# Approximating $E[g(\overline X_n)]$ and want to bound the remainder using some form of CLT or Berry-Essen Theorem

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 $$g(\overline X_n) = |\overline X_n| \mathbb{1}_{\{|\overline X_n| > c\}}$$.

• 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.
– TMat
Mar 30 at 8:33
• The function $g$ would generally be something like $g(\overline X_n) = |\overline X_n| \mathbb{1}_{\{|\overline X_n| > c\}}$. Mar 30 at 8:46

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 $$c$$.

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.)