and since $\alpha > \frac{1}{2}$ and $\lim_{N \to \infty} \int_{\mathbb{R}^d} |y|^\ell \bar{\nu}_N(dy) = \int_{\mathbb{R}^d} |y|^\ell \mu(dy)$, we can conclude that the left-hand side goes to 0 as $N$ goes to infinity and so the third step is concluded.

**FOURTH STEP** We are now going to prove the convergence in distribution of $\sqrt{N}Q_N$ to a Gaussian random variable. In this case, we can immediately apply the Central Limit Theorem for martingales while in Theorem 25 we had to introduce first the Poisson equation before applying it to $K_{3,N}$. For $1 \le i \le N$ let

$$Y_{N,i} = \frac{1}{\sqrt{N}} \left( \frac{\delta U}{\delta m} \left( \mu_{N}^{i \wedge I_{N}, 0}, X_{i} \right) - \int_{\mathbb{R}^{d}} \frac{\delta U}{\delta m} \left( \mu_{N}^{i \wedge I_{N}, 0}, x \right) \nu_{i}(dx) \right).$$

Let us start by **checking that** $\mathbb{E}(Y_{N,i}|\mathcal{F}_{i-1}) = 0$.

Since $X_i$ is independent of $\mathcal{F}_{i-1}$ and $\mu_N^{i \wedge I_N, 0} = \sum_{j=1}^{i-1} 1_{\{I_N=j\}} \mu_N^{j, 0} + 1_{\{I_N>i-1\}} \mu_N^{i, 0}$ is $\mathcal{F}_{i-1}$-measurable, applying the Freezing Lemma one has

$$\begin{aligned} \mathbb{E}(Y_{N,i}|\mathcal{F}_{i-1}) &= \mathbb{E}\left(\frac{1}{\sqrt{N}}\left(\frac{\delta U}{\delta m}\left(\mu_N^{i \wedge I_N, 0}, X_i\right) - \int_{\mathbb{R}^d} \frac{\delta U}{\delta m}\left(\mu_N^{i \wedge I_N, 0}, x\right) \nu_i(dx)\right)\bigg|\mathcal{F}_{i-1}\right) \\ &= \frac{1}{\sqrt{N}}\left(\mathbb{E}\left(\frac{\delta U}{\delta m}(t, X_i)\right)_{t=\mu_N^{i \wedge I_N, 0}} - \int_{\mathbb{R}^d} \frac{\delta U}{\delta m}\left(\mu_N^{i \wedge I_N, 0}, x\right) \nu_i(dx)\right) = 0. \end{aligned}$$

For what concerns the **square integrability**, let us check that both the components are squared integrable by applying Hypothesis **RU2** and the fact that $\nu_i \in \mathbb{P}_\ell(\mathbb{R}^d)$.

$$\bullet \quad \mathbb{E}\left(\frac{\delta U}{\delta m}\left(\mu_N^{i \wedge I_N, 0}, X_i\right)^2\right) \le C^2 \mathbb{E}\left(\left(1 + |X_i|^{\frac{\ell}{2}}\right)^2\right) \le 2C^2 \mathbb{E}\left(1 + |X_i|^{\ell}\right) < \infty$$

$$\bullet \quad \mathbb{E}\left(\left(\int_{\mathbb{R}^d} \frac{\delta U}{\delta m}(\mu_N^{i \wedge I_N, 0}, x) \nu_i(dx)\right)^2\right) \le \mathbb{E}\left(\int_{\mathbb{R}^d} \frac{\delta U}{\delta m}(\mu_N^{i \wedge I_N, 0}, x)^2 \nu_i(dx)\right) \\ \le 2C^2 \left(1 + \int_{\mathbb{R}^d} |x|^{\ell} \nu_i(dx)\right) < \infty.$$

Let us now study the **convergence of** $\sum_{i=1}^N \mathbb{E}(Y_{N,i}^2|\mathcal{F}_{i-1})$:

$$\begin{aligned} \sum_{i=1}^{N} \mathbb{E}(Y_{N,i}^2 | \mathcal{F}_{i-1}) &= \frac{1}{N} \sum_{i=1}^{N} \mathbb{E} \left( \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, X_i) - \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) \nu_i(dx) \right)^2 | \mathcal{F}_{i-1} \right) \\ &= \frac{1}{N} \sum_{i=1}^{N} \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x)^2 \nu_i(dx) - \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) \nu_i(dx) \right)^2 \right). \end{aligned}$$

As in the fourth step of the proof of the Theorem 25, thanks to Hypothesis **RU3** and thanks to the a.s. convergence of $\max_{1 \le i \le N} W_\ell(\mu_N^{i, 0}, \mu)$ to 0, we obtain the existence of a sequence of random variables $(\epsilon_N)_{N \ge 0}$ converging a.s. to 0 such that

$$\forall 1 \le i \le N \quad \left| \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) - \frac{\delta U}{\delta m} (\mu, x) \right| \le (1 + |x|^{\ell}) \epsilon_N. \quad (2.54)$$
**First Component** Let us first show that

$$ \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x)^2 \nu_i(dx) - \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x)^2 \mu(dx) = 0 \quad \text{a.s.} $$

We can rewrite the difference in the following way

$$ \begin{aligned} & \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x)^2 \nu_i(dx) - \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x)^2 \nu_i(dx) \\ & + \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x)^2 \nu_i(dx) - \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x)^2 \mu(dx) \\ & = \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x)^2 - \frac{\delta U}{\delta m} (\mu, x)^2 \right) \nu_i(dx) \\ & + \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x)^2 (\bar{\nu}_N - \mu)(dx). \end{aligned} $$

Since $\lim_{N \to \infty} W_\ell(\bar{\nu}_N, \mu) = 0$, by the characterization (2.7), we can deduce that the second term of the sum goes to 0 thanks to Assumption **RU2** and Assumption **RU4**. For what concerns the first term for $i = 1, \dots, N$, using Assumption **RU2** and (2.54), one has

$$ \begin{aligned} \left| \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x)^2 - \frac{\delta U}{\delta m} (\mu, x)^2 \right| & \le \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x) - \frac{\delta U}{\delta m} (\mu, x) \right)^2 \\ & \quad + 2 \left| \frac{\delta U}{\delta m} (\mu, x) \right| \left| \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x) - \frac{\delta U}{\delta m} (\mu, x) \right| \\ & \le (1 + |x|^{\frac{\ell}{2}})^2 \epsilon_N^2 + 2C (1 + |x|^{\frac{\ell}{2}})^2 \epsilon_N \\ & \le 2 (1 + |x|^{\ell}) \epsilon_N (\epsilon_N + 2C). \end{aligned} $$

Therefore

$$ \begin{aligned} \left| \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x)^2 - \frac{\delta U}{\delta m} (\mu, x)^2 \right) \nu_i(dx) \right| & \le 2\epsilon_N (\epsilon_N + 2C) \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} (1 + |x|^{\ell}) \nu_i(dx) \\ & = 2\epsilon_N (\epsilon_N + 2C) \left( 1 + \int_{\mathbb{R}^d} |x|^{\ell} \bar{\nu}_N(dx) \right) \end{aligned} $$

where the right-hand side goes to 0 a.s. since $\lim_{N \to \infty} \int_{\mathbb{R}^d} |x|^{\ell} \bar{\nu}_N(dx) = \int_{\mathbb{R}^d} |x|^{\ell} \mu(dx)$.

**Second Component** We are going to prove that

$$ \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^{N} \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x) \nu_i(dx) \right)^2 - \int_{\mathbb{R}^d \times \mathbb{R}^d} \frac{\delta U}{\delta m}(\mu, x) \frac{\delta U}{\delta m}(\mu, y) \eta(dx, dy) = 0 \quad \text{a.s.} $$

Let us rewrite the difference in the following way

$$ \frac{1}{N} \sum_{i=1}^{N} \left( \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_{N,0}}, x) \nu_i(dx) \right)^2 - \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x) \nu_i(dx) \right)^2 \right) \\ + \left( \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d \times \mathbb{R}^d} \frac{\delta U}{\delta m}(\mu, x) \frac{\delta U}{\delta m}(\mu, y) \nu_i(dx) \nu_i(dy) - \int_{\mathbb{R}^d \times \mathbb{R}^d} \frac{\delta U}{\delta m}(\mu, x) \frac{\delta U}{\delta m}(\mu, y) \eta(dx, dy) \right) $$
Since Hypothesis 1 holds, again by the characterization (2.7), we can deduce that the second term of the sum goes to 0 a.s. thanks to Hypothesis **RU2** and Hypothesis **RU4**. As far as the first term is concerned, using Hypothesis **RU2** and (2.54), one has that for $i = 1, \dots, N$

$$
\begin{aligned}
& \left| \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) \nu_i(dx) \right)^2 - \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x) \nu_i(dx) \right)^2 \right| \\
& \le \left( \int \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) - \frac{\delta U}{\delta m} (\mu, x) \right) \nu_i(dx) \right)^2 \\
& \quad + 2 \left| \int \frac{\delta U}{\delta m} (\mu, x) \nu_i(dx) \right| \left| \int \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) - \frac{\delta U}{\delta m} (\mu, x) \right) \nu_i(dx) \right| \\
& \le \int \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) - \frac{\delta U}{\delta m} (\mu, x) \right)^2 \nu_i(dx) \\
& \quad + 2 \int \left| \frac{\delta U}{\delta m} (\mu, x) \right| \nu_i(dx) \int_{\mathbb{R}^d} \left| \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) - \frac{\delta U}{\delta m} (\mu, x) \right) \right| \nu_i(dx) \\
& \le 2 \epsilon_N^2 \int_{\mathbb{R}^d} (1 + |x|^\ell) \nu_i(dx) + 2C \epsilon_N \int_{\mathbb{R}^d} (1 + |x|^{\ell/2}) \nu_i(dx) \int_{\mathbb{R}^d} (1 + |x|^{\ell/2}) \nu_i(dx) \\
& \le 2 \epsilon_N^2 \int_{\mathbb{R}^d} (1 + |x|^\ell) \nu_i(dx) + 4C \epsilon_N \int_{\mathbb{R}^d} (1 + |x|^\ell) \nu_i(dx) = (2 \epsilon_N^2 + 4C \epsilon_N) \left( 1 + \int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) \right).
\end{aligned}
$$

Therefore

$$
\begin{aligned}
& \frac{1}{N} \sum_{i=1}^{N} \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) \nu_i(dx) \right)^2 - \frac{1}{N} \sum_{i=1}^{N} \left( \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x) \nu_i(dx) \right)^2 \\
& \le (2\epsilon_N^2 + 4C\epsilon_N) \left( 1 + \int_{\mathbb{R}^d} |x|^\ell \bar{\nu}_N(dx) \right)
\end{aligned}
$$

and the left-hand side goes to 0 since $\lim_{N \to \infty} \int_{\mathbb{R}^d} |x|^\ell \bar{\nu}_N(dx) = \int_{\mathbb{R}^d} |x|^\ell \mu(dx)$.

To finally apply the Central Limit Theorem for martingales (see Corollary 3.1 [31]) to conclude that

$$
\sqrt{N} Q_N = \sum_{i=1}^{N} Y_{N,i} \underset{d}{\implies} \mathcal{N} \left( 0, \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x)^2 \mu(dx) - \int_{\mathbb{R}^d \times \mathbb{R}^d} \frac{\delta U}{\delta m} (\mu, x) \frac{\delta U}{\delta m} (\mu, y) \eta(dx, dy) \right),
$$

it remains just to verify that the **Lindeberg condition** holds.

We need to check that for $\epsilon > 0$, $\sum_{i=1}^N \mathbb{E}(Y_{N,i}^2 1_{\{Y_{N,i}^2 > \epsilon\}} | \mathcal{F}_{i-1})$ goes to 0 in probability as $N \to \infty$. By Jensen's inequality and Hypothesis **RU2**, one has

$$
\begin{aligned}
NY_{N,i}^2 &= \left( \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, X_i) - \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x) \nu_i(dx) \right)^2 \\
&\le 2 \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, X_i)^2 + 2 \int_{\mathbb{R}^d} \frac{\delta U}{\delta m} (\mu_N^{i \wedge I_N, 0}, x)^2 \nu_i(dx) \\
&\le 4C^2 (1 + |X_i|^\ell) + 4C^2 \int_{\mathbb{R}^d} (1 + |x|^\ell) \nu_i(dx) = 4C^2 \left( 2 + |X_i|^\ell + \int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) \right).
\end{aligned}
$$

Therefore, as in the fourth step of the proof of the Theorem 25, it is enough to check that for each $\epsilon > 0$
$$2 \times 1_{\{\frac{2}{N} > \epsilon\}} + \sum_{i=1}^{N} \mathbb{E} \left( \frac{|X_i|^\ell}{N} 1_{\{|X_i|^\ell > N\epsilon\}} | \mathcal{F}_{i-1} \right) + \sum_{i=1}^{N} \frac{\int_{\mathbb{R}^d} |x|^\ell \nu_i(dx)}{N} 1_{\{\int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) > N\epsilon\}}$$

goes to 0 as $N$ goes to infinity (a.s.).

It is immediate to prove that the **first component** goes to 0 while the **second component** goes to 0 by the independence of $X_i$ by $\mathcal{F}_{i-1}$ and (2.49). For what concerns the **third component**, we have that $\bar{y}_N := \frac{1}{N} \sum_{i=1}^N \int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) \to \int_{\mathbb{R}^d} |x|^\ell \mu(dx)$. Therefore

$$\frac{\int_{\mathbb{R}^d} |x|^\ell \nu_N(dx)}{N} = \bar{y}_N - \frac{N-1}{N} \bar{y}_{N-1} \underset{N \to \infty}{\longrightarrow} 0$$

that implies $\forall \epsilon > 0$

$$\int_{\mathbb{R}^d} |x|^\ell \nu_N(dx) 1_{\{\int_{\mathbb{R}^d} |x|^\ell \nu_N(dx) > N\epsilon\}} \underset{N \to \infty}{\longrightarrow} 0$$

and taking the Cesaro mean we deduce that

$$\lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) 1_{\{\int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) > N\epsilon\}} \le \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^{N} \int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) 1_{\{\int_{\mathbb{R}^d} |x|^\ell \nu_i(dx) > i\epsilon\}} = 0$$

and so the proof is concluded. □
<figure><img src="image_7.png" /></figure>
# Chapter 3

## Convergence to the uniform distribution of vectors of partial sums modulo one with a common factor

The material for this chapter has been released in [27].

In this work, we prove the joint convergence in distribution of $q$ variables modulo one obtained as partial sums of a sequence of i.i.d. square integrable random variables multiplied by a common factor given by some function of an empirical mean of the same sequence. The limit is uniformy distributed over $[0, 1]^q$. To deal with the coupling introduced by the common factor, we assume that the joint distribution of the random variables has a non zero component absolutely continuous with respect to the Lebesgue measure, so that the convergence in the central limit theorem for this sequence holds in total variation distance. While our result provides a generalization of Benford's law to a data adapted mantissa, our main motivation is the derivation of a central limit theorem for the stratified resampling mechanism, which is performed in the companion paper [25].

### 3.1 Introduction

Given $(Y_i)_{i \ge 1}$ a sequence of i.i.d. square integrable random variables and a measurable real valued function $\phi$, we are going to give sufficient conditions for the convergence in distribution to the uniform law on $[0, 1]$ of

$$ \left\{ \phi \left( \frac{1}{M} \sum_{i=1}^{\beta_M} Y_i \right) (Y_1 + \cdots + Y_M) \right\}, \qquad (3.1) $$

for $(\beta_M)_{M \ge 1} \subseteq \mathbb{N}^*$ such that $\lim_{M \to \infty} \sqrt{M} \left( \frac{\beta_M}{M} - \beta \right) = 0$ with $\beta > 1$. Here $\{x\}$ denotes the fractional part of the real number $x$ given by $\{x\} = x - \lfloor x \rfloor$ where $\lfloor x \rfloor$ is the integer such that $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$ and we also define $\lfloor x \rfloor$ as the integer such that $\lfloor x \rfloor - 1 < x \le \lfloor x \rfloor$. Our main motivation for considering (3.1) comes from the derivation of a central limit theorem for the stratified resampling mechanism. Before giving more details about this particular application, let us review the existing literature which addresses the case of a constant function $\phi$ with the derivation of Benford's law as a common motivation.

The convergence in distribution of the sequence $(V_M = \{Y_1 + \cdots + Y_M\})_{M \ge 1}$ of sums of random variables defined modulo 1 to the uniform distribution on $[0, 1]$ has been studied by many researchers using Fourier analysis. In 1939, Lévy [39] gave necessary and sufficient conditions for this convergence when the $Y_i$ are i.i.d.. In 1986, Störmer [58] provided sufficient conditions in terms of the distribution functions of the $Y_i$ for the convergence to hold when the $Y_i$ are merely independent. In
2007, under the assumption of independent absolutely continuous $Y_i$, Miller and Nigrini [44] char-
acterized the convergence of $(V_M)_{M \ge 1}$ in $L^1([0, 1])$. In 2010, Szewczak [59] generalized the above
results by getting rid of the hypothesis of independence of the $Y_i$. In particular he proved that, if
the absolute value of the characteristic function of $Y_1 + \dots + Y_M$ satisfies a certain growth condition
and if the set $\{n \in \mathbb{N} \setminus \{0\} : |\mathbb{E}(e^{2\pi i n Y_1})| = 1\}$ is empty, then $(V_M)_{M \ge 1}$ converges in distribution
to a uniform random variable on $[0, 1]$.

Let us now briefly introduce Benford's law and the problem of the distribution of the leading digits of products of random variables (see for instance [18]). Benford's law in base $b > 1$ is the probability measure $\mu_b$ on the interval $[1, b)$ defined by

$$
\mu_b([1, a)) = \log_b a, \quad \forall a \in [1, b).
$$

The mantissa in base $b$ of a positive real number $x$ is the unique number $M_b(x)$ in $[1, b)$ such that
$x = M_b(x) \times b^{\lfloor \log_b(x) \rfloor}$. Given a sequence of positive random variables $(X_i)_{i \ge 1}$, many researchers
([54], [9], [55], [40]) were interested in studying the weak convergence as $M \to \infty$ of the law of
$M_b\left(\prod_{i=1}^{M} X_i\right)$ to $\mu_b$. Since $\log_b M_b(x) = \{\log_b(x)\}$ for each positive real number $x$, one has

$$
\log_b M_b \left( \prod_{i=1}^{M} X_i \right) = \left\{ \sum_{i=1}^{M} \log_b (X_i) \right\},
$$

and this convergence is equivalent to the weak convergence of the partial sums modulo 1 of the
random variables $(\log_b (X_i))_{i \ge 1}$ by continuity of $[1, b) \ni z \mapsto \log_b(z)$ and its inverse.

The introduction of a non-constant function $\phi$ in our work permits to address the choice of a
data dependent mantissa for instance given by the geometric mean $\hat{b}_M = \exp\left(\frac{1}{\beta_M} \sum_{i=1}^{\beta_M} \ln(X_i)\right)$.
Indeed,

$$
\log_{\hat{b}_M} \mathcal{M}_{\hat{b}_M} \left( \prod_{i=1}^{M} X_i \right) = \left\{ \frac{\sum_{i=1}^{M} \ln(X_i)}{\frac{1}{\beta_M} \sum_{i=1}^{\beta_M} \ln(X_i)} \right\} = \left\{ \phi \left( \frac{1}{M} \sum_{i=1}^{\beta_M} \ln(X_i) \right) \sum_{i=1}^{M} \ln(X_i) \right\}
$$

where $\phi(x) = \frac{1}{\beta x}$. Compared to previous works, the main difficulty that we have to address comes
from the coupling between the variables introduced through the common factor $\phi\left(\frac{1}{M} \sum_{i=1}^{\beta_M} Y_i\right)$. To
overcome this difficulty, we assume that the $Y_i$ are i.i.d. according to a common distribution with
a non-zero component absolutely continuous with respect to the Lebesgue measure. This allows
us to apply one of the key results for our proof, the convergence in total variation in the central
limit theorem, that we now recall. Let $F$ be a centered square-integrable random vector in $\mathbb{R}^n$ with
identity covariance matrix and let $(F_i)_{i \ge 1}$ independent copies of $F$. Under the assumption that the
law of $F$ has an absolutely continuous component, Prohorov [50] in the one-dimensional case $n = 1$
and Mamatov et Halikov [42] in the multidimensional case, proved that the total variation distance
between the distribution of $\frac{1}{\sqrt{M}} \sum_{i=1}^{M} F_i$ and the standard Gaussian law in $\mathbb{R}^n$ goes to 0 as $M \to \infty$.

The study of the joint convergence of several partial sums modulo 1 with a common factor and an additional component satisfying a central limit theorem does not add further significant difficulties and is useful in the derivation of the central limit theorem for the stratified resampling mechanism. That is why we address the convergence in distribution of

$$
\left( \left( \left\{ \phi \left( \frac{1}{M} \sum_{m=1}^{\beta_M^{q+1}} Y_m \right) (Y_1 + \cdots + Y_{\beta_M^i}) \right\} \right)_{1 \le i \le q} \right) , \sqrt{M} \left( \phi \left( \frac{1}{M} \sum_{m=1}^{\beta_M^{q+1}} Y_m \right) \frac{1}{M} \sum_{i=1}^{\beta_M^{q+1}} Z_i - \theta \right) \right) \quad (3.2)
$$
for $(\beta_M^1, \dots, \beta_M^{q+1})_{M \ge 1} \subseteq \mathbb{N}^{q+1}$ such that $\lim_{M \to \infty} \sqrt{M} \left(\frac{\beta_M^i}{M} - \beta^i\right) = 0$ with $0 < \beta^1 < \dots < \beta^{q+1}$ and $(Z_i)_{i \ge 1}$ such that the sequence $((Y_i, Z_i))_{i \ge 1}$ also is i.i.d. and the last component where $\theta$ is a constant (typically equal to $\beta^{q+1} \phi(\beta^{q+1} \mathbb{E}(Y_1)) \mathbb{E}(Z_1)$) converges in distribution to $T$ as $M \to \infty$. We give mild conditions ensuring that the full vectors converge in distribution to $(U, T)$ where $U$ is uniformly distributed on $[0, 1]^q$ and independent of $T$.

## Main motivation

Let us now come back to our main motivation. We are interested in providing a central limit theorem for the stratified resampling scheme under the simplifying assumption that the $\mathbb{R}^d$-valued initial draws $(X_m)_{m \ge 1}$ are independent and identically distributed and weighted proportionally to their image by some measurable function $g : \mathbb{R}^d \to (0, +\infty)$. For $M \ge 1$, resampling schemes (see for instance [19]) permit to replace the probability measure $\frac{\sum_{m=1}^M g(X_m) \delta_{X_m}}{\sum_{m=1}^M g(X_m)}$ with non equal weights by some empirical measure $\frac{1}{M} \sum_{m=1}^M \delta_{\xi_m^M}$ with the same conditional expectation given $\mathcal{F} = \sigma((X_m)_{m \ge 1})$. For $(U_m)_{m \ge 1}$ an independent sequence of independent random variables uniformly distributed on $(0, 1)$, the stratified resampling scheme consists in setting

$$ \xi_m^M = \sum_{i=1}^{M} 1_{\{\bar{S}_{i-1}^M < m - U_m \le \bar{S}_i^M\}} X_i \quad \text{for } m \in \{1, \dots, M\} $$

with $\bar{S}_j^M = \frac{M \sum_{m=1}^j g(X_m)}{\sum_{m=1}^M g(X_m)}$ for $j \in \{1, \dots, M\}$, under the convention $\bar{S}_0^M = 0$. For $f : \mathbb{R}^d \to \mathbb{R}$ measurable, we have

$$ \mathbb{E}(f(\xi_m^M) | \mathcal{F}) = \sum_{i=1}^{M} f(X_i) \int_{m-1}^{m} 1_{\{\bar{S}_{i-1}^M < u \le \bar{S}_i^M\}} du. \quad (3.3) $$

The central limit theorem deals with the asymptotic behaviour of $\frac{1}{\sqrt{M}} \sum_{m=1}^M \left( f(\xi_m^M) - \frac{\mathbb{E}(f(X_1)g(X_1))}{\mathbb{E}(g(X_1))} \right)$ as $M \to \infty$. Let us explain how the computation of the asymptotic variance is related to our main result. By a standard decomposition of the variance and the conditional independence of $(\xi_m^M)_{1 \le m \le M}$ given $\mathcal{F}$,

$$ \begin{aligned} \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(\xi_m^M) \right) &= \mathrm{Var} \left( \mathbb{E} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(\xi_m^M) \mid \mathcal{F} \right) \right) + \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(\xi_m^M) \mid \mathcal{F} \right) \right) \\ &= \mathrm{Var} \left( \sqrt{M} \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{m=1}^{M} g(X_m)} \right) + \mathbb{E} \left( \frac{\sum_{m=1}^{M} g(X_m) f^2(X_m)}{\sum_{m=1}^{M} g(X_m)} \right) - \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} \left( f(\xi_m^M) \mid \mathcal{F} \right)^2 \right). \end{aligned} $$

Since the asymptotic behaviour as $M \to \infty$ of the first two terms in the right-hand side can be analysed using standard arguments, we focus on that of the third term. Using (3.3) and $\bar{S}_M^M = M$, we get that $\sum_{m=1}^M \mathbb{E}(f(\xi_m^M) | \mathcal{F})^2$ is equal to

$$ \sum_{k=0}^{M-1} (1+1_{\{k \ge 1\}}) \sum_{i=1}^{M-k} f(X_i) f(X_{i+k}) \sum_{m \ge \lceil \bar{S}_{i-1}^M \rceil+1} \int_{m-1}^{m} 1_{\{\bar{S}_{i-1}^M < u \le \bar{S}_i^M\}} du \int_{m-1}^{m} 1_{\{\bar{S}_{i+k-1}^M < u \le \bar{S}_{i+k}^M\}} du. $$

Since $\frac{1}{M} \sum_{i=1}^M \gamma_i = \int_0^1 \gamma_{\lceil \alpha M \rceil} d\alpha$, we deduce that $\frac{1}{M} \sum_{m=1}^M \mathbb{E}(f(\xi_m^M) | \mathcal{F})^2$ is equal to the sum over $k \in \mathbb{N}$ of

$$ (1+1_{\{k \ge 1\}}) \int_0^1 1_{\{\lceil \alpha M \rceil \le M-k\}} f(X_{\lceil \alpha M \rceil}) f(X_{\lceil \alpha M \rceil+k}) \psi_k \left( \left\{ \bar{S}_{\lceil \alpha M \rceil-1}^M \right\}, \frac{Mg(X_{\lceil \alpha M \rceil})}{\sum_{m=1}^M g(X_m)}, \dots, \frac{Mg(X_{\lceil \alpha M \rceil+k})}{\sum_{m=1}^M g(X_m)} \right) d\alpha $$
where $\psi_k(u_0, w_1, \dots, w_{k+1}) = \sum_{m \ge 1} \int_{m-1}^{m} 1_{\{u_0 < u \le u_0 + w_1\}} du \int_{m-1}^{m} 1_{\{u_0 + \sum_{\ell=1}^{k} w_\ell < u \le u_0 + \sum_{\ell=1}^{k+1} w_\ell\}} du$. That is why, in order to compute the asymptotic variance of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(\xi_m^M)$, it is very useful to understand the behaviour as $M \to \infty$ of $\{\bar{S}_{[\alpha M]-1}^M\}$ which is equal to the first component in (3.2) when $q=1$, $Y_i = g(X_i)$, $\phi(x) = \frac{1}{x}$, $\beta_M^1 = [\alpha M] - 1$ and $\beta_M^2 = M$.

Regarding the Central Limit Theorem, the characteristic function of

$$ \frac{1}{\sqrt{M}} \sum_{m=1}^{M} \left( f(\xi_m^M) - \frac{\mathbb{E}(f(X_1)g(X_1))}{\mathbb{E}(g(X_1))} \right) $$

writes for $u \in \mathbb{R}$

$$ \mathbb{E} \left( e^{i u \sqrt{M} \left( \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{m=1}^{M} g(X_m)} - \frac{\mathbb{E}(f(X_1)g(X_1))}{\mathbb{E}(g(X_1))} \right)} \prod_{m=1}^{M} \mathbb{E} \left( e^{\frac{i u}{\sqrt{M}} \left( f(\xi_m^M) - \mathbb{E}(f(\xi_m^M | \mathcal{F})) \right)} \mid \mathcal{F} \right) \right). $$

Using some Taylor expansion of $e^{\frac{i u}{\sqrt{M}} (f(\xi_m^M) - \mathbb{E}(f(\xi_m^M | \mathcal{F}))}$, one may approximate the product by a sum of integrals of functions of

$$ \left( \{ \bar{S}_{[\alpha_1 M]-1}^M \}, \{ \bar{S}_{[\alpha_2 M]-1}^M \}, \dots, \{ \bar{S}_{[\alpha_q M]-1}^M \} \right) $$

with respect to $1_{\{0 < \alpha_1 < \alpha_2 < \dots < \alpha_q < 1\}} d\alpha_1 d\alpha_2 \dots d\alpha_q$. This closely relates the asymptotic behaviour of the characteristic function to the one of the vectors (3.2) for the choice $Y_i = g(X_i)$, $\phi(x) = \frac{1}{x}$, $Z_i = g(X_i)f(X_i)$, $\beta_M^i = [\alpha_i M] - 1$ for $i \in \{1, \dots, q\}$ and $\beta_M^{q+1} = M$. In the companion paper [25], using the main result of the present paper, we compute the asymptotic variance and prove the associated central limit theorem.

## 3.2 Notation

We denote by $\mathbb{N}^*$ the set of natural numbers without 0 and by $\mathbb{R}_+$ the set of non negative real numbers. We denote by $\lfloor x \rfloor$ the integer $j$ such that $j \le x < j+1$ and by $\{x\} = x - \lfloor x \rfloor$ the fractional part of $x \in \mathbb{R}$.

We denote by $\mu_Y$ the law of a $\mathbb{R}^d$-valued random vector $Y = (Y^1, \dots, Y^d)$ where $d \in \mathbb{N}^*$ and by $\phi_Y$ its characteristic function given by $\phi_Y(u) = \mathbb{E}(e^{iu^T Y}) = \int_{\mathbb{R}^d} e^{iu^T Y} \mu_Y(dy)$, $u \in \mathbb{R}^d$. The convergence in distribution is denoted by $\stackrel{d}{\Rightarrow}$. Moreover let $\mu_Y = \mu_{Y,c} + \mu_{Y,s}$ denote the decomposition of $\mu_Y$ into a part $\mu_{Y,c}$ absolutely continuous with respect to the Lebesgue measure and a singular part $\mu_{Y,s}$. Notice that there exists $A$, a Borel subset of $\mathbb{R}^d$, such that $\mu_{Y,s}(A) = 0$ and $\int_{\mathbb{R}^d} 1_{\{x \notin A\}} dx = 0$. Let $p_Y$ denote a density of $\mu_{Y,c}$ with respect to the Lebesgue measure. In what follows, we will always consider absolute continuity with respect to the Lebesgue measure and so we avoid to write it everytime. Moreover, when we write that the law of a random variable has an absolutely continuous component we mean a non zero absolutely continuous component. We denote by $m_Y = (\mathbb{E}(Y^1), \dots, \mathbb{E}(Y^d))$ the expected value vector of $Y$.

For the total variation distance between the measures $\mu_1$ and $\mu_2$, we write

$$ d_{TV}(\mu_1, \mu_2) = \sup_{A \in \mathcal{B}(\mathbb{R}^d)} |\mu_1(A) - \mu_2(A)| = \frac{1}{2} \sup_{\|f\|_{\infty, \mathbb{R}} \le 1} \left| \int_{\mathbb{R}^d} f(x) (\mu_1(dx) - \mu_2(dx)) \right| \quad (3.4) $$

$$ = \frac{1}{2} \sup_{\|f\|_{\infty, \mathbb{C}} \le 1} \left| \int_{\mathbb{R}^d} f(x) (\mu_1(dx) - \mu_2(dx)) \right| \quad (3.5) $$
where the first supremum is taken over the real-valued measurable functions and the second one is taken over the complex-valued measurable functions. The latter formulation is less usual so we will provide the proof of the third equality in the Appendix (see Lemma 10).

