For the following PDF of the Gamma distribution,
$ f(x)= \frac{1}{\Gamma{(k) \theta^k}} x^{k-1} e^{-\frac{x}{\theta}}$
with mean= $k \theta$ and variance = $k \theta^2$ , and if I assume $k \rightarrow \infty$,
then the random variable x should have approximately a normal distribution with the same mean and variance. Is this correct?
Can anyone please check whether the PDF below is the appropriate representation for the case of $k \rightarrow \infty$? Does it look like the following PDF?
$ f(x)= \frac{1}{\theta \sqrt{2\pi}} e^{ – {\frac{1}{2}} ( \frac{x- \theta}{\theta})^2}$
Thank you so much.
Best Answer
Yes, same variance and mean, say $N(k\theta;k\theta^2)$
Thus
$$f_X(x)=\frac{1}{\theta\sqrt{2\pi k}}e^{-\frac{1}{2k}(\frac{x-k\theta}{\theta})^2 }$$
It is an application of CLT because $Gamma(k;\theta)$ is the sum of $k$ iid $Exp(\theta)$ rv's. Note that, for example $(k=30; \theta=0,5)=\infty$
Take a look here at the densities' graphics when k increases... with $k=9$ the shape is already symmetric around $k\theta$