Finding PDF of $X^2$ when $X$ is standard normal

Question

I'm given a standard normal $X$, and $Y=X^2$. To find the PDF of $Y$, we have $$F_Y(a)=P(X^2\le a)=P(X \le a^{1/2})=F_X(a^{1/2}),$$

so $$f_Y(a)=f_X(a^{1/2})*\frac{1}{2a^{1/2}}=\frac{1}{\sqrt{2\pi}}e^{-a/2}\frac{1}{2a^{1/2}}.$$

The latter isn't a PDF since its integral from 0 to infinity (the range of $Y$) is 1/2. What is my mistake?

Answer 1

If $a\leqslant 0$ then $F_Y(a)=0$. If $a>0$, then $F_Y(a)=\mathbb P(-\sqrt{a}\leqslant X\leqslant \sqrt{a})=F_X(\sqrt{a})-F_X(-\sqrt{a})$.