Change of variable problems in probability

Question

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.

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

Answer 1

$$P(\sqrt{X}\leq t)=P(X\leq t^2)=F(t^2)$$ 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.

Answer 2

Suppose that the CDF for $X$ is denoted by $F_X(x)$ and similarly for $Y$. The you have:

$$ F_Y(t)=\Pr(Y\leq t)=\Pr(X\leq t^2)= F_X(t^2) $$

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

Answer 3

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