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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.

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    $\begingroup$ 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 $\endgroup$
    – Brennan
    Commented Nov 27, 2019 at 3:01
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    $\begingroup$ @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 $\frac{1}{V}-1$ $\endgroup$
    – Glen_b
    Commented Nov 27, 2019 at 4:03
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    $\begingroup$ 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. $\endgroup$
    – jbowman
    Commented Nov 27, 2019 at 4:48
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    $\begingroup$ In other words, we can exploit the observation that $U = 1 - V$. $\endgroup$
    – Sycorax ♦
    Commented Nov 27, 2019 at 5:05
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    $\begingroup$ Nothing Beta than being pointed in the right direction... $\endgroup$
    – statsplease
    Commented Nov 27, 2019 at 7:31

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