If Z has a normal distribution with mean 0 and variance $\sigma^{2}$, and $Y=Z^{2}$, what would the density function of Y be?

Question

How would I go about finding this density function? Thanks

Answer 1

The cumulative distribution function for $Y$ is: $$ F_Y(y)=\operatorname{Prob}(Y<y)=\operatorname{Prob}(-\sqrt{y}<z<\sqrt{y})=F_Z(\sqrt{y})-F_Z(-\sqrt{y})=2F_Z(\sqrt{y})-1 $$ Now to get the density differentiate with respect to $y$, and of course to differentiate $2F_Z(\sqrt{y})$ you use the chain rule and the fact that the derivative of $F_Z(x)$ with respect to $x$ is $f_Z(x)$ the density of $Z$, which you know as $Z\sim N(0,1)$.