{c\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right).$ (31) $$

Combining ([G.7](#A7.Ex290)) and ([G.7](#A7.Ex306)), we have that

$$ $s_{T}\leq\underbrace{\left(\frac{1}{p}(1-\delta)^{p}e^{-p\delta\log(1/\delta)}+(1-\delta)^{p}\frac{e^{-p\delta\log(1/\delta)}-1}{1-\delta p}\right)}_{=:s_{\infty}}\left(1+O\left(\frac{\log T}{T}\right)\right).$ $$

Also, for all large $T$’s, since $s_{\infty}\to\frac{1}{p}$ and $\delta\log(1/\delta)\to 0$ as $\delta\to 0$, when $\delta\ll 1$,

$$ $s_{T}\leq\frac{1}{p}-\left(1+\frac{p+1}{2p}\right)\frac{c\log T}{T}+\tilde{O}\left(\frac{1}{T^{2}}\right).$ $$

The proof is now complete.
∎

###### Lemma 7 .

###### Proof.

## Appendix H Proofs in Section 4

### H.1 Proof of Theorem 3

Fix $t\geq 1$. Using the forward model in ([5](#S2.E5)), we have that $q_{t|0,y}$ is the p.d.f. of $\mathcal{N}(\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)x_{0}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y,\Sigma_{t|0,y})$. Thus,

$$ $\displaystyle\nabla\log q_{t|y}(x)=\frac{1}{q_{t|y}(x)}\int_{x_{0}\in\mathbb{R}^{d}}\nabla q_{t|0,y}(x|x_{0})\mathrm{d}Q_{0|y}(x_{0})$ $\displaystyle=-\frac{1}{q_{t|y}(x)}\Sigma_{t|0,y}^{-1}\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0,y}(x|x_{0})(x-\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)x_{0}-\sqrt{\bar{\alpha}_{t}}H^{\dagger}y)\mathrm{d}Q_{0|y}(x_{0})$ $\displaystyle=-\Sigma_{t|0,y}^{-1}(x-\sqrt{\bar{\alpha}_{t}}H^{\dagger}y)$ $\displaystyle\qquad+\frac{\sqrt{\bar{\alpha}_{t}}}{q_{t|y}(x)}\Sigma_{t|0,y}^{-1}(I_{d}-H^{\dagger}H)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0,y}(x|x_{0})x_{0}\mathrm{d}Q_{0|y}(x_{0}).$ $$

Thus, the equality for $\nabla\log q_{t|y}$ is established because by [Lemma 9](#Thmlemma9),

$$ $(\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}+(1-\bar{\alpha}_{t})I_{d})^{-1}(I_{d}-H^{\dagger}H)=(1-\bar{\alpha}_{t})^{-1}(I_{d}-H^{\dagger}H).$ $$

To see the optimality with $f_{t,y}^{*}$, fix $t\geq 1$ and $x\in\mathbb{R}^{d}$. First note that $(I_{d}-H^{\dagger}H)f_{t,y}^{*}(x)=(I_{d}-H^{\dagger}H)\Sigma_{t|0,y}^{-1}(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx)=0$ by [Lemma 9](#Thmlemma9). Now, suppose that $f_{t,y}=f_{t,y}^{*}+v$ such that $(I_{d}-H^{\dagger}H)f_{t,y}=0\implies(I_{d}-H^{\dagger}H)v=0$. From the definition of $\Delta_{t,y}$ in ([7](#S2.E7)),

$$ $\Delta_{t,y}(x)=(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}(x)-\nabla\log q_{t}(x))+(H^{\dagger}H)\nabla\log q_{t|y}(x)-f_{t,y}(x)$ $$

where

$$ $\displaystyle(H^{\dagger}H)\nabla\log q_{t|y}(x)-f_{t,y}(x)$ $\displaystyle=(H^{\dagger}H)\Sigma_{t|0,y}^{-1}(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-x)-\Sigma_{t|0,y}^{-1}(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx)-v$ $\displaystyle=-(H^{\dagger}H)\Sigma_{t|0,y}^{-1}(I_{d}-H^{\dagger}H)x-(I_{d}-H^{\dagger}H)\Sigma_{t|0,y}^{-1}(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx)-v$ $\displaystyle=-v$ $$

where the last line follows from [Lemma 9](#Thmlemma9).

Thus, if $v=0$, then $f_{t,y}=f_{t,y}^{*}$, and we have

$$ $\Delta_{t,y}=(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}(x)-\nabla\log q_{t}(x)).$ (32) $$

Also, if $v\neq 0$, since $v$ is orthogonal to the space induced by $(I_{d}-H^{\dagger}H)$, we have

$$ $\left\lVert\Delta_{t,y}(x)\right\rVert^{2}=\left\lVert(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}(x)-\nabla\log q_{t}(x))\right\rVert^{2}+\left\lVert v\right\rVert^{2}$ (33) $$

which is minimized at $v=0$. The proof is now complete.

### H.2 Proof of Theorem 4

Fix $t\geq 2$. Recall that the unconditional score $\nabla\log q_{t}(x)$ is

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

since $q_{t|0}$ is the p.d.f. of $\mathcal{N}(\sqrt{\bar{\alpha}_{t}}x_{0},(1-\bar{\alpha}_{t})I_{d})$.

In the first half, we consider the case where $\sigma_{y}^{2}$ is known, and thus $f_{t,y}=f_{t,y}^{*}$ in ([10](#S4.E10)). Note that from [Theorem 3](#Thmtheorem3),

$$ $\displaystyle\nabla\log q_{t|y}(x)$ $\displaystyle=\Sigma_{t|0,y}^{-1}(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-x)$ $\displaystyle\qquad+\frac{\sqrt{\bar{\alpha}_{t}}}{q_{t|y}(x)}\Sigma_{t|0,y}^{-1}(I_{d}-H^{\dagger}H)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0,y}(x|x_{0})x_{0}\mathrm{d}Q_{0|y}(x_{0}).$ $$

Here we also recall from [Theorem 3](#Thmtheorem3) that

$$ $\Sigma_{t|0,y}:=\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}+(1-\bar{\alpha}_{t})I_{d}.$ $$

Since $H^{\dagger}(H^{\dagger})^{\intercal}$ is positive semi-definite, all its eigenvalues are non-negative. Write the eigen-decomposition as $H^{\dagger}(H^{\dagger})^{\intercal}=P\mathrm{diag}(D_{1},\dots,D_{d})P^{\intercal}$ where $D_{1}\geq\dots\geq D_{d}\geq 0,\leavevmode\nobreak\ \forall i\in[d]$. Then, $\lambda_{min}(\Sigma_{t|0,y})\geq\bar{\alpha}_{t}\sigma_{y}^{2}D_{d}+1-\bar{\alpha}_{t}\geq 1-\bar{\alpha}_{t}$, and we get

$$ $\left\lVert\Sigma_{t|0,y}^{-1}\right\rVert\leq\frac{1}{1-\bar{\alpha}_{t}}.$ (34) $$

Also, from ([6](#S2.E6)), with the $f_{t,y}^{*}$ in ([10](#S4.E10)),

$$ $\displaystyle g_{t,y}(x)$ $\displaystyle=f_{t,y}^{*}(x)+(I_{d}-H^{\dagger}H)\nabla\log q_{t}(x)$ $\displaystyle=\Sigma_{t|0,y}^{-1}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx\right)-\frac{1}{(1-\bar{\alpha}_{t})}(I_{d}-H^{\dagger}H)x$ $\displaystyle\qquad+\frac{\sqrt{\bar{\alpha}_{t}}}{(1-\bar{\alpha}_{t})q_{t}(x)}(I_{d}-H^{\dagger}H)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})x_{0}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\Sigma_{t|0,y}^{-1}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx\right)-\Sigma_{t|0,y}^{-1}(I_{d}-H^{\dagger}H)x$ $\displaystyle\qquad+\frac{\sqrt{\bar{\alpha}_{t}}}{q_{t}(x)}\Sigma_{t|0,y}^{-1}(I_{d}-H^{\dagger}H)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})x_{0}\mathrm{d}Q_{0}(x_{0})$ $\displaystyle=\Sigma_{t|0,y}^{-1}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-x\right)+\frac{\sqrt{\bar{\alpha}_{t}}}{q_{t}(x)}\Sigma_{t|0,y}^{-1}(I_{d}-H^{\dagger}H)\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})x_{0}\mathrm{d}Q_{0}(x_{0})$ $$

where $(i)$ follows from [Lemma 9](#Thmlemma9).
Then, the norm-squared of the score mismatch at time $t\geq 2$ is

$$ $\displaystyle\left\lVert\Delta_{t,y}\right\rVert^{2}=\left\lVert\nabla\log q_{t|y}-g_{t,y}\right\rVert^{2}=\left\lVert(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}-\nabla\log q_{t})\right\rVert^{2}$ $\displaystyle\leq\bar{\alpha}_{t}\left\lVert\Sigma_{t|0,y}^{-1}\right\rVert^{2}\left\lVert\frac{\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0,y}(x|x_{0})x_{0}\mathrm{d}Q_{0|y}(x_{0})}{q_{t|y}(x)}-\frac{\int_{x_{0}\in\mathbb{R}^{d}}q_{t|0}(x|x_{0})x_{0}\mathrm{d}Q_{0}(x_{0})}{q_{t}(x)}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}\bar{\alpha}_{t}\left\lVert\Sigma_{t|0,y}^{-1}\right\rVert^{2}\int_{x_{a},x_{b}\in\mathbb{R}^{d}}\left\lVert x_{a}-x_{b}\right\rVert^{2}\mathrm{d}Q_{0|t,y}(x_{a})\mathrm{d}Q_{0|t}(x_{b})$ $\displaystyle\leq\bar{\alpha}_{t}\left\lVert\Sigma_{t|0,y}^{-1}\right\rVert^{2}\max_{\begin{subarray}{c}x_{a}\in\mathrm{supp}(Q_{0|y})\\ x_{b}\in\mathrm{supp}(Q_{0})\end{subarray}}\left\lVert x_{a}-x_{b}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\lesssim}}\frac{\bar{\alpha}_{t}}{(1-\bar{\alpha}_{t})^{2}}d.$ (35) $$

Here $(ii)$ follows from Jensen’s inequality, and $(iii)$ follows by ([34](#A8.E34)) and from the assumption that $Q_{0}$ has bounded support (and thus also for both $Q_{0|t}$ and $Q_{0|t,y}$). Therefore, with the $\alpha_{t}$ in ([8](#S3.E8)) (cf. [Lemma 6](#Thmlemma6)), [Assumption 4](#Thmassumption4) is satisfied for all $\sigma_{y}^{2}\geq 0$.
Thus, [Theorem 2](#Thmtheorem2) holds with $\gamma=1$ and $r=2$.

Now, we consider the case where $\sigma_{y}^{2}$ is unknown, and the conditional sampler of interest is $g_{t,y}^{N}(x)=f_{t,y}^{N}(x)+(I_{d}-H^{\dagger}H)\nabla\log q_{t}(x)$ where $f_{t,y}^{N}(x)=(1-\bar{\alpha}_{t})^{-1}\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx\right)$. With the same notation as in the proof of [Theorem 3](#Thmtheorem3), we can write $v=f_{t,y}^{N}-f_{t,y}^{*}=((1-\bar{\alpha}_{t})^{-1}I_{d}-\Sigma_{t|0,y}^{-1})\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx\right)$. Note that $v$ still satisfies that $(I_{d}-H^{\dagger}H)v=0$.
Using the result in ([33](#A8.E33)), we have

$$ $\left\lVert\Delta_{t,y}^{N}\right\rVert^{2}=\left\lVert(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}(x)-\nabla\log q_{t}(x))\right\rVert^{2}\\ +\left\lVert(\Sigma_{t|0,y}^{-1}-(1-\bar{\alpha}_{t})^{-1}I_{d})\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}Hx\right)\right\rVert^{2}$ $$

where the first term is the same as in ([H.2](#A8.Ex333)) which can be upper-bounded in a similar way. To upper-bound the second term, note that by Woodbury matrix identity,

$$ $\displaystyle\left\lVert\Sigma_{t|0,y}^{-1}-\frac{1}{1-\bar{\alpha}_{t}}I_{d}\right\rVert$ $\displaystyle=\frac{\bar{\alpha}_{t}\sigma_{y}^{2}}{(1-\bar{\alpha}_{t})^{2}}\left\lVert H^{\dagger}\left(I_{p}+\frac{\sigma_{y}^{2}}{1-\bar{\alpha}_{t}}(H^{\dagger})^{\intercal}H^{\dagger}\right)^{-1}(H^{\dagger})^{\intercal}\right\rVert$ $\displaystyle\lesssim\frac{\bar{\alpha}_{t}\sigma_{y}^{2}}{(1-\bar{\alpha}_{t})^{2}}$ (36) $$

where the inequality follows because $\left\lVert H^{\dagger}\right\rVert\lesssim 1$ is a constant and the minimum eigenvalue of $(I_{p}+\frac{\sigma_{y}^{2}}{1-\bar{\alpha}_{t}}(H^{\dagger})^{\intercal}H^{\dagger})$ is at least 1.
Thus,

$$ $\displaystyle\mathbb{E}_{Q_{t|y}}\left\lVert(\Sigma_{t|0,y}^{-1}-(1-\bar{\alpha}_{t})^{-1}I_{d})\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}HX_{t}\right)\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{\leq}}\frac{\bar{\alpha}_{t}^{2}\sigma_{y}^{4}}{(1-\bar{\alpha}_{t})^{4}}\mathbb{E}_{Q_{t|y}}\left\lVert\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}HX_{t}\right\rVert^{2}$ $\displaystyle=\frac{\bar{\alpha}_{t}^{2}\sigma_{y}^{4}}{(1-\bar{\alpha}_{t})^{4}}\mathbb{E}_{Q_{0|y}}\mathbb{E}_{Q_{t|0,y}}\left\lVert\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}HX_{t}\right\rVert^{2}$ $\displaystyle=\frac{\bar{\alpha}_{t}^{2}\sigma_{y}^{4}}{(1-\bar{\alpha}_{t})^{4}}\mathbb{E}_{Q_{0|y}}\mathbb{E}_{Q_{t|0}}\left\lVert\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}HX_{t}\right\rVert^{2}$ $\displaystyle\leq\frac{2\bar{\alpha}_{t}^{2}\sigma_{y}^{4}}{(1-\bar{\alpha}_{t})^{4}}\mathbb{E}_{Q_{0|y}}\left[\bar{\alpha}_{t}\left\lVert H^{\dagger}y-H^{\dagger}HX_{0}\right\rVert^{2}+\mathbb{E}_{Q_{t|0}}\left\lVert H^{\dagger}H(X_{t}-\sqrt{\bar{\alpha}_{t}}X_{0})\right\rVert^{2}\right]$ $\displaystyle\leq\frac{2\bar{\alpha}_{t}^{2}\sigma_{y}^{4}}{(1-\bar{\alpha}_{t})^{4}}\mathbb{E}_{Q_{0|y}}\left[\bar{\alpha}_{t}\left\lVert H^{\dagger}y\right\rVert^{2}+\bar{\alpha}_{t}\left\lVert X_{0}\right\rVert^{2}+d(1-\bar{\alpha}_{t})\right]$ $\displaystyle\stackrel{{\scriptstyle(v)}}{{\lesssim}}\frac{\bar{\alpha}_{t}^{2}\sigma_{y}^{4}}{(1-\bar{\alpha}_{t})^{4}}d$ $$

where $(iv)$ follows from ([H.2](#A8.Ex337)), and $(v)$ follows from the fact that $Q_{0|y}$ has bounded support. Similarly, for general moments $\ell\geq 2$,

$$ $\mathbb{E}_{Q_{t|y}}\left\lVert(\Sigma_{t|0,y}^{-1}-(1-\bar{\alpha}_{t})^{-1}I_{d})\left(\sqrt{\bar{\alpha}_{t}}H^{\dagger}y-H^{\dagger}HX_{t}\right)\right\rVert^{\ell}\lesssim\left(\frac{\bar{\alpha}_{t}\sigma_{y}^{2}}{(1-\bar{\alpha}_{t})^{2}}d\right)^{\ell/2}.$ $$

Therefore, with the $\alpha_{t}$ in ([8](#S3.E8)), we still have that [Assumption 4](#Thmassumption4) is satisfied (see [Lemma 6](#Thmlemma6)), and [Theorem 2](#Thmtheorem2) still holds with $\gamma=1$ and $r=4$. The proof is now complete.

