Gamma Family Density Function of Y

Question

I have tried to think of how to prove this but at a loss. I keep getting it squared so not sure what I'm doing wrong.

Answer 1

You would just use the following formula for a one-to-one transformation of a random variable: if $Y = g(X)$ where $g$ is invertible, then $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right|.$$ So with $Y = g(X) = X^{1/\gamma}$, we have $X = g^{-1}(Y) = Y^\gamma$, and the rest is straightforward.

Answer 2

\begin{align*} \mathbb{P}(Y \le y) &= \mathbb{P}(X^{1/\gamma} \le y)\\ &= \mathbb{P}(X \le y^\gamma)\\ &= 1-e^{- y^{\gamma}/\beta} \end{align*}

\begin{align*} f_Y(y) &= \frac{d}{dy} \mathbb{P}(Y \le y)\\ &= \frac{\gamma}{\beta} y^{\gamma-1} e^{-y^\gamma/\beta} \end{align*}