Question

If X1 and X2 are independent exponential random variables, each having parameter $\lambda$, find the joint density function of $Y_1=X_1+X_2$ and $Y_2=e^{X_1}$ .

Short answer

The joint probability density function of $Y_{1}and{Y}_2$is${\lambda }^2{e}^{-\lambda n}\times \frac{1}{y_2}$.

Step-by-step solution

Joint density function :  

The joint probability density function (joint pdf) is a function that is used to characterize a continuous random vector's probability distribution.

Explanation : 

If $X_{1}and{X}_2$are two independent random variables,

The joint probability density function is equal to the product of the marginal distribution functions of two random variables if they are independent.

The joint probability density function of $X_{1}and{X}_2$is,

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

Consider $y_1=x_1+x_{2}and{y}_2=e^{x_1}$

The joint probability density function of $Y_1$and $Y_2$is,

$f(y_1,y_2)=f\left(x_1,x_2\right)·{\left|J\left(x_1,x_2\right)\right|}^{-1}$

The Jacobean transformation is $J\left(x_1,x_2\right)$.

$$\begin{gathered} J\left(x_1,x_2\right)=\left|\begin{array}{cc}\frac{\partial y_1}{\partial x_1}& \frac{\partial y_1}{\partial x_2}\\ \frac{\partial y_2}{\partial x_1}& \frac{\partial y_2}{\partial x_2}\end{array}\right| \\ =\left|\begin{array}{cc}1& 1\\ e^{x_1}& 0\end{array}\right| \\ =1\left(0\right)-1\left(e^{x_1}\right) \\ =-\left(e^{x_1}\right) \end{gathered}$$

Substitute the values of $Y_{1}and{Y}_2$.

$$\begin{gathered} f\left(y_1,y_2\right)=\left\{{\lambda }^2{e}^{-\lambda (x_1+x_2)}\right\}·\left\{\frac{1}{e^{x_1}}\right\} \\ ={\lambda }^2{e}^{-\lambda n}\times \frac{1}{y^2}\left\{\sin ce{y}_1=x_1+x_{2}and{y}_2=e^x\right\} \\ ={\lambda }^2{e}^{-\lambda n}\times \frac{1}{y_2} \end{gathered}$$

Hence,${\lambda }^2{e}^{-\lambda n}\times \frac{1}{y_2}$is the joint probability density function of$Y_{1}and{Y}_2$.