# Normal approximation for partial sums: general convex costs

Jérôme Dedecker, Florence Merlevède, Emmanuel Rio

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Jérôme Dedecker, Florence Merlevède, Emmanuel Rio. Normal approximation for partial sums: general convex costs. 2026. hal-05528218

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Normal approximation for partial sums: general
convex costs

Jérôme Dedecker,\* Florence Merlevède†and Emmanuel Rio ‡

Abstract

We provide non-asymptotic bounds and asymptotic limits for convex trans-
port costs between the distribution of partial sums of independent and iden-
tically distributed square integrable and centered random variables and the
normal distribution with mean zero and the same variance. The proof relies
on controlling the transport cost by an appropriate ideal distance, combined
with an adaptation of Lindeberg’s method. The numerical constants and the
asymptotic constants are explicit.

**Mathematics subject classifications.** (2020) - 60 F 05, 60 E 15.

**Keywords :** Transportation cost, Normal approximation.

# 1 Introduction

Let $\psi$ be a non-negative, even and convex function defined on $\mathbb{R}$. Define then the
following transport cost between two probability laws $\mu$ and $\nu$ on $\mathbb{R}$:

$$
\kappa_{\psi}(\mu, \nu) = \inf\{\mathbb{E}(\psi(X - Y)) : X \sim \mu, Y \sim \nu\}
$$

According to Theorem (8.1) in Major [13],

$$
\kappa_{\psi}(\mu, \nu) = \int_{0}^{1} \psi(F^{-1}(u) - G^{-1}(u))du. \quad (1.1)
$$

<footnote>*Jérôme Dedecker, Université Paris Cité, MAP5, UMR 8145 CNRS, 45 rue des Saints-Pères, F-75006 Paris, France.</footnote>

<footnote>†Florence Merlevède, LAMA, Univ Gustave Eiffel, Univ Paris Est Créteil, UMR 8050 CNRS, F-77454 Marne-La-Vallée, France.</footnote>

<footnote>‡Emmanuel Rio, Université de Versailles, LMV, UMR 8100 CNRS, 45 avenue des Etats-Unis, F-78035 Versailles, France.</footnote>
where $F$ and $G$ are the distribution functions of $\mu$ and $\nu$ respectively, and $F^{-1}$ and $G^{-1}$ their generalized inverses. In this paper, we are interested in transport costs from the law of a sum of independent and identically distributed (iid) centered real-valued random variables with finite variance to the Gaussian law with mean zero and the same variance. More precisely our aim is to find conditions ensuring that the transport cost $\kappa_\psi$ between these two distributions remains bounded as $n$ tends to $\infty$.

Let us now recall the known results in this area. In order to shorten the notations, we denote by $\kappa_p$ the transport cost in the case $\psi(x) = |x|^p$ (here $p \ge 1$) and $W_p = \kappa_p^{1/p}$ the associated Wasserstein distance. In the case $p = 1$, (1.1) implies that

$$ \kappa_1(\mu, \nu) = \int_{\mathbb{R}} |F(x) - G(x)|dx. $$

Consequently the results of Esseen [8] on the $L^1$-norm of the difference between the distribution functions apply to $\kappa_1$ (see also Ibragimov-Linnik [12], Section 5.3). It follows from Esseen's results that the costs $\kappa_1$ between the distribution of partial sums and the corresponding Gaussian distribution remain bounded as soon as the random variables have a finite absolute moment of order three. More precisely, let $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued centered random variables in $L^3$, with positive variance $\sigma^2$. Set $S_n = X_1 + X_2 + \dots + X_n$ and let $G_{n\sigma^2}$ denote a Gaussian random variable with mean 0 and variance $n\sigma^2$. For a random variable $X$ denote by $P_X$ its distribution. Esseen [8] proved that

$$ \sup_{n>0} \kappa_1(P_{S_n}, P_{G_{n\sigma^2}}) \le c_1 \sigma^{-2} \mathbb{E}(|X_1|^3) \quad (1.2) $$

for some positive universal constant $c_1$. According to Goldstein [9], $c_1 \le 1$. Furthermore Esseen proved that the above costs converge to some asymptotic constant (depending on the law of $X_1$) as $n$ tends to $\infty$ (see Esseen [8], Theorem 4.2).

Let us now discuss the case $p > 1$. In this case, one cannot derive asymptotic results for the cost $\kappa_p$ in the central limit theorem from results on the $L^p$-norm of the difference between the distribution functions. Bártfai [1] proved that, if the law of $X_1$ has a finite Laplace transform in a neighborhood of 0, then, for any positive $\varepsilon$,

$$ \lim_{n \to \infty} n^{-\varepsilon} \kappa_2(P_{S_n}, P_{G_{n\sigma^2}}) = 0. $$

Later, Rio [20] extended (1.2) to the costs $\kappa_p$ for $p$ in $]1, 2]$; he proved that, for any $p$ in $]1, 2]$, there exists a positive constant $c_p$, depending only on $p$, such that

$$ \sup_{n>0} \kappa_p(P_{S_n}, P_{G_{n\sigma^2}}) \le c_p \sigma^{-2} \mathbb{E}(|X_1|^{2+p}). \quad (1.3) $$
Next Bobkov [3] proved that (1.3) also holds true for $p > 2$. On another hand, Rio [21] obtained asymptotic constants for the costs $\kappa_p$ in the case $p$ in $]1, 2]$ under the moment condition $\mathbb{E}(|X_1|^{2+p}) < \infty$ (without conditions on the smoothness of the law of $X_1$). For $p > 2$, Bobkov [3] obtained asymptotic expansions for the costs $\kappa_p$ for random variables satisfying $\mathbb{E}(|X_1|^{2+p}) < \infty$, under the Crámer condition.

In this paper, we are interested in giving upper bounds for costs associated with general even and convex functions $\psi$ such that $x^{-2}\psi(x) \to 0$, as $x \to \infty$ (the precise definition of the class of costs is given in Definition 2.1). In particular this class of functions includes the fonction $x \mapsto h(|x|)$ where $h(x) = (x+1)\ln(x+1) - x$. Our aim is to give sufficient conditions ensuring that the costs $\kappa_\psi(P_{S_n}, P_{G_{n\sigma^2}})$ are uniformly bounded. As we shall see in Theorem 2.2, this will be the case as soon as $\mathbb{E}(X_1^2\psi(|X_1|)) < \infty$. Also, as a consequence of the introduced techniques, we will be in position to get asymptotic constants for general convex costs (see Corollary 2.1) and to provide tail inequalities for $|S_n - G_{n\sigma^2}|$ with $G_{n\sigma^2}$ defined from $S_n$ via the quantile transformation, as soon as the underlying random variables have a weak moment of order $2+p$ with $p \in ]1, 2[$ (see Theorem 2.3). We shall also give upper bounds for the conditional value at risk associated with the partial sum $S_n$ (see Corollary 2.3 and Remark 2.6). We refer to Section 2, which is devoted to our main results, for more details about the theoretical results.

We now give some insights on the proofs of our main results, which are given in Section 3, in the case $\sigma = 1$. The main tools of the proofs are Proposition 2.1 and a suitable version of the Lindeberg method combined with a technique of acceleration of the convergence. As in Rio [20], if $\beta_3 := \mathbb{E}(X_1^3) \neq 0$, we replace step by step the initial random variables $X_k$ by iid random variables $Y_k$ with a finite Laplace transform such that $\mathbb{E}(Y_1^3) = \beta_3$. More precisely, if $\Pi(\lambda)$ denotes a random variable with Poisson distribution of parameter $\lambda$,

$$Y_1 = B_1 + N_1, \qquad (1.4)$$

where $B_1$ and $N_1$ are independent, $N_1$ is normally distributed with mean 0 and variance 1/2, and

$$B_1 = 2\beta_3 \left( \Pi(1/(8\beta_3)^2) - 1/(8\beta_3^2) \right). \qquad (1.5)$$

Our methods allow us to get some numerical constants for this first step. In order to get general results with numerical constants, we then need to give estimates of the quadratic transportation cost between a Poisson distribution of parameter $\lambda$ and the normal distribution with mean $\lambda$ and variance $\lambda$ with a suitable numerical constant. Although it is known since a long time that the transportation costs $\kappa_p$ for the normal approximation of the Poisson distribution remain bounded as $\lambda$ tends to
$\infty$, up to our knowledge, numerical bounds do not exist, except in the case $p = 1$. In the case $p = 1$, the general results of Peccati et al. ([15], Example 3.5) provide the upper bound

$$W_1(\mu_\lambda, \nu_\lambda) \le 1 \text{ for any } \lambda > 0, \qquad (1.6)$$

where $\mu_\lambda$ denotes the distribution of $\Pi(\lambda) - \lambda$ and $\nu_\lambda$ the normal distribution with mean 0 and variance $\lambda$. In Proposition 3.1 of Section 3, we give the more efficient upper bound

$$W_2(\mu_\lambda, \nu_\lambda) \le \sqrt{0.937} \le 0.968 \text{ for any } \lambda > 0, \qquad (1.7)$$

which implies that $W_1(\mu_\lambda, \nu_\lambda) \le 0.968$, since $W_1(\mu, \nu) \le W_2(\mu, \nu)$. The proof of this result, given in Section 4, is based on a Tusnády type Lemma proved in Massart [14] together with a dyadic decomposition of a Poisson random variable. Notice that the asymptotic constants are much better: indeed, from the results of Esseen [8] in the case $p = 1$ and Rio [21] for $p$ in $]1, 2]$,

$$\lim_{n \to \infty} W_p(\mu_n, \nu_n) = \|(G^2 - 1)/6 + U\|_p,$$

where $G$ is a standard normal and $U$ is a random variable with uniform distribution over $[-1/2, 1/2]$, independent of $G$. In the case $p = 2$, the above limit is equal to $\sqrt{5}/6 = 0.37267...$ (see Corollary 1.3 in Rio [21]).

Concerning arbitrary laws, the numerical constants in (1.3) have not been studied, except in the case $p = 1$. In this paper, as a consequence of more general results, we get the upper bound $W_2(P_{S_n}, G_{n\sigma^2}) \le 6.825 \sigma^{-1} \sqrt{\mathbb{E}(X_1^4)}$ (see Remark 2.1 for more details). One can conjecture that the constant in (1.3) is less than 1: most probably, to prove this conjecture, more sophisticated methods, such as the operator methods given in Bonis [4], should be used. Our techniques of proofs are more elementary. Therefore they can be adapted to generalize the results of this paper to the class of weakly dependent sequences considered in Dedecker et al. [6].

# 2 Main results

Throughout this paper, $(X_i)_{i \in \mathbb{Z}}$ is a sequence of iid real-valued random variables, $X$ denotes a real-valued random variable with the same law as $X_1$ and $\Phi$ denotes the cumulative distribution function (c.d.f.) of a standard Normal distribution.

## 2.1 Normal approximation for convex costs

To handle the class of general convex costs we consider in this paper, we first introduce the following convex functions $\psi_x$, which generate all the convex functions
in the class Ψ defined in Definition 2.1 below (see Remark 2.2). Some additional
results are stated and proved in Appendix.

For $x > 0$, define the convex and even function $\psi_x$ by

$$
\psi_x(t) = \begin{cases} t^2/4 & \text{if } |t| \le 2x, \\ |tx| - x^2 & \text{if } |t| > 2x. \end{cases} \tag{2.1}
$$

The next proposition allows to compare the convex cost associated with $\psi_x$ with an
appropriate ideal metric.

**Proposition 2.1.** Let $\mathcal{F}_x := \{f : \mathbb{R} \to \mathbb{R} : \|f'\|_{\infty} \le x, f' \text{ is 1-Lipschitz}\}$. Then

$$
\kappa_{\psi_x}(\mu, \nu) \le \sup\{\mu(f) - \nu(f) : f \in \mathcal{F}_x\}.
$$

Proposition 2.1 and a Lindeberg-type method lead to the theorem below.

**Theorem 2.1.** Let $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables that are centered and in $\mathbb{L}^3$, with positive variance $\sigma^2$. Let $\mu_3 = \mathbb{E}(X^3)$ and $\lambda_3 = \mathbb{E}(|X|^3)$. Then there exist positive universal constants $\gamma_0$, $\gamma_1$, $\gamma_2$ and $\gamma_3$ such that, setting $C_0(X) = \gamma_1\sigma^2 + \sigma^{-4}(\gamma_0\lambda_3^2 + \gamma_3\mu_3^2)$, for any $n \ge 1$ and any $x > 0$,

$$
\sqrt{\kappa_{\psi_x}(P_{S_n}, P_{G_{n\sigma^2}})} \le \sqrt{C_0(X) + \gamma_2\sigma^{-2}\mathbb{E}(|X|^3 \min(|X|, 8x))} + 0.968\sigma^{-2}|\mu_3|,
$$

where $G_{n\sigma^2} \sim \mathcal{N}(0, n\sigma^2)$. The constants $\gamma_0, \gamma_1, \gamma_2$ and $\gamma_3$ can be chosen as follows:
$\gamma_0 = 3/5$, $\gamma_1 = 3.0376$, $\gamma_2 = 0.64584$ and $\gamma_3 = 1.6917$.

**Remark 2.1.** When the variables are in $\mathbb{L}^4$, taking the limit as $x$ tend to $\infty$ gives

$$
W_2(P_{S_n}, P_{G_{n\sigma^2}}) \le 2\sqrt{C_0(X) + \gamma_2\sigma^{-2}\mu_4} + 1.936\sigma^{-2}|\mu_3|
$$

for any $n \ge 1$, where $\mu_4 = \mathbb{E}(X^4)$. In particular, since $\sigma^4 \le \mu_4$ and $\mu_3^2 \le \lambda_3^2 \le \sigma^2 \mu_4$,

$$
W_2(P_{S_n}, P_{G_{n\sigma^2}}) \le 2\sqrt{4.2835\sigma^{-2}\mu_4 + 1.6917\sigma^{-4}\mu_3^2 + 1.936\sigma^{-2}|\mu_3|} \le 6.825\sigma^{-1}\sqrt{\mu_4}.
$$

In the case $\mathbb{E}(X^3) = 0$, the above upper bound yields

$$
W_2(P_{S_n}, P_{G_{n\sigma^2}}) \le 4.140\sigma^{-1}\sqrt{\mu_4}.
$$

Conversely, in the case of Rademacher random variables with parameter 1/2 (for
which $\mu_3 = 0$ and $\sigma^{-1}\sqrt{\mu_4} = 1$), (4.14) gives

$$
\sup_{n \ge 1} W_2(P_{S_n}, P_{G_{n\sigma^2}}) \ge \sqrt{2(1 - \sqrt{2/\pi})} \ge 0.63579.
$$

We now give applications of Theorem 2.1 to a class of convex costs whose defi-
nition is given below.
**Definition 2.1.** Let $\Psi$ be the class of functions $\varphi$ defined on $\mathbb{R}$, even, convex, $C^1$ and such that $\varphi(0) = \varphi'(0) = 0$, $\varphi'$ is concave, $\varphi'$ is derivable at 0, $\varphi''(0) = 1$ and $\lim_{x\to\infty} x^{-1}\varphi'(x) = 0$.

