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Random Variables $X,Y$

Density function $f(x,y)=\left\{\begin{array}{l}2e^{−2x}/xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 0≤x<\mathrm{∞},0≤y≤x\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{otherwise}\end{array}\right.$

$Cov(X,Y)=?$

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:

$f(x,y)=\left\{\begin{array}{l}2e^{−2x}/xâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 0≤x<\mathrm{∞},0≤y≤x\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{otherwise}\end{array}\right.$

Calculate$E\left(XY\right)$

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

$={∫}_0^{∞}{∫}_0^{x}xy\frac{2}{x}{e}^{-2x}dydx$

$={∫}_0^{∞}{∫}_0^{x}2y{e}^{-2x}dydx$

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

By integrating the function,

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

$={∫}_0^{∞}{x}^2{e}^{-2x}dx$

Integration by parts:

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

$=-\frac{1}{2}\left(∞-0×e^{-0}\right)+{∫}_0^{∞}x{e}^{-2x}dx$

$=0+{∫}_0^{∞}x{e}^{-2x}dx$

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

$=0+\frac{1}{2}{∫}_0^{∞}{e}^{-2x}dx$

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

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

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

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

$=\frac{1}{4}$

Calculate $E\left(X\right)$

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

$={∫}_0^{∞}{∫}_0^{x}x\frac{2}{x}{e}^{-2x}dydx$

$={∫}_0^{∞}{∫}_0^{x}2e^{-2x}dydx$

$=2{∫}_0^{∞}{e}^{-2x}(y{)}_0^{x}dx$

$=2{∫}_0^{∞}{e}^{-2x}(x-0)dx$

$=2{∫}_0^{∞}x{e}^{-2x}dx$

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

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

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

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

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

$=\frac{1}{2}$

Calculate $E\left(Y\right)$

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

$={∫}_0^{∞}{∫}_0^{x}y\frac{2}{x}{e}^{-2x}dydx$

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

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

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

$={∫}_0^{∞}x{e}^{-2x}dx$

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

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

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

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

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

$=\frac{1}{4}$

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

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

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

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

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

$=\frac{1}{8}$

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