Given two $\mathbb{R}^d$-valued random vectors $X$ and $Y$ and a measurable function $g : \mathbb{R}^d \to \mathbb{R}^{d'}$ where $d' \in \mathbb{N}^*$, the following inequality holds

$$d_{TV}(\mu_{g(X)}, \mu_{g(Y)}) \le d_{TV}(\mu_X, \mu_Y). \qquad (3.6)$$

In particular,

$$d_{TV}(\mu_{(0,X)}, \mu_{(0,Y)}) = d_{TV}(\mu_X, \mu_Y) \qquad (3.7)$$

and given $A \in \mathbb{R}^{d \times d}$ an invertible matrix

$$d_{TV}(\mu_{AX}, \mu_{AY}) = d_{TV}(\mu_X, \mu_Y). \qquad (3.8)$$

Additionally we introduce the following notation: given a real sequence $(x_j)_{j \ge 1}$ and a real number $c \in \mathbb{R}$ we define for any integer $M \ge 1$

$$\bar{x}_M^c := \frac{x_1 + \cdots + x_M + c}{M}.$$

## 3.3 Main Result

Let $(Y_i, Z_i)_{i \ge 1}$ be a sequence of square-integrable i.i.d. random vectors in $\mathbb{R}^2$ where $Y_i$ is not constant. Moreover let $q \in \mathbb{N}^*$ and consider a sequence of vectors of integers $(\beta_M^1, \cdots, \beta_M^{q+1})_{M \ge 1} \subseteq \mathbb{N}^{q+1}$ such that

$$\lim_{M \to \infty} \sqrt{M} \left( \frac{\beta_M^i}{M} - \beta^i \right) = 0$$

with

$$0 < \beta^1 < \cdots < \beta^{q+1}.$$

Let us observe that for $M$ big enough $0 < \beta_M^1 < \cdots < \beta_M^{q+1}$. Given $(x, z, y_1, \cdots, y_q) \in \mathbb{R}^{q+2}$ and $\phi$ a measurable real-valued function, we are interested in studying the convergence in distribution as $M \to \infty$ of the following random vector

$$\left( \left\{ R_{\beta_M}^{1, y_1} \right\}, \cdots, \left\{ R_{\beta_M}^{q, y_q} \right\}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}}^z - \theta \right) \right) \qquad (3.9)$$

where

$$R_{\beta_M}^{i, y_i} := \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) (Y_1 + \cdots + Y_{\beta_M^i} + y_i) \qquad (3.10)$$

and

$$\theta = \phi(\beta^{q+1}m_Y) \beta^{q+1}m_Z. \qquad (3.11)$$

**Remark 26.** Let us observe that if $\phi$ is differentiable at $\beta^{q+1}m_Y$, the application of the delta method provides the convergence in distribution of the last component of (3.9):

$$\sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}}^z - \theta \right) \stackrel{d}{\Longrightarrow} T \sim \mathcal{N}(0, \sigma_T^2) \qquad (3.12)$$

where $\sigma_T^2 = \beta^{q+1} (\phi'(\beta^{q+1}m_Y)\beta^{q+1}m_Z, \phi(\beta^{q+1}m_Y))^T \Sigma_{(Y_1, Z_1)} (\phi'(\beta^{q+1}m_Y)\beta^{q+1}m_Z, \phi(\beta^{q+1}m_Y))$ being $\Sigma_{(Y_1, Z_1)}$ the covariance matrix of $(Y_1, Z_1)$.

We are now ready to state the main result of this work.
**Theorem 27.** Let $(Y_i, Z_i)_{i \ge 1}$ be a sequence of square-integrable i.i.d. random vectors in $\mathbb{R}^2$ such that the law of $Y_i$ has an absolutely continuous component. Moreover let $(x, z, y_1, \dots, y_q) \in \mathbb{R}^{q+2}$ and let $\phi: \mathbb{R} \to \mathbb{R}$ be a measurable function differentiable at $\beta^{q+1}m_Y$ such that $\phi(\beta^{q+1}m_Y) \neq 0$. If there exists $\tilde{M} \in \mathbb{N}$ such that

$$ \int_{\mathbb{R}} \frac{1}{|\phi(\beta^{q+1}m_Y + y)|^{q+1}} e^{-\tilde{M}y^2} dy < \infty, \quad (3.13) $$

then the random vector

$$ \left( \left\{ R_{\beta_M}^{1,y_1} \right\}, \dots, \left\{ R_{\beta_M}^{q,y_q} \right\}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M}^{x} \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M}^{z} - \theta \right) \right) $$

converges in distribution as $M \to \infty$ to $(U, T)$ where $T$ has been introduced in (3.12) and where $U$ is a uniform random variable on $[0, 1]^q$ independent of $T$.

**Remark 28.** Let us observe that we can apply Theorem 27 to $(Y_i, 0)_{i \ge 1}$ where $(Y_i)_{i \ge 1}$ is a sequence of square-integrable i.i.d. real-valued random variables such that the law of $Y_i$ has an absolutely continuous component. In particular we have

$$ \left( \left\{ R_{\beta_M}^{1,y_1} \right\}, \dots, \left\{ R_{\beta_M}^{q,y_q} \right\} \right) \stackrel{d}{\Longrightarrow} U $$

where $U$ is a uniform random variable on $[0, 1]^q$. It is possible to prove that in this case the hypothesis (3.13) can be replaced by the following slightly weaker hypothesis: $\exists \tilde{M} \in \mathbb{N}$ such that

$$ \int_{\mathbb{R}} \frac{1}{|\phi(\beta^{q+1}m_Y + y)|^q} e^{-\tilde{M}y^2} dy < \infty. \quad (3.14) $$

We are now going to provide the statement of Theorem 27 in the particular case where the law of $(Y_1, Z_1)$ has an absolutely continuous component and $(z, y_1, \dots, y_q) = (0, \dots, 0)$. This, together with the lemma that immediately follows, will allow to prove Theorem 27.

**Proposition 5.** Theorem 27 holds under the reinforced hypotheses that the law of $(Y_i, Z_i)$ has an absolutely continuous component and $(z, y_1, \dots, y_q) = (0, \dots, 0)$.

**Lemma 6.** Let $Y$ and $Z$ be two real-valued random variables such that the law of $Y$ has an absolutely continuous component. If $\xi$ is an absolutely continuous real-valued random variable independent of $(Y, Z)$, the law of $(Y, Z + \xi)$ has an absolutely continuous component.

We provide the proof of Proposition 5 in Section 3.4 and the proof of Lemma 6 in the Appendix.

*Proof of Theorem 27.* The proof consists of two steps. In step (i) we are going to check that the conclusion still holds when the hypothesis of existence of an absolutely continuous component for $(Y_1, Z_1)$ made in Proposition 5 is weakened to the existence of an absolutely continuous component for $Y_1$. Moreover we suppose that $(z, y_1, \dots, y_q) = (0, \dots, 0)$. In step (ii) we deal with the case when the vector $(z, y_1, \dots, y_q) \neq (0, \dots, 0)$.

(i) To simplify the notation in what follows we write $R_{\beta_M}^i$ instead of $R_{\beta_M}^{i,0}$ and $\bar{Z}_{\beta_M}^{q+1}$ instead of $\bar{Z}_{\beta_M}^0$.

Let $(\xi_i)_{i \ge 1}$ be a sequence of zero-mean absolutely continuous square-integrable i.i.d. real-valued random variables independent of $(Y_i, Z_i)_{i \ge 1}$. For each $n \ge 1$, let $\xi_i^n = \frac{\xi_i}{n}$ and let us consider the sequence $(Y_i, Z_i + \xi_i^n)_{i \ge 1}$ of square-integrable i.i.d. random vectors in $\mathbb{R}^2$ such that by Lemma 6 the law of $(Y_i, Z_i + \xi_i^n)$ has an absolutely continuous component. Thus we can apply Proposition 5 and obtain that for each $n \ge 1$ the random vector
$$ \left( \{R_{\beta_M}^1\}, \dots, \{R_{\beta_M}^q\}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \left( \bar{Z}_{\beta_M^{q+1}} + \bar{\xi}^n_{\beta_M^{q+1}} \right) - \theta \right) \right) \stackrel{d}{\Longrightarrow} (U, T_n) \quad (3.15) $$

where $\theta = \phi(\beta^{q+1}m_Y)\beta^{q+1}m_Z$, $U$ is a uniform random variable on $[0, 1]^q$ independent of $T_n$ and by Remark 26, $T_n \sim \mathcal{N}(0, \sigma_{T_n}^2)$ with

$$ \sigma_{T_n}^2 = \beta^{q+1} (\phi'(\beta^{q+1}m_Y)\beta^{q+1}m_Z, \phi(\beta^{q+1}m_Y))^T \Sigma_{(Y_1, Z_1+\xi_1^n)} (\phi'(\beta^{q+1}m_Y)\beta^{q+1}m_Z, \phi(\beta^{q+1}m_Y)). $$

Since $(Y_1, Z_1 + \xi_1^n)$ converges in $L^2$ to $(Y_1, Z_1)$ as $n$ goes to infinity, $\Sigma_{(Y_1, Z_1+\xi_1^n)} \xrightarrow{n \to \infty} \Sigma_{(Y_1, Z_1)}$ and $\sigma_{T_n}^2$ converges as $n \to \infty$ to

$$ \sigma_T^2 = \beta^{q+1} (\phi'(\beta^{q+1}m_Y)\beta^{q+1}m_Z, \phi(\beta^{q+1}m_Y))^T \Sigma_{(Y_1, Z_1)} (\phi'(\beta^{q+1}m_Y)\beta^{q+1}m_Z, \phi(\beta^{q+1}m_Y)). $$

Let us now prove that this implies the convergence in distribution as $M \to \infty$ of (3.9) to (U, T) where $T \sim \mathcal{N}(0, \sigma_T^2)$ independent of $U = (U_1, \dots, U_q)$.

Let $(u_1, u_2) \in \mathbb{R}^q \times \mathbb{R}$ and let us denote the random vector $(\{R_{\beta_M}^1\}, \dots, \{R_{\beta_M}^q\})$ by $\{R_{\beta_M}^i\}_{i \le q}$. For $n \ge 1$ one has

$$ \begin{aligned} & \left| \mathbb{E} \left( e^{iu_1^T \left( \{R_{\beta_M}^i\}_{i \le q} \right)} + iu_2 \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}} - \theta \right) \right) - \mathbb{E} \left( e^{iu_1^T U} \right) e^{-\frac{1}{2} \sigma_T^2 u_2^2} \right| \\ & \le \left| \mathbb{E} \left( e^{iu_1^T \left( \{R_{\beta_M}^i\}_{i \le q} \right)} e^{iu_2 \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}} - \theta \right) \left( 1 - e^{iu_2 \sqrt{M} \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \frac{\beta_M^{q+1}}{M} \bar{\xi}^n_{\beta_M^{q+1}} } \right) \right) \right| \\ & \quad + \left| \mathbb{E} \left( e^{iu_1^T \left( \{R_{\beta_M}^i\}_{i \le q} \right)} + iu_2 \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \left( \bar{Z}_{\beta_M^{q+1}} + \bar{\xi}^n_{\beta_M^{q+1}} \right) - \theta \right) \right) - \mathbb{E} \left( e^{iu_1^T U} \right) e^{-\frac{1}{2} \sigma_{T_n}^2 u_2^2} \right| \\ & \quad + \left| \mathbb{E} \left( e^{iu_1^T U} \right) e^{-\frac{1}{2} \sigma_{T_n}^2 u_2^2} - \mathbb{E} \left( e^{iu_1^T U} \right) e^{-\frac{1}{2} \sigma_T^2 u_2^2} \right| \\ & \le \mathbb{E} \left( \left| 1 - e^{iu_2 \sqrt{M} \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \frac{\beta_M^{q+1}}{M} \bar{\xi}^n_{\beta_M^{q+1}} } \right| \right) \\ & \quad + \left| \mathbb{E} \left( e^{iu_1^T \left( \{R_{\beta_M}^i\}_{i \le q} \right)} + iu_2 \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \left( \bar{Z}_{\beta_M^{q+1}} + \bar{\xi}^n_{\beta_M^{q+1}} \right) - \theta \right) \right) - \mathbb{E} \left( e^{iu_1^T U} \right) e^{-\frac{1}{2} \sigma_{T_n}^2 u_2^2} \right| \\ & \quad + \left| e^{-\frac{1}{2} \sigma_{T_n}^2 u_2^2} - e^{-\frac{1}{2} \sigma_T^2 u_2^2} \right|. \end{aligned} $$

Let us now study the first term of the right-hand side.

By observing that $H_M = \sqrt{M}\phi\left(\frac{\beta_M^{q+1}}{M}\bar{Y}_{\beta_M^{q+1}}^x\right)\frac{\beta_M^{q+1}}{M}\bar{\xi}_{\beta_M^{q+1}} \stackrel{d}{\Longrightarrow} \mathcal{N}(0, \beta^{q+1}\phi^2(\beta^{q+1}m_Y)\text{Var}(\xi))$, this in particular implies that the sequence $H_M$ is tight: $\forall \epsilon > 0$ there exists $K_\epsilon > 0$ such that $\sup_{M \ge 1} \mathbb{P}(|H_M| > K_\epsilon) \le \epsilon$. One has
$$
\begin{align*}
\mathbb{E} \left( \left| 1 - e^{i u_2 \sqrt{M} \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \frac{\beta_M^{q+1}}{M} \bar{\xi}^n_{\beta_M^{q+1}} } \right| \right) &= \mathbb{E} \left( \left| 1 - e^{i \frac{u_2}{n} H_M } \right| \right) \\
&= \mathbb{E} \left( \left| 1 - e^{i \frac{u_2}{n} H_M } \right| \mathbf{1}_{|H_M| \le K_\epsilon} \right) + \mathbb{E} \left( \left| 1 - e^{i \frac{u_2}{n} H_M } \right| \mathbf{1}_{|H_M| > K_\epsilon} \right) \\
&\le \frac{|u_2|}{n} K_\epsilon + 2\epsilon
\end{align*}
$$

where to obtain the last inequality we use that $|1 - e^{ix}| \le |x|$, $\forall x \in \mathbb{R}$.
Therefore we have obtained that $\forall \epsilon > 0$, $\forall n > 0$

$$
\left| \mathbb{E} \left( e^{i u_1^T \left( \{R_{\beta_M}^i\}_{i \le q} \right)} + i u_2 \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}} - \theta \right) \right) - \mathbb{E} \left( e^{i u_1^T U} \right) e^{-\frac{1}{2} \sigma_T^2 u_2^2} \right| \\
\le \frac{|u_2|}{n} K_\epsilon + 2\epsilon + \left| e^{-\frac{1}{2} \sigma_T^2 u_2^2} - e^{-\frac{1}{2} \sigma_T^2 u_2^2} \right| \\
+ \left| \mathbb{E} \left( e^{i u_1^T \left( \{R_{\beta_M}^i\}_{i \le q} \right)} + i u_2 \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \left( \bar{Z}_{\beta_M^{q+1}} + \bar{\xi}^n_{\beta_M^{q+1}} \right) - \theta \right) \right) - \mathbb{E} \left( e^{i u_1^T U} \right) e^{-\frac{1}{2} \sigma_T^2 u_2^2} \right|.
$$

To conclude this part of the proof it is sufficient to take in the above expression first the superior limit as $M \to \infty$ so that the last term converges to 0 by (3.15), then the superior limit as $n \to \infty$ so that $|u_2|/n K_\epsilon \to 0$ and $|e^{-1/2 \sigma_T^2 u_2^2} - e^{-1/2 \sigma_T^2 u_2^2}| \to 0$ and finally the superior limit as $\epsilon \to 0$.

(ii) Let us now take $(z, y_1, \dots, y_q) \in \mathbb{R}^{q+2}$ different from the zero vector. One has

$$
\begin{align}
& \left( \left\{ R_{\beta_M}^{1,y_1} \right\}, \dots, \left\{ R_{\beta_M}^{1,y_q} \right\}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}}^z - \theta \right) \right) \tag{3.16} \\
&= \left( \left( \left\{ \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) y_i + \left\{ R_{\beta_M}^i \right\} \right\} \right)_{1 \le i \le q}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}} - \theta \right) \right) \nonumber \\
&\quad + \left( 0, \dots, 0, \frac{z}{\sqrt{M}} \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \right). \nonumber
\end{align}
$$

By the Strong Law of Large Numbers and the hypothesis of continuity of $\phi$ at $\beta^{q+1}m_Y$,
$$
\lim_{M \to \infty} \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) y_i = \phi (\beta^{q+1}m_Y) y_i. \text{ We can therefore apply the previous step and Slutsky's theorem to deduce the following convergence in distribution}
$$

$$
\left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^{x_2} \right) y_1 + \{R_{\beta_M}^1\}, \dots, \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) y_q + \{R_{\beta_M}^q\}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}} - \theta \right) \right) \\
\stackrel{d}{\Rightarrow} \left( \phi (\beta^{q+1} m_Y) y_1 + U_1, \dots, \phi (\beta^{q+1} m_Y) y_q + U_q, T \right).
$$

Now by observing that the set of the points of discontinuity of the function $(x_1, \dots, x_q, y) \mapsto (\{x_1\}, \dots, \{x_q\}, y)$ has a zero measure with respect to the law of

$$
(\phi (\beta^{q+1} m_Y) y_1 + U_1, \dots, \phi (\beta^{q+1} m_Y) y_q + U_q, T)
$$

and applying the continuous mapping theorem we can deduce that

$$
\left( \left( \left\{ \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) y_i + \{R_{\beta_M}^i\} \right\} \right)_{1 \le i \le q}, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}} - \theta \right) \right) \\
\stackrel{d}{\Rightarrow} \left( \left\{ \phi (\beta^{q+1} m_Y) y_1 + U_1 \right\}, \dots, \left\{ \phi (\beta^{q+1} m_Y) y_q + U_q \right\}, T \right) \stackrel{d}{=} (U_1, \dots, U_q, T).
$$
This combined with the fact that $(0, \dots, 0, \frac{z}{\sqrt{M}}\phi\left(\frac{\beta_M^{q+1}}{M}\overline{Y}\beta_M^{q+1}\right))$ converges as $M$ goes to infinity to the zero vector allows to conclude that (3.16) converges in distribution to $(U, T)$. $\square$

## 3.4 Proof of Proposition 5

Before proving Proposition 5, we need some preliminary results. As done in the proof of Theorem 27, in what follows we write $R_{\beta_M}^i$ instead of $R_{\beta_M}^{i,0}$ and $\overline{Z}_{\beta_M^{q+1}}$ instead of $\overline{Z}_{\beta_M^{q+1}}^0$. Let us observe that for each $j = 1, \dots, q$

$$R_{\beta_M}^j = \sum_{\ell=1}^{j} R_{\beta_M}^{\ell-1:\ell}$$

where we have introduced the notation $R_{\beta_M}^{\ell-1:\ell} := \phi\left(\frac{\beta_M^{q+1}}{M}\overline{Y}\beta_M^{q+1}\right) (Y_{\beta_M^{\ell-1}+1} + \dots + Y_{\beta_M^{\ell}})$ with $\beta_M^0 := 0$ for each $M$ by convention. Let us therefore study the asymptotic behaviour of the vector

$$(R_M, K_M) \qquad (3.17)$$

where

$$R_M := (R_{\beta_M}^{0:1}, R_{\beta_M}^{1:2}, \dots, R_{\beta_M}^{q-1:q})$$

and

$$K_M := \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \overline{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \overline{Z}_{\beta_M^{q+1}} - \theta \right) = \frac{\phi \left( \frac{\beta_M^{q+1}}{M} \overline{Y}_{\beta_M^{q+1}}^x \right) \beta_M^{q+1} \overline{Z}_{\beta_M^{q+1}} - M\theta}{\sqrt{M}}$$

with $\theta$ as in (3.11).

The proof of the following proposition is given in Section 3.5.

**Proposition 6.** Let $\gamma = (\beta^1, \beta^2 - \beta^1, \dots, \beta^q - \beta^{q-1})$. Under the assumptions of Proposition 5 the following convergence in total variation holds

$$d_{TV}\left(\mu\left(\frac{R_{M-m_Y\phi(\beta^{q+1}m_Y)M\gamma}}{\sqrt{M}}, K_M\right), \mathcal{N}(0, \Gamma)\right) \underset{M \to \infty}{\longrightarrow} 0 \qquad (3.18)$$

where $\Gamma \in \mathbb{R}^{(q+1) \times (q+1)}$ is a positive definite covariance matrix.

Let us now state the Weyl criterion concerning the convergence in distribution to a vector composed of a uniform random variable on $[0, 1]^q$ and an independent vector.

**Theorem 29 (Weyl criterion).** Let $(B_M)_{M \ge 1}$ be a sequence of $\mathbb{R}^q$-valued random vectors and let $(H_M)_{M \ge 1}$ be a sequence of $\mathbb{R}^{q'}$-valued random vectors that converges in distribution to a $\mathbb{R}^{q'}$-valued random vector $H$. Then as $M \to \infty$ the sequence $(\{B_M^1\}, \dots, \{B_M^q\}, H_M)$ converges in distribution to $(U, H)$ where $U$ is independent of $H$ and uniformly distributed on $[0, 1]^q$ if and only if for every $k \in \mathbb{Z}^q \setminus \{0\}$ and $u \in \mathbb{R}^{q'}$

$$\lim_{M \to \infty} \phi_{(B_M, H_M)} (2k\pi, u) = 0. \qquad (3.19)$$

For the sake of completeness, the proof of Theorem 29 is provided in Section 3.5. We are now ready to prove Proposition 5.
Proof of Proposition 5. By Theorem 29, it is sufficient to prove that for each $k \in \mathbb{Z}^q \setminus \{0\}$, $u \in \mathbb{R}$

$$ \lim_{M \to \infty} \phi \left( R_{\beta_M}^1, \dots, R_{\beta_M}^q, \sqrt{M} \left( \phi \left( \frac{\beta_M^{q+1}}{M} \bar{Y}_{\beta_M^{q+1}}^x \right) \times \frac{\beta_M^{q+1}}{M} \bar{Z}_{\beta_M^{q+1}-\theta} \right) \right) (2k\pi, u) = 0. $$

Let us observe that it is equivalent to prove that for each $k \in \mathbb{Z}^q \setminus \{0\}$, $u \in \mathbb{R}$

$$ \lim_{M \to \infty} \phi_{(R_M, K_M)}(2k\pi, u) = 0 \qquad (3.20) $$

where $(R_M, K_M)$ has been introduced in (3.17).

By Proposition 6,

$$ \lim_{M \to \infty} d_{TV}(\mu_{(a_M R_M + b_M, K_M)}, \mu_Y) = 0 \qquad (3.21) $$

where $a_M = \frac{1}{\sqrt{M}}$, $b_M = -m_Y \phi(\beta^{q+1} m_Y) \sqrt{M} \gamma$ and $Y \sim \mathcal{N}(0, \Gamma)$, $\Gamma$ positive definite.

For $u := (u_1, u_2) \in \{\mathbb{R}^q \setminus \{0\}\} \times \mathbb{R}$

$$ \begin{aligned} |\phi_{(R_M, K_M)}(u_1, u_2)| &= \left| \phi_{(a_M R_M + b_M, K_M)} \left( \frac{u_1}{a_M}, u_2 \right) e^{-\frac{i u_1^T b_M}{a_M}} \right| = \left| \phi_{(a_M R_M + b_M, K_M)} \left( \frac{u_1}{a_M}, u_2 \right) \right| \\ &\le \left| \phi_{(a_M R_M + b_M, K_M)} \left( \frac{u_1}{a_M}, u_2 \right) - \phi_Y \left( \frac{u_1}{a_M}, u_2 \right) \right| + \left| \phi_Y \left( \frac{u_1}{a_M}, u_2 \right) \right|. \end{aligned} $$

Since by (3.5) the first term of the right-hand side can be bounded from above by $2d_{TV}(\mu_{(a_M R_M + b_M, K_M)}, \mu_Y)$, we deduce that

$$ \limsup_{M \to \infty} |\phi_{(R_M, K_M)}(u_1, u_2)| \le \limsup_{M \to \infty} \left| \phi_Y \left( \frac{u_1}{a_M}, u_2 \right) \right|. $$

Since the law of $Y$ is absolutely continuous and $\lim_{M \to \infty} a_M = 0$, the right-hand side goes to 0 as $M \to \infty$ by the Riemann Lebesgue lemma. In particular (3.20) is true.

□

## 3.5 Proof of Proposition 6 and Theorem 29

In this section we are first going to prove Proposition 6. Let $(Y_i, Z_i)_{i \ge 1}$ be a sequence of square-integrable i.i.d. random vectors in $\mathbb{R}^2$ such that the law of $(Y_i, Z_i)$ has an absolutely continuous component. We denote by $\Sigma$ its covariance matrix which has rank 2.

The proof of Proposition 6 strongly relies on the following result.

**Theorem 30.** Let $(Y_i, Z_i)_{i \ge 1}$ be a sequence of square-integrable i.i.d. random vectors in $\mathbb{R}^2$ such that the law of $(Y_i, Z_i)$ has an absolutely continuous component. Under the notations introduced above

$$ \lim_{M \to \infty} d_{TV} \left( \mu_{\frac{1}{\sqrt{M}} \sum_{k=1}^{M} (Y_k - m_Y, Z_k - m_Z)}, \mu_G \right) = 0 $$

where $G \sim \mathcal{N}(0, \Sigma)$.

Let $F$ be a centered square-integrable random variable in $\mathbb{R}^n$ with identity covariance matrix and let $F_k$, $k \in \mathbb{N}^*$, independent copies of $F$. The main instrument that we will use in the proof of Theorem 30 is the result about the convergence in total variation for the CLT that is $\lim_{M \to \infty} d_{TV} \left( \mu_{\frac{1}{\sqrt{M}} \sum_{k=1}^{M} F_k}, \mathcal{N}(0, I_{n \times n}) \right) = 0$ where $I_{n \times n}$ denotes the identity matrix of size $n$. Prohorov [50] in 1952 was the first to give his contribution to the problem: he proved that, in dimension 1, a necessary and sufficient condition in order to get the result is that there exists
$M_0$ such that the law of $\sum_{k=1}^{M_0} F_k$ has an absolutely continuous component. In 1964, Mamatov et Halikov [42] contribute in the same direction by extending the result to any dimension. They prove that under the assumption that the law of $F$ has an absolutely continuous component, $\lim_{M \to \infty} d_{TV}\left(\mu_{\frac{1}{\sqrt{M}}\sum_{k=1}^{M} F_k}, \mathcal{N}(0, I_{n \times n})\right) = 0$. See also [3].

We are now ready to prove Theorem 30.

*Proof of Theorem 30.* Let $O \in \mathbb{R}^{2 \times 2}$ be an orthogonal matrix ($O^T O = I$) that diagonalizes $\Sigma$ that is $\Sigma = O^T D O$ where $D$ is the diagonal matrix containing the eigenvalues $\lambda_1, \lambda_2$ of $\Sigma$. Given $G \sim \mathcal{N}(0, \Sigma)$ and by introducing the following notation $D^{-\frac{1}{2}} = \text{diag}(\frac{1}{\sqrt{\lambda_1}}, \frac{1}{\sqrt{\lambda_2}})$, we have

$$d_{TV}\left(\mu_{\frac{1}{\sqrt{M}}\sum_{k=1}^{M}(Y_k-m_Y, Z_k-m_Z)}, \mu_G\right) = d_{TV}\left(\mu_{\frac{D^{-\frac{1}{2}}}{\sqrt{M}}\sum_{k=1}^{M}O(Y_k-m_Y, Z_k-m_Z)}, \mu_{D^{-\frac{1}{2}}OG}\right)$$

where to obtain the above equality we apply (3.8) with $A = D^{-\frac{1}{2}}O$. By the hypothesis that the law of $(Y_1, Z_1)$ has an absolutely continuous component, the right-hand side converges to 0 as $M \to \infty$ by the result of convergence in total variation mentioned above. $\square$

We now recall a result obtained by Parthasarathy and Steerneman in the second section of [49] regarding the behavior of the total variation convergence with respect to the sum and the multiplication by a real sequence.

**Lemma 7.** Let $(B_M)_{M \ge 1}$ and $(T_M)_{M \ge 1}$ be two independent sequences of $\mathbb{R}^d$-valued random variables such that $\lim_{M \to \infty} d_{TV}(\mu_{B_M}, \mu_B) = 0$ and $T_M \stackrel{d}{\Rightarrow} T$ for $\mathbb{R}^d$-valued random variables $B, T$. If $\mu_B$ is absolutely continuous with respect to the Lebesgue measure, then $\lim_{M \to \infty} d_{TV}(\mu_{B_M+T_M}, \mu_{B+T}) = 0$. Moreover under the same condition, $\lim_{M \to \infty} d_{TV}(\mu_{c_M B_M}, \mu_{c_0 B}) = 0$ if $(c_M)_{M \ge 1}$ is a deterministic real sequence converging to $c_0 \in \mathbb{R}^*$.

The first step in the proof of Proposition 6 consists in applying Theorem 30 so that we can work with gaussian random vectors. Thus the new problem becomes to study the convergence in total variation of the law of a given function of a normal random vector. The following lemma deals with this problem and its proof is given after the proof of Proposition 6.

**Lemma 8.** Let

* $(\eta_{M,1}, \dots, \eta_{M,q+2})_{M \ge 1} \subseteq \mathbb{R}^{q+2}$ such that for $i = 1, \dots, q+2$ $\lim_{M \to \infty} \sqrt{M} \left(\frac{\eta_{M,i}}{M} - \eta_i\right) = 0$ with $(\eta_1, \dots, \eta_{q+2}) \in \mathbb{R}^{q+2}$

* $(W_1, \dots, W_{q+2})$ a zero-mean normal vector with a positive definite covariance matrix $\Sigma_1 \in \mathbb{R}^{(q+2) \times (q+2)}$

* $\phi: \mathbb{R} \to \mathbb{R}$ be a measurable function differentiable at $\eta_{q+2}$ and such that $\phi(\eta_{q+2}) \neq 0$.

If there exists $\tilde{M} \in \mathbb{N}$ such that

$$\int_{\mathbb{R}} \frac{1}{|\phi(\eta_{q+2} + y)|^{q+1}} e^{-\tilde{M}y^2} dy < \infty, \quad (3.22)$$

the law of the following random vector

$$\left( \frac{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{W_{q+2}}{\sqrt{M}}\right) \left(\sqrt{M}W_i + \eta_{M,i}\right) - M\phi(\eta_{q+2})\eta_i}{\sqrt{M}} \right)_{1 \le i \le q+1} \quad (3.23)$$

converges in total variation as $M \to \infty$ to the law of $(\phi(\eta_{q+2})W_i + \eta_i\phi'(\eta_{q+2})W_{q+2})_{1 \le i \le q+1}$.
**Remark 31.** Let us observe that it is not difficult to prove the pointwise convergence of (3.23) to

$$ (\phi(\eta_{q+2})W_i + \eta_i\phi'(\eta_{q+2})W_{q+2})_{1\le i\le q+1}. $$

Indeed for $i = 1, \dots, q+1$, using that $\lim_{M \to \infty} \sqrt{M} \left(\frac{\eta_{M,j}}{M} - \eta_j\right) = 0$ and the hypothesis of differentiability of $\phi$ at $\eta_{q+2}$, one has

$$ \frac{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{W_{q+2}}{\sqrt{M}}\right)\left(\sqrt{M}W_i + \eta_{M,i}\right) - M\phi(\eta_{q+2})\eta_i}{\sqrt{M}} \\ = \phi(\eta_{q+2})\left(W_i + \sqrt{M}\left(\frac{\eta_{M,i}}{M} - \eta_i\right)\right) + \phi'(\eta_{q+2})\left(W_{q+2} + \sqrt{M}\left(\frac{\eta_{M,q+2}}{M} - \eta_{q+2}\right)\right)\left(\frac{W_i}{\sqrt{M}} + \frac{\eta_{M,i}}{M}\right) + o(1) \\ \xrightarrow{M \to \infty} \phi(\eta_{q+2})W_i + \eta_i\phi'(\eta_{q+2})W_{q+2}. $$

The proof of the convergence in total variation will require more effort.

**Remark 32.** Let us observe that the covariance matrix of the zero-mean normal random vector $(\phi(\eta_{q+2})W_i + \eta_i\phi'(\eta_{q+2})W_{q+2})_{1\le i\le q+1}$ is $A_1\Sigma_1 A_1^T$ with $A_1 \in \mathbb{R}^{q+1 \times (q+2)}$ the $q+1$-rank matrix given by

$$ A_1 = \begin{pmatrix} \phi(\eta_{q+2}) & 0 & 0 & 0 & \cdots & 0 & \eta_1\phi'(\eta_{q+2}) \\ 0 & \phi(\eta_{q+2}) & 0 & \cdots & 0 & 0 & \eta_2\phi'(\eta_{q+2}) \\ 0 & 0 & \phi(\eta_{q+2}) & 0 & 0 & 0 & \cdots \\ \cdots & \cdots & \cdots & \cdots & 0 & \cdots & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots & \cdots & \phi(\eta_{q+2}) & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & \phi(\eta_{q+2}) & \eta_{q+1}\phi'(\eta_{q+2}) \end{pmatrix}. \quad (3.24) $$

Therefore the covariance matrix is positive definite.

