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If we have a set $X_1,\dots,X_n$ of iid random variables with finite mean $\mu$ and variance $\sigma$, the CLT says that $\sqrt{n}(\overline X_n - \mu) \stackrel{d}{\to} \mathcal{N}(0,\sigma^2)$. If we also have a finite third moment, then the Berry-Essen theorem says that $$ |F_n(x) - \Phi(x)| \le \frac{C E[|X_1|^3]}{\sigma^3 \sqrt{n}}. $$ where $F_n$ is the CDF of $\sqrt{n} \overline X_n /\sigma$.

Do either the CLT or the Berry-Essen theorem allow us to say something about the convergence of the pdf of $\overline X_n$ to the pdf of the normal distribution?

My motivation is that I would like to approximate an expectation of a function of the sample mean for large $n$. So I write $$ \begin{align} E[g(\overline X_n)] & = \int g(x) p_{\overline X_n}(x) dx \\ & = \int g(x) (\varphi(x) - \varphi(x) + p_{\overline X_n}(x)) dx \\ & = \int g(x) \varphi(x)dx + R, \end{align} $$ where the remainder $R$ is $$ R = \int g(x) (p_{\overline X_n}(x) - \varphi(x)) dx. $$ where $\varphi(x)$ is the pdf of the (properly scaled) normal distribution. So I would like to bound the remainder as $n \to \infty$ to show this is a valid approximation. If we were talking about the cdf instead of the pdf, the Berry-Essen inequality could be used to bound the remainder. So is it possible to bound the remainder given that it is in terms of the pdf?

The function $g$ that I am most interested is of the form $g(\overline X_n) = |\overline X_n| \mathbb{1}_{\{|\overline X_n| > c\}}$.

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    $\begingroup$ Is $g$ (almost surely) differentiable ? If yes, I would use an integration by part to say that $\int g(y)p(y)dy=\int g'(y)F(y)dy$ and then we can use Berry-Esseen inequality. $\endgroup$ – TMat Mar 30 at 8:33
  • $\begingroup$ The function $g$ would generally be something like $g(\overline X_n) = |\overline X_n| \mathbb{1}_{\{|\overline X_n| > c\}}$. $\endgroup$ – ManUtdBloke Mar 30 at 8:46
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Let us begin with the example $g(x)=|x| 1_{|x|>c}$.

Suppose that $X_1,\dots, X_n$ are centered and have finite second moment $\sigma^2$. Denote $F_n(x)=\mathbb{P}\left(\frac{1}{\sigma\sqrt{n}}\overline{X}_n\le x\right)$. We work with $\frac{1}{\sigma \sqrt{n}}\overline{X}_n$ instead of $\overline{X}_n$ as it simplifies the computations, please change $c$ accordingly to recover a result on you original question. We have the following observation: by integration by part, \begin{align*} \mathbb{E}\left[g\left(\frac{1}{\sigma \sqrt{n}}\overline{X}_n\right)\right]&=\int_{-\infty}^{-c} -x p_{\frac{1}{\sigma \sqrt{n}}\overline{X_n}}(x)dx+\int_{c}^{\infty} x p_{\frac{1}{\sigma \sqrt{n}}\overline{X_n}}(x)dx\\ &=cF_n(-c)+\int_{-\infty}^{-c} F_n(x)dx+c(1-F_n(c))+\int_{c}^{\infty} (1-F_n(x))dx. \end{align*} Then, apply the following local Berry-Esseen inequality (ref Theorem 14 Petrov's book)

Theorem

Let $X_1,\dots,X_n$ be i.i.d random variables with $\mathbb{E}[X]=0$, $\mathbb{E}[X^2]=\sigma^2<\infty$ and $\mathbb{E}[|X|^3]=\sigma^3 \rho <\infty$. Then, there exists an absolute constant $A>0$ such that $$|F_n(x)-\Phi(x)|\le A\frac{\rho}{\sqrt{n}(1+|x|)^3}.$$

Hence, denoting by $A$ a generic $>0$ constant that can change at each apparition, \begin{align*} \mathbb{E}\left[g\left(\frac{1}{\sigma \sqrt{n}}\overline{X}_n\right)\right]\le& c\Phi(-c)+A\frac{\rho}{\sqrt{n}(1+c)^3} +\int_{-\infty}^{-c} \Phi(x) + A\frac{\rho}{\sqrt{n}(1-x)^3}dx\\ &+c(1-\Phi(c))+\int_{c}^{\infty} (1-\Phi(x)) +A\frac{\rho}{\sqrt{n}(1+x)^3} dx. \end{align*} then, having $\int_{c}^{\infty}\frac{dx}{(1+x)^3}=\frac{1}{2(1+c)^2}$ and denoting $Z\sim \mathcal{N}(0,1)$, \begin{align*} \mathbb{E}\left[g(\overline{X}_n)\right]\le& \mathbb{E}[g(Z)]+A\frac{\rho}{\sqrt{n}(1+c)^2}. \end{align*}

This method is kind of inspired by Stein Method, read about Stein Method if you are interested. This can be generalized to any $g$ function that is derivable in the sense of distribution.

EDIT: Another less complicated method is to use Edgeworth expansion but this is asymptotic whether my method is not asymptotic.

EDIT 2: I corrected the proof to have a better dependency in $c$.

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Note that this expectation does not exist for all $g$, because if $g$ is sufficiently fast-growing then $E[g(\bar X_n)]$ may not be finite for any $n$. For example, this happens if each $X_i$ follows a standard normal distribution, so that $\bar X_n \sim N(0, \frac 1 {\sqrt{n}})$, and $g(x) = e^{x^8}$. (Though the specific $g$ that you are most interested in doesn't have this problem.)

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