### H.3 Theorem 6 and its proof

Before we enter the proof of [Propositions 1](#Thmproposition1) and [5](#Thmtheorem5), we first state a similar set of results for Gaussian $Q_{0}$, which turns out to be useful for analyzing Gaussian mixture $Q_{0}$’s. To begin, the following lemma investigates $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}$ when $Q_{0}$ is Gaussian. This quantity is proportional to the asymptotic bias $\mathcal{W}_{\text{bias}}$.

###### Proposition 2 .

For $Q_{0}=\mathcal{N}(\mu_{0},\Sigma_{0})$, if $f_{t,y}=f_{t,y}^{*}$ in ([10](#S4.E10)) and $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$, with the $\alpha_{t}$’s according to [Definition 1](#Thmdefinition1), [Assumption 4](#Thmassumption4) is satisfied, and

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}$ $\displaystyle\leq\bar{\alpha}_{t}^{2}\frac{\max\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\}}{\min\{\lambda_{d},1\}^{2}\min\{\tilde{\lambda}_{d-p},1\}^{2}}\left\lVert[\Sigma_{0}]_{y\bar{y}}[\Sigma_{0}]_{\bar{y}y}\right\rVert$ $\displaystyle\lesssim\bar{\alpha}_{t}^{2}\cdot(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d)$ $$

where $\lambda_{1}$ is the largest eigenvalue of $\Sigma_{0}$, and $\lambda_{d}$ and $\tilde{\lambda}_{d-p}$ are the smallest eigenvalues of $\Sigma_{0}$ and $[\Sigma_{0}]_{\bar{y}\bar{y}}$, respectively.

###### Proof.

See [Section I.1](#A9.SS1).
∎

With this lemma, the following theorem characterizes the conditional KL divergence when $Q_{0}$ is Gaussian.

###### Theorem 6 .

Suppose that $\sigma_{y}^{2}>0$. Suppose that [Assumptions 1](#Thmassumption1) and [5](#Thmassumption5) hold.
Under the same conditions as in [Proposition 2](#Thmproposition2),
if $\alpha_{t}$ further satisfies $\sum_{t=1}^{T}(1-\alpha_{t})\bar{\alpha}_{t}=1+o(1)$, we have

$$ $\displaystyle\mathrm{KL}(Q_{0|y}\|\widehat{P}_{0|y})\lesssim(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d)$ $\displaystyle\qquad+(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d)\frac{(\log T)^{2}}{T}+\sqrt{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d}\cdot(\log T)\varepsilon.$ $$

Note that a similar result can be obtained for $\mathrm{KL}(Q_{1|y}\|\widehat{P}_{1|y})$ (where $\mathrm{W}_{2}(Q_{1|y},Q_{0|y})^{2}\lesssim\delta d$) for any general $\sigma_{y}^{2}\geq 0$ using the $\alpha_{t}$ in ([8](#S3.E8)) (see [Remark 1](#Thmremark1)).

###### Proposition 2 .

###### Proof.

###### Theorem 6 .

#### H.3.1 Proof of Theorem 6

Throughout the proof we use the same notations as in ([I.1](#A9.Ex597)). Since [Assumption 4](#Thmassumption4) is satisfied from [Proposition 2](#Thmproposition2), in order to invoke [Theorem 1](#Thmtheorem1), we still need to check [Assumption 3](#Thmassumption3). Since each $Q_{t|y}\leavevmode\nobreak\ (\forall t\geq 0)$ is Gaussian, all partial derivatives of its log-p.d.f. higher than third-order equal zero. For the first and second-order, note that $\Sigma_{t|y}=\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+(1-\bar{\alpha}_{t})I_{d}+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}H$. Thus, when $\sigma_{y}^{2}>0$, $\lambda_{min}(\Sigma_{t-1|y})\geq\min\{1-\bar{\alpha}_{t}+\bar{\alpha}_{t}\sigma_{y}^{2},1-\bar{\alpha}_{t}+\bar{\alpha}_{t}\tilde{\lambda}_{d}\}\geq\min\{1,\sigma_{y}^{2},\tilde{\lambda}_{d}\}>0$, which yields

$$ $\left\lVert\Sigma_{t-1|y}^{-1}\right\rVert\lesssim 1,\quad\forall t\geq 1.$ (37) $$

Thus, we have, $\forall\ell\geq 1$,

$$ $\displaystyle\mathbb{E}_{Q_{t|y}}\left\lVert\nabla\log q_{t|y}(X_{t})\right\rVert^{\ell}$ $\displaystyle\quad=\mathbb{E}_{Q_{t|y}}\left\lVert\Sigma_{t|y}^{-1}(X_{t}-\mu_{t|y})\right\rVert^{\ell}\leq\left\lVert\Sigma_{t|y}^{-\frac{1}{2}}\right\rVert^{\ell}\mathbb{E}_{Q_{t|y}}\left\lVert\Sigma_{t|y}^{-\frac{1}{2}}(X_{t}-\mu_{t|y})\right\rVert^{\ell}$ $\displaystyle\quad\lesssim d^{\ell/2}=O(1),$ $\displaystyle\mathbb{E}_{Q_{t|y}}\left\lVert\nabla\log q_{t-1|y}(m_{t,y}(X_{t}))\right\rVert^{\ell}$ $\displaystyle\quad=\mathbb{E}_{Q_{t|y}}\left\lVert\Sigma_{t|y}^{-1}(m_{t,y}(X_{t})-\mu_{t|y})\right\rVert^{\ell}$ $\displaystyle\quad\lesssim\left\lVert\Sigma_{t|y}^{-\frac{1}{2}}\right\rVert^{\ell}\mathbb{E}_{Q_{t|y}}\left\lVert\Sigma_{t|y}^{-\frac{1}{2}}(X_{t}-\mu_{t|y})\right\rVert^{\ell}+\left\lVert\Sigma_{t|y}^{-1}\right\rVert^{\ell}\mathbb{E}_{Q_{t|y}}\left\lVert\nabla\log q_{t|y}(X_{t})\right\rVert^{\ell}$ $\displaystyle\quad\lesssim d^{\ell/2}=O(1),$ $\displaystyle\mathbb{E}_{Q_{t|y}}\left\lVert\nabla^{2}\log q_{t|y}(X_{t})\right\rVert^{\ell}=\left\lVert\Sigma_{t|y}^{-1}\right\rVert^{\ell}=O(1),$ $\displaystyle\mathbb{E}_{Q_{t|y}}\left\lVert\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\right\rVert^{\ell}=\left\lVert\Sigma_{t-1|y}^{-1}\right\rVert^{\ell}=O(1).$ $$

Thus, [Assumption 3](#Thmassumption3) holds when $1-\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1).

Now, we can invoke [Theorem 1](#Thmtheorem1) and get $\mathrm{KL}(Q_{0|y}\|\widehat{P}_{0|y})\lesssim\mathcal{W}_{\text{oracle}}+\mathcal{W}_{\text{bias}}+\mathcal{W}_{\text{vanish}}$,
where

$$ $\displaystyle\mathcal{W}_{\text{oracle}}$ $\displaystyle=\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\nabla^{2}\log q_{t|y}(X_{t})\Big{)}\right]+(\log T)\varepsilon^{2}$ $\displaystyle\mathcal{W}_{\text{bias}}$ $\displaystyle=\sum_{t=1}^{T}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}$ $\displaystyle\mathcal{W}_{\text{vanish}}$ $\displaystyle=\sum_{t=1}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(\nabla\log q_{t-1|y}(m_{t,y}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(X_{t}))^{\intercal}\Delta_{t,y}(X_{t})\bigg{]}$ $\displaystyle-\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\Delta_{t,y}(X_{t})^{\intercal}\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})\right]$ $\displaystyle+\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}3\sum_{i=1}^{d}\partial^{3}_{iii}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{i}$ $\displaystyle\qquad+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{j}\bigg{]}$ $\displaystyle+\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}(\log T)\varepsilon.$ $$

We first consider the estimation error (in both $\mathcal{W}_{\text{oracle}}$ and $\mathcal{W}_{\text{vanish}}$), which can be upper-bounded as

$$ $\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}(\log T)\varepsilon+(\log T)\varepsilon^{2}\lesssim\left(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d\right)^{\frac{1}{2}}(\log T)\varepsilon$ $$

from [Proposition 2](#Thmproposition2).
Also, when $Q_{t|y}$ is Gaussian, we can calculate, for any $x_{t}\in\mathbb{R}^{d}$,

$$ $\displaystyle\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1|y}(m_{t,y}(x_{t}))\nabla^{2}\log q_{t|y}(x_{t})\Big{)}=\mathrm{Tr}(\Sigma_{t-1|y}^{-1}\Sigma_{t|y}^{-1})$ $\displaystyle=\mathrm{Tr}(\Sigma_{t-1|y}^{-1}(\alpha_{t}\Sigma_{t-1|y}+(1-\alpha_{t})I_{d})^{-1})$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathrm{Tr}(\Sigma_{t-1|y}^{-1}(\alpha_{t}^{-1}\Sigma_{t-1|y}^{-1}-\frac{1-\alpha_{t}}{\alpha_{t}^{2}}\Sigma_{t-1|y}^{-2}+O((1-\alpha_{t})^{2})))$ $\displaystyle\lesssim\frac{1}{\alpha_{t}}\mathrm{Tr}(\Sigma_{t-1|y}^{-2})$ $$

where $(i)$ follows from Taylor expansion when $1-\alpha_{t}$ is small. Using ([37](#A8.E37)), this implies that

$$ $\mathcal{W}_{\text{oracle}}\lesssim\frac{d(\log T)^{2}}{T}+(\log T)\varepsilon^{2}.$ $$

Also, from the condition on $\alpha_{t}$,

$$ $\sum_{t=1}^{T}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}\lesssim(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d).$ $$

Now we focus on $\mathcal{W}_{\text{vanish}}$ (except the estimation error).
Since $Q_{t|y}$ is Gaussian, all third-order partial derivatives are zero, and only the first two terms in $\mathcal{W}_{\text{vanish}}$ remain.
In the following we fix $t\geq 1$. Also recall from ([I.1](#A9.Ex609)) that when $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$,

$$ $\displaystyle\Delta_{t,y}$ $\displaystyle=-\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle=-\bar{\alpha}_{t}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0}).$ $$

For the first term of $\mathcal{W}_{\text{vanish}}$, we first calculate for each $x_{t}$ that

$$ $\displaystyle\nabla\log q_{t-1|y}(m_{t,y})-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(x_{t})$ $\displaystyle=\sqrt{\alpha_{t}}\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})-\Sigma_{t-1|y}^{-1}(m_{t,y}-\sqrt{\bar{\alpha}_{t-1}}\mu_{0|y})$ $$

Recall that

$$ $\displaystyle m_{t,y}$ $\displaystyle=\frac{1}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t|y}(x_{t})=\frac{1}{\sqrt{\alpha_{t}}}x_{t}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y}),$ $\displaystyle\Sigma_{t|y}^{-1}$ $\displaystyle=(\alpha_{t}\Sigma_{t-1|y}+(1-\alpha_{t})I_{d})^{-1}=\frac{1}{\alpha_{t}}\Sigma_{t-1|y}^{-1}-\frac{1-\alpha_{t}}{\alpha_{t}^{2}}\Sigma_{t-1|y}^{-2}+O((1-\alpha_{t})^{2}).$ $$

Thus,

$$ $\displaystyle\nabla\log q_{t-1|y}(m_{t,y})-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(x_{t})$ $\displaystyle=\sqrt{\alpha_{t}}\left(\frac{1}{\alpha_{t}}\Sigma_{t-1|y}^{-1}-\frac{1-\alpha_{t}}{\alpha_{t}^{2}}\Sigma_{t-1|y}^{-2}\right)(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})$ $\displaystyle\qquad-\Sigma_{t-1|y}^{-1}\left(\frac{1}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t|y}(x_{t})-\sqrt{\bar{\alpha}_{t-1}}\mu_{0|y}\right)+O((1-\alpha_{t})^{2})$ $\displaystyle=-\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}\Sigma_{t-1|y}^{-2}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Sigma_{t-1|y}^{-1}\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})+O((1-\alpha_{t})^{2}).$ $$

Combining with the definition for $\Delta_{t,y}$ in ([I.1](#A9.Ex609)) and using [Lemma 9](#Thmlemma9), we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\Delta_{t,y}(X_{t})^{\intercal}(\nabla\log q_{t-1|y}(m_{t,y}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(X_{t}))\right]$ $\displaystyle=\bar{\alpha}_{t}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}\Sigma_{t}^{-1}(H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)$ $\displaystyle\qquad\left(\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}\Sigma_{t-1|y}^{-2}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Sigma_{t-1|y}^{-1}\Sigma_{t|y}^{-1}\right)(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})\bigg{]}+O((1-\alpha_{t})^{2})$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\bar{\alpha}_{t}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}\Sigma_{t}^{-1}(H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)$ $\displaystyle\qquad\left(\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}\Sigma_{t-1|y}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-1}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Sigma_{t-1|y}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{t|y}^{-1}\right)$ $\displaystyle\qquad(I_{d}-H^{\dagger}H)(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\bigg{]}+O((1-\alpha_{t})^{2})$ $\displaystyle=\bar{\alpha}_{t}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}[\Sigma_{t}^{-1}]_{:y}[\Sigma_{0}]_{y\bar{y}}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}$ $\displaystyle\qquad\left(\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}^{2}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}[\Sigma_{t|y}^{-1}]_{\bar{y}\bar{y}}\right)\begin{pmatrix}0&I_{d-p}\end{pmatrix}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\bigg{]}$ $\displaystyle\qquad+O((1-\alpha_{t})^{2})$ $\displaystyle=\bar{\alpha}_{t}\mathrm{Tr}\bigg{(}[\Sigma_{t}^{-1}]_{:y}[\Sigma_{0}]_{y\bar{y}}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\left(\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}^{2}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}[\Sigma_{t|y}^{-1}]_{\bar{y}\bar{y}}\right)$ $\displaystyle\qquad\begin{pmatrix}0&I_{d-p}\end{pmatrix}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}\right]\bigg{)}$ $\displaystyle\qquad+O((1-\alpha_{t})^{2})$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\leq}}\left\lVert[\Sigma_{t}^{-1}]_{:y}\right\rVert\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert\times$ $\displaystyle\qquad\left(\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}\left\lVert[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}\right\rVert^{2}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\left\lVert[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}\right\rVert\left\lVert[\Sigma_{t|y}^{-1}]_{\bar{y}\bar{y}}\right\rVert\right)\times$ $\displaystyle\qquad\mathrm{Tr}\left(\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}\right]\right)$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{\lesssim}}\left(\frac{1-\alpha_{t}}{\alpha_{t}^{3/2}}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\right)\max\left\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\right\}$ $\displaystyle\lesssim(1-\alpha_{t})\left(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d\right)$ $$

where $(ii)$ follows by [Lemma 9](#Thmlemma9) and from definition that $(I_{d}-H^{\dagger}H)\mu_{0|y}=(I_{d}-H^{\dagger}H)\mu_{0}$, $(iii)$ follows because $\left|\mathrm{Tr}(UV)\right|\leq\left\lVert U\right\rVert\mathrm{Tr}(V)$ if $V$ is positive semi-definite, and $(iv)$ follows from ([I.1](#A9.Ex630)) and the same reasons for ([I.1](#A9.Ex623)). In particular, we note that $[\Sigma_{t|y}^{-1}]_{\bar{y}\bar{y}}=[\Sigma_{t,sig}^{-1}]_{\bar{y}\bar{y}}=(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}$ and $\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert\leq\frac{1}{\min\{\tilde{\lambda}_{d-p},1\}}<\infty$.

