I'm trying to show that as $\alpha$ tends to 0, the gamma distribution $$\Gamma(\lambda,\alpha),$$ is properly standardised, tends to the standard normal distribution. I have figured out that the moment generating function for the gamma distribution is $$\left(\frac{\lambda}{\lambda-t}\right)^\alpha.$$ Also, I've worked out that the mean and variance of a gamma random variable is $$\frac{\alpha}{\lambda}$$ and $$\frac{\alpha}{\lambda^2}$$ respectively.
However, I am not sure how to proceed further. I tried by defining $$Z=\frac{X-\frac{\alpha}{\lambda}}{\frac{\alpha^0.5}{\lambda}}$$ and using the fact that $M_{_Z}(t) = e^{bt}M_{X}(at)$
However, I can't show that $$M_{_Z}(t)=e^{t^2/2}$$ which is the moment generating function of a standard normal random variable. Is this the correct way to proceed?
Best Answer
First of all, you seem to be using $t$ for two different purposes: a parameter of the Gamma distribution and the variable in the moment generating function. These should be completely different. The Gamma distribution with shape parameter $k$ and rate parameter $r$ has mean $\mu = k/r$, variance $\sigma^2 = k/r^2$, and moment generating function $M_X(t) = \left(\frac{r}{r-t}\right)^k$. The limit you should be taking is $k \to \infty$ with $r$ fixed. The MGF of the scaled and translated variable $Y = (X-\mu)/\sigma$ is then $M_Y(t) = \left( 1 - \frac{t}{\sqrt{k}}\right)^{-k} e^{-\sqrt{k} t}$. I suggest taking logarithms, and using a degree-2 Taylor expansion of $\ln(1 - t/\sqrt{k})$.