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I was studying probability distribution functions and had a particular question about an exercise problem. The problem goes:

Let $Z$ ~ $\mathcal{N}(0, 1)$. Find the PDF for $Z^4$.

I've solved the problem, but missed defining the support, and frankly am not sure how the support came to be. Allow me to elaborate.


My Solution

Let $Y = Z^4$. We cannot use the change of variables formula due to the fact that $y = z^4$ is not strictly increasing or decreasing. We will find the CDF for $Y$ first, then differentiate to find the PDF.

\begin{align} F_Y(y) & = P(Y\le y) \\ & = P(Z^4 \le y) \\ & = P(|Z| \le y^{\frac{1}{4}}) \\ & = P(-y^{\frac{1}{4}} \le Z \le y^{\frac{1}{4}}) \\ & = P(Z \le y^{\frac{1}{4}})\ -\ P(Z\le -y^{\frac{1}{4}}) \\ & = \Phi(y^{\frac{1}{4}})\ -\ \Phi(-y^{\frac{1}{4}}) \\ & = \Phi(y^{\frac{1}{4}})\ -\ (1\ -\ \Phi(y^{\frac{1}{4}})) \\ & = 2\Phi(y^{\frac{1}{4}})\ -\ 1 \end{align}

Now that we've found the CDF of $Y$, we can find the PDF of $Y$ by differentiating the CDF and using the chain rule.

\begin{align} \frac{d}{dy}F_Y(y) & = \frac{d}{dy}(2\Phi(y^{\frac{1}{4}})\ -\ 1) \\ & = 2\frac{d}{dy}\Phi(y^{\frac{1}{4}})\ -\ \frac{d}{dy}1 \\ & = 2\phi(y^{\frac{1}{4}})\times \frac{d}{dy}(y^{\frac{1}{4}})\ - 0\\ & = \frac{1}{4\sqrt{2\pi}}e^{-\frac{y^{1/2}}{2}}y^{-\frac{3}{4}} \end{align}


Apparently the equation itself is correct, but I got it wrong because I forgot to specify that it only holds when $y \gt 0$ and is $0$ elsewhere.


I'm having trouble how this support came to be. I know that a Normal distribution's support is defined on $(-\infty,\ \infty)$, and I guess I was under the assumption that the same would hold for $\mathcal{N}^4$. Where did the support for $y$, $(0,\ \infty)$, come from?


Any feedback is appreciated. Thank you.

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    $\begingroup$ $z\in\mathbb R\implies z^2\ge 0\implies (z^2)^2\ge 0$. $\endgroup$ – StubbornAtom Nov 26 '18 at 14:58
  • $\begingroup$ Thank you! You may also post an answer if you'd like. :) $\endgroup$ – Seankala Nov 26 '18 at 15:40

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