# Joint distrib of gamma distribs

• Oct 31st 2009, 09:48 AM
isarutas
Joint distrib of gamma distribs
Let X_i, with i \epsilon {1,2,3} be independant r.v. and suppose that X_i \epsilon Gamma(r_i,1).
Now set:
Y_1= X_1/(X_1+X_2)
Y_2= (X_1+X_2)/(X_1+X_2+X_3)
Y_3= X_1+X_2+X_3

Determine the joint distrib of Y_1, Y_2, Y_3.

What is the best strategy to tackle this problem? rewrite X_i in Y_i and try to find a jacobian? I tried that but i got some divided by 0 in it.

(answer: f_\vec{y}(\vec{y})=1/(gamma(r_1)*gamma(r_2)*gamma(r_3))*y_1^{r_1-1}*y_2^{r_1+r_2-1}*y_3^{r_1+r_2+r_3-1}*(1-y_1)^{r_2-1}*(1-y_2)^{r_3-1}*exp(-y_3) with 0<y_1<1, 0<y_2<1 and y_3>0 )

im sorry about the notation but i did not get the mathmode to work...
• Oct 31st 2009, 12:28 PM
theodds
Yeah, the Jacobian should work. I don't see where a divided by 0 has an opportunity to occur, but man the det of the Jacobian matrix doesn't look very fun to compute or work with. You should end up with

$\displaystyle X_1 = Y_1 Y_2 Y_3$
$\displaystyle X_2 = Y_2 Y_3 - Y_1 Y_2 Y_3$
$\displaystyle X_3 = Y_3 - Y_2 Y_3$

[assuming I did this right]

Update: Yep, just finished it and this works. It works out super nice, particularly the Jacobian simplifies to $\displaystyle Y_2 Y_3 ^2$ and it happens to be exactly what you need. They're independent, weird o.O I'm guessing you can extend this result to an arbitrary number of independent gammas.
• Oct 31st 2009, 03:25 PM
Laurent
Quote:

Originally Posted by theodds
They're independent, weird o.O I'm guessing you can extend this result to an arbitrary number of independent gammas.

If I may add a piece of explanation.

This, and its generalization, are consequences of two simpler and widely known facts (perhaps previous questions?):
If $\displaystyle X,Y$ are independent random variables respectively distributed according to $\displaystyle \Gamma(\alpha,1)$ and $\displaystyle \Gamma(\beta,1)$, then
a) $\displaystyle X+Y$ has $\displaystyle \Gamma(\alpha+\beta,1)$ distribution
b) $\displaystyle \frac{X}{X+Y}$ has $\displaystyle B(\alpha,\beta)$ distribution and is independent of $\displaystyle X+Y$.

Then you use these properties several times: if $\displaystyle X_i$, $\displaystyle 1\leq i\leq n$ are independent random variables with respective distributions $\displaystyle \Gamma(r_i,1)$, $\displaystyle 1\leq i\leq n$, then:
- $\displaystyle \frac{X_1}{X_1+X_2}$ is $\displaystyle B(r_1,r_2)$ distributed and independent of $\displaystyle X_1+X_2$ (and of $\displaystyle X_3,\ldots,X_n$)
- $\displaystyle Y_2=X_1+X_2$ is $\displaystyle \Gamma(r_1+r_2,1)$ distributed, hence you can repeat the first step to the sequence $\displaystyle (Y_2,X_3,X_4,\ldots,X_n)$: $\displaystyle \frac{Y_2}{Y_2+X_3}$ is $\displaystyle B(r_1+r_2,r_3)$ distributed and independent of $\displaystyle Y_3=Y_2+X_3=X_1+X_2+X_3$, which is $\displaystyle \Gamma(r_1+r_2+r_3,1)$ distributed.
- and so on.
• Oct 31st 2009, 04:00 PM
theodds
That also would make the problem much easier. I saw that the 2nd and 3rd were independant gamma and beta based on (a) and (b) above, but I figured you'd only be able to get pairwise independance of those two.