Independence of $X_1+X_2$ and $\frac{X_1}{X_1+X_2}$ where $X_1,X_2$ are gamma distributed

Question

So in my textbook it states $$\text{Let $X_1$ and $X_2$ be independent and }X_i\sim \Gamma(\alpha_i,\beta),i=1,2.\text{ Then }\\ (i)\frac{X_1}{X_1+X_2}\sim \operatorname{Beta}(\alpha_1,\beta)\\(ii) X_1+X_2\text{ and }\frac{X_1}{X_1+X_2}\text{ is independent}$$ There is a proof for $(i)$ but not a proof for $(ii)$ and I'm having trouble proving it. A link or an answer or maybe just a hint would be appreciated!

Answer 1

Standard proof:

Let $Y_1 = X_1 + X_2 \sim Gamma(\alpha_1+\alpha_2,\beta)$ and $Y_2=\frac{X_1}{X_1+X_2} \sim Beta(\alpha_1, \beta)$.

Then

$X_1 = Y_1\cdot Y_2 = v_1(Y_1,Y_2)$,

$X_2 = Y_1 - X_1 = Y_1 - Y_1\cdot Y_2 = Y_1(1-Y_2) = v_2(X_1,X_2)$.

By the change-of-variables theorem, we have that $$ f_Y(y_1,y_2) = |J_v| f_X\circ v(y_1,y_2) = |J_v| f_{X_1}(y_1\cdot y_2) f_{X_2}(y_1\cdot(1-y_2)) $$

Computing the jacobian is straightforward:

$$ |J_v| = \begin{vmatrix} \frac{d v_1}{dy_1} & \frac{d v_1}{dy_2} \\ \frac{d v_2}{dy_1} & \frac{d v_2}{dy_2} \end{vmatrix} = \begin{vmatrix} y_2 & y_1 \\ 1-y_2 & -y_1 \end{vmatrix} = |-y_1 y_2 - y_1(1-y_2)| = |-y_1| \stackrel{(1)}{=} y_1 $$

(1) since the support of a Gamma is positive

so we have that

$$ f_Y(y_1,y_2) = y_1 f_{X_1}(y_1\cdot y_2) f_{X_2}(y_1\cdot(1-y_2)) = y_1 \frac{1}{\Gamma(\alpha_1)} \frac{1}{\beta^{\alpha_1}} (y_1\cdot y_2)^{\alpha_1-1} e^{-y_1\cdot y_2 / \beta} \frac{1}{\Gamma(\alpha_2)} \frac{1}{\beta^{\alpha_2}} (y_1\cdot(1-y_2))^{\alpha_2-1} e^{- (y_1\cdot(1-y_2)) / \beta} \stackrel{(2)}{\approx} y_1^{1 + (\alpha_1 - 1) + (\alpha_2 - 1)} y_2^{\alpha_1-1} (1-y_2)^{\alpha_2-1} e^{\frac{y_1y_2}{\beta} + \frac{y_1(1-y_2)}{\beta}} = (y_1^{\alpha_1 + \alpha_2 - 1} e^{\frac{y_1}{\beta}}) (y_2^{\alpha_1-1} (1-y_2)^{\alpha_2-1}) \approx f_{Y_1}(y_1) f_{Y_2}(y_2) $$

(2) we can forget about the constants - everything must integrate to one so they will coincide.

So $f_Y(y_1,y_2) = f_{Y_1}(y_1) f_{Y_2}(y_2)$ ie they are independent.