For the second term of $\mathcal{W}_{\text{vanish}}$, we use the fact that $[\Sigma_{t-1|y}^{-1}]_{\bar{y}\bar{y}}=(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-1}$ and have

$$ $\displaystyle-\mathbb{E}_{X_{t}\sim Q_{t|y}}\Delta_{t,y}(X_{t})^{\intercal}\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})$ $\displaystyle=\bar{\alpha}_{t}^{2}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}\Sigma_{t}^{-1}(H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-1}$ $\displaystyle\qquad(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\bigg{]}$ $\displaystyle=\bar{\alpha}_{t}^{2}\mathrm{Tr}\bigg{(}[\Sigma_{t}^{-1}]_{:y}[\Sigma_{0}]_{y\bar{y}}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-1}$ $\displaystyle\qquad(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}\mathbb{E}_{X_{t}\sim Q_{t|y}}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}\bigg{)}$ $\displaystyle\stackrel{{\scriptstyle(v)}}{{\leq}}\max\left\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\right\}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert^{2}$ $\displaystyle\qquad\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-1}\right\rVert\left\lVert\Sigma_{t}^{-1}\right\rVert^{2}\left\lVert[\Sigma_{0}]_{y\bar{y}}[\Sigma_{0}]_{\bar{y}y}\right\rVert$ $\displaystyle\lesssim\max\left\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\right\}$ $\displaystyle\lesssim\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d.$ $$

Here $(v)$ follows from the fact that $\left|\mathrm{Tr}(UV)\right|\leq\left\lVert U\right\rVert\mathrm{Tr}(V)$ if $V$ is positive semi-definite and ([I.1](#A9.Ex630)).
The proof is complete by plugging all the results above into [Theorem 1](#Thmtheorem1).

###### Remark 1 .

Before we end the proof, we leave a note for the case of $\sigma_{y}^{2}=0$ (indeed, for any general $\sigma_{y}^{2}\geq 0$). The only difference is how to upper-bound $\mathcal{W}_{\text{oracle}}$. In particular, if $\sigma_{y}^{2}=0$, ([37](#A8.E37)) no longer holds (i.e., we can no longer upper-bound $\left\lVert\Sigma_{t-1|y}^{-1}\right\rVert$ as a constant). Instead, we can obtain an upper bound as $\left\lVert\Sigma_{t-1|y}^{-1}\right\rVert\lesssim(1-\bar{\alpha}_{t-1})^{-1}$. Then, with the $\alpha_{t}$ in ([8](#S3.E8)), we have

$$ $\mathcal{W}_{\text{oracle}}\lesssim\frac{d(\log T)^{2}\log(1/\delta)^{2}}{T}+(\log T)\varepsilon^{2}.$ $$

The rest of the proof still follows because the $\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1) when $t\geq 2$.
Combining with [Lemma 7](#Thmlemma7), we would finally obtain

$$ $\displaystyle\mathrm{KL}(Q_{1|y}\|\widehat{P}_{1|y})\lesssim(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d)\left(1-\frac{3\log(1/\delta)\log T}{T}\right)$ $\displaystyle\qquad+(\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d)\frac{(\log T)^{2}\log(1/\delta)^{2}}{T}+\sqrt{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d}\cdot(\log T)\varepsilon.$ $$

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

###### Remark 1 .

### H.4 Proof of Proposition 1

We first introduce some useful notations for this subsection.
Recall that $Q_{0}$ has mixture p.d.f. in which the mixture prior $\pi_{n}$ is independent of $y\ (=Hx_{0}+n)$. Thus, using the fact that $x_{0}=(I_{d}-H^{\dagger}H)x_{0}+H^{\dagger}y-H^{\dagger}n$, we can define $Q_{0,n|y}$ as (cf. [])

$$ $\displaystyle Q_{0|y}$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}Q_{0,n|y}$ $\displaystyle:=\sum_{n=1}^{N}\pi_{n}\mathcal{N}((I_{d}-H^{\dagger}H)\mu_{0,n}+H^{\dagger}y,(I_{d}-H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}).$ $$

Note that when $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$ and $\sigma_{y}^{2}>0$, $q_{0|y}$ exists.
From the conditional forward model in ([5](#S2.E5)), we further define

$$ $\displaystyle Q_{t}$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}Q_{t,n},\leavevmode\nobreak\ \leavevmode\nobreak\ Q_{t|y}=\sum_{n=1}^{N}\pi_{n}Q_{t,n|y},\leavevmode\nobreak\ \leavevmode\nobreak\ Q_{t,n}:=\mathcal{N}(\mu_{t,n},\Sigma_{t}),\leavevmode\nobreak\ \leavevmode\nobreak\ Q_{t,n|y}:=\mathcal{N}(\mu_{t,n|y},\Sigma_{t|y})$ $\displaystyle\mu_{t,n}$ $\displaystyle:=\sqrt{\bar{\alpha}_{t}}\mu_{0,n},\leavevmode\nobreak\ \leavevmode\nobreak\ \Sigma_{t}:=\bar{\alpha}_{t}\Sigma_{0}+(1-\bar{\alpha}_{t})I_{d},\leavevmode\nobreak\ \leavevmode\nobreak\ \mu_{t,n|y}:=\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mu_{0,n}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y$ $\displaystyle\Sigma_{t|y}$ $\displaystyle:=\Sigma_{t,sig}+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal},\leavevmode\nobreak\ \leavevmode\nobreak\ \Sigma_{t,sig}:=\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+(1-\bar{\alpha}_{t})I_{d}.$ (38) $$

Similar to ([37](#A8.E37)), we still have

$$ $\left\lVert\Sigma_{t-1|y}^{-1}\right\rVert\lesssim 1,\quad\forall t\geq 1.$ $$

We can also calculate the scores of $Q_{t}$ and $Q_{t|y}$ in as follows.

$$ $\displaystyle\nabla\log q_{t}(x_{t})$ $\displaystyle=-\frac{1}{q_{t}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t}^{-1}(x_{t}-\mu_{t,n}),$ $\displaystyle\nabla\log q_{t|y}(x_{t})$ $\displaystyle=-\frac{1}{q_{t|y}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n|y}(x_{t})\Sigma_{t|y}^{-1}(x_{t}-\mu_{t,n|y}).$ (39) $$

Now, with $f_{t,y}=f_{t,y}^{*}$ (in ([10](#S4.E10))), from the expression of $\Delta_{t,y}$ in ([32](#A8.E32)), under the assumption $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$, the score mismatch at each diffusion step is equal to

$$ $\displaystyle\Delta_{t,y}$ $\displaystyle=(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}-\nabla\log q_{t})$ $\displaystyle=\frac{1}{q_{t}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})(I_{d}-H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})$ $\displaystyle\quad-\frac{1}{q_{t|y}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n|y}(x_{t})(I_{d}-H^{\dagger}H)\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n|y})$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}\left(\frac{q_{t,n}(x_{t})}{q_{t}(x_{t})}-\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\right)(I_{d}-H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})$ $\displaystyle\quad+\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}(I_{d}-H^{\dagger}H)\left(\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})-\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n|y})\right)$ $\displaystyle=-\sqrt{\bar{\alpha}_{t}}\sum_{n=1}^{N}\pi_{n}\left(\frac{q_{t,n}(x_{t})}{q_{t}(x_{t})}-\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\right)(I_{d}-H^{\dagger}H)\Sigma_{t}^{-1}\mu_{0,n}$ $\displaystyle\quad+\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}(I_{d}-H^{\dagger}H)\left(\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})-\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n|y})\right)$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}-\sqrt{\bar{\alpha}_{t}}\sum_{n=1}^{N}\pi_{n}\left(\frac{q_{t,n}(x_{t})}{q_{t}(x_{t})}-\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\right)(I_{d}-H^{\dagger}H)\Sigma_{t}^{-1}\mu_{0,n}$ $\displaystyle\quad-\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}A_{t}\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})$ (40) $$

where $A_{t}:=(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)$. Here $(i)$ follows from similar arguments as in ([I.1](#A9.Ex609)). Note that since $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$, we have equivalently $A_{t}=(I_{d}-H^{\dagger}H)\Sigma_{t|y}^{-1}(I_{d}-H^{\dagger}H)$.
Since $H^{\dagger}H=\begin{pmatrix}I_{p}&0\\
0&0\end{pmatrix}$, we can also re-express the second term in $\Delta_{t,y}$ such that $[\Delta_{t,y}]_{y}=0$ and

$$ $\displaystyle[\Delta_{t,y}]_{\bar{y}}$ $\displaystyle=-\sqrt{\bar{\alpha}_{t}}\sum_{n=1}^{N}\pi_{n}\left(\frac{q_{t,n}(x_{t})}{q_{t}(x_{t})}-\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\right)[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}$ $\displaystyle\quad-\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})$ (41) $$

since when $H^{\dagger}H=\begin{pmatrix}I_{p}&0\\
0&0\end{pmatrix}$, $A_{t}=\begin{pmatrix}0&0\\
0&(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\end{pmatrix}$.

Now, for the second moment, we follow similar analyses in ([I.1](#A9.Ex623)) and get

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{2}$ $\displaystyle\leq 4\bar{\alpha}_{t}\mathbb{E}_{X_{t}\sim Q_{t|y}}\max_{n\in[N]}\left\lVert\Sigma_{t}^{-1}\mu_{0,n}\right\rVert^{2}$ $\displaystyle\quad+2\bar{\alpha}_{t}^{2}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert^{2}\left\lVert[\Sigma_{t}^{-1}]_{y:}\right\rVert^{2}\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}\times$ $\displaystyle\qquad\quad\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}\right]$ $$

where

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}\right]$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\mathbb{E}_{N\sim\Pi_{\cdot|t,y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{2}$ $\displaystyle=\mathbb{E}_{N\sim\Pi_{\cdot|y}}\mathbb{E}_{X_{t}\sim Q_{t,N|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\mathbb{E}_{N\sim\Pi_{\cdot|y}}\left[\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,N}\right\rVert^{2}\right]$ $\displaystyle=\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ $$

where $(ii)$ follows from ([I.1](#A9.Ex630)) and note that $Q_{t,N|y}$ is Gaussian for each $N=n$.
Denote $\lambda_{1}\geq\dots\geq\lambda_{d}>0$ and $\tilde{\lambda}_{1}\geq\dots\geq\tilde{\lambda}_{d-p}>0$ to be the eigenvalues of $\Sigma_{0}$ and $[\Sigma_{0}]_{\bar{y}\bar{y}}$, respectively. Similarly as the proof of [Proposition 2](#Thmproposition2), we have $\left\lVert[\Sigma_{0}]_{\bar{y}y}\right\rVert\leq\left\lVert\Sigma_{0}\right\rVert=\lambda_{1}$, $\left\lVert[\Sigma_{t}^{-1}]_{y:}\right\rVert\leq\left\lVert\Sigma_{t}^{-1}\right\rVert\leq(\bar{\alpha}_{t}\lambda_{d}+(1-\bar{\alpha}_{t}))^{-1}\leq\frac{1}{\min\{\lambda_{d},1\}}$, $\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert\leq\frac{1}{\bar{\alpha}_{t}\tilde{\lambda}_{d-p}+(1-\bar{\alpha}_{t})}\leq\frac{1}{\min\{\tilde{\lambda}_{d-p},1\}}$, and $\left\lVert\Sigma_{t|y}\right\rVert\leq\bar{\alpha}_{t}(\lambda_{1}+\sigma_{y}^{2})+(1-\bar{\alpha}_{t})$.
Therefore,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{2}$ $\displaystyle\lesssim\bar{\alpha}_{t}d+\bar{\alpha}_{t}^{2}\frac{\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}}{\min\{\tilde{\lambda}_{d-p},1\}^{2}\min\{\lambda_{d},1\}^{2}}\times$ $\displaystyle\qquad\left(d(1-\bar{\alpha}_{t})+\bar{\alpha}_{t}d(\lambda_{1}+\sigma_{y}^{2})+\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)$ $\displaystyle\lesssim\bar{\alpha}_{t}d+\bar{\alpha}_{t}^{2}\frac{\left\lVert[\Sigma_{0}]_{y\bar{y}}\right\rVert^{2}}{\min\{\tilde{\lambda}_{d-p},1\}^{2}\min\{\lambda_{d},1\}^{2}}\max\left\{d(\lambda_{1}+\sigma_{y}^{2})+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2},d\right\}.$ $$

The proof is complete.