Note that the class $\Psi$ includes the function $x \mapsto (|x|+1)\ln(|x|+1) - |x|$. Next, for $p \in [1, 2[$, let $g_p$ be the even function defined on $\mathbb{R}^+$ by

$$g_p(x) = \begin{cases} x^2/2 & \text{if } x \in [0, 1] \\ x^p/p + 1/2 - 1/p & \text{if } x > 1. \end{cases} \qquad (2.2)$$

It is easy to prove that $g_p$ is in the class $\Psi$, and that for any $x$ in $\mathbb{R}^+$,

$$x^p \le pg_p(x) + 1 - p/2. \qquad (2.3)$$

**Remark 2.2.** Note that if $\varphi$ is in $\Psi$ then there exists a probability measure $\nu$ on $\mathbb{R}_+$ such that

$$\varphi(z) = \frac{1}{2} \int_{\mathbb{R}_+} \psi_t(2z) d\nu(t) = 2 \int_{\mathbb{R}_+} \psi_{t/2}(z) d\nu(t)$$

where $\psi_t$ is defined in (2.1). This fact will be proved in Section 3.3.

Let $\mathcal{P}_\varphi(\mathbb{R})$ be the set of probability laws $\mu$ on the real line such that $\mu(\varphi) < \infty$. For two probability laws $\mu$ and $\nu$ in $\mathcal{P}_\varphi(\mathbb{R})$, define now

$$W_\varphi(\mu, \nu) = (\kappa_\varphi(\mu, \nu))^{1/2}.$$

$W_\varphi$ is a distance between probability laws on $\mathcal{P}_\varphi(\mathbb{R})$ (see Lemma 3.2).

**Theorem 2.2.** Let $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables that are centered and in $\mathbb{L}^3$, with positive variance $\sigma^2$. Let $\varphi$ be a function in $\Psi$ and $G_{n\sigma^2}$ be a centered normal random variable with variance $n\sigma^2$. Then, with the same notations and the same constants as in Theorem 2.1, for any $n \ge 1$,

$$W_\varphi(P_{S_n}, P_{G_{n\sigma^2}}) \le \sqrt{2C_0(X) + 8\gamma_2\sigma^{-2}\mathbb{E}(|X|^3\varphi'(|X|/4))} + 1.369\sigma^{-2}|\mu_3|.$$

**Remark 2.3.** As we shall prove in Section 3.5,

$$\mathbb{E}(|X|^3\varphi'(|X|/4)) \le 8\mathbb{E}(|X|^2\varphi(|X|/4)).$$

It follows that $W_\varphi(P_{S_n}, P_{G_{n\sigma^2}})$ is uniformly bounded in $n$ provided that

$$\mathbb{E}(|X|^2\varphi(|X|)) < \infty. \qquad (2.4)$$

Concerning the Wasserstein distances of order $p$, we obtain the constants below in (1.3).
**Proposition 2.2.** Let $p \in ]1, 2[$ and $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables that are centered and in $\mathbb{L}^{2+p}$. Then, for any $n \ge 1$,

$$W_p(P_{S_n}, P_{G_{n\sigma^2}}) \le 2((C_0(X))^{p/2} + p2^{5-3p}\gamma_2\sigma^{-2}\mathbb{E}(|X|^{p+2}))^{1/p} + 1.936\sigma^{-2}|\mu_3|.$$

Starting from Theorem 2.2 and using the Cornish-Fisher expansion, one obtains the following extension of the results of Rio [21] on asymptotic constants for general convex costs.

**Corollary 2.1.** Let $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables that are centered and in $\mathbb{L}^3$, with variance 1. Let $G_n$ be a centered normal random variable with variance $n$ and $G$ be a random variable with standard normal law. Let $\varphi$ be any function in $\Psi$ such that $\mathbb{E}(|X|^2\varphi(|X|)) < \infty$.

(a) If the distribution of $X$ is not a lattice distribution, then

$$\lim_{n \to \infty} \kappa_{\varphi}(P_{S_n}, P_{G_n}) = \mathbb{E}(\varphi(\mathbb{E}(X^3)(G^2 - 1)/6)).$$

(b) If $X$ takes its values in the arithmetic progression $\{a + kh : k \in \mathbb{Z}\}$ ($h$ being maximal) and $V$ is a random variable with uniform law over $[-1/2, 1/2]$, independent of $(X_i)_{i \in \mathbb{Z}}$, then

$$\lim_{n \to \infty} \kappa_{\varphi}(P_{S_n+hV}, P_{G_n}) = \mathbb{E}(\varphi(\mathbb{E}(X^3)(G^2 - 1)/6)).$$

**Remark 2.4.** Note that the above limits are equal to 0 if and only if $\mathbb{E}(X^3) = 0$. In addition, as in Rio [21], one can also prove that if $X$ takes its values in the arithmetic progression $\{a + kh : k \in \mathbb{Z}\}$ ($h$ being maximal) and $U$ is a random variable with uniform law over $[0, 1]$, independent of $G$, then

$$\lim_{n \to \infty} \kappa_{\varphi}(P_{S_n}, P_{G_n}) = \mathbb{E}(\varphi(\mathbb{E}(X^3)(G^2 - 1)/6 + h(U - 1/2))).$$

Let $U$ denote a random variable with uniform law over $[0, 1]$. From (1.1),

$$\kappa_{\varphi}(P_{S_n}, P_{G_n}) = \mathbb{E}(\varphi(Z_n)), \text{ with } Z_n = |F_{S_n}^{-1}(U) - \sqrt{n}\Phi^{-1}(U)|. \quad (2.5)$$

The proof of Corollary 2.1 is based on the uniform integrability of the sequence $(\varphi(Z_n))_{n \ge 1}$, proved in Lemma 3.3. Using this lemma, one can also obtain asymptotic constants for the weighted costs defined below. Let $g: [0, 1] \to \mathbb{R}$ be a positive and measurable function. Define the weighted cost $\kappa_{1,g}$ by

$$\kappa_{1,g}(P_X, P_Y) = \mathbb{E}(g(U)|F_X^{-1}(U) - F_Y^{-1}(U)|). \quad (2.6)$$
For $\varphi$ in $\Psi$, the Young dual $\varphi^*$ of $\varphi$ is defined by $\varphi^*(x) = \sup_{t \in \mathbb{R}} (xt - \varphi(t))$. We denote by $L_{\varphi^*}$ the space of real-valued random variables $X$ such that $\mathbb{E}(\varphi^*(aX)) < \infty$ for some positive $a$. The corollary below provides asymptotic constants for the costs $\kappa_{1,g}$ when $X$ satisfies the integrability condition of Corollary 2.1 and $g(U)$ belongs to the space $L_{\varphi^*}$.

**Corollary 2.2.** Let $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables that are centered and in $\mathbb{L}^3$, with variance 1. Let $G_n$ be a centered normal random variable with variance $n$ and $G$ be a random variable with standard normal distribution and $U = \Phi(G)$. Let $\varphi$ be any function in $\Psi$ such that $\mathbb{E}(|X|^2\varphi(|X|)) < \infty$ and $g:]0, 1[\mapsto \mathbb{R}$ be a positive and measurable function such that $g(U)$ belongs to $L_{\varphi^*}$.

(a) If the distribution of $X$ is not a lattice distribution, then

$$ \lim_{n \to \infty} \kappa_{1,g}(P_{S_n}, P_{G_n}) = |\mathbb{E}(X^3)| \mathbb{E}(g(\Phi(G))|G^2 - 1|/6). $$

(b) If $X$ takes its values in the arithmetic progression $\{a + kh : k \in \mathbb{Z}\}$ ($h$ being maximal) and $V$ is a random variable with uniform law over $[-1/2, 1/2]$, independent of $(X_i)_{i \in \mathbb{Z}}$, then

$$ \lim_{n \to \infty} \kappa_{1,g}(P_{S_n+hV}, P_{G_n}) = |\mathbb{E}(X^3)| \mathbb{E}(g(\Phi(G))|G^2 - 1|/6). $$

let us illustrate Corollary 2.2 when $\varphi$ is defined by $\varphi(x) = (1+x)\log(1+x) - x$ for $x \ge 0$. Since $\varphi^*(x) = e^x - 1 - x$ for $x \ge 0$, the random variable $(\Phi^{-1}(U))^2$ belongs to $L_{\varphi^*}$. Consequently, in the non lattice case, if $\mathbb{E}(|X|^3 \log(1+|X|)) < \infty$,

$$ \lim_{n \to \infty} \int_0^1 (\Phi^{-1}(u))^2 |F_{S_n}^{-1}(u) - \sqrt{n}\Phi^{-1}(u)| du = |\mathbb{E}(X^3)| \mathbb{E}(G^2|G^2 - 1|/6). $$

$$ \text{Moreover } \mathbb{E}(G^2|G^2 - 1|/6) = 2(2\pi e)^{-1/2} + 1 - (4/3)\Phi(1) \approx 0.3622. $$

**Remark 2.5.** Using the same line of proof, one can easily extend the results of Corollary 2.2 to signed functionals as follows. Let $\ell: \mathbb{R} \mapsto \mathbb{R}$ be a Lipschitz function and $g:]0, 1[\mapsto \mathbb{R}$ be a measurable function such that $g(U)$ belongs to $L_{\varphi^*}$. Then, in the non lattice case

$$ \lim_{n \to \infty} \int_0^1 g(u) \ell(F_{S_n}^{-1}(u) - \sqrt{n}\Phi^{-1}(u)) du = \mathbb{E}(g(\Phi(G)) \ell(\mathbb{E}(X^3)(G^2 - 1)/6)) \quad (2.7) $$

In the lattice case the same result holds with $F_{S_n+hV}^{-1}(u)$ instead of $F_{S_n}^{-1}(u)$ on left hand. Applying (2.7) with $\ell(x) = x$ and $g = (\Phi^{-1})^2$, one obtains that, in the non lattice case, if $\mathbb{E}(|X|^3 \log(1+|X|)) < \infty$,

$$ \lim_{n \to \infty} \int_0^1 (\Phi^{-1}(u))^2 (F_{S_n}^{-1}(u) - \sqrt{n}\Phi^{-1}(u)) du = \mathbb{E}(X^3) \mathbb{E}(G^2(G^2 - 1)/6) = \mathbb{E}(X^3)/3. $$
## 2.2 Coupling inequalities

Enlarging the probability space if necessary, recall that, for any $n \ge 1$,

$$ \kappa_p(P_{S_n}, P_{G_{n\sigma^2}}) = \mathbb{E}(|S_n - G_{n\sigma^2}|^p) $$

where the random variable $G_{n\sigma^2}$ is a centered normal random variable (r.v.) with variance $n\sigma^2$, defined from $S_n$ via the quantile transformation as follows:

$$ G_{n\sigma^2} = \sigma\sqrt{n}\Phi^{-1}(F_n(S_n - 0) + \delta(F_n(S_n) - F_n(S_n - 0))), \quad (2.8) $$

where $F_n$ is the c.d.f. of $S_n$, $\delta$ is a r.v. with uniform distribution over $[0, 1]$ independent of $(X_i)_{i \in \mathbb{Z}}$ and $F_n(x - 0)$ means the left limit of $F_n$ at $x$.

Recall that Inequality (1.3) ensures that if the random variables have a strong moment of order $2 + p$ with $p \in ]1, 2]$, $\kappa_p(P_{S_n}, P_{G_{n\sigma^2}})$ is uniformly bounded with respect to $n$. Now, starting from the tools introduced in the proofs of the above results, we give upper bounds on the weak norm of order $p$ of the random variable $Z_n$ defined by

$$ Z_n = |S_n - G_{n\sigma^2}|, \quad (2.9) $$

where $G_{n\sigma^2}$ is defined by (2.8), under suitable conditions on the tail of the random variables $X_k$. In order to get more general results, we will introduce the conditional value at risk of a real valued random variable and its generalized inverse. For a real-valued random variable $Z$ we define its conditional value at risk by: for any $u \in ]0, 1[$

$$ \mathrm{CVar}(Z)(u) = \inf\{t + u^{-1}\mathbb{E}((Z - t)_+) : t > 0\}. \quad (2.10) $$

According to Proposition 5.4 in Pinelis [17],

$$ \mathrm{CVar}(Z)(u) = \tilde{Q}_Z(u) = \frac{1}{u} \int_0^u Q_Z(v)dv, $$

where $Q_Z$ is the tail quantile function of $Z$, i.e. $Q_Z(v) = F_Z^{-1}(1 - v)$. Note that $\tilde{Q}_Z$ was introduced by Hardy and Littlewood [11]. We also define $\tilde{H}_Z$ by

$$ \tilde{H}_Z(x) = \inf\{(x - t)^{-1}\mathbb{E}((Z - t)_+) : t < x\} $$

Define also the tail function $H_Z$ by $H_Z(x) = \mathbb{P}(Z > x)$. By the Markov inequality $\tilde{H}_Z \ge H_Z$. From the variational formula (2.10), $\tilde{H}_Z$ is the generalized inverse of $\tilde{Q}_Z$. Indeed, for any $u$ in $]0, 1]$,

$$ \tilde{H}_Z(x) < u \text{ if and only if } x > \tilde{Q}_Z(u). \quad (2.11) $$

It follows that $\tilde{H}_Z$ is the tail function of $\tilde{Q}_Z$. Hence an upper bound on $\tilde{H}_Z$ provides immediately an upper bound on $\tilde{Q}_Z$. Below we give an upper bound on $\tilde{H}_{Z_n}$, which will allow us to extend (1.3) to weak moments.
**Theorem 2.3.** Under the conditions of Theorem 2.1 and with the same notations, for any positive t,

$$t^2 \tilde{H}_{Z_n}(t) \leq 42.943 \sigma^{-4} \lambda_3^2 + 5.2041 \sigma^{-2} \mathbb{E}(|X|^3 \min(|X|, 2.8183t)).$$

From Theorem 2.3, we now derive upper bounds on the tail of $Z_n$ in case of weak moment of order $2+p$ with $p \in ]1, 2[$. For any $q \ge 1$ and a real-valued random variable $Z$, define the following weak moments of order $q$ of $Z$:

$$\Lambda_q(Z) = \sup_{x>0} x^q H_{|Z|}(x) \quad \text{and} \quad \tilde{\Lambda}_q(Z) = \sup_{x>0} x^q \tilde{H}_{|Z|}(x). \qquad (2.12)$$

Clearly $\Lambda_q(Z) \le \tilde{\Lambda}_q(Z)$. Starting from Theorem 2.3, we obtain the following upper bound on the weak moments of $Z_n$.

