This is from a proof in my monte carlo course. let $h$ be a smooth function, $T_n = h(\bar{X})$ $\mu = E(X)$ then by taylor expansion
$E(T_n -\tau) = E[h(\bar{X} -h(\mu)] = E[\bar{X} - \mu]h'(\mu)+ \frac{(E[\bar{X} - \mu]]) ^2}{2} h''(\mu) = \frac{Var(X)}{2n}h''(\mu)$
The second equality is due to the taylor expansion but I fail to se why there is no $E[h'(\mu)]$ term. and why there is a expected value around the $\bar{X}−\mu$? Also why is the third exuality true? what is happening to all the other temrs? Thanks
edit: I think the first term goes away since the definition of $\mu = E(X)$ and $E[\bar{X} - \mu]]^2 = E[\bar{X} - E(\bar{X})]]^2 $ gives the last equality. But what about the first question about the expected value?
The Taylor expansion of $h$ around $x=\mu$ yields $$ h(x)=h(\mu)+h'(\mu)(x-\mu)+\frac12 h''(\mu)(x-\mu)^2+R(x) $$ for all $x\in\mathbb{R}$, where the $R$ is a function such that $R(x)\to 0$ when $x\to\mu$. If we disregard this "error term" $R$ then we may write that $$ h(x)-h(\mu)=h'(\mu)(x-\mu)+\frac12 h''(\mu)(x-\mu)^2,\quad x\in\mathbb{R},\tag{1} $$ where the equality actually means approximately equal when $x$ is close to $\mu$ (writing $\approx$ would have been a better notation). Assuming that $(1)$ holds for all $x$, then also $$ h(\bar{X})-h(\mu)=h'(\mu)(\bar{X}-\mu)+\frac12 h''(\mu)(\bar{X}-\mu)^2 $$ so by taking expectation on both sides we get $$ \begin{align} {\rm E}[h(\bar{X})-\tau]&={\rm E}\Big[h'(\mu)(\bar{X}-\mu)+\frac12 h''(\mu)(\bar{X}-\mu)^2\Big]\\ &=h'(\mu){\rm E}\big[\bar{X}-\mu\big]+\frac12 h''(\mu){\rm E}\big[(\bar{X}-\mu)^2\big] \end{align} $$ with $\tau=h(\mu)$. All we have used here is that ${\rm E}[aX+b]=a{\rm E}[X]+b$ for constants $a$ and $b$. All that is left is note that ${\rm E}[(\bar{X}-\mu)^2]={\rm Var}(\bar{X})/n$ which you probably know.