The combination of Theorem 30, Lemma 7 and Lemma 8 allows to prove Proposition 6.

**Proof of Proposition 6.** Let

$$ \begin{aligned} \bar{\gamma} &:= (\beta^1, \beta^2 - \beta^1, \dots, \beta^q - \beta^{q-1}, \beta^{q+1} - \beta^q) \\ &= (\beta^1 - \beta^0, \beta^2 - \beta^1, \dots, \beta^q - \beta^{q-1}, \beta^{q+1} - \beta^q) = (\gamma, \beta^{q+1} - \beta^q) \end{aligned} $$

with $\beta^0 := 0$ by convention. For $j = 1, \dots, q+1$ let us consider the following independent $\mathbb{R}^2$-valued random vectors

$$ V_{M,j} := (V_{M,j}^1, V_{M,j}^2) = \frac{1}{\sqrt{\bar{\gamma}_j M}} \sum_{i=\beta_M^{j-1}+1}^{\beta_M^j} (Y_i - m_Y, Z_i - m_Z) $$

and let us rewrite $(\frac{R_M - m_Y \phi(\beta^{q+1} m_Y) M \gamma}{\sqrt{M}}, K_M)$ in terms of $V_M := (V_{M,j})_{1\le j\le q+1}$. One has

$$ \left( \frac{R_M - m_Y \phi(\beta^{q+1} m_Y) M \gamma}{\sqrt{M}}, K_M \right) = \left( \left( \frac{R_{\beta_M^{\ell-1:\ell}} - m_Y \phi(\beta^{q+1} m_Y) \bar{\gamma}_\ell M}{\sqrt{M}} \right)_{1\le \ell \le q}, K_M \right) = g_M(V_M) $$
with the component $g_{M,\ell}$ of $g_M: \mathbb{R}^{2(q+1)} \to \mathbb{R}^{q+1}$ for $\ell = 1, \dots, q$ given by: $\forall c = (c_j^1, c_j^2)_{1\le j\le q+1} \in \mathbb{R}^{2(q+1)}$

$$ g_{M,\ell}(c) = \frac{\phi\left(\frac{m_Y \beta_M^{q+1}}{M} + \frac{x}{M} + \frac{\sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} c_j^1}{\sqrt{M}}\right) \left(\sqrt{\bar{\gamma}_\ell M} c_\ell^1 + m_Y (\beta_M^\ell - \beta_M^{\ell-1})\right) - m_Y \phi(\beta^{q+1} m_Y) \bar{\gamma}_\ell M}{\sqrt{M}}, \quad (3.25) $$

and $\forall c = (c_j^1, c_j^2)_{1\le j\le q+1} \in \mathbb{R}^{2(q+1)}$

$$ g_{M,q+1}(c) = \frac{\phi\left(\frac{m_Y \beta_M^{q+1}}{M} + \frac{x}{M} + \frac{\sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} c_j^1}{\sqrt{M}}\right) \left(\sqrt{M} \sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} c_j^2 + m_Z \beta_M^{q+1}\right) - m_Z \phi(\beta^{q+1} m_Y) \beta^{q+1} M}{\sqrt{M}}. \quad (3.26) $$

The purpose now is to “asymptotically rewrite” the vector $(\frac{R_M - m_Y \phi(\beta^{q+1} m_Y) \gamma_M}{\sqrt{M}}, K_M) = g_M(V_M)$ in terms of a normal random vector so to apply Lemma 8. By recalling that we denote by $\Sigma$ the covariance matrix of $(Y_i, Z_i)$, let $G = (G_1, \dots, G_{q+1})$ where the two dimensional vectors $G_j = (G_j^1, G_j^2)$ $1 \le j \le q+1$ are i.i.d. according to $\mathcal{N}(0, \Sigma)$. By the independence of the random vectors $V_{M,j}$, the independence of the random vectors $G_j$ and the well known fact that $d_{TV}(\prod_{i=1}^\ell \tilde{\nu}_i, \prod_{i=1}^\ell \nu_i) \le \sum_{i=1}^\ell d_{TV}(\tilde{\nu}_i, \nu_i)$ with $\tilde{\nu}_i, \nu_i$ probability measures for $i = 1, \dots, \ell$, we have

$$ d_{TV}(\mu_{V_M}, \mu_G) \le \sum_{j=1}^{q+1} d_{TV}(\mu_{V_{M,j}}, \mu_{G_j}). \quad (3.27) $$

We are now going to prove that the right-hand side converges to 0 as $M \to \infty$. Let us observe that for $j = 1, \dots, q+1$

$$ V_{M,j} = \frac{\sqrt{\beta_M^j - \beta_M^{j-1}}}{\sqrt{\bar{\gamma}_j M}} \times \frac{1}{\sqrt{\beta_M^j - \beta_M^{j-1}}} \sum_{i=\beta_M^{j-1}+1}^{\beta_M^j} (Y_i - m_Y, Z_i - m_Z) \quad (3.28) $$

$$ \text{with } \lim_{M \to \infty} \frac{\sqrt{\beta_M^j - \beta_M^{j-1}}}{\sqrt{\bar{\gamma}_j M}} = 1. $$

Thanks to Theorem 30 and the hypothesis that the law of $(Y_i, Z_i)$ has an absolutely continuous component, $\forall j = 1, \dots, q+1$ we have that the law of $\frac{1}{\sqrt{\beta_M^j - \beta_M^{j-1}}} \sum_{i=\beta_M^{j-1}+1}^{\beta_M^j} (Y_i - m_Y, Z_i - m_Z)$ converges in total variation as $M \to \infty$ to the law of $G_j$ and, by Lemma 7 and (3.28), this implies

$$ \lim_{M \to \infty} d_{TV}(\mu_{V_{M,j}}, \mu_{G_j}) = 0, \quad j = 1, \dots, q+1. $$

Hence we have proved the right-hand side of (3.27) converges to 0 as $M \to \infty$. Applying (3.6) with $g = g_M$, we deduce that

$$ d_{TV}(\mu_{g_M(V_M)}, \mu_{g_M(G)}) \le d_{TV}(\mu_{V_M}, \mu_G) \underset{M \to \infty}{\longrightarrow} 0. $$
We are now going to prove the convergence in total variation of the law of $g_M(G)$ and to do so we apply Lemma 8. Recalling that $G_j \sim \mathcal{N}(0, \Sigma)$ where $\Sigma$ is positive definite, $G_j$ i.i.d., it is possible to prove that the random variables $\sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} G_j^1$, $\sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} G_j^2$, $\sqrt{\bar{\gamma}_1} G_1^1$, ..., $\sqrt{\bar{\gamma}_q} G_q^1$ are linearly independent.

Thus by using that $\lim_{M \to \infty} \sqrt{M} \left( \frac{\beta_M^i}{M} - \beta^i \right) = 0$ for $i = 1, \dots, q+1$ and (3.13), we can apply Lemma 8 with

• $(\eta_{M,1}, \dots, \eta_{M,q}, \eta_{M,q+1}, \eta_{M,q+2}) = \left( m_Y(\beta_M^1 - \beta_M^0), \dots, m_Y(\beta_M^q - \beta_M^{q-1}), m_Z \beta_M^{q+1}, \frac{m_Y \beta_M^{q+1}}{M} + \frac{x}{M} \right)$ and $(\eta_1, \dots, \eta_q, \eta_{q+1}, \eta_{q+2}) = (m_Y \bar{\gamma}_1, \dots, m_Y \bar{\gamma}_q, m_Z \beta^{q+1}, m_Y \beta^{q+1})$

• $(W_1, \dots, W_{q+2}) = \left( \sqrt{\bar{\gamma}_1} G_1^1, \dots, \sqrt{\bar{\gamma}_q} G_q^1, \sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} G_j^2, \sum_{j=1}^{q+1} \sqrt{\bar{\gamma}_j} G_j^1 \right)$

to conclude that the law of $g_M(G)$ converges in total variation as $M \to \infty$ to the law of $Y \sim \mathcal{N}(0, \Gamma)$ where $\Gamma \in \mathbb{R}^{(q+1) \times (q+1)}$ is positive definite by Remark 32. $\square$

The following lemma contains the key result to prove Lemma 8. Its proof is provided in the Appendix.

**Lemma 9.** Let $(Y_M)_{M \ge 1}$ be a sequence of $\mathbb{R}^d$-valued random vectors such that

$$ \liminf_{M \to \infty} p_{Y_M}(x) \ge p(x) \quad \text{d}x \text{ a.e.} \qquad (3.29) $$

with $p$ the density of an absolutely continuous $\mathbb{R}^d$-valued random variable $Y$. Then

$$ 1. \lim_{M \to \infty} \int_{\mathbb{R}^d} |p_{Y_M}(x) - p(x)| \, dx = 0 $$

$$ 2. \lim_{M \to \infty} d_{TV}(\mu_{Y_M}, \mu_Y) = 0. $$

*Proof of Lemma 8.* According to Lemma 9, to prove the convergence in total variation of the law of the random vector (3.23) it is enough to check that its density with respect to the Lebesgue measure converges pointwise to that of its limit.

Let us preliminary observe that the density of the zero-mean normal vector $(W_1, \dots, W_{q+2})$ is given by

$$ p(x_{1:q+2}) = \frac{1}{(2\pi)^{\frac{q+2}{2}} \times \det(\Sigma_1)^{1/2}} e^{-\frac{1}{2} x_{1:q+2}^T \Sigma_1^{-1} x_{1:q+2}}, \quad x_{1:q+2} \in \mathbb{R}^{q+2} \qquad (3.30) $$

where we denote the vector $(x_1, \dots, x_{q+2})$ by $x_{1:q+2}$. Let us moreover observe that by the property of positive definiteness of $\Sigma_1^{-1}$, the smallest eigenvalue $\lambda_1$ of $\Sigma_1^{-1}$ is positive and we have

$$ p(x_{1:q+2}) \le \frac{1}{(2\pi)^{\frac{q+2}{2}} \times \det(\Sigma_1)^{1/2}} e^{-\frac{1}{2} \lambda_1 \left( \sum_{i=1}^{q+2} x_i^2 \right)}, \quad x_{1:q+2} \in \mathbb{R}^{q+2}. \qquad (3.31) $$

Let us now compute the density of (3.23). Let $f: \mathbb{R}^{q+1} \to \mathbb{R}_+$ be a non-negative measurable function. Writing the expectation

$$ \mathbb{E} \left( f \left( \left( \frac{\phi \left( \frac{\eta_{M,q+2}}{M} + \frac{W_{q+2}}{\sqrt{M}} \right) \left( \sqrt{M} W_i + \eta_{M,i} \right) - M \phi(\eta_{q+2}) \eta_i}{\sqrt{M}} \right)_{1 \le i \le q+1} \right) \right) \qquad (3.32) $$
as an integral with respect to the density p and then applying the change of variable

$$ \xi_i = \frac{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}}\right) \left(\sqrt{M}x_i + \eta_{M,i}\right) - M\phi(\eta_{q+2})\eta_i}{\sqrt{M}} \quad i = 1, \dots, q+1 $$

with inverse

$$ x_{M,i} := \frac{1}{\sqrt{M}} \left( \frac{\sqrt{M}\xi_i + M\phi(\eta_{q+2})\eta_i}{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}}\right)} - \eta_{M,i} \right) \quad i = 1, \dots, q+1 \quad (3.33) $$

for $x_{q+2} \in \mathbb{R}$ outside the set $\{t \in \mathbb{R} : \phi\left(\frac{\eta_{M,q+2}}{M} + \frac{t}{\sqrt{M}}\right) = 0\}$ which is Lebesgue negligible since by a change of variable and (3.22)

$$ \int_{\mathbb{R}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{t}{\sqrt{M}} \right) \right|^q + 1} e^{-\tilde{M} \left( \frac{\eta_{M,q+2}}{M} - \eta_{q+2} + \frac{t}{\sqrt{M}} \right)^2} dt = \sqrt{M} \int_{\mathbb{R}} \frac{1}{\left| \phi(\eta_{q+2} + y) \right|^{q+1}} e^{-\tilde{M}y^2} dy < \infty, $$

we get

$$ \int_{\mathbb{R}} dx_{q+2} \int_{\mathbb{R}^{q+1}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}} \right) \right|^{q+1}} f(\xi_{1:q+1}) p(x_{M,1}, \dots, x_{M,q+1}, x_{q+2}) d\xi_{1:q+1}. $$

We have therefore obtained that the density of $\left( \frac{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{W_{q+2}}{\sqrt{M}}\right)\left(\sqrt{M}W_i + \eta_{M,i}\right) - M\phi(\eta_{q+2})\eta_i}{\sqrt{M}} \right)_{1 \le i \le q+1}$ at $\xi \in \mathbb{R}^{q+1}$ is given by

$$ \int_{\mathbb{R}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}} \right) \right|^{q+1}} p(x_{M,1}, \dots, x_{M,q+1}, x_{q+2}) dx_{q+2}. \quad (3.34) $$

Since, by hypothesis, $\phi$ is continuous at $\eta_{q+2}$ with $\phi(\eta_{q+2}) \ne 0$, there exists $\delta > 0$ such that $\forall t \in \mathbb{R} : |t| \le \delta$

$$ |\phi(\eta_{q+2} + t)| \ge \frac{|\phi(\eta_{q+2})|}{2}. \quad (3.35) $$

Let us now study the pointwise convergence of (3.34). Let $M \ge 1$ and let us rewrite the integral as

$$ \int_{\mathbb{R}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}} \right) \right|^{q+1}} p(x_{M,1:q+1}, x_{q+2}) dx_{q+2} \quad (3.36) $$

$$ = \int_{|x_{q+2}| \le \frac{\delta}{2} \sqrt{M}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}} \right) \right|^{q+1}} p(x_{M,1:q+1}, x_{q+2}) dx_{q+2} \quad (3.37) $$

$$ + \int_{|x_{q+2}| > \frac{\delta}{2} \sqrt{M}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}} \right) \right|^{q+1}} p(x_{M,1:q+1}, x_{q+2}) dx_{q+2}. \quad (3.38) $$

Let us start studying the convergence as $M \to \infty$ of (3.37). Thanks to (3.35) and (3.31) and for $M$ big enough so that $|\frac{\eta_{M,q+2}}{M} - \eta_{q+2}| \le \frac{\delta}{2}$, we have

$$ 1\{|x_{q+2}|\le \frac{\delta}{2}\sqrt{M}\} \frac{1}{\left|\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}}\right)\right|^{q+1}} p(x_{M,1:q+1}, x_{q+2}) \le \frac{2^{q+1}}{(2\pi)^{\frac{q+2}{2}} \times \det(\Sigma_1)^{\frac{1}{2}} |\phi(\eta_{q+2})|^{q+1}} e^{-\frac{1}{2}\lambda_1 x_{q+2}^2} $$
and since by hypothesis $\phi$ is continuous at $\eta_{q+2}$, for each $x_{q+2} \in \mathbb{R}$ we have

$$ \phi\left(\frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}}\right) \underset{M \to \infty}{\longrightarrow} \phi(\eta_{q+2}). \quad (3.39) $$

Moreover for each $x_{q+2} \in \mathbb{R}$, using that $\lim_{M \to \infty} \sqrt{M} \left(\frac{\eta_{M,i}}{M} - \eta_i\right) = 0$ for $i = 1, \dots, q+2$ and the hypothesis of differentiability of $\phi$ one has

$$ \begin{aligned} x_{M,i} &= \frac{1}{\sqrt{M}} \left( \frac{\sqrt{M}\xi_i + M\phi(\eta_{q+2})\eta_i}{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}}\right) - \eta_{M,i}} \right) \\ &= \left( \frac{\xi_i + \phi(\eta_{q+2})\sqrt{M}\left(\eta_i - \frac{\eta_{M,i}}{M}\right) - \phi'(\eta_{q+2})\frac{\eta_{M,i}}{\sqrt{M}}\left(\frac{x_{q+2}}{\sqrt{M}} + \frac{\eta_{M,q+2}}{M} - \eta_{q+2}\right) + o(1)}{\phi(\eta_{q+2}) + \phi'(\eta_{q+2})\left(\frac{x_{q+2}}{\sqrt{M}} + \frac{\eta_{M,q+2}}{M} - \eta_{q+2}\right) + o\left(\frac{1}{\sqrt{M}}\right)} \right) \\ &\underset{M \to \infty}{\longrightarrow} \frac{\xi_i - \phi'(\eta_{q+2})\eta_i x_{q+2}}{\phi(\eta_{q+2})}. \end{aligned} \quad (3.40) $$

We can therefore apply Lebesgue's theorem to (3.37) and obtain that for each $\xi \in \mathbb{R}^{q+1}$

$$ \begin{aligned} & \lim_{M \to \infty} \int_{|x_{q+2}| \le \frac{\delta}{2} \sqrt{M}} \frac{1}{\left|\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}}\right)\right|^{q+1}} p(x_{M,1:q+1}, x_{q+2}) dx_{q+2} \\ &= \frac{1}{|\phi(\eta_{q+2})|^{q+1}} \int_{\mathbb{R}} p\left(\left(\frac{\xi_i - \phi'(\eta_{q+2})\eta_i x_{q+2}}{\phi(\eta_{q+2})}\right)_{1 \le i \le q+1}, x_{q+2}\right) dx_{q+2} \\ &= \int_{\mathbb{R}} p\left((\phi(\eta_{q+2})W_i + \phi'(\eta_{q+2})\eta_i W_{q+2})_{1 \le i \le q+1}, W_{q+2}\right) (\xi_{1:q+1}, x_{q+2}) dx_{q+2} \\ &= p\left((\phi(\eta_{q+2})W_i + \phi'(\eta_{q+2})\eta_i W_{q+2})_{1 \le i \le q+1}\right) (\xi_{1:q+1}). \end{aligned} $$

Let us now prove that (3.38) converges to 0 as $M \to \infty$. By applying a change of variable and by choosing $M$ big enough so that $|\frac{\eta_{M,q+2}}{M} - \eta_{q+2}| \le \frac{\delta}{4}$, we obtain

$$ \begin{aligned} & \int_{|x_{q+2}| > \frac{\delta}{2} \sqrt{M}} \frac{1}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + \frac{x_{q+2}}{\sqrt{M}} \right) \right|^{q+1}} p(x_{M,1:q+1}, x_{q+2}) dx_{q+2} \\ &= \int_{|y| > \frac{\delta}{2}} \frac{\sqrt{M}}{\left| \phi \left( \frac{\eta_{M,q+2}}{M} + y \right) \right|^{q+1}} p(x_{M,1:q+1}, \sqrt{M}y) dy \\ &\le \int_{|z| > \frac{\delta}{4}} \frac{\sqrt{M}}{\left| \phi (\eta_{q+2} + z) \right|^{q+1}} p(x_{M,1:q+1}, \sqrt{M}z + \sqrt{M} \left( \eta_{q+2} - \frac{\eta_{M,q+2}}{M} \right)) dz. \end{aligned} $$

Therefore by (3.31) and by using that $\forall x_1, x_2 \in \mathbb{R} (x_1 - x_2)^2 \ge \frac{x_1^2}{2} - x_2^2$, one has

$$ \begin{aligned} & 1_{\{|z| > \frac{\delta}{4}\}} \frac{\sqrt{M}}{|\phi(\eta_{q+2} + z)|^{q+1}} p(x_{M,1:q+1}, \sqrt{M}z + \sqrt{M} \left(\eta_{q+2} - \frac{\eta_{M,q+2}}{M}\right)) \\ & \le 1_{\{|z| > \frac{\delta}{4}\}} \frac{\sqrt{M}}{(2\pi)^{\frac{q+2}{2}} \times \det(\Sigma_1)^{\frac{1}{2}} |\phi(\eta_{q+2} + z)|^{q+1}} e^{-\frac{1}{2}\lambda_1 \left(\sqrt{M}z - \sqrt{M}\left(\frac{\eta_{M,q+2}}{M} - \eta_{q+2}\right)\right)^2} \\ & \le 1_{\{|z| > \frac{\delta}{4}\}} \frac{\sqrt{M}}{(2\pi)^{\frac{q+2}{2}} \times \det(\Sigma_1)^{\frac{1}{2}} |\phi(\eta_{q+2} + z)|^{q+1}} e^{-\frac{1}{4}\lambda_1 M z^2 e^{\frac{1}{2}\lambda_1 \left(\sqrt{M}\left(\frac{\eta_{M,q+2}}{M} - \eta_{q+2}\right)\right)^2}} \\ & \le C 1_{\{|z| > \frac{\delta}{4}\}} \frac{\sqrt{M}}{(2\pi)^{\frac{q+2}{2}} \times \det(\Sigma_1)^{\frac{1}{2}} |\phi(\eta_{q+2} + z)|^{q+1}} e^{-\frac{1}{4}\lambda_1 M z^2} \end{aligned} $$
where for the last inequality we use that the sequence $(\sqrt{M} (\frac{\eta_{M,q+2}}{M} - \eta_{q+2}))_{M \ge 1}$ converges to 0 and so in particular $e^{\frac{1}{2}\lambda_1}(\sqrt{M}(\frac{\eta_{M,q+2}}{M} - \eta_{q+2}))^2$ is bounded by a positive constant $C < \infty$. The right-hand side converges to 0 as $M \to \infty$. Let us now observe that for $|z| > \frac{\delta}{4}$ and $t \ge \frac{32}{\delta^2\lambda_1}$,

$$ \partial_t(\sqrt{t}e^{-\frac{1}{4}\lambda_1 t z^2}) = \frac{e^{-\frac{1}{4}\lambda_1 z^2 t}}{4} \left( \frac{2}{\sqrt{t}} - z^2 \sqrt{t} \lambda_1 \right) \le 0. $$

Therefore if $M \ge \bar{M} := \max(\lfloor \frac{4\bar{M}}{\lambda_1} \rfloor, \lfloor \frac{32}{\delta^2\lambda_1} \rfloor)$, where $\tilde{M}$ is defined in the statement of the Lemma 8,

$$ 1_{\{|z|>\frac{\delta}{4}\}} \frac{\sqrt{M}}{|\phi(\eta_{q+2}+z)|^{q+1}} e^{-\frac{1}{4}\lambda_1 M z^2} \le 1_{\{|z|>\frac{\delta}{4}\}} \frac{\sqrt{M}}{|\phi(\eta_{q+2}+z)|^{q+1}} e^{-\frac{1}{4}\lambda_1 \bar{M} z^2} \quad (3.41) $$

$$ \le 1_{\{|z|>\frac{\delta}{4}\}} \frac{\sqrt{M}}{|\phi(\eta_{q+2}+z)|^{q+1}} e^{-\tilde{M}z^2} \quad (3.42) $$

where the last term is integrable by hypothesis (3.22). By Lebesgue's theorem we can therefore conclude that $\int_{|z|>\frac{\delta}{4}} \frac{\sqrt{M}}{|\phi(\eta_{q+2}+z)|^{q+1}} p(x_{M,1:q+1}, \sqrt{M}z + \sqrt{M} (\eta_{q+2} - \frac{\eta_{M,q+2}}{M})) dz$ tends to 0 as $M \to \infty$.

In conclusion, we have obtained that the density of $\left( \frac{\phi\left(\frac{\eta_{M,q+2}}{M} + \frac{W_{q+2}}{\sqrt{M}}\right)\left(\sqrt{M}W_i + \eta_{M,i}\right) - M\phi(\eta_{q+2})\eta_i}{\sqrt{M}} \right)_{1\le i\le q+1}$ converges pointwise as $M \to \infty$ to the density of $((\phi(\eta_{q+2})W_i + \phi'(\eta_{q+2})\eta_i W_{q+2}))_{1\le i\le q+1}$. $\square$

We are now going to prove Theorem 29.

*Proof of Theorem 29.* We denote $(\{B_M\}, H_M) := (\{B_M^1\}, \dots, \{B_M^q\}, H_M)$. Let us first observe that for every $k \in \mathbb{Z}^q, u \in \mathbb{R}^{q'}$ one has

$$ \begin{aligned} \phi(\{B_M\}, H_M)(2\pi k, u) &= \mathbb{E} \left( e^{i2\pi k^T \{B_M\}} e^{iu^T H_M} \right) = \mathbb{E} \left( e^{i2\pi k^T B_M} e^{-i2\pi \sum_{j=1}^q k_j \lfloor B_M^j \rfloor} e^{iu^T H_M} \right) \\ &= \phi(\{B_M, H_M\}) (2\pi k, u). \end{aligned} $$

If $(\{B_M\}, H_M)$ converges in distribution as $M \to \infty$ to $(U, H)$ where $U$ is independent of $H$ and uniformly distributed on $[0, 1]^q, \forall (k, u) := (k_1, \dots, k_q, u) \in \{\mathbb{Z}^q \setminus \{0\}\} \times \mathbb{R}^{q'}$ one has

$$ \begin{aligned} \lim_{M \to \infty} \phi_{(B_M, H_M)} (2\pi k, u) &= \lim_{M \to \infty} \phi_{(\{B_M\}, H_M)} (2\pi k, u) = \phi_{(U, H)} (2k\pi, u) \\ &= \mathbb{E} \left( e^{i2\pi kU} \right) \mathbb{E} \left( e^{iuH} \right) = 0. \end{aligned} $$

Conversely, assume that (3.19) holds for every $(k, u) \in \{\mathbb{Z}^q \setminus \{0\}\} \times \mathbb{R}^{q'}$ and let us prove that $\forall (u_1, u_2) \in \mathbb{R}^q \times \mathbb{R}^{q'}$,

$$ \lim_{M \to \infty} \phi_{(\{B_M\}, H_M)}(u_1, u_2) = \phi_U(u_1)\phi_H(u_2) $$

with $(U, H)$ as above.

Fix $(u_1, u_2) \in \{\mathbb{R}^q \setminus \{0\}\} \times \mathbb{R}^{q'}$. Given $0 < \epsilon < \frac{1}{4}$, let us define for $j = 1, \dots, q$ a complex $\lceil \frac{q+1}{2} \rceil$-times continuously differentiable function $\phi_{\epsilon,j}$ defined on $[0, 1]$ that coincides with $x_j \mapsto e^{iu_1^j x_j}$ on $[0, 1 - \epsilon]$ and such that $\phi_{\epsilon,j}(1) = \phi_{\epsilon,j}(0)$. It is possible to choose $\phi_{\epsilon,j}$ such that $\sup_{x \in [0,1]} |\phi_{\epsilon,j}(x)| \le C^{\frac{1}{q}}$, with $C < \infty$ not depending on $\epsilon$ and $j$. If we now consider the function $\phi_{\epsilon}(x_1, \dots, x_q) := \phi_{\epsilon,1}(x_1)$.
$\cdots \phi_{\epsilon,q}(x_q) \forall (x_1, \dots, x_q) \in [0,1]^q$, then it is $\lceil \frac{q+1}{2} \rceil$-times continuously differentiable, coincides with $(x_1, \dots, x_q) \mapsto e^{iu_1^1 x_1} \cdots e^{iu_1^q x_q}$ on $[0,1-\epsilon]^q$ and $\forall j, \forall (x_1, \dots, x_{j-1}, x_{j+1}, \dots, x_q) \in [0,1]^{q-1}$

$$
\phi_{\epsilon}(x_1, \dots, x_{j-1}, 0, x_{j+1}, \dots, x_q) = \phi_{\epsilon}(x_1, \dots, x_{j-1}, 1, x_{j+1}, \dots, x_q).
$$

By the theory of Fourier's series (see for instance the Section Sobolev Spaces in Chapter 5 of [53]),
there exists $M^\epsilon \in \mathbb{N}$ such that

$$
\sup_{x \in [0,1]^q} \left| \phi_{\epsilon}(x) - \sum_{\max_{j=1,\dots,q} |k_j| \le M^{\epsilon}} c_k^{\epsilon} e^{i2\pi k^T x} \right| = \sup_{x \in [0,1]^q} \left| \phi_{\epsilon}(x) - \sum_{|k_1| \le M^{\epsilon}} \dots \sum_{|k_q| \le M^{\epsilon}} c_k^{\epsilon} e^{i2\pi k^T x} \right| \le \epsilon \quad (3.43)
$$

where for $k \in \mathbb{Z}^q$

$$
c_k^\epsilon = \int_{[0,1]^q} \phi_\epsilon(x) e^{-i2\pi k^T x} dx
$$

satisfies $|c_k^\epsilon| \le C$. Let $\nu_M$ denote the law of $(\{B_M\}, H_M)$ (having support in $[0,1]^q \times \mathbb{R}^{q'}$).
Given $M \ge 1$ one has

$$
\begin{align*}
& \left| \phi(\{B_M\}, H_M)(u_1, u_2) - \phi_U(u_1)\phi_H(u_2) \right| \\
& \le \left| \int_{[0,1]^q \times \mathbb{R}^{q'}} \left( e^{iu_1^T x} - \phi_\epsilon(x) \right) e^{iu_2^T y} \nu_M(dx, dy) \right| \\
& \quad + \left| \int_{[0,1]^q \times \mathbb{R}^{q'}} \phi_\epsilon(x) e^{iu_2^T y} \nu_M(dx, dy) - \sum_{\max_{j=1,\dots,q} |k_j| \le M^\epsilon} c_k^\epsilon \int_{[0,1]^q \times \mathbb{R}^{q'}} e^{i2\pi k^T x} e^{iu_2^T y} \nu_M(dx, dy) \right| \\
& \quad + \left| \sum_{\max_{j=1,\dots,q} |k_j| \le M^\epsilon, k \ne 0} c_k^\epsilon \int_{[0,1]^q \times \mathbb{R}^{q'}} e^{i2\pi k^T x} e^{iu_2^T y} \nu_M(dx, dy) \right| + |c_0^\epsilon \phi_{H_M}(u_2) - c_0^\epsilon \phi_H(u_2)| \\
& \quad + |c_0^\epsilon \phi_H(u_2) - \phi_U(u_1)\phi_H(u_2)| \\
& \le (1+C) \int_{[0,1]^q \setminus [0,1-\epsilon]^q} \nu_M(dx, \mathbb{R}^{q'}) + \epsilon + \left| \sum_{\max_{j=1,\dots,q} |k_j| \le M^\epsilon, k \ne 0} c_k^\epsilon \phi_{(B_M,H_M)}(2k\pi, u_2) \right| \\
& \quad + C |\phi_{H_M}(u_2) - \phi_H(u_2)| + |\phi_U(u_1) - c_0^\epsilon|.
\end{align*}
$$

Let us look in more detail at the first term of the right-hand side. Let $\bar{\phi}_{\epsilon}$ be a real continuously
differentiable function defined on $[0, 1]$ that is equal to 0 on $[2\epsilon, 1 - 2\epsilon]$ and equal to 1 on $[0, \epsilon]$ and
$[1 - \epsilon, 1]$. By the theory of Fourier's series, there exists $\bar{M}^{\epsilon} \in \mathbb{N}$ such that

$$
\sup_{x \in [0,1]} \left| \bar{\phi}_{\epsilon}(x) - \sum_{|\ell| \le \bar{M}^{\epsilon}} d_{\ell}^{\epsilon} e^{i2\pi \ell x} \right| \le \epsilon \quad (3.44)
$$

where for $\ell \in \mathbb{Z}$

$$
d_{\ell}^{\epsilon} = \int_{[0,1]} \bar{\phi}_{\epsilon}(x) e^{-i2\pi \ell x} dx.
$$
One has

$$
\begin{align*}
\int_{[0,1]^q \setminus [0,1-\epsilon]^q} \nu_M(dx, \mathbb{R}^{q'}) &\le \sum_{j=1}^q \int_{[1-\epsilon,1]} \mu_{\{B_M^j\}}(dx) \le \sum_{j=1}^q \int_{[0,1]} \bar{\phi}_\epsilon(x) \mu_{\{B_M^j\}}(dx) \\
&\le \sum_{j=1}^q \int_{[0,1]} \left| \bar{\phi}_\epsilon(x) - \sum_{|\ell| \le \bar{M}^\epsilon} d_\ell^\epsilon e^{i2\pi\ell x} \right| \mu_{\{B_M^j\}}(dx) \\
&\quad + \sum_{j=1}^q \left| \sum_{|\ell| \le \bar{M}^\epsilon, \ell \ne 0} d_\ell^\epsilon \int_{[0,1]} e^{i2\pi\ell x} \mu_{\{B_M^j\}}(dx) \right| + q |d_0^\epsilon| \\
&\le q\epsilon + \sum_{j=1}^q \left| \sum_{|\ell| \le \bar{M}^\epsilon, \ell \ne 0} d_\ell^\epsilon \phi_{B_M^j}(2\pi\ell) \right| + q |d_0^\epsilon|.
\end{align*}
$$

Thus we have obtained that

$$
\begin{align*}
& \left| \phi_{(\{B_M\}, H_M)}(u_1, u_2) - \phi_U(u_1)\phi_H(u_2) \right| \\
& \le (1+C) \left( q\epsilon + \sum_{j=1}^q \left| \sum_{|\ell| \le \bar{M}^\epsilon, \ell \ne 0} d_\ell^\epsilon \phi_{B_M^j}(2\pi\ell) \right| + q |d_0^\epsilon| \right) + \epsilon \\
& \quad + \left| \sum_{\max_{j=1,\dots,q} |k_j| \le M^\epsilon, k \ne 0} c_k^\epsilon \phi_{(B_M,H_M)}(2\pi k, u_2) \right| + C |\phi_{H_M}(u_2) - \phi_H(u_2)| + |\phi_U(u_1) - c_0^\epsilon|.
\end{align*}
$$

Given $0 < \epsilon < \frac{1}{4}$, we can first take the superior limit as $M \to \infty$ of the right-hand side.
By (3.19) we have

$$
\lim_{M \to \infty} \sum_{j=1}^{q} \left| \sum_{|\ell| \le \bar{M}^{\epsilon}, \ell \ne 0} d_{\ell}^{\epsilon} \phi_{B_{M}^{j}}(2\pi\ell) \right| = 0 = \lim_{M \to \infty} \left| \sum_{\max_{j=1, \dots, q} |k_j| \le M^{\epsilon}, k \ne 0} c_{k}^{\epsilon} \phi_{(B_{M}, H_{M})}(2\pi k, u_2) \right|
$$

and by the hypothesis that $H_M \stackrel{d}{\Rightarrow} H$ we have $\lim_{M \to \infty} |\phi_{H_M}(u_2) - \phi_H(u_2)| = 0$. We therefore obtain

$$
\limsup_{M \to \infty} |\phi_{(\{B_M\}, H_M)}(u_1, u_2) - \phi_U(u_1)\phi_H(u_2)| \le q(1+C)(\epsilon + |d_0^\epsilon|) + \epsilon + |\phi_U(u_1) - c_0^\epsilon|. \quad (3.45)
$$

Now since

$$
|d_0^\epsilon| = d_0^\epsilon = \int_{[0,1]} \bar{\phi}_\epsilon(x) dx \le 4\epsilon
$$

and

$$
|\phi_U(u_1) - c_0^\epsilon| = \left| \int_{[0,1]^q} (e^{iu_1^T x} - \phi_\epsilon(x)) dx \right| \le (C+1) \int_{[0,1]^q \setminus [0,1-\epsilon]^q} dx
$$

with $\int_{[0,1]^q \setminus [0,1-\epsilon]^q} dx = 1 - (1-\epsilon)^q \underset{\epsilon \to 0}{\longrightarrow} 0$, the limit as $\epsilon \to 0$ of the right-hand side of (3.45) is 0.
If $u_1 = 0$ and $u_2 \in \mathbb{R}$, by the hypothesis that $H_M \stackrel{d}{\Rightarrow} H$, we directly obtain

$$
\phi_{(\{B_M\}, H_M)}(0, u_2) = \phi_{H_M}(u_2) \underset{M \to \infty}{\longrightarrow} \phi_H(u_2) = \phi_U(0)\phi_H(u_2).
$$

□
3.6 Appendix

**Lemma 10.** Given the measures $\mu_1$ and $\mu_2$ on $\mathbb{R}^d$, one has

$$
\sup_{\|f\|_{\infty,\mathbb{C}} \le 1} \left| \int_{\mathbb{R}^d} f(x) (\mu_1(dx) - \mu_2(dx)) \right| = 2d_{TV}(\mu_1, \mu_2) \quad (3.46)
$$

where the supremum is taken over the complex-valued measurable functions.

