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If random variables $X \sim \mathrm{Gamma}(\frac{n}{2}, \frac{1}{2})$ and $Y \sim \mathrm{Gamma}(\frac{m}{2}, \frac{1}{2})$, where $m$ and $n$ are constants, why does $\frac{X}{X + Y} \sim \mathrm{Beta}(\frac{n}{2}, \frac{m}{2})$?

In general, can we just combine Gamma-distributed variables like this into Beta-distributed ones?

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    Sorry, you are right. I must proofread in the future.2012-12-05

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David. The equality in law that you mention is a special case of the following

Lemma. If $X \sim \Gamma(\alpha,\lambda)$ and $Y \sim \Gamma(\beta,\lambda)$ are independent, then $Z = \frac{X}{X+Y}\sim B(\alpha,\beta)$ and is independent of $X+Y \sim \Gamma(\alpha+\beta,\lambda)$.

This can be shown by computing the law of the couple $(Z,X+Y)$ using a change of variables.

A nice application : if $X \sim B(\alpha,\beta)$ and $Y \sim \Gamma(\alpha+\beta,\lambda)$ are independent, then $XY \sim \Gamma(\alpha,\lambda)$

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    According to the lemma $(X,Y) \overset{d}{=} (\frac{Z_\alpha}{Z_\alpha+Z_\beta},Z_\alpha+Z_\beta)$ as a couple, where $Z_\alpha \sim \Gamma(\alpha,\lambda)$ and $Z_\beta \sim \Gamma(\beta,\lambda)$ are independent. Taking the product, we get $XY \overset{d}{=} \frac{Z_\alpha(Z_\alpha+Z_\beta)}{Z_\alpha + Z_\beta}=Z_\alpha$.2012-12-05