Show that as $d$ goes to $\infty$, a standardized version of $X$ has the STD Normal Dist I am currently stuck on this problem and I would greatly appreciate some help. The problem is as follows:
Let $X$ have a chi-square with $d$ degrees of freedom. Show that a standardized version of $X$ has a limiting standard normal distribution as $d$ goes to $\infty$.
Here is my current work/take on the problem.
Since $X$ is a chi-square distribution with $d$ degrees of freedom, we know that 
$$X = \sum_1^d{Z_i^2}$$
Now of course, before we apply the Central Limit Theorem, we need to standardize $X$. So we need to find the $E(X)$ and $Var(X)$.
$E(X)=E(\sum_1^d{Z_i^2})=d$
$Var(X)=Var(\sum_1^d{Z_i^2})=2d$
Is this correct?
If so, then I would standardize like so:
$$ \frac{X-d}{\sqrt{2d}} = \frac{\sum_1^d{Z_i^2}-d}{\sqrt{2d}} = \frac{\sum_1^d{(Z_i^2-1)}}{\sqrt{2d}}$$
Then what I would do is get the MGF of just ${(Z_i^2-1)}$ and then the MGF of the whole numerator then the whole thing. Is that correct?
Help would be greatly appreciated in showing me where I went wrong or if the methodology is correct? thanks!
 A: I believe there is a simpler way. To avoid confusion with squares, set $Z_i^2 \equiv Y_i$.
Then note that
$$\operatorname{Var}(Y_i) \equiv \sigma^2_y= E(Z_i^4) - \Big[E(Z_i^2)\Big]^2 = 3 - 1 = 2$$
and so $\sigma_y = \sqrt2$. Also $E(Y) \equiv \mu_y =1$.
Therefore
$$\frac{X-d}{\sqrt{2d}} =  \frac 1{\sqrt d}\sum_1^d\frac {Y_i-\mu_y}{\sigma_y}$$
Then you can easily verify that, say, the Lyapunov condition is satisfied, and so that the Central Limit Theorem holds.
A: I will present an mgf approach as it sometimes makes convergence a little easier to see.
Let's start with your RV $X \sim \chi ^2 (d)$. The mgf of $X$ is then $(1-2t)^{-d/2}$. Now we are interested in the limiting mgf of the standardized variable, let's call it $Y$
$$Y=\frac{X-d}{\sqrt{2d}}$$
Based on the mgf of $X$ we now find the mgf of $Y$
$$M_Y =  E \left\{ \exp \left[ t \left( \frac{X-d}{\sqrt{2d}} \right) \right] \right\}=e^{-td/\sqrt{2d}} E \left( e^{tX/\sqrt{2d}} \right)= \exp \left[ - \left(t \sqrt{\frac{2}{d}} \right) \frac{d}{2} \right] \left( 1-2\frac{t}{\sqrt{2d}}\right)^{-d/2} \quad,t<\frac{\sqrt{2d}}{2} $$
Now we can rewrite the last line as:
$$M_Y = \left( e^{t\sqrt{2/d}} -t \sqrt{\frac{2}{d}} e^{t\sqrt{2/d}} \right)^{-d/2}, t<\sqrt{\frac{d}{2}}$$
We now use Taylor's formula to make the above a little more tractable. Expanding $e^{t\sqrt{2/d}}$ around $0$,
$$e^{t\sqrt{2/d}}=1+t\sqrt{\frac{2}{d}}+\frac{1}{2} \left( t \sqrt{\frac{2}{d}} \right)^2+\frac{e^{\xi}}{6} \left( t \sqrt{\frac{2}{d}} \right)^3 $$
where $\xi$ is as usual between $0$ and $t\sqrt{2/d}$.
If we now substitute the above into the last expression for $M_Y$, it is seen that:
$$M_Y=\left(1-\frac{t^2}{d}+\frac{\psi}{d} \right)^{-d/2}$$
where 
$$\psi= \frac{\sqrt{2}t^3 e^{\xi}}{3\sqrt{d}}-\frac{\sqrt{2}t^3}{\sqrt{d}}-\frac{2t^{4}e^{\xi}}{3d}$$
Now there is nothing left for us to do but take the  limit as $ d \to \infty $. Now, as $d \to \infty$ $\xi \to 0$ which implies that $\psi \to 0$ as well.
Bearing that in mind, let us compute the limit of the mgf:
$$ \lim_{d \to \infty}M_Y= \lim_{d \to \infty} \left(1-\frac{t^2}{d}+\frac{\psi}{d} \right)^{-d/2}=\lim_{d \to \infty} \left(1-\frac{t^2}{d} \right)^{-d/2}=e^{t^2 /2}$$
the mgf of a standard normal distribution as required $\blacksquare$
The proof is long but instructive. You can bypass it by invoking the CLT but in that case you would miss several important results concerning mgf's ; ). In any case, since you put the mgf tag in your question, here you go!