Proof of Lemma 10. By observing that the set of complex-valued measurable functions contains the set of real-valued measurable functions and that $\|f\|_{\infty,\mathbb{C}} = \|f\|_{\infty,\mathbb{R}}$ when $f$ is real-valued, thanks to (3.4) we can easily obtain that

$$
\sup_{\|f\|_{\infty,\mathbb{C}} \le 1} \left| \int_{\mathbb{R}^d} f(x) (\mu_1(dx) - \mu_2(dx)) \right| \ge 2d_{TV}(\mu_1, \mu_2).
$$

For what concerns the other inequality, let $f$ be a complex-valued measurable function such that
$\|f\|_{\infty,\mathbb{C}} \le 1$. Then there exist $\psi \ge 0$ and $\rho \in [0, 2\pi)$ such that

$$
\int_{\mathbb{R}^d} f(x) (\mu_1(dx) - \mu_2(dx)) = \psi e^{i\rho}.
$$

Note that $\|\mathfrak{Re}(e^{-i\rho}f)\|_{\infty,\mathbb{R}} \le \|e^{-i\rho}f\|_{\infty,\mathbb{C}} = \|f\|_{\infty,\mathbb{C}} \le 1$. Hence

$$
\begin{align*}
\left| \int_{\mathbb{R}^d} f(x) (\mu_1(dx) - \mu_2(dx)) \right| &= \int_{\mathbb{R}^d} \mathfrak{Re}(e^{-i\rho} f(x)) (\mu_1(dx) - \mu_2(dx)) \\
&\le \sup_{\|g\|_{\infty,\mathbb{R}} \le 1} \left| \int_{\mathbb{R}^d} g(x) (\mu_1(dx) - \mu_2(dx)) \right| = 2d_{TV}(\mu_1, \mu_2)
\end{align*}
$$

where for the last equality we use (3.4). Since *f* is arbitrary, we can conclude the proof. □

Proof of Lemma 6. Let $\xi$ be an absolutely continuous random variable independent of (Y, Z) where the law of Y has an absolutely continuous component and let us observe that $\mu_{(Y,Z)}(dy,dz) = \mu_{Z|Y=y}(dz)\mu_Y(dy)$ where $\mu_{Z|Y=y}(dz)$ denotes the conditional law of Z given $Y=y$. Given $f: \mathbb{R}^2 \to \mathbb{R}_+$ a non-negative measurable function, one has

$$
\begin{align*}
\mathbb{E} (f (Y, Z + \xi)) &= \int_{\mathbb{R}^3} f(y, z + \epsilon) \mu_{(Y,Z,\xi)} (dy, dz, d\epsilon) = \int_{\mathbb{R}^3} f(y, z + \epsilon) p_{\xi} (\epsilon) \mu_{(Y,Z)} (dy, dz) d\epsilon \\
&= \int_{\mathbb{R}^3} f(y, z + \epsilon) p_{\xi} (\epsilon) \mu_{Z|Y=y} (dz) \mu_Y (dy) d\epsilon = \int_{\mathbb{R}^3} f(y, x) p_{\xi} (x - z) \mu_{Z|Y=y} (dz) \mu_Y (dy) dx \\
&= \int_{\mathbb{R}^2} f(y, x) \int_{\mathbb{R}} p_{\xi} (x - z) \mu_{Z|Y=y} (dz) p_Y (y) dy dx + \int_{\mathbb{R}^2} f(y, x) \int_{\mathbb{R}} p_{\xi} (x - z) \mu_{Z|Y=y} (dz) \mu_{Y,s} (dy) dx
\end{align*}
$$

where $\int_{\mathbb{R}^2} \int_{\mathbb{R}} p_{\xi}(x-z) \mu_{Z|Y=y}(dz)p_Y(y)dydx$ is positive since $Y$ has an absolutely continuous component and $\xi$ is an absolutely continuous random variable. Thus the law of $(Y, Z + \xi)$ has an absolutely continuous component. $\square$

*Proof of Lemma 9.* Let us first observe that by Fatou’s lemma, (3.29) and the fact that *p* is a probability density, one has

$$
\liminf_{M \to \infty} \int_{\mathbb{R}^d} p_{Y_M}(x) dx \ge \int_{\mathbb{R}^d} \liminf_{M \to \infty} p_{Y_M}(x) dx \ge \int_{\mathbb{R}^d} p(x) dx = 1.
$$

Moreover $\int_{\mathbb{R}^d} p_{Y_M}(x)dx \le 1$ for each $M \in \mathbb{N}^*$. Therefore we can conclude that

$$
\lim_{M \to \infty} \int_{\mathbb{R}^d} p_{Y_M}(x) dx = 1. \qquad (3.47)
$$
As an immediate consequence of (3.47) we have that

$$ \lim_{M \to \infty} \mu_{Y_M, s}(\mathbb{R}^d) = 1 - \lim_{M \to \infty} \int_{\mathbb{R}^d} p_{Y_M}(x)dx = 0. \quad (3.48) $$

Let us now prove the $L^1$ convergence. For each $M \in \mathbb{N}^*$,

$$ \int_{\mathbb{R}^d} |p(x) - p_{Y_M}(x)| dx = 2 \int_{\mathbb{R}^d} (p(x) - p_{Y_M}(x))^+ dx - \int_{\mathbb{R}^d} (p(x) - p_{Y_M}(x)) dx $$

where the second component of the right-hand side goes to 0 as $M \to \infty$ by (3.47) and the fact that $p$ is a probability density while the first component goes to 0 as $M \to \infty$ by Lebesgue theorem and the fact that $0 \le \liminf_{M \to \infty} (p(x) - p_{Y_M}(x))^+ \le \limsup_{M \to \infty} (p(x) - p_{Y_M}(x))^+ = (\limsup_{M \to \infty} (p(x) - p_{Y_M}(x)))^+ = 0$.

Hence $\lim_{M \to \infty} \int_{\mathbb{R}^d} |p(x) - p_{Y_M}(x)| dx = 0$.

Let us now prove the second point of the Lemma. For a fixed $M \in \mathbb{N}^*$, one has

$$ \begin{aligned} d_{TV}(\mu_{Y_M}, \mu_Y) &= \sup_{A \in \mathcal{B}(\mathbb{R}^d)} |\mu_{Y_M}(A) - \mu_Y(A)| = \sup_{A \in \mathcal{B}(\mathbb{R}^d)} |\mu_{Y_M,c}(A) + \mu_{Y_M,s}(A) - \mu_Y(A)| \\ &\le \sup_{A \in \mathcal{B}(\mathbb{R}^d)} |\mu_{Y_M,c}(A) - \mu_Y(A)| + \mu_{Y_M,s}(\mathbb{R}^d) \le \int_{\mathbb{R}^d} |p_{Y_M}(x) - p(x)| dx + \mu_{Y_M,s}(\mathbb{R}^d). \end{aligned} $$

and the right-hand side tends to 0 as $M \to \infty$ thanks to what has been proved in the previous steps. $\square$
<figure><img src="image_8.png" /></figure>
# Chapter 4

## Central Limit Theorem for the stratified resampling mechanism

The material for this chapter has been released in [25].

The stratified resampling mechanism is one of the resampling schemes commonly used in the resampling steps of particle filters. In the present paper, we prove a central limit theorem for this mechanism under the assumption that the initial positions are independent and identically distributed and the weights proportional to a positive function of the positions such that the image of their common distribution by this function has a non zero component absolutely continuous with respect to the Lebesgue measure. This result relies on the convergence in distribution of the fractional part of partial sums of the normalized weights to some random variable uniformly distributed on $[0, 1]$, which is established in the companion paper [27] by overcoming the difficulty raised by the coupling through the normalization. Under the conjecture that a similar convergence in distribution remains valid at the next steps of a particle filter which alternates selections according to the stratified resampling mechanism and mutations according to Markov kernels, we provide an inductive formula for the asymptotic variance of the resampled population after $n$ steps. We perform numerical experiments which support the validity of this formula.

### 4.1 Introduction

Particle filtering, also known as Sequential Monte Carlo methods (see Chapter 4 in [10] and [15] for a general introduction), is a powerful method to estimate the evolving state of a system over time, even when the state cannot be directly observed but can only be inferred through noisy measurements or observations. It has become a very popular class of numerical methods for the solution of optimal estimation problems in non-linear non-Gaussian scenarios. This kind of method is used in real-time applications appearing in fields such as chemical engineering, computer vision, financial econometrics, target tracking, robotics and statistics (see among others [51] and [20]). The use of Monte Carlo methods for nonlinear filtering problems can be traced back to Handschin [32] and Mayne [33]. They introduced a sequential version of the importance sampling method (see [22], [52] as references for the importance sampling) and the corresponding algorithm is known as sequential importance sampling, often abbreviated SIS. A drawback of the latter was identified by Gordon et al. in [29]: as the number of iterations increases, the importance weights tend to degenerate (a phenomenon usually known as *weight degeneracy*). This means that after a certain number of iterations, some weights tend to become very small so that the corresponding positions no longer contribute to the estimation. Thus, Gordon et al. introduced the resampling step where, in view of stabilizing the Monte Carlo error over time, the key idea is to eliminate the particles having low weights and to replicate the particles having high weights. Therefore, by propagating $M$ particles through weighting, resampling and mutation steps (each particle evolves randomly according to a given transition probability kernel), particle filters can be used to numerically estimate the state of
the system given the observations (see also [20], [15]).

Let us henceforth concentrate on the resampling step. Let $M \ge 1$. Given a sequence of $\mathbb{R}^d$-valued random vectors $(X_m)_{1\le m\le M}$ with associated random weights $(w_m^M)_{1\le m\le M}$ such that $w_m^M > 0$ and $\sum_{m=1}^M w_m^M = 1$, a resampling scheme defines the resampled sequence $(Y_m)_{1\le m\le M}$ such that

$$ \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \delta_{Y_m} \mid \mathcal{G} \right) = \sum_{m=1}^{M} w_m^M \delta_{X_m}. $$

where $\mathcal{G} := \sigma((X_m, w_m^M)_{1\le m\le M})$. Resampling schemes [19] permit to replace the probability measure $\sum_{m=1}^M w_m^M \delta_{X_m}$ with non equal weights by some empirical measure $\frac{1}{M} \sum_{m=1}^M \delta_{Y_m}$ with the same conditional expectation given $\mathcal{G}$. Depending on the definition used for the random variables $Y_m$, several resampling schemes can be considered. The most common resampling techniques are of the following types: multinomial, residual, stratified [37] and systematic [12]. See [19] for a brief description of such methods. The simplest approach is the multinomial resampling. It consists in drawing, conditionally upon $\mathcal{G}$, the new positions $(Y_m)_{1\le m\le M}$ independently from the common distribution $\sum_{m=1}^M w_m^M \delta_{X_m}$. While the residual resampling consists in replicating $[Mw_m^M]$-times $X_m$ for $m \in \{1, \dots, M\}$ and the remaining $M - \sum_{m=1}^M [Mw_m^M]$ variables $Y_m$ are drawn, conditionally upon $\mathcal{G}$, independently from the common distribution $\frac{1}{M - \sum_{m=1}^M [Mw_m^M]} \sum_{\ell=1}^M \{Mw_\ell^M\} \delta_{X_\ell}$. Concerning the stratified resampling, it is not straightforward to understand its behaviour, notably due to the complicated structure in the definition of the $Y_m$ which, however, continue to be conditionally independent given $\mathcal{G}$. The systematic resampling is even more complicated to understand since the $Y_m$ are no longer conditionally independent given $\mathcal{G}$.

Resampling schemes have been largely studied in the literature, we now present a selection of such results. The asymptotic behaviour for the multinomial resampling scheme has been extensively studied in [47] (see Corollary 7.4.2 and Section 9.4.2). Douc et al. in [19] showed that residual and stratified resampling improve over multinomial resampling in the sense that they have a lower conditional variance (with respect to the $\sigma$-algebra generated by $(X_m)_{1\le m\le M}$). They also proved, by means of a counter-example, that the same property does not hold for systematic resampling. Furthermore, they established a central limit theorem for the residual resampling approach suggesting that a similar result should be obtained for the stratified resampling scheme. One of the last contributions concerning resampling schemes, is given by Gerber et al. [28]. Using the notion of negative association [35], they first provided a general consistency result for resampling. An application of this theorem gives the proof of almost sure weak convergence of $\frac{1}{M} \sum_{m=1}^M \delta_{Y_m}$ in the stratified resampling method. Moreover they provide a counter-example to almost sure weak convergence for the systematic resampling method. More recently, Chopin et al. [16] studied the resampling schemes for particle filters with weakly informative observations. Empirical evidence indicates that when the weights used in resampling exhibit high variability, the selection of the resampling strategy tends to have a weak impact. However, in cases where the weights are close to being uniform, the performance differences between the different resampling methods can be substantial. By keeping $M$ fixed, they also considered the asymptotic behaviour of the resampling schemes as the weights become less and less informative. See also [41], [23], [38], [14] for additional references.

In this paper, we focus on the stratified resampling scheme for the weights $w_m^M = g(X_m)/\sum_{\ell=1}^M g(X_\ell)$ where $g$ is a positive measurable function. Our purpose is to study the asymptotic behaviour of the method as the number of particles $M$ goes to $\infty$. Let $f$ be a real-valued measurable function. To
compute the asymptotic variance of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m)$, it turns out to be essential to understand the behaviour as $M \to \infty$ of

$$ \left\{ \frac{M \sum_{\ell=1}^{[\alpha M]-1} g(X_\ell)}{\sum_{\ell=1}^{M} g(X_\ell)} \right\} \qquad (4.1) $$

where $0 < \alpha < 1$ and where $\{x\}$ denotes the fractional part of $x \in \mathbb{R}$. Under the assumption that the $X_i$ are i.i.d. such that the law of $g(X_i)$ has an absolutely continuous component, we prove in the companion paper [27] the convergence in distribution of (4.1) to a random variable uniformly distributed on $[0, 1]$. Under this assumption, we explicit the limit of $\text{Var}\left(\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m)\right)$ and prove that $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} \left(f(Y_m^M) - \frac{\mathbb{E}(f(X_1)g(X_1))}{\mathbb{E}(g(X_1))}\right)$ converges in distribution to a centered Gaussian random variable with variance equal to the limit. The proof of this Central Limit Theorem relies on the asymptotic behaviour as $M \to \infty$ of a vector composed of $\sqrt{M} \left(\frac{\sum_{m=1}^{M} g(X_m)f(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)} - \mathbb{E}(f(X_1)g(X_1))\right)$ and the fractional parts (4.1) for $\alpha \in \{\alpha_1, \dots, \alpha_s\}$ with $0 < \alpha_1 < \dots < \alpha_s < 1$. In the companion paper [27], we check that this vector converges in distribution to a vector with centered Gaussian first component and independent $s$ last components uniformly distributed on $[0, 1]^s$. Under the conjecture that a similar convergence in distribution remains valid at the next steps of a particle filter which alternates selections according to the stratified resampling mechanism and mutations according to Markov kernels, we provide an inductive formula for the asymptotic variance of the resampled population after $n$ steps.

The paper is organized as follows. In Section 4.2, we recall the definition of the stratified sampling scheme and the statement of the main result (Theorem 34) is given. In Section 4.3, the asymptotic variance is derived and in Section 4.4, the proof of the Central Limit Theorem is given. In Section 4.6, we consider a particle filter which alternates selections according to the stratified resampling mechanism and mutations according to Markov kernels and we provide an inductive formula for the asymptotic variance of the resampled sequence. We perform numerical experiments which support the validity of this formula.

**Notation** We denote by $\lfloor x \rfloor$ the integer $j$ such that $j \le x < j + 1$ and by $\{x\} = x - \lfloor x \rfloor$ the fractional part of $x \in \mathbb{R}$. We denote the set of real-valued bounded measurable functions on $\mathbb{R}^d$ by $B_b(\mathbb{R}^d)$. Given $\mu$ a positive measure on $\mathbb{R}^d$ endowed with the Borel sigma algebra and $\phi: \mathbb{R}^d \to \mathbb{R}$ a measurable function that is either positive or such that $\int_{\mathbb{R}^d} |\phi(x)| \mu(dx) < \infty$, we denote $\mu(\phi) = \int_{\mathbb{R}^d} \phi(x) \mu(dx)$.

## 4.2 Statement of the Main Result

Given $M > 0$, let $(X_m)_{1 \le m \le M}$ be a sequence of i.i.d. $\mathbb{R}^d$-valued random vectors following the law $\eta$ and let $g: \mathbb{R}^d \to (0, \infty)$ be a measurable function such that $0 < \inf_{x \in \mathbb{R}^d} g(x) \le \sup_{x \in \mathbb{R}^d} g(x) < \infty$. In what follows we denote $\bar{g} := \sup_{x \in \mathbb{R}^d} g(x)$ and $g := \inf_{x \in \mathbb{R}^d} g(x)$.

We now generate the sequence $(Y_m^M)_{1 \le m \le M}$ according to the selection step of the stratified sampling. We recall that starting from $M$ random variables $(U_m)_{1 \le m \le M}$ i.i.d. distributed according to the uniform law on $(0, 1)$ and independent of $(X_m)_{1 \le m \le M}$, the sequence $(Y_m^M)_{1 \le m \le M}$ is defined in the following way

$$ Y_m^M = \sum_{\ell=1}^{M} \left\{ \sum_{j=1}^{\ell-1} w_j^M < m - U_m \le \sum_{j=1}^{\ell} w_j^M \right\} X_\ell \quad \text{for } m \in \{1, \dots, M\} \qquad (4.2) $$
where $w_m^M = \frac{Mg(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)}$ for $m = 1, \dots, M$. Since the weights are preserved when multiplied by a positive constant, up to dividing $g$ by $\mathbb{E}(g(X_1))$, without loss of generality we may suppose that

$$ \mathbb{E}(g(X_1)) = 1. $$

Let us observe that the random vectors $(Y_m^M)_{1 \le m \le M}$ are conditionally independent given $\mathcal{F}$, the $\sigma$-algebra generated by the sequence $(X_m)_{1 \le m \le M}$. Moreover one has

$$ \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \delta_{Y_m^M} | \mathcal{F} \right) = \frac{1}{\sum_{m=1}^{M} g(X_m)} \sum_{m=1}^{M} g(X_m) \delta_{X_m}. \quad (4.3) $$

Given $f: \mathbb{R}^d \to \mathbb{R}$ a measurable function, our purpose is to provide a central limit theorem for $\frac{1}{M} \sum_{m=1}^{M} f(Y_m^M)$. If we start looking at the variance of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M)$, using (4.3) one has

$$ \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \right) \quad (4.4) $$

$$ = \mathrm{Var} \left( \mathbb{E} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) \right) + \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) \right) \quad (4.5) $$

$$ = \mathrm{Var} \left( \sqrt{M} \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)} \right) + \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) \right). \quad (4.6) $$

Thus, in particular, we will prove respectively the convergence of the first and second term in (4.6). The first term is common to all the resampling schemes while the second one really depends on the considered resampling scheme.

Let us consider the following hypotheses:

$$ \mathbf{I1} \quad \sup_{x \in \mathbb{R}^d} |f(x)| < \infty $$

$$ \mathbf{I2} \quad \text{the law of } g(X_1) \text{ has an absolutely continuous component with respect to the Lebesgue measure on } \mathbb{R} $$

Before providing the statement of the central limit theorem, to express the asymptotic variance we introduce the real-valued functions $\beta_0$ and $\beta_1$ respectively given by

$$ \beta_0(x, y_1) = \{x + y_1\} (1 - \{x + y_1\}) + x(1 - x) - 2x(1 - x - y_1)1_{\{y_1 < 1 - x\}} \quad (4.7) $$

$$ \begin{aligned} \beta_1(x, y_1, y_2, y_3) = & 2 \left( \{x + y_1\} (1 - \{x + y_1\} - y_2) 1_{\{y_2 < 1 - \{x + y_1\}\}} \right. \\ & \left. - \{x + y_1\} (1 - \{x + y_1\} - y_2 - y_3) 1_{\{y_2 + y_3 < 1 - \{x + y_1\}\}} \right. \\ & \left. - x (1 - x - y_1 - y_2) 1_{\{y_1 + y_2 < 1 - x\}} \right. \\ & \left. + x (1 - x - y_1 - y_2 - y_3) 1_{\{y_1 + y_2 + y_3 < 1 - x\}} \right). \end{aligned} \quad (4.8) $$
**Remark 33.** Let us observe that $\beta_0$ is continuous. Indeed, the fractional part is continuous apart from the integers where its left-hand limit is equal to 1 and its right-hand limit is equal to 0. The composition with the function $f(z) = z(1-z)$ which is such that $f(0) = f(1) = 0$, allows us to conclude that $(x, y_1) \mapsto \{x+y_1\}(1-\{x+y_1\})$ is continuous. Moreover $(x, y_1) \mapsto x(1-x-y_1)1_{\{y_1<1-x\}}$ is continuous since the function $\bar{f}(x, y_1) = 1-x-y_1$ is equal to 0 on the set $\{(x, y_1) \in \mathbb{R}^2 : y_1 = 1-x\}$ of discontinuity points of the indicator function. Similarly it is possible to prove that also $\beta_1$ is a continuous function on $\mathbb{R} \times \mathbb{R} \times \mathbb{R}^+ \times \mathbb{R}^+$.

The following result holds.

**Theorem 34.** Under the notation introduced above and under **I1-2** we have

$$ \lim_{M \to \infty} \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \right) = \sigma_1^2(f) + \sigma_2^2(f) \quad (4.9) $$

where

$$ \sigma_1^2(f) = \eta \left( (g(f - \eta(fg)))^2 \right) \quad (4.10) $$

and

$$ \sigma_2^2(f) := \sum_{k=0}^{\lfloor \frac{g}{g} \rfloor} \mathbb{E}(F_k) \quad (4.11) $$

with $(F_k)_{k \in \mathbb{N}}$ given by

$$ F_k = \begin{cases} f^2(X_1)\beta_0(U_1, g(X_1)) & k=0 \\ -f(X_1)f(X_{k+1})\beta_1\left(U_1, g(X_1), \sum_{\ell=2}^k g(X_\ell), g(X_{k+1})\right) & k>0 \end{cases} \quad (4.12) $$

where $U_1 \sim \mathcal{U}(0, 1)$ is independent of $X_1, \dots, X_{k+1}$.

Moreover the following convergence in distribution holds

$$ \sqrt{M} \left( \frac{1}{M} \sum_{m=1}^{M} f(Y_m^M) - \eta(fg) \right) \stackrel{d}{\Longrightarrow} \mathcal{N} (0, \sigma_1^2(f) + \sigma_2^2(f)) \quad (4.13) $$

We will split the proof of the theorem into two parts: in Section 4.3, we are going to prove the result about the asymptotic variance and in Section 4.4, the proof of the Central Limit Theorem is provided.

## 4.3 Asymptotic Variance

The following proposition provides the asymptotic behaviour of the first term in (4.6). We provide its proof for the sake of completeness.

**Proposition 7.** Under **I1**, the following convergence in distribution holds

$$ \sqrt{M} \left( \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)} - \eta(fg) \right) \stackrel{d}{\Longrightarrow} \mathcal{N} (0, \sigma_1^2(f)) \quad (4.14) $$

where $\sigma_1^2(f)$ has been defined in (4.10). Moreover

$$ \lim_{M \to \infty} \mathrm{Var} \left( \sqrt{M} \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)} \right) = \sigma_1^2(f). \quad (4.15) $$
Proof. We recall that we suppose that $\eta(g) = 1$. Let us define $h := g(f - \eta(fg))$. By observing that $\eta(h) = 0$ and

$$ \sqrt{M} \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)} - \sqrt{M} \eta(fg) = \frac{M}{\sum_{m=1}^{M} g(X_m)} \times \frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(X_m), \quad (4.16) $$

by Slutsky's theorem we have

$$ \sqrt{M} \left( \frac{\sum_{m=1}^{M} g(X_m) f(X_m)}{\sum_{\ell=1}^{M} g(X_\ell)} - \eta(fg) \right) \stackrel{d}{=} \mathcal{N}(0, \eta(h^2)). $$

Let us now prove (4.15). Let us preliminary study the following quantity

$$ \mathbb{E} \left( \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(X_m) \right)^4 \right) = \frac{1}{M^2} \sum_{m_1, m_2, m_3, m_4=1}^{M} \mathbb{E} \left( \prod_{i=1}^{4} h(X_{m_i}) \right). $$

Among the $M^4$ expectations appearing in the sum on the right-hand side, the ones where one index is different from the other three are equal to 0 by the independence of the $X_i$ and the fact that $\eta(h) = 0$. Therefore

$$ \mathbb{E} \left( \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(X_m) \right)^4 \right) = \frac{1}{M} \eta(h^4) + \frac{3(M-1)}{M} \eta(h^2)^2 \le D, \quad (4.17) $$

where $D$ is a finite constant depending on $g$ and $f$ but not on $M$. Since (4.16) holds, to prove (4.15) we can study the convergence as $M$ goes to infinity of the variance of $\sqrt{M} \frac{1}{\bar{g}_M} \bar{h}_M$ given by

$$ \mathbb{E} \left( \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} \right)^2 \right) - \mathbb{E} \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} \right)^2 $$

where we denote by $\bar{g}_M$ the empirical mean $\frac{1}{M} \sum_{m=1}^{M} g(X_m)$ and by $\bar{h}_M$ the empirical mean $\frac{1}{M} \sum_{m=1}^{M} h(X_m)$.

By the Cauchy-Schwarz inequality we have

$$ \begin{aligned} \mathbb{E} \left( \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} - \sqrt{M} \bar{h}_M \right)^2 \right) &= \mathbb{E} \left( \left( \sqrt{M} \bar{h}_M \right)^2 \left( \frac{1}{\bar{g}_M} - 1 \right)^2 \right) \\ &\le \mathbb{E}^{\frac{1}{2}} \left( \left( \sqrt{M} \bar{h}_M \right)^4 \right) \mathbb{E}^{\frac{1}{2}} \left( \left( \frac{1}{\bar{g}_M} - 1 \right)^4 \right) \\ &\le D^{\frac{1}{2}} \cdot \mathbb{E}^{\frac{1}{2}} \left( \left( \frac{1}{\bar{g}_M} - 1 \right)^4 \right), \end{aligned} $$

where the right-hand side converges to 0 as $M$ goes to infinity by the Strong Law of Large Numbers and by Lebesgue's theorem. With $\eta(h) = 0$, this in particular implies that

$$ 0 = \lim_{M \to \infty} \left| \mathbb{E}^{\frac{1}{2}} \left( \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} \right)^2 \right) - \mathbb{E}^{\frac{1}{2}} \left( \left( \sqrt{M} \bar{h}_M \right)^2 \right) \right| = \lim_{M \to \infty} \left| \mathbb{E}^{\frac{1}{2}} \left( \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} \right)^2 \right) - \eta(h^2)^{\frac{1}{2}} \right| $$

and

$$ 0 = \lim_{M \to \infty} \left| \mathbb{E} \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} \right) - \mathbb{E} \left( \sqrt{M} \bar{h}_M \right) \right| = \lim_{M \to \infty} \left| \mathbb{E} \left( \sqrt{M} \frac{\bar{h}_M}{\bar{g}_M} \right) \right| $$

and so the proof is complete. □
Let us now study the second term in the expression (4.6). Our purpose is to prove the following result:

**Theorem 35.** Under **I1-2**, the following convergence holds

$$ \lim_{M \to \infty} \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) | \mathcal{F} \right) \right) = \sigma_2^2(f) \quad (4.18) $$

where $\sigma_2^2(f)$ has been defined in (4.11).

To prove Theorem 35 we need some preliminary results the proofs of which are given in Section 4.5. By the conditional independence of the random vectors $(Y_m^M)_{1\le m\le M}$ with respect to the $\sigma$-algebra $\mathcal{F}$, we have

$$ \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) = \frac{1}{M} \sum_{m=1}^{M} \mathrm{Var} \left( f(Y_m^M) \mid \mathcal{F} \right) \quad (4.19) $$

$$ = \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} \left( f^2(Y_m^M) \mid \mathcal{F} \right) - \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} \left( f(Y_m^M) \mid \mathcal{F} \right)^2. \quad (4.20) $$

We introduce the following notation that will be useful in what follows: for $i = 1, \dots, M$, let us denote $u_i^M := \{w_1^M + \dots + w_i^M\}$, $\mu_i^M := \lfloor w_1^M + \dots + w_i^M \rfloor + 1$ where $w_i^M = \frac{Mg(X_i)}{\sum_{\ell=1}^{M} g(X_\ell)}$ has been introduced in (4.2) and by convention $u_0^M = 0$, $\mu_0^M = 1$.

We now rewrite in a more explicit way the conditional variance $\mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right)$. The proof is given in the Section 4.5.

**Proposition 8.** Under **I1**, we have

$$ \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) = \frac{1}{M} \sum_{i=1}^{M} f^2(X_i) \beta_0 (u_{i-1}^M, w_i^M) \quad (4.21) $$

$$ - \frac{1}{M} \sum_{k=1}^{(M-1)\lfloor\frac{\bar{g}}{g}\rfloor} \sum_{i=1}^{M-k} f(X_i) f(X_{i+k}) \beta_1(u_{i-1}^M, w_i^M, \sum_{\ell=2}^{k} w_{i+\ell-1}^M, w_{i+k}^M) \quad (4.22) $$

where $\beta_0$ and $\beta_1$ are respectively defined in (4.7) and (4.8).