### H.5 Proof of Theorem 5

We first recall all the notations in ([H.4](#A8.Ex414)) under Gaussian mixture.
We also recall the scores from ([H.4](#A8.Ex417)):

$$ $\displaystyle\nabla\log q_{t}(x_{t})$ $\displaystyle=-\frac{1}{q_{t}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t}^{-1}(x_{t}-\mu_{t,n})$ $\displaystyle\qquad=-\Sigma_{t}^{-1}x_{t}+\frac{1}{q_{t}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n}(x_{t})\Sigma_{t}^{-1}\mu_{t,n}$ $\displaystyle\nabla\log q_{t|y}(x_{t})$ $\displaystyle=-\frac{1}{q_{t|y}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n|y}(x_{t})\Sigma_{t|y}^{-1}(x_{t}-\mu_{t,n|y})$ $\displaystyle\qquad=-\Sigma_{t|y}^{-1}x_{t}+\frac{1}{q_{t|y}(x_{t})}\sum_{n=1}^{N}\pi_{n}q_{t,n|y}(x_{t})\Sigma_{t|y}^{-1}\mu_{t,n|y}$ $$

Also, we recall the explicit expression of $\Delta_{t,y}$ from ([H.4](#A8.Ex426)), such that $[\Delta_{t,y}]_{y}=0$ and

$$ $\displaystyle[\Delta_{t,y}]_{\bar{y}}$ $\displaystyle=-\sqrt{\bar{\alpha}_{t}}\sum_{n=1}^{N}\pi_{n}\left(\frac{q_{t,n}(x_{t})}{q_{t}(x_{t})}-\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\right)[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}$ $\displaystyle\quad-\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}).$ $$

In order to invoke [Theorem 1](#Thmtheorem1), we need to check [Assumptions 3](#Thmassumption3) and [4](#Thmassumption4). From [, Lemmas 13 and 14], since $\left\lVert\Sigma_{t|y}^{-1}\right\rVert\lesssim 1$ for all $t\geq 0$, the absolute values of any-order partial derivative are bounded by $O(1)$ in expectation, and thus [Assumption 3](#Thmassumption3) is satisfied.
The following lemma verifies [Assumption 4](#Thmassumption4) using the $\alpha_{t}$ in [Definition 1](#Thmdefinition1).

###### Lemma 8 .

Under the same condition of [Theorem 5](#Thmtheorem5), [Assumption 4](#Thmassumption4) holds if the $\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1).

###### Proof.

See [Section I.2](#A9.SS2).
∎

Now we start to upper-bound the conditional KL-divergence of interest.
Recall that from [Theorem 1](#Thmtheorem1), $\mathrm{KL}(Q_{0|y}\|\widehat{P}_{0|y})\lesssim\mathcal{W}_{\text{oracle}}+\mathcal{W}_{\text{bias}}+\mathcal{W}_{\text{vanish}}$,
where

$$ $\displaystyle\mathcal{W}_{\text{oracle}}$ $\displaystyle=\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\mathrm{Tr}\Big{(}\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\nabla^{2}\log q_{t|y}(X_{t})\Big{)}\right]+(\log T)\varepsilon^{2}$ $\displaystyle\mathcal{W}_{\text{bias}}$ $\displaystyle=\sum_{t=1}^{T}(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}$ $\displaystyle\mathcal{W}_{\text{vanish}}$ $\displaystyle=\sum_{t=1}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(\nabla\log q_{t-1|y}(m_{t,y}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(X_{t}))^{\intercal}\Delta_{t,y}(X_{t})\bigg{]}$ $\displaystyle-\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\Delta_{t,y}(X_{t})^{\intercal}\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})\right]$ $\displaystyle+\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}3\sum_{i=1}^{d}\partial^{3}_{iii}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{i}$ $\displaystyle\qquad+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{j}\bigg{]}$ $\displaystyle+\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}(\log T)\varepsilon.$ $$

From [, Theorem 2] (and by assumption $N\leq d$), if the $\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1),

$$ $\mathcal{W}_{\text{oracle}}\lesssim\frac{d^{2}(\log T)^{2}}{T}+(\log T)\varepsilon^{2}.$ $$

Also, from [Proposition 1](#Thmproposition1), under the assumption on $\alpha_{t}$,
$\mathcal{W}_{\text{bias}}$ can be upper-bounded as

$$ $\mathcal{W}_{\text{bias}}\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}.$ $$

Among the terms in $\mathcal{W}_{\text{vanish}}$, the last estimation error term can be upper-bounded using [Proposition 1](#Thmproposition1) as

$$ $\max_{t\geq 1}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}(\log T)\varepsilon\lesssim\sqrt{d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}}(\log T)\varepsilon.$ $$

It remains to analyze the rest of the terms in $\mathcal{W}_{\text{vanish}}$.
In the following we fix $t\geq 1$.
We remind readers of the notations in ([H.4](#A8.Ex414)).
For the first term in $\mathcal{W}_{\text{vanish}}$, we first provide the following useful calculations. Note that by exchanging the order of expectation, for any function $\mathrm{fn}$ we have

$$ $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\mathrm{fn}(X_{t},n)\right]=\sum_{n=1}^{N}\pi_{n}\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\mathrm{fn}(X_{t},n).$ $$

Thus,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\left|(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})^{\intercal}X_{t}\right|$ $\displaystyle=\sum_{n=1}^{N}\pi_{n}\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left|(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})^{\intercal}X_{t}\right|$ $\displaystyle\leq\sum_{n=1}^{N}\pi_{n}\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}+\sqrt{\bar{\alpha}_{t}}\sum_{n=1}^{N}\pi_{n}\left\lVert\mu_{0,n}\right\rVert\sqrt{\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ $\displaystyle\qquad+\sqrt{\bar{\alpha}_{t}}\sum_{n=1}^{N}\pi_{n}\left\lVert\mu_{0,n}\right\rVert\sqrt{\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}}$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ (42) $$

where $(i)$ follows from ([I.1](#A9.Ex630)).

Also, note that $(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-r}(H^{\dagger}H)=0$ using the following simple induction argument. For the base case, we have $(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-1}(H^{\dagger}H)=0$ from [Lemma 9](#Thmlemma9). Then, suppose $(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-(r-1)}(H^{\dagger}H)=0$, we have $(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-r}(H^{\dagger}H)=(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-(r-1)}(I_{d}-H^{\dagger}H+H^{\dagger}H)\Sigma_{t-1|y}^{-1}(H^{\dagger}H)=(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-(r-1)}(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-1}(H^{\dagger}H)+(I_{d}-H^{\dagger}H)\Sigma_{t-1|y}^{-(r-1)}(H^{\dagger}H)\Sigma_{t-1|y}^{-1}(H^{\dagger}H)=0$.
Thus, for all $r\geq 1$ and any fixed vector $v$, with the definition of $\Delta_{t,y}$ in ([H.4](#A8.Ex418)) and ([H.4](#A8.Ex426)),

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t-1|y}^{-r}v)^{\intercal}\Delta_{t,y}\right|$ $\displaystyle\leq\left\lVert v\right\rVert\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Sigma_{t-1|y}^{-r}\Delta_{t,y}\right\rVert=\left\lVert v\right\rVert\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert A_{t-1}^{r}\Delta_{t,y}\right\rVert$ $\displaystyle\leq 4\left\lVert v\right\rVert\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-r}\right\rVert\max_{n\in[N]}\left\lVert[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}\right\rVert$ $\displaystyle\quad+2\left\lVert v\right\rVert\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-r}\right\rVert\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}\right\rVert\times$ $\displaystyle\qquad\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}}$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ (43) $$

where the last line follows from ([I.1](#A9.Ex630)). Similarly,

$$ $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t|y}^{-r}v)^{\intercal}\Delta_{t,y}\right|\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}.$ $$

Also, for all $r\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t-1|y}^{-r}X_{t})^{\intercal}\Delta_{t,y}\right|=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|X_{t}^{\intercal}\Sigma_{t-1|y}^{-r}(I_{d}-H^{\dagger}H)\Delta_{t,y}\right|$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|X_{t}^{\intercal}A_{t-1}^{r}\Delta_{t,y}\right|$ $\displaystyle\leq 4\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-r}\right\rVert\max_{n\in[N]}\left\lVert[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}\right\rVert\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}\right\rVert$ $\displaystyle\quad+2\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-r}\right\rVert\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}\right\rVert\times$ $\displaystyle\qquad\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\left|(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})^{\intercal}X_{t}\right|$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ (44) $$

where the last line follows from ([H.5](#A8.Ex457)) and the fact that, from ([I.1](#A9.Ex630)),

$$ $\displaystyle\sqrt{d}\cdot\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}\right\rVert$ $\displaystyle\leq\sqrt{d}\sum_{n=1}^{N}\pi_{n}\sqrt{2\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\mu_{t,n}\right\rVert^{2}+2\left\lVert\mu_{t,n}\right\rVert^{2}}$ $\displaystyle\lesssim\sqrt{d}\sum_{n=1}^{N}\pi_{n}\sqrt{\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}+d}$ $\displaystyle\lesssim d+\sqrt{d}\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}.$ $$

Similarly, we also have $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t|y}^{-r}X_{t})^{\intercal}\Delta_{t,y}\right|\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$.

Also, for all $r\geq 1$, using the expression of $\nabla\log q_{t|y}$ in ([H.4](#A8.Ex417)), and noting that by definition $(I_{d}-H^{\dagger}H)(x_{t}-\mu_{t,n|y})=(I_{d}-H^{\dagger}H)(x_{t}-\mu_{t,n})$, we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t-1|y}^{-r}\nabla\log q_{t|y}(X_{t}))^{\intercal}\Delta_{t,y}\right|$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\left(A_{t-1}^{r}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}A_{t}(I_{d}-H^{\dagger}H)(X_{t}-\mu_{t,n|y})\right)^{\intercal}\Delta_{t,y}\right|$ $\displaystyle\leq 4\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-r}\right\rVert\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert\times$ $\displaystyle\quad\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\sum_{\ell=1}^{N}\pi_{\ell}\frac{q_{t,\ell|y}(X_{t})}{q_{t|y}(X_{t})}[X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,\ell}]_{\bar{y}}\right\rVert\times\max_{n\in[N]}\left\lVert[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}\right\rVert$ $\displaystyle\quad+2\left\lVert(\bar{\alpha}_{t-1}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t-1})I_{d-p})^{-r}\right\rVert\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert^{2}\times$ $\displaystyle\quad\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N,L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\bigg{[}[X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,L}]_{\bar{y}}^{\intercal}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N})\bigg{]}$ $\displaystyle\lesssim\sqrt{\sum_{n=1}^{N}\pi_{n}\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}}\times\sqrt{d}$ $\displaystyle\quad+\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N,L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,L}\right\rVert\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ (45) $$

where the last line follows because, from ([I.1](#A9.Ex630)),

$$ $\displaystyle\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N,L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left[\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,L}\right\rVert\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert\right]$ $\displaystyle\leq\sqrt{\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,L}\right\rVert^{2}}\times\sqrt{\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{2}}$ $\displaystyle=\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{2}$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}.$ $$

Now, we start to analyze the first term of $\mathcal{W}_{\text{vanish}}$. Recall that $m_{t,y}(x_{t})=\mathbb{E}_{X_{t-1}\sim Q_{t-1|t,y}}[X_{t-1}]=\frac{1}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t|y}(x_{t})$. Using the score expressions in ([H.4](#A8.Ex417)), we can calculate that given $x_{t}$ (and thus $m_{t,y}=m_{t,y}(x_{t})$),

$$ $\displaystyle\nabla\log q_{t-1|y}(m_{t,y})-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(x_{t})$ $\displaystyle=\sqrt{\alpha_{t}}\Sigma_{t|y}^{-1}x_{t}-\Sigma_{t-1|y}^{-1}m_{t,y}$ $\displaystyle\qquad-\sqrt{\alpha_{t}}\Sigma_{t|y}^{-1}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(x_{t})}{q_{t|y}(x_{t})}\mu_{t,n|y}+\Sigma_{t-1|y}^{-1}\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\mu_{t-1,n|y}$ $\displaystyle=\left(\Sigma_{t|y}^{-1}-\Sigma_{t-1|y}^{-1}\right)\sqrt{\alpha_{t}}x_{t}-\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\Sigma_{t-1|y}^{-1}(x_{t}+\nabla\log q_{t|y}(x_{t}))$ $\displaystyle\qquad+(1-\alpha_{t})\Sigma_{t-1|y}^{-1}\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\mu_{t-1,n|y}$ $\displaystyle\qquad+\alpha_{t}\frac{\sum_{n,\ell=1}^{N}\pi_{n}\pi_{\ell}\left(q_{t-1,n|y}(m_{t,y})q_{t,\ell|y}(x_{t})\Sigma_{t-1|y}^{-1}-q_{t-1,\ell|y}(m_{t,y})q_{t,n|y}(x_{t})\Sigma_{t|y}^{-1}\right)\mu_{t-1,n|y}}{q_{t-1|y}(m_{t,y})q_{t|y}(x_{t})}.$ $$

Here, using similar analyses as in the proof of [Lemma 5](#Thmlemma5), we get

$$ $\displaystyle(m_{t,y}-\mu_{t-1,n|y})-(x_{t}-\mu_{t,n|y})=\frac{1-\sqrt{\alpha_{t}}}{\sqrt{\alpha_{t}}}x_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t|y}(x_{t})-(1-\sqrt{\alpha_{t}})\mu_{t-1,n|y}$ $\displaystyle\Sigma_{t|y}^{-1}-\Sigma_{t-1|y}^{-1}=\frac{1-\alpha_{t}}{\alpha_{t}}\Sigma_{t-1,n}^{-1}+\frac{1-\alpha_{t}}{\alpha_{t}^{2}}\Sigma_{t-1,n}^{-2}+O((1-\alpha_{t})^{2})$ $\displaystyle q_{t-1,n|y}(m_{t,y})q_{t,\ell|y}(x_{t})\Sigma_{t-1|y}^{-1}-q_{t-1,\ell|y}(m_{t,y})q_{t,n|y}(x_{t})\Sigma_{t|y}^{-1}$ $\displaystyle=q_{t-1,n|y}(m_{t,y})q_{t,\ell|y}(x_{t})(\Sigma_{t-1|y}^{-1}-\Sigma_{t|y}^{-1})$ $\displaystyle\qquad+(q_{t-1,n|y}(m_{t,y})q_{t,\ell|y}(x_{t})-q_{t-1,\ell|y}(m_{t,y})q_{t,n|y}(x_{t}))\Sigma_{t|y}^{-1}$ $\displaystyle=q_{t-1,n|y}(m_{t,y})q_{t,\ell|y}(x_{t})(\Sigma_{t-1|y}^{-1}-\Sigma_{t|y}^{-1})$ $\displaystyle\qquad+\bigg{(}\frac{1}{2}((m_{t,y}-\mu_{t-1,\ell|y})-(x_{t}-\mu_{t,\ell|y}))^{\intercal}\Sigma_{t-1|y}^{-1}(m_{t,y}-\mu_{t-1,\ell|y})$ $\displaystyle\quad\qquad+\frac{1}{2}(x_{t}-\mu_{t,\ell|y})^{\intercal}(\Sigma_{t-1|y}^{-1}-\Sigma_{t|y}^{-1})(m_{t,y}-\mu_{t-1,\ell|y})$ $\displaystyle\quad\qquad+\frac{1}{2}(x_{t}-\mu_{t,\ell|y})^{\intercal}\Sigma_{t|y}^{-1}((m_{t,y}-\mu_{t-1,\ell|y})-(x_{t}-\mu_{t,\ell|y}))$ $\displaystyle\quad\qquad-\frac{1}{2}((m_{t,y}-\mu_{t-1,n|y})-(x_{t}-\mu_{t,n|y}))^{\intercal}\Sigma_{t-1|y}^{-1}(m_{t,y}-\mu_{t-1,n|y})$ $\displaystyle\quad\qquad-\frac{1}{2}(x_{t}-\mu_{t,n|y})^{\intercal}(\Sigma_{t-1|y}^{-1}-\Sigma_{t|y}^{-1})(m_{t,y}-\mu_{t-1,n|y})$ $\displaystyle\quad\qquad-\frac{1}{2}(x_{t}-\mu_{t,n|y})^{\intercal}\Sigma_{t|y}^{-1}((m_{t,y}-\mu_{t-1,n|y})-(x_{t}-\mu_{t,n|y}))\bigg{)}\Sigma_{t|y}^{-1}$ $\displaystyle\qquad+O((1-\alpha_{t})^{2})$ $$

