CDF RV Let Y = X^2. Compute the pdf fY (y). Since X = pY ,

Question

Let $X$ be a continuous random variable with pdf $$f_X(x)=\begin{cases}6x^{-7}&\text{if }x\ge1\\0&\text{otherwise.}\end{cases}$$

Let $Y=X^2$. Compute the pdf $f_Y(y)$. Since $X=\sqrt Y$, $f_Y(y)=f_X(\sqrt y)/|dy/dx|=f_X(\sqrt y)/2\sqrt y=3y^{-4}$

This is for my review. However, I don't get this part: Why is $dy/dx$ absolute? Is it inverse?

Answer 1

If $F_{Y}\left(x\right)=F_{X}\left(g\left(x\right)\right)$ for some differentiable function $g$ and $F_X$ is differentiable then differentiating (wich application of chainrule) leads to: $$f_{Y}\left(x\right)=g'\left(x\right)f_{X}\left(g\left(x\right)\right)$$

In your case $g\left(x\right)=\sqrt{x}$ and $x\geq1$.