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 one year ago
Probability and Statistics: Let X and Y be independent random variables, and E(A) be the expectation of any random variable A. Simplify this expression: E(2XY  2XE(Y)  2YE(X) + 2E(X)E(Y))
I think it's supposed to be 0, by the way, but I just don't know how this turns out to be 0.
 one year ago
Probability and Statistics: Let X and Y be independent random variables, and E(A) be the expectation of any random variable A. Simplify this expression: E(2XY  2XE(Y)  2YE(X) + 2E(X)E(Y)) I think it's supposed to be 0, by the way, but I just don't know how this turns out to be 0.

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windsylph
 one year ago
Best ResponseYou've already chosen the best response.1*please note the correction: E(2XY... instead of E(4XY...

wio
 one year ago
Best ResponseYou've already chosen the best response.1Do you know about Covariance? Somehow feel this plays a part.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Well, what formula do you know about covariance that involve the expected value and independent random variables?

windsylph
 one year ago
Best ResponseYou've already chosen the best response.1Thank you, I just figured out the solution (through looking deeper into covariance). This question is actually part of a larger proof that I'm trying to do, which is the variance of the sum of two independent random variables. I attached the proof, please review it if you wish, for correctness.
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