**Corollary 2.3.** Let $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables that are centered, with variance $\sigma^2$ and such that $\Lambda_{p+2}(X) < \infty$ for some $p \in ]1, 2[$. Assume that the probability space is large enough and, for any $n \ge 1$, define $G_{n\sigma^2}$ by (2.8) and let $Z_n = |S_n - G_{n\sigma^2}|$. Then

$$\tilde{\Lambda}_p(Z_n) \le (a_1 \sigma^{-2} \lambda_3)^p + a_2 ((p-1)(2-p))^{-1} (p+2) \sigma^{-2} \Lambda_{p+2}(X),$$

with $\lambda_3 = \mathbb{E}(|X|^3)$, $a_1 = 6.5531$ and $a_2 = 5.2041(2.8183)^{2-p}$. In addition, if $\mathbb{E}(|X|^{p+2}) < \infty$,

$$\tilde{\Lambda}_p(Z_n) \le (a_1 \sigma^{-2} \lambda_3)^p + a_2 \sigma^{-2} \mathbb{E}(|X|^{p+2}).$$

Now, define the Calderon weak norm of order $p$ (see DeVore and Lorentz [7], page 26) of a random variable $Z$ by

$$\|Z\|_{w,p} = \sup_{u \in [0,1]} u^{1/p} \tilde{Q}_{|Z|}(u).$$

One can easily deduce from (2.11) that

$$\tilde{\Lambda}_p(Z_n) = \|Z_n\|_{w,p}^p.$$

Hence Corollary 2.3 provides immediately the upper bound below:

$$\|Z_n\|_{w,p} \le \tilde{\kappa}_p^{1/p} \quad \text{with} \quad \tilde{\kappa}_p = (a_1 \sigma^{-2} \lambda_3)^p + a_2 ((p-1)(2-p))^{-1} (p+2) \sigma^{-2} \Lambda_{p+2}(X).$$

**Remark 2.6.** By subadditivity and monotonicity of the conditional value at risk (see Theorem 3.4 in [17]), one gets that for real-valued random variables $X$ and $Y$,

$$|\tilde{Q}_X(u) - \tilde{Q}_Y(u)| \le \tilde{Q}_{|X-Y|}(u).$$
Hence, under the conditions of Corollary 2.3, for any $n \ge 1$ and any $u \in ]0, 1[$,

$$|\tilde{Q}_{S_n}(u) - \tilde{Q}_{G_{n\sigma^2}}(u)| \le (\tilde{\kappa}_p/u)^{1/p}. \qquad (2.13)$$

Consequently

$$\tilde{Q}_{S_n}(u) \le \left(\frac{\tilde{\kappa}_p}{u}\right)^{1/p} + \frac{\sigma\sqrt{n}}{u\sqrt{2\pi}} \exp\left(-\frac{(\Phi^{-1}(u))^2}{2}\right).$$

Note that, if the random variables admit a strong moment of order $p+2$ for $p \in ]1, 2]$, the upper bound (2.13) can be improved. Indeed, Corollary 3.1 in Rio [22] asserts that in this case there exists a positive constant $C_p$ such that for any $u$ in $]0, 1/2]$,

$$|\tilde{Q}_{S_n}(u) - \tilde{Q}_{G_{n\sigma^2}}(u)| \le C_p(u|\ln(u)|)^{-1/p}.$$

We now give asymptotic constants for $\Lambda_p(Z_n)$ and $\tilde{\Lambda}_p(Z_n)$ under an additional condition on the tail of $|X|$.

**Corollary 2.4.** Let $p \in [1, 2[$ and $(X_i)_{i \in \mathbb{Z}}$ be a sequence of iid real-valued random variables in $\mathbb{L}^3$ that are centered, with variance 1. If $p > 1$, assume furthermore that

$$\lim_{t \to +\infty} t^{p+2} \mathbb{P}(|X| > t) = 0. \qquad (2.14)$$

(a) Assume that the distribution of $X$ is not a lattice distribution. Then

$$\lim_{n \to \infty} \Lambda_p^{1/p}(Z_n) = \frac{|\mathbb{E}(X^3)|}{6} \Lambda_p^{1/p}(G^2 - 1) \text{ and } \lim_{n \to \infty} \tilde{\Lambda}_p^{1/p}(Z_n) = \frac{|\mathbb{E}(X^3)|}{6} \tilde{\Lambda}_p^{1/p}(G^2 - 1).$$

(b) Assume that $X$ takes its values in the arithmetic progression $\{a + kh : k \in \mathbb{Z}\}$ ($h$ being maximal). Let $V$ be a random variable with uniform law over $[-1/2, 1/2]$, independent of $(X_i)_{i \in \mathbb{Z}}$ and $\bar{S}_n = S_n + hV$. Define $\bar{G}_n = \sqrt{n}\Phi^{-1}(F_{\bar{S}_n}(\bar{S}_n))$ and $\bar{Z}_n = |\bar{S}_n - \bar{G}_n|$. Then

$$\lim_{n \to \infty} \Lambda_p^{1/p}(\bar{Z}_n) = \frac{|\mathbb{E}(X^3)|}{6} \Lambda_p^{1/p}(G^2 - 1) \text{ and } \lim_{n \to \infty} \tilde{\Lambda}_p^{1/p}(\bar{Z}_n) = \frac{|\mathbb{E}(X^3)|}{6} \tilde{\Lambda}_p^{1/p}(G^2 - 1).$$

**Remark 2.7.** When $\mathbb{E}(X^3) = 0$, the asymptotic constants are equal to zero.

# 3 Proofs

## 3.1 Preliminary considerations

We start with a smoothing lemma which is a consequence of Lemma 6.1 in [5]. Below $\mathcal{F}_x$ is the class of functions defined in Proposition 2.1.
**Lemma 3.1.** Let $f$ in $\mathcal{F}_x$, $N$ be a standard normal and $B$ be an integrable real-valued random variable independent of $N$. Let $c_j = \|\phi^{(j)}\|_1$ where $\phi$ denotes the density of $N$. Then, for any integer $i \ge 3$ and any $t > 0$,

$$ \left| \frac{d^i}{du^i} \mathbb{E} f(u + tN + B) \right| \le \min \left( c_{i-1} x t^{1-i}, c_{i-2} t^{2-i} \right). $$

Elementary computations lead to

$$ c_1 = \sqrt{2/\pi}, c_2 = \sqrt{8/(\pi e)} \le 0.9679, c_3 = \sqrt{2/\pi} (1 + 4e^{-3/2}). $$

In addition

$$ c_4 = 4\sqrt{3/(\pi e^3)}((3 - \sqrt{6})^{1/2}e^{\sqrt{3}/2} + (3 + \sqrt{6})^{1/2}e^{-\sqrt{3}/2}), $$

$$ c_5 = \frac{4}{\sqrt{2\pi}} \left( \frac{3}{2} + 4(2 + \sqrt{10})e^{-(5+\sqrt{10})/2} + 4(\sqrt{10} - 2)e^{-(5-\sqrt{10})/2} \right) \le 5.9101. $$

Recall that the class $\Psi$ has been defined in Definition 2.1.

**Lemma 3.2.** Let $\varphi$ in $\Psi$. Then $W_\varphi$ is a distance on the set $\mathcal{P}_\varphi(\mathbb{R})$ of probability laws $\mu$ on the real line such that $\mu(\varphi) < \infty$.

**Proof of Lemma 3.2.** Let $\mu_1, \mu_2$ and $\mu_3$ in $\mathcal{P}_\varphi(\mathbb{R})$ with respective distribution functions $F_1, F_2$ and $F_3$.

Clearly $W_\varphi(\mu_1, \mu_2) \ge 0$ and $W_\varphi(\mu_1, \mu_2) = 0$ iff $\mu_1 = \mu_2$. We now prove the triangular inequality. Let $U \sim \mathcal{U}([0, 1])$. Set $A = F_1^{-1}(U) - F_2^{-1}(U)$ and $B = F_2^{-1}(U) - F_3^{-1}(U)$. With these notations, we have

$$ \kappa_\varphi(\mu_1, \mu_3) = \mathbb{E}(\varphi(A + B)), \kappa_\varphi(\mu_1, \mu_2) = \mathbb{E}(\varphi(A)), \kappa_\varphi(\mu_2, \mu_3) = \mathbb{E}(\varphi(B)). $$

Hence proving the triangular inequality is equivalent to prove that

$$ \sqrt{\mathbb{E}(\varphi(A + B))} \le \sqrt{\mathbb{E}(\varphi(A))} + \sqrt{\mathbb{E}(\varphi(B))}. \quad (3.1) $$

To prove (3.1), we first note that $\sqrt{\varphi}$ is subadditive on $\mathbb{R}_+$. Indeed, since $\varphi'$ is concave and $\varphi'(0) = 0$, $x \mapsto \frac{\varphi'(x)}{x}$ is nonincreasing on $]0, \infty[$ and so $x \mapsto \frac{\varphi(x)}{x^2}$ also since $\varphi(0) = 0$, by l'Hospital rule for monotonicity (see Corollary 1.3 and Remark 1.2 in Pinelis [16]). Hence, since $x \mapsto x^{-1}\sqrt{\varphi(x)}$ is nonincreasing on $]0, \infty[$, for any positive reals $a$ and $b$,

$$ \frac{\sqrt{\varphi(a+b)}}{a+b} \le \min \left( \frac{\sqrt{\varphi(a)}}{a}, \frac{\sqrt{\varphi(b)}}{b} \right) \le \frac{a}{a+b} \frac{\sqrt{\varphi(a)}}{a} + \frac{b}{a+b} \frac{\sqrt{\varphi(b)}}{b}. $$

This implies that, for any positive reals $a$ and $b$,

$$ \sqrt{\varphi(a+b)} \le \sqrt{\varphi(a)} + \sqrt{\varphi(b)}. \quad (3.2) $$
We go back to the proof of (3.1). Using (3.2), we get

$$ \mathbb{E}(\varphi(A+B)) \leq \mathbb{E}((\sqrt{\varphi(A)} + \sqrt{\varphi(B)})^2). $$

Hence, by the Schwarz inequality,

$$ \begin{aligned} \mathbb{E}(\varphi(A+B)) &\leq \mathbb{E}(\varphi(A)) + \mathbb{E}(\varphi(B)) + 2\mathbb{E}(\sqrt{\varphi(A)}\sqrt{\varphi(B)}) \\ &\leq \left(\sqrt{\mathbb{E}(\varphi(A))} + \sqrt{\mathbb{E}(\varphi(B))}\right)^2, \end{aligned} $$

which ends the proof of (3.1). □

The next result gives a quantitative upper bound for the Wasserstein distance of order 2 between the Poisson distribution and the normal distribution with the same mean and same variance. The proof will be given in Appendix.

**Proposition 3.1.** Let $(\Pi(t))_{t \geq 0}$ be a Poisson Process with parameter 1. For any $m > 0$ and any $\alpha \neq 0$, denote by $\mu_{m,\alpha}$ the law of $\alpha(\Pi(\alpha^{-2}m) - \alpha^{-2}m)$ and $\nu_m$ the $\mathcal{N}(0, m)$ law. Then $W_2^2(\mu_{m,\alpha}, \nu_m) \leq 0.937 \alpha^2$.

## 3.2 Proof of Theorem 2.1

Let $\sigma^2 = \mathbb{E}(X^2) > 0$. Note that, for any positive $a$,

$$ \psi_{ax}(az) = a^2 \psi_x(z). \qquad (3.3) $$

This leads to

$$ \kappa_{\psi_x}(P_A, P_B) = a^2 \kappa_{\psi_{x/a}}(P_{a^{-1}A}, P_{a^{-1}B}). \qquad (3.4) $$

Therefore we shall prove the result in case $\sigma^2 = 1$ and use the above equality with $A = S_n$, $B = G_{n\sigma^2}$ and $a = \sigma$ to get the general case. So, from now, we assume that $\sigma^2 = 1$.

Let $\beta_3 = \mathbb{E}(X^3)$. Assume first that $\beta_3 \neq 0$.

Let $(N_k)_{k \geq 1}$ be iid Gaussian random variables that are independent of $(X_k)_{k \geq 1}$, centered and with variance $1/2$. Let $(B_k)_{k \geq 1}$ be a sequence of iid random variables, independent of $(N_k, X_k)_{k \geq 1}$ and with the same law as

$$ B = 2\beta_3 \left( \Pi\left(\frac{1}{(8\beta_3^2)}\right) - \frac{1}{(8\beta_3^2)} \right), $$

where $\Pi(\lambda)$ is a r.v. with Poisson distribution of parameter $\lambda$. For any $k \geq 1$, let $Y_k = B_k + N_k$. The random variables $Y_k$ are iid and their first moments satisfy

$$ \mathbb{E}(Y_1) = 0, \mathbb{E}(Y_1^2) = 1, \mathbb{E}(Y_1^3) = \beta_3 \text{ and } \mathbb{E}(Y_1^4) = 3 + 2\beta_3^2. \qquad (3.5) $$
We now define set $T_n = \sum_{k=1}^n Y_k$. Note now that $2\psi_x$ belongs to $\Psi$. Hence, by Lemma 3.2, $\sqrt{\kappa_{\psi_x}}$ is a distance between probability laws, from which

$$ \sqrt{\kappa_{\psi_x}(P_{S_n}, P_{G_n})} \le \sqrt{\kappa_{\psi_x}(P_{S_n}, P_{T_n})} + \sqrt{\kappa_{\psi_x}(P_{T_n}, P_{G_n})}. \quad (3.6) $$

Next, using the fact that $\psi_x(t) \le t^2/4$ and Proposition 3.1 with $\alpha = 2\beta_3$ and $m = n/2$, we get

$$ \kappa_{\psi_x}(P_{T_n}, P_{G_n}) \le \frac{1}{4} W_2^2(P_{T_n}, P_{G_n}) \le \frac{1}{4} W_2^2(\mu_{n/2,2\beta_3}, P_{G_{n/2}}) \le 0.937\beta_3^2. \quad (3.7) $$

The above inequalities together with Proposition 2.1 imply that

$$ \sqrt{\kappa_{\psi_x}(P_{S_n}, P_{G_n})} \le \sqrt{\sup\{\mathbb{E}(f(S_n) - f(T_n)) : f \in \mathcal{F}_x\} + 0.968|\beta_3|}. \quad (3.8) $$

It remains to give an upper bound on $\mathbb{E}(f(S_n) - f(T_n))$ for $f$ in $\mathcal{F}_x$. Let $v^2 > 1/2$ to be chosen later and $Z$ be a random variable with normal distribution $N(0, v^2)$ such that $Z$ is independent of $(X_k, Y_k)_{k \in \mathbb{Z}}$. We first write

$$ \mathbb{E}(f(S_n)) - f(T_n) = \mathbb{E}(f(S_n + Z)) - f(T_n + Z) + R_{n,2} - R_{n,1}, \quad (3.9) $$

with $R_{n,1} = \mathbb{E}(f(S_n + Z) - f(S_n))$ and $R_{n,2} = \mathbb{E}(f(T_n + Z) - f(T_n))$. Now, since $Z$ is independent of $(S_n, T_n)$, $\mathbb{E}(f'(S_n)Z) = \mathbb{E}(f'(T_n)Z) = 0$. It follows that $R_{n,1} = \mathbb{E}(f(S_n + Z) - f(S_n) - f'(S_n)Z)$, which together with the fact that $f'$ is 1-Lipschitz implies that $|R_{n,1}| \le \mathbb{E}(Z^2/2) = v^2/2$. Similarly $|R_{n,2}| \le v^2/2$, whence

$$ |R_{n,2} - R_{n,1}| \le v^2. \quad (3.10) $$

From now on, let $\Delta_{n,v}(f) = \mathbb{E}(f(S_n + Z)) - f(T_n + Z)$. Setting

$$ f_{n-k}(x) = \mathbb{E}\left(f\left(x + \sum_{u=k+1}^{n} Y_u + Z\right)\right), $$

and taking into account the independence between the sequences, we get that

$$ \begin{aligned} \Delta_{n,v}(f) &= \sum_{k=1}^{n} \mathbb{E}(f(S_{k-1} + X_k + T_n - T_k + Z)) - f(S_{k-1} + Y_k + T_n - T_k + Z)) \\ &= \sum_{k=1}^{n} \left( \mathbb{E}(f'_{n-k}(S_{k-1})(X_k - Y_k)) + \frac{1}{2}\mathbb{E}(f''_{n-k}(S_{k-1})(X_k^2 - Y_k^2)) \right) \\ &\quad + \int_{0}^{1} \frac{(1-s)^2}{2} \sum_{k=1}^{n} \left\{ \mathbb{E}(f^{(3)}_{n-k}(S_{k-1} + sX_k)X_k^3) - \mathbb{E}(f^{(3)}_{n-k}(S_{k-1} + sY_k)Y_k^3) \right\} ds. \end{aligned} $$
By independence and since $\mathbb{E}(X_k) = \mathbb{E}(Y_k)$ and $\mathbb{E}(X_k^2) = \mathbb{E}(Y_k^2)$, it follows that

$$ \Delta_{n,v}(f) = \int_0^1 \frac{(1-s)^2}{2} \sum_{k=1}^n \{ \mathbb{E}(f_{n-k}^{(3)}(S_{k-1} + sX_k)X_k^3) - \mathbb{E}(f_{n-k}^{(3)}(S_{k-1} + sY_k)Y_k^3) \} ds. \quad (3.11) $$

