Limiting distribution. 
Let $Y_n \sim \chi^2(n) $. Find the limiting distribution, $(Y_n-n)/ \sqrt{2n}$ as $n\rightarrow \infty $, using moment generating functions.

I don't know how to properly calculate the moment generating function. Or to calculate the limit. 
I'll be grateful for the help and advices. 
 A: The MGF of a chi-square distribution with $n$ degrees of freedom is $$M_Y(t) = {\rm E}[e^{t Y}] = (1-2t)^{-n/2}, \quad t < \tfrac{1}{2}.$$  (Throughout, I have used $Y$ instead of $Y_n$ for simplicity of notation.)  Now let $$X = \frac{Y - n}{\sqrt{2n}}.$$  Use the above to compute the MGF of $X$ $$M_X(t) = {\rm E}[e^{tX}].$$  Then let $X_\infty$ be the limiting distribution of $X$ as $n \to \infty$; then the MGF of $X_\infty$ is simply $$M_{X_\infty}(t) = \lim_{n \to \infty} M_X(t).$$  What is this limit, and what is the distribution that has this MGF?
A: Given
$X\sim\chi^2(n)$ then,
$$f(x)=\frac{1}{2^{\frac{n}{2}}\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{\frac{n}{2}},x>0$$
$$Y_n=\frac{X-n}{\sqrt{2n}}$$
Let start,
$$E_{Y_n}(t)=E(e^{Y_nt})=E(e^{\frac{X-n}{\sqrt{2n}}t})=E\bigg(e^{\frac{X}{\sqrt{2n}}t-\frac{n}{\sqrt{2n}}t}\bigg)$$
$$=e^{-\frac{n}{\sqrt{2n}}t}E\bigg(e^{\frac{X}{\sqrt{2n}}t}\bigg)=e^{-\frac{n}{\sqrt{2n}}t}\bigg(1-2\frac{t}{\sqrt{2n}}\bigg)^{-\frac{n}{2}}$$
The limiting distribution,
$$\lim_{n \to \infty} E_{Y_n}(t)=\lim_{n \to \infty} e^{-\frac{n}{\sqrt{2n}}t}\bigg(1-2\frac{t}{\sqrt{2n}}\bigg)^{-\frac{n}{2}}=\lim_{n \to \infty} e^{-t\sqrt{\frac{n}{2}}}\bigg(1-t\sqrt{\frac{2}{n}}\bigg)^{-(\frac{n}{2})}$$
$$=\lim_{k \to \infty} e^{-tk}\bigg(1-\frac{t}{k}\bigg)^{-k^2}=\lim_{k \to \infty} e^{\ln{\bigg[e^{-tk}\bigg(1-\frac{t}{k}\bigg)^{-k^2}}\bigg]}=\lim_{k \to \infty} e^{-tk-k^2\ln{(1-\frac{t}{k}})}$$
Nearly there, let expand the $\ln{(1-\frac{t}{k}})$,
$$=\lim_{k \to \infty} e^{-tk-k^2(-\frac{t}{k}-\frac{1}{2}(\frac{t}{k})^2-\frac{1}{3}(\frac{t}{k})^3-\frac{1}{4}(\frac{t}{k})^4-...)}$$
$$=\lim_{k \to \infty} e^{-tk+tk+\frac{t^2}{2}+\frac{t^3}{3k}+\frac{t^4}{4k^2}+...}$$
$$=\lim_{k \to \infty} e^{\frac{t^2}{2}+\frac{t^3}{3k}+\frac{t^4}{4k^2}+...}$$
$$=e^{\frac{t^2}{2}}$$
The rest I think you can done by youself.
A: I know this has been answered, but I think this is a slightly quicker answer.
The mgf of $\chi^2_{\nu}$ is 
    $$\left(\frac{1}{1-2t}\right)^{\frac{\nu}{2}}$$
        The mgf of a linear combination is, in general,
    $$M_{a\chi^2_\nu+b}(t)=e^{bt}M_{\chi^2_\nu}(at)$$
    In our case, $b=-\sqrt{\frac{\nu}{2}}$ and $a=\frac{1}{\sqrt{2\nu}}$. We find that, essentially incorporating the $1/\sqrt{2\nu}$ into $t$ that:
    $$ E\left(e^{t\chi^2_{\nu}/\sqrt{2\nu}}\right)=\left(\frac{1}{1-2t/\sqrt{2\nu}}\right)^{\nu/2}$$
    So, the mgf of $\frac{\chi^2_{\nu}-\nu}{\sqrt{2\nu}}$ is 
        $$e^{-\frac{\sqrt{\nu} t}{\sqrt{2}}}\left(\frac{1}{1-2t/\sqrt{2\nu}}\right)^{\nu/2}$$
    Now, we take the log:
    \begin{align*}
 &-\sqrt{\frac{\nu}{2}}\cdot t+\frac{\nu}{2}\ln(\frac{1}{1-2t/\sqrt{2\nu}})\\
 &=-\sqrt{\frac{\nu}{2}}\cdot t-\frac{\nu}{2}\ln(1-2t/\sqrt{2\nu})\\
% &=-\sqrt{\frac{\nu}{2}}\cdot t-\frac{\nu}{2}\ln(1-2t/\sqrt{2\nu})
 \end{align*}
    Taylor expand $\ln(1-2t/\sqrt{2\nu})$.
In general, the Taylor expansion for $\ln(1-x)$ around 0 is 
$$\ln(1-a=0)=\ln(1-0)-\frac{1}{1-0}\frac{x-0}{1!}+\frac{1}{(1+0)^2}\frac{(x-0)^2}{2!}-\frac{2}{(1+0)^3}\frac{(x-0)^3}{3!}$$
With the change of variable $x=\frac{2t}{\sqrt{2\nu}}$.
$$-\sqrt{\frac{\nu}{2}}\cdot t-\frac{\nu}{2}\left(-\frac{2t}{\sqrt{2\nu}}+                    \frac{4t^2}{2\cdot2\nu}-\frac{8t^3}{3!\cdot2\sqrt{2}\cdot\nu^{3/2}}\right)=\frac{t^2}{2}+\frac{\nu t^3}{3\nu^{3/2}}+\ldots$$
As $\nu\rightarrow\infty$, the only term left is $\frac{t^2}{2}$.  Since we took the log, this means the mgf converges to $e^{\frac{t^2}{2}}$, the mgf of the standard normal.  
We can visualize this too: $\nu=100000$ data generated from sample of 100000 points
$\nu=100000$ data generated from sample of 10000 points">
Here is some code for the graph, in R.
  library(tidyverse)
  library(dplyr)
  library(reshape2)
  nu=100000
  nhere=100000
  datamerge930b=data.frame(cbind((rchisq(nhere,nu)-  nu)/sqrt(2*nu),rnorm(nhere, mean=0, sd=1)))
 colnames(datamerge930b)=c('chisquared', 'stdnorm')
 datamerge930b=melt(datamerge930b)

 datamerge930b%>%
 ggplot(aes(x = value, fill = variable)) + geom_density(alpha=0.69)+
 scale_fill_manual( values =c("dodgerblue4",'firebrick4'),
               name="distributions",
                    breaks=c("chisquared","stdnorm"),
                     labels=c("chi-squared","standard norm"))+
 ggtitle('\u03bd=100000') +theme_minimal()+ theme(plot.title =    element_text(hjust = 0.5), text = element_text(size=16))