In what follows, without specifying it every time, we will suppose that $M \ge 1 + \lfloor\frac{\bar{g}}{g}\rfloor$ so that $(M-1)\lfloor\frac{\bar{g}}{g}\rfloor = \lfloor\frac{\bar{g}}{g}\rfloor$. Let us observe that it is possible to rewrite (4.21) and (4.22) in the following way

$$ \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) = \sum_{k=0}^{\lfloor\frac{\bar{g}}{g}\rfloor} \int_0^1 F_{k, \alpha_1}^M d\alpha_1 \quad (4.23) $$

where $F_{k,\alpha_1}^M$ is given by $f^2(X_{\lfloor\alpha_1 M\rfloor})\beta_0(u_{\lfloor\alpha_1 M\rfloor-1}^M, w_{\lfloor\alpha_1 M\rfloor}^M)$ if $k=0$ and by

$$ -1_{\{\lfloor\alpha_1 M\rfloor \le M-k\}} f(X_{\lfloor\alpha_1 M\rfloor}) f(X_{\lfloor\alpha_1 M\rfloor+k}) \beta_1(u_{\lfloor\alpha_1 M\rfloor-1}^M, w_{\lfloor\alpha_1 M\rfloor}^M, \sum_{\ell=2}^{k} w_{\lfloor\alpha_1 M\rfloor+\ell-1}^M, w_{\lfloor\alpha_1 M\rfloor+k}^M) $$
if $k > 0$. Given $0 < \alpha_1 < 1$ and $k \in \mathbb{N}$, $F_{k,\alpha_1}^M$ is a function of the vector

$$T_{\alpha_1, k}^M := \left( f(X_{\lfloor \alpha_1 M \rfloor}), f(X_{\lfloor \alpha_1 M \rfloor + k}), u_{\lfloor \alpha_1 M \rfloor - 1}^M, w_{\lfloor \alpha_1 M \rfloor}, \sum_{\ell=2}^k w_{\lfloor \alpha_1 M \rfloor + \ell - 1}^M, w_{\lfloor \alpha_1 M \rfloor + k}^M \right)$$

which is well defined for $M$ big enough. Therefore to study the limit behavior as $M \to \infty$ of the expectation of $F_{k,\alpha_1}^M$, we are going to study the limit behavior of the vector $T_{\alpha_1,k}^M$. To obtain a much more general formulation useful in the next section, we introduce also the following family of vectors well defined for $M$ big enough. Let $s \ge 1$ and $0 < \alpha_1 < \dots < \alpha_s < 1$. For $k := (k_1, \dots, k_s) \in \mathbb{N}^s$, let us define $\forall i = 1, \dots, s$

$$T_{\alpha_i, k_i}^M := \left( f(X_{\lfloor \alpha_i M \rfloor}), f(X_{\lfloor \alpha_i M \rfloor + k_i}), u_{\lfloor \alpha_i M \rfloor - 1}^M, w_{\lfloor \alpha_i M \rfloor}, \sum_{\ell=2}^{k_i} w_{\lfloor \alpha_i M \rfloor + \ell - 1}^M, w_{\lfloor \alpha_i M \rfloor + k_i}^M \right). \quad (4.24)$$

Moreover let

$$H^M := \sqrt{M} \left( \frac{\sum_{m=1}^{M} (fg)(X_m)}{\sum_{m=1}^{M} g(X_m)} - \eta(fg) \right) \quad (4.25)$$

that thanks to (4.14) converges in distribution to a centered Gaussian random variable with variance $\sigma_1^2(f)$. The purpose of the next proposition is therefore to study the convergence in distribution of $(H^M, T_{\alpha_1,k_1}^M, \dots, T_{\alpha_s,k_s}^M)$.

**Proposition 9.** Assume I1-2. Let $s \in \mathbb{N}^*$ and $k := (k_1, \dots, k_s) \in \mathbb{N}^s$. We set $K_i := k_1 + \dots + k_{i-1} + i - 1$ for $i \in \{2, \dots, s\}$ and $K_1 := 0$. Let $(U_j)_{1 \le j \le s}$ be a sequence of i.i.d random variables distributed according to the uniform law on $(0,1)$ and independent of $H \sim \mathcal{N}(0, \sigma_1^2(f))$ with $(U_1, \dots, U_s, H)$ independent from $(X_j)_{1 \le j \le K_{s+1}}$. For $0 < \alpha_1 < \dots < \alpha_s < 1$, as $M$ goes to infinity, the following convergence in distribution holds

$$(H^M, T_{\alpha_1, k_1}^M, \dots, T_{\alpha_s, k_s}^M) \stackrel{d}{\Longrightarrow} (H, T_{1,k}, \dots, T_{s,k}) \quad (4.26)$$

where

$$T_{i,k} := \left( f(X_{K_i+1}), f(X_{K_{i+1}}), U_i, g(X_{K_i+1}), \sum_{\ell=K_i+2}^{K_{i+1}-1} g(X_\ell), g(X_{K_{i+1}}) \right) \quad \forall i = 1, \dots, s. \quad (4.27)$$

*Proof of Proposition 9.* Let $M \ge \max\left(\frac{2}{\alpha_1}, \frac{k_s}{1-\alpha_s}, \max_{1 \le i \le s-1} \frac{k_{i+1}}{\alpha_{i+1}-\alpha_i}\right)$ so that $2 \le \lceil \alpha_i M \rceil \le M - k_i$ $\forall i = 1, \dots, s$ and $k_i < \lceil \alpha_{i+1} M \rceil - \lceil \alpha_i M \rceil$ $\forall i = 1, \dots, s-1$. The last condition allows us to separate the variables: $(X_{\lfloor \alpha_i M \rfloor}, \dots, X_{\lfloor \alpha_i M \rfloor + k_i})$ is independent of $(X_{\lfloor \alpha_j M \rfloor}, \dots, X_{\lfloor \alpha_j M \rfloor + k_j})$ for $i, j$ distinct elements of $\{1, \dots, s\}$.

By defining $J_{\alpha_i, k_i}^M := \{\lceil \alpha_i M \rceil, \dots, \lceil \alpha_i M \rceil + k_i\}$ for $i = 1, \dots, s$, let $J^M := \bigcup_{i=1}^s J_{\alpha_i, k_i}^M$.

Moreover let $(\tilde{X}_m)_{m \ge 1}$ be a sequence of i.i.d. $\mathbb{R}^d$-valued random vectors following the law $\eta$ independent of $(X_m)_{1 \le m \le M}$. Given $x := (x_j)_{1 \le j \le K_{s+1}} \in \mathbb{R}^{d(K_{s+1})}$, we define

$$H_x^M := \sqrt{M} \left( \frac{\sum_{j=1}^{K_{s+1}} (fg)(x_j) + \sum_{j \notin J_s^M} (fg)(X_j)}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j \notin J_s^M} g(X_j)} - \eta(fg) \right)$$
and

$$
\tilde{H}_x^M := \sqrt{M} \left( \frac{\sum_{j=1}^{K_{s+1}} (fg)(x_j) + \sum_{j=1}^{M-K_{s+1}} (fg)(\tilde{X}_j)}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} - \eta(fg) \right).
$$

Similarly for $i = 1, \dots, s$ we define the vector $T_{\alpha_i, k_i, x}^M$ by

$$
\left(
\begin{array}{@{}l@{}}
  \left(
    f(x_{K_i+1}), f(x_{K_i+1}),
    \begin{array}{@{}c@{}}
      \left( \sum_{j=1}^{[\alpha_i M]-1} g(x_j) + \sum_{j=1}^{K_i} g(x_j) \right) \\
      \times \frac{M}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j \notin J_s^M} g(x_j)}
    \end{array}
  \right)
  \left\{
    \begin{array}{@{}c@{}}
      \left( \sum_{j \notin J_s^M}^{[\alpha_i M]-1} g(x_j) + \sum_{j=1}^{K_i} g(x_j) \right) \\
      \times \frac{Mg(x_{K_i+1})}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j \notin J_s^M} g(x_j)}
    \end{array}
  \right\}
  \left(
    \begin{array}{@{}c@{}}
      \sum_{\ell=K_i+2}^{K_{i+1}-1} g(x_\ell) \\
      \times \frac{M}{\sum_{\ell=K_i+2}^{K_{i+1}-1} g(x_\ell)}
    \end{array}
  \right)
  \left(
    \begin{array}{@{}c@{}}
      \frac{Mg(x_{K_{i+1}})}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j \notin J_s^M} g(x_j)}
    \end{array}
  \right)
\end{array}
\right)
$$

and the vector $\tilde{T}_{\alpha_i, k_i, x}$ by $(f(x_{K_i+1}), f(x_{K_i+1}), \tilde{T}_{\alpha_i, k_i, x, f}^M)$ with

$$
\tilde{T}_{\alpha_i, k_i, x, f}^M = \left( \left\{ M \times \frac{\sum_{j=1}^{[\alpha_i M] - K_i - 1} g(\tilde{X}_j) + \sum_{j=1}^{K_i} g(x_j)}{K_{s+1} \sum_{j=1}^{[\alpha_i M] - K_i - 1} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} \right\}, \frac{Mg(x_{K_i+1})}{K_{s+1} \sum_{j=1}^{[\alpha_i M] - K_i - 1} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)}, \frac{Mg(x_{K_i+1})}{K_{s+1} \sum_{j=1}^{[\alpha_i M] - K_i - 1} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)}, \frac{Mg(x_{K_i+1})}{K_{s+1} \sum_{j=1}^{[\alpha_i M] - K_i - 1} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} \right).
$$

We notice that

$$
(H_x^M, T_{\alpha_1, k_1, x}^M, \dots, T_{\alpha_s, k_s, x}^M) \stackrel{\mathcal{L}}{=} (\tilde{H}_x^M, \tilde{T}_{\alpha_1, k_1, x}^M, \dots, \tilde{T}_{\alpha_s, k_s, x}^M). \quad (4.28)
$$

Let us observe that by hypothesis **I2** the law of $g(\tilde{X}_i)$ has a component absolutely continuous.
Moreover since $\int_{\mathbb{R}} |1+y|^{s+1} e^{-y^2} dy < \infty \quad \forall s \ge 1$, we can apply Theorem 1 of [27] with $q=s$ and

• $(Y_i, Z_i)$ equal to $(g(\tilde{X}_i), (fg)(\tilde{X}_i))$

• $(\beta_M^1, \dots, \beta_M^s, \beta_M^{s+1})$ equal to $(\lceil \alpha_1 M \rceil - K_1 - 1, \dots, \lceil \alpha_s M \rceil - K_s - 1, M - K_{s+1})$

• $\phi(x) = \frac{1}{x}$

• $x$ equal to $\sum_{j=1}^{K_{s+1}} g(x_j)$, $z$ equal to $\sum_{j=1}^{K_{s+1}} (fg)(x_j)$, $\theta$ equal to $\eta(fg)$ and $(y_1, \dots, y_s)$ equal to
$$ \left( \sum_{j=1}^{K_1} g(x_j), \dots, \sum_{j=1}^{K_s} g(x_j) \right) $$

• $T := H \sim \mathcal{N}(0, \sigma_1^2(f))$ by (4.14)

and deduce the following convergence in distribution as $M \to \infty$

$$
\left( \tilde{H}_x^M, \left( \left\{ M \times \frac{\sum_{j=1}^{[\alpha_i M] - K_i - 1} g(\tilde{X}_j) + \sum_{j=1}^{K_i} g(x_j)}{K_{s+1} \sum_{j=1}^{[\alpha_i M] - K_i - 1} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} \right\} \right)_{1 \le i \le s} \right) \underset{d}{\Longrightarrow} (H, U_1, \dots, U_s). \quad (4.29)
$$

Moreover, by the Strong Law of Large Numbers

$$
\lim_{M \to \infty} \frac{Mg(x_{K_i+1})}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} = g(x_{K_i+1}),
$$
$$
\lim_{M \to \infty} \frac{M \sum_{\ell=K_i+2}^{K_{i+1}-1} g(x_\ell)}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} = \sum_{\ell=K_i+2}^{K_{i+1}-1} g(x_\ell)
$$
and

$$ \lim_{M \to \infty} \frac{Mg(x_{K_{i+1}})}{\sum_{j=1}^{K_{s+1}} g(x_j) + \sum_{j=1}^{M-K_{s+1}} g(\tilde{X}_j)} = g(x_{K_{i+1}}). $$

Therefore by Slutsky's theorem, for each $x := (x_j)_{1 \le j \le K_{s+1}} \in \mathbb{R}^{d(K_{s+1})}$ we can conclude that the following random vector

$$ (\tilde{H}_x^M, \tilde{T}_{\alpha_1,k_1,x}^M, \dots, \tilde{T}_{\alpha_s,k_s,x}^M) $$

converges in distribution as $M$ goes to infinity to

$$ (H, T_{1,k,x}, \dots, T_{s,k,x}) \qquad (4.30) $$

where the vector $T_{i,k,x}$ is given by

$$ \left( f(x_{K_{i+1}}), f(x_{K_{i+1}}), U_i, g(x_{K_{i+1}}), \sum_{\ell=K_{i+2}}^{K_{i+1}-1} g(x_\ell), g(x_{K_{i+1}}) \right). $$

In particular, given $\phi: \mathbb{R}^{1+6s} \to \mathbb{R}$ a continuous bounded function one has

$$ \lim_{M \to \infty} \mathbb{E} \left( \phi \left( \tilde{H}_x^M, \tilde{T}_{\alpha_1,k_1,x}^M, \dots, \tilde{T}_{\alpha_s,k_s,x}^M \right) \right) = \mathbb{E} \left( \phi \left( H, T_{1,k,x}, \dots, T_{s,k,x} \right) \right). \quad (4.31) $$

We are now ready to prove (4.26). By conditioning with respect to $\sigma((X_i)_{i \in J_s^M})$ and by applying the Freezing Lemma, (4.28) and (4.31), one has

$$ \begin{aligned} & \mathbb{E} \left( \phi \left( H^M, T_{\alpha_1,k_1}^M, \dots, T_{\alpha_s,k_s}^M \right) \right) \\ &= \int_{\mathbb{R}^{d(k_1+\dots+k_{s+s})}} \mathbb{E} \left( \phi \left( H_x^M, T_{\alpha_1,k_1,x}^M, \dots, T_{\alpha_s,k_s,x}^M \right) \right) \prod_{i=1}^{K_{s+1}} \eta(dx_i) \\ &= \int_{\mathbb{R}^{d(k_1+\dots+k_{s+s})}} \mathbb{E} \left( \phi \left( \tilde{H}_x^M, \tilde{T}_{\alpha_1,k_1,x}^M, \dots, \tilde{T}_{\alpha_s,k_s,x}^M \right) \right) \prod_{i=1}^{K_{s+1}} \eta(dx_i) \\ &\xrightarrow{M \to \infty} \mathbb{E} \left( \phi \left( H, T_{1,k}, \dots, T_{s,k} \right) \right) \end{aligned} $$

where $T_{i,k}$, $i = 1, \dots, s$ is defined in (4.27) and so the proof is complete. $\square$

We are now ready to prove Theorem 35.

*Proof of Theorem 35.* By (4.23), it is enough to study the convergence as $M$ goes to infinity of

$$ \sum_{k=0}^{\lfloor \frac{g}{g} \rfloor} \int_{0}^{1} \mathbb{E} (F_{k, \alpha_1}^M) d\alpha_1. \qquad (4.32) $$

Let us observe that we can replace $\beta_0$ and $\beta_1$ (introduced in (4.7) and (4.8)) in the definition of $F_{k,\alpha_1}^M$ respectively by the bounded functions

$$ \beta_0 (0 \lor x \land 1, 0 \lor y_1), \beta_1 (0 \lor x \land 1, 0 \lor y_1, 0 \lor y_2, 0 \lor y_3) $$

that to simplify the notation we will keep calling $\beta_0$ and $\beta_1$.

For what has been said there exists a constant $\tilde{C} < \infty$ such that
$$ \sup_{(x,y_1) \in \mathbb{R}^2} |\beta_0(x, y_1)| \lor \sup_{(x,y_1,y_2,y_3) \in \mathbb{R}^4} |\beta_1(x, y_1, y_2, y_3)| \le \tilde{C}. \quad (4.33) $$

By Proposition 9 applied with $s = 1$ and by Remark 33, we have that for each $\alpha_1 \in (0, 1)$

$$ F_{0, \alpha_1}^{M} \stackrel{d}{\Rightarrow} F_0 $$

with $F_0$ defined in (4.12). Moreover by the hypothesis of boundedness of $f$ and (4.33) we have

$$ \lim_{M \to \infty} \mathbb{E} (F_{0, \alpha_1}^M) = \mathbb{E} (F_0). $$

By Lebesgue's theorem, $\int_0^1 \mathbb{E} (F_{0, \alpha_1}^M) d\alpha_1$ will converge to the same limit.

Let us now study the convergence of $\sum_{k=1}^M \int_0^1 \mathbb{E} (F_{k, \alpha_1}^M) d\alpha_1$. By Proposition 9 applied with $s = 1$ and by Remark 33, we have that for each $\alpha_1 \in (0, 1)$ and $k \ge 1$

$$ F_{k, \alpha_1}^M \stackrel{d}{\Rightarrow} F_k $$

with $F_k$ defined in (4.12). Moreover by the hypothesis of boundedness of $f$ and (4.33),

$$ \lim_{M \to \infty} \mathbb{E} (F_{k, \alpha_1}^M) = \mathbb{E} (F_k). $$

By Lebesgue's theorem, $\lim_{M \to \infty} \int_0^1 \mathbb{E} (F_{k, \alpha_1}^M) d\alpha_1 = \mathbb{E} (F_k)$ for each $k = 1, \dots, \lfloor \frac{g}{g} \rfloor$ and so the proof is complete. $\square$

We are now ready to provide a central limit theorem for $\frac{1}{M} \sum_{m=1}^M f(Y_m^M)$ knowing its asymptotic variance.

## 4.4 Central Limit Theorem

**Theorem 36.** Let assume **I1-2**. The following convergence in distribution holds

$$ \sqrt{M} \left( \frac{1}{M} \sum_{m=1}^{M} f(Y_m^M) - \eta(fg) \right) \stackrel{d}{\Rightarrow} \mathcal{N} \left( 0, \sigma_1^2(f) + \sigma_2^2(f) \right) \quad (4.34) $$

where $\sigma_1^2(f)$ and $\sigma_2^2(f)$ are respectively defined in (4.10) and (4.11).

*Proof of Theorem 36.* Let $u \in \mathbb{R}$. By introducing the notation

$$ a_m^M := e^{\frac{iu}{\sqrt{M}} (f(Y_m^M) - \mathbb{E}(f(Y_m^M) | \mathcal{F}))} \quad \text{for } m = 1, \dots, M \quad (4.35) $$

and by applying (4.3), we have

$$ \begin{aligned} \mathbb{E} \left( e^{\frac{iu}{\sqrt{M}} \sum_{m=1}^{M} (f(Y_m^M) - \eta(fg))} \right) &= \mathbb{E} \left( e^{i u \sqrt{M} \left( \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) - \eta(fg) \right)} \right) \mathbb{E} \left( \prod_{m=1}^{M} a_m^M \mid \mathcal{F} \right) \\ &= \mathbb{E} \left( e^{iuH^M} \mathbb{E} \left( \prod_{m=1}^{M} a_m^M \mid \mathcal{F} \right) \right) \end{aligned} \quad (4.36) $$
where $H^M$ is defined in (4.25). The purpose of what follows is therefore to study the convergence
as $M$ goes to infinity of (4.36).

First step

We are first going to study the conditional expectation $\mathbb{E} \left( \prod_{m=1}^{M} a_m^M \mid \mathcal{F} \right)$.

By using that for each $u, y \in \mathbb{R} \mid e^{iuy} - 1 - iuy + \frac{(uy)^2}{2} \mid \le \frac{|uy|^3}{6}$, one has

$$
\begin{aligned}
& \left| a_m^M - 1 - \frac{iu}{\sqrt{M}} (f(Y_m^M) - \mathbb{E}(f(Y_m^M) \mid \mathcal{F})) + \frac{u^2 (f(Y_m^M) - \mathbb{E}(f(Y_m^M) \mid \mathcal{F}))^2}{2M} \right| \\
& \le \frac{|u|^3 |f(Y_m^M) - \mathbb{E}(f(Y_m^M) \mid \mathcal{F})|^3}{6M^{\frac{3}{2}}} \le \frac{4 \|f\|_{\infty}^3 |u|^3}{3M^{\frac{3}{2}}}.
\end{aligned}
$$

Therefore for $m = 1, \dots, M$, $a_m^M$ can be rewritten as

$$
a_m^M = 1 + \frac{iu}{\sqrt{M}} (f(Y_m^M) - \mathbb{E}(f(Y_m^M) | \mathcal{F})) - \frac{u^2 (f(Y_m^M) - \mathbb{E}(f(Y_m^M) | \mathcal{F}))^2}{2M} + r_m^M
$$

with $|r_m^M| \le \frac{4\|f\|_\infty^3|u|^3}{3M^{\frac{3}{2}}}$.

By defining for $m = 1, \dots, M$

$$
b_m^M := 1 - \frac{u^2 (f(Y_m^M) - \mathbb{E}(f(Y_m^M) | \mathcal{F}))^2}{2M}, \quad (4.37)
$$

let us now study the difference

$$
\left| \mathbb{E} \left( \prod_{m=1}^{M} a_m^M \mid \mathcal{F} \right) - \mathbb{E} \left( \prod_{m=1}^{M} b_m^M \mid \mathcal{F} \right) \right|. \qquad (4.38)
$$

By rewriting $\prod_{m=1}^{M} a_m^M - \prod_{m=1}^{M} b_m^M = \sum_{m=1}^{M} \prod_{\ell=1}^{m-1} a_\ell^M (a_m^M - b_m^M) \prod_{\ell=m+1}^{M} b_\ell^M$, where by convention the empty product is equal to one, by using that the random vectors $(Y_m^M)_{1\le m \le M}$ are conditionally independent given $\mathcal{F}$ and that $|r_m^M| \le \frac{4\|f\|_{\infty}^3|u|^3}{3M^{\frac{3}{2}}}$, $|a_\ell^M| \le 1$ and $|b_\ell^M| \le 1 + \frac{2\|f\|_{\infty}^2 u^2}{M}$, (4.38) becomes

$$
\begin{align*}
& \left| \mathbb{E} \left( \prod_{m=1}^{M} a_m^M \mid \mathcal{F} \right) - \mathbb{E} \left( \prod_{m=1}^{M} b_m^M \mid \mathcal{F} \right) \right| \\
&= \left| \sum_{m=1}^{M} \mathbb{E} \left( \prod_{\ell=1}^{m-1} a_\ell^M \left( \frac{iu}{\sqrt{M}} (f(Y_m^M) - \mathbb{E}(f(Y_m^M) \mid \mathcal{F})) + r_m^M \right) \prod_{\ell=m+1}^{M} b_\ell^M \mid \mathcal{F} \right) \right| \\
&= \left| \sum_{m=1}^{M} \mathbb{E} \left( \prod_{\ell=1}^{m-1} a_\ell^M \mid \mathcal{F} \right) \mathbb{E} \left( \frac{iu}{\sqrt{M}} (f(Y_m^M) - \mathbb{E}(f(Y_m^M) \mid \mathcal{F})) + r_m^M \mid \mathcal{F} \right) \mathbb{E} \left( \prod_{\ell=m+1}^{M} b_\ell^M \mid \mathcal{F} \right) \right| \\
&= \left| \sum_{m=1}^{M} \mathbb{E} \left( \prod_{\ell=1}^{m-1} a_\ell^M \mid \mathcal{F} \right) \mathbb{E}(r_m^M \mid \mathcal{F}) \mathbb{E} \left( \prod_{\ell=m+1}^{M} b_\ell^M \mid \mathcal{F} \right) \right| \\
&\le M \times \frac{4 \|f\|_{\infty}^3 |u|^3}{3M^{\frac{3}{2}}} \times \left( 1 + \frac{2 \|f\|_{\infty}^2 u^2}{M} \right)^M \xrightarrow{M \to \infty} 0.
\end{align*}
$$
This implies that

$$ \limsup_{M \to \infty} \left| \mathbb{E} \left( e^{iuH^M} \mathbb{E} \left( \prod_{m=1}^{M} a_m^M \middle| \mathcal{F} \right) \right) - \mathbb{E} \left( e^{iuH^M} \mathbb{E} \left( \prod_{m=1}^{M} b_m^M \middle| \mathcal{F} \right) \right) \right| = 0. \quad (4.39) $$

## Second step

With $b_m^M$ defined in (4.37), by the previous step, it is enough to study the limit behaviour as $M \to \infty$ of $\mathbb{E} \left( e^{iuH^M} \mathbb{E} \left( \prod_{m=1}^{M} b_m^M \middle| \mathcal{F} \right) \right)$. By using the general fact that

$$ \prod_{m=1}^{M} (1 - c_m) = \sum_{s=0}^{M} (-1)^s \sum_{1 \le m_1 < \dots < m_s \le M} \prod_{j=1}^{s} c_{m_j} $$

for any real sequence $(c_m)_{1 \le m \le M}$, we have

$$ \begin{aligned} \mathbb{E} \left( \prod_{m=1}^{M} b_m^M \mid \mathcal{F} \right) &= \sum_{s=0}^{M} \left( \frac{-u^2}{2M} \right)^s \sum_{1 \le m_1 < \dots < m_s \le M} \mathbb{E} \left( \prod_{j=1}^{s} \left( f(Y_{m_j}^M) - \mathbb{E} \left( f(Y_{m_j}^M) \mid \mathcal{F} \right) \right)^2 \mid \mathcal{F} \right) \\ &= \sum_{s=0}^{M} \left( \frac{-u^2}{2M} \right)^s \sum_{1 \le m_1 < \dots < m_s \le M} \prod_{j=1}^{s} \mathbb{E} \left( \left( f(Y_{m_j}^M) - \mathbb{E} \left( f(Y_{m_j}^M) \mid \mathcal{F} \right) \right)^2 \mid \mathcal{F} \right) \end{aligned} \quad (4.40) $$

where for the last equality we use that random vectors $(Y_{m_j}^M)_{1 \le m_j \le M}$ are conditionally independent given $\mathcal{F}$. Let us introduce $B_{m_1, \dots, m_s} := \prod_{j=1}^s \mathbb{E} \left( \left( f(Y_{m_j}^M) - \mathbb{E} \left( f(Y_{m_j}^M) \mid \mathcal{F} \right) \right)^2 \mid \mathcal{F} \right)$. Since $B_{m_1, \dots, m_s}$ is symmetric in its indexes and satisfies $|B_{m_1, \dots, m_s}| \le (4 \|f\|_{\infty}^2)^s$, for each $s \ge 1$ we have

$$ \left| \frac{1}{M^s} \left( \frac{1}{s!} \sum_{m_1, \dots, m_s=1}^{M} B_{m_1, \dots, m_s} - \sum_{1 \le m_1 < \dots < m_s \le M} B_{m_1, \dots, m_s} \right) \right| \le \frac{(4 \|f\|_{\infty}^2)^s}{M^s} \left( \frac{M^s}{s!} - \binom{M}{s} \right). $$

It is therefore possible to rewrite (4.40) as

$$ \begin{aligned} \mathbb{E} \left( \prod_{m=1}^{M} b_m^M \mid \mathcal{F} \right) &= \sum_{s=0}^{M} \left( -\frac{u^2}{2M} \right)^s \sum_{1 \le m_1 < \dots < m_s \le M} \prod_{j=1}^{s} \mathbb{E} \left( \left( f(Y_{m_j}^M) - \mathbb{E} \left( f(Y_{m_j}^M) \mid \mathcal{F} \right) \right)^2 \mid \mathcal{F} \right) \\ &= \sum_{s=0}^{M} \left( -\frac{u^2}{2M} \right)^s \frac{1}{s!} \sum_{m_1, \dots, m_s=1}^{M} \prod_{j=1}^{s} \mathbb{E} \left( \left( f(Y_{m_j}^M) - \mathbb{E} \left( f(Y_{m_j}^M) \mid \mathcal{F} \right) \right)^2 \mid \mathcal{F} \right) + \bar{r}^M \\ &= \sum_{s=0}^{M} \left( -\frac{u^2}{2} \right)^s \frac{1}{s!} \left( \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} \left( \left( f(Y_m^M) - \mathbb{E} \left( f(Y_m^M) \mid \mathcal{F} \right) \right)^2 \mid \mathcal{F} \right) \right)^s + \bar{r}^M \\ &= \sum_{s=0}^{M} \left( -\frac{u^2}{2} \right)^s \frac{1}{s!} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) \right)^s + \bar{r}^M \end{aligned} $$

with $\bar{r}^M$ such that

$$ |\bar{r}^M| \le \sum_{s \ge 1} \frac{(2u^2 \|f\|_{\infty}^2)^s}{s!} \left( 1 - \frac{M(M-1)\cdots(M-s+1)}{M^s} \right) \underset{M \to \infty}{\longrightarrow} 0. $$
With $|e^{iuH^M}| \le 1$, (4.36) and (4.39), this implies that

$$ \left| \mathbb{E} \left( e^{\frac{i u}{\sqrt{M}} \sum_{m=1}^{M} (f(Y_m^M) - \eta(fg))} \right) - \mathbb{E} \left( e^{i u H^M} \sum_{s=0}^{M} \left(-\frac{u^2}{2}\right)^s \frac{1}{s!} \left( \text{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) \right)^s \right) \right| \underset{M \to \infty}{\longrightarrow} 0. $$

### Third step

The purpose of what follows is therefore to study the convergence as $M$ goes to infinity of

$$ \mathbb{E} \left( e^{i u H^M} \sum_{s=0}^{M} \left(-\frac{u^2}{2}\right)^s \frac{1}{s!} \left( \text{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_m^M) \mid \mathcal{F} \right) \right)^s \right). \quad (4.41) $$

By (4.23), (4.41) can be rewritten as

$$ \begin{aligned} & \mathbb{E} \left( e^{i u H^M} \right) + \sum_{s=1}^{M} \left(-\frac{u^2}{2}\right)^s \frac{1}{s!} \sum_{k_1, \dots, k_s = 0}^{\lfloor \frac{\bar{g}}{g} \rfloor} \int_{[0,1]^s} \mathbb{E} \left( e^{i u H^M} F_{k_1, \alpha_1}^M \cdots F_{k_s, \alpha_s}^M \right) d\alpha_1 \cdots d\alpha_s \\ & = \mathbb{E} \left( e^{i u H^M} \right) + \sum_{s \ge 1} \mathbb{1}_{\{s \le M\}} z(s, M) \end{aligned} \quad (4.42) $$

where for $s \ge 1$ and $M \ge 1$

$$ z(s, M) = \left(-\frac{u^2}{2}\right)^s \sum_{k_1, \dots, k_s = 0}^{\lfloor \frac{\bar{g}}{g} \rfloor} \int_{[0,1]^s} \mathbb{1}_{\{0 < \alpha_1 < \dots < \alpha_s < 1\}} \mathbb{E} \left( e^{i u H^M} F_{k_1, \alpha_1}^M \cdots F_{k_s, \alpha_s}^M \right) d\alpha_1 \cdots d\alpha_s $$

and where we recall that $F_{k_i, \alpha_i}^M$ for $i = 1, \dots, s$ is given by $f^2(X_{\lceil \alpha_i M \rceil}) \beta_0(u_{\lceil \alpha_i M \rceil-1}^M, w_{\lceil \alpha_i M \rceil}^M)$ if $k_i = 0$ and by

$$ -1_{\{\lceil \alpha_i M \rceil \le M - k_i\}} f(X_{\lceil \alpha_i M \rceil}) f(X_{\lceil \alpha_i M \rceil + k_i}) \beta_1(u_{\lceil \alpha_i M \rceil-1}^M, w_{\lceil \alpha_i M \rceil}^M, \sum_{\ell=2}^{k_i} w_{\lceil \alpha_i M \rceil+\ell-1}^M, w_{\lceil \alpha_i M \rceil+k_i}^M) $$

if $k_i > 0$. Let us now study the convergence as $M \to \infty$ of (4.42): $\lim_{M \to \infty} \mathbb{E} \left( e^{i u H^M} \right) = e^{-\frac{u^2 \sigma_1^2 (f)}{2}}$ by (4.14) and to study the convergence of the sum we will apply Lebesgue's theorem.