Thus,

$$ $\displaystyle\left|\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(\nabla\log q_{t-1|y}(m_{t,y})-\sqrt{\alpha_{t}}\nabla\log q_{t|y})^{\intercal}\Delta_{t,y}\bigg{]}\right|$ $\displaystyle\leq\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}\left|X_{t}^{\intercal}(\Sigma_{t|y}^{-1}-\Sigma_{t-1|y}^{-1})\Delta_{t,y}\right|\bigg{]}$ $\displaystyle\quad+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}\left|(X_{t}+\nabla\log q_{t|y})^{\intercal}(\Sigma_{t-1|y}^{-1})\Delta_{t,y}\right|\bigg{]}$ $\displaystyle\quad+(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}\left|\left(\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\mu_{t-1,n|y}\right)^{\intercal}\Sigma_{t-1|y}^{-1}\Delta_{t,y}\right|\bigg{]}$ $\displaystyle\quad+\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}^{\intercal}\sum_{n,\ell=1}^{N}\pi_{n}\pi_{\ell}\left(\frac{q_{t-1,n|y}(m_{t,y})q_{t,\ell|y}(X_{t})}{q_{t-1|y}(m_{t,y})q_{t|y}(X_{t})}\Sigma_{t-1|y}^{-1}-\frac{q_{t-1,\ell|y}(m_{t,y})q_{t,n|y}(X_{t})}{q_{t-1|y}(m_{t,y})q_{t|y}(X_{t})}\Sigma_{t|y}^{-1}\right)\mu_{t-1,n|y}\right|.$ $$

Among the four terms above, the first term $\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ from ([H.5](#A8.Ex468)) (along with the similar result for $\Sigma_{t|y}^{-1}$), the second term $\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ from ([H.5](#A8.Ex468)) and ([H.5](#A8.Ex477)), and both the third and the fourth term $\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ from ([H.5](#A8.Ex462)) (along with the similar result for $\Sigma_{t|y}^{-1}$). Thus,

$$ $\left|\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(\nabla\log q_{t-1|y}(m_{t,y})-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(X_{t}))^{\intercal}\Delta_{t,y}\bigg{]}\right|\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2},$ $$

and the first term in $\mathcal{W}_{\text{vanish}}$ satisfies that

$$ $\sum_{t=1}^{T}\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}(\nabla\log q_{t-1|y}(m_{t,y}(X_{t}))-\sqrt{\alpha_{t}}\nabla\log q_{t|y}(X_{t}))^{\intercal}\Delta_{t,y}(X_{t})\bigg{]}\\ \lesssim\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)\frac{\log(1/\delta)^{2}(\log T)^{2}}{T}.$ $$

For the second term in $\mathcal{W}_{\text{vanish}}$, we first provide the following useful calculation.
Similar to ([H.5](#A8.Ex462)), for all $r\geq 1$ and any fixed vector $v$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t|y}^{-r}v)^{\intercal}\Delta_{t,y}\right|^{2}$ $\displaystyle\leq\left\lVert v\right\rVert^{2}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Sigma_{t|y}^{-r}\Delta_{t,y}\right\rVert^{2}=\left\lVert v\right\rVert^{2}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert A_{t}^{r}\Delta_{t,y}\right\rVert^{2}$ $\displaystyle\lesssim\left\lVert v\right\rVert^{2}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-r}\right\rVert^{2}\max_{n\in[N]}\left\lVert[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}\right\rVert^{2}$ $\displaystyle\quad+\left\lVert v\right\rVert^{2}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-r}\right\rVert^{2}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}\right\rVert^{2}\times$ $\displaystyle\qquad\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{2}$ $\displaystyle\lesssim d^{2}+d\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ (46) $$

where the last line follows from ([I.1](#A9.Ex630)).

Also, similar to ([H.5](#A8.Ex468)), for all $r\geq 1$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\Sigma_{t|y}^{-r}m_{t,y})^{\intercal}\Delta_{t,y}\right|^{2}$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|X_{t}^{\intercal}\Sigma_{t|y}^{-r}\Delta_{t,y}\right|^{2}+(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\nabla\log q_{t|y}(X_{t}))^{\intercal}\Sigma_{t|y}^{-r}\Delta_{t,y}\right|^{2}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|X_{t}^{\intercal}\Sigma_{t|y}^{-r}(I_{d}-H^{\dagger}H)\Delta_{t,y}\right|^{2}=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|X_{t}^{\intercal}A_{t}^{r}\Delta_{t,y}\right|^{2}$ $\displaystyle\lesssim\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-r}\right\rVert^{2}\max_{n\in[N]}\left\lVert[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,n}\right\rVert^{2}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}\right\rVert^{2}$ $\displaystyle\quad+\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-r}\right\rVert^{2}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}\right\rVert^{2}\times$ $\displaystyle\qquad\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\left|(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})^{\intercal}X_{t}\right|^{2}$ $\displaystyle\lesssim d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}$ (47) $$

where $(ii)$ follows from the fact that $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(\nabla\log q_{t|y}(X_{t}))^{\intercal}\Sigma_{t|y}^{-r}\Delta_{t,y}\right|^{2}\lesssim d^{2}$ (using a similar argument for deriving ([H.5](#A8.Ex477))), and the last line follows because

$$ $\displaystyle d\cdot\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}\right\rVert^{2}$ $\displaystyle\leq d\sum_{n=1}^{N}\pi_{n}\left(2\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\mu_{t,n}\right\rVert^{2}+2\left\lVert\mu_{t,n}\right\rVert^{2}\right)$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\lesssim}}d\sum_{n=1}^{N}\pi_{n}\left(\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}+d\right)$ $\displaystyle\lesssim d^{2}+d\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}$ $$

where $(iii)$ follows from ([I.1](#A9.Ex630)), and also

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t,n|y}(X_{t})}{q_{t|y}(X_{t})}\left|(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n})^{\intercal}X_{t}\right|^{2}$ $\displaystyle=\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left((X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N})^{\intercal}X_{t}\right)^{2}$ $\displaystyle\leq 2\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{4}+2\sqrt{\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{4}}\sqrt{\mathbb{E}_{N\sim\Pi}\left\lVert\mu_{0,N}\right\rVert^{4}}$ $\displaystyle\lesssim d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}$ $$

where the last line above follows because for all $r\geq 1$ and each $n\in[N]$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{r}$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n|y}\right\rVert^{r}+\left\lVert\sqrt{\bar{\alpha}_{t}}\mu_{0,n|y}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n}\right\rVert^{r}$ $\displaystyle\leq\left\lVert\Sigma_{t|y}^{\frac{1}{2}}\right\rVert^{r}\mathbb{E}_{X_{t}\sim Q_{t,n|y}}\left\lVert\Sigma_{t|y}^{-\frac{1}{2}}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,n|y})\right\rVert^{r}+(\bar{\alpha}_{t})^{r/2}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{r}$ $\displaystyle\lesssim d^{r/2}+\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{r}.$ (48) $$

Now we are ready to analyze the second term of $\mathcal{W}_{\text{vanish}}$. Note that

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

Thus,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}^{\intercal}\nabla^{2}\log q_{t-1|y}(m_{t,y})\Delta_{t,y}\right|$ $\displaystyle\leq 3\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\left(\Delta_{t,y}^{\intercal}\Sigma_{t|y}^{-1}(m_{t,y}-\mu_{t,n|y})\right)^{2}\right]$ $\displaystyle\quad+3\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|$ $\displaystyle\quad+3\mathbb{E}_{X_{t}\sim Q_{t|y}}\left(\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\Delta_{t,y}^{\intercal}\Sigma_{t|y}^{-1}(m_{t,y}-\mu_{t,n|y})\right)^{2}$ $\displaystyle\leq 3\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|$ $\displaystyle\quad+6\mathbb{E}_{X_{t}\sim Q_{t|y}}\left[\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\left(\Delta_{t,y}^{\intercal}\Sigma_{t|y}^{-1}(m_{t,y}-\mu_{t,n|y})\right)^{2}\right].$ $$

To determine the rate of these two terms, we get

$$ $\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}^{\intercal}A_{t}\Delta_{t,y}\right|\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2},$ $$

and

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{n=1}^{N}\pi_{n}\frac{q_{t-1,n|y}(m_{t,y})}{q_{t-1|y}(m_{t,y})}\left|(m_{t,y}-\mu_{t,n|y})^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|^{2}$ $\displaystyle\qquad\leq\mathbb{E}_{X_{t}\sim Q_{t|y}}\max_{n\in[N]}\left|(m_{t,y}-\mu_{t,n|y})^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|^{2}$ $\displaystyle\qquad\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|m_{t,y}^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|^{2}+\mathbb{E}_{X_{t}\sim Q_{t|y}}\max_{n\in[N]}\left|\mu_{t,n|y}^{\intercal}\Sigma_{t|y}^{-1}\Delta_{t,y}\right|^{2}$ $\displaystyle\qquad\lesssim N\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)$ $$

where the last line follows from ([H.5](#A8.Ex514)) and ([H.5](#A8.Ex519)).
Thus, the second term of $\mathcal{W}_{\text{vanish}}$ satisfies that

$$ $\displaystyle\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{2\alpha_{t}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\Delta_{t,y}(X_{t})^{\intercal}\nabla^{2}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})\right|$ $\displaystyle\lesssim N\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)\frac{c^{2}(\log T)^{2}}{T}.$ $$

For the third term of $\mathcal{W}_{\text{vanish}}$, we provide the following useful calculations. Denote $v^{\circ 3}$ as the element-wise (Hadamard) third power of a vector $v$. For each $n\in[N]$, we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(m_{t,y}-\mu_{t,n|y})^{\circ 3}(I_{d}-H^{\dagger}H)\Delta_{t,y}\right|=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|[\Delta_{t,y}]_{\bar{y}}^{\intercal}[m_{t,y}-\mu_{t,n}]^{\circ 3}_{\bar{y}}\right|$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|[\Delta_{t,y}]_{\bar{y}}^{\intercal}[X_{t}-\mu_{t,n}]^{\circ 3}_{\bar{y}}\right|+(1-\alpha_{t})\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{2}}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\nabla\log q_{t|y}(X_{t})\right\rVert_{6}^{6}}$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{\lesssim}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|[\Delta_{t,y}]_{\bar{y}}^{\intercal}[X_{t}-\mu_{t,n}]^{\circ 3}_{\bar{y}}\right|$ $\displaystyle\lesssim\max_{\ell}\left\lVert[\Sigma_{t}^{-1}]_{\bar{y}:}\mu_{0,\ell}\right\rVert\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}-\mu_{t,n}\right\rVert_{6}^{6}}+\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\mu_{t,L}\right\rVert_{4}^{4}$ $\displaystyle\lesssim\sqrt{\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\mu_{t,L}\right\rVert^{6}}+\sqrt{\mathbb{E}_{L\sim\Pi}\left\lVert\mu_{t,n}-\mu_{t,L}\right\rVert^{6}}+\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\mu_{t,L}\right\rVert^{4}$ $\displaystyle\lesssim d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}$ (49) $$

where $(iv)$ follows from [Lemma 6](#Thmlemma6) (using the $\alpha_{t}$ in ([8](#S3.E8))) and [, Lemma 15], and the last line follows from ([H.5](#A8.Ex532)). With a similar argument,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(m_{t,y}-\mu_{t,n|y})(I_{d}-H^{\dagger}H)\Delta_{t,y}\right|$ $\displaystyle\lesssim\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\mu_{t,L}\right\rVert^{2}$ $\displaystyle\lesssim d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}.$ (50) $$

Now, employing the notations from [, Section G.1], we define

$$ $z_{t,n}(x):=\Sigma_{t|y}^{-1}(x-\mu_{t,n|y}),\quad\xi_{t}(x,i):=\max_{n}\left|z_{t,n}^{i}(x)\right|,\quad\bar{\Sigma}_{t}^{ij}:=\max_{n}\left|[\Sigma_{t|y}^{-1}]^{ij}\right|.$ $$

When $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$, we note that

$$ $\Sigma_{t|y}^{-1}=\begin{pmatrix}(1-\bar{\alpha}_{t}+\bar{\alpha}_{t}\sigma_{y}^{2})^{-1}I_{p}&0\\ 0&(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\end{pmatrix}.$ $$

Thus, we have $\bar{\Sigma}_{t}^{ij}\equiv 0$ whenever $(i,j)\in[1,p]\times[p+1,d]$ or $(i,j)\in[p+1,d]\times[1,p]$ and $\max_{i,j\in[p+1,d]}\bar{\Sigma}_{t}^{ij}=O(1)$.
Since $\sigma_{y}^{2}>0$, we also have $\left\lVert\Sigma_{t|y}^{-1}\right\rVert\lesssim 1$.

From [, Section G.1.2], an upper bound for third-order partial derivatives is

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

We also remind readers that $\Delta_{t,y}$ is supported on $\mathrm{range}(I_{d}-H^{\dagger}H)$, namely that $[\Delta_{t,y}]_{y}\equiv 0$.

Now, the third term of $\mathcal{W}_{\text{vanish}}$ can be upper-bounded as

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=1}^{d}\partial^{3}_{iii}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{i}\right|$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=p+1}^{d}\partial^{3}_{iii}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{i}\right|$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=p+1}^{d}\xi_{t}(m_{t,y}(X_{t}),i)^{3}\Delta_{t,y}(X_{t})^{i}\right|+\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=p+1}^{d}\bar{\Sigma}_{t}^{ii}\xi_{t}(m_{t,y}(X_{t}),i)\Delta_{t,y}(X_{t})^{i}\right|$ $\displaystyle\lesssim\sum_{n=1}^{N}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert^{3}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=p+1}^{d}(m_{t,y}(X_{t})^{i}-\mu_{t,n|y}^{i})^{3}\Delta_{t,y}(X_{t})^{i}\right|$ $\displaystyle\quad+\sum_{n=1}^{N}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=p+1}^{d}(m_{t,y}(X_{t})^{i}-\mu_{t,n|y}^{i})\Delta_{t,y}(X_{t})^{i}\right|$ $\displaystyle\lesssim N\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right).$ $$

Here the last line follows from ([H.5](#A8.Ex551)) and ([H.5](#A8.Ex556)).

