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anonymous
 3 years 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.
anonymous
 3 years 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0*please note the correction: E(2XY... instead of E(4XY...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know about Covariance? Somehow feel this plays a part.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, what formula do you know about covariance that involve the expected value and independent random variables?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank 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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