Question

The joint density function of$X$and$Y$is given by

$f(x,y)=\frac{1}{y}{e}^{-(y+x/y)},x>0,y>0$

Find $E\left[X\right],E\left[Y\right]$and show that $Cov(X,Y)=1$

Short answer

The value of $E\left[X\right]=1$

The value of $E\left[Y\right]=1$

The value of$Cov(X,Y)=1$

Step-by-step solution

Given Information

The density of $X$and $Y$is $f(x,y)=\frac{1}{y}{e}^{-(y+x/y)},x>0,y>0$

$E\left[X\right]=?$

$E\left[Y\right]=?$

$Cov(X,Y)=?$

Explanation

Calculate the value of$E\left[X\right],$

$\mathrm{E}\left[\mathrm{X}\right]={\int }_0^{\infty }{\int }_0^{\infty }x·\frac{1}{y}{e}^{-\left(y+\frac{x}{y}\right)}dydx$

$={\int }_0^{\infty }{e}^{-y}\left[{\int }_0^{\infty }\frac{x}{y}{e}^{-\frac{x}{y}}dx\right]dy$

$={\int }_0^{\infty }y{e}^{-y}dy$

$={\left.y\frac{e^{-y}}{-1}\right|}_0^{\infty }-{\int }_0^{\infty }\frac{e^{-y}}{-1}dy$

$=0+1$

$=1$

Explanation

Calculate the value of $E\left[Y\right],$

$\mathrm{E}\left[\mathrm{Y}\right]={\int }_0^{\infty }{\int }_0^{\infty }y·\frac{1}{y}{e}^{-\left(y+\frac{x}{y}\right)}dxdy$

$={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-y}{e}^{-\frac{x}{y}}dxdy$

$={\int }_0^{\infty }{e}^{-y}\left[{\int }_0^{\infty }{e}^{-\frac{x}{y}}dx\right]dy$

$={\int }_0^{\infty }{e}^{-y}ydy$

$=1$

Explanation

Calculate the value of $Cov(X,Y),$

$E\left(XY\right)={\int }_0^{\infty }{\int }_0^{\infty }xy·\frac{1}{y}{e}^{-\left(y+\frac{x}{y}\right)}dxdy$

$={\int }_0^{\infty }{e}^{-y}\left[{\int }_0^{\infty }x{e}^{\frac{-x}{y}}dx\right]dy$

$={\int }_0^{\infty }{e}^{-y}\left[{\left.\frac{x{e}^{\frac{-x}{y}}}{\frac{-1}{y}}\right|}_0^{\infty }-{\int }_0^{\infty }\frac{e^{\frac{-x}{y}}}{\frac{-1}{y}}dx\right]dy$

$={\int }_0^{\infty }{e}^{-y}\left[0+y^2\right]dy$

$={\int }_0^{\infty }{y}^2{e}^{-y}dy$

$=-{\left.y^2{e}^{-y}\right|}_0^{\infty }+{\int }_0^{\infty }2y{e}^{-y}dy$

$=0+2=2$

Final Answer

Hence$Cov(X,Y)=E\left(XY\right)-E\left(X\right)E\left(Y\right)=2-1.1$

$=2-1$

$=1$

So, the value of$Cov(X,Y)=1$