Hello, I put together the following by trying to make sense of my notes. I discussed this with a classmate and we had some disagreement, so I would like to know if I'm doing it correctly. Assume that X is normally distributed with mean mu m and variance sigma s squared. So Based on the above assumption, this implies X's pdf is: (1/(sqrt(2 pi) s))*exp((-1/2)((x-m)/s)^2) So far okay? Ok so now the problem is: Find the PDF of Y = X^2. I have a relationship P(X <= t) = integral( f(x), x, -inf, t) where f(x) is the pdf of x. So I do, P(Y <= t) = P( -sqrt(t) <= X <= sqrt(t)) = ugly integral. Then I differentiate using the Fundamental Theorem of Calculus to get rid of the integral, and that result is my PDF for X^2. The work is pretty messy so I won't post it, and I'm confident I did that part correctly. I'm just wondering if the process outlined above is correct.