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So, I have that a joint probability density function is given by the formula: $$ 5e^{-5x} / x, \quad 0 < y < x < \infty $$ and I have to find the $\operatorname{Cov}(X,Y)$.

I know that $\operatorname{Cov}(X,Y) = {\bf E}[XY] - {\bf E}[X]{\bf E}[Y]$. I've been able to find ${\bf E}[XY]$ and ${\bf E}[X] $ ($1/25$ and $1/5$ respectively, hopefully I'm correct there), but I've been unable to find the ${\bf E}[Y]$... Can someone help?

user642796
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Shell
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1 Answers1

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$E[Y]=\int_0^\infty \int_0^x y\cdot \frac{5}{x}e^{-5x} dy dx$. This formula is obtained after inserting into general form $E[Y]=\int_{\mathbb{R}^2} y \cdot pdf(x,y) dydx$.

Jacek Podlewski
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  • Of course... I kept trying to do dxdy, since I had done the other two that way. Thank you so much! – Shell Apr 17 '14 at 00:24