I have to prove this variation of the Central Limit Theorem:
Let $(Z_n)_{n \in \mathbb{N}}$ be i.i.d. random variables with existing variance $\sigma^2 = \mathrm{Var}(Z_1) > 0$ and $\mu = \mathrm{E}(Z_1)$. Let $I \subset \mathbb{R}$ be an interval and $f: I \rightarrow \mathbb{R}$ continuously differentiable twice with a bounded second derivative and $f'(\mu) ≠ 0$. Show the following:
$$\frac{\sqrt{n}}{\sigma f'(\mu)}\left(f(\bar{Z_n}) - f(\mu)\right) \overset{d}{\rightarrow} \mathcal{N}(0, 1),$$ where $\bar{Z_n} := \frac{1}{n}\sum_{i = 1}^n Z_i$.
My first idea was to maybe look at a taylor expansion for $f$, but I honestly did not get very far with this approach. Does someone have a better way for solving this?
Using Taylor expansion of $f$ is the right idea. Taylor expansion of $f$ about $\mu$: $$f(\bar{Z}_n)-f(\mu) = f'(\mu)(\bar{Z}_n-\mu)+\frac{f''(\xi_n)}{2}(\bar{Z}_n-\mu)^2,$$ for some $\xi_n$ between $\bar{Z}_n$ and $\mu$. As $f'(\mu)$ is non-zero and $f''$ is bounded, $$\frac{\sqrt{n}}{f'(\mu)}\left(f(\bar{Z}_n)-f(\mu)\right) = \sqrt{n}(\bar{Z}_n-\mu)+\color{red}{\frac{f''(\xi_n)}{2f'(\mu)}}\cdot\color{red}{\sqrt{n}(\bar{Z}_n-\mu)}\cdot\color{blue}{(\bar{Z}_n-\mu)}.$$ The red terms are $O_p(1)$, and the blue term is $o_p(1)$, so the second term on the RHS is $o_p(1)$. As $\sqrt{n}(\bar{Z}_n-\mu)$ goes in distribution to $\mathcal{N}(0,\sigma^2)$, the LHS converges in distribution to $\mathcal{N}(0,\sigma^2)$ as well (by Slutsky's theorem).