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Let $X_1 \sim Gamma(\alpha_1,1)$ and $X_2 \sim Gamma(\alpha_2,1)$ be independent random variables. How can I find the marginal distributions of $\frac{X_1}{X_1+X_2}$ and $\frac{X_2}{X_1+X_2}$?

By setting $U=\frac{X_1}{X_1+X_2}$ and $V=\frac{X_2}{X_1+X_2}$, I am having a hard time trying to isolate $X_1$ and $X_2$. Once I can isolate these variables, I can complete the problem. However, I was wondering if there was a better way to solve for those marginal distributions.

Stephan Kolassa
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Jen Snow
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    Not sure if its helpful but you can rearrange $U=\frac{X_1}{X_1 + X_2} = 1 + \frac{X_1}{X_2}$ which may make it easier – Brennan Nov 27 '19 at 03:01
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    @Brennan I think you made a small mistake there. But you can get somewhere with $1/U$ and thereby $\frac{1}{U} -1$ (which is what I expect you were looking to get to); similarly $\frac{1}{V}-1$ – Glen_b Nov 27 '19 at 04:03
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    Note that if you solve for the marginal distribution of $U$, it will necessarily have $\alpha_1$ and $\alpha_2$ as parameters somewhere in the expression, in which case, by a symmetry argument, the marginal distribution of $V$ will be the same except with $\alpha_1$ and $\alpha_2$ switched. So you only need to solve for one of them. – jbowman Nov 27 '19 at 04:48
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    In other words, we can exploit the observation that $U = 1 - V$. – Sycorax Nov 27 '19 at 05:05
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    Nothing Beta than being pointed in the right direction... – epp Nov 27 '19 at 07:31
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    Check Theorem 4.1, p.430, in [Devroye's Non-Uniform Random Variate Generation.](https://amzn.to/2DlXmPg) – Xi'an Nov 27 '19 at 09:37

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