Find the distribution of $Y=\sqrt X$ when $X$~$N(0,1)$ PDF of $X$ is $f_x(t)=\frac{1}{\sqrt{2pi}}e^{\frac{-x^2}{2}}$.
I tried to found the CDF of $Y$: $F_Y(t)=P(Y \leq t)=P(\sqrt X \leq t)=P(X \leq t^2)=F_X(t^2)$
How do i get distrbution of Y from this? If I were to find the $P(X\leq t^2)$ I would get error function and i don't have the CDF of X to find $F_X(t^2)$
I tried to find the PDF of $Y$: $f_Y(t)=F'_Y(t)=F'_X(t^2)=2tf_X(t^2)$
Does PDF define the distribution of random variable? If not how to i find it?
 A: Here is a solution to finding the distribution of $Y=\sqrt{X}$ conditional on $X>0$ where unconditionally $X\sim \mathcal{N}(0,1)$. In this case, $Y=\sqrt{X}$, given $X>0$ is properly defined, because the square-root is undefined for negative reals. The conditional distribution of $X$ is given by the basic definition of conditional probability conditional on events:
$$P(X\leq x| X>0)=P(0<X\leq x)/P(X>0)=\frac{1}{1-\Phi(0)}[\Phi(x)-\Phi(0)],$$
so that then
$$P(Y\leq y|X>0)=P(\sqrt{X} \leq y |X >0)$$
$$=P(X\leq y^2 |X>0)=\frac{1}{1-\Phi(0)}[\Phi(y^2)-\Phi(0)],$$
now differentiate with respect to $y$ to find that the conditional probability-density-function of $Y$, conditional on $X>0$, is given by
$$f_Y(y|x>0)=\frac{2y \cdot \phi(y^2)}{1-\Phi(0)}$$
Here is a visual/numerical verification in R (black is the empirical density, blue is the exact density derived above):

Please comment for further clarifications/questions/corrections.
A: If you consider the absolute value of $X$ instead (half-normal distribution), you should be able to derive the pdf of $Y:=\sqrt{|X|}$ via the density transformation formula.