Given $s \ge 1$, we first compute the pointwise convergence as $M \to \infty$ of $z(s, M)$ and then we show the existence of a function $\tilde{z}$ independent of $M$ that dominates $|z|$ and such that $\sum_{s \ge 1} \tilde{z}(s) < \infty$. Let us

start by computing the pointwise convergence. Let $s \ge 1$ and $0 \le k_1, k_2, \dots, k_s \le \lfloor \frac{\bar{g}}{g} \rfloor$. By applying the reasoning based on Proposition 9 done in the proof of Theorem 35, $\forall 0 < \alpha_1 < \dots < \alpha_s < 1$ we check that

$$ e^{i u H^M} F_{k_1, \alpha_1}^M \cdots F_{k_s, \alpha_s}^M \stackrel{d}{\Longrightarrow} e^{i u H} F_{1, k} \cdots F_{s, k} $$

where, recalling that $k$ denotes the multiindex $(k_1, \dots, k_s)$ and $K_i := k_1 + \dots + k_{i-1} + i - 1$ $\forall i = 1, \dots, s$ with the convention $K_1 := 0$, $F_{i,k}$ for $i = 1, \dots, s$ is given by

$$ F_{i,k} = \begin{cases} f^2(X_{K_i+1}) \beta_0 (U_i, g(X_{K_i+1})) & k_i = 0 \\ -f(X_{K_i+1}) f(X_{K_i+1}) \beta_1 \left( U_i, g(X_{K_i+1}), \sum_{\ell=K_i+2}^{K_i+1-1} g(X_\ell), g(X_{K_i+1}) \right) & k_i > 0 \end{cases} $$
and $H \sim \mathcal{N}(0, \sigma_1^2(f))$ independent of $(U_j)_{1\le j\le s}$ and $(X_j)_{1\le j\le K_{s+1}}$.
Moreover by the hypothesis of boundedness of $f$, (4.33) and the fact that $|e^{iuH^M}| \le 1$,

$$
\lim_{M \to \infty} \mathbb{E} \left( e^{iuH^M} F_{k_1, \alpha_1}^M \cdots F_{k_s, \alpha_s}^M \right) = \mathbb{E} \left( e^{iuH} F_{1,k} \cdots F_{s,k} \right).
$$

By Lebesgue's theorem and by observing that $F_{1,k}, \dots, F_{s,k}$ are independent, respectively dis-
tributed as $F_{k_1}, \dots, F_{k_s}$ and independent of $H$, $z(s, M)$ converges to

$$
\begin{align*}
\left(-\frac{u^2}{2}\right)^s \sum_{k_1, \dots, k_s = 0}^{\lfloor \frac{\bar{g}}{g} \rfloor} \int_{[0,1]^s} \mathbb{E} \left( e^{iuH} F_{1,k} \cdots F_{s,k} \right) d\alpha_1 \cdots d\alpha_s \\
&= e^{-\frac{u^2 \sigma_1^2(f)}{2}} \frac{1}{s!} \left(-\frac{u^2}{2}\right)^s \sum_{k_1, \dots, k_s = 0}^{\lfloor \frac{\bar{g}}{g} \rfloor} \mathbb{E}(F_{k_1}) \cdots \mathbb{E}(F_{k_s}) \\
&= e^{-\frac{u^2 \sigma_1^2(f)}{2}} \frac{1}{s!} \left(-\frac{u^2}{2}\right)^s \left( \sum_{k_1=0}^{\lfloor \frac{\bar{g}}{g} \rfloor} \mathbb{E}(F_{k_1}) \right)^s = e^{-\frac{u^2 \sigma_1^2(f)}{2}} \frac{1}{s!} \left(-\frac{u^2}{2}\right)^s (\sigma_2^2(f))^s.
\end{align*}
$$

Let us now show that $z$ is dominated by a summable function $\tilde{z}$. By using that $f$, $\beta_0$ and $\beta_1$
are bounded functions, for $s \ge 1$ one has

$$
|z(s, M)| = \left| \left(-\frac{u^2}{2}\right)^s \sum_{k_1, \dots, k_s = 0}^{\lfloor \frac{\bar{g}}{g} \rfloor} \int_{[0,1]^s} \mathbb{E} \left( e^{iuH} F_{1,k} \cdots F_{s,k} \right) d\alpha_1 \cdots d\alpha_s \right| \\
\le C \frac{1}{s!} \left( \frac{u^2 \left( \lfloor \frac{\bar{g}}{g} \rfloor + 1 \right)}{2} \right)^s
$$

for a given finite constant $C$. It is now sufficient to observe that $\sum_{s \ge 1} \frac{1}{s!} \left( \left( \lfloor \frac{\bar{g}}{g} \rfloor + 1 \right) \times \frac{u^2}{2} \right)^s = e^{\left( \lfloor \frac{\bar{g}}{g} \rfloor + 1 \right) \frac{u^2}{2}} < \infty$. In conclusion we have proved that (4.42) converges as $M \to \infty$ to

$$
e^{-\frac{u^2 \sigma_1^2(f)}{2}} \sum_{s \ge 0} \frac{1}{s!} \left(-\frac{u^2}{2}\right)^s (\sigma_2^2(f))^s = e^{-\frac{u^2 (\sigma_1^2(f) + \sigma_2^2(f))}{2}}.
$$

□

4.5 Proof of Proposition 8

We recall that for $i = 1, \dots, M$ we denote $u_i^M := \{w_1^M + \dots + w_i^M\}$ and $\mu_i^M := \lfloor w_1^M + \dots + w_i^M \rfloor + 1$,
where by convention $u_0^M = 0, \mu_0^M = 1$. The following technical result holds.

**Lemma 11.** If $1 \le p < \ell \le M$ then

1. $\mu_p^M < \mu_\ell^M$ if and only if $w_{p+1}^M + \dots + w_\ell^M \ge 1 - u_p^M$

2. $\mu_p^M = \mu_\ell^M$ if and only if $w_{p+1}^M + \dots + w_\ell^M < 1 - u_p^M$

3. $\mu_p^M = \mu_\ell^M$ if and only if $u_\ell^M = u_p^M + w_{p+1}^M + \dots + w_\ell^M$.
Proof of Lemma 11. Let us observe that 2. follows directly from 1. and 3. is a direct consequence of the definition of the integer and fractional part. Therefore it is enough to prove 1.
If $\mu_p^M < \mu_\ell^M$, then

$$w_1^M + \cdots + w_p^M - u_p^M + 1 = \mu_p^M \le \mu_\ell^M - 1 \le w_1^M + \cdots + w_\ell^M$$

that implies

$$1 - u_p^M \le w_{p+1}^M + \cdots + w_\ell^M.$$

Let us now prove the other implication. If $1 - u_p^M \le w_{p+1}^M + \cdots + w_\ell^M$, then

$$\mu_p^M = w_1^M + \cdots + w_p^M + 1 - u_p^M \le w_1^M + \cdots + w_\ell^M < \mu_\ell^M.$$

□

The following lemma provides an explicit expression for the conditional expectation $\mathbb{E}(f(Y_m^M) | \mathcal{F})$ for $m = 1, \cdots, M$ which appears in (4.20) and so it allows to prove Proposition 8.

**Lemma 12.** Given $\varrho \in \mathcal{B}_b(\mathbb{R}^d)$, for $m = 1, \cdots, M$ we have

$$\mathbb{E}(\varrho(Y_m^M) | \mathcal{F}) = \sum_{i=1}^{M} \varrho(X_i) q_{m,i}^M 1\{\mu_{i-1}^M \le m \le \mu_i^M\}$$

where for $i = 1, \cdots, M$

$$q_{m,i}^M = 1_{\{\mu_{i-1}^M < \mu_i^M\}} (1_{\{\mu_{i-1}^M < m < \mu_i^M\}} + 1_{\{m = \mu_{i-1}^M\}} (1 - u_{i-1}^M) + 1_{\{m = \mu_i^M\}} u_i^M) + 1_{\{\mu_{i-1}^M = m = \mu_i^M\}} w_i^M. \quad (4.43)$$

Proof of Lemma 12. For $m = 1, \cdots, M$, by (1.17), by observing that $m - U_m$ is uniformly distributed on $[m-1, m]$ and by the Freezing Lemma, we have

$$\mathbb{E}(\varrho(Y_m^M) | \mathcal{F}) = \sum_{i=1}^{M} \varrho(X_i) \int_{m-1}^{m} 1\left\{\sum_{j=1}^{i-1} w_j^M < u \le \sum_{j=1}^{i} w_j^M\right\} du. \quad (4.44)$$

By observing that for $i = 1, \cdots, M$

$$1 = 1_{\{m < \mu_{i-1}^M\}} + 1_{\{m > \mu_i^M\}} + 1_{\{\mu_{i-1}^M < \mu_i^M\}} (1_{\{m = \mu_{i-1}^M\}} + 1_{\{m = \mu_i^M\}} + 1_{\{\mu_{i-1}^M < m < \mu_i^M\}}) + 1_{\{m = \mu_{i-1}^M = \mu_i^M\}},$$

let us study the value of the integral in (4.44) according to this partition. If $m < \mu_{i-1}^M$ then $m \le \mu_{i-1}^M - 1 \le w_1^M + \cdots + w_{i-1}^M$ and if $m > \mu_i^M$ then $m \ge \mu_i^M + 1 > w_1^M + \cdots + w_i^M + 1$ and so in both cases the integral is zero. Let us now suppose that $\mu_{i-1}^M < \mu_i^M$. If $\mu_{i-1}^M < m < \mu_i^M$, then $m \ge \mu_{i-1}^M + 1 > w_1^M + \cdots + w_{i-1}^M + 1$ and $m \le w_1^M + \cdots + w_i^M$. Therefore

$$\int_{m-1}^{m} 1 \left\{ \sum_{j=1}^{i-1} w_j^M < u \le \sum_{j=1}^{i} w_j^M \right\} du = 1.$$

If $m = \mu_{i-1}^M$:

$$\int_{m-1}^{m} 1 \left\{ \sum_{j=1}^{i-1} w_j^M < u \le \sum_{j=1}^{i} w_j^M \right\} du = \int_{w_1^M + \cdots + w_{i-1}^M}^{\mu_{i-1}^M} du = 1 - u_{i-1}^M.$$
If $m = \mu_i^M$:

$$ \int_{m-1}^{m} 1 \left\{ \sum_{j=1}^{i-1} w_j^M < u \le \sum_{j=1}^{i} w_j^M \right\} du = \int_{\mu_i^M - 1}^{w_1^M + \dots + w_i^M} du = w_i^M. $$

Finally if $m = \mu_i^M = \mu_{i-1}^M$:

$$ \int_{m-1}^{m} 1 \left\{ \sum_{j=1}^{i-1} w_j^M < u \le \sum_{j=1}^{i} w_j^M \right\} du = \int_{w_1^M + \dots + w_{i-1}^M}^{w_1^M + \dots + w_i^M} du = w_i^M. $$

To sum up we have obtained that for $i = 1, \dots, M$

$$ \int_{m-1}^{m} 1 \left\{ \sum_{j=1}^{i-1} w_j^M < u \le \sum_{j=1}^{i} w_j^M \right\} du \\ = 1_{\{\mu_{i-1}^M < \mu_i^M\}} (1_{\{\mu_{i-1}^M < m < \mu_i^M\}} + 1_{\{m = \mu_{i-1}^M\}} (1 - u_{i-1}^M) + 1_{\{m = \mu_i^M\}} u_i^M) + 1_{\{\mu_{i-1}^M = m = \mu_i^M\}} w_i^M. $$

<figure><img src="image_9.png" /></figure>

*Proof of Proposition 8.* Let us rewrite (4.20). By Lemma 12 we have

$$ \begin{aligned} & \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} (f^2(Y_m^M) | \mathcal{F}) - \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} (f(Y_m^M) | \mathcal{F})^2 \\ &= \frac{1}{M} \sum_{m=1}^{M} \sum_{i=1}^{M} f^2(X_i) q_{m,i}^M 1_{\{\mu_{i-1}^M \le m \le \mu_i^M\}} - \frac{1}{M} \sum_{m=1}^{M} \sum_{i=1}^{M} f^2(X_i) (q_{m,i}^M)^2 1_{\{\mu_{i-1}^M \le m \le \mu_i^M\}} \\ &\quad - \frac{2}{M} \sum_{m=1}^{M} \sum_{1 \le i < j \le M} f(X_i) f(X_j) q_{m,i}^M q_{m,j}^M 1_{\{\mu_{i-1}^M \le m \le \mu_i^M\}} 1_{\{\mu_{j-1}^M \le m \le \mu_j^M\}} \\ &= \frac{1}{M} \sum_{m=1}^{M} \sum_{i=1}^{M} f^2(X_i) (q_{m,i}^M - (q_{m,i}^M)^2) 1_{\{\mu_{i-1}^M \le m \le \mu_i^M\}} - \frac{2}{M} \sum_{m=1}^{M} \sum_{1 \le i < j \le M} f(X_i) f(X_j) q_{m,i}^M q_{m,j}^M 1_{\{m = \mu_i^M = \mu_{j-1}^M\}}. \end{aligned} \quad (4.45) $$

We are now going first to rewrite the first component of the right-hand side and then the second one.

**First term:**

Let us observe that by Lemma 11 applied to the couple $(p, l)$ equal to $(i-1, i)$, we have $1_{\{\mu_i^M > \mu_{i-1}^M\}} = 1_{\{w_i^M \ge 1 - u_{i-1}^M\}}$ and $1_{\{\mu_i^M = \mu_{i-1}^M\}} = 1_{\{w_i^M < 1 - u_{i-1}^M\}}$ for $i = 1, \dots, M$. Therefore by Lemma 12 and by using that $\mu_i^M - \mu_{i-1}^M = u_{i-1}^M - u_i^M + w_i^M$, we have

$$ \begin{aligned} & \sum_{m=1}^{M} q_{m,i}^M 1_{\{\mu_{i-1}^M \le m \le \mu_i^M\}} \\ &= \sum_{m=1}^{M} (1_{\{w_i^M \ge 1 - u_{i-1}^M\}} (1_{\{\mu_{i-1}^M < m < \mu_i^M\}} + 1_{\{m = \mu_{i-1}^M\}} (1 - u_{i-1}^M) \\ &\quad + 1_{\{m = \mu_i^M\}} u_i^M) + 1_{\{w_i^M < 1 - u_{i-1}^M\}} 1_{\{\mu_{i-1}^M = m = \mu_i^M\}} w_i^M) \\ &= 1_{\{w_i^M \ge 1 - u_{i-1}^M\}} (\mu_i^M - \mu_{i-1}^M - u_{i-1}^M + u_i^M) + 1_{\{w_i^M < 1 - u_{i-1}^M\}} w_i^M \\ &= w_i^M 1_{\{w_i^M \ge 1 - u_{i-1}^M\}} + w_i^M 1_{\{w_i^M < 1 - u_{i-1}^M\}} = w_i^M. \end{aligned} $$
Let us observe that since, by (4.3), $\frac{1}{M} \sum_{m=1}^{M} \mathbb{E}(f^2(Y_m^M) | \mathcal{F}) = \frac{1}{M} \sum_{i=1}^{M} f^2(X_i)w_i^M$, we got the expected result. Similarly, by using that $\mu_i^M - \mu_{i-1}^M = u_{i-1}^M - u_i^M + w_i^M$, we have

$$ 
\begin{aligned}
& \sum_{m=1}^{M} (q_{m,i}^M)^2 1\{\mu_{i-1}^M \le m \le \mu_i^M\} \\
&= 1\{w_i^M \ge 1 - u_{i-1}^M\} (\mu_i^M - \mu_{i-1}^M - 1 + (1 - u_{i-1}^M)^2 + (u_i^M)^2) + 1\{w_i^M < 1 - u_{i-1}^M\} (w_i^M)^2 \\
&= (w_i^M - u_i^M(1 - u_i^M) - u_{i-1}^M(1 - u_{i-1}^M))1\{w_i^M \ge 1 - u_{i-1}^M\} + (w_i^M)^2 1\{w_i^M < 1 - u_{i-1}^M\} \\
&= w_i^M - u_i^M(1 - u_i^M) - u_{i-1}^M(1 - u_{i-1}^M) - (w_i^M(1 - w_i^M) - u_i^M(1 - u_i^M) - u_{i-1}^M(1 - u_{i-1}^M))1\{w_i^M < 1 - u_{i-1}^M\}.
\end{aligned}
$$

Substrating the two quantities and using Lemma 11 to rewrite $u_i^M$ as $w_i^M + u_{i-1}^M$ if $w_i^M < 1 - u_{i-1}^M$, we have

$$ 
\begin{aligned}
& \sum_{m=1}^{M} q_{m,i}^M 1\{\mu_{i-1}^M \le m \le \mu_i^M\} - \sum_{m=1}^{M} (q_{m,i}^M)^2 1\{\mu_{i-1}^M \le m \le \mu_i^M\} \\
&= u_i^M(1 - u_i^M) + u_{i-1}^M(1 - u_{i-1}^M) + (w_i^M(1 - w_i^M) - u_i^M(1 - u_i^M) - u_{i-1}^M(1 - u_{i-1}^M))1\{w_i^M < 1 - u_{i-1}^M\} \\
&= u_i^M(1 - u_i^M) + u_{i-1}^M(1 - u_{i-1}^M) + (w_i^M(1 - w_i^M) - (w_i^M + u_{i-1}^M)(1 - w_i^M - u_{i-1}^M) - u_{i-1}^M(1 - u_{i-1}^M))1\{w_i^M < 1 - u_{i-1}^M\} \\
&= u_i^M(1 - u_i^M) + u_{i-1}^M(1 - u_{i-1}^M) - (2u_{i-1}^M(1 - u_{i-1}^M) - 2w_i^M u_{i-1}^M)1\{w_i^M < 1 - u_{i-1}^M\}.
\end{aligned}
$$

Therefore by observing that for $i = 1, \dots, M$, $u_i^M = \{u_{i-1}^M + w_i^M\}$, the first term can be rewritten as

$$ \frac{1}{M} \sum_{m=1}^{M} \sum_{i=1}^{M} f^2(X_i) (q_{m,i}^M - (q_{m,i}^M)^2) 1\{\mu_{i-1}^M \le m \le \mu_i^M\} = \frac{1}{M} \sum_{i=1}^{M} f^2(X_i) \beta_0(u_{i-1}^M, w_i^M) \quad (4.46) $$

with $\beta_0$ defined in (4.7).

## Second term:

By Lemma 12 and by rewriting $1\{\mu_{i-1}^M < \mu_i^M\} = 1 - 1\{\omega_i^M < 1 - u_{i-1}^M\}$, we can rewrite the right-hand side of (4.45) in the following way

$$ 
\begin{aligned}
& \frac{2}{M} \sum_{m=1}^{M} \sum_{1 \le i < j \le M} f(X_i) f(X_j) q_{m,i}^M q_{m,j}^M 1\{m = \mu_i^M = \mu_{j-1}^M\} \\
&= \frac{2}{M} \sum_{1 \le i < j \le M} f(X_i) f(X_j) \left(1\{\mu_{i-1}^M < \mu_i^M\} u_i^M + 1\{\mu_{i-1}^M = \mu_i^M\} w_i^M\right) \\
& \qquad \times \left(1\{\mu_{j-1}^M < \mu_j^M\} (1 - u_{j-1}^M) + 1\{\mu_{j-1}^M = \mu_j^M\} w_j^M\right) 1\{\mu_i^M = \mu_{j-1}^M\} \\
&= \frac{2}{M} \sum_{k=1}^{M-1} \sum_{i=1}^{M-k} f(X_i) f(X_{i+k}) (\rho_{i,k}^M + \tilde{\rho}_{i,k}^M)
\end{aligned}
$$

where for $k = 1, \dots, M-1$ and $i = 1, \dots, M-k$

$$ \rho_{i,k}^M = u_i^M \left(1\{\mu_{i+k-1}^M < \mu_{i+k}^M\} (1 - u_{i+k-1}^M) + 1\{\mu_{i+k-1}^M = \mu_{i+k}^M\} w_{i+k}^M\right) 1\{\mu_i^M = \mu_{i+k-1}^M\} $$

$$ \tilde{\rho}_{i,k}^M = (w_i^M - u_i^M) \left(1\{\mu_{i+k-1}^M < \mu_{i+k}^M\} (1 - u_{i+k-1}^M) + 1\{\mu_{i+k-1}^M = \mu_{i+k}^M\} w_{i+k}^M\right) 1\{\mu_{i-1}^M = \mu_i^M\} 1\{\mu_i^M = \mu_{i+k-1}^M\}. $$
Let us now rewrite $\rho_{i,k}^M$ and $\tilde{\rho}_{i,k}^M$. To simplify the notation, we will denote $\sum_{\ell=j_1}^{j_2} w_\ell^M$ for $1 \le j_1 \le j_2 \le M$ by $s_{j_1}^{j_2}$. We can apply Lemma 11 to the couples

1. $(p, \ell) = (i, i + k - 1)$ so that $1_{\{\mu_i^M = \mu_{i+k-1}^M\}} = 1_{\{s_{i+1}^{i+k-1} < 1 - u_i^M\}}$

2. $(p, \ell) = (i - 1, i)$ so that $1_{\{\mu_{i-1}^M = \mu_i^M\}} = 1_{\{\omega_i^M < 1 - u_{i-1}^M\}}$

3. $(p, \ell) = (i + k - 1, i + k)$ so that $1_{\{\mu_{i+k-1}^M = \mu_{i+k}^M\}} = 1_{\{\omega_{i+k}^M < 1 - u_{i+k-1}^M\}}$ and $1_{\{\mu_{i+k-1}^M < \mu_{i+k}^M\}} = 1_{\{\omega_{i+k}^M \ge 1 - u_{i+k-1}^M\}}$

so that, by observing that when $s_{i+1}^{i+k-1} < 1 - u_i^M$ we can rewrite $u_{i+k-1}^M$ as $u_i^M + s_{i+1}^{i+k-1}$, one has

$$
\begin{align*}
\rho_{i,k}^{M} &= u_{i}^{M} \left( 1_{\{\omega_{i+k}^{M} \ge 1 - u_{i+k-1}^{M}\}} (1 - u_{i+k-1}^{M}) + w_{i+k}^{M} 1_{\{\omega_{i+k}^{M} < 1 - u_{i+k-1}^{M}\}} \right) 1_{\{s_{i+1}^{i+k-1} < 1 - u_{i}^{M}\}} \\
&= u_{i}^{M} \left( 1_{\{\omega_{i+k}^{M} \ge 1 - u_{i}^{M} - s_{i+1}^{i+k-1}\}} (1 - u_{i}^{M} - s_{i+1}^{i+k-1}) + w_{i+k}^{M} 1_{\{\omega_{i+k}^{M} < 1 - u_{i}^{M} - s_{i+1}^{i+k-1}\}} \right) 1_{\{s_{i+1}^{i+k-1} < 1 - u_{i}^{M}\}} \\
&= u_{i}^{M} (1 - u_{i}^{M} - s_{i+1}^{i+k-1}) 1_{\{s_{i+1}^{i+k-1} < 1 - u_{i}^{M} \le s_{i+1}^{i+k}\}} + u_{i}^{M} w_{i+k}^{M} 1_{\{s_{i+1}^{i+k} < 1 - u_{i}^{M}\}} \\
&= u_{i}^{M} (1 - u_{i}^{M} - s_{i+1}^{i+k-1}) 1_{\{s_{i+1}^{i+k-1} < 1 - u_{i}^{M}\}} - u_{i}^{M} (1 - u_{i}^{M} - s_{i+1}^{i+k}) 1_{\{s_{i+1}^{i+k} < 1 - u_{i}^{M}\}}
\end{align*}
$$

and, by observing that when $w_i^M < 1 - u_{i-1}^M$ and $s_{i+1}^{i+k-1} < 1 - u_i^M$ we can rewrite $u_i^M = u_{i-1}^M + w_i^M$ and $u_{i+k-1}^M = u_{i-1}^M + s_i^{i+k-1}$, one has

$$
\begin{align*}
\tilde{\rho}_{i,k}^M &= (w_i^M - u_i^M) \left( 1_{\{\omega_{i+k}^M \ge 1 - u_{i+k-1}^M\}} (1 - u_{i+k-1}^M) + w_{i+k}^M 1_{\{\omega_{i+k}^M < 1 - u_{i+k-1}^M\}} \right) 1_{\{\omega_i^M < 1 - u_{i-1}^M\}} 1_{\{s_{i+1}^{i+k-1} < 1 - u_i^M\}} \\
&= -(u_{i-1}^M + w_i^M) \left( 1_{\{s_i^{i+k} \ge 1 - u_{i-1}^M\}} (1 - u_{i-1}^M - s_i^{i+k-1}) + w_{i+k}^M 1_{\{s_i^{i+k} < 1 - u_{i-1}^M\}} \right) 1_{\{s_i^{i+k-1} < 1 - u_{i-1}^M\}} \\
&\quad + w_i^M \left( 1_{\{s_i^{i+k} \ge 1 - u_{i-1}^M\}} (1 - u_{i-1}^M - s_i^{i+k-1}) + w_{i+k}^M 1_{\{s_i^{i+k} < 1 - u_{i-1}^M\}} \right) 1_{\{s_i^{i+k-1} < 1 - u_{i-1}^M\}} \\
&= -u_{i-1}^M (1 - u_{i-1}^M - s_i^{i+k-1}) 1_{\{s_i^{i+k-1} < 1 - u_{i-1}^M \le s_i^{i+k}\}} - u_{i-1}^M w_{i+k}^M 1_{\{s_i^{i+k} < 1 - u_{i-1}^M\}} \\
&= -u_{i-1}^M (1 - u_{i-1}^M - s_i^{i+k-1}) 1_{\{s_i^{i+k-1} < 1 - u_{i-1}^M\}} + u_{i-1}^M (1 - u_{i-1}^M - s_i^{i+k}) 1_{\{s_i^{i+k} < 1 - u_{i-1}^M\}}.
\end{align*}
$$

Therefore we have obtained that

$$
\begin{align*}
& \frac{1}{M} \sum_{m=1}^{M} \sum_{1 \le i < j \le M} f(X_i) f(X_j) q_{m,i}^{M} q_{m,j}^{M} 1_{\{m = \mu_i^M = \mu_{j-1}^M\}} \\
& = \frac{1}{M} \sum_{k=1}^{M-1} \sum_{i=1}^{M-k} f(X_i) f(X_{i+k}) \beta_1 \left( u_{i-1}^{M}, w_i^M, \sum_{\ell=2}^{k} w_{i+\ell-1}^{M}, w_{i+k}^{M} \right)
\end{align*}
$$

with $\beta_1$ defined as in (4.8). Let us now observe that there exists a finite constant $C_1$ such that for $k = 1, \dots, M-1$, $i = 1, \dots, M-k$

$$
\left| \beta_1 \left( u_{i-1}^M, w_i^M, \sum_{\ell=2}^k w_{i+\ell-1}^M, w_{i+k}^M \right) \right| \le C_{11} \left\{ \sum_{\ell=2}^k w_{i+\ell-1}^M \right\} \le C_{11} \left\{ k < 1 + \frac{\bar{g}}{g} \right\}. \quad (4.47)
$$
Therefore

$$ 
\begin{aligned}
& \frac{1}{M} \sum_{k=1}^{M-1} \sum_{i=1}^{M-k} f(X_i) f(X_{i+k}) \beta_1 \left( u_{i-1}^M, w_i^M, \sum_{\ell=2}^k w_{i+\ell-1}^M, w_{i+k}^M \right) \\
& = \frac{1}{M} \sum_{k=1}^{(M-1)\wedge\lceil\frac{\bar{g}}{g}\rceil} \sum_{i=1}^{M-k} f(X_i) f(X_{i+k}) \beta_1 \left( u_{i-1}^M, w_i^M, \sum_{\ell=2}^k w_{i+\ell-1}^M, w_{i+k}^M \right).
\end{aligned}
$$

□

## 4.6 Asymptotic Variance for the Next Steps

What we have seen so far is the study of the stratified sampling selection step: it is actually part of a more general algorithm where each step consists of a selection part and a mutation part that we are now going to describe more in details (see for instance [47] for a more general version of the algorithm).

For each step $n \ge 0$, we are going to recursively define the selection sequence $(Y_n^{M,m})_{1\le m \le M}$ and the mutation sequence $(X_n^{M,m})_{1\le m \le M}$. Let $g_n : \mathbb{R}^d \to (0, \infty), n \ge 0$ be a family of measurable functions such that $0 < \inf_{x \in \mathbb{R}^d} g_n(x) \le \sup_{x \in \mathbb{R}^d} g_n(x) < \infty$. In what follows we denote $\bar{g}_n := \sup_{x \in \mathbb{R}^d} g_n(x)$ and $g_n := \inf_{x \in \mathbb{R}^d} g_n(x)$. Let for $n \ge 0$ and $k \ge 0$

$$ \phi_n(k) := \left\lceil \frac{\bar{g}_n}{g_n} (1+k) \right\rceil. \qquad (4.48) $$

Moreover let $(Z_k)_{k \ge 0}$ be a $\mathbb{R}^d$-valued time-inhomogeneous Markov chain that is for each bounded measurable function $h : \mathbb{R}^d \to \mathbb{R}, \mathbb{E}(h(Z_k) | \sigma(Z_0, \dots, Z_{k-1})) = \int_{\mathbb{R}^d} h(x) P_k(Z_{k-1}, dx)$ for a transition kernel $P_k$ possibly depending on $k$. We denote the law of $Z_0$ by $\eta$.

**Initialization:** we generate a sequence $(X_0^{M,m})_{1\le m \le M}$ of i.i.d. $\mathbb{R}^d$-valued random vectors following the law $\eta$. By convention $Y_0^{M,m} := X_0^{M,m}$ for $1 \le m \le M$.

**n → n+1:** the transition from $(Y_n^{M,m}, X_n^{M,m})_{1\le m \le M}$ to $(Y_{n+1}^{M,m}, X_{n+1}^{M,m})_{1\le m \le M}$ for $n \ge 0$ consists of two steps.