We provide the following useful calculations to upper-bound the fourth term of $\mathcal{W}_{\text{vanish}}$. First, for all $r\geq 1$ and any fixed vector $v$,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert m_{t,y}-v\right\rVert^{r}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\frac{1}{\sqrt{\alpha_{t}}}X_{t}+\frac{1-\alpha_{t}}{\sqrt{\alpha_{t}}}\nabla\log q_{t|y}(X_{t})-v\right\rVert^{r}$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}-v\right\rVert^{r}+(1-\alpha_{t})\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\nabla\log q_{t|y}(X_{t})\right\rVert^{r}$ $\displaystyle\stackrel{{\scriptstyle(v)}}{{\lesssim}}\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ L\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\mu_{t,L}\right\rVert^{r}+\mathbb{E}_{L\sim\Pi}\left\lVert\mu_{t,L}-v\right\rVert^{r}$ $\displaystyle\lesssim d^{r/2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{r}$ (51) $$

where $(v)$ follows from [Lemma 6](#Thmlemma6) (using the $\alpha_{t}$ in ([8](#S3.E8))) and [, Lemma 15], and the last line follows from ([H.5](#A8.Ex532)).
Now,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{i=1}^{d}\xi_{t}(m_{t,y},i)^{2}\left|\sum_{j=p+1}^{d}\xi_{t}(m_{t,y},j)\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\lesssim\sum_{n,\ell=1}^{N}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert m_{t,y}-\mu_{t,n|y}\right\rVert^{2}\left|(m_{t,y}-\mu_{t,\ell|y})^{\intercal}\Delta_{t,y}(X_{t})\right|$ $\displaystyle\leq\sum_{n,\ell=1}^{N}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert m_{t,y}-\mu_{t,n|y}\right\rVert^{4}}\left(\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert m_{t,y}-\mu_{t,\ell|y}\right\rVert^{4}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{4}\right)^{1/4}$ $\displaystyle\lesssim N^{2}\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)$ (52) $$

where the first inequality follows from $\left\lVert\Sigma_{t|y}^{-1}\right\rVert\lesssim 1$ and the last line follows from ([H.5](#A8.Ex566)) and ([I.2](#A9.Ex642)).
Also,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i,j=p+1}^{d}\bar{\Sigma}_{t}^{ii}\xi_{t}(m_{t,y},j)\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle=\left|\sum_{i=p+1}^{d}\bar{\Sigma}_{t}^{ii}\right|\cdot\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{j=p+1}^{d}\xi_{t}(m_{t,y},j)\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\lesssim\left|\sum_{i=p+1}^{d}\bar{\Sigma}_{t}^{ii}\right|\cdot\sum_{n=1}^{N}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|(m_{t,y}-\mu_{t,n})^{\intercal}\Delta_{t,y}(X_{t})\right|$ $\displaystyle\lesssim\left|\sum_{i=p+1}^{d}\bar{\Sigma}_{t}^{ii}\right|\cdot\sum_{n=1}^{N}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert m_{t,y}-\mu_{t,n}\right\rVert^{2}}\sqrt{\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}$ $\displaystyle\lesssim Nd\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)$ (53) $$

where the last line follows from [Proposition 1](#Thmproposition1) and ([H.5](#A8.Ex566)).
Also,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i,j=p+1}^{d}\bar{\Sigma}_{t}^{ij}\xi_{t}(m_{t,y},i)\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\left(\sum_{i=p+1}^{d}\xi_{t}(m_{t,y},i)\right)\left|\sum_{j=p+1}^{d}\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\leq\sqrt{d\cdot\mathbb{E}_{X_{t}\sim Q_{t|y}}\left(\sum_{i=p+1}^{d}\xi_{t}(m_{t,y},i)^{2}\right)}\sqrt{d\cdot\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}$ $\displaystyle\lesssim\sqrt{d\cdot\sum_{n=1}^{N}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert m_{t,y}-\mu_{t,n}\right\rVert^{2}}\sqrt{d\cdot\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}(X_{t})\right\rVert^{2}}$ $\displaystyle\lesssim\sqrt{N}d\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)$ (54) $$

where the last line follows from [Proposition 1](#Thmproposition1) and ([H.5](#A8.Ex566)).

Now, the fourth term of $\mathcal{W}_{\text{vanish}}$ can be upper-bounded as

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i,j=1}^{d}\partial^{3}_{iij}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i=1}^{d}\sum_{j=p+1}^{d}\partial^{3}_{iij}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\lesssim\mathbb{E}_{X_{t}\sim Q_{t|y}}\sum_{i=1}^{d}\xi_{t}(m_{t,y}(X_{t}),i)^{2}\left|\sum_{j=p+1}^{d}\xi_{t}(m_{t,y}(X_{t}),j)\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\quad+\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i,j=p+1}^{d}\bar{\Sigma}_{t}^{ii}\xi_{t}(m_{t,y}(X_{t}),j)\Delta_{t,y}(X_{t})^{j}\right|$ $\displaystyle\quad+\mathbb{E}_{X_{t}\sim Q_{t|y}}\left|\sum_{i,j=p+1}^{d}\bar{\Sigma}_{t}^{ij}\xi_{t}(m_{t,y}(X_{t}),i)\Delta_{t,y}(X_{t})^{j}\right|.$ $$

For the three terms above, the first term $\lesssim N^{2}\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)$ from ([H.5](#A8.Ex569)), the second term $\lesssim Nd\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)$ from ([H.5](#A8.Ex572)), and the last term $\lesssim\sqrt{N}d\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)$ from ([H.5](#A8.Ex576)).

Thus, overall, with the $\alpha_{t}$ in ([8](#S3.E8)) (cf. [Lemma 6](#Thmlemma6)), the third and fourth terms give us

$$ $\displaystyle\sum_{t=1}^{T}\frac{(1-\alpha_{t})^{2}}{3!{\alpha_{t}}^{3/2}}\mathbb{E}_{X_{t}\sim Q_{t|y}}\bigg{[}3\sum_{i=1}^{d}\partial^{3}_{iii}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{i}$ $\displaystyle\qquad+\sum_{\begin{subarray}{c}i,j=1\\ i\neq j\end{subarray}}^{d}\partial^{3}_{iij}\log q_{t-1|y}(m_{t,y}(X_{t}))\Delta_{t,y}(X_{t})^{j}\bigg{]}$ $\displaystyle\lesssim N^{2}\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)\frac{(\log T)^{2}}{T}.$ $$

Therefore, combining all the above, since $N$ is constant,

$$ $\displaystyle\mathrm{KL}(Q_{0|y}\|\widehat{P}_{0|y})$ $\displaystyle\lesssim\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)$ $\displaystyle\quad+\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)\frac{(\log T)^{2}}{T}$ $\displaystyle\quad+\sqrt{d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}}(\log T)\varepsilon.$ $$

###### Remark 2 .

In case of general $\sigma_{y}\geq 0$, the same upper bound can be applied to $\left\lVert\Sigma_{t-1|y}^{-1}\right\rVert$ as detailed in [Remark 1](#Thmremark1). Thus, with the $\alpha_{t}$ in ([8](#S3.E8)), we similarly have

$$ $\mathcal{W}_{\text{oracle}}\lesssim\frac{d^{2}(\log T)^{2}\log(1/\delta)^{2}}{T}+(\log T)\varepsilon^{2}.$ $$

The rest of the proof is similar.
Combining with [Lemma 7](#Thmlemma7), we would finally obtain

$$ $\displaystyle\mathrm{KL}(Q_{1|y}\|\widehat{P}_{1|y})\lesssim\left(d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}\right)\left(1-\frac{2\log(1/\delta)\log T}{T}\right)$ $\displaystyle+\left(d^{2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{4}\right)\frac{(\log T)^{2}\log(1/\delta)^{2}}{T}+\sqrt{d+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{2}}(\log T)\varepsilon.$ $$

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

###### Lemma 8 .

###### Proof.

###### Remark 2 .

## Appendix I Auxiliary Lemmas and Proofs in Section 4

### I.1 Proof of Proposition 2

Given $Q_{0}=\mathcal{N}(\mu_{0},\Sigma_{0})$, from the conditional forward model in ([5](#S2.E5)), we can calculate

$$ $\displaystyle Q_{t}$ $\displaystyle=\mathcal{N}(\sqrt{\bar{\alpha}_{t}}\mu_{0},\bar{\alpha}_{t}\Sigma_{0}+(1-\bar{\alpha}_{t})I_{d})=:\mathcal{N}(\mu_{t},\Sigma_{t})$ $\displaystyle Q_{t|y}$ $\displaystyle=\mathcal{N}(\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mu_{0}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y,$ $\displaystyle\qquad\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}+(1-\bar{\alpha}_{t})I_{d}).$ $$

Note that when $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$ and $\sigma_{y}^{2}>0$, $q_{0|y}$ exists. Define

$$ $\displaystyle\mu_{t|y}$ $\displaystyle:=\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mu_{0}+\sqrt{\bar{\alpha}_{t}}H^{\dagger}y$ $\displaystyle\Sigma_{t,sig}$ $\displaystyle:=\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+(1-\bar{\alpha}_{t})I_{d}$ $\displaystyle\Sigma_{t|y}$ $\displaystyle:=\Sigma_{t,sig}+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}.$ (55) $$

Here $\Sigma_{t,sig}$ is the signal variance at time $t$, and $\Sigma_{t|y}$ is the total variance of the signal and the measurement noise. Note that when $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$, $[\Sigma_{t|y}^{-1}]_{\bar{y}\bar{y}}=[\Sigma_{t,sig}^{-1}]_{\bar{y}\bar{y}}$.
We also calculate the respective scores of $Q_{t}$ and $Q_{t|y}$:

$$ $\nabla\log q_{t}(x_{t})=-\Sigma_{t}^{-1}(x_{t}-\mu_{t}),\quad\nabla\log q_{t|y}(x_{t})=-\Sigma_{t|y}^{-1}(x_{t}-\mu_{t|y}).$ $$

Since $f_{t,y}=f_{t,y}^{*}$ (defined in ([10](#S4.E10))), from ([32](#A8.E32)), the bias at each time is equal to

$$ $\displaystyle\Delta_{t,y}$ $\displaystyle=(I_{d}-H^{\dagger}H)(\nabla\log q_{t|y}(x_{t})-\nabla\log q_{t}(x_{t}))$ $\displaystyle=(I_{d}-H^{\dagger}H)\left(\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})-\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})\right)$ $\displaystyle=(I_{d}-H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle\quad-(I_{d}-H^{\dagger}H)\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mu_{0}-\sqrt{\bar{\alpha}_{t}}H^{\dagger}y).$ $$

Now, define

$$ $\displaystyle V_{t}$ $\displaystyle:=(H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+(I_{d}-H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)+(H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)$ $\displaystyle A_{t}$ $\displaystyle:=(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)$ $$

Thus, we have $\Sigma_{t}=\Sigma_{t,sig}+\bar{\alpha}_{t}V_{t}$ and $\Sigma_{t|y}=\Sigma_{t,sig}+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}$. By Woodbury matrix identity, for any two matrices $A$ and $B$, their sum can be inversed as $(A+B)^{-1}=A^{-1}-A^{-1}B(A+B)^{-1}$. Thus, we get

$$ $\displaystyle\Sigma_{t}^{-1}$ $\displaystyle=(\Sigma_{t,sig}+\bar{\alpha}_{t}V_{t})^{-1}=\Sigma_{t,sig}^{-1}-\bar{\alpha}_{t}\Sigma_{t,sig}^{-1}V_{t}\Sigma_{t}^{-1}$ $\displaystyle\Sigma_{t|y}^{-1}$ $\displaystyle=(\Sigma_{t,sig}+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal})^{-1}=\Sigma_{t,sig}^{-1}-\bar{\alpha}_{t}\sigma_{y}^{2}\Sigma_{t,sig}^{-1}H^{\dagger}(H^{\dagger})^{\intercal}\Sigma_{t|y}^{-1}$ $\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\Sigma_{t,sig}^{-1}-\bar{\alpha}_{t}\sigma_{y}^{2}\Sigma_{t,sig}^{-1}(H^{\dagger}H)\Sigma_{t|y}^{-1}$ $$

where $(i)$ holds under assumption $H=\begin{pmatrix}I_{p}&0\end{pmatrix}$.
Thus,

$$ $\displaystyle\Delta_{t,y}$ $\displaystyle=(I_{d}-H^{\dagger}H)\left(\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})-\Sigma_{t|y}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0|y})\right)$ $\displaystyle=(I_{d}-H^{\dagger}H)\left(\Sigma_{t,sig}^{-1}-\bar{\alpha}_{t}\Sigma_{t,sig}^{-1}V_{t}\Sigma_{t}^{-1}\right)(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle\quad-(I_{d}-H^{\dagger}H)\left(\Sigma_{t,sig}^{-1}-\bar{\alpha}_{t}\sigma_{y}^{2}\Sigma_{t,sig}^{-1}(H^{\dagger}H)\Sigma_{t|y}^{-1}\right)(x_{t}-\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mu_{0}-\sqrt{\bar{\alpha}_{t}}H^{\dagger}y)$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}(I_{d}-H^{\dagger}H)\left(\Sigma_{t,sig}^{-1}-\bar{\alpha}_{t}\Sigma_{t,sig}^{-1}V_{t}\Sigma_{t}^{-1}\right)(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle\quad-(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}(I_{d}-H^{\dagger}H)\mu_{0}-\sqrt{\bar{\alpha}_{t}}H^{\dagger}y)$ $\displaystyle=-\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}V_{t}\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle\quad+(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}\Big{(}(I_{d}-H^{\dagger}H)(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})+H^{\dagger}H(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\Big{)}$ $\displaystyle\quad-(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}\Big{(}(I_{d}-H^{\dagger}H)(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})+((H^{\dagger}H)x_{t}-\sqrt{\bar{\alpha}_{t}}H^{\dagger}y)\Big{)}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{=}}-\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}V_{t}\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle\stackrel{{\scriptstyle(iv)}}{{=}}-\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ $\displaystyle=-\bar{\alpha}_{t}A_{t}\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})$ (56) $$

where $(ii)$–$(iv)$ hold because $(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(H^{\dagger}H)=0$ by [Lemma 9](#Thmlemma9).

