# Gamma distribution PDF

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.

Yes, same variance and mean, say $$N(k\theta;k\theta^2)$$
$$f_X(x)=\frac{1}{\theta\sqrt{2\pi k}}e^{-\frac{1}{2k}(\frac{x-k\theta}{\theta})^2 }$$
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 $$k\theta$$