Let us first handle the quantity

$$ A(n) := \int_0^1 \frac{(1-s)^2}{2} \sum_{k=1}^n \{ \mathbb{E}(f_{n-k}^{(3)}(T_{k-1} + sX_k)X_k^3) - \mathbb{E}(f_{n-k}^{(3)}(T_{k-1} + sY_k)Y_k^3) \} ds. $$

By independence and the fact that $\mathbb{E}(X_k^3) = \mathbb{E}(Y_k^3) = \beta_3$, we have

$$ \mathbb{E}(f_{n-k}^{(3)}(T_{k-1})X_k^3) = \mathbb{E}(f_{n-k}^{(3)}(T_{k-1}))\beta_3 = \mathbb{E}(f_{n-k}^{(3)}(T_{k-1})Y_k^3). $$

It follows that $A(n) = A_1(n) - A_2(n)$, where

$$ A_1(n) := \int_0^1 \frac{(1-s)^2}{2} \sum_{k=1}^n \{ \mathbb{E}(f_{n-k}^{(3)}(T_{k-1} + sX_k)X_k^3) - \mathbb{E}(f_{n-k}^{(3)}(T_{k-1})X_k^3) \} ds $$

$$ A_2(n) := \int_0^1 \frac{(1-s)^2}{2} \sum_{k=1}^n \{ \mathbb{E}(f_{n-k}^{(3)}(T_{k-1} + sY_k)Y_k^3) - \mathbb{E}(f_{n-k}^{(3)}(T_{k-1})Y_k^3) \} ds. $$

Let $u_n := (n-1)/2 + v^2$. Since $v^2 \ge 1/2$, we have that $u_n \ge n/2$. For any $s \in [0, 1]$, taking into account Lemma 3.1,

$$ \sum_{k=1}^{n} \mathbb{E}\left(|\left(f_{n-k}^{(3)}(T_{k-1} + sX_k) - f_{n-k}^{(3)}(T_{k-1})\right)X_k^3|\right) \\ \le n \mathbb{E}\left(|X|^3 \min\left(2\|f_n^{(3)}\|_{\infty}, s|X|\|f_n^{(4)}\|_{\infty}\right)\right) \le n c_2 \mathbb{E}\left(|X|^3 \min\left\{\frac{2x}{u_n}, |X|\frac{s}{u_n}\right\}\right). $$

Since $(n/u_n) \le 2$, it follows that

$$ |A_1(n)| \le \frac{c_2}{12} \mathbb{E}(|X|^3 (|X| \wedge 8x)). \quad (3.12) $$

On another hand, for any $s$ in $[0, 1]$, taking into account Lemma 3.1,

$$ \sum_{k=1}^{n} \mathbb{E}\left(\left|\left(f_{n-k}^{(3)}(T_{k-1} + sY_k) - f_{n-k}^{(3)}(T_{k-1})\right)Y_k^3\right|\right) \le n s \|f_n^{(4)}\|_{\infty} \mathbb{E}(Y^4) \le c_2 \frac{ns}{u_n}(3 + 2\beta_3^2). $$

Therefore

$$ |A_2(n)| \le \frac{c_2}{12} \times (3 + 2\beta_3^2). \quad (3.13) $$

Taking into account (3.12) and (3.13), we finally get that

$$ |A(n)| \le \frac{c_2}{12} \left( 3 + 2\beta_3^2 + \mathbb{E}(|X|^3(|X| \wedge 8x)) \right). \quad (3.14) $$
Next, we write

$$ \Delta_{n,v}(f) - A(n) := B_1(n) - B_2(n), \qquad (3.15) $$

where

$$ B_1(n) := \int_0^1 \frac{(1-s)^2}{2} \sum_{k=1}^n \{ \mathbb{E}(f_{n-k}^{(3)}(S_{k-1} + sX_k)X_k^3) - \mathbb{E}(f_{n-k}^{(3)}(T_{k-1} + sX_k)X_k^3) \} ds $$

$$ B_2(n) := \int_0^1 \frac{(1-s)^2}{2} \sum_{k=1}^n \{ \mathbb{E}(f_{n-k}^{(3)}(S_{k-1} + sY_k)Y_k^3) - \mathbb{E}(f_{n-k}^{(3)}(T_{k-1} + sY_k)Y_k^3) \} ds. $$

We first handle $B_1(n)$. For any integer $k$ in $[1, n]$ and any $s \in [0, 1]$,

$$ \begin{aligned} & \mathbb{E}(X_k^3 \{ f_{n-k}^{(3)}(S_{k-1} + sX_k) - (f_{n-k}^{(3)}(T_{k-1} + sX_k)) \}) \\ &= \sum_{\ell=1}^{k-1} \mathbb{E} \left( X_k^3 \left\{ f_{n-k}^{(3)}(S_{\ell-1} + X_\ell + T_{k-1} - T_\ell + sX_k) - f_{n-k}^{(3)}(S_{\ell-1} + Y_\ell + T_{k-1} - T_\ell + sX_k) \right\} \right) \\ &= \sum_{\ell=1}^{k-1} \mathbb{E} \left( X_k^3 \left\{ f_{n-\ell-1}^{(3)}(S_{\ell-1} + X_\ell + sX_k) - f_{n-\ell-1}^{(3)}(S_{\ell-1} + Y_\ell + sX_k) \right\} \right). \end{aligned} $$

By independence, since $\mathbb{E}(X_\ell) = \mathbb{E}(Y_\ell)$ and $\mathbb{E}(X_\ell^2) = \mathbb{E}(Y_\ell^2)$ for any $\ell \le k-1$,

$$ \mathbb{E}(X_k^3(X_\ell - Y_\ell) f_{n-\ell-1}^{(4)}(S_{\ell-1} + sX_k)) = 0, \quad \mathbb{E}(X_k^3(X_\ell^2 - Y_\ell^2) f_{n-\ell-1}^{(5)}(S_{\ell-1} + sX_k)) = 0. $$

Hence, by the Taylor integral formula,

$$ \begin{aligned} & \mathbb{E}(X_k^3 \{ f_{n-k}^{(3)}(S_{k-1} + sX_k) - (f_{n-k}^{(3)}(T_{k-1} + sX_k)) \}) \\ &= \int_0^1 \frac{(1-t)^2}{2} \sum_{\ell=1}^{k-1} \mathbb{E} \left( X_k^3 \left\{ X_\ell^3 f_{n-\ell-1}^{(6)}(S_{\ell-1} + tX_\ell + sX_k) - Y_\ell^3 f_{n-\ell-1}^{(6)}(S_{\ell-1} + tY_\ell + sX_k) \right\} \right) dt. \end{aligned} $$

Since $\mathbb{E}(X_\ell^3) = \mathbb{E}(Y_\ell^3)$, it follows that

$$ \begin{aligned} & \mathbb{E}(X_k^3 \{ f_{n-k}^{(3)}(S_{k-1} + sX_k) - (f_{n-k}^{(3)}(T_{k-1} + sX_k)) \}) \\ &= \int_0^1 \frac{(1-t)^2}{2} \sum_{\ell=1}^{k-1} (C_{n,k,\ell}^X(s,t) - D_{n,k,\ell}^X(s,t)) dt, \quad (3.16) \end{aligned} $$

where

$$ C_{n,k,\ell}^X(s,t) = \mathbb{E} \left( X_k^3 X_\ell^3 \left\{ f_{n-\ell-1}^{(6)}(S_{\ell-1} + tX_\ell + sX_k) - f_{n-\ell-1}^{(6)}(S_{\ell-1} + sX_k) \right\} \right), $$

$$ D_{n,k,\ell}^X(s,t) = \mathbb{E} \left( X_k^3 Y_\ell^3 \left\{ f_{n-\ell-1}^{(6)}(S_{\ell-1} + tY_\ell + sX_k) - f_{n-\ell-1}^{(6)}(S_{\ell-1} + sX_k) \right\} \right). $$
For $\ell < k$, we have

$$ 
\begin{aligned}
|C_{n,k,\ell}^X(s,t)| & \le \mathbb{E} \left( |X_k^3 X_\ell^3| \min \left( 2 \|f_{n-\ell-1}^{(6)}\|_\infty, t |X_\ell| \|f_{n-\ell-1}^{(7)}\|_\infty \right) \right) \\
& \le \mathbb{E}(|X|^3) \mathbb{E} \left( |X|^3 \min \left( 2 \|f_{n-\ell-1}^{(6)}\|_\infty, t |X| \|f_{n-\ell-1}^{(7)}\|_\infty \right) \right).
\end{aligned}
$$

Using the notation

$$ u_{n,\ell} = (n - \ell - 1)/2 + v^2 \quad (3.17) $$

and Lemma 3.1, we derive

$$ 
\begin{aligned}
& \int_0^1 \int_0^1 \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} \sum_{k=2}^n \sum_{\ell=1}^{k-1} |C_{n,k,\ell}^X(s,t)| dsdt \\
& \le \mathbb{E}(|X|^3) \int_0^1 \int_0^1 \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} \sum_{k=2}^n \sum_{\ell=1}^{k-1} \mathbb{E} \left( |X|^3 \min \left( \frac{2c_5x}{u_{n,\ell}^{5/2}}, \frac{c_5t|X|}{u_{n,\ell}^{5/2}} \right) \right) dsdt \\
& \le \frac{c_5}{144} \mathbb{E}(|X|^3) \mathbb{E}(|X|^3 (|X| \wedge 8x)) \sum_{k=0}^{n-2} \sum_{\ell=k}^{n-2} \frac{2^{5/2}}{(\ell + 2v^2)^{5/2}}.
\end{aligned}
$$

Next, we note that, by convexity, the following upper bound holds: for any $a \ge 1/2$, any $m \in \mathbb{N}$ and any $p > 1$,

$$ \sum_{k \ge m} (k+a)^{-p} \le \int_{m+a-1/2}^{+\infty} x^{-p} dx. $$

Applying twice the above inequality, we obtain that

$$ \sum_{k=0}^{n-2} \sum_{\ell=k}^{n-2} \frac{1}{(\ell + 2v^2)^{5/2}} \le \frac{2}{3} \int_{2v^2-1}^{+\infty} \frac{1}{x^{3/2}} dx = \frac{4}{3} (2v^2 - 1)^{-1/2}. \quad (3.18) $$

From the above considerations, it follows that

$$ \int_0^1 \int_0^1 \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} \sum_{k=2}^n \sum_{\ell=1}^{k-1} |C_{n,k,\ell}^X(s,t)| dsdt \le \frac{\sqrt{2c_5}\mathbb{E}(|X^3|)}{27\sqrt{2v^2-1}} \mathbb{E}(|X|^3(|X| \wedge 8x)). \quad (3.19) $$

On another hand, taking into account Lemma 3.1 and the notation (3.17), we derive that

$$ |D_{n,k,\ell}^X(s,t)| \le t c_5 \mathbb{E}(|X|^3) \mathbb{E}(Y^4) u_{n,\ell}^{-5/2}. $$

Therefore, taking into account the upper bound (3.18), we get

$$ \int_0^1 \int_0^1 \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} \sum_{k=2}^n \sum_{\ell=1}^{k-1} |D_{n,k,\ell}^X(s,t)| dsdt \le \frac{\sqrt{2c_5}\mathbb{E}(|X^3|)}{27\sqrt{2v^2-1}} (3+2\beta_3^2). \quad (3.20) $$
Starting from (3.16) and considering (3.19) and (3.20), we derive that

$$|B_1(n)| \le \frac{\sqrt{2}c_5\mathbb{E}(|X^3|)}{27\sqrt{2v^2-1}} (3 + 2\beta_3^2 + \mathbb{E}(|X|^3(|X| \wedge 8x))). \quad (3.21)$$

To handle the quantity $B_2(n)$ we proceed as above. We first write

$$B_2(n) = \sum_{k=1}^{n} \sum_{\ell=1}^{k-1} \int_{0}^{1} \int_{0}^{1} \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} (C_{n,k,\ell}^{Y}(s,t) - D_{n,k,\ell}^{Y}(s,t)) ds dt,$$

where

$$C_{n,k,\ell}^{Y}(s,t) = \mathbb{E}\left(Y_{k}^{3}X_{\ell}^{3}\left\{f_{n-\ell-1}^{(6)}(S_{\ell-1} + tX_{\ell} + sY_{k}) - f_{n-\ell-1}^{(6)}(S_{\ell-1} + sY_{k})\right\}\right),$$

$$D_{n,k,\ell}^{Y}(s,t) = \mathbb{E}\left(Y_{k}^{3}Y_{\ell}^{3}\left\{f_{n-\ell-1}^{(6)}(S_{\ell-1} + tY_{\ell} + sY_{k}) - f_{n-\ell-1}^{(6)}(S_{\ell-1} + sY_{k})\right\}\right).$$

Taking into account the previous computations, we infer that

$$|C_{n,k,\ell}^{Y}(s,t)| \le \mathbb{E}(|Y|^3) \mathbb{E}\left(|X|^3 \min\left(2\|f_{n-\ell-1}^{(6)}\|_{\infty}, t|X||f_{n-\ell-1}^{(7)}\|_{\infty}\right)\right).$$

Therefore

$$\int_{0}^{1} \int_{0}^{1} \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} \sum_{k=2}^{n} \sum_{\ell=1}^{k-1} |C_{n,k,\ell}^{Y}(s,t)| ds dt \le \frac{\sqrt{2}c_5 \mathbb{E}(|Y|^3)}{27\sqrt{2v^2-1}} \mathbb{E}(|X|^3(|X| \wedge 8x)).$$

On another hand

$$|D_{n,k,\ell}^{Y}(s,t)| \le t c_5 \mathbb{E}(|Y|^3) \mathbb{E}(Y^4) u_{n,\ell}^{-5/2},$$

implying that

$$\int_{0}^{1} \int_{0}^{1} \frac{(1-s)^2}{2} \frac{(1-t)^2}{2} \sum_{k=2}^{n} \sum_{\ell=1}^{k-1} |D_{n,k,\ell}^{Y}(s,t)| ds dt \le \frac{\sqrt{2}c_5 \mathbb{E}(|Y|^3)}{27\sqrt{2v^2-1}} (3 + 2\beta_3^2).$$

So, overall,

$$|B_2(n)| \le \frac{\sqrt{2}c_5\mathbb{E}(|Y|^3)}{27\sqrt{2v^2-1}} (3 + 2\beta_3^2 + \mathbb{E}(|X|^3(|X| \wedge 8x))). \quad (3.22)$$

Starting from (3.15) and considering the upper bounds (3.14), (3.21) and (3.22), it follows that

$$|\Delta_{n,v}(f)| \le \left( \frac{\sqrt{2}c_5\mathbb{E}(|X|^3 + |Y|^3)}{27\sqrt{2v^2-1}} + \frac{c_2}{12} \right) \left( 3 + 2\beta_3^2 + \mathbb{E}(|X|^3(|X| \wedge 8x)) \right). \quad (3.23)$$

In order to minimize the numerical constants, we now set

$$v^2 = 0.50 + 0.15 (\mathbb{E}(|X|^3) + \mathbb{E}(|Y|^3))^2. \quad (3.24)$$

This choice, combined with (3.9), (3.10), (3.23) leads to the following proposition:
**Proposition 3.2.** Let

$$B(n, x) := 0.50 + 0.15(\mathbb{E}(|X|^3) + \mathbb{E}(|Y|^3))^2 + \alpha_1(3 + 2\beta_3^2 + \mathbb{E}(|X|^3(|X| \wedge 8x))),$$
with $\alpha_1 = \sqrt{20/3}(c_5/27) + (c_2/12) \le \gamma_2 := 0.64584$. Then

$$\sup_{f \in \mathcal{F}_x} |\mathbb{E}(f(S_n)) - f(T_n)| \le B(n, x).$$

In order to complete the proof of Theorem 2.1, we now give an upper bound on $B(n, x)$ depending more explicitly on the law of $X$. First

$$(\mathbb{E}(|X|^3) + \mathbb{E}(|Y|^3))^2 \le 4(\mathbb{E}(|X|^3))^2 + (4/3)(\mathbb{E}(|Y|^3))^2,$$
and second, since $\mathbb{E}(Y^2) = 1$, $(\mathbb{E}(|Y|^3))^2 \le \mathbb{E}(Y^4) = 3 + 2\beta_3^2$. Hence

$$B(n, x) \le 1.1 + 0.6(\mathbb{E}(|X|^3))^2 + 0.4\beta_3^2 + \alpha_1(3 + 2\beta_3^2 + \mathbb{E}(|X|^3(|X| \wedge 8x))). \quad (3.25)$$

When $\sigma^2 = 1$ and $\beta_3 \ne 0$, Theorem 2.1 follows by considering (3.6) together with Proposition 3.2, (3.25) and inequality (3.8).

