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windsylph
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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
 one year ago
windsylph Group Title
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
 one year ago

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

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

windsylph Group TitleBest ResponseYou've already chosen the best response.1
yes, but how..?
 one year ago

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

windsylph Group TitleBest ResponseYou've already chosen the best response.1
Thank 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.
 one year ago
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