**Selection:** we generate the random vectors $(Y_{n+1}^{M,m})_{1\le m \le M}$ conditionally independent given $\mathcal{F}^n := \sigma((X_i^{M,m}, Y_i^{M,m})_{1\le m \le M, 0 \le i \le n})$ by

$$ Y_{n+1}^{M,m} = \sum_{\ell=1}^{M} \left\{ \frac{\sum_{j=1}^{\ell-1} g_n(x_n^{M,j})}{\sum_{j=1}^{M} g_n(x_n^{M,j})} \right\}^{\ell} X_n^{M,\ell} \quad m = 1, \dots, M \quad (4.49) $$

where $(U_n^m)_{1\le m \le M}$ is a sequence of i.i.d. random variables independent of $\mathcal{F}^n$ and distributed according to the uniform law on $(0, 1)$.
Mutation: given $\mathcal{G}^{n+1} := \sigma\left(\left(X_i^{M,m}, Y_i^{M,m}\right)_{1\le m\le M, 0\le i\le n}, \left(Y_{n+1}^{M,m}\right)_{1\le m\le M}\right)$ we generate the ran-
dom vectors $X_{n+1}^{M,m}, 1 \le m \le M$ conditionally independent and respectively distributed
according to the probability measure $P_{n+1}(Y_{n+1}^{M,m}, \cdot)$

For each $n \ge 0$ we denote the weights by

$$
w_n^{M,m} = \frac{M g_n(X_n^{M,m})}{\sum_{\ell=1}^{M} g_n(X_n^{M,\ell})} \quad m = 1, \dots, M.
$$

Moreover let us observe that the selection property for $n = 0$ (4.3) remains valid for each $n$:
$\forall n \ge 0$

$$
\mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \delta_{Y_{n+1}^{M,m}} |\mathcal{F}^n\rangle \right) = \sum_{m=1}^{M} \frac{w_n^{M,m}}{M} \delta_{X_n^{M,m}}. \quad (4.50)
$$

The study of the asymptotic variance of the selection part of the first step that is $\mathrm{Var}\left(\frac{1}{\sqrt{M}}\sum_{m=1}^{M} f\left(Y_{1}^{M,m}\right)\right)$ for a given bounded measurable function $f: \mathbb{R}^d \to \mathbb{R}$ was the object of Section 4.3. Our purpose now is to generalize this result by studying the asymptotic variance of $\frac{1}{\sqrt{M}}\sum_{m=1}^{M} f\left(Y_{n+1}^{M,m}\right)$ for $n \ge 1$. Before doing it let us begin with some notation: for $n \ge 0$, $M \ge 1$ and $k = 0, \dots, M-1$ let

$$
\tilde{\eta}_n^{k,M} := \frac{1}{M} \sum_{i=1}^{M-k} \delta(Y_n^{M,i}, \dots, Y_n^{M,i+k})
$$

$$
\bar{\eta}_n^{k,M} := \frac{1}{M} \sum_{i=1}^{M-k} \delta(X_n^{M,i}, \dots, X_n^{M,i+k}).
$$

**Remark 37.** Let $f \in B_b(\mathbb{R}^d)$. If $k=0$ and $n=0$, by the classical Strong Law of Large Numbers we have

$$
\lim_{M \to \infty} \tilde{\eta}_0^{0,M}(f) = \lim_{M \to \infty} \bar{\eta}_0^{0,M}(f) = \bar{\eta}_0^0(f) := \eta(f) = \mathbb{E}(f(Z_0)) \quad \text{a.s.}
$$

It is possible to prove that a Strong Law of Large Numbers holds (see [47, Corollary 7.4.2]) also
in the case $k = 0$ and $n \ge 1$:

$$
\lim_{M \to \infty} \tilde{\eta}_n^{0,M}(f) = \lim_{M \to \infty} \frac{1}{M} \sum_{i=1}^{M} f(Y_n^{M,i}) = \tilde{\eta}_n^0(f) := \frac{\mathbb{E}\left(f(Z_{n-1}) \prod_{p=0}^{n-1} g_p(Z_p)\right)}{\mathbb{E}\left(\prod_{p=0}^{n-1} g_p(Z_p)\right)} \quad \text{a.s.}
$$

$$
\lim_{M \to \infty} \bar{\eta}_n^{0,M}(f) = \lim_{M \to \infty} \frac{1}{M} \sum_{i=1}^{M} f(X_n^{M,i}) = \bar{\eta}_n^0(f) := \frac{\mathbb{E}\left(f(Z_n) \prod_{p=0}^{n-1} g_p(Z_p)\right)}{\mathbb{E}\left(\prod_{p=0}^{n-1} g_p(Z_p)\right)} \quad \text{a.s.}
$$

We are now going to prove that given $h: \mathbb{R}^{2d} \to \mathbb{R}$ a bounded measurable function, the study of the asymptotic variance of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(Y_{n+1}^{M,m}, X_{n+1}^{M,m})$ depends on the study of the asymptotic variance of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_{n+1}^{M,m})$ for $f: \mathbb{R}^d \to \mathbb{R}$ a bounded measurable function.
**Proposition 10.** Let $h \in \mathcal{B}_b(\mathbb{R}^{2d})$. The following convergence holds

$$ \left| \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \right) - \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} P_{n+1} h(Y_{n+1}^{M,m}) \right) - \frac{\bar{\eta}_n^0 \left( g_n \left( P_{n+1} h^2 - (P_{n+1} h)^2 \right) \right)}{\bar{\eta}_n^0(g_n)} \right| \underset{M \to \infty}{\longrightarrow} 0. $$

*Proof of Proposition 10.* By using that the $X_{n+1}^{M,m}$ are conditionally independent given $\mathcal{G}^{n+1}$ and that the $Y_{n+1}^{M,m}$ are $\mathcal{G}^{n+1}$ measurable, one has

$$ \begin{aligned} & \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \right) \\ &= \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} \mathbb{E} \left( h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \mid \mathcal{G}^{n+1} \right) \right) + \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \mathrm{Var} \left( h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \mid \mathcal{G}^{n+1} \right) \right) \\ &= \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} \mathbb{E} \left( h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \mid \mathcal{G}^{n+1} \right) \right) + \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} \left( h^2(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \mid \mathcal{G}^{n+1} \right) \right) \\ &\quad - \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \mathbb{E} \left( h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \mid \mathcal{G}^{n+1} \right)^2 \right). \end{aligned} $$

Since $\mathcal{L}(X_{n+1}^{M,m} | \mathcal{G}^{n+1}) \sim P_{n+1}(Y_{n+1}^{M,m}, \cdot)$, for each bounded measurable function $\tilde{h}: \mathbb{R}^{2d} \to \mathbb{R}$ one has

$$ \mathbb{E} \left( \tilde{h}(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \mid \mathcal{G}^{n+1} \right) = P_{n+1} \tilde{h} \left( Y_{n+1}^{M,m} \right) := \int_{\mathbb{R}^d} \tilde{h}(Y_{n+1}^{M,m}, x) P_{n+1} \left( Y_{n+1}^{M,m}, dx \right). \quad (4.51) $$

Therefore we obtain that

$$ \begin{aligned} & \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} h(Y_{n+1}^{M,m}, X_{n+1}^{M,m}) \right) \\ &= \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} P_{n+1} h(Y_{n+1}^{M,m}) \right) + \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} P_{n+1} h^2(Y_{n+1}^{M,m}) \right) - \mathbb{E} \left( \frac{1}{M} \sum_{m=1}^{M} \left( P_{n+1} h(Y_{n+1}^{M,m}) \right)^2 \right) \\ &= \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} P_{n+1} h(Y_{n+1}^{M,m}) \right) + \mathbb{E} \left( \sum_{m=1}^{M} \frac{w_n^{M,m}}{M} P_{n+1} h^2(X_n^{M,m}) \right) - \mathbb{E} \left( \sum_{m=1}^{M} \frac{w_n^{M,m}}{M} \left( \mathrm{Ph}(X_n^{M,m}) \right)^2 \right) \end{aligned} $$

where to obtain the last equality we use the selection property (4.50). By Remark 37,

$$ \lim_{M \to \infty} \frac{1}{M} \sum_{m=1}^{M} g_n(X_n^{M,m}) P_{n+1} h^2(X_n^{M,m}) = \bar{\eta}_n^0(g_n P_{n+1} h^2) \quad \mathrm{a.s.,} $$

$$ \lim_{M \to \infty} \frac{1}{M} \sum_{m=1}^{M} g_n(X_n^{M,m}) (P_{n+1}h(X_n^{M,m}))^2 = \bar{\eta}_n^0(g_n(P_{n+1}h)^2) \quad \mathrm{a.s.,} $$

$$ \lim_{M \to \infty} \frac{1}{M} \sum_{m=1}^{M} g_n(X_n^{M,m}) = \bar{\eta}_n^0(g_n) \quad \mathrm{a.s.} $$
Thus by Lebesgue theorem,

$$
\lim_{M \to \infty} \mathbb{E} \left( \sum_{m=1}^{M} \frac{w_n^{M,m}}{M} P_{n+1} h^2 (X_n^{M,m}) \right) = \frac{\bar{\eta}_n^0(g_n P_{n+1} h^2)}{\bar{\eta}_n^0(g_n)},
$$
$$
\lim_{M \to \infty} \mathbb{E} \left( \sum_{m=1}^{M} \frac{w_n^{M,m}}{M} (P_{n+1} h (X_n^{M,m}))^2 \right) = \frac{\bar{\eta}_n^0(g_n (P_{n+1} h)^2)}{\bar{\eta}_n^0(g_n)}
$$

and this concludes the proof.

<figure><img src="image_10.png" /></figure>

We are now ready to study for $n \ge 1$ and $f \in \mathcal{B}_b(\mathbb{R}^d)$ the convergence as $M \to \infty$ of

$$
V_{n+1}^{M}(f) := \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_{n+1}^{M,m}) \right).
$$

With the aim of studying $\lim_{M \to \infty} V_{n+1}^M (f)$, we will assume the following conjectures:

**Conjecture 1.** For each $f \in \mathcal{B}_b(\mathbb{R}^d)$ one has

$$
\left| \mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_n^{0,M}(g_n f)}{\bar{\eta}_n^{0,M}(g_n)} \right) - \frac{V_n^M(P_n f_n)}{(\bar{\eta}_n^0(g_n))^4} - \frac{\bar{\eta}_{n-1}^0 \left( g_{n-1} \left( P_n f_n^2 - (P_n f_n)^2 \right) \right)}{(\bar{\eta}_n^0(g_n))^4 \bar{\eta}_{n-1}^0(g_{n-1})} \right| \underset{M \to \infty}{\longrightarrow} 0
$$

where

$$
f_n := g_n (\bar{\eta}_n^0(g_n) f - \bar{\eta}_n^0(g_n f)).
$$

**Remark 38.** In Section 4.3 we proved that in the case $n = 0$, by observing that $\mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_0^{0,M}(g_0 f)}{\bar{\eta}_0^{0,M}(g_0)} \right) =$

$$
\mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_0^{0,M} \left( \frac{g_0}{\eta(g_0)} f \right)}{\bar{\eta}_0^{0,M} \left( \frac{g_0}{\eta(g_0)} \right)} \right), \text{ one has}
$$

$$
\left| \mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_0^{0,M}(g_0 f)}{\bar{\eta}_0^{0,M}(g_0)} \right) - \frac{\eta(f_0^2)}{\eta(g_0)^4} \right| \underset{M \to \infty}{\longrightarrow} 0.
$$

For $n \ge 1$ the idea underlying the Conjecture 1 is the following: by considering

$$
\mathrm{Var} \left( \sqrt{M} \left( \frac{\bar{\eta}_n^{0,M}(g_n f)}{\bar{\eta}_n^{0,M}(g_n)} - \frac{\bar{\eta}_n^0(g_n f)}{\bar{\eta}_n^0(g_n)} \right) \right) = \mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_n^{0,M}(f_n)}{\bar{\eta}_n^{0,M}(g_n) \bar{\eta}_n^0(g_n)} \right),
$$

we asymptotically replace in the denominator of the right-hand side $\bar{\eta}_n^{0,M}(g_n)$ by $\bar{\eta}_n^0(g_n)$. Thus we have

$$
\left| \mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_n^{0,M}(f_n)}{\bar{\eta}_n^{0,M}(g_n) \bar{\eta}_n^0(g_n)} \right) - \frac{1}{(\bar{\eta}_n^0(g_n))^4} \mathrm{Var} \left( \sqrt{M} \bar{\eta}_n^{0,M}(f_n) \right) \right| \underset{M \to \infty}{\longrightarrow} 0.
$$

We then rewrite $\mathrm{Var}(\sqrt{M}\bar{\eta}_n^{0,M}(f_n))$ following the same reasoning used in the proof of Proposition 10.
**Conjecture 2.** Let $t \in \mathbb{N}$. Given $\psi \in \mathcal{B}_b(\mathbb{R}^{t+2})$ continuous and $h \in \mathcal{B}_b(\mathbb{R}^{d(t+1)})$, the following convergence holds

$$ \left| \frac{1}{M} \sum_{m=1}^{M-t} \mathbb{E} \left( h(X_n^{M,m}, \dots, X_n^{M,m+t}) \psi(u_n^{M,m-1}, w_n^{M,m}, \dots, w_n^{M,m+t}) \right) - \frac{1}{M} \sum_{m=1}^{M-t} \mathbb{E} \left( h(X_n^{M,m}, \dots, X_n^{M,m+t}) \int_0^1 \psi(u, \tilde{g}_n(X_n^{M,m}), \dots, \tilde{g}_n(X_n^{M,m+t})) du \right) \right| \underset{M \to \infty}{\longrightarrow} 0 \\ \text{where } u_n^{M,m-1} = \left\{ \frac{\sum_{i=1}^{m-1} g_n(X_n^{M,i})}{\sum_{i=1}^{M} g_n(X_n^{M,i})} \right\} \text{ and } \tilde{g}_n(x) = \frac{g_n(x)}{\bar{\eta}_n^0(g_n)} $$

**Remark 39.** In Section 4.3 we have provided a formal proof of Conjecture 2 in the case $n=0$. In that case we strongly used the fact that the random variables we are working with $\left(\left(X_0^{M,m}\right)_{m \ge 1}\right)$ are i.i.d. and the fact that the law of $g_0(X_0^{M,m})$ has an absolutely continuous component to prove that is possible to asymptotically replace $u_0^{M,i-1}$ with a uniformly distributed random variable independent of $\mathcal{F}^0$ and $w_0^{M,i}$ with $\frac{g_0(X_0^{M,i})}{\bar{\eta}_0^0(g_0)}$.

Before providing the main result of this section, let us introduce some notation. For $k \ge 0$ let $P_n^{\otimes(k+1)} : \mathcal{B}_b(\mathbb{R}^{d(k+1)}) \to \mathcal{B}_b(\mathbb{R}^{d(k+1)})$ be defined by

$$ P_n^{\otimes(k+1)} h(x_0, \dots, x_k) := \int_{\mathbb{R}^{d(k+1)}} h(y_0, \dots, y_k) P_n(x_0, dy_0) \cdots P_n(x_k, dy_k) $$

and let for $s_k = 0, \dots, \phi_{n-1}(k)$, $T_{n-1}^{k \to s_k} : \mathcal{B}_b(\mathbb{R}^{d(k+1)}) \to \mathcal{B}_b(\mathbb{R}^{d(s_k+1)})$ given by

$$ T_{n-1}^{k \to s_k} f(x_0, x_1, \dots, x_{s_k}) = \sum_{0 \le s_1 \le s_2 \le \dots \le s_k} f(x_0, x_{s_1}, \dots, x_{s_k}) \int_0^1 \psi_{s_1:s_k}(u, \tilde{g}_{n-1}(x_0), \tilde{g}_{n-1}(x_1), \dots, \tilde{g}_{n-1}(x_{s_k})) du $$

and where given $0 \le s_k \le \phi_{n-1}(k)$ and $s_0 := 0 \le s_1 \le s_2 \le \dots \le s_k$:

$$ \psi_{s_1:s_k}(u, y_0, y_1, \dots, y_{s_k}) = \sum_{i=1}^{\lfloor 1+\bar{g}_n/\underline{g}_n \rfloor} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} \mathbf{1}\left\{u+\sum_{j=0}^{s_q-1} y_j < u' \le u+\sum_{j=0}^{s_q} y_j\right\} du' $$

Finally let $\mathcal{T}_{n-1}^k : \mathcal{B}_b(\mathbb{R}^{d(k+1)}) \to \mathcal{B}_b(\mathbb{R}^{d(\phi_{n-1}(k)+1)})$ defined by

$$ \mathcal{T}_{n-1}^k h(x_0, \dots, x_{\phi_{n-1}(k)}) = \sum_{s_k=0}^{\phi_{n-1}(k)} T_{n-1}^{k \to s_k} (P_n^{\otimes(k+1)} h)(x_0, x_1, \dots, x_{s_k}). $$

**Theorem 40.** Let us assume Conjecture 1 and Conjecture 2. Then $\forall n \ge 1$ and $\forall f \in \mathcal{B}(\mathbb{R}^d)$, $V_n^M(f)$ converges as $M$ goes to infinity and we denote its limit by $V_n(f)$. Moreover, by defining $\phi_{m:n}(0) := \phi_m(\phi_{m+1}(\dots(\phi_n(0)))))$ for $m \le n$, $m, n \in \mathbb{N}$, $V_n(f)$ is defined by the following recursive formula:

$$ V_{n+1}(f) = \frac{V_n(P_n f_n)}{(\bar{\eta}_n^0(g_n))^4} + \frac{\bar{\eta}_{n-1}^0 \left( g_{n-1} \left( P_n f_n^2 - (P_n f_n)^2 \right) \right)}{(\bar{\eta}_n^0(g_n))^4 \bar{\eta}_{n-1}^0(g_{n-1})} \\ + \int \mathcal{T}_0^{\phi_{1:n}(0)} \cdots \mathcal{T}_{n-2}^{\phi_{n-1:n}(0)} \mathcal{T}_{n-1}^{\phi_{n:n}(0)} \bar{f}_n(x) \eta^{\otimes \phi_{0:n}(0)}(dx) $$
where $\bar{f}_n \in \mathcal{B}_b(\mathbb{R}^{d(1+\phi_n(0))})$ is given by

$$ \bar{f}_n(x_0, \dots, x_{\phi_n(0)}) := \sum_{k=0}^{\phi_n(0)} f(x_0) f(x_k) \int_0^1 \bar{\beta}_k(u, \tilde{g}_n(x_0), \dots, \tilde{g}_n(x_k)) du $$

with $\bar{\beta}_k(u, y_0, \dots, y_k) = \beta_0(u, y_0)1_{\{k=0\}} - \beta_1 \left(u, y_0, \sum_{\ell=1}^{k-1} y_\ell, y_k\right) 1_{\{k \neq 0\}}$ where $\beta_0$ and $\beta_1$ are respectively defined in (4.7) and (4.8).

**Remark 41.** Let us observe that by Theorem 34, $\lim_{M \to \infty} V_1^M(h) = V_1(h) = \sigma_1^2(h) + \sigma_2^2(h) \forall h \in \mathcal{B}_b(\mathbb{R}^d)$.

The proof of Theorem 40 relies on the following proposition the proof of which is provided after the proof of the theorem.

**Proposition 11.** Let us assume Conjecture 2. Given $n \ge 1$ and $h \in \mathcal{B}_b(\mathbb{R}^{d(\phi_n(0)+1)})$ one has

$$ \left| \mathbb{E} \left( \bar{\eta}_n^{\phi_n(0),M}(h) \right) - \int \mathcal{T}_0^{\phi_1:n(0)} \cdots \mathcal{T}_{n-2}^{\phi_{n-1:n(0)}} \mathcal{T}_{n-1}^{\phi_{n:n(0)}} h(x) \eta^{\otimes \phi_{0:n(0)}}(dx) \right| \underset{M \to \infty}{\longrightarrow} 0. $$

*Proof of Theorem 40.* Similarly to what we have done in (4.5) and (4.6), we can rewrite the variance in the following way

$$ \begin{aligned} V_{n+1}^M(f) &= \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^M f(Y_{n+1}^{M,m}) \right) \\ &= \mathrm{Var} \left( \mathbb{E} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^M f(Y_{n+1}^{M,m}) \mid \mathcal{F}^n \right) \right) + \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^M f(Y_{n+1}^{M,m}) \mid \mathcal{F}^n \right) \right) \\ &= \mathrm{Var} \left( \sqrt{M} \frac{\sum_{m=1}^M g_n(X_n^{M,m}) f(X_n^{M,m})}{\sum_{m=1}^M g_n(X_n^{M,m})} \right) + \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^M f(Y_{n+1}^{M,m}) \mid \mathcal{F}^n \right) \right) \end{aligned} \quad (4.52) $$

where to obtain the last equality we use the selection property (4.50). Conjecture 1 gives the asymptotic behavior of the first term of (4.52).

Let us then study the second term of (4.52). Using the definition of $Y_{n+1}^{M,m}$ and following the same reasoning done to prove Proposition 8, we can rewrite the expression inside the expectation: for $M \ge 1 + \phi_n(0)$

$$ \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_{n+1}^{M,m}) \mid \mathcal{F}^n \right) \quad (4.53) $$

$$ = \frac{1}{M} \sum_{k=0}^{\phi_n(0)} \sum_{i=1}^{M-k} f(X_n^{M,i}) f(X_n^{M,i+k}) \bar{\beta}_k (u_n^{M,i-1}, w_n^{M,i}, \dots, w_n^{M,i+k}) \quad (4.54) $$

with $\bar{\beta}_k(u, y_0, \dots, y_k) := \beta_0(u, y_0)1_{\{k=0\}} - \beta_1 \left(u, y_0, \sum_{\ell=1}^{k-1} y_\ell, y_k\right) 1_{\{k \neq 0\}}$ where we recall that $\beta_0$ and $\beta_1$ are respectively defined in (4.7) and (4.8). Given $k = 0, \dots, \phi_n(0)$, we can apply Conjecture 2 with $t = k$, $h(x_0, \dots, x_k) = f(x_0) f(x_k)$ and $\psi(u, y_0, \dots, y_k) = \bar{\beta}_k(u, y_0, \dots, y_k)$ and obtain that

$$ \left| \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_{n+1}^{M,m}) \mid \mathcal{F}^n \right) \right) - \sum_{k=0}^{\phi_n(0)} \mathbb{E} \left( \bar{\eta}_n^{k,M} (\tilde{f}_k) \right) \right| \underset{M \to \infty}{\longrightarrow} 0 \quad (4.55) $$
where for $k = 0, \dots, \phi_n(0)$, $\tilde{f}_k(x_0, \dots, x_k) = f(x_0) f(x_k) \int_0^1 \bar{\beta}_k(u, \tilde{g}_n(x_0), \dots, \tilde{g}_n(x_k)) du$.

We now observe that by defining $\bar{f}_n(x_0, \dots, x_{\phi_n(0)}) = \sum_{k=0}^{\phi_n(0)} \tilde{f}_k(x_0, \dots, x_k)$, one has

$$ 
\begin{aligned}
& \left| \sum_{k=0}^{\phi_n(0)} \bar{\eta}_n^{k,M}(\tilde{f}_k) - \bar{\eta}_n^{\phi_n(0),M}(\bar{f}_n) \right| \\
&= \frac{1}{M} \left| \sum_{k=0}^{\phi_n(0)} \left( \sum_{i=1}^{M-k} \tilde{f}_k(X_n^{M,i}, \dots, X_n^{M,i+k}) - \sum_{i=1}^{M-\phi_n(0)} \tilde{f}_k(X_n^{M,i}, \dots, X_n^{M,i+k}) \right) \right| \\
&\le \frac{C}{M} \sum_{k=0}^{\phi_n(0)} (\phi_n(0) - k) \xrightarrow{M \to \infty} 0
\end{aligned}
$$

for a finite constant $C$. Thus we can combine it with (4.55) and obtain that

$$ \left| \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_{n+1}^{M,m}) \mid \mathcal{F}^n \right) \right) - \mathbb{E} \left( \bar{\eta}_n^{\phi_n(0),M}(\bar{f}_n) \right) \right| \xrightarrow{M \to \infty} 0. $$

We can now apply Proposition 11 with $h = \bar{f}_n$ and obtain that

$$ \left| \mathbb{E} \left( \bar{\eta}_n^{\phi_n(0),M}(\bar{f}_n) \right) - \int \mathcal{T}_0^{\phi_{1:n}(0)} \cdots \mathcal{T}_{n-2}^{\phi_{n-1:n}(0)} \mathcal{T}_{n-1}^{\phi_{n:n}(0)} \bar{f}_n(x) \eta^{\otimes \phi_{0:n}(0)}(dx) \right| \xrightarrow{M \to \infty} 0. $$

Thus we have obtained that $\forall n \ge 1$, $\forall f \in \mathcal{B}_b(\mathbb{R}^d)$

$$ \left| V_{n+1}^{M}(f) - \frac{V_{n}^{M}(P_{n}f_{n})}{(\bar{\eta}_{n}^{0}(g_{n}))^{4}} - \frac{\bar{\eta}_{n-1}^{0}(g_{n-1}(P_{n}f_{n}^{2} - (P_{n}f_{n})^{2}))}{(\bar{\eta}_{n}^{0}(g_{n}))^{4}\bar{\eta}_{n-1}^{0}(g_{n-1})} - \int \mathcal{T}_{0}^{\phi_{1:n}(0)} \cdots \mathcal{T}_{n-2}^{\phi_{n-1:n}(0)} \mathcal{T}_{n-1}^{\phi_{n:n}(0)} \bar{f}_{n}(x) \eta^{\otimes \phi_{0:n}(0)}(dx) \right| \xrightarrow{M \to \infty} 0. $$

To conclude the proof, it is now sufficient to observe that by Theorem 34, $\forall h \in \mathcal{B}_b(\mathbb{R}^d) V_1^M(h)$ converges as $M$ goes to infinity and its limit is given by $V_1(h) = \sigma_1^2(h) + \sigma_2^2(h)$. $\square$

The proof of Proposition 11 is a direct consequence of the following Lemma.

**Lemma 13.** Let $n \ge 0$ and $k \ge 0$. Given $h \in \mathcal{B}_b(\mathbb{R}^{d(k+1)})$ one has

$$ \left| \mathbb{E} \left( \bar{\eta}_{n+1}^{k,M}(h) \right) - \mathbb{E} \left( \bar{\eta}_n^{\phi_n(k),M}(\mathcal{T}_n^k h) \right) \right| \xrightarrow{M \to \infty} 0. $$

*Proof of Lemma 13.* Since the $X_{n+1}^{M,m}$ are conditionally independent given $\mathcal{G}^{n+1}$ and $\mathcal{L}(X_{n+1}^{M,m} | \mathcal{G}^{n+1}) \sim P_{n+1}(Y_{n+1}^{M,m}, \cdot)$, for each bounded measurable function $h: \mathbb{R}^{d(k+1)} \to \mathbb{R}$ one has

$$ \mathbb{E} \left( h(X_{n+1}^{M,i}, \dots, X_{n+1}^{M,i+k}) \mid \mathcal{G}^{n+1} \right) = P_{n+1}^{\otimes(k+1)} h \left( Y_{n+1}^{M,i}, \dots, Y_{n+1}^{M,i+k} \right) \quad (4.56) $$

$$ := \int_{\mathbb{R}^{d(k+1)}} h(x_0, \dots, x_k) P_{n+1} \left( Y_{n+1}^{M,i}, dx_0 \right) \cdots P_{n+1} \left( Y_{n+1}^{M,i+k}, dx_k \right). \quad (4.57) $$

Therefore for $M \ge k + 1$
$$ \mathbb{E}(\tilde{\eta}_{n+1}^{k,M}(h)) = \frac{1}{M} \sum_{i=1}^{M-k} \mathbb{E}(h(X_{n+1}^{M,i}, \dots, X_{n+1}^{M,i+k})) \quad (4.58) $$

$$ = \frac{1}{M} \sum_{i=1}^{M-k} \mathbb{E} \left( P_{n+1}^{\otimes(k+1)} h \left( Y_{n+1}^{M,i}, \dots, Y_{n+1}^{M,i+k} \right) \right) = \mathbb{E} \left( \tilde{\eta}_{n+1}^{k,M} (P_{n+1}^{\otimes(k+1)} h) \right). \quad (4.59) $$

Thus our purpose now becomes to study the asymptotic behaviour of $\mathbb{E}(\tilde{\eta}_{n+1}^{k,M}(f))$ for a given $f \in B_b(\mathbb{R}^{d(k+1)})$. Recalling the definition (4.49) of $Y_{n+1}^{M,m}$, let us observe that if we denote by $\ell_m$ the random index in $\{1, \dots, M\}$ such that $1_{\left\{ \sum_{j=1}^{\ell_m-1} w_n^{M,j} < m - U_m^{m} \le \sum_{j=1}^{\ell_m} w_n^{M,j} \right\}} = 1$ so that $Y_{n+1}^{M,m} = X_n^{M,\ell_m}$, one has that $m \mapsto \ell_m$ is non decreasing.
Therefore given $k \ge 0$, $M \ge k+1$ and $1 \le i \le M-k$ one has:

$$ \begin{aligned} & (Y_{n+1}^{M,i}, \dots, Y_{n+1}^{M,i+k}) \\ &= \sum_{\ell_0=1}^{M} \dots \sum_{\ell_k=\ell_{k-1}}^{M} (X_n^{M,\ell_0}, \dots, X_n^{M,\ell_k}) \prod_{q=0}^{k} 1 \left\{ \sum_{j=1}^{\ell_q-1} w_n^{M,j} < i+q - U_n^{i+q} \le \sum_{j=1}^{\ell_q} w_n^{M,j} \right\}. \end{aligned} $$

We can now apply the following change of variables $m = \ell_0, s_1 = \ell_1 - \ell_0, s_2 = \ell_2 - \ell_0, \dots, s_k = \ell_k - \ell_0$ and set $s_0 := 0$ so that the above expression becomes

$$ \sum_{0 \le s_k \le M-1} \sum_{0 \le s_1 \le s_2 \le \dots \le s_k} \sum_{m=1}^{M-s_k} (X_n^{M,m}, X_n^{M,m+s_1}, \dots, X_n^{M,m+s_k}) \prod_{q=0}^{k} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < i+q - U_n^{i+q} \le \sum_{j=1}^{m+sq} w_n^{M,j} \right\}. $$

Therefore one has

$$ \begin{aligned} \mathbb{E}(\tilde{\eta}_{n+1}^{k,M}(f)) &= \frac{1}{M} \sum_{i=1}^{M-k} \mathbb{E} \left( f \left( Y_{n+1}^{M,i}, \dots, Y_{n+1}^{M,i+k} \right) \right) = \frac{1}{M} \sum_{i=1}^{M-k} \mathbb{E} \left( \mathbb{E} \left( f \left( Y_{n+1}^{M,i}, \dots, Y_{n+1}^{M,i+k} \right) \mid \mathcal{F}^n \right) \right) \\ &= \frac{1}{M} \sum_{0 \le s_k \le M-1} \sum_{0 \le s_1 \le s_2 \le \dots \le s_k} \sum_{m=1}^{M-s_k} \mathbb{E} \left( f \left( X_n^{M,m}, X_n^{M,m+s_1}, \dots, X_n^{M,m+s_k} \right) \right. \\ & \qquad \left. \times \sum_{i=1}^{M-k} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u \le \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du \right) \end{aligned} $$

where to obtain the last equality we apply the Freezing Lemma and the fact that the sequence $(U_n^m)_{1 \le m \le M}$ is independent of $\mathcal{F}^n$ and the sequence $(X_n^{M,m})_{1 \le m \le M}$ is $\mathcal{F}^n$-measurable.
Given $k \ge 0$, $M \ge k+1$, $0 \le s_k \le M-1$, $0 \le s_1 \le s_2 \le \dots \le s_k$ and $1 \le m \le M-s_k$, let us now focus on the sum over $i$ appearing in the above expression:

$$ \sum_{i=1}^{M-k} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u \le \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du. $$

Let us first observe that if the quantity inside the sum is different from zero, then in particular $\int_{i-1}^{i} 1 \left\{ \sum_{j=1}^{m-1} w_n^{M,j} < u \le \sum_{j=1}^{m} w_n^{M,j} \right\} du \ne 0$ and $\int_{i+k-1}^{i+k} 1 \left\{ \sum_{j=1}^{m+s_k-1} w_n^{M,j} < u \le \sum_{j=1}^{m+s_k} w_n^{M,j} \right\} du \ne 0$. This implies that $\sum_{j=1}^{m} w_n^{M,j} > i-1$ and $\sum_{j=1}^{m+s_k-1} w_n^{M,j} < i+k$. Thus
$$ \frac{g_n}{\bar{g}_n} (s_k - 1) \leq \sum_{j=m+1}^{m+s_k-1} w_n^{M,j} = \sum_{j=1}^{m+s_k-1} w_n^{M,j} - \sum_{j=1}^{m} w_n^{M,j} < i + k - i + 1 = k + 1 $$

and so we have obtained an upper bound for $s_k$: $s_k \leq \left\lfloor \frac{\bar{g}_n}{g_n} (k+1) \right\rfloor = \phi_n(k)$ where we use the notation introduced in (4.48). Moreover we observe that we can replace the finite sum over $i$ with an infinite sum: if $i \geq M-k+1$

$$ \int_{i+k-1}^{i+k} 1 \left\{ \sum_{j=1}^{m+s_k-1} w_n^{M,j} < u \leq \sum_{j=1}^{m+s_k} w_n^{M,j} \right\} du = 0 $$

since $\sum_{j=1}^{m+s_k} w_n^{M,j} \leq \sum_{j=1}^{M} w_n^{M,j} = M$. Therefore after all one has

$$ \sum_{i=1}^{M-k} \int_{i-1}^{i} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u \leq \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du \\ = 1\{s_k \leq \phi_n(k)\} \sum_{i \geq 1} \prod_{q=1}^{k} \int_{i+q-1}^{i+q} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u \leq \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du. $$

We can now apply the change of variable $u := u' + \left\lfloor \sum_{j=1}^{m-1} w_n^{M,j} \right\rfloor$ in each of the above integrals so to obtain

$$ \begin{aligned} & \sum_{i \ge 1} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u \le \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du \\ &= \sum_{i \ge 1} \prod_{q=0}^{k} \int_{i+q-1}^{i+q-1} \left\{ \sum_{j=1}^{m-1} w_n^{M,j} \right\} \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u' \le \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du' \\ &= \sum_{i \ge 1} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} 1 \left\{ \sum_{j=1}^{m+sq-1} w_n^{M,j} < u' \le \sum_{j=1}^{m+sq} w_n^{M,j} \right\} du' \end{aligned} \quad (4.60) $$

where to obtain the last equality we use the fact that if $i < 1$,

$$ \int_{i-1}^{i} 1 \left\{ u_n^{M,m-1} + \sum_{j=m}^{m-1} w_n^{M,j} < u' \le u_n^{M,m-1} + \sum_{j=m}^{m} w_n^{M,j} \right\} du' = 0. $$

We observe that if the quantity inside the sum is different from zero, then in particular

$$ \int_{i-1}^{i} 1 \left\{ u_n^{M,m-1} < u' \le u_n^{M,m-1} + w_n^{M,m} \right\} du' \neq 0. $$

