# If $Y\sim\operatorname{Beta}(a,1-a)$ and $Z\sim\operatorname{Exp}(1)$, then $YZ\sim\operatorname{Gamma}(0,1)$?

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 $$YZ \sim \operatorname{Gamma}(a, 1)$$ ?

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_{00}\quad,\,0

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 $$00\implies 0.

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

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 $$f_U(u)=\frac{1}{\Gamma(a)}e^{-u}u^{a-1}\mathbf1_{u>0}$$

• Pretty interesting, I thought making joint PDF would be more complicated. Did it via Mellin though... – Severin Pappadeux May 21 at 17:10

Use the following result:

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

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

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

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