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.
Best Answer
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 $$ \lim_{n\to +\infty}n\mathbb E\left[\int_\mu^{\overline{X_n}}g^{(3)}(t)\left(\overline{X_n}-t\right)^2\mathrm dt\right]=0. $$