Let $Y = X^2$. Find the pdf of Y when the distribution is $N(0,1)$.

Question

I've performed a change of variable: $$X = \sqrt{y}$$ $$X'=\frac{1}{2}Y^{-\frac{1}{2}}$$ Thus: $$f(\sqrt{y})*X'=f(y)=\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}$$ However the book gives: $$f(y)=\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}}$$ Where did I go wrong?

Answer 1

HINT: You should consider $P(-\sqrt{x}<y<\sqrt{x})$.

Answer 2

Thanks for the hint. I computed only half the result. To complete the solution:

$$f(y)=\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}+\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}=\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}},0<y<\infty$$