# Transformation theorem

Given $$X_1$$ is $$\Gamma(\alpha,1)$$ distributed and $$X_2$$ is $$\Gamma(\beta,1)$$ distributed and set

$$Y=\frac{X_1}{X_1+X_2}.$$ The task is to show that $$Y$$ is $$\operatorname{Beta}(\alpha,\beta)$$ distributed.

I use the following for hint trying to solve the problem: set $$Y_1=X_1$$ and $$Y_2=\frac{X_1}{X_1+X_2}$$.

So I solve for $$X_1=Y_1$$ and $$X_2=\frac{Y_1(1-Y_2)}{Y_2}$$. By the transformation theorem, I need to find the Jacobian for $$X_1$$ and $$X_2$$, that is $$|J| = -\frac{Y_1}{Y_2}$$. Now from here I am quite confused. Should I use the probability density functions $$\Gamma(a,1)$$ and $$\Gamma(b,1)$$ and insert the expressions of $$X_1$$ and $$X_2$$ in then and multiply with the Jacobian? Because when I do so, I really get messed up calculations that does not make any sense. Would appreciate for help.

• $X_1,X_2$ must be independently distributed. Yes the transformation that you are using will be painful to deal with, instead take $U=X_1+X_2$ and $V=\frac{X_1}{X_1+X_2}$. This transformation will also give you $U$ and $V$ to be independent as a by product!
– Saty
Jul 22 '15 at 23:56
• How does the joint density then look like? I get $\frac{1}{\Gamma(a,1)\Gamma(b,1)}(vu)^{a-1}(u(1-v))^{b-1}e^{-u}$ which I cannot see as a Beta distribution Jul 23 '15 at 0:11
• I am not a fan of using the Jacobian unless when doing a two-variable to two-variable transformation. When I'm doing a two-variable to one-variable transformation, I do the CDF method. $$\Pr\left(Y \leq y\right) = \Pr\left(\dfrac{X_1}{X_1 + X_2} \leq y\right) = \Pr\left((1-y)X_1 \leq yX_2\right) = \Pr\left(X_2 \geq \dfrac{(1-y)X_1}{y}\right)\text{.}$$ Now find the appropriate region in $\mathbb{R}^2$ and fix $y \in (0, 1)$. Jul 23 '15 at 0:47
• @Elekko just integrate $u$ out of. Besides $f(u,v)$ can be written as a product of $g(u)$ and $h(v)$ which implies $U$ and $V$ are independent.
– Saty
Jul 23 '15 at 0:51
• Well by writing $f(u,v)$ as a $g(u)h(v)$ then $g(u)h(v)\mid J \mid$ is the expression I showed above, which is not Beta distributed... :/ Jul 23 '15 at 13:32

I assume $X_1,X_2$ are independent. As Saty suggested, it's an easier transformation of variables if you set

$$Y_1 = X_1+X_2\\ Y_2 = \dfrac{X_1}{X_1+X_2}.$$

Then $X_1 = Y_1Y_2$ and $X_2 = Y_1-Y_1Y_2$ and the Jacobian is

$$J = \begin{vmatrix} y_2 & 1-y_2 \\ y_1 & -y_1 \\ \end{vmatrix} = -y_1.$$ Then,

\begin{eqnarray*} f_{Y_1,Y_2}(y_1,y_2) &=& f_{X_1,X_2}(x_1(y_1,y_2),\;x_2(y_1,y_2))\vert J\vert \\ &=& \dfrac{1}{\Gamma(\alpha)}(y_1y_2)^{\alpha-1}e^{-y_1y_2}\;\dfrac{1}{\Gamma(\beta)}y_1^{\beta-1}(1-y_2)^{\beta-1}e^{y_1y_2-y_1}y_1 \\ &=& \dfrac{1}{\Gamma(\alpha)\Gamma(\beta)}y_1^{\alpha+\beta-1}e^{-y_1}\;y_2^{\alpha-1}(1-y_2)^{\beta-1} \\ &=& \left( \dfrac{1}{\Gamma(\alpha+\beta)}y_1^{\alpha+\beta-1}e^{-y_1} \right) \left( \dfrac{1}{B(\alpha,\beta)}y_2^{\alpha-1}(1-y_2)^{\beta-1} \right) \quad\text{using $B(\alpha,\beta)=\dfrac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$.} \\ \end{eqnarray*}

Since $f_{Y_1,Y_2}(y_1,y_2)$ has form $f_{Y_1}(y_1)f_{Y_2}(y_2)$ and $f_{Y_2}(y_2)$ is a Beta density function with paramaters $\alpha,\beta,$ we are done.

• How does the product show that $f_{Y_1,Y_2}(y_1.y_2)$ has Beta density function? I know that $\frac{1}{B(\alpha,\beta)}=\frac{\Gamma (\alpha + \beta)}{\Gamma(\alpha) \Gamma(\beta)}$, but how the product show that? Jul 24 '15 at 22:31
• @Elekko It's not $f_{Y_1,Y_2}$ that has Beta pdf. $f_{Y_2}$ does. If a joint pdf $f_{Y_1,Y_2}$ can be expressed as the product of 2 functions with $y_1$ in one and $y_2$ in the other then $Y_1,Y_2$ are independent and the joint pdf is the product of their respective pdfs. (Obviously, you can multiply one pdf by a constant and divide the other by same constant and they'd no longer be pdfs: but there's a unique one where they are pdfs.) Second one is Beta pdf: $\dfrac{1}{B(\alpha,\beta)}y_2^{\alpha-1}(1-y_2)^{\beta-1}$. Jul 24 '15 at 23:34