# 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? [duplicate]

How would I go about finding this density function? Thanks

• (1) Do you know how to find the partial distribution function for $Z$? (2) Do you know how to find the partial distribution function for $Y$, if you have $Y = f(Z)$ and you know the PDF of $Z$? Jul 8 '16 at 11:45
• See here. $Z^2$ is distributed as Chi-Square. So possible duplicate Jul 8 '16 at 12:07

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 $Z\sim N(0,1)$.
• @ConradTurner Thanks for the answer! I understand the answer but just had one more query. I've got the formula for the normal probability density function but to differentiate the cumulative distribution function for Y stated above, I would need to find cumulative distribution function for Z up to $\sqrt{y}$, but how could I do that? Jul 8 '16 at 13:31
• @SamuelOpiyo 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 $Z$. Jul 8 '16 at 15:59