Find the pdf of $X^2$ when $X$ is standard normal

Question

I have to calculate the pdf of $Y=X^2$ and I know that $X=N(0;1)$. I want to use the following theorem: $f_{Y}(y)=f_{X}(g^{-1}(y))∣\frac{\partial}{\partial y}(g^{-1}(y)∣$ However, I seem to be doing something wrong because if I integrate the pdf that I get, the result is $\frac{1}{2}$ when it should be $1$. Could someone please help me?

Answer 1

I assume what happened is you obtained the PDF as $\frac{1}{\sqrt{8\pi}}y^{-1/2}\exp-\frac{y}{2}$, which integrates on $[0,\,\infty)$ to $\frac{2^{1/2}\Gamma(1/2)}{\sqrt{8\pi}}=\frac12$. The problem is that because $x^2$ doesn't have a unique inverse, you should really sum over all $y$ consistent with a given $x$, which introduces a factor of $2$. In particular,$$x=\pm y^{1/2}\implies\frac{1}{\sqrt{2\pi}}\exp-\frac{x^2}{2}\cdot\left|\frac{dx}{dy}\right|=\frac{1}{\sqrt{8\pi}}y^{-1/2}\exp-\frac{y}{2},$$so now we just need to sum over the $\pm$.