Question

$$\begin{gathered} ThejointprobabilitydensityfunctionofXandYisgivenby \\ f(x,y)=e-(x+y)0\dots x<q,0\dots y<q \\ Find \\ \left(a\right)PX<Yand \\ \left(b\right)PX<a \end{gathered}$$

Short answer

(a) $P(X<Y)=\frac{1}{2}$

$\left(b\right)P(X<a)=1-e^{-a}$

Step-by-step solution

Introduction

The joint probability density function of X and Y$f(x,y)=e-(x+y)0\dots x<q,0\dots y<q$

Explanation of (a).

$$\begin{gathered} Giveninformation: \\ Thejointprobabilitydensityfunctionis \\ f(x,y) \\ =e^{-(x+y)} \\ \le x\le \infty ,0\le y\le \infty \\ \text{formula used:} \\ P(X<Y)={\int }_0^{\infty }{\int }_0^{y}f(x,y)dxdy \\ {\int }_0^{\infty }{\int }_0^y{e}^{-x}\times e^{-y}dxdy \\ {\int }_0^{\infty }{\left[-e^{-x}\right]}_0^y\times e^{-y}dy \\ {\int }_0^{\infty }\left(-e^{-y}+1\right)\times e^{-y}dy \\ {\int }_0^{\infty }\left(e^{-y}-e^{-2y}\right)dy \\ Solvingitfurther \\ {\left.\left[-e^{-y}-\frac{e^{-2y}}{-2}\right]\right|}_0^{\infty }=1-\frac{1}{2}=\frac{1}{2} \end{gathered}$$

Explanation of (b).

Formula used:

$$\begin{gathered} P(X<a)={\int }_0^{\infty }{\int }_0^{a}f(x,y)dxdy \\ Calculation: \\ Findtherequiredprobabilityu\sin gtheaboveformula \\ {\int }_0^{\infty }{\int }_0^a{e}^{-x}\times e^{-y}dxdy \\ {\int }_0^{\infty }{\left[-e^{-x}\right]}_0^a\times e^{-y}dy \\ {\int }_0^{\infty }\left(-e^{-a}+1\right)\times e^{-y}dy \\ \left(-e^{-a}+1\right)\times {\left.\frac{e^{-y}}{-1}\right|}_0^{\infty }=1-e^{-a} \end{gathered}$$