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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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    $\begingroup$ 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, $[0,\infty)$. $\endgroup$
    – hardmath
    May 28, 2013 at 12:34
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    $\begingroup$ 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. $\endgroup$ May 28, 2013 at 12:38

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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*}

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  • $\begingroup$ @hardmath What are you talking about? The density of $Z^2$ is infinite at zero. $\endgroup$
    – Did
    May 28, 2013 at 13:05
  • $\begingroup$ @Did: Thanks for the correction. The two terms of the formula obviously blow up, but I managed to convince myself the signs could be rearranged to cancel. $\endgroup$
    – hardmath
    May 28, 2013 at 14:06
  • $\begingroup$ @hardmath And I am quite curious to see how you managed to convince yourself of this. Once again, the density at zero is infinite. $\endgroup$
    – Did
    May 28, 2013 at 14:20
  • $\begingroup$ @Did: With a sign mistake: $$\frac{f_Z(\sqrt{x})-f_Z(-\sqrt{x})}{\sqrt{x} - (-\sqrt{x})}$$ $\endgroup$
    – hardmath
    May 28, 2013 at 14:23
  • $\begingroup$ @hardmath Indeed. (I thought my first comment was enough to make you realize this, but apparently not...) $\endgroup$
    – Did
    May 28, 2013 at 14:28
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If we make use of the following property for a random variable transformation

$f_Y(y) = f_Z(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|$.

Replacing $Y=g(Z)=Z^2$ and $f_Z(z)=\frac{1}{\sqrt{2 \, \pi}} \exp \left(-\frac{1}{2}(z- \mu)^2 \right)$ yields

$f_Y(y) = \sqrt{\frac{y}{2 \, \pi}} \exp\left( -\frac{(\sqrt{y}-\mu)^2}{2} \right)$

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