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I am having problem with a probablity theory question.

Given that $X$ is a $N(0,1)$-distributed (normal) random variable. Find the density funciton of $X^2$.

So, i put $X^2 = Y$.

$F_Y(y) = P(Y \le y) = P(X^2 \le y) = P(X \le\sqrt y) = F_X(\sqrt y)$

This is where i get stuck. How do i differentiate $F_X(\sqrt y)$ in order to find $f_Y(y)$?

Thanks and best regards!

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    $\begingroup$ $P(X^2 \leq y) = 2 P(X \leq \sqrt{y})$. Note $\frac{d}{dy} F_X(\sqrt(y)) = F'_X(\sqrt{y}) \frac{1}{2 \sqrt{y}}$. $\endgroup$ – user438618 Jul 2 '17 at 13:03
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$X$~$N (0,1)$ this implies $X^2$~$\chi^2 (1) $ What is $F_X (\sqrt (y))$= the probability that a standard normal variable is less that or equal to $\sqrt (y)$. Now just differentiate under the integration using Leibnitz rule. What you end up with will be your pdf of $X^2$

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