Change of variable problems in probability If $X$ is a continuous random variable with positive values,find it's Cumulative distribution function (CDF) and it's probability density function (PDF) for $$Y = \sqrt{X}$$
This was a detable question on another topic so I made it a seperate one.
Can someone explain how do we solve this problem to it's end? (putting some theory/ explanations in between would be appreciated. My goal is to understand the concepts that are behind this and be prepared for problems of equal difficulty). Thank you.


Added by A.F. This was originally the "third pattern" of this question.  Possible solutions may exist on the linked question.

 A: $$P(\sqrt{X}\leq t)=P(X\leq t^2)=F(t^2)$$
where $F$ is the cumulative distribution function of $X$. Therefore pdf of $Y$ is 
$$\frac{dF(t^2)}{dt}=f(t^2)\times2t$$
where $f$ is the pdf of $X$.
See any elementary book on probability and statistics, for a better understanding of transformation of random variables.
A: Suppose that the CDF for $X$ is denoted by $F_X(x)$ and similarly for $Y$. The you have:
$$
F_Y(t)=\Pr(Y\leq t)=\Pr(X\leq t^2)= F_X(t^2)
$$

$$
f_Y(t)=2t\times\frac{d F_X}{dt}(t^2)=2t\times f_X(t^2)
$$

A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\expo}{{\rm e}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}$
$$
{\rm P}_{Y}\left(Y\right)\,{\rm d}Y
=
{\rm P}_{X}\left(X\right)\,{\rm d}X
\quad\imp\quad
\color{#ff0000}{{\rm P}_{Y}\left(Y\right)}
=
{\rm P}_{X}\left(X\right)\,{{\rm d}X \over {\rm d}Y}
=
{\rm P}_{X}\left(X\right)\,2Y
=
\color{#ff0000}{2{\rm P}_{X}\left(Y^{2}\right)Y}
$$
