Question

The random variables X and Y have a joint density function is given by

$f(x,y)=\{\begin{array}{ll}2e^{-2x}/x& 0\le x<\mathrm{\infty },0\le y\le x\\ 0& \text{otherwise}\end{array}$

Compute$Cov(X,Y)$

Short answer

The computation of$Cov(X,Y)$is$\frac{1}{8}.$

Step-by-step solution

Given Information

Random Variables $X,Y$

Density function $f(x,y)=\left\{\begin{array}{l}2e^{-2x}/x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le x<\mathrm{\infty },0\le y\le x\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{otherwise}\end{array}\right.$

$Cov(X,Y)=?$

Explanation

From the information, observe that the random variable $X$and $Y$are the random variables that have the joint probability density function is as follows:

$f(x,y)=\left\{\begin{array}{l}2e^{-2x}/x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le x<\mathrm{\infty },0\le y\le x\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{otherwise}\end{array}\right.$

Calculate$E\left(XY\right)$

$E\left(XY\right)={\int }_0^{\infty }{\int }_0^{x}xyf(x,y)dydx$

$={\int }_0^{\infty }{\int }_0^{x}xy\frac{2}{x}{e}^{-2x}dydx$

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

$=2{\int }_0^{\infty }{e}^{-2x}{\left(\frac{y^2}{2}\right)}_0^{x}dx$

Explanation

By integrating the function,

$={\int }_0^{\infty }{e}^{-2x}\left(x^2\right)dx$

$={\int }_0^{\infty }{x}^2{e}^{-2x}dx$

Integration by parts:

$={\left(x^2\frac{e^{-2x}}{-2}\right)}_0^{\infty }-{\int }_0^{\infty }2x\left(\frac{e^{-2x}}{-2}\right)dx$

$=-\frac{1}{2}\left(\infty -0\times e^{-0}\right)+{\int }_0^{\infty }x{e}^{-2x}dx$

$=0+{\int }_0^{\infty }x{e}^{-2x}dx$

$={\left(x\frac{e^{-2x}}{-2}\right)}_0^{\infty }-{\int }_0^{\infty }\frac{e^{-2x}}{-2}dx$$={\left(x\frac{e^{-2x}}{-2}\right)}_0^{\infty }-{\int }_0^{\infty }\frac{e^{-2x}}{-2}dx$

$=0+\frac{1}{2}{\int }_0^{\infty }{e}^{-2x}dx$

$=\frac{1}{2}{\left[\frac{e^{-2x}}{-2}\right]}_0^{\infty }$

$=-\frac{1}{4}\left[e^{-\infty }-e^{-0}\right]$

$=-\frac{1}{4}(0-1)$

$=\frac{1}{4}\left(1\right)$

$=\frac{1}{4}$

Explanation

Calculate $E\left(X\right)$

$E\left(X\right)={\int }_0^{\infty }{\int }_0^{x}xf(x,y)dydx$

$={\int }_0^{\infty }{\int }_0^{x}x\frac{2}{x}{e}^{-2x}dydx$

$={\int }_0^{\infty }{\int }_0^{x}2e^{-2x}dydx$

$=2{\int }_0^{\infty }{e}^{-2x}(y{)}_0^{x}dx$

$=2{\int }_0^{\infty }{e}^{-2x}(x-0)dx$

$=2{\int }_0^{\infty }x{e}^{-2x}dx$

$=2\left[{\left(x\frac{e^{-2x}}{-2}\right)}_0^{\infty }-{\int }_0^{\infty }{e}^{-2x}dx\right]$

$=\left[0-{\left(\frac{e^{-2x}}{-2}\right)}_0^{\infty }\right]$

$=-\frac{1}{2}\left(e^{-\infty }-e^{-0}\right)$

$=-\frac{1}{2}(0-1)$

$=\frac{1}{2}\left(1\right)$

$=\frac{1}{2}$

Explanation

Calculate $E\left(Y\right)$

$E\left(Y\right)={\int }_0^{\infty }{\int }_0^{x}yf(x,y)dydx$

$={\int }_0^{\infty }{\int }_0^{x}y\frac{2}{x}{e}^{-2x}dydx$

$={\int }_0^{\infty }{\int }_0^x\frac{2}{x}{e}^{-2x}\left(ydy\right)dx$

$={\int }_0^{\infty }\frac{2}{x}{e}^{-2x}{\left(\frac{y^2}{2}\right)}_0^{x}dx$

$={\int }_0^{\infty }\frac{1}{x}{e}^{-2x}\left(x^2\right)dx$

$={\int }_0^{\infty }x{e}^{-2x}dx$

$=\left[{\left(x\frac{e^{-2x}}{-2}\right)}_0^{\infty }-{\int }_0^{\infty }\frac{e^{-2x}}{-2}dx\right]$

$=\left[0+\frac{1}{2}{\left(\frac{e^{-2x}}{-2}\right)}_0^{\infty }\right]$

$=-\frac{1}{4}\left(e^{-\infty }-e^{-0}\right)$

$=-\frac{1}{4}(0-1)$

$=\frac{1}{4}\left(1\right)$

$=\frac{1}{4}$

Explanation

Therefore, the calculation of $Cov(X,Y)$

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

$=\frac{1}{4}-\frac{1}{2}\times \frac{1}{4}$

$=\frac{1}{4}-\frac{1}{8}$

$=\frac{2-1}{8}$

$=\frac{1}{8}$

Final Answer

Hence, the computation of$Cov(X,Y)$is$\frac{1}{8}.$