I want to prove that $\frac{(X_n-n)}{\sqrt{2n}}\xrightarrow{{L}}Z\sim N(0,1)$ as $n\rightarrow\infty$ where $X_n\sim\chi^2_n$.
I don't want to use the CLL directly. So far this is what I got, but I am stuck at the end:
Considering sequence of rvs of $Z_1,Z_2,...\sim iid N(0,1)$, $\chi^2_n\equiv\Sigma_{i=1}^nZ_i^2\sim Chisq(n)$. We know that $E(\chi^2_n)=n$ and $Var(\chi^2_n)=2n$.
Let $Y_i=(X_i-n)/\sqrt{2n}$ with mean $n$ and variance of $2n$. (Should I consider n and 2n or just 1 and 2 (assume n=1) for mean and var?)
So $(X_i-n)/\sqrt{2n}=\Sigma_{i=1}^nY_i=Z_n$.
By Taylor series $Y_1$ has mgf:
$$M_{Y_1}(s)=E[e^{sY_1}]=\int_Re^{sY}dP(Y_1\leq y)=\int_R(1+sy+\frac{1}{2!}s^2y^2+o(s^2y^2))dF_{Y_1}(y)$$ $$=1+n+s^2n+o(2s^2n)$$
Since $X_i$s are iid: $$E[e^{sZ_n}]=E[e^{s\Sigma_{i=1}^nY_i}]=(E[e^{sY_1}])^n=(1+n+s^2n+o(2s^2n))^n$$
To prove the statement mentioned, this should converge to $e^{s^2/2}$ as $n\rightarrow\infty$ which is the mgf of $Z$, but it doesn't! Can someone please help me find out which part I am doing wrong and how to do this correctly?
Thank you in advance
The way you approximated the mgf of $Y_1$ is not quite right. Moreover, you wrote $Z_n = \sum_{i=1}^n (X_i - n)/\sqrt{2n},$ which should actually be $Z_n = \sum_{i=1}^n (X_i - 1)/\sqrt{2n}.$
An easier way is to use the usual mgf of $X_1$, which is given by $$\mathbb{E}[e^{tX_1}] = (1-2t)^{-1/2}, \text{ for }t < 1/2.$$ This gives $$\mathbb{E}[e^{sY_1}] = \mathbb{E}[e^{s(X_1 - 1)/\sqrt{2n}}] = e^{-s/\sqrt{2n}} (1-2s/\sqrt{2n})^{-1/2}.$$ Note that this is valid only for $s/\sqrt{2n}< 1/2.$ This is why it is better to use characteristic functions instead of moment generating functions. The characteristic function of $Z_n$ is given by $$\mathbb{E}[e^{isZ_n}] = e^{-is\sqrt{n/2}} (1-2is/\sqrt{2n})^{-n/2}.$$ The advantage here is that it is defined for all $s\in \mathbb{R},$ and now you should show that it converges to the required characteristic function, which is $e^{-s^2/2}.$ This can be shown by taking log of the characteristic function of $Z_n,$ as shown below. $$-is\sqrt{\frac{n}{2}} -\frac{n}{2}\log \left(1 - \frac{2is}{\sqrt{2n}}\right) = -is\sqrt{\frac{n}{2}} -\frac{n}{2} \left(- \frac{2is}{\sqrt{2n}} + \frac{2s^2}{n} + o\left(\frac{1}{n}\right)\right) = - \frac{s^2}{2} + o(1).$$