Characteristic function in central limit theorem I have difficulties following the proof of the CLT. I know that there are essentially three steps:


*

*Using Characteristic functions,

*Using Taylor's Theorem,

*Using Lévy's continuity theorem.


In the proof of the classical CLT on Wikipedia they state that $\varphi_{Y_1} = 1 - \frac{t^2}{2n} + o\left(\frac{t^2}{n}\right), \frac{t}{\sqrt n}\to0$. $\frac{t}{\sqrt n}\to0$ as $n$ tends to $\infty$ is obvious to me. But how do I arrive at

$\varphi_{Y_1} = 1 - \frac{t^2}{2n} + o\left(\frac{t^2}{n}\right)$

and what is the characteristic function before using Taylor's theorem. Where do we use the fact, that the variance and mean of

$\frac{S_n - \mathbb E(S_n)}{\mathbb V(S_n)}$

are $1$ and $0$ respectively.
Thank you
 A: (1). Characteristic functions of random variables always exist: indeed if $X$ is a $\mathbb{R}$-valued random variable, then $|e^{i\xi X}|\leq 1,\,\forall \xi \in \mathbb{R}$, which makes $e^{i \xi  X}\in L^1(\Omega,\mathscr{F},P)$. We have $(X_1,X_2,...)$ IID with $E[X_1^2]<\infty$ so that $X_1 \in L^1\cap L^2$ and $V[X_1]=E[X_1^2]-E[X_1]^2<\infty$. If we define
$$Z_n:=\frac{(X_1+...+X_n)-nE[X_1]}{\sqrt{nV[X_1]}}=\sum_{k\leq n}\frac{X_k-E[X_1]}{\sqrt{nV[X_1]}}$$
We have that $Z_n$ is the (scaled by $\sqrt{n}$) sum of IID rvs defined as $Y_k:=(X_k-E[X_1])/\sqrt{V[X_1]}$. Of course, $E[Y_1]=0$ and $V[Y_1]=E[Y_1^2]=1$. The characteristic function of $Z_n$ is
$$\begin{aligned}E[e^{i\xi Z_n}]&=E[e^{(i\xi/\sqrt{n})(Y_1+...+Y_n)}]=\\
&\stackrel{(a)}{=}E[e^{(i\xi/\sqrt{n})Y_1}]E[e^{(i\xi/\sqrt{n})Y_2}](...)E[e^{(i\xi/\sqrt{n})Y_n}]=\\
&\stackrel{(b)}{=}(E[e^{(i\xi/\sqrt{n})Y_1}])^n
\end{aligned}$$
where $(a)$ holds by independence and $(b)$ by identical distribution of $(Y_k)_{k\leq n}$.
(2). Here I provide references for the analytical framework needed to prove that the limit works. We can prove (Klenke; Th. 15.31) that
$$E[e^{(i\xi/\sqrt{n})Y_1}]=1+\frac{i\xi}{\sqrt{n}}E[Y_1]-\frac{\xi^2}{2n}E[Y_1^2]+\frac{\xi^2}{n}\varepsilon\bigg(\frac{\xi}{\sqrt{n}}\bigg)$$
where $\lim_{x \to 0}\varepsilon(x)=0$. It follows that $E[e^{(i\xi/\sqrt{n})Y_1}]=1-\frac{\xi^2}{2n}+\frac{\xi^2}{n}\varepsilon(\frac{\xi}{\sqrt{n}})$ and that (Klenke; Lemma 15.36) we ultimately have $(E[e^{(i\xi/\sqrt{n})Y_1}])^n\to e^{-\xi^2/2}$.
(3). We use Lévy continuity to claim $Z_n \stackrel{\textrm{d}}{\to}\mathcal{N}(0,1)$.
