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Suppose that $X_1 \sim \Gamma(\alpha_1,\beta)$ and $X_2 \sim \Gamma(\alpha_2,\beta)$ and let $Z = \frac{X_1}{X_1 + X_2}$ ($X_1$ and $X_2$ are assumed to be independent).

I want to prove that $Z$ is beta distributed as follows: $Z \sim Beta(\alpha_1, \alpha_2)$.

I don't want to use the Jacobian as shown there: http://www.math.wm.edu/~leemis/chart/UDR/PDFs/GammaBeta.pdf.

Instead, I know that $Z = \frac{X_1}{X_1 + X_2} \Rightarrow X_2 = X_1\frac{1-Z}{Z}$ and the joint of $X_1$ and $X_2$ is:

$$P(X_1,X_2) = \frac{\beta^{\alpha_1 +\alpha_2}}{\Gamma(\alpha_1)\Gamma(\alpha_2)}X_1^{\alpha_1-1}X_2^{\alpha_2-1} \exp[-\beta(X_1 + X_2)]$$

By integrating the joint I get:

$$P(Z) = \int_0^{+\infty} P(X_1,X_1\frac{1-Z}{Z}) dX_1 = \int_0^{+\infty} \frac{\beta^{\alpha_1 +\alpha_2}}{\Gamma(\alpha_1)\Gamma(\alpha_2)}X_1^{\alpha_1-1}\Bigg(X_1\frac{1-Z}{Z}\Bigg)^{\alpha_2-1} \exp[-\beta X_1/Z]dX_1$$

Using that $Beta(\alpha_1,\alpha_2)\Gamma(\alpha_1 + \alpha_2) = \Gamma(\alpha_1)\Gamma(\alpha_2)$ and pulling the constant outside of the integral:

$$P(Z) = \frac{Z^{\alpha_1 - 1} (1-Z)^{\alpha_2 - 1}}{Beta(\alpha_1,\alpha_2)} \int_0^{+\infty} \frac{\beta^{\alpha_1 + \alpha_2}}{\Gamma(\alpha_1+\alpha_2)}\Bigg( \frac{X_1}{Z} \Bigg)^{\alpha_1-1}\Bigg(\frac{X_1}{Z}\Bigg)^{\alpha_2-1} \exp[-\beta X_1/Z]dX_1\\ = \underbrace{\frac{Z^{\alpha_1 - 1} (1-Z)^{\alpha_2 - 1}}{Beta(\alpha_1,\alpha_2)}}_{\text{Beta distribution}} \underbrace{\int_0^{+\infty} \frac{\beta^{\alpha_1 + \alpha_2}}{\Gamma(\alpha_1+\alpha_2)}\Bigg( \frac{X_1}{Z} \Bigg)^{\alpha_1 + \alpha_2 - 2} \exp[-\beta X_1/Z]dX_1}_{\text{Almost a gamma distribution that should integrate to 1}}$$

How do I get the Beta distribution from there? I have tried a change of variable $Y = \frac{X_1}{Z}$ but it does not seems to lead to the result.

Theophile Champion
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  • Hint: How can you possibly determine $P(Z)$ without using a Jacobian? – whuber Sep 26 '20 at 16:39
  • Xi'an: That's the post I am trying to understand at the end but I am starting with something simpler, i.e. Beta instead of dirichlet. – Theophile Champion Sep 27 '20 at 08:15
  • whuber: Well, I guess I won't be asking if I knew the answer. There is the beginning of a prove in my post, but there is a power of $\alpha_1 + \alpha_2 - 2$ that should be $\alpha_1 + \alpha_2 - 1$ and the integration over $X_1$ that should be over $X_1/Z$. – Theophile Champion Sep 27 '20 at 08:19

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