Assume now that $\beta_3 = 0$. In this case, first we add an independent Gaussian r.v. $Z$ centered and with variance $v^2$, and then we take $(Y_k)_{k \ge 1}$ iid standard Gaussian random variables that are independent of $(X_k)_{k \ge 1}$. We proceed as for the proof of the case $\beta_3 \ne 0$, but with $u_n = n - 1 + v^2$ and $u_{n,l} = n - l - 1 + v^2$. So, overall, we infer that when $\sigma^2 = 1$ and $\beta_3 = 0$,

$$\kappa_{\psi_x}(P_{S_n}, P_{G_n}) \le \tilde{\gamma}_1 + 0.6(\mathbb{E}(|X^3|))^2 + \tilde{\gamma}_2 \mathbb{E}(|X|^3(|X| \wedge 8x)),$$

where $\tilde{\gamma}_1 \le \gamma_1$ and $\tilde{\gamma}_2 \le \gamma_2$. This ends the proof of the theorem. $\square$

**3.3 Proof of Remark 2.2**

Since $\varphi'$ is concave and then absolutely continuous, there exists a positive measure $\nu$ on $\mathbb{R}_+$ such that $\varphi''(0) - \varphi''(z) = \nu([0, z])$ almost surely. Since $\varphi''(0) = 1$ and $\varphi'(0) = 0$, $x - \varphi'(x) = \int_0^x \nu([0, t])dt$. It follows that

$$1 - \lim_{x \to \infty} x^{-1} \varphi'(x) = \lim_{x \to \infty} x^{-1} \int_{0}^{x} \nu([0, t]) dt.$$

Since $\lim_{x \to \infty} x^{-1} \varphi'(x) = 0$, it follows that $\nu$ is a probability measure. Hence

$$\varphi'(x) = \int_{0}^{x} \nu([y, +\infty]) dy = \int_{0}^{\infty} (x \wedge y) d\nu(y) = \frac{1}{2} \int_{0}^{\infty} \psi'_{y}(2x) d\nu(y), \quad (3.26)$$

where $\psi_y$ is defined in (2.1). Since $\varphi'(0) = 0$, it implies that

$$\varphi(z) = \frac{1}{2} \int_{\mathbb{R}_+} \psi_t(2z) d\nu(t) = 2 \int_{\mathbb{R}_+} \psi_{t/2}(z) d\nu(t),$$

where the second equality comes from (3.3). $\square$
## 3.4 Proof of Theorem 2.2

By Lemma 3.2,

$$W_{\varphi}(P_{S_n}, P_{G_{n\sigma^2}}) \le W_{\varphi}(P_{S_n}, P_{\sigma T_n}) + W_{\varphi}(P_{\sigma T_n}, P_{G_{n\sigma^2}}), \quad (3.27)$$

where $T_n$ is defined as in the proof of Theorem 2.1 but with $\beta_3 = \mathbb{E}(X^3)/\sigma^3$ in (3.5).

Since $\varphi$ is in $\Psi$, $\varphi(x) \le x^2/2$. Whence, taking into account (3.7),

$$W_{\varphi}^2(P_{\sigma T_n}, P_{G_{n\sigma^2}}) \le \frac{\sigma^2}{2} W_2^2(P_{T_n}, P_{G_n}) \le 1.874 \frac{\mu_3^2}{\sigma^4} \le (1.369\sigma^{-2}\mu_3)^2. \quad (3.28)$$

We now handle the quantity $W_{\varphi}(P_{S_n}, P_{\sigma T_n})$. Taking into account Remark 2.2,

$$\kappa_{\varphi}(P_{S_n}, P_{\sigma T_n}) = 2 \int_{\mathbb{R}_+} \kappa_{\psi_{t/2}}(P_{S_n}, P_{\sigma T_n}) d\nu(t).$$

Applying inequality (3.4) with $a = \sigma$, we get that

$$\kappa_{\psi_{t/2}}(P_{S_n}, P_{\sigma T_n}) = \sigma^2 \kappa_{\psi_{t/(2\sigma)}}(P_{\sigma^{-1}S_n}, P_{T_n}).$$

Hence

$$\kappa_{\varphi}(P_{\sigma^{-1}S_n}, P_{\sigma T_n}) = 2\sigma^2 \int_{\mathbb{R}_+} \kappa_{\psi_{t/(2\sigma)}}(P_{\sigma^{-1}S_n}, P_{T_n}) d\nu(t).$$

Now, using Proposition 3.2 and (3.25),

$$\kappa_{\varphi}(P_{S_n}, P_{\sigma T_n}) \le 2C_0(X) + 8\gamma_2\sigma^{-2} \int_{\mathbb{R}_+} \mathbb{E}(|X|^3 \min(|X|/4, t)) d\nu(t). \quad (3.29)$$

Recall that $\varphi'(|z|) = \int_{\mathbb{R}_+} (|z| \wedge t) d\nu(t)$ (see (3.26)). Hence

$$\int_{\mathbb{R}_+} \mathbb{E}(|X|^3 \min(|X|/4, t)) d\nu(t) \le \mathbb{E}(|X|^3 \varphi'(|X|/4)). \quad (3.30)$$

Starting from (3.27), the theorem follows by taking into account (3.28), (3.29) and (3.30). $\square$

## 3.5 Proof of Remark 2.3

As quoted in the proof of Lemma 3.2, $x \mapsto x^{-2}\varphi(x)$ is nonincreasing on $]0, +\infty[$. Hence the derivative of $x \mapsto x^{-2}\varphi(x)$ is nonpositive which ensures that $\varphi'(x) \le 2x^{-1}\varphi(x)$. Therefrom

$$\mathbb{E}(|X|^3 \varphi'(|X|/4)) \le 8\mathbb{E}(|X|^2 \varphi(|X|/4)). \quad \square$$
## 3.6 Proof of Proposition 2.2

We first notice that

$$W_p(P_{S_n}, P_{G_{n\sigma^2}}) = \sigma W_p(P_{S_n/\sigma}, P_{G_n}) \le \sigma W_p(P_{S_n/\sigma}, P_{T_n}) + \sigma W_p(P_{T_n}, P_{G_n}), \quad (3.31)$$

where $T_n$ has been defined in the proof of Theorem 2.1. Using the fact that $p \in ]1, 2[$ and the upper bound (3.7), we get

$$\sigma W_p(P_{T_n}, P_{G_n}) \le \sigma W_2(P_{T_n}, P_{G_n}) \le 1.936\sigma^{-2}|\mu_3|. \quad (3.32)$$

In order to bound up the first term on right hand in (3.31), we define the rescaled functions $g_{p,a}$ from the function $g_p$ defined in (2.2) by $g_{p,a}(x) = a^2 g_p(x/a)$ for any $a > 0$. Note that the functions $g_{p,a}$ are in the class $\Psi$. In addition, using inequality (2.3), we get that

$$x^p \le a^{p-2} p g_{p,a}(x) + a^p (1 - p/2). \quad (3.33)$$

Taking into account the upper bounds (3.29) and (3.30), it follows that

$$\kappa_p(P_{S_n/\sigma}, P_{T_n}) \le (1-p/2)a^p + 2pa^{p-2}C_0\left(\frac{X}{\sigma}\right) + 8\gamma_2 pa^{p-2}\mathbb{E}\left(\left|\frac{X}{\sigma}\right|^3 g'_{p,a}\left(\frac{|X|}{4\sigma}\right)\right). \quad (3.34)$$

Next, since $g'_p(z) = \min(z, z^{p-1}) \le z^{p-1}$ for any $z \ge 0$, we have that

$$g'_{p,a}(z) = a g'_p(z/a) \le a^{2-p} z^{p-1}$$

for any $z \ge 0$. Hence

$$\kappa_p(P_{S_n/\sigma}, P_{T_n}) \le (1-p/2)a^p + 2pa^{p-2}C_0(X/\sigma) + 8p4^{1-p}\gamma_2\mathbb{E}(|X/\sigma|^{2+p}). \quad (3.35)$$

In order to minimize the above upper bound, we then choose $a = \sqrt{4C_0(X/\sigma)}$, which gives

$$\kappa_p(P_{S_n/\sigma}, P_{T_n}) \le 2^p (C_0(X/\sigma))^{p/2} + p 2^{5-2p} \gamma_2 \mathbb{E}(|X/\sigma|^{2+p}). \quad (3.36)$$

Starting from (3.31) and considering the upper bounds (3.32) and (3.36), the proposition follows by noticing that $\sigma^2 C_0(X/\sigma) = C_0(X)$. $\square$

## 3.7 Proof of Corollaries 2.1 and 2.2

In this subsection, we prove Corollary 2.1 and item (a) of Corollary 2.2. The proof of item (b) of Corollary 2.2, being similar, will be omitted. We start by proving the following lemma.

**Lemma 3.3.** Let $(Z_n)_{n \ge 1}$ be defined in (2.9). Under the conditions of Corollary 2.1, $(\varphi(Z_n))_{n \ge 1}$ is uniformly integrable.
**Proof of Lemma 3.3.** We have to prove that

$$ \lim_{M \to \infty} \sup_{n \ge 1} \mathbb{E}(\varphi(Z_n) \mathbf{1}_{Z_n \ge M}) = 0. \qquad (3.37) $$

Let $M > 0$ and define the function $\psi_M$ on $\mathbb{R}$ by the following conditions: $\psi_M$ is even, $\psi_M(0) = 0$ and its derivative on $\mathbb{R}_+$ is defined by

$$ \psi'_{M}(x) = \begin{cases} x & \text{if } x \le M, \\ \frac{M}{\varphi'(M)} \varphi'(x) & \text{if } x \ge M. \end{cases} $$

Clearly $\psi_M$ belongs to the class $\Psi$. Now applying Theorem 2.2 to the function $\psi_M$, it follows that there exists a universal constant $C$ such that for any $n \ge 1$,

$$ \mathbb{E}(\psi_M(Z_n)) \le C \left( \mathbb{E}(|X|^3) + \mathbb{E}(|X|^3 \psi'_M(|X|)) \right). \qquad (3.38) $$

On another hand, since $\varphi$ is convex and $\varphi(0) = 0$, $\varphi(x) \le x\varphi'(x)$ for $x \ge 0$. This implies that

$$ \mathbb{E}(\varphi(Z_n) \mathbf{1}_{Z_n \ge M}) \le \mathbb{E}(Z_n \varphi'(Z_n) \mathbf{1}_{Z_n \ge M}) \le \frac{\varphi'(M)}{M} \mathbb{E}(Z_n \psi'_M(Z_n)). $$

Now, as quoted in Section 3.5, since $\psi_M$ belongs to the class $\Psi$, $x\psi'_M(x) \le 2\psi_M(x)$. Hence

$$ \mathbb{E}(\varphi(Z_n) \mathbf{1}_{Z_n \ge M}) \le 2 \frac{\varphi'(M)}{M} \mathbb{E}(\psi_M(Z_n)). \qquad (3.39) $$

Combining (3.38) and (3.39), it follows that

$$ \mathbb{E}(\varphi(Z_n) \mathbf{1}_{Z_n \ge M}) \le 2C \left( \frac{\varphi'(M)}{M} \mathbb{E}(|X|^3) + \frac{\varphi'(M)}{M} \mathbb{E}(|X|^3 \psi'_M(|X|)) \right). \qquad (3.40) $$

Note that

$$ \frac{\varphi'(M)}{M} |X|^3 \psi'_M(|X|) = \frac{\varphi'(M)}{M} |X|^4 \mathbf{1}_{|X| \le M} + |X|^3 \varphi'(|X|) \mathbf{1}_{|X| > M}. $$

Now, since $x \mapsto x^{-1}\varphi'(x)$ is nonincreasing,

$$ \frac{\varphi'(M)}{M} |X|^4 \mathbf{1}_{|X| \le M} + |X|^3 \varphi'(|X|) \mathbf{1}_{|X| > M} \le |X|^3 \varphi'(|X|). $$

Next, since $\lim_{M \to +\infty} M^{-1}\varphi'(M) = 0$, the term of the left hand side of the above inequality tends to 0 as $M \to +\infty$. Hence, since $\mathbb{E}(|X|^3\varphi'(|X|)) < \infty$, by the dominated convergence theorem, we get that

$$ \lim_{M \to +\infty} \frac{\varphi'(M)}{M} \mathbb{E}(|X|^3 \psi'_M(|X|)) = 0. \qquad (3.41) $$

Starting from (3.40) and considering (3.41), the convergence (3.37) follows. $\square$
**Proof of Item (a) of Corollary 2.1.** Let $G = n^{-1/2}G_n$ and $U = \Phi(G)$ where $\Phi$ is the c.d.f. of a standard normal. Then $U$ has the uniform distribution over $[0, 1]$ and

$$Z_n = |F_n^{-1}(U) - \sqrt{n}\Phi^{-1}(U)|.$$

Set

$$Z = \left| \frac{\mu_3}{6} \left( (\Phi^{-1}(U))^2 - 1 \right) \right|. \qquad (3.42)$$

According to the Cornish-Fisher expansion (see for instance Lemma 2.1 (a) in Rio [21] or [10, Theorem 2.4] and final comments to Chapter 2), we get that

$$Z_n \to^{\mathcal{L}} Z, \text{ as } n \to \infty,$$

where $\to^{\mathcal{L}}$ means convergence in distribution. Since $\varphi$ is a continuous function, by the continuous mapping theorem, we also have that $\varphi(Z_n) \to^{\mathcal{L}} \varphi(Z)$, as $n \to \infty$. Together with Lemma 3.3 and the convergence of moments Theorem (see Theorem 3.5 in [2]), this convergence in distribution implies that $\mathbb{E}(\varphi(Z_n)) \to \mathbb{E}(\varphi(Z))$, as $n \to \infty$. This ends the proof of Item (a) of Corollary 2.1.

