Q.6.58 - A First Course in Probability

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 $Y_2=e^{X_1}$ . 

Step-by-Step Solution

Answer

The joint probability density function of $Y_1 and Y_2$ is${\lambda }^2{e}^{-\lambda n}\times \frac{1}{y_2}$.

Step 1 : 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. 

Step 2 : 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 $Y_1 and Y_2$.