Discussion
Suppose X and Y are independent r.v.'s such that :
\[ \begin{aligned} f(x) &= \frac{e^{-x} x^{\alpha - 1}}{\Gamma(\alpha)}; \ x \gt 0, \ \alpha \gt 0 \\ f(y) &= \frac{e^{-y} y^{\beta - 1}}{\Gamma(\beta)}; \ y \gt 0, \ \beta \gt 0 \end{aligned} \]
The distribution of \(\dfrac{X}{X + Y}\) is :
Answer: A
\(X/(X + Y)\) has the Beta \((\alpha, \beta)\) distribution.
X and Y are independent Gamma variables with shapes \(\alpha\) and \(\beta\) and a common scale 1. Put \(U = X/(X + Y)\) and \(V = X + Y\), so \(X = UV\), \(Y = (1 - U)V\), with Jacobian \(V\). The joint density factorises:
\[ \begin{aligned} g(u, v) &= \frac{(uv)^{\alpha - 1}((1 - u)v)^{\beta - 1} e^{-v}}{\Gamma(\alpha)\Gamma(\beta)} v \\ &= u^{\alpha - 1}(1 - u)^{\beta - 1} \cdot v^{\alpha + \beta - 1}e^{-v} \cdot \frac{1}{\Gamma(\alpha)\Gamma(\beta)} \end{aligned} \]
So U and V are independent, \(V \sim\) Gamma \((\alpha + \beta, 1)\) and \(U \sim\) Beta \((\alpha, \beta)\) on (0, 1).
Gamma \((\alpha + \beta, 1)\) is the distribution of the sum X + Y, not of the ratio, which lies in (0, 1) and so cannot be Gamma at all. Beta \((\alpha + \beta, \alpha - \beta)\) is not even defined when \(\alpha \le \beta\).
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