Given random variables like below:
$Y \sim \operatorname{Gamma}(a + 1, 1)$
$U_0 \sim \operatorname{Unif}(0,1)$
$U = U_0^\frac1a$
If $Y$ and $U_0$ is independent,
How can I proof $X=YU \sim \operatorname{Gamma}(a, 1)$ ?
I tried to solve this problem with theorem $f_{U,V}(u,v) = f_{X,Y}(h_1(u,v), h_2(u,v))|J|$
But I'm confusing what should be preimage.
Find the moment generating function of $YU$. Use independence to make it equal to MGF of $Y$ times MGF of $U$. Look up the gamma MGF and uniform MGF and then times them together, and see that it is also gamma.
Again, like in If $Y\sim\operatorname{Beta}(a,1-a)$ and $Z\sim\operatorname{Exp}(1)$, then $YZ\sim\operatorname{Gamma}(0,1)$? we will be using Mellin transform to get image of the product, using the fact, that image of the product is the product of Mellin images. For $U(0,1)^{\frac{1}{a}}$ we could easily reconstruct PDF
$$ PDF_X(x) = a x^{a-1} 1_{0<x<1} $$
Its Mellin image is
$$ M(X) = \frac{a}{s+a-1} $$
For Gamma-distribution
$$ M(Y) = \frac{\Gamma(s+a-1+1)}{\Gamma(a+1)} $$
Therefore
$$ M(XY) = \frac{\Gamma(s+a-1+1)}{s+a-1} \frac{a}{\Gamma(a+1)} = \frac{\Gamma(s+a-1)}{\Gamma(a)} $$
using well-known property of Gamma-function. Which is obviously could be transformed back to
$$ PDF(x|XY) = M^{-1}\{\frac{ \Gamma(s+a-1) }{\Gamma(a)} \} = \frac{ \exp(-x) x^{a-1} }{\Gamma(a)} 1_{x>0} $$
which is obviously Gamma distribution
