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I have two random variables

$Y \sim \operatorname{Beta}(a, 1 - a)$

$Z \sim \operatorname{Exp}(1)$

If $Y$ and $Z$ are independent, why is the distribution of $X = YZ \sim \operatorname{Gamma}(a, 1)$?

$f_X(x) = \int_0^\infty|\frac{1}{y}|f_Y(y)f_Z(\frac xy)dy$

$f_X(x) = \int_0^\infty \frac{1}{y}\frac{1}{\Gamma(\alpha)\Gamma(1-\alpha)}y^{\alpha-1}(1-y)^{-\alpha}e^{-\frac{x}{y}}dy$

but I can't derive more than it.

How can I proof $YZ \sim \operatorname{Gamma}(a, 1)$ ?

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Here is a very familiar approach; nothing special about it.

Joint pdf of $(Y,Z)$ is $$f_{Y,Z}(y,z)=\frac{e^{-z}y^{a-1}(1-y)^{-a}}{\Gamma(a)\Gamma(1-a)}\mathbf1_{0<y<1,z>0}\quad,\,0<a<1$$

You can use a change of variables $(Y,Z)\to (U,V)$ such that $U=YZ$ and $V=Z$.

So the preimages are $z=v$ and $y=u/v$, and $0<y<1,z>0\implies 0<u<v$.

Absolute value of jacobian of transformation is $1/v$.

This gives the joint pdf of $(U,V)$:

$$f_{U,V}(u,v)=\frac{e^{-v}u^{a-1}(v-u)^{-a}}{\Gamma(a)\Gamma(1-a)}\mathbf1_{0<u<v}$$

Therefore, marginal pdf of $U$ is $$f_U(u)=\frac{u^{a-1}}{\Gamma(a)\Gamma(1-a)}\int_u^\infty e^{-v}(v-u)^{-a}\,dv\,\mathbf1_{u>0}$$

Substitute $v-u=t$, which converts the integral to a Gamma function, ultimately giving the answer $$f_U(u)=\frac{1}{\Gamma(a)}e^{-u}u^{a-1}\mathbf1_{u>0}$$

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  • $\begingroup$ Pretty interesting, I thought making joint PDF would be more complicated. Did it via Mellin though... $\endgroup$ – Severin Pappadeux May 21 at 17:10
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Use the following result:

Assuming $Y$ and $Z$ are independent, the PDF of $X = YZ$ is given by:

$$f_X(x) = \int_{-\infty}^{\infty} \frac{1}{|u|} f_{Y}(u) f_Z\left(\frac{x}{u}\right) du$$

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You could use Mellin transform to get distribution for $YZ$. For a product of two RVs, there is a very simple theorem, which states that Mellin transform of product distribution is the product of Mellin transform of the constituent RVs. So there is simple algorithm - Mellin transform of $Y$ multiplied by Mellin transform of $Z$, and then do inverse Mellin transform to get final PDF. It is just like using Fourier transform for sum of two RVs.

$$ M(YZ) = M(Y) M(Z) $$

For exponential distribution $Y = \exp(-x)$

$$ M(Y) = \Gamma(s) $$

For $Z = B(a, 1-a)$ one could easily get

$$ M(Z) = \frac{ \Gamma(s+a-1) }{\Gamma(s) \Gamma(a)} $$

Therefore, for product

$$ M(YZ) = \frac{ \Gamma(s+a-1) }{\Gamma(a)} $$

which gives us pretty obvious inverse transform in the form of

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

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