**Proof of Item (b) of Corollary 2.1.** Let

$$\bar{Z}_n = |\bar{F}_n^{-1}(U) - \sqrt{n}\Phi^{-1}(U)|$$

where $\bar{F}_n$ is the cdf of $\bar{S}_n = S_n+hV$. Recalling the notation (3.42), we have $\bar{Z}_n \to^{\mathcal{L}} Z$, as $n \to \infty$ (see for instance Lemma 2.1 (b) in Rio [21]). Whence, following the proof of Item (a), it is enough to prove the uniform integrability of $(\varphi(\bar{Z}_n))_{n \ge 1}$. With this aim, we first notice that for any $n \ge 1$,

$$\bar{Z}_n \le Z_n + |\bar{F}_n^{-1}(U) - F_n^{-1}(U)|.$$

Next, since $V$ is a random variable with uniform distribution over $[-1/2, 1/2]$ and $\bar{S}_n = S_n + hV$,

$$S_n - h/2 \le \bar{S}_n \le S_n + h/2 \text{ almost surely,}$$

which implies that $F_n^{-1}(u) - (h/2) \le \bar{F}_n^{-1}(u) \le F_n^{-1}(u) + (h/2)$ for any $u$ in $]0, 1[$. Hence $|\bar{F}_n^{-1}(U) - F_n^{-1}(U)| \le h/2$ almost surely, which ensures that

$$\bar{Z}_n \le Z_n + h/2 \text{ almost surely.} \qquad (3.43)$$

Now, using both the above inequality and (3.2), $\varphi(\bar{Z}_n) \le 2\varphi(Z_n) + 2\varphi(h/2)$ almost surely. Since $(\varphi(Z_n))_{n \ge 1}$ is uniformy integrable, this proves the uniform integrability of $(\varphi(\bar{Z}_n))_{n \ge 1}$ and ends the proof of Item (b) of Corollary 2.1.

**Proof of Item (a) of Corollary 2.2.** By Lemma 2.1(a) in Rio [21], for any positive $\varepsilon$, the sequence of functions $(F_n^{-1} - \sqrt{n}\Phi^{-1})_{n \ge 1}$ converges uniformly to
$\mu_3((\Phi^{-1})^2 - 1)/6$ on $[\varepsilon, 1 - \varepsilon]$. This implies that the above sequence of functions is uniformly bounded over $[\varepsilon, 1 - \varepsilon]$. Next $g(U)$ belongs to $L_{\varphi}^*$ and $L_{\varphi^*} \subset L^1$, which ensures the $g$ is integrable over $[0, 1]$. Hence, by the dominated convergence theorem,

$$ \lim_{n \to \infty} \int_{\varepsilon}^{1-\varepsilon} g(u) |F_n^{-1}(u) - \sqrt{n}\Phi^{-1}(u)| du = \frac{|\mu_3|}{6} \int_{\varepsilon}^{1-\varepsilon} g(u) |(\Phi^{-1}(u))^2 - 1| du. \quad (3.44) $$

It remains to prove that

$$ \limsup_{\varepsilon \to 0} \int_{[0,1[ \setminus [\varepsilon, 1-\varepsilon]} g(u) |F_n^{-1}(u) - \sqrt{n}\Phi^{-1}(u)| du = 0. \quad (3.45) $$

Let $a$ be a positive real such that $E(\varphi^*(ag(U))) < \infty$. By the Young inequality $xy \le \varphi^*(x) + \varphi(y)$ applied to $x = ag(U)$ and $y = |F_n^{-1}(u) - \sqrt{n}\Phi^{-1}(u)|$,

$$ a \int_{[0,1[ \setminus [\varepsilon, 1-\varepsilon]} g(u) |F_n^{-1}(u) - \sqrt{n}\Phi^{-1}(u)| du \le I(\varepsilon) + J_n(\varepsilon), $$

with

$$ I(\varepsilon) = \int_{[0,1[ \setminus [\varepsilon, 1-\varepsilon]} \varphi^*(ag(u))du \quad \text{and} \quad J_n(\varepsilon) = \int_{[0,1[ \setminus [\varepsilon, 1-\varepsilon]} \varphi(F_n^{-1}(u) - \sqrt{n}\Phi^{-1}(u))du. $$

Now, Lemma 3.3 ensures that $\lim_{\varepsilon \to 0} \sup_{n \ge 1} J_n(\varepsilon) = 0$. Next, since $u \mapsto \varphi^*(ag(u))$ is integrable over $]0, 1[$, $\lim_{\varepsilon \to 0} I(\varepsilon) = 0$, which completes the proof of (3.45). Now (3.44) and (3.45) imply Item (a) of Corollary 2.2. $\square$

## 3.8 Proof of Theorem 2.3

We start with the following lemma, which is of independent interest. The proof of this lemma is postponed to Appendix.

**Lemma 3.4.** Let A and B be two real-valued random variables. Then for any real numbers $x$ and $t$, $\tilde{H}_{A+B}(x) \le \max(\tilde{H}_A(t), \tilde{H}_B(x-t))$.

**Remark 3.1.** The classical inequality $P(A+B \ge x) \le P(A \ge t) + P(B > x-t)$ for usual tail functions cannot be improved, according to Proposition 1 in Pinelis [18]. The above lemma proves that for sums of r.v.’s. $\tilde{H}$ has a better behavior than the tail function.

We go back to the proof of Theorem 2.3. We first note that

$$ \tilde{H}_{Z_n}(t) = \tilde{H}_{Z_n/\sigma}(t/\sigma). $$

We shall apply Lemma 3.4 with $A = |\sigma^{-1}S_n - T_n|$ and $B = |T_n - G_n|$ where $T_n$ be the random variable defined in the proof of Theorem 2.1. For any $\alpha \in ]0, 1[$ and any positive $t$, we then get

$$ \tilde{H}_{Z_n}(t) \le \max(\tilde{H}_X(|\sigma^{-1}S_n - T_n|)(\alpha t/\sigma), \tilde{H}_{|T_n - G_n|}((1 - \alpha)t/\sigma)). \quad (3.46) $$
Next, for any $s > 0$, note that $(x - s)_+ \le s^{-1}\psi_s(x)$ where $\psi_s$ is defined in (2.1).
Hence, for any real-valued random variable $Z$ and any positive real $u$,

$$ \tilde{H}_{|Z|}(u) \le \inf_{s<u} \frac{1}{(u-s)s} \mathbb{E}(\psi_s(|Z|)) \le \frac{4}{u^2} \mathbb{E}(\psi_{u/2}(|Z|)). \quad (3.47) $$

On one hand, taking into account (3.47) and (3.7), we get

$$ \tilde{H}_{|T_n-G_n|} \left( \frac{(1-\alpha)t}{\sigma} \right) \le \frac{4\sigma^2}{(1-\alpha)^2 t^2} \kappa_{\psi_{u/2}}(P_{T_n}, P_{G_n}) \le \frac{4 \times 0.937 \lambda_3^2}{(1-\alpha)^2 t^2 \sigma^4}. \quad (3.48) $$

On another hand, taking into account (3.47) together with Proposition (3.2), the upper bound (3.25) and the fact that $|\mathbb{E}(X^3)| \le \lambda_3$ and $\sigma^3 \le \lambda_3$, we infer that

$$ \tilde{H}_{|\sigma^{-1}S_n - T_n|} \left( \frac{\alpha t}{\sigma} \right) \le \frac{4}{\alpha^2 t^2} \left( \left( \gamma_1 + \frac{3}{5} + \gamma_3 \right) \frac{\lambda_3^2}{\sigma^4} + \frac{\gamma_2}{\sigma^2} \mathbb{E}(|X|^3 (|X| \wedge 4\alpha t)) \right). \quad (3.49) $$

In the inequality above the constants $\gamma_1$, $\gamma_2$ and $\gamma_3$ are those involved in the statement of Theorem 2.1. In particular the constant $a := \gamma_1 + 3/5 + \gamma_3$ can be chosen equal to 5.3294. We now choose $\alpha = \tilde{\alpha}$ such that

$$ \frac{a}{\tilde{\alpha}^2 t^2} \ge \frac{0.937}{(1 - \tilde{\alpha})^2 t^2}. \quad (3.50) $$

Numerical computation gives that $\tilde{\alpha} = 0.7045699$ satisfies (3.50). Then starting from (3.46) and taking into account (3.48) and (3.49), we derive that, for any positive real $t$,

$$ \tilde{H}_{Z_n}(t) \le \frac{4a}{\tilde{\alpha}^2 t^2} \frac{\lambda_3^2}{\sigma^4} + \frac{4\gamma_2}{\tilde{\alpha}^2 \sigma^2} \mathbb{E}(|X|^3 (|X| \wedge 4\tilde{\alpha}t)). $$

This ends the proof by taking into account the values of $a$, $\tilde{\alpha}$ and $\gamma_2$. $\square$

## 3.9 Proof of Corollary 2.3

According to Theorem 2.3 and the fact that $\tilde{H}_{Z_n}(t) \le 1$, we have

$$ t^p \tilde{H}_{Z_n}(t) \le t^p \min(1, 42.943 \sigma^{-4} \lambda_3^2 t^{-2}) + 5.2041 t^{p-2} \sigma^{-2} \mathbb{E}(|X|^3 \min(|X|, 2.8183t)) $$

since $\min(a+b, 1) \le \min(a, 1) + b$ for any nonnegative reals $a$ and $b$. Next, for $p \in ]1, 2[$, note that

$$ \mathbb{E}(|X|^3 \min(|X|, u)) \le (p+2)((2-p)(p-1))^{-1} \Lambda_{p+2}(X) u^{2-p}. \quad (3.51) $$

Whence

$$ t^p \tilde{H}_{Z_n}(t) \le t^p \min(1, 42.943 \sigma^{-4} \lambda_3^2 t^{-2}) + a_2(p+2)((2-p)(p-1))^{-1} \sigma^{-2} \Lambda_{p+2}(X). $$

The supremum over $t > 0$ is reached at $t_0$ such that $1 = 42.943 \sigma^{-4} \lambda_3^2 t_0^{-2}$. Numerical computation shows that $t_0 \le 6.5531 \sigma^{-2} \lambda_3$. This proves the first part of the corollary. When the random variables have strong moment of order $p+2$, we use the fact that

$$ \mathbb{E}(|X|^3 \min(|X|, 2.8183t)) \le (2.8183t)^{2-p} \mathbb{E}(|X|^{p+2}), $$

and the same arguments as before. $\square$
## 3.10 Proof of Corollary 2.4

The proof is based on the following lemma.

**Lemma 3.5.** Under the conditions of Corollary 2.4 and if the distribution of X is not a lattice distribution, then:

(a) for any $\varepsilon > 0$, the sequence of functions $(Q_{Z_n})_{n \ge 1}$ converges uniformly to $Q_{|G^2-1|} \times \frac{|\mathbb{E}(X^3)|}{6}$ over $[\varepsilon, 1]$.

(b) $\lim_{u \to 0} \sup_{n \ge 1} u^{1/p} \tilde{Q}_{Z_n}(u) = 0$.

**Proof of Lemma 3.5.** We start with the proof of Item (a). Recall that by the Cornish-Fisher expansion, $Z_n \to^L Z$ as $n \to \infty$, where $Z = |G^2 - 1| \times \frac{|\mathbb{E}(X^3)|}{6}$. Since $Z$ is a continuous r.v., it follows that, for any $u \in [0, 1]$, $Q_{Z_n}(u) \to Q_Z(u) = Q_{|G^2-1|}(u) \times \frac{|\mathbb{E}(X^3)|}{6}$, as $n \to \infty$ (see for instance Lemma 21.2 in [24]). Note now that for any positive integer $n$, $u \mapsto Q_{Z_n}(u)$ is nonincreasing. Whence, by the second Dini's Theorem, the convergence is also uniform over all intervals of the form $[\varepsilon, 1]$ for any $\varepsilon > 0$ and Item (a) is proved.

We turn now to Item (b). We first prove that

$$ \lim_{t \to \infty} t^p \sup_{n \ge 1} \tilde{H}_{Z_n}(t) = 0. \qquad (3.52) $$

By Theorem 2.3, there exists a positive constant $C$ such that for any positive $t$,

$$ t^2 \sup_{n \ge 1} \tilde{H}_{Z_n}(t) \le C \left( 1 + \mathbb{E}(|X|^3 \min(|X|, t)) \right). $$

Next

$$ \lim_{t \to \infty} t^{p-2} \mathbb{E}(|X|^3 \min(|X|, t)) = 0. $$

For $p \in [1, 2[$, this follows from the fact that $H_{|X|}(x) = o(x^{-p-2})$, and for $p = 1$ this follows from the dominated convergence theorem and the fact that $\mathbb{E}(|X|^3) < \infty$. Now the above convergence implies (3.52). Next, for all $\eta > 0$ there exists $A > 0$ such that for any $n \ge 1$ and any $x \ge A$, $\tilde{H}_{Z_n}(x) < \eta^p x^{-p}$. Hence, according to inequality (2.11), $\tilde{Q}_{Z_n}(\eta^p x^{-p}) < x$ for any $x \ge A$. This implies that $\eta x^{-1} \tilde{Q}_{Z_n}(\eta^p x^{-p}) < \eta$ for any $x \ge A$. It follows that for any $u < \eta^p A^{-p}$, $u^{1/p} \tilde{Q}_{Z_n}(u) < \eta$, for any $n \ge 1$. This ends the proof of Item (b) of Lemma 3.5. $\square$

We now return to the proof of Corollary 2.4. Let us start by Item (a). According to (2.12) and the fact that $H_Z(x) < u$ iff $x > \tilde{Q}_Z(u)$, we have $\Lambda_p(Z_n) = \sup_{u \in [0,1]} u Q_{Z_n}^p(u)$. By Item (a) of Lemma 3.5, for any $\varepsilon > 0$,

$$ \lim_{n \to \infty} \sup_{u \in [\varepsilon, 1]} u^{1/p} Q_{Z_n}(u) = \frac{|\mathbb{E}(X^3)|}{6} \sup_{u \in [\varepsilon, 1]} u^{1/p} Q_{|G^2-1|}(u). $$
Moreover, by Item (b) of Lemma 3.5,

$$ \limsup_{\epsilon \to 0} \sup_{n \ge 1} u^{1/p} Q_{Z_n}(u) = 0 \quad \text{and} \quad \limsup_{\epsilon \to 0} \sup_{u \in [0, \epsilon]} u^{1/p} Q_{|G^2-1|}(u) = 0. $$

This ends the proof of the first part of Item (a) of Corollary 2.4.

We turn now to the proof of the second part of Item (a) of Corollary 2.4. With this aim, we first prove that Item (a) of Lemma 3.5 also holds if one replaces $Q_{Z_n}$ by $\tilde{Q}_{Z_n}$ and $Q_{|G^2-1|}$ by $\tilde{Q}_{|G^2-1|}$. First, Item (a) of Lemma 3.5 implies the point-wise convergence of $Q_{Z_n}$ to $Q_{|G^2-1|} \times \frac{|\mathbb{E}(X^3)|}{6}$ over $]0, 1]$. Next, by Items (a) and (b) of Lemma 3.5, there exists a positive constant $C$, such that for any $u \in ]0, 1]$, $\sup_{n \ge 1} Q_{Z_n}(u) \le Cu^{-1/p}$. Hence, by the dominated convergence theorem, for any $u \in [0, 1]$,

$$ \lim_{n \to \infty} u \tilde{Q}_{Z_n}(u) = \lim_{n \to \infty} \int_0^u Q_{Z_n}(v) dv = u \tilde{Q}_{|G^2-1|}(u) \times \frac{|\mathbb{E}(X^3)|}{6}. $$

Since, for any $n$, $u \to u \tilde{Q}_{Z_n}(u)$ is nondecreasing, by the second Dini's theorem, the above convergence is uniform over $[0, 1]$. It follows that, for any $\epsilon > 0$, $\tilde{Q}_{Z_n}$ converges uniformly to $\tilde{Q}_{|G^2-1|} \times \frac{|\mathbb{E}(X^3)|}{6}$ over $[\epsilon, 1]$, which implies that

$$ \limsup_{n \to \infty} \sup_{u \in [\epsilon, 1]} u^{1/p} \tilde{Q}_{Z_n}(u) = \frac{|\mathbb{E}(X^3)|}{6} \sup_{u \in [\epsilon, 1]} u^{1/p} \tilde{Q}_{|G^2-1|}(u). $$

Finally

$$ \limsup_{\epsilon \to 0} \sup_{n \ge 1} u^{1/p} \tilde{Q}_{Z_n}(u) = 0 \quad \text{and} \quad \limsup_{\epsilon \to 0} \sup_{u \in [0, \epsilon]} u^{1/p} \tilde{Q}_{|G^2-1|}(u) = 0. $$

This ends the proof of the second part of Item (a) of Corollary 2.4.

