How $X=y^b$ distributes if $f(y)=aby^{b-1}e^{-ay^{b}}$ , $y\geq0$. How $X=y^b$ distributes if $f(y)=aby^{b-1}e^{-ay^{b}}$ , $y\geq0$. (Weibull distribution) It would be great if you can elaborate on the calculation. Thanks :D

As $X=y^b$ is an monotonically increasing function for $y\geq 0$, therefore,

pdf of $X$ is $$g(x)=f(y)|\frac{dy}{dx}|=f(x^{1/b})\frac{x^{(\frac{1-b}{b})}}{b}, x\geq 0$$

See Here