# Square of a normal distribution

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.

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Working with the CDF is a usual approach. Notice that while the normal distribution has support on the whole real line, no matter what mean $\mu$ is, the square $Z^2$ will have have support only on the nonnegative half, $[0,\infty)$. – hardmath May 28 '13 at 12:34
Hint: For $\alpha \geq 0$, $F_Z(\alpha) = P\{Z \leq \alpha\} = P\{-\sqrt{\alpha} \leq X \leq \sqrt{\alpha}\}$. Now express the right side in terms of $F_X(\cdot)$ and differentiate w.r.t $\alpha$, remembering to use the chain rule. – Dilip Sarwate May 28 '13 at 12:38
thanks, sorted it now – Tom May 28 '13 at 12:49

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*}
@hardmath What are you talking about? The density of $Z^2$ is infinite at zero. – Did May 28 '13 at 13:05
@Did: With a sign mistake: $$\frac{f_Z(\sqrt{x})-f_Z(-\sqrt{x})}{\sqrt{x} - (-\sqrt{x})}$$ – hardmath May 28 '13 at 14:23