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If $X_i$, $i=1,2$ are independent gamma$(\alpha_i,1)$ random variables, find the distribution of $\frac{X_1}{X_1+X_2}$ and $\frac{X_2}{X_1+X_2}$.

Attempt: Let $Y_1 = \frac{X_1}{X_1+X_2}$ and $Y_2=X_1+X_2$.

Then $X_1 = Y_1Y_2$ and $X_2 = Y_2-Y_1Y_2$.

The Jacobian matrix is :

$\hspace{15mm}J = \left| \begin{array}{ccc} y_2 & y_1 \\ -y_2& 1-y_1 \\ \end{array} \right| = y_2$

I'm having trouble plugging the values back in to the gamma equation. Since we want to know the distribution of $Y_1$, do we first find the joint distribution of $f_{y_1,y_2}(y_1,y_2)$ and then integrate $y_2$ out?

StubbornAtom
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statsguyz
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2 Answers2

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Here How exactly are the beta and gamma distributions related? you can find in one of the solutions posted. X1/X1+X2 has betta distribution

Community
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Javi
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We have that $X_1$ and $X_2$ are independent $\Gamma(\alpha_i,1)$ random variables. Note that $X_1$ and $X_2$ are non-negative with values in $[0,\infty).$

Given the transformation, the joint density of $Y_1$ and $Y_2$ is:

$$ f_{Y_1Y_2}(y_1,y_2)=f_{X_1X_2}[X_1(y_1,y_2),X_2(y_1,y_2)]|J|= f_{X_1}(y_1y_2)f_{X_2}(y_2-y_1y_2)y_2= \frac1{\Gamma(\alpha_1)}[y_1y_2]^{\alpha_1-1}e^{-y_1y_2}\frac1{\Gamma(\alpha_2)}[y_2-y_1y_2]^{\alpha_2-1}e^{-y_2+y_1y_2}y_2=\frac1{\Gamma(\alpha_1)\Gamma(\alpha_2)}y_1^{\alpha_1-1}(1-y_1)^{\alpha_2-1}y_2^{\alpha_1+\alpha_2-1}e^{-y_2}$$

The marginal density of $Y_1$ is

$$f_{Y_1}(y_1) = \frac{y_1^{\alpha_1-1}(1-y_1)^{\alpha_2-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)}\int_{0}^{\infty}y_2^{\alpha_1+\alpha_2-1}e^{-y_2}dy_2=\frac{\Gamma(\alpha_1+\alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)}y_1^{\alpha_1-1}(1-y_1)^{\alpha_2-1}$$

RRL
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