Consider $G = \operatorname{Gamma}(p)$. As $p$ goes to $\infty$, the Gamma becomes more and more bell-shaped. How do I show that $\frac{G - p}{\sqrt{p}} \to Z \sim N(0,1)$ as $p \to \infty$?

I started with the CDF of the Gamma and began taking the limit, but it got very messy.

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    $\begingroup$ Have you considered using the MGF? (or the CF) . It's often a convenient strategy. Perhaps consider a Taylor-type expansion. $\endgroup$ – Glen_b Oct 25 '18 at 5:23
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    $\begingroup$ I have not. My instructor suggested this as a fun practice problem using only the CDF and PDF. $\endgroup$ – purpleostrich Oct 25 '18 at 5:44
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    $\begingroup$ @StubbornAtom it doesn't help that Z is used to represent two distinct things in the question. It would be necessary to fix that first $\endgroup$ – Glen_b Oct 25 '18 at 6:00
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    $\begingroup$ Alternative to using MGF, you can write $G_p$ as being equal in distribution to the sum of $p$ i.i.d $exp(1)$ random variables. The result is then immediate by CLT. $\endgroup$ – Xiaomi Oct 25 '18 at 10:46
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    $\begingroup$ The brute-force analysis isn't that difficult if you plan it out. Expand the log of the (unnormalized) PDF of $Z$ in a Maclaurin series. It will equal $$f_Z(z) = -\frac{1}{\sqrt p} + \left(\frac{1}{2p} - \frac{1}{2} \right)z^2 + O(p^{-1/2})O(z^3).$$ Thus its exponential is $e^{-z^2/2}$ times an expression that is very close to $1.$ Justify taking the limit under the integral sign and you're done. $\endgroup$ – whuber Oct 25 '18 at 14:24

This answer is part of a previous answer with a link here. That portion of the previous answer is copied over here so that one can see that the question above has been answered, however, as the answer here formed only part of an answer to a different question, it might not have been noticed in the different context of the question above. Text as follows:

A more direct relationship between the gamma distribution (GD) and the normal distribution (ND) with mean zero follows. Simply put, the GD becomes normal in shape as its shape parameter is allowed to increase. Proving that that is the case is more difficult. For the GD, $$\text{GD}(z;a,b)=\begin{array}{cc} & \begin{cases} \dfrac{b^{-a} z^{a-1} e^{-\dfrac{z}{b}}}{\Gamma (a)} & z>0 \\ 0 & \text{other} \\ \end{cases} \,. \\ \end{array}$$

As the GD shape parameter $a\rightarrow \infty$, the GD shape becomes more symmetric and normal, however, as the mean increases with increasing $a$, we have to left shift the GD by $(a-1) \sqrt{\dfrac{1}{a}} k$ to hold it stationary, and finally, if we wish to maintain the same standard deviation for our shifted GD, we have to decrease the scale parameter ($b$) proportional to $\sqrt{\dfrac{1}{a}}$.

To wit, to transform a GD to a limiting case ND we set the standard deviation to be a constant ($k$) by letting $b=\sqrt{\dfrac{1}{a}} k$ and shift the GD to the left to have a mode of zero by substituting $z=(a-1) \sqrt{\dfrac{1}{a}} k+x\ .$ Then $$\text{GD}\left((a-1) \sqrt{\frac{1}{a}} k+x;\ a,\ \sqrt{\frac{1}{a}} k\right)=\begin{array}{cc} & \begin{cases} \dfrac{\left(\dfrac{k}{\sqrt{a}}\right)^{-a} e^{-\dfrac{\sqrt{a} x}{k}-a+1} \left(\dfrac{(a-1) k}{\sqrt{a}}+x\right)^{a-1}}{\Gamma (a)} & x>\dfrac{k(1-a)}{\sqrt{a}} \\ 0 & \text{other} \\ \end{cases} \\ \end{array}\,.$$

Note that in the limit as $a\rightarrow\infty$ the most negative value of $x$ for which this GD is nonzero $\rightarrow -\infty$. That is, the semi-infinite GD support becomes infinite. Taking the limit as $a\rightarrow \infty$ of the reparameterized GD, we find

$$\lim_{a\to \infty } \, \frac{\left(\frac{k}{\sqrt{a}}\right)^{-a} e^{-\frac{\sqrt{a} x}{k}-a+1} \left(\frac{(a-1) k}{\sqrt{a}}+x\right)^{a-1}}{\Gamma (a)}=\dfrac{e^{-\dfrac{x^2}{2 k^2}}}{\sqrt{2 \pi } k}=\text{ND}\left(x;0,k^2\right)$$

Graphically for $k=2$ and $a=1,2,4,8,16,32,64$ the GD is in blue and the limiting $\text{ND}\left(x;0,\ 2^2\right)$ is in orange, below

enter image description here


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