Now, since $H^{\dagger}H=\begin{pmatrix}I_{p}&0\\
0&0\end{pmatrix}$, we can re-express $A_{t}$ and $\Delta_{t,y}$ as follows.

$$ $\displaystyle A_{t}$ $\displaystyle=(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)$ $\displaystyle=\begin{pmatrix}0&0\\ 0&I_{d-p}\end{pmatrix}\begin{pmatrix}(1-\bar{\alpha}_{t})I_{p}&0\\ 0&\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p}\end{pmatrix}^{-1}\begin{pmatrix}0&0\\ 0&I_{d-p}\end{pmatrix}$ $\displaystyle=\begin{pmatrix}0&0\\ 0&(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\end{pmatrix}$ $\displaystyle\Delta_{t,y}$ $\displaystyle=\begin{pmatrix}0\\ -\bar{\alpha}_{t}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\end{pmatrix},$ $$

and we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{2}=\bar{\alpha}_{t}^{2}\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\right\rVert^{2}$ $\displaystyle=\bar{\alpha}_{t}^{2}\mathbb{E}_{X_{t}\sim Q_{t|y}}\mathrm{Tr}\bigg{(}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}$ $\displaystyle\qquad[\Sigma_{t}^{-1}]_{:y}[\Sigma_{0}]_{y\bar{y}}(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\bigg{)}$ $\displaystyle\stackrel{{\scriptstyle(v)}}{{\leq}}\bar{\alpha}_{t}^{2}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert^{2}\left\lVert[\Sigma_{t}^{-1}]_{y:}\right\rVert^{2}\left\lVert[\Sigma_{0}]_{y\bar{y}}[\Sigma_{0}]_{\bar{y}y}\right\rVert\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0}\right\rVert^{2}$ (57) $$

where $(v)$ follows from the fact that $\left|\mathrm{Tr}(UV)\right|\leq\left\lVert U\right\rVert\mathrm{Tr}(V)$ if $V$ is positive semi-definite.
To analyze each norm above, denote $\lambda_{1}\geq\dots\geq\lambda_{d}>0$ to be the eigenvalues of $\Sigma_{0}$, and note that $\left\lVert[\Sigma_{0}]_{\bar{y}y}\right\rVert\leq\left\lVert\Sigma_{0}\right\rVert=\lambda_{1}$. The largest eigenvalue of $\Sigma_{t}^{-1}$ is $(\bar{\alpha}_{t}\lambda_{d}+(1-\bar{\alpha}_{t}))^{-1}$, and note that $\left\lVert[\Sigma_{t}^{-1}]_{y:}\right\rVert\leq\left\lVert\Sigma_{t}^{-1}\right\rVert$.
Also, since $[\Sigma_{0}]_{\bar{y}\bar{y}}$ is positive semi-definite, denote $\tilde{\lambda}_{1}\geq\dots\geq\tilde{\lambda}_{d-p}>0$ to be its eigenvalues, and thus

$$ $\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert\leq\frac{1}{\bar{\alpha}_{t}\tilde{\lambda}_{d-p}+(1-\bar{\alpha}_{t})}\leq\frac{1}{\min\{\tilde{\lambda}_{d-p},1\}}<\infty.$ $$

Also, since

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})^{\intercal}$ $\displaystyle=\mathbb{E}_{X_{t}\sim Q_{t|y}}(X_{t}-\mu_{t|y}+\sqrt{\bar{\alpha}_{t}}(H^{\dagger}y-H^{\dagger}H\mu_{0}))(X_{t}-\mu_{t|y}+\sqrt{\bar{\alpha}_{t}}(H^{\dagger}y-H^{\dagger}H\mu_{0}))^{\intercal}$ $\displaystyle=\Sigma_{t|y}+\bar{\alpha}_{t}(H^{\dagger}y-H^{\dagger}H\mu_{0})(H^{\dagger}y-H^{\dagger}H\mu_{0})^{\intercal},$ $$

we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0}\right\rVert^{2}=\mathrm{Tr}(\Sigma_{t|y})+\bar{\alpha}_{t}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}$ $\displaystyle\stackrel{{\scriptstyle(vi)}}{{\leq}}\bar{\alpha}_{t}(\lambda_{1}+\sigma_{y}^{2})+(1-\bar{\alpha}_{t})+\bar{\alpha}_{t}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}$ $\displaystyle\leq\max\left\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\right\}$ (58) $$

where $(vi)$ is because $\Sigma_{t|y}$ is positive definite with

$$ $\displaystyle\left\lVert\Sigma_{t|y}\right\rVert$ $\displaystyle=\left\lVert\bar{\alpha}_{t}(I_{d}-H^{\dagger}H)\Sigma_{0}(I_{d}-H^{\dagger}H)+\bar{\alpha}_{t}\sigma_{y}^{2}H^{\dagger}(H^{\dagger})^{\intercal}+(1-\bar{\alpha}_{t})I_{d}\right\rVert$ $\displaystyle\leq\bar{\alpha}_{t}\left\lVert\Sigma_{0}\right\rVert+\bar{\alpha}_{t}\sigma_{y}^{2}+(1-\bar{\alpha}_{t})$ $\displaystyle=\bar{\alpha}_{t}(\lambda_{1}+\sigma_{y}^{2})+(1-\bar{\alpha}_{t}).$ $$

Therefore, for the second moment,

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{2}$ $\displaystyle\leq\bar{\alpha}_{t}^{2}\frac{\max\left\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\right\}}{\min\{\lambda_{d},1\}^{2}\min\{\tilde{\lambda}_{d-p},1\}^{2}}\left\lVert[\Sigma_{0}]_{y\bar{y}}[\Sigma_{0}]_{\bar{y}y}\right\rVert$ $\displaystyle\lesssim\bar{\alpha}_{t}^{2}\cdot\max\left\{\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0}\right\rVert^{2}+d(\lambda_{1}+\sigma_{y}^{2}),d\right\}$ $$

since $\left\lVert[\Sigma_{0}]_{y\bar{y}}[\Sigma_{0}]_{\bar{y}y}\right\rVert\leq\left\lVert\Sigma_{0}\right\rVert^{2}=\lambda_{1}^{2}$.
Also, for general moments with $m\geq 2$,

$$ $\displaystyle\left\lVert\Delta_{t,y}\right\rVert^{m}$ $\displaystyle\leq\bar{\alpha}_{t}^{m}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}(x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0})\right\rVert^{m}$ $\displaystyle\leq\bar{\alpha}_{t}^{m}\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}\right\rVert^{m}\left\lVert\Sigma_{0}\right\rVert^{m}\left\lVert\Sigma_{t}^{-1}\right\rVert^{m}\left\lVert x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0}\right\rVert^{m}$ $\displaystyle\lesssim\bar{\alpha}_{t}^{m}\left\lVert x_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0}\right\rVert^{m}$ $$

and thus

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{m}$ $\displaystyle\lesssim\bar{\alpha}_{t}^{m}\left(\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Sigma_{t|y}^{-\frac{1}{2}}(X_{t}-\mu_{t|y})\right\rVert^{m}+\left\lVert\sqrt{\bar{\alpha}_{t}}(H^{\dagger}y-H^{\dagger}H\mu_{0})\right\rVert^{m}\right)$ $\displaystyle\leq\bar{\alpha}_{t}^{m}\left((m-1)!!\cdot d^{m/2-1}+\left\lVert\sqrt{\bar{\alpha}_{t}}(H^{\dagger}y-H^{\dagger}H\mu_{0})\right\rVert^{m}\right)=O(\bar{\alpha}_{t}).$ $$

Therefore, [Assumption 4](#Thmassumption4) is satisfied if the $\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1). The proof is now complete.

### I.2 Proof of Lemma 8

We continue from the expression of $\Delta_{t,y}$ in ([H.4](#A8.Ex418)) when $Q_{0}$ is Gaussian mixture. For $m\geq 2$, we have

$$ $\displaystyle\mathbb{E}_{X_{t}\sim Q_{t|y}}\left\lVert\Delta_{t,y}\right\rVert^{m}$ $\displaystyle\leq 2^{m-1}(\bar{\alpha}_{t})^{m/2}\max_{n\in[N]}\left\lVert\Sigma_{t}^{-1}\mu_{0,n}\right\rVert^{m}$ $\displaystyle\quad+2^{m-1}(\bar{\alpha}_{t})^{m}\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}(X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N})\right\rVert^{m}$ $\displaystyle\stackrel{{\scriptstyle(ii)}}{{\lesssim}}(\bar{\alpha}_{t})^{m/2}d^{m/2}+(\bar{\alpha}_{t})^{m}\mathbb{E}_{\begin{subarray}{c}X_{t}\sim Q_{t|y}\\ N\sim\Pi_{\cdot|t,y}(\cdot|X_{t})\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{m}$ $\displaystyle=(\bar{\alpha}_{t})^{m/2}d^{m/2}+(\bar{\alpha}_{t})^{m}\mathbb{E}_{\begin{subarray}{c}N\sim\Pi\\ X_{t}\sim Q_{t,N|y}\end{subarray}}\left\lVert X_{t}-\sqrt{\bar{\alpha}_{t}}\mu_{0,N}\right\rVert^{m}$ $\displaystyle\stackrel{{\scriptstyle(iii)}}{{\lesssim}}(\bar{\alpha}_{t})^{m/2}d^{m/2}+(\bar{\alpha}_{t})^{m}\left(d^{m/2}+\sum_{n=1}^{N}\pi_{n}\left\lVert H^{\dagger}y-H^{\dagger}H\mu_{0,n}\right\rVert^{m}\right)$ (59) $\displaystyle=O(\bar{\alpha}_{t}).$ $$

Here $(ii)$ follows because

$$ $\displaystyle\left\lVert(I_{d}-H^{\dagger}H)\Sigma_{t,sig}^{-1}(I_{d}-H^{\dagger}H)\Sigma_{0}(H^{\dagger}H)\Sigma_{t}^{-1}\right\rVert$ $\displaystyle=\left\lVert(\bar{\alpha}_{t}[\Sigma_{0}]_{\bar{y}\bar{y}}+(1-\bar{\alpha}_{t})I_{d-p})^{-1}[\Sigma_{0}]_{\bar{y}y}[\Sigma_{t}^{-1}]_{y:}\right\rVert$ $\displaystyle\leq\frac{\lambda_{1}}{\min\{\tilde{\lambda}_{d-p},1\}\min\{\lambda_{d},1\}}=O(1),$ $$

and $(iii)$ follows from ([H.5](#A8.Ex532)).
Therefore, this verifies [Assumption 4](#Thmassumption4) when the $\alpha_{t}$ satisfies [Definition 1](#Thmdefinition1). The proof is now complete.

### I.3 Lemma 9 and its proof

###### Lemma 9 .

Given a positive semi-definite matrix $\Sigma$, $\sigma\geq 0$, and $\alpha\in(0,1)$,

$$ $\displaystyle(\alpha\sigma H^{\dagger}(H^{\dagger})^{\intercal}+(1-\alpha)I_{d})^{-1}(I_{d}-H^{\dagger}H)=\frac{1}{1-\alpha}(I_{d}-H^{\dagger}H)$ $\displaystyle(I_{d}-H^{\dagger}H)(\alpha(I_{d}-H^{\dagger}H)\Sigma(I_{d}-H^{\dagger}H)+\alpha\sigma H^{\dagger}H+(1-\alpha)I_{d})^{-1}(H^{\dagger}H)=0.$ $$

###### Proof.

The key of the proof is the Woodbury matrix identity, which states that for any matrices $U\in\mathbb{R}^{d\times p},V\in\mathbb{R}^{p\times d}$,

$$ $(I_{d}+UV)^{-1}=I_{d}-U(I_{p}+VU)^{-1}V.$ $$

For the first equality, we apply Woodbury with $U=\sqrt{\frac{\alpha\sigma}{1-\alpha}}H^{\dagger}$ and $V=\sqrt{\frac{\alpha\sigma}{1-\alpha}}(H^{\dagger})^{\intercal}$ and we get

$$ $(\alpha\sigma H^{\dagger}(H^{\dagger})^{\intercal}+(1-\alpha)I_{d})^{-1}=\frac{1}{1-\alpha}\left(I_{d}-\frac{\alpha\sigma}{1-\alpha}H^{\dagger}\left(I_{p}+\frac{\alpha\sigma}{1-\alpha}(H^{\dagger})^{\intercal}H^{\dagger}\right)^{-1}(H^{\dagger})^{\intercal}\right).$ $$

Since $p\leq d$, the pseudo-inverse equals $H^{\dagger}=H^{\intercal}(HH^{\intercal})^{-1}$, and by the orthogonal property we have

$$ $(H^{\dagger})^{\intercal}(I_{d}-H^{\dagger}H)=(HH^{\intercal})^{-1}H(I_{d}-H^{\dagger}H)=0.$ $$

We have thus shown the first equality.

For the second equality, we first consider the case where $\sigma=0$. Write $S=(1-\alpha)I_{d}+\alpha(I_{d}-H^{\dagger}H)\Sigma(I_{d}-H^{\dagger}H)$. Since $\Sigma$ is positive semi-definite, there exists matrix $L$ such that $\Sigma=LL^{\intercal}$. Thus,

$$ $\displaystyle S^{-1}$ $\displaystyle=\bigg{(}(1-\alpha)I_{d}+\alpha(I_{d}-H^{\dagger}H)\Sigma(I_{d}-H^{\dagger}H)\bigg{)}^{-1}$ $\displaystyle=\bigg{(}(1-\alpha)I_{d}+\alpha\Big{(}(I_{d}-H^{\dagger}H)L\Big{)}\Big{(}(I_{d}-H^{\dagger}H)^{\intercal}L\Big{)}^{\intercal}\bigg{)}^{-1}$ $\displaystyle=\frac{1}{1-\alpha}I_{d}-\frac{\alpha}{1-\alpha}((I_{d}-H^{\dagger}H)L)\left(I_{d}+\frac{\alpha}{1-\alpha}L^{\intercal}(I_{d}-H^{\dagger}H)L\right)^{-1}L^{\intercal}(I_{d}-H^{\dagger}H),$ $$

where in the last line we have applied Woodbury with $U=V^{\intercal}=\sqrt{\frac{\alpha}{1-\alpha}}(I_{d}-H^{\dagger}H)L$.
The equality is achieved because

$$ $\displaystyle(I_{d}-H^{\dagger}H)S^{-1}(H^{\dagger}H)$ $\displaystyle=\frac{1}{1-\alpha}\underbrace{(I_{d}-H^{\dagger}H)(H^{\dagger}H)}_{=0}$ $\displaystyle\quad-\frac{\alpha}{1-\alpha}(I_{d}-H^{\dagger}H)L\left(I_{d}+\frac{\alpha}{1-\alpha}L^{\intercal}(I_{d}-H^{\dagger}H)L\right)^{-1}L^{\intercal}\underbrace{(I_{d}-H^{\dagger}H)(H^{\dagger}H)}_{=0}$ $\displaystyle=0.\qed$ $$

When $\sigma>0$, we can apply Woodbury identity to sum of matrices $A$ and $B$ and get $(A+B)^{-1}=A^{-1}-A^{-1}B(A+B)^{-1}$. Thus,

$$ $(S+\alpha\sigma H^{\dagger}H)^{-1}=S^{-1}-\alpha\sigma S^{-1}(H^{\dagger}H)(S+\alpha\sigma H^{\dagger}H)^{-1}$ $$

and

$$ $\displaystyle(I_{d}-H^{\dagger}H)(S+\alpha\sigma H^{\dagger}H)^{-1}(H^{\dagger}H)$ $\displaystyle=\underbrace{(I_{d}-H^{\dagger}H)S^{-1}(H^{\dagger}H)}_{=0}$ $\displaystyle\qquad-\alpha\sigma\underbrace{(I_{d}-H^{\dagger}H)S^{-1}(H^{\dagger}H)}_{=0}(S+\alpha\sigma H^{\dagger}H)^{-1}(H^{\dagger}H)$ $\displaystyle=0.$ $$

The proof is now complete.

