Question

If $X1/X2$are independent exponential random variables with respective parameters $\lambda 1$and $\lambda 2$, find the distribution of $Z=X1/X2$. Also compute $P\left\{X1<X2\right\}$.

Short answer

The distribution of Z is $f_z\left(z\right)=\frac{{\lambda }_1{\lambda }_2}{{({\lambda }_{1}z+{\lambda }_2)}^2}$

$P(X_1<X_2)=\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2}$

Step-by-step solution

Given information

X₁ and X₂ are independent exponential random variables.

Parameters ${\lambda }_1\text{and}{\lambda }_2$

The joint density function is

$f_{x_1{x}_2}\left(x,y\right)={\lambda }_1{e}^{-4x}{\lambda }_2{e}^{-{\lambda }_{2}x}$ for x, y$>0$

Explanation

The joint density function:

$$\begin{gathered} F_Z\left(z\right)=P(Z<z) \\ =P\left(\frac{X_1}{X_2}<z\right) \\ =P\left(X_1<z{X}_2\right) \\ ={\int }_0^{\infty }{\int }_0^{2y}{\lambda }_1{e}^{-\lambda x}{\lambda }_2{e}^{-iy}dxdy \\ ={\int }_0^{\infty }(1-e^{-4z}){\lambda }_2{e}^{-\lambda y}dy \\ =1-\frac{{\lambda }_2}{{\lambda }_{1}z+{\lambda }_2} \\ =\frac{{\lambda }_{1}z+{\lambda }_2-{\lambda }_2}{{\lambda }_{1}z+{\lambda }_2} \\ =\frac{{\lambda }_{1}z}{{\lambda }_{1}z+{\lambda }_2} \end{gathered}$$

The distribution of Z is

$$\begin{gathered} f_2\left(z\right)=\frac{d}{dz}{F}_z\left(z\right) \\ =\frac{d}{dz}\frac{{\lambda }_{1}z}{{\lambda }_{1}z+{\lambda }_2} \\ =\frac{{\lambda }_1{\lambda }_2}{{({\lambda }_{1}z+{\lambda }_2)}^2} \end{gathered}$$

hence the distribution of z is

$f_z\left(z\right)=\frac{{\lambda }_1{\lambda }_2}{{({\lambda }_{1}z+{\lambda }_2)}^2}$

Compute P (X₁ < X₂)

$$\begin{gathered} P(X_1<X_2)=P\left(\frac{X_1}{X_2}<1\right) \\ =P(Z<1) \\ =\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2} \end{gathered}$$