Square of a normal distribution

Question

Let $Z$ have a normal distribution with mean $\mu$ and variance $1$. Show that $Z^2$ is a continuous random variable and find its p.d.f.

I really don't know what to do with this... I tried working out the CDF but it didn't get me anywhere.

Answer 1

Note as we have $Z^2\ge 0$, we have $P(Z^2 \le x) = 0$ for $x < 0$. Now let $x \ge 0$, then $$ F_{Z^2}(x) = P(Z^2 \le x) = P(Z \in [-\sqrt x,\sqrt x]) = F_Z(\sqrt x) - F_Z(-\sqrt x) $$ where $F_Z$ is the cdf of $Z$. Now take the derivative with respect to $x$, we get for the pdf that \begin{align*} f_{Z^2}(x) &= \frac 1{2\sqrt x}f_Z(\sqrt x) + \frac 1{2\sqrt x} f_Z(-\sqrt x)\\ \end{align*}