This implies that $i-1 < u_n^{M,m-1} + w_n^{M,m} \leq 1 + \frac{\bar{g}_n}{g_n}$ and so $i \leq \left\lfloor 1 + \frac{\bar{g}_n}{g_n} \right\rfloor$. In conclusion, (4.60) can be rewritten as

$$ \sum_{i=1}^{\lfloor 1 + \frac{\bar{g}_n}{g_n} \rfloor} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} 1 \left\{ u_n^{M,m-1} + \sum_{j=m}^{m+sq-1} w_n^{M,j} < u' \le u_n^{M,m-1} + \sum_{j=m}^{m+sq} w_n^{M,j} \right\} du'. \quad (4.61) $$
We denote (4.61) by $\psi_{s_1:s_k}(u_n^{M,m-1}, w_n^{M,m}, w_n^{M,m+1}, \dots, w_n^{M,m+s_k})$ where $\psi_{s_1:s_k}: \mathbb{R}^{s_k+2} \to \mathbb{R}$.
Therefore if $M \ge \max(1+k, 1+\phi_n(k))$ we have obtained that

$$ \mathbb{E}(\tilde{\eta}_{n+1}^{k,M}(f)) \qquad (4.62) $$

$$ = \sum_{s_k=0}^{\phi_n(k)} \sum_{0 \le s_1 \le s_2 \le \dots \le s_k} \frac{1}{M} \sum_{m=1}^{M-s_k} \mathbb{E} \left( f(X_n^{M,m}, X_n^{M,m+s_1}, \dots, X_n^{M,m+s_k}) \right) \quad (4.63) $$

$$ \times \psi_{s_1:s_k}(u_n^{M,m-1}, w_n^{M,m}, w_n^{M,m+1}, \dots, w_n^{M,m+s_k}) \quad (4.64) $$

We can therefore apply Conjecture 2 with $t = s_k$, $h(x_0, x_1, \dots, x_{s_k}) = f(x_0, x_{s_1}, x_{s_2}, \dots, x_{s_k})$, $\psi(u, y_0, \dots, y_{s_k}) = \psi_{s_1:s_k}(u, y_0, \dots, y_{s_k})$.

$$ \left| \mathbb{E} \left( \tilde{\eta}_{n+1}^{k,M}(f) \right) - \sum_{s_k=0}^{\phi_n(k)} \mathbb{E} \left( \bar{\eta}_n^{s_k,M} (T_n^{k \to s_k} f) \right) \right| \underset{M \to \infty}{\longrightarrow} 0 \qquad (4.65) $$

where $T_n^{k \to s_k} : \mathcal{B}_b(\mathbb{R}^{d(k+1)}) \to \mathcal{B}_b(\mathbb{R}^{d(s_k+1)})$ is defined by

$$ T_n^{k \to s_k} f(x_0, x_1, \dots, x_{s_k}) = \sum_{0 \le s_1 \le s_2 \le \dots \le s_k} f(x_0, x_{s_1}, \dots, x_{s_k}) \int_0^1 \psi_{s_1:s_k}(u, \tilde{g}_n(x_0), \tilde{g}_n(x_1), \dots, \tilde{g}_n(x_{s_k})) du $$

and where given $0 \le s_k \le \phi_n(k)$ and $0 \le s_1 \le s_2 \le \dots \le s_k$:

$$ \psi_{s_1:s_k}(u, y_0, y_1, \dots, y_{s_k}) = \sum_{i=1}^{\lfloor 1+\bar{g}_n/\underline{g}_n \rfloor} \prod_{q=0}^{k} \int_{i+q-1}^{i+q} \mathbb{1}\left\{u+\sum_{j=0}^{s_q-1} y_j < u' \le u+\sum_{j=0}^{s_q} y_j\right\} du'. $$

We now observe that by defining $T_n^k : \mathcal{B}_b(\mathbb{R}^{d(k+1)}) \to \mathcal{B}_b(\mathbb{R}^{d(\phi_n(k)+1)})$ by

$$ T_n^k f(x_0, \dots, x_{\phi_n(k)}) = \sum_{s_k=0}^{\phi_n(k)} T_n^{k \to s_k} f(x_0, \dots, x_{s_k}), $$

one has

$$ \left| \sum_{s_k=0}^{\phi_n(k)} \bar{\eta}_n^{s_k,M} (T_n^{k \to s_k} f) - \bar{\eta}_n^{\phi_n(k),M} (T_n^k f) \right| \\ = \frac{1}{M} \left| \sum_{s_k=0}^{\phi_n(k)} \left( \sum_{i=1}^{M-s_k} (T_n^{k \to s_k} f) (X_n^{M,i}, \dots, X_n^{M,i+s_k}) - \sum_{i=1}^{M-\phi_n(k)} (T_n^{k \to s_k} f) (X_n^{M,i}, \dots, X_n^{M,i+s_k}) \right) \right| \\ \le \frac{1}{M} \sum_{s_k=0}^{\phi_n(k)} \| T_n^{k \to s_k} f \|_{\infty} (\phi_n(k) - s_k) \underset{M \to \infty}{\longrightarrow} 0. $$

Combining the above estimate with (4.65) we obtain that

$$ \left| \mathbb{E} \left( \tilde{\eta}_{n+1}^{k,M}(f) \right) - \mathbb{E} \left( \bar{\eta}_n^{\phi_n(k),M} (T_n^k f) \right) \right| \underset{M \to \infty}{\longrightarrow} 0. \qquad (4.66) $$

In conclusion, combining (4.59) with (4.66) by taking $f = P_{n+1}^{\otimes(k+1)}h$, we obtain that

$$ \left| \mathbb{E} \left( \tilde{\eta}_{n+1}^{k,M}(h) \right) - \mathbb{E} \left( \bar{\eta}_n^{\phi_n(k),M} (\mathcal{T}_n^k h) \right) \right| \underset{M \to \infty}{\longrightarrow} 0 $$

where $\mathcal{T}_n^k = T_n^k P_{n+1}^{\otimes(k+1)}$ and this concludes the proof. $\square$
## 4.7 Numerical Results

In what follows we fix $d = 1$ and given a sequence $(W_n)_{n \ge 1}$ of independent real-valued random variables distributed according to the uniform law on $(0, 1)$, let $Z_{n+1} = Z_n + W_{n+1}$ for $n \ge 0$ with $Z_0$ distributed according to the uniform law on $(0, 1)$ $(\eta(dx) = 1_{[0,1]}(x) dx)$. Thus in this case the transition kernel is given by $P(x, dy) = 1_{[x,x+1]}(y)dy$. Moreover we will fix $g_n(x) = f(x) = e^x$ $\forall n \ge 0$, $\forall x \in \mathbb{R}$.

In this section, we are first going to numerically verify the Conjecture 1 in the case $n = 1$ and the Conjecture 2 in the case $n = 1$ and $n = 2$. In the second place, we will test the two conjectures together by directly studying the asymptotic variance. As done in the theory, we study numerically the asymptotic behaviour of following expression:

$$V_{n+1}^{M}(f) = \mathrm{Var} \left( \sqrt{M} \frac{\bar{\eta}_{n}^{0,M}(g_{n}f)}{\bar{\eta}_{n}^{0,M}(g_{n})} \right) + \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_{n+1}^{M,m}) \mid \mathcal{F}^{n} \right) \right). \quad (4.67)$$

We will consider separately the case $n = 0$ and the case $n = 1$. We recall that the first case has been fully studied in Section 4.3 without the need to introduce any conjecture.

### 4.7.1 Verification of the Conjectures

#### Conjecture 1

We recall the notation

$$f_1 := g_1 (\bar{\eta}_1^0(g_1) f - \bar{\eta}_1^0(g_1 f)).$$

Since we have fixed $g_n(x) = f(x) = e^x$ $\forall n \ge 0$, $\forall x \in \mathbb{R}$, one has $\bar{\eta}_0^0(g_0) = e - 1$, $\bar{\eta}_1^0(g_1) = \frac{e^2 - 1}{2}$, $\bar{\eta}_1^0(g_1 f) = \frac{1}{6}(e^3 - 1)(e + 1)$. Moreover $Pf_1(x) = \bar{\eta}_1^0(g_1) \frac{e^{2(x+1)} - e^{2x}}{2} - \bar{\eta}_1^0(fg_1) (e^{x+1} - e^x)$ and

$$Pf_1^2(x) = (\bar{\eta}_1^0(g_1))^2 \frac{e^{4(x+1)} - e^{4x}}{4} + (\bar{\eta}_1^0(fg_1))^2 \frac{e^{2(x+1)} - e^{2x}}{2} - 2\bar{\eta}_1^0(g_1)\bar{\eta}_1^0(fg_1) \frac{e^{3(x+1)} - e^{3x}}{3}$$

and

$$\bar{\eta}_0^0(g_0(Pf_1^2 - (Pf_1)^2)) = (\bar{\eta}_1^0(g_1))^2 \left(\frac{e^5-1}{5}\right) \left(\frac{e^2-1}{2}\right) \\ + (\bar{\eta}_1^0(g_1 f))^2 \left(\frac{e^3-1}{3}\right) \left(\frac{e^2-1}{2} - (e-1)^2\right) \\ - \bar{\eta}_1^0(g_1)\bar{\eta}_1^0(g_1 f) \left(\frac{e^4-1}{4}\right) \left(\frac{2}{3}(e^3-1) - (e^2-1)(e-1)\right).$$

We first observe that given a sequence $(T_i)_{i \ge 1}$ of square integrable i.i.d. random variables, by using the delta method it is possible to prove that as $M$ goes to infinity the following convergence in distribution holds:

$$\sqrt{M} \left( \left( \frac{1}{M} \sum_{i=1}^{M} T_i^2 - \left( \frac{1}{M} \sum_{i=1}^{M} T_i \right)^2 \right) - \mathrm{Var}(T_1) \right) \stackrel{d}{\Rightarrow} \mathcal{N} \left( 0, \mathrm{Var} \left( (T_1 - \mathbb{E}(T_1))^2 \right) \right). \quad (4.68)$$

The general strategy will be the following: we fix $M = 10000$ and
1. we simulate $n_1 = 10^7$ independent samples $T_j$ of $\sqrt{M}\frac{\bar{\eta}_1^{0,M}(g_1f)}{\bar{\eta}_1^{0,M}(g_1)} = \sqrt{M}\frac{\sum_{i=1}^M(g_1f)(X_1^{M,i})}{\sum_{i=1}^M g_1(X_1^{M,i})}$ and we compute, by using (4.68), the estimator

$$ \hat{v}_1 := \frac{1}{n_1} \sum_{i=1}^{n_1} T_i^2 - \left( \frac{1}{n_1} \sum_{i=1}^{n_1} T_i \right)^2 \quad (4.69) $$

of $\mathrm{Var}\left(\sqrt{M}\frac{\bar{\eta}_1^{0,M}(g_1f)}{\bar{\eta}_1^{0,M}(g_1)}\right)$ with relative 95% confidence interval $[a_1, b_1]$ given by

$$ a_1 := \hat{v}_1 - \frac{1.96}{\sqrt{n_1}} \sqrt{ \frac{1}{n_1} \sum_{i=1}^{n_1} \left( T_i - \frac{1}{n_1} \sum_{j=1}^{n_1} T_j \right)^4 - \left( \frac{1}{n_1} \sum_{i=1}^{n_1} \left( T_i - \frac{1}{n_1} \sum_{j=1}^{n_1} T_j \right)^2 \right)^2 } \quad (4.70) $$

$$ b_1 := \hat{v}_1 + \frac{1.96}{\sqrt{n_1}} \sqrt{ \frac{1}{n_1} \sum_{i=1}^{n_1} \left( T_i - \frac{1}{n_1} \sum_{j=1}^{n_1} T_j \right)^4 - \left( \frac{1}{n_1} \sum_{i=1}^{n_1} \left( T_i - \frac{1}{n_1} \sum_{j=1}^{n_1} T_j \right)^2 \right)^2 } \quad (4.71) $$

2. we simulate $n_1 = 10^7$ independent samples $H_j$ of $\frac{1}{\sqrt{M}}\sum_{m=1}^M Pf_1(Y_1^{M,m})$ and we compute the estimator $\hat{v}_2$ of $V_1^M(Pf_1) = \mathrm{Var}\left(\frac{1}{\sqrt{M}}\sum_{m=1}^M Pf_1(Y_1^{M,m})\right)$ with relative 95% confidence interval $[a_2, b_2]$ with $\hat{v}_2$, $a_2$, and $b_2$ respectively defined as in (4.69), (4.70) and (4.71). Therefore

$$ \frac{\hat{v}_2}{(\bar{\eta}_1^0(g_1))^4} + \frac{\bar{\eta}_0^0(g_0(P_1f_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)} $$

is an estimator of $\frac{V_1^M(Pf_1)}{(\bar{\eta}_1^0(g_1))^4} + \frac{\bar{\eta}_0^0(g_0(P_1f_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$ with relative 95% confidence interval

$$ \left[ \frac{a_2}{(\bar{\eta}_1^0(g_1))^4} + \frac{\bar{\eta}_0^0(g_0(P_1f_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}, \frac{b_2}{(\bar{\eta}_1^0(g_1))^4} + \frac{\bar{\eta}_0^0(g_0(P_1f_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)} \right] $$

3. we check that $\hat{v}_1$ is close to $\frac{\hat{v}_2}{(\bar{\eta}_1^0(g_1))^4} + \frac{\bar{\eta}_0^0(g_0(P_1f_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$ as expected

<table><thead><tr><th>n=1</th><th>value</th><th>CI (95%)</th></tr></thead><tbody><tr><td>hat<sub>1</sub></td><td>2.8021446</td><td>[ 2.7775619 , 2.8267273 ]</td></tr><tr><td>(hat<sub>2</sub> / (eta<sub>1</sub><sup>0</sup>(g<sub>1</sub>))) + (eta<sub>0</sub><sup>0</sup>(g<sub>0</sub>(P<sub>1</sub>f<sub>1</sub><sup>2</sup> - (Pf<sub>1</sub>)<sup>2</sup>)) / (eta<sub>1</sub><sup>0</sup>(g<sub>1</sub>)))<sup>4</sup> (eta<sub>0</sub><sup>0</sup>(g<sub>0</sub>))</td><td>2.7932862</td><td>[ 2.7831461 , 2.8034263 ]</td></tr></tbody></table>

### Conjecture 2

Let $n \in \{1, 2\}$ and $t \in \{1, 2\}$. Moreover let $h(x_0, \dots, x_t) = x_0 + \dots + x_t$ and $\psi(u_0, w_0, \dots, w_{t+1}) = u_0 + w_0 + \dots + w_{t+1}$. The general strategy will be the following: we fix $M = 10000$ and
1. we simulate $n_1 = 10^5$ independent samples $T_j$ of

$$ \frac{1}{M} \sum_{m=1}^{M-t} h(X_n^{M,m}, \dots, X_n^{M,m+t}) \psi(u_n^{M,m-1}, w_n^{M,m}, \dots, w_n^{M,m+t}) $$

and we compute the standard estimator

$$ \hat{v}_1 := \frac{1}{n_1} \sum_{j=1}^{n_1} T_j \qquad (4.72) $$

of $\frac{1}{M} \sum_{m=1}^{M-t} \mathbb{E}(h(X_n^{M,m}, \dots, X_n^{M,m+t}) \psi(u_n^{M,m-1}, w_n^{M,m}, \dots, w_n^{M,m+t}))$ with relative 95% confidence interval $[a_1, b_1]$ given by

$$ a_1 := \hat{v}_1 - \frac{1.96}{\sqrt{n_1}} * \sqrt{\frac{1}{n_1} \sum_{j=1}^{n_1} (T_j - \hat{v}_1)^2} \qquad (4.73) $$

$$ b_1 := \hat{v}_1 + \frac{1.96}{\sqrt{n_1}} * \sqrt{\frac{1}{n_1} \sum_{j=1}^{n_1} (T_j - \hat{v}_1)^2} \qquad (4.74) $$

2. we simulate $n_1 = 10^5$ independent samples $H_j$ of

$$ \frac{1}{M} \sum_{m=1}^{M-t} h(X_n^{M,m}, \dots, X_n^{M,m+t}) \psi(U, \tilde{g}_n(X_n^{M,m}), \dots, \tilde{g}_n(X_n^{M,m+t})) $$

where $U$ is a random variable uniformly distributed on $(0, 1)$ independent of $\mathcal{F}^n$ and $\tilde{g}_n = \frac{g_n}{\eta_n^0(g_n)} (\eta_1^0(g_1) = \frac{e^2-1}{2}, \eta_2^0(g_2) = \frac{e^3-1}{3})$. We then compute the standard estimator $\hat{v}_2$ of

$$ \frac{1}{M} \sum_{m=1}^{M-t} \mathbb{E} \left( h(X_n^{M,m}, \dots, X_n^{M,m+t}) \psi(U, \tilde{g}_n(X_n^{M,m}), \dots, \tilde{g}_n(X_n^{M,m+t})) \right) $$

with relative 95% confidence interval $[a_2, b_2]$ with $\hat{v}_2, a_2, b_2$ respectively defined as in (4.72), (4.73) and (4.74).

3. we check that $\hat{v}_1$ is close to $\hat{v}_2$ as expected

### 4.7.2 Focus on the Variance

case $n = 0$

We recall the notation $\bar{\eta}_0^0(h) := \eta(h) \forall h \in \mathcal{B}_b(\mathbb{R})$ and $f_0 := g_0(\bar{\eta}_0^0(g_0) f - \bar{\eta}_0^0(g_0 f))$.

By (4.15) and by observing that $\mathrm{Var}\left(\sqrt{M}\frac{\bar{\eta}_0^{0,M}(g_0 f)}{\bar{\eta}_0^{0,M}(g_0)}\right) = \mathrm{Var}\left(\sqrt{M}\frac{\bar{\eta}_0^{0,M}\left(\frac{g_0}{\eta(g_0)}f\right)}{\bar{\eta}_0^{0,M}\left(\frac{g_0}{\eta(g_0)}\right)}\right)$, one has

$$ \left| \mathrm{Var}\left(\sqrt{M} \frac{\bar{\eta}_0^{0,M}(g_0 f)}{\bar{\eta}_0^{0,M}(g_0)}\right) - \frac{\eta(f_0^2)}{\eta(g_0)^4} \right| \underset{M \to \infty}{\longrightarrow} 0 $$
<table><thead><tr><th>n=1,t=1</th><th>value</th><th>CI (95%)</th></tr></thead><tbody><tr><td>ŷ<sub>1</sub></td><td>5.7510732</td><td>[5.7509322, 5.7512143]</td></tr><tr><td>ŷ<sub>2</sub></td><td>5.7509738</td><td>[5.7470892, 5.7548583]</td></tr></tbody><thead><tr><th>n=1,t=2</th><th></th><th></th></tr></thead><tbody><tr><td>ŷ<sub>1</sub></td><td>11.8853516</td><td>[11.8850608, 11.8856425]</td></tr><tr><td>ŷ<sub>2</sub></td><td>11.8835799</td><td>[11.8777417, 11.8894182]</td></tr></tbody><thead><tr><th>n=2,t=1</th><th></th><th></th></tr></thead><tbody><tr><td>ŷ<sub>1</sub></td><td>9.2154319</td><td>[9.2152408, 9.215623]</td></tr><tr><td>ŷ<sub>2</sub></td><td>9.2150201</td><td>[9.2087834, 9.2212568]</td></tr></tbody><thead><tr><th>n=2,t=2</th><th></th><th></th></tr></thead><tbody><tr><td>ŷ<sub>1</sub></td><td>19.0901421</td><td>[19.0897513, 19.0905329]</td></tr><tr><td>ŷ<sub>2</sub></td><td>19.0895089</td><td>[19.0801501, 19.0988676]</td></tr></tbody></table>

where with the choices made $\bar{\eta}_0^0(g_0) = e-1$, $\bar{\eta}_0^0(g_0 f) = \frac{e^2-1}{2}$, $\eta(f_0^2) = (\bar{\eta}_0^0(g_0))^2 \frac{e^4-1}{4} + (\bar{\eta}_0^0(g_0 f))^2 \frac{e^2-1}{2} - 2\bar{\eta}_0^0(g_0) \bar{\eta}_0^0(g_0 f) \frac{e^3-1}{3}$. The application of Theorem 35 with $g = \tilde{g}_0 = \frac{g_0}{\bar{\eta}_0^0(g_0)}$ gives

$$ \left| \mathbb{E} \left( \mathrm{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_1^{M,m}) \mid \mathcal{F}^0 \right) \right) - \sum_{k=0}^{\phi_0(0)} \mathbb{E} (F_k) \right| \underset{M \to \infty}{\longrightarrow} 0 $$

with $(F_k)_{k \in \mathbb{N}}$ given by $f^2(X_0^{M,1})\beta_0(U_1, \tilde{g}_0(X_0^{M,1}))$ if $k=0$ and by

$$ -f(X_0^{M,1})f(X_0^{M,k+1})\beta_1 \left( U_1, \tilde{g}_0(X_0^{M,1}), \sum_{\ell=2}^{k} \tilde{g}_0(X_0^{M,\ell}), \tilde{g}_0(X_0^{M,k+1}) \right) $$

if $k > 0$ with $U_1 \sim \mathcal{U}(0,1)$ independent of $X_0^{M,1}, \dots, X_0^{M,k+1}$ and $\beta_0$ and $\beta_1$ respectively defined in (4.7) and (4.8). By using that when $U_1$ is uniformly distributed on $(0,1)$, $\{U_1+r\}$ is uniformly distributed on $(0,1)$ for each $r \ge 0$, we can apply the Freezing Lemma to rewrite $\mathbb{E}(F_k) \forall k = 0, \dots, \phi_0(0)$:

$$ \sum_{k=0}^{\phi_0(0)} \mathbb{E}(F_k) = \sum_{k=0}^{\phi_0(0)} \mathbb{E} \left( \phi_{k, \tilde{g}_0} \left( X_0^{M,1}, \dots, X_0^{M,k+1} \right) \right) \quad (4.75) $$

for measurable functions $\phi_{k, \tilde{g}_0} : \mathbb{R}^{(k+1)} \to \mathbb{R}$ given by

$$ \phi_{0, \tilde{g}_0}(x) = \frac{1}{3} \left( f^2(x) \left( 1 - \mathbb{1}_{\{\tilde{g}_0(x) < 1\}} \right) (1 - \tilde{g}_0(x))^3 \right) \quad (4.76) $$

and for each $k \ge 1$,

$$ \phi_{k, \tilde{g}_0}(x) = f(x_1)f(x_{k+1})\psi_{k, \tilde{g}_0}(x) \quad (4.77) $$
where $\psi_{k,\tilde{g}_0}(x)$ is given by

$$ \frac{1}{2} \left( \left( 1_{\left\{ \sum_{i=2}^{k} \tilde{g}_0(x_i) < 1 \right\}} \right) \left( 1_{\left\{ \sum_{i=1}^{k+1} \tilde{g}_0(x_i) < 1 \right\}} \right) \tilde{g}_0(x_1) \tilde{g}_0(x_{k+1}) \left( 2 - 2 \sum_{i=2}^{k} \tilde{g}_0(x_i) - (\tilde{g}_0(x_1) + \tilde{g}_0(x_{k+1})) \right) \right) \\ + \frac{1}{6} \left( 1_{\left\{ \sum_{i=2}^{k} \tilde{g}_0(x_i) < 1 \right\}} \right) \left( 1_{\left\{ \sum_{i=1}^{k+1} \tilde{g}_0(x_i) \ge 1 \right\}} \right) \left( 1 - \sum_{i=2}^{k} \tilde{g}_0(x_i) \right)^3 \\ - \frac{1}{6} \left( 1_{\left\{ \sum_{i=1}^{k} \tilde{g}_0(x_i) < 1 \right\}} \right) \left( 1_{\left\{ \sum_{i=1}^{k+1} \tilde{g}_0(x_i) \ge 1 \right\}} \right) \left( 1 - \sum_{i=1}^{k} \tilde{g}_0(x_i) \right)^3 \\ - \frac{1}{6} \left( 1_{\left\{ \sum_{i=2}^{k+1} \tilde{g}_0(x_i) < 1 \right\}} \right) \left( 1_{\left\{ \sum_{i=1}^{k+1} \tilde{g}_0(x_i) \ge 1 \right\}} \right) \left( 1 - \sum_{i=2}^{k+1} \tilde{g}_0(x_i) \right)^3. $$

Thus, using numerical methods we are going to check what we already know theoretically that is

$$ \left| V_1^M(f) - \frac{\eta(f_0^2)}{\eta(g_0)^4} - \sum_{k=0}^{\phi_0(0)} \mathbb{E} \left( \phi_{k, \tilde{g}_0} \left( X_0^{M,1}, \dots, X_0^{M,k+1} \right) \right) \right| \underset{M \to \infty}{\longrightarrow} 0. $$

The general strategy will be the following: we fix $M = 10000$ and

1. we simulate $n_1 = 10^7$ independent samples $T_j$ of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_1^{M,m})$ and we compute the estimator $\hat{v}_1$ of $V_1^M(f) = \text{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_1^{M,m}) \right)$ with relative 95% confidence interval $[a_1, b_1]$ with $\hat{v}_1, a_1, b_1$ respectively defined as in (4.69), (4.70) and (4.71). Therefore $\hat{v}_1 - \frac{\eta(f_0^2)}{\eta(g_0)^4}$ is an estimator of $V_1^M(f) - \frac{\eta(f_0^2)}{\eta(g_0)^4}$ with relative 95% confidence interval $[a_1 - \frac{\eta(f_0^2)}{\eta(g_0)^4}, b_1 - \frac{\eta(f_0^2)}{\eta(g_0)^4}]$.

2. We simulate $n_2 = 10^5$ independent samples $Z_j$ of $\sum_{k=0}^{\phi_0(0)} \phi_{k, \tilde{g}_0} (X_0^{M,1}, \dots, X_0^{M,k+1})$ and we compute the standard estimator $\hat{v}_2$ of $\sum_{k=0}^{\phi_0(0)} \mathbb{E} \left( \phi_{k, \tilde{g}_0} \left( X_0^{M,1}, \dots, X_0^{M,k+1} \right) \right)$ with relative 95% confidence interval $[a_2, b_2]$ with $\hat{v}_2, a_2, b_2$ respectively defined as in (4.72), (4.73) and (4.74).

3. We check that $\hat{v}_1 - \frac{\eta(f_0^2)}{\eta(g_0)^4}$ is close to $\hat{v}_2$ as expected.

<table><thead><tr><th>n=0</th><th>value</th><th>CI (95%)</th></tr></thead><tbody><tr><td>\hat{v}_1 - \frac{\eta(f_0^2)}{\eta(g_0)^4}</td><td>0.07943</td><td>[ 0.079127 , 0.079733 ]</td></tr><tr><td>\hat{v}_2</td><td>0.0793412</td><td>[ 0.0790773 , 0.0796051 ]</td></tr></tbody></table>

### case n = 1

In this case the asymptotic behaviour of the first term of the right-hand side of (4.67) is given by Conjecture 1

$$ \left| \text{Var} \left( \sqrt{M} \frac{\bar{\eta}_1^{0,M}(g_1 f)}{\bar{\eta}_1^{0,M}(g_1)} \right) - \frac{V_1^M(Pf_1)}{(\bar{\eta}_1^0(g_1))^4} - \frac{\bar{\eta}_0^0(g_0(Pf_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)} \right| \underset{M \to \infty}{\longrightarrow} 0 $$
and by (4.55) the asymptotic behaviour of the second term of the right-hand side of (4.67) is given by

$$ \left| \mathbb{E} \left( \operatorname{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_2^{M,m}) \mid \mathcal{F}^1 \right) \right) - \sum_{k=0}^{\phi_1(0)} \mathbb{E} \left( \bar{\eta}_1^{k,M} (\tilde{f}_k) \right) \right| \underset{M \to \infty}{\longrightarrow} 0 \quad (4.78) $$

where for $k = 0, \dots, \phi_1(0)$,

$$ \tilde{f}_k(x_0, \dots, x_k) = f(x_0) f(x_k) \int_0^1 \bar{\beta}_k(u, \tilde{g}_1(x_0), \dots, \tilde{g}_1(x_k)) du $$

with $\bar{\beta}_k(u, y_0, \dots, y_k) := \beta_0(u, y_0) \mathbf{1}_{\{k=0\}} - \beta_1 \left( u, y_0, \sum_{\ell=1}^{k-1} y_\ell, y_k \right) \mathbf{1}_{\{k \neq 0\}}$. We recall that $\beta_0$ and $\beta_1$ are respectively defined in (4.7) and (4.8) and $\tilde{g}_1 = \frac{g_1}{\bar{\eta}_1^0(g_1)}$. We now observe that

$$ \sum_{k=0}^{\phi_1(0)} \mathbb{E} \left( \bar{\eta}_1^{k,M} (\tilde{f}_k) \right) = \sum_{k=0}^{\phi_1(0)} \mathbb{E} \left( \bar{\eta}_1^{k,M} (\phi_{k,\tilde{g}_1}) \right) \quad (4.79) $$

where the $\phi_{k,\tilde{g}_1} : \mathbb{R}^{(k+1)} \to \mathbb{R}$ are measurable functions defined in (4.76) and (4.77).

Thus using numerical methods we are going to check that

$$ \left| V_2^M (f) - \frac{V_1^M (P f_1)}{(\bar{\eta}_1^0 (g_1))^4} - \frac{\bar{\eta}_0^0 (g_0 (P f_1^2 - (P f_1)^2))}{(\bar{\eta}_1^0 (g_1))^4 \bar{\eta}_0^0 (g_0)} - \sum_{k=0}^{\phi_1(0)} \mathbb{E} \left( \bar{\eta}_1^{k,M} (\phi_{k,\tilde{g}_1}) \right) \right| \underset{M \to \infty}{\longrightarrow} 0 $$

so that Conjecture 1 and Conjecture 2 are numerically verified at the same time. The general strategy will be the following:

1. (a) we simulate $n_1 = 10^7$ independent samples $T_j$ of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_2^{M,m})$ and we compute the estimator $\hat{v}_{1,1}$ of $\operatorname{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} f(Y_2^{M,m}) \right)$ with relative 95% confidence interval $[a_{1,1}, b_{1,1}]$ with $\hat{v}_{1,1}, a_{1,1}$, and $b_{1,1}$ respectively defined as in (4.69), (4.70) and (4.71).

(b) we simulate $n_1 = 10^7$ independent samples $H_j$ of $\frac{1}{\sqrt{M}} \sum_{m=1}^{M} P f_1(Y_1^{M,m})$ independent of $(T_j)_{1 \le j \le n_1}$. We compute the estimator $\hat{v}_{1,2}$ of $\operatorname{Var} \left( \frac{1}{\sqrt{M}} \sum_{m=1}^{M} P f_1(Y_1^{M,m}) \right)$ with relative 95% confidence interval $[a_{1,2}, b_{1,2}]$ with $\hat{v}_{1,2}, a_{1,2}$, and $b_{1,2}$ respectively defined as in (4.69), (4.70) and (4.71).

(c) we compute the estimator $\hat{v}_{1,1} - \frac{\hat{v}_{1,2}}{(\bar{\eta}_1^0(g_1))^4} - \frac{\bar{\eta}_0^0(g_0(Pf_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$ of

$$ V_2^M (f) - \frac{V_1^M (P f_1)}{(\bar{\eta}_1^0 (g_1))^4} - \frac{\bar{\eta}_0^0 (g_0 (P f_1^2 - (P f_1)^2))}{(\bar{\eta}_1^0 (g_1))^4 \bar{\eta}_0^0 (g_0)} $$

with 90% confidence interval $[A, B]$ with $A = a_{1,1} - \frac{b_{1,2}}{(\bar{\eta}_1^0(g_1))^4} - \frac{\bar{\eta}_0^0(g_0(Pf_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$ and $B = b_{1,1} - \frac{a_{1,2}}{(\bar{\eta}_1^0(g_1))^4} - \frac{\bar{\eta}_0^0(g_0(Pf_1^2 - (Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$.
2. We simulate $n_2 = 10^5$ independent samples $Z_j$ of $\sum_{k=0}^{\phi_1(0)} \bar{\eta}_1^{k,M}(\phi_{k,\bar{g}_1})$ and we compute the standard estimator $\hat{v}_2$ of $\sum_{k=0}^{\phi_1(0)} \mathbb{E}(\bar{\eta}_1^{k,M}(\phi_{k,\bar{g}_1}))$ with relative 95% confidence interval $[a_2, b_2]$ with $\hat{v}_2, a_2, b_2$ respectively defined as in (4.72), (4.73) and (4.74).

3. We check that $\hat{v}_{1,1} - \frac{\hat{v}_{1,2}}{(\bar{\eta}_1^0(g_1))^4} - \frac{\bar{\eta}_0^0(g_0(Pf_1^2-(Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$ is close to $\hat{v}_2$ as expected.

<table><thead><tr><th>n=1</th><th>value</th><th>CI</th></tr></thead><tbody><tr><td>$\hat{v}_{1,1} - \frac{\hat{v}_{1,2}}{(\bar{\eta}_1^0(g_1))^4} - \frac{\bar{\eta}_0^0(g_0(Pf_1^2-(Pf_1)^2))}{(\bar{\eta}_1^0(g_1))^4 \bar{\eta}_0^0(g_0)}$</td><td>0.4729737</td><td>$[0.4690806, 0.4768669]$</td></tr><tr><td>$\hat{v}_2$</td><td>0.4725217</td><td>$[0.4724719, 0.4725714]$</td></tr></tbody></table>
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