If I have a normal distribution $X$ with mean 0 and variance $\sigma^2$ for $\sigma>0$, how would I find the moment generating function of $Y=X^2$?
I can find the moment generating function of a normal distribution. But I'm not sure how that changes if I'm squaring the distribution.
Thank you in advance!
The square of a standard normal distribution is a chi-squared distribution, and its moment generating function can be looked up from here:
http://en.wikipedia.org/wiki/Chi-squared_distribution
I think you can proceed formally and compute it by $E(e^{tX^{2}})$ by definition as well. This might be easier in practice.
To make matter simple I will do this for the standard normal, we have $$ f_{X}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}} $$ and we want to integrate $$ \frac{1}{\sqrt{2\pi}}e^{(t-\frac{1}{2})x^{2}} $$ We can use a scale transformation $u=\sqrt{1-2t}x$. Then we change the integral to be $$ \int^{\infty}_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{u^{2}}{2}}du*\frac{1}{\sqrt{(1-2t)}}=\frac{1}{\sqrt{(1-2t)}} $$ as desired. I think the general case is similar.