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I am given n independent Bernoulli trials, prob success = p. If X is # of successes, and Y is the number of failures, what is E(XY) and Cov(X,Y)? I was trying to use E(XY) - E(X)E(Y), the second component is easy, but I can't think of an easy way of finding E(XY) (maybe there isn't any?) Thanks!

problab
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1 Answers1

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The number of successes and failures have to add up to $n$, i.e. $X+Y=n$, so you can write

$$E[XY] = E[X(n-X)] = E[nX - X^2] = nE[X] - E[X^2]$$

Does that help?

Chris Taylor
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  • Ahh brilliant! I did try to use the sum in calculation, but didn't think that substitution will make it that much easier... Thanks! – problab Nov 18 '11 at 08:10