# $X\sim\mathcal N(0,1);\ Y = \sqrt{|X|}$; find $f_Y(y)$

Let $X\sim\mathcal N(0,1)$ be a normal distribution and $Y = \sqrt{|X|}$. Find $f_Y(y)$, the probability density function of $Y$.

I figure trying to find the expected value of $Y$ and the variance is a good start, but I have no idea how to integrate $\frac{2}{\sqrt{2 \pi}}\int^\infty_0 x^{\frac{1}{2}}e^{\frac{-x^2}{2}}$. Is there a good way to integrate this or am I missing an easier method to solving this?

• Hint: $f_Y(y) = \frac{d}{dy} P(Y \le y) = \frac{d}{dy} P(-y^2 \le x \le y^2)$ for $y > 0$. Sep 24, 2016 at 0:47

Differeniate with respect to $y$.