# A question about the central limit theorem

The question is:
$$g:R\rightarrow R$$ has at least three bounded continuous derivatives and let $$X_i$$ be $$iid$$ and in $$L^2$$. Prove that:

i) $$\sqrt{n}[g(\overline{X_n}) - g(\mu)]\xrightarrow{w} N(0,g^{'}(\mu)^{2} \sigma ^2)$$ and that

ii) $$E[g(\overline{X_n})-g(\mu)] = \frac{\sigma^2g''(\mu)}{2n} + o(n^{-1})$$ as $$n\rightarrow \infty$$

where $$\overline{X_n} = \frac{\sum X_n}{n}$$, $$\mu = EX_1$$, $$\sigma^2=Var(X_1)$$

I have proved i) using CLT but for ii) $$g(\overline{X_n}) - g(\mu)\approx N(0,g^{'}(\mu)^{2} \sigma ^2/n)$$ as $$n\rightarrow \infty$$. Since $$RHS$$ has $$g''(\mu)$$, I was thinking of expanding $$e^{\frac{-x^2}{2g'(\mu)\sigma^2/n}}$$using Taylor's series at $$\mu$$ but it already has $$g'(\mu)$$ in it which is a constant. If it had a $$g'(x)$$, I would get a $$2^{nd}$$ derivative, so not sure how to approach the problem. Thanks and appreciate a hint.

• Are you tacitly assuming the $X_i$ are iid, so the $\overline X_n$ obeys the CLT? – kimchi lover Feb 26 at 19:48
• Yes, $X_i$ are $iid$. – manifolded Feb 26 at 19:49
• Are you also tacitly assuming the $X_i$ are in $L^3$? – kimchi lover Feb 27 at 0:19
• No, $X_{i}\in L^2$ but $g$ has at least $3$ bounded continuous derivatives as stated. Thanks. – manifolded Feb 27 at 0:26

Here is a variation on the other answer. It has two parts.

First, writing the Taylor approximation as $$g(x)=g(\mu)+g'(\mu)\,(x-\mu) + g''(\mu)\,(x-\mu)^2/2 + (x-\mu)^2h(x-\mu)$$ where $$h$$ is continuous and bounded, and vanishes at $$0$$. (This should be clear from the usual Taylor expansion error bound for $$x$$ near $$\mu$$; and from the boundedness of $$g$$: one has $$h(x)=(g(x)-(g(\mu)+g'(\mu)\,(x-\mu) + g''(\mu)\,(x-\mu)^2/2)/(x-\mu)^2$$, which is clearly bounded when $$x$$ is far from $$\mu$$.)

Second, write $$Z_n = \sqrt n(\overline X_n-\mu)$$, so $$Z_n\to Z$$ in distribution, where $$Z\sim N(0,1)$$, by the usual CLT.

Now the error term in the original problem's (ii) is $$E\left( Z_n^2 h(Z_n/\sqrt n)\right) / n$$ and the problem boils down to showing that $$\lim_{n\to\infty} E\left( Z_n^2 h(Z_n/\sqrt n) \right)= 0$$.

Since the $$Z_n$$ converge in distribution to $$Z$$ there exist random variables $$Y_n$$ with the same distribution as $$Z_n$$ such that $$Y_n\to Z$$ with probability 1. (Such as, $$Y_n = F_n^{-1}(\Phi(Z))$$ where $$F_n$$ is the cdf of $$Z_n$$ and $$\Phi$$ is the cdf of $$Z$$, or by Skorohod's theorem.) Since $$Y_n\to Z$$ w.p.1 and $$EY_n^2 = EZ^2$$, the sequence $$Y_n^2$$ is uniformly integrable. This, plus $$h$$ being bounded, implies the sequence $$Y_n^2 h(Y_n/\sqrt n)$$ is uniformly integrable.

(For details of the uniform integability: Hewitt and Stromberg theorem 13.47, for instance implies the convergence $$Y_n^2\to Z^2$$ holds in $$L^1$$, and exercise 13.39 then gives uniform integrability of the $$Y_n^2$$. See also Theorem 21 in Dellacherie and Meyer, Probability and Potential, 1978 edition, p."23-11".)

So $$\lim_{n\to\infty}E(Y_n^2 h(Y_n/\sqrt n)) = E (\lim_{n\to\infty} Y_n^2 h(Y_n/\sqrt n)) = E Z^2 h(0) = E(Z^20) = 0.$$ But $$Z_n$$ and $$Y_n$$ have the same distribution, so $$\lim_{n\to\infty}E(Z_n^2 h(Z_n/\sqrt n)) = \lim_{n\to\infty}E(Y_n^2 h(Y_n/\sqrt n)) = 0,$$ as desired.

• Thanks a lot, let me take some time to read and get back. – manifolded Feb 28 at 19:28
• Fair enough. I always find uniform integrability arguments tricky. – kimchi lover Feb 28 at 19:44
• Hi, sorry for a late reply! Just making sure if $h$ being bounded follows from $g$ being bounded and also I didn't understand why $h(0)=0$. Invoking UI is very interesting. Is it also a known result that if $EY_n^{2} = EZ^2$ and $Yn\xrightarrow{w.p.1} Z$, then $Y_n$ is UI? Because I am just aware of standard definition, you know if $Sup_n E(|Y_n|1_{|Y_n>y|})\rightarrow 0$ as $y\rightarrow 0$. Thanks and appreciate your time. – manifolded Mar 1 at 18:40
• @manifolded I have tried to answer these questions in an edit to my answer . I think UI gets short shrift in analysis classes, as it seems technical and boringly nerdy. I remembered that it is the big brother of the dominated convergence theorem: you don't actually need a common dominating function but only just UI. Then I scoured old textbooks to see how to fill in the details of the argument. – kimchi lover Mar 1 at 19:04
• Brilliant, thanks, yes UI is definitely powerful and I should look for more opportunities to apply it. I will look at those references too, appreciate it. – manifolded Mar 1 at 19:10

This is more a comment than an answer, but at least an idea.

Using Taylor's formula with integral remainder gives $$g\left(\overline{X_n}\right)-g(\mu)=\left(\overline{X_n}-\mu\right)g'(\mu) +\frac{\left(\overline{X_n}-\mu\right)^2}2g''(\mu)+\frac 12\int_\mu^{\overline{X_n}}g^{(3)}(t)\left(\overline{X_n}-t\right)^2\mathrm dt.$$ Therefore, taking the expectations reduces us to show that $$\lim_{n\to +\infty}n\mathbb E\left[\int_\mu^{\overline{X_n}}g^{(3)}(t)\left(\overline{X_n}-t\right)^2\mathrm dt\right]=0.$$

• Interesting, thanks. I will try to prove the $limit$ now. – manifolded Feb 27 at 19:21
• It has to be subtle, because if you replace $g^{(3)}(t)$ in the integral with a constant upper bound, as you may, and as you might be tempted, you get a cubic polynomial in $\overline X_n$, whose expectation is not guaranteed to exist. – kimchi lover Feb 27 at 21:20
• Hit the nail on the head, I was stuck there because I felt obligated to use $g^{(3)}(t)$ is bounded. – manifolded Feb 28 at 19:26