Find distribution $Y=X^2$

Question

X~N(0,1). Find distribution $Y=X^2$

Can someone help me? I have no idea how to do it. I could try to start like this:

$F_Y(t)=P(X^2<t)=P(-\sqrt(t)<X<\sqrt{t})$

Answer 1

Going off what you have, we have $$P(-\sqrt{t}<X<\sqrt{t})=P(X<\sqrt{t})-P(X<-\sqrt{t})$$ but by symmetry we have $$P(<\sqrt{t})-\left(1-P(<\sqrt{t})\right)=2P(X<\sqrt{t})-1=2F(\sqrt(t))-1$$ Now to find distribution we will find pdf with what we have above thus we have $$f_{y}(t)=\frac{d}{dt}\left(F_{Y}(t)\right)=\frac{d}{dt}\left(2F_{X}(\sqrt{t})-1\right)=2\frac{1}{2\sqrt{t}}f_{X}(\sqrt{t})=\frac{1}{\sqrt{t}}f_{X}(\sqrt{t})$$

Now from here just use pdf of $X$ (i.e. pdf of $N(0,1)$) and you should get pdf for $Gamma(\frac{1}{2},\frac{1}{2})$ (i.e chi square with df =1). Recall also that $\Gamma(\frac{1}{2})=\sqrt{\pi}$

Answer 2

You are almost here, but you should consider two cases: $$F_Y(t) = P(X^2 < t) = \begin{cases} 0,& t \leqslant 0\\ P(-\sqrt t < X < \sqrt t), & t > 0.\end{cases} = \\ \begin{cases} 0,& t \leqslant 0,\\ \displaystyle \frac{1}{\sqrt{2\pi}}\int\limits_{-\sqrt t}^{\sqrt t}e^{-\frac{x^2}{2}}dx, & t > 0.\end{cases} = \begin{cases} 0,& t \leqslant 0,\\ \Phi(\sqrt t) - \Phi(- \sqrt t), & t > 0.\end{cases}$$ where $\Phi(t)$ is the cdf of standard normal distribution. The distribution of $Y$ is also known as $\chi^2(1)$, mentioned by Troy Woo. It is called Chi-squared distribution with 1 degree of freedom.

Updated: To find pdf instead of cdf you can use the formula: $$f_{g(X)}(y) = \sum_{i = 1}^{n(y)}\Big|\frac{d}{dy}\big(g_k^{-1}(y)\big)\Big|f_X\big(g_k^{-1}(y)\big),$$ where $n(y)$ is the number of solutions to $y = g(x)$ and $g_k^{-1}(y)$ are these solutions.

From the above considerations it is clear that $f_{X^2}(y) = 0, \text{for }y \leqslant 0$. In case of $y = x^2$ the solutions are $g_1^{-1}(y) = -\sqrt y$ and $g_2^{-1}(y) = \sqrt y$. Applying the formula above we get: $$ f_{X^2}(y) = \Big|\frac{d}{dy}\big(-\!\sqrt y\big)\Big|f_X\big(-\!\sqrt y\big) + \Big|\frac{d}{dy}\big(\sqrt y\big)\Big|f_X\big(\sqrt y\big) = \frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}},\ \text{for } y >0.$$