If $Y \sim \operatorname{Gamma}(a+1, 1), U_0 \sim \operatorname{Unif}(0, 1), U = U_0^\frac1a$, why $YU \sim \operatorname{Gamma} (a, 1)$?

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.

• 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 $Y$. – Severin Pappadeux May 21 at 22:36

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

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

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