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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.

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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.

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  • $\begingroup$ Are you sure $MGF(XY) = MGF(X)MGF(Y)$? I cannot see how you could untangle exponents under integral even in the case of independent $Y$ and $Y$. It is usually stated as a fact for sum of $X$ and $Y$. $\endgroup$ – Severin Pappadeux May 21 at 22:36
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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

$$ PDF(x|XY) = M^{-1}\{\frac{ \Gamma(s+a-1) }{\Gamma(a)} \} = \frac{ \exp(-x) x^{a-1} }{\Gamma(a)} 1_{x>0} $$

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