The proof of Item (b) of Corollary 2.4 follows the same path as that for Item (a). Using the fact that $\bar{Z}_n \to^L Z$, as $n \to \infty$ (see for instance Lemma 2.1 (b) in Rio [21]), one obtains that Item (a) of Lemma 3.5 also holds for the sequence $(\tilde{Q}_{\bar{Z}_n})_{n \ge 1}$. Next, from Inequality (3.43), $\tilde{Q}_{\bar{Z}_n} \le \tilde{Q}_{Z_n} + (h/2)$, which ensures that Item (b) of Lemma 3.5 also holds for the sequence $(\tilde{Q}_{\bar{Z}_n})_{n \ge 1}$. Now the end of the proof of Item (b) of Corollary 2.4 is exactly the same as for Item (a). $\square$

# 4 Appendix

## 4.1 Proof of Proposition 3.1

By homogeneity, it suffices to prove the result for $\alpha = 1$. For any $m > 0$, note that

$$ \Pi(m) - m = \sum_{\ell \ge 0} (2\Pi(2^{\ell}m) - \Pi(2^{\ell+1}m)) 2^{-\ell-1} \quad \text{in } \mathbb{L}^2. \quad (4.1) $$
Indeed, for any positive integer $N$,

$$ \sum_{\ell=0}^{N} (2\Pi(2^{\ell}m) - \Pi(2^{\ell+1}m)) 2^{-\ell-1} = \Pi(m) - 2^{-N-1}\Pi(2^{N+1}m) $$

and $\|2^{-N-1}\Pi(2^{N+1}m) - m\|_2^2 = 2^{-N-1}m$, which converges to zero as $N \to \infty$.

Let $\tilde{U}_{\ell+1,m} := 2\Pi(2^{\ell}m) - \Pi(2^{\ell+1}m)$ and $U_{\ell+1,m} := \Pi(2^{\ell+1}m)$. Let $\Phi$ be the distribution function of a standard real-valued Gaussian random variable and, for any $\ell \in \mathbb{N}$, let $\tilde{F}_{\ell+1,m}$ be the distribution function of the conditional law of $\tilde{U}_{\ell+1,m}$ given $U_{\ell+1,m}$. Let $(\delta_{\ell})_{\ell \ge 0}$ be a sequence of i.i.d. random variables with uniform distribution on $[0, 1]$, independent of the Poisson process $(\Pi(t))_{t \ge 0}$. For any $\ell \ge 0$, let

$$ \xi_{\ell,m} = \Phi^{-1} \left( \tilde{F}_{\ell+1,m}(\tilde{U}_{\ell+1,m} - 0) + \delta_{\ell} \left( \tilde{F}_{\ell+1,m}(\tilde{U}_{\ell+1,m}) - \tilde{F}_{\ell+1,m}(\tilde{U}_{\ell+1,m} - 0) \right) \right). $$

By the properties of the conditional quantile transform, $\xi_{\ell,m}$ is independent of $U_{\ell+1,m}$, $\sigma(\delta_{\ell}, \tilde{U}_{\ell+1,m}, U_{\ell+1,m})$-measurable and $\mathcal{N}(0, 1)$ distributed (for more details see Lemma F1 in [23]). Setting $\mathcal{G}_{\ell+1} = \sigma((U_{\ell+k,m}, \delta_{\ell+k}), k \ge 1)$, we note that $\xi_{\ell,m}$ is independent of $\mathcal{G}_{\ell+1}$. By induction, it follows that the random variables $(\xi_{\ell,m})_{\ell \ge 0}$ are independent and $\mathcal{N}(0, 1)$ distributed.

Next, let

$$ W(m) := \sum_{\ell \ge 0} \xi_{\ell,m} \sqrt{2^{\ell+1}m} 2^{-\ell-1}. $$

From the above facts, $W(m)$ is $\mathcal{N}(0, m)$ distributed.

Next, for any $\ell \ge 0$, let $V_{\ell,m} = \xi_{\ell,m}\sqrt{2^{\ell+1}m} - \tilde{U}_{\ell+1,m}$. Notice that $\mathcal{L}(\tilde{U}_{\ell+1,m}|\mathcal{G}_{\ell+1})$ is the law of $2B - U_{\ell+1,m}$ with $B \sim \mathcal{B}(U_{\ell+1,m}, 1/2)$. Hence $\mathbb{E}(\tilde{U}_{\ell+1,m}|\mathcal{G}_{\ell+1}) = 0$ implying that the random variables $(V_{\ell,m})_{\ell \ge 0}$ are orthogonal in $\mathbb{L}^2$. Therefore it follows that

$$ \|\Pi(m) - m - W(m)\|_2^2 = \left\| \sum_{\ell \ge 0} V_{\ell,m} 2^{-\ell-1} \right\|_2^2 = \sum_{\ell \ge 0} 2^{-2(\ell+1)} \|V_{\ell,m}\|_2^2, $$

from which

$$ \|\Pi(m) - m - W(m)\|_2^2 \le \frac{1}{3} \sup_{\ell \ge 0} \|V_{\ell,m}\|_2^2. \quad (4.2) $$

We now take care of the quantities $\|V_{\ell,m}\|_2^2$. Set $\tilde{W}_{\ell+1,m} = \xi_{\ell,m}\sqrt{2^{\ell+1}m}$. Then

$$ \|V_{\ell,m}\|_2^2 = \|\tilde{U}_{\ell+1,m} - \tilde{W}_{\ell+1,m}\|_2^2. $$

Let $\mu_0 = 4/3$. We first consider the case $\mu \le \mu_0$. In order to shorten the notations, we set $\mu = 2^{\ell+1}m$, $U = U_{\ell+1,m}$, $\tilde{U} = \tilde{U}_{\ell+1,m}$ and $\tilde{W} = \tilde{W}_{\ell+1,m}$. We first write

$$ \mathbb{E}(\tilde{U}\tilde{W}) = \sum_{n>0} \mathbb{P}(U = n) \mathbb{E}(\tilde{U}\tilde{W} | U = n), \quad (4.3) $$
since $\tilde{U} = 0$ if $U = 0$. Since $\xi_{\ell,m}$ is the conditional quantile transform of $\tilde{U}$ conditionally to $U$, according to Exercise 3)-a) page 30 in Rio [23], for any $n > 0$,

$$ \mathbb{E}(\tilde{U}\xi_{\ell,m} | U = n) \geq \mathbb{E}(\tilde{U} | U = n)\mathbb{E}(\xi_{\ell,m} | U = n) = 0. $$

This implies that $\mathbb{E}(\tilde{U}\tilde{W}) \geq 0$. Hence

$$ \mathbb{E}((\tilde{U} - \tilde{W})^2) \leq \mathbb{E}(\tilde{U}^2 + \tilde{W}^2) = 2\mu \leq 2\mu_0 = 8/3. \quad (4.4) $$

We turn now to the case $\mu > \mu_0$. In order to bound up the $L^2$ distance between $\tilde{U}$ and $\tilde{W}$, we will give a lower bound on $\mathbb{E}(\tilde{U}\tilde{W})$. Indeed

$$ \|V_{\ell,m}\|_2^2 = \mathbb{E}((\tilde{U} - \tilde{W})^2) = 2\mu - 2\mathbb{E}(\tilde{U}\tilde{W}), \quad (4.5) $$

since $\mathbb{E}(\tilde{W}^2) = \mu$, $\mathbb{E}(\tilde{U}^2 | \mathcal{G}_{\ell+1}) = U_{\ell+1,m}$ and $\mathbb{E}(U_{\ell+1,m}) = 2^{\ell+1}m$.

We rewrite (4.3) as follows:

$$ \mathbb{E}(\tilde{U}\tilde{W}) = \sum_{n>0} \mathbb{P}(U = n)\mathbb{E}(n^{-1/2}\tilde{U}\mu^{-1/2}\tilde{W} | U = n)\sqrt{n\mu}. \quad (4.6) $$

Now

$$ \mathbb{E}(n^{-1/2}\tilde{U}\mu^{-1/2}\tilde{W} | U = n) = 1 - \frac{1}{2}\mathbb{E}((n^{-1/2}\tilde{U} - \mu^{-1/2}\tilde{W})^2 | U = n). \quad (4.7) $$

Note now that $\mathcal{L}(\tilde{U}|U = n)$ is the law of $2B - n$ with $B \sim \mathcal{B}(n, 1/2)$. Hence $\mathcal{L}(n^{-1/2}\tilde{U}|U = n)$ is the law of $n^{-1/2}(\varepsilon_1 + \cdots + \varepsilon_n)$ where $(\varepsilon_i)_{1\leq i\leq n}$ are iid r.v.'s with law $\mathbb{P}(\varepsilon_1 = 1) = \mathbb{P}(\varepsilon_1 = -1) = 1/2$. In addition, the conditional law of $\mu^{-1/2}\tilde{W}$ given $U = n$ is a $\mathcal{N}(0, 1)$. Therefore, by Lemma 4.1 below,

$$ \mathbb{E}((n^{-1/2}\tilde{U} - \mu^{-1/2}\tilde{W})^2 | U = n) \leq \min\left(\frac{33}{16n}, 2(1 - \sqrt{2/\pi})\right). \quad (4.8) $$

Starting from (4.6) and considering (4.7) and (4.8), it follows that

$$ \mathbb{E}(\tilde{U}\tilde{W}) \geq \sum_{n>0} \mathbb{P}(U = n)\sqrt{n\mu}\left(1 - \min\left(\frac{33}{32n}, 1 - \sqrt{2/\pi}\right)\right). $$

The last inequality is equivalent to

$$ \mu^{-1/2}\mathbb{E}(\tilde{U}\tilde{W}) \geq \mathbb{E}(\sqrt{U}) - \mathbb{E}\left(\min\left(\frac{33}{32\sqrt{U}}, (1 - \sqrt{2/\pi})\sqrt{U}\right)\right). $$

We now claim that

$$ \min\left(\frac{33}{32\sqrt{U}}, (1 - \sqrt{2/\pi})\sqrt{U}\right) \leq c_0(U + 1)^{-1/2} \text{ with } c_0 = 1.1139. $$
Indeed, for $U \le 5$, $(1 - \sqrt{2/\pi})\sqrt{U} \le 1.108(U + 1)^{-1/2}$, and, for $U \ge 6$,

$$ \frac{33}{32\sqrt{U}} \le \frac{33}{32\sqrt{U} + 1} \sqrt{7/6} \le c_0(U + 1)^{-1/2}. $$

Hence

$$ \mu^{-1/2} \mathbb{E}(\tilde{U}\tilde{W}) \ge \mathbb{E}(\sqrt{U}) - c_0 \mathbb{E}((U+1)^{-1/2}). $$

Now, recall that $U$ has the Poisson distribution with parameter $\mu$. Therefrom

$$ \mathbb{E}((U + 1)^{-1/2}) = e^{-\mu} \sum_{n \ge 0} \frac{1}{n!} (n + 1)^{-1/2} \mu^n = e^{-\mu} \sum_{n \ge 0} \frac{(n + 1)^{1/2}}{(n + 1)!} \mu^n = \mu^{-1} \mathbb{E}(\sqrt{U}). $$

The two last inequalities ensure that

$$ \mathbb{E}(\tilde{U}\tilde{W}) \ge (\mu - c_0)\mathbb{E}(\sqrt{\mu^{-1}U}). \quad (4.9) $$

We now provide a lower bound on $\mathbb{E}(\sqrt{\mu^{-1}U})$. Let $Z = \mu^{-1}(U - \mu)$. Then $\sqrt{\mu^{-1}U} = \sqrt{1 + Z}$. Now, by the Taylor integral formula at order 4, for any real $z \ge -1$,

$$ \sqrt{1+z} = 1 + (z/2) - (z^2/8) + (z^3/16) - (5/32)z^4 \int_0^1 (1-t)^3(1+tz)^{-7/2}dt. $$

Next, for $z \ge -1$, $(1 + tz)^{-7/2} \le (1 - t)^{-7/2}$, from which

$$ -(5/32)z^4 \int_0^1 (1-t)^3(1+tz)^{-7/2}dt \ge -(5/32)z^4 \int_0^1 (1-t)^{-1/2}dt = -(5/16)z^4. $$

It follows that

$$ \sqrt{\mu^{-1}U} \ge 1 + (Z/2) - (Z^2/8) + (Z^3/16) - (5/16)Z^4 $$

Recall now that $U$ has the Poisson distribution with parameter $\mu$. Therefrom $\mathbb{E}(Z) = 0$, $\mathbb{E}(Z^2) = \mu^{-1}$, $\mathbb{E}(Z^3) = \mu^{-2}$ and $\mathbb{E}(Z^4) = \mu^{-3} + 3\mu^{-2}$. Together with the above inequality, it implies that

$$ \mathbb{E}(\sqrt{\mu^{-1}U}) \ge 1 - \frac{1}{8\mu} - \frac{7}{8\mu^2} - \frac{5}{16\mu^3}. \quad (4.10) $$

Combining this lower bound with (4.9) we finally obtain that

$$ \mathbb{E}(\tilde{U}\tilde{W}) \ge \mu - 1.2389 - 0.7358\mu^{-1} + 0.6621\mu^{-2} + 0.3480\mu^{-3} \ge \mu - 1.4055 $$

for any $\mu \ge \mu_0$. The above inequality together with (4.5) imply that, for any $\mu \ge \mu_0$,

$$ \mathbb{E}((\tilde{U} - \tilde{W})^2) \le 2 \times 1.4055 = 2.8110. \quad (4.11) $$
From (4.4), the above inequality also holds true for $\mu \le \mu_0$. Finally, combining this upper bound with (4.2), we get that

$$ \|\Pi(m) - m - W(m)\|_2^2 \le 0.9370. \quad (4.12) $$

Since $W_2^2(\mu_{m,1}, \nu_m) \le \|\Pi(m) - m - W(m)\|_2^2$, the upper bound (4.12) implies Proposition 3.1. To complete the proof of the proposition, we state and prove Lemma 4.1.

