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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?

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

2 Answers 2

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\begin{align} f_Y(y) & = \frac d {dy} \Pr(Y\le y) = \frac d{dy} \Pr( -y^2 \le X \le y^2) \\[10pt] & = \frac d {dy}\int_{-y^2}^{y^2} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \,dx = 2 \frac d {dy} \int_0^{y^2} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \,dx \\[10pt] & = 2 \frac d {dy} \int_0^u \cdots = 2 \frac{du}{dy} \cdot \frac d {du} \int_0^u \cdots =2 (2y) \cdot \frac d {du} \int_0^u \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \, dx \\[10pt] & = 4y \cdot \frac 1 {\sqrt{2\pi}} e^{-u^2/2} = 4y \cdot \frac 1 {\sqrt{2\pi}} e^{-y^4/2} \text{ for } y\ge 0. \end{align}

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  • $\begingroup$ didn't you loose somewhere -4y? $\endgroup$
    – Kran
    May 10, 2019 at 14:31
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Hint:

\begin{align} \Pr(Y \leq y) & = \Pr(\sqrt{|X|} \leq y) \\ &= \Pr( |X| \leq y^2) \\ &= \Pr(-y^2 \leq X \leq y^2)\\ &=\Pr(X \leq y^2)-Pr(X < -y^2) \end{align}

Differeniate with respect to $y$.

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