Product of two densities

Question

Let $X \sim \Gamma(\alpha, \lambda)$ and $Y \sim \Gamma(\beta, \lambda)$. I denoty by $f_X$ the density of X and by $f_Y$ the density of Y. Additionally, I assume that the density of (X, Y) is $f_X \cdot f_Y$.

Let $Z = X+Y$ and $U = \frac{X}{X+Y}$.

Why can the density of (Z, U) be written as the product of two densities?

Because U and Z are independent. – Vons Apr 29 '21 at 16:34 — Vons, Apr 29 '21 at 16:34
https://math.stackexchange.com/a/190695/321264 – StubbornAtom Apr 29 '21 at 19:08 — StubbornAtom, Apr 29 '21 at 19:08
How to prove these two random variables are independent? – StubbornAtom Nov 26 '21 at 19:23 — StubbornAtom, Nov 26 '21 at 19:23

score 2 · Accepted Answer · answered Apr 29 '21 at 16:46

2

This is actually Example 7c on page 284 of Sheldon Ross' A First Course in Probability, fifth edition. Quoting from that source:

If $g_1(x,y) = x+y$ and $g_2(x,y) = \frac{x}{x+y}$, then

$$\frac{\partial g_1}{\partial x} = 1, \frac{\partial g_1}{\partial y} = 1, \frac{\partial g_2}{\partial x} = \frac{y}{(x+y)^2}, \frac{\partial g_2}{\partial y} = -\frac{x}{(x+y)^2}$$ and so $$J(x,y) = \begin{vmatrix} 1 & 1 \\ \frac{y}{(x+y)^2} & -\frac{x}{(x+y)^2} \end{vmatrix} = -\frac{1}{x+y}$$

It turns out that $X+Y$ is distributed Gamma$(\alpha + \beta, \lambda)$ while $\frac{X}{X+Y}$ is distributed Beta$(\alpha, \beta)$ and they are independent.

answered Apr 29 '21 at 16:46

RobertTheTutor

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Thanks a lot for the answer and the reference. – StMan Apr 29 '21 at 17:21
Sorry, just to clarify: So the density of X/(X+Y) is...? – StMan Apr 29 '21 at 17:43
It's a beta distribution, with parameters $\alpha$ and $\beta$. The density is $cy^{\alpha-1}(1-y)^{\beta-1}$, for $0 < y < 1$,where $c$ is a normalizing constant. – RobertTheTutor Apr 29 '21 at 18:22
Great, thanks a lot! =) – StMan Apr 29 '21 at 20:00

Product of two densities

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