**Lemma 4.1.** Let $(\varepsilon_i)_{1\le i\le n}$ be iid r.v.'s with law $P(\varepsilon_1 = 1) = P(\varepsilon_1 = -1) = 1/2$, and $\nu_n$ be the $\mathcal{N}(0, n)$ probability measure. Then

$$ W_2^2(P_{\varepsilon_1+\dots+\varepsilon_n}, \nu_n) \le \min(33/16, 2n(1-\sqrt{2/\pi})). $$

*Proof.* Let $G_n \sim \mathcal{N}(0, n)$ (hence $P_{G_n} = \nu_n$) and $F_n$ be the c.d.f. of $\varepsilon_1 + \dots + \varepsilon_n$, and $\Phi$ be the c.d.f. of a $\mathcal{N}(0, 1)$. Define

$$ S_n = F_n^{-1}(\Phi(n^{-1/2}G_n)). $$

Then $P_{S_n} = P_{\varepsilon_1+\dots+\varepsilon_n}$ and $\mathbb{E}((S_n - G_n)^2) = W_2^2(P_{S_n}, P_{G_n})$. Let $Y = n^{-1/2}G_n$. According to Theorem 1.1 in [14],

$$ |S_n - G_n| \le (3/2) + Y^2/4. $$

Now, using the elementary inequality $x^2 \le \frac{3}{2}|x| + |x|\max(|x| - \frac{3}{2}, 0)$, we derive that

$$ \mathbb{E}((S_n - G_n)^2) \le \frac{3}{2}\mathbb{E}(|S_n - G_n|) + \mathbb{E}\left(\left[\frac{3}{2} + \frac{Y^2}{4}\right]\frac{Y^2}{4}\right) \le \frac{3}{2}\mathbb{E}(|S_n - G_n|) + \frac{9}{16}. $$

Next $\mathbb{E}(|S_n - G_n|) = W_1(P_{S_n}, P_{G_n})$. Since $\mathbb{E}(\varepsilon_1^2) = \mathbb{E}(|\varepsilon_1|^3) = 1$, according to Theorem 1.1 in [9], it follows that $\mathbb{E}(|S_n - G_n|) \le 1$. So, overall,

$$ \mathbb{E}((S_n - G_n)^2) \le (3/2) + (9/16) = 33/16. $$

It remains to prove that we also have

$$ W_2^2(P_{\varepsilon_1+\dots+\varepsilon_n}, \nu_n) \le 2n(1 - \sqrt{2/\pi}). \quad (4.13) $$

Let $Y_1, \dots, Y_n$ be iid r.v.'s with law $\mathcal{N}(0, 1)$. Define $\eta_k$ by $\eta_k = \mathbf{1}_{Y_k \ge 0} - \mathbf{1}_{Y_k < 0}$ for $k$ in $[1, n]$. Then $\eta_1, \dots, \eta_n$ are iid r.v.'s with law $P(\eta_1 = 1) = P(\eta_1 = -1) = 1/2$. Therefrom

$$ W_2^2(P_{\varepsilon_1+\dots+\varepsilon_n}, \nu_n) \le \mathbb{E}\left(\left(\sum_{k=1}^n (\eta_k - Y_k)\right)^2\right). $$

Since the r.v.'s $((\eta_k - Y_k)_{1\le k\le n})$ are iid,

$$ \mathbb{E}\left(\left(\sum_{k=1}^n (\eta_k - Y_k)\right)^2\right) = n\mathbb{E}((\eta_1 - Y_1)^2). $$

Next

$$ \mathbb{E}((\eta_1 - Y_1)^2) = \mathbb{E}((|Y_1| - 1)^2) = 2 - 2\mathbb{E}(|Y_1|) = 2(1 - \sqrt{2/\pi}). \quad (4.14) $$

This ends the proof of (4.13) and completes the proof of Lemma 4.1. $\square$
## 4.2 Proof of Proposition 2.1

The proof follows immediately from the general proposition below applied to $\varphi = \psi_x$ where $\psi_x$ is defined in (2.1).

**Proposition 4.1.** Let $\varphi$ be a $C^1$, non-negative, even and convex function defined on $\mathbb{R}$ such that $\varphi(0) = 0$. Assume in addition that $\varphi'$ is a concave function. Let $\mu$ and $\nu$ two probability laws on the real line with mean zero and respective cumulative distribution functions $F$ and $G$. Let $\mathcal{F}_\varphi$ be the class of continuously differentiable functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $f(0) = 0$, $\|f'\|_\infty \le \|\varphi'\|_\infty$ and

$$|f'(x) - f'(y)| \le 2\varphi'(|x - y|) \text{ for any } (x, y) \in \mathbb{R}^2.$$

Then

$$\kappa_{\varphi}(\mu, \nu) \le \zeta_{\varphi}(\mu, \nu) := \sup\{\mu(f) - \nu(f) : f \in \mathcal{F}_{\varphi}\}.$$

**Remark 4.1.** If $\varphi(x) = |x|^r$ with $r \in ]1, 2]$, we derive that $\kappa_r(\mu, \nu) \le 2r\zeta_r(\mu, \nu)$, where $\zeta_r$ is the ideal distance of order $r$ (see for instance [19, eq. (9.1.2)] for its definition). In this particular case we recover [20, Theorem 3.1].

**Proof of Proposition 4.1.** Using the arguments developed in the beginning of the proof of [20, Theorem 3.1], we start by noting that it is enough to prove the proposition for probability laws with strictly positive and smooth densities. Indeed,

$$\kappa_{\varphi}(\mu, \nu) = \lim_{\sigma \to 0} \kappa_{\varphi}(\mu_{\sigma}, \nu_{\sigma}) \text{ and } \zeta_{\varphi}(\mu, \nu) = \lim_{\sigma \to 0} \zeta_{\varphi}(\mu_{\sigma}, \nu_{\sigma}).$$

where $\mu_\sigma = \mu * \phi_\sigma * \lambda$ with $\phi_\sigma$ the density of the normal law $N(0, \sigma^2)$.

So, from now, $\mu$ and $\nu$ denote two probability laws with distributions function $F$ and $G$, respectively, and we assume that $F$ and $G$ are continuous and strictly increasing from $\mathbb{R}$ to $]0, 1[$. For $U$ a r.v. with the uniform distribution over $[0, 1]$, the random vector $(F^{-1}(U), G^{-1}(U))$ has respective marginal distributions $\mu$ and $\nu$. Hence, to prove the proposition, it is enough to prove that there exists some function $h$ in $\mathcal{F}_\varphi$ such that, for any $u$ in $]0, 1[$,

$$h(F^{-1}(u)) - h(G^{-1}(u)) \ge \varphi(F^{-1}(u) - G^{-1}(u)). \quad (4.15)$$

With this aim, let

$$H = G - F \text{ and } A = H^{-1}(\{0\}).$$

Since $H$ is continuous, $A$ is a closed set. Let $h$ be a function from $\mathbb{R}$ to $\mathbb{R}$ that is derivable and such that

$$h'(t) = \operatorname{sign} H(t) \varphi'(2d(t, A)).$$
Notice first that $A$ cannot be equal to $\emptyset$. Indeed, if $A = \emptyset$, then either $F > G$ or $G > F$, which entails that $\mu$ and $\nu$ cannot have the same mean.

Next, if $[s, t] \cap A = \emptyset$, then

$$|h'(s) - h'(t)| = |\varphi'(2d(s, A)) - \varphi'(2d(t, A))| \leq \varphi'(2|t-s|) \leq 2\varphi'(|t-s|),$$

where the first and the second inequalities come from the fact that $\varphi'$ is a concave function, that $\varphi'$ is nondecreasing and $\varphi'(0) = 0$.

Now assume that $[s, t] \cap A \neq \emptyset$. Let $\alpha$ and $\beta$ such that $d(s, A) = |\alpha - s|$ and $d(t, A) = |t - \beta|$. Clearly

$$|h'(s) - h'(t)| = \varphi'(2|t - \beta|) + \varphi'(2|\alpha - s|).$$

Since $\varphi'$ is a concave and nondecreasing function, for any $x$ and $y$ such that $x + y \leq t - s$, the following inequalities hold:

$$\varphi'(x) + \varphi'(y) \leq 2\varphi'((x+y)/2) \leq 2\varphi'((t-s)/2).$$

This implies that

$$|h'(s) - h'(t)| \leq 2\varphi'(|t-s|).$$

All these considerations show that $h$ belongs to $\mathcal{F}_{\varphi}$. The rest of the proof consists of showing that $h$ satisfies the inequality (4.15).

With this aim, we start by noticing that $F^{-1}(u) \in A$ is equivalent to $G^{-1}(u) = F^{-1}(u)$. Indeed, $F^{-1}(u) \in A$ is equivalent to $G(F^{-1}(u)) = F(F^{-1}(u)) = u$ which in turn is equivalent to $G^{-1}(u) = F^{-1}(u)$. Consequently, if $F^{-1}(u) \in A$, since $\varphi(0) = 0$,

$$h(F^{-1}(u)) - h(G^{-1}(u)) = \varphi(F^{-1}(u) - G^{-1}(u)) = 0.$$

Assume from now that $F^{-1}(u) \notin A$. There are different cases.

**Case 1.** $F^{-1}(u) \notin \inf A, \sup A$. For instance assume that $F^{-1}(u) > \sup A$ and that $\sup A < +\infty$ (the case where $F^{-1}(u) < \inf A$ can be handled similarly). Let $z = \sup A$. Since $A$ is closed, $z \in A$. Now $F^{-1}(u) > z$ if and only if $u > F(z)$. Hence, since $z \in A$,

$$G^{-1}(u) > G^{-1}(F(z)) = G^{-1}(G(z)) = z.$$

Note that for any $z \in A$,

$$h(F^{-1}(u)) - h(G^{-1}(u)) = \int_{z}^{F^{-1}(u)} h'(t)dt - \int_{z}^{G^{-1}(u)} h'(t)dt.$$

Therefore, if $F^{-1}(u) > G^{-1}(u)$,

$$h(F^{-1}(u)) - h(G^{-1}(u)) = \int_{G^{-1}(u)}^{F^{-1}(u)} h'(t)dt,$$
and we note that $G^{-1}(u) < t < F^{-1}(u)$ if and only if $F(t) < u < G(t)$ implying that
$\mathrm{sign}(H(t)) = 1$. Next, if $F^{-1}(u) < G^{-1}(u)$,

$$
h(F^{-1}(u)) - h(G^{-1}(u)) = - \int_{F^{-1}(u)}^{G^{-1}(u)} h'(t) dt,
$$

and $\mathrm{sign}(H(t)) = -1$. By definition of $h$, if $t > z$, $d(t, A) = t - z$ implying that
$h'(t) = \mathrm{sign}H(t)\varphi'(2(t-z))$. So, overall,

$$
D(u) := h(F^{-1}(u)) - h(G^{-1}(u)) = \int_{\min(F^{-1}(u), G^{-1}(u))}^{\max(F^{-1}(u), G^{-1}(u))} \varphi'(2(t-z))dt.
$$

Assume now that $F^{-1}(u) > G^{-1}(u)$ (the case $F^{-1}(u) < G^{-1}(u)$ can be handled similarly). Then, since $\varphi'$ is nondecreasing and $\varphi'(0) = 0$,

$$
D(u) = \int_{G^{-1}(u)-z}^{F^{-1}(u)-z} \varphi'(2s)ds \geq \int_{0}^{F^{-1}(u)-G^{-1}(u)} \varphi'(2s)ds \geq \varphi(F^{-1}(u)-G^{-1}(u)).
$$

We consider now the complementary case of case 1.
Case 2. $F^{-1}(u) \in] \inf A, \sup A[$. Let

$$
a = \sup(A \cap [-\infty, F^{-1}(u)]) \text{ and } b = \inf(A \cap [F^{-1}(u), +\infty)).
$$

Therefore $a < F^{-1}(u) < b$ which implies that $a < G^{-1}(u) < b$. Indeed $a < F^{-1}(u)$
entails that $u > F(a)$. Hence, since $a \in A$,

$$
G^{-1}(u) > G^{-1}(F(a)) = G^{-1}(G(a)) = a.
$$

We proceed similarly to prove that $G^{-1}(u) < b$. Let then

$$
I =]F(a), F(b)[=]G(a), G(b)[.
$$

For any $v \in I$, $F^{-1}(v) \notin A$, from which $F^{-1}(v) \neq G^{-1}(v)$. Hence the sign of
$F^{-1}(v) - G^{-1}(v)$ is constant on $I$.

Without loss of generality, we assume in what follows that $F^{-1}(u) > G^{-1}(u)$ and
we set $\delta(u) = F^{-1}(u) - G^{-1}(u)$. We have

$$
D(u) = h(F^{-1}(u)) - h(G^{-1}(u)) = \int_{G^{-1}(u)}^{F^{-1}(u)} \varphi'\left(2\left((t-a) \wedge (b-t)\right)\right)dt := f(G^{-1}(u)),
$$

where

$$
f(y) = \int_{y}^{y+\delta(u)} \varphi'\left(2\left((t-a) \wedge (b-t)\right)\right) dt.
$$
By analyzing the derivative of $f$, one can see that the minimum of $f$ on $[a, b]$ is achieved in $y = a$ or in $y = b - \delta(u)$. Therefore

$$D(u) \geq \int_{0}^{\delta(u)} \varphi'(2(s \wedge (b-a-s)))ds. \qquad (4.16)$$

Since $\delta(u) \leq b - a$,

$$D(u) \geq \int_{0}^{\delta(u)/2} \varphi'(2s)ds + \int_{\delta(u)/2}^{\delta(u)} \varphi'(2(s \wedge (b-a-s)))ds.$$

We argue now as to get the inequality (4.16). Hence we analyze the derivative of the function $g$ defined by $g(y) = \int_{y}^{y+\delta(u)/2} \varphi'(2(s \wedge (b-a-s)))ds$. One can see that the minimum of $g$ is achieved in $y = 0$ or in $y = b - a - \delta(u)/2$. Therefore

$$\int_{\delta(u)/2}^{\delta(u)} \varphi'(2(s \wedge (b-a-s)))ds \geq \int_{0}^{\delta(u)/2} \varphi'(2s)ds.$$

Hence

$$D(u) \geq 2 \int_{0}^{\delta(u)/2} \varphi'(2s)ds = \varphi(\delta(u)).$$

This ends the proof of the proposition. $\square$

## 4.3 Proof of Lemma 3.4

Let $C = B - x$. Clearly $\tilde{H}_{A+B}(x) = \tilde{H}_{A+C}(0)$. Hence it is enough to prove the lemma in case $x = 0$. Note that

$$\tilde{H}_{A+C}(0) = \inf_{s>0} s^{-1}\mathbb{E}((A + C + s)_+) = \inf_{\substack{v>0 \\ w>0}} (v+w)^{-1}\mathbb{E}((A + C + v + w)_+).$$

Similarly

$$\tilde{H}_A(t) = \inf_{v>0} v^{-1} \mathbb{E}((A - t + v)_+) \quad \text{and} \quad \tilde{H}_C(-t) = \inf_{w>0} w^{-1} \mathbb{E}((C + t + w)_+).$$

Now $(A + C + v + w)_+ \leq (A - t + v)_+ + (C + t + w)_+$. Hence, for any $v, w > 0$,

$$\tilde{H}_{A+C}(0) \leq \frac{\mathbb{E}((A - t + v)_+)}{v} \frac{v}{v + w} + \frac{\mathbb{E}((C + t + w)_+)}{w} \frac{w}{v + w}.$$

It follows that

$$\tilde{H}_{A+C}(0) \leq \max(v^{-1}\mathbb{E}((A - t + v)_+), w^{-1}\mathbb{E}((C + t + w)_+)).$$

Let $\varepsilon > 0$ and $v, w$ be chosen in such a way that $v^{-1}\mathbb{E}((A - t + v)_+) \leq \tilde{H}_A(t) + \varepsilon$ and $w^{-1}\mathbb{E}((C + t + w)_+) \leq \tilde{H}_C(-t) + \varepsilon$. Then $\tilde{H}_{A+C}(0) \leq \max(\tilde{H}_A(t), \tilde{H}_C(-t)) + \varepsilon$. The result follows by letting $\varepsilon$ tend to 0. $\square$
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