###### Lemma 9 .

###### Proof.

### I.4 Lemma 10 and its proof

###### Lemma 10 .

With $1-\alpha_{t}\equiv\frac{\log T}{T},\leavevmode\nobreak\ \forall t\geq 0$ (which satisfies [Definition 1](#Thmdefinition1)), given any $p>0$,

$$ $\sum_{t=2}^{T}(1-\alpha_{t})\bar{\alpha}_{t}^{p}=\frac{1}{p}\left(1-\frac{2pc\log T}{T}\right)+\tilde{O}\left(\frac{1}{T^{2}}\right).$ $$

###### Proof.

Define the sum as $s_{T}$. Then,

$$ $\displaystyle s_{T}$ $\displaystyle=\sum_{t=2}^{T}\frac{c\log T}{T}\left(1-\frac{c\log T}{T}\right)^{pt}=\frac{c\log T}{T}\left(1-\frac{c\log T}{T}\right)^{2p}\frac{1-\left(1-\frac{c\log T}{T}\right)^{p(T-1)}}{1-\left(1-\frac{c\log T}{T}\right)^{p}}$ $\displaystyle=\frac{c\log T}{T}\left(1-\frac{c\log T}{T}\right)^{2p}\frac{1}{1-\left(1-\frac{c\log T}{T}\right)^{p}}(1+O(T^{-cp}))$ $\displaystyle=\frac{1}{p}\left(1-\frac{2pc\log T}{T}\right)+\tilde{O}(T^{-2}).\qed$ $$

###### Lemma 10 .

###### Proof.

## Peer Reviews

confidence: 3
contribution: 2
initial_rating: 6
presentation: 2
questions: Q1. How does Assumption 3 relate to the standard smoothness and regularity conditions typically required for Langevin dynamics sampling, such as Lipschitz continuity and bounds on the gradient and Hessian of the log-density?

Q2. Considering high-dimensional data, how does Assumption 3 ensure that the expectations involving higher-order derivatives remain well-controlled?

Q3. Are there any connections between Assumption 3 and functional inequalities like log-Sobolev or Poincaré inequalities?
soundness: 3
strengths: S1. The paper is well-organized and easy to follow.

S2. The convergence results for zero-shot conditional diffusion models are new and address an area that remains largely unexplored in the literature.

S3. The authors propose a new zero-shot conditional sampler that accelerates the convergence rates of the zero-shot conditional DDPM sampler by reducing the asymptotic distributional bias in the obtained bounds.
summary: This paper studies the convergence guarantees of zero-shot conditional diffusion models under certain regularity assumptions on the target distributions and assuming small score estimation errors. The authors provide distribution estimation error bounds in KL divergence for score-mismatched DDPM and establish convergence rates under several settings. They also introduce a new zero-shot conditional sampler designed to accelerate the convergence rates by reducing the asymptotic distributional bias in the obtained bounds.
weaknesses: My main concern revolves around Assumption 3, which appears to be relatively strong and may limit the applicability of the results. Specifically:

C1. **Smoothness Requirements:** The assumption requires the existence and control of higher-order derivatives of the log densities. Distributions that are not sufficiently smooth (e.g., those with discontinuities or singularities) may not satisfy this assumption.

C2. **Scaling with $(1−\alpha_t)^m$:** When $\alpha_t$ approaches 1, the factor $(1-\alpha_t)^m$  becomes very small. If the expectations do not decrease at the same rate, the assumption may not hold. This can occur in practice when $\alpha_t$ is close to 1, as in the noise schedule Equation (8) in the paper.

C3. **High Dimensionality Issues:** In high-dimensional settings (large $d$), the behavior of the derivatives can be more complex. The curse of dimensionality may make the assumption unrealistic because the number of derivatives grows exponentially with $d$.

C4. **Applicability to other distributions**: The assumption may not be valid for distributions outside certain classes, such as those with multimodal densities where modes are not well-separated, or distributions with skewness and kurtosis that affect the derivatives.

Additional concerns include:

C5. **Avoiding overstatements of contributions:** In Section 1.1, the paper claims to provide convergence bounds for general target distributions with finite second moments (e.g., Line 92-93). However, their analysis requires regular high-order derivatives for the score functions, which significantly restricts their results to smooth distributions. Adjusting the claim accordingly would provide a more accurate representation of the scope of the contributions.

C6. **Insufficient Discussion on the obtained bounds:** The paper lacks sufficient discussions on the practical implications of the bounds. For example, how do the obtained bounds compare to conventional conditional sampling methods, such as Langevin dynamics? How can the bounds explain the success of conditional diffusion models over other sampling methods?

C7. **Clarification on early stopping in Theorem 2:** The bound in Theorem 2 is given in terms of the KL divergence between the target and learned distribution at time step 1 rather than at time step 0, implying that early stopping has been imposed. Clarification on this point would be helpful.

---

confidence: 2
contribution: 2
initial_rating: 5
presentation: 2
questions: What does the final line in theorem 2 signify: "Here $W_2(\tilde{Q}_1, \tilde{Q}_0) < \delta d$"?  Is it an assumption or a result of the theorem ? It is unclear from the proof as well.
soundness: 2
strengths: This paper participates in the effort to provide theoretical characterizations of guidance and conditional sampling in diffusion models, an area which has not yet been well explored.
summary: The paper derives KL convergence bounds for diffusion models when the model is not trained on the target distribution but on a mismatched one. Under some assumptions, the authors exhibit a bound with a term depending on the mismatch. Building on this result, the authors propose an algorithm for conditional sampling that minimizes the mismatch term.
weaknesses: 1. Assumption 2 is not realistic: Assumption 2 measures how much the approximated conditional expectation deviates from the true $\mu_t$. The expectation in assumption 2 is taken with respect to $\tilde{Q}$ the target distribution, which is unknown during training. Assumption 2 can thus not be guaranteed to hold. Moreover, the authors state mistakenly, as far as I understand, that Assumption 5 is the same as Assumption 4 when is taken under different distributions. Assumption 5, under $Q_t$, is a classic assumption and, as it relates closely to the training loss, is a realistic assumption. Assumption 4 on the other hand needs more justification.
2. A minor modification of prior work with assumptions to control the additional terms: Theorem 1 is obtained through modification Liang et al 2024 by assuming the target distribution is different from the training one. The simple additional terms that appear due to the modification are controlled using a series of assumptions. This limits the technical novelty over Liang et al 2024.
3. The paper is not easy to read. This can be fixed by reducing the large number of inline equations (paragraphs line 513-523 or paragraph A in the appendix). Unnecessary factors $(1-\alpha_t)$ appear on both sides of an equation in crucial assumptions 3 and 4 which I found to be confusing.

---

confidence: 2
contribution: 2
initial_rating: 5
presentation: 2
questions: 1. Theorem 4 is somewhat confusing. Specifically, it claims the establishment of Theorem 2 under the specific choice of $f_{t,y}$, while Theorem 2 has additional assumptions for $\E[\|\Delta_t(X_t)\|]$. The problem is whether the upper bound of the mismatch-score, i.e., $\E[\|\Delta_t(X_t)\|]$, is required for Theorem 4 when the conditional target distribution is general.
2. The BO-DDNM is too dependent on the linear structure of $y=Hx+n$. Considering a general mismatch-score setting, i.e., $y=h(x_0)$. How can we adapt BO-DDNM to the general case except for a Taylor extension?

I am willing to raise my score if the authors solve my concerns.
soundness: 3
strengths: 1. The problem setting considered in this paper is interesting, especially regarding the effectiveness of the zero-shot sampler in the score-mismatched diffusion.
2. Authors denote the gap between ground truth and generated samples lies in the accumulative mismatch between the target and training distributions. This qualitative result is useful and aligns with their sampler design.
summary: This paper mainly focuses on the convergence analysis of zero-shot samplers when training scores are based on the unconditional distribution while inference samples for a conditional distribution. Under these conditions, the inference required scores vary from the predicted one, which is considered as the score-mismatched diffusion.
Although various practical works have appeared, from a theoretical perspective, this is the first work to analyze the inference convergence under score-mismatched settings. The core step of this paper is to recognize and minimize (Theorem 3) the mismatch scores (Eq.(7)) by considering the conditional reverse process (Eq.(5)). The conditional reverse process is deduced by a known connection (e.g., linear connection) between the random variables and the conditions (Eq.(4)).
weaknesses: 1. I am a bit afraid that this paper overclaims the problem settings it actually solved. Specifically, this paper claims to consider a general score-mismatched setting. Although it provides a convergence analysis (Theorem 1) for such a setting, the assumption, e.g., (Assumption 4) required, is highly artificial. Besides, it can hardly find an algorithm to achieve Theorem 1, since authors do not provide any approximation for $\grad \ln \tilde{q}_t(x_t)$ under a general score-mismatched setting ($y=h(x_0)$). Instead, I think this paper more focused on the linear setting, i.e., $y=Hx_0+n$ (Eq.(4))
2. I am curious about whether the problem setting is really useful beyond the super-resolution problem. Since the conditioning (y) is usually correlated with the random variable (x) implicitly. I suggest authors provide more examples of utilizing zero-shot samplers to solve score-mismatch diffusions.
3. The establishment of Assumption 4 requires strict assumptions. It not only constrains the unconditional distribution $Q_0$ but also constrains the $H=[I_p, 0]$. I hope the authors can demonstrate whether the results can extend to a general $H$ and how.
4. The final KL divergence bound between ground truth and generated samples is unsatisfactory. Although it is an upper bound, it contains a factor O(d). Such a result will make me concern whether a zero-shot sampler is really effective for mismatch-score settings.

---

confidence: 2
contribution: 2
initial_rating: 5
presentation: 3
questions: 1. I am slightly confused by theorem 2 following from theorem 1 (more specifically, Corollary 1 following from theorem 1).  Theorem 2 bounds the KL between Q_1 (the marginal distribution of the zero-shot sampler at time 1) and P_1 (e.g. the time 1 marginal of the VP-SDE at time 1 where P_0 is the conditional distribution). Typically for all practical purposes in diffusion models Q_1 is simply a Gaussian and this bound can be obtained from exploring the ergodicity of the OU-process and prior work on the convergence of diffusion models. So two questions 

     * Why is this bound for a general Q_1 relevant? We typically fix/choose Q_1 , so Im a bit confused about what specific aspect of the zero-shot diffusion models considered in this work induces/requires a particular Q_1 which needs to be analysed 
     * How does corollary 1 follow from theorem 2 , do you simply just preform the same analysis in reverse time ?  (e.g. flip the time indices) more detail here would be useful 

2. How realistic is assumption 4 for zero-shot models such as Chung 2022a-b, Chung 2023 where the score mismatch cant really be controlled as it makes rather unrealistic Gaussian approximations of the transition density in the backwards process.  Assumption 4 is a rather specific bound on the mismatch it seems its form is required to make everything else work so discussion on its relevance outside of a Gaussian setting for the prior (data distribution) would be extremely helpful.
soundness: 3
strengths: 1. The paper is clear and well written
2. The short summary sketches of the proofs allow us to for a high-level verification in the soundness of the results
3. The work is novel and clearly relevant to the community; with some small modifications and clarifications, this work can be impactful.
summary: This work provides the first guarantees for conditional sampling (within the inverse problems setting) via diffusion models, which approximate the guidance term via techniques that do not require additional supervision (e.g., x,y pairs such as in CG and CFG), which the authors coin zero-shot as they do not require to observe any additional data to fine-tune. 

The authors then provide a preconditioned linear-approximation-based approach to their proposed decomposition of the posterior score, which minimises the mismatch; the proposed estimator is very cheap and could be useful to the community. Furthermore, the authors provide a very modest numerical benchmark showing significant improvements to denoising diffusion null space models. In general, the strategy/sketch used in Theorem 3 can be of great use as a template to practitioners looking to derive quick yet accurate estimators.
weaknesses: 1. The reviewer found the presentation of the paper a bit confusing. Calling methodologies such as reconstruction guidance (Chung 2023 in the paper) zero-shot (whilst potentially accurate) can be confusing and misleading for the following reasons: 


      *  Whilst many of these methodologies do not require additional supervision (e.g. observing x,y pairs such as in CFG or classifier guidance) they do often still approximate the same quantity e.g. the h-transform / guidance term [1], ultimately with the goal of approximating the posterior score.

     *  Many of these zero shot approximations induce costs that are comparable to training when one considers (see wallclock times in [1]) , furthermore with methods like DPS / reconstruction guidance we can see that they are effectively fine-tuning/minimising a reconstruction loss across timesteps, whilst there's no extra data (thus arguably zero shot) , there is certainly a significant extra cost arising from this "online-finetuning". 

In short, I think the paper would improve its reach/impact if it revised its presentation of what it considers to be zero-shot, or make it much more precise from earlier in the manuscript. 


2. Assumption 4 seems quite strong and rather non-practical for some of these zero-shot methodologies, which make Gaussian approximations on the reverse process and thus do not induce a mismatch that can be controlled. 


[1] Denker, A., Vargas, F., Padhy, S., Didi, K., Mathis, S., Dutordoir, V., Barbano, R., Mathieu, E., Komorowska, U.J. and Lio, P., 2024. DEFT: Efficient Finetuning of Conditional Diffusion Models by Learning the Generalised $ h $-transform. arXiv preprint arXiv:2406.01781.

---

confidence: 3
contribution: 3
initial_rating: 5
presentation: 2
questions: - The big-O notation in this work is confusing. To name a few, in Assumptions 3 and 4, should $(1-\alpha_t)^m \mathbb{E}[\cdot] = O((1-\alpha_t)^m)$ be equivalent to $\mathbb{E}[\cdot] = O(1)$? Or do the hidden constants in these assumptions implicitly depend on $(1-\alpha_t)^m$? In Definition 1, the noise schedule needs to satisfy $\bar{\alpha}\_T = o (1/T)$ which is defined as an asymptotic bound $\lim\sup_{T\to\infty} |\bar{\alpha}_T / (1/T)| \to 0$, while Theorem 1 presents a non-asymptotic analysis. How does the asymptotic assumption apply to the non-asymptotic analysis? I suggest the authors clarify the notations or even explicitly write the constants' dependency on the hyperparameters if possible. 
- Does $1-\alpha_1=\delta$ in equation (8) contradict with $1-\alpha_1 \lesssim \frac{\log T}{T}$ in Definition 1? What can we obtain from Theorem 1 if $\alpha_t$ is chosen as (8)?
soundness: 2
strengths: - The convergence bound of DDPM under score mismatch applies to a variety of target distributions without restrictive assumptions. 
- A bias-optimal zero-shot sampler is proposed, and its convergence rate is analyzed.
summary: This paper provides a non-asymptotic convergence theory for diffusion models with a mismatched score function. The results can be applied to zero-shot conditional samplers, especially linear conditional models. The authors also propose the Bias-Optimal (BO) DDNM sampler and provide a comparison with the previous algorithms.
weaknesses: - Assumption 2 requires an upper bound on the score estimation error $\epsilon^2 = \tilde{\mathcal{O}}(T^{-2})$, which could be restrictive compared to the previous literature (e.g., Li et al. 2024b) which applies to any $\epsilon$. 
- The in-line equations in Section 2.1 and Section 5 are hard to follow. I suggest the authors re-organize the equations for better readability, especially by highlighting the definitions and differences of $q$, $p$, and $\hat{p}$.

Meta-review: The reviews are all generally on the positive side, though usually not enthusiastically so.  The contributions are generally deemed to be valuable, giving convergence analyses for  zero-shot conditional diffusion models under suitable regularity conditions, and introducing a zero-shot sampler to accelerate convergence.  The opinions on the paper writing are mixed (some positive and some negative).  Maybe most significantly, several of the assumptions were raised into question or clarifications sought, but the rebuttal seems to have addressed these sufficiently.

Decision: Accept (Poster)