Let X, and X, be independent exponential distributions with the same parameter 1. What is the joint distribution of Y, = X, + X2 and Y, X1 ? What are the distributions of Y, and X1+X2 Y,?

Given:  X₁ & X₂ are independent random variables with pdfs

            $f\left(x_1\right)=\lambda e^{-\lambda x_1}; x_1>0$         &      $f\left(x_2\right)=\lambda e^{-\lambda x_2}; x_2>0$

Then their joint pdf f(x₁, x₂) (say) is given by

       $f(x_1,x_2)= f\left(x_1\right)f\left(x_2\right)= {\lambda }^2{e}^{-\lambda (x_1+x_2)}$

We are given

       $Y_1=X_1+X_2 \& Y_2=\frac{X_1}{X_1+X_2}$

First, we find the joint density of $Y_1 \& Y_2$

Let $T(x_1,x_2)=\left(x_1+x_2, \frac{x_1}{x_1+x_2}\right)$, then T is a  bijection from $[0,\infty )\times [0,\infty ) onto [0,\infty )\times [0,1]$. Now we need to find $(x_1,x_2)$ in terms of $(y_1,y_2)$.

Multiplying Y₁ & Y₂ we get

           $x_1=y_1{y}_2$  then    $x_2=y_1-x_1=y_1-y_1{y}_2$

So,         $T^{-1}(y_1,y_2)=\left(y_1{y}_2, y_1- y_1{y}_2\right)$

The Jacobian of transformation is

          $J=\frac{\delta (x_1,x_2)}{\delta y_1\delta y_2}=\left|\begin{array}{cc}\frac{\delta x_1}{\delta y_1}& \frac{\delta x_2}{\delta y_1}\\ \frac{\delta x_1}{\delta y_2}& \frac{\delta x_2}{\delta y_2}\end{array}\right|=\left|\begin{array}{cc}{y}_2& 1-y_2\\ y_1& -y_1\end{array}\right|=-y_1$

Thus the joint pdf of Y₁ and Y₂, g(y_1, y₂) is 

       $$\begin{gathered} g(y_1,y_2) = f_{X_1{X}_2}(x_1,x_2)\times \left|J\right| \\ = f_{X_1}\left(x_1\right)f_{X_2}\left(x_2\right)\times y_1 \\ = f_{X_1}\left(y_1{y}_2\right)f_{X_2}(y_1-y_1{y}_2)y_1 \\ = \lambda e^{-\lambda y_1{y}_2}·\lambda e^{-\lambda (y_1-y_1{y}_2)}{y}_1 \\ = {\lambda }^2{y}_1{e}^{-\lambda y_1} \\ \therefore g(y_1,y_2)=\left\{\begin{array}{l} {\lambda }^2{y}_1{e}^{-\lambda y_1}; y_1\ge 0, 0\le y_2\le 1\\ 0; otherwise\end{array}\right. \end{gathered}$$

 Therefore, the marginal pdf of Y₁ is 

         $$\begin{gathered} g\left(y_1\right)={\int }_0^{1}g(y_1,y_2)d{y}_2 \\ = {\int }_0^1{\lambda }^2{y}_1{e}^{-\lambda y_1}d{y}_2 \\ = {\lambda }^2{y}_1{e}^{-\lambda y_1}{\left|y_2\right|}_0^1 \\ = {\lambda }^2{y}_1{e}^{-\lambda y_1} \\ \therefore g\left(y_1\right) =\left\{\begin{array}{l}{\lambda }^2{y}_1{e}^{-\lambda y_1}; y_1\ge 0\\ 0; otherwise\end{array}\right. \end{gathered}$$