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I know this is an easy question but I'm having problems solving it

I sorta thought that you'd get $E(X^2) + 2E(XY) + E(Y^2)$ and that'd add up to: $3+4+4=11$ but my answer isn't correct. I'm guessing my intuition is definitely wrong

Thanks a lot.

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    $\begingroup$ I believe the answer is 11. What makes you say it's incorrect? $\endgroup$ Oct 22, 2019 at 1:31
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    $\begingroup$ The first term in your title is: $[E(X+Y)]^2$ or $E[(X+Y)^2]$? $\endgroup$ Oct 22, 2019 at 2:24
  • $\begingroup$ Oh thank you so much. I must've slipped up $\endgroup$
    – Howell Lu
    Oct 22, 2019 at 2:48

1 Answer 1

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First, observe that we have $E((X+Y)^2) = E(X^2 + 2XY + Y^2)$

The key property that we will use is Linearity of expectation. This says for any random variables $X$ and $Y$, and any constants $a$ and $b$, we have $$E(aX + bY) = aE(X) + bE(Y)$$ Applying this to what we did above, we get $$E(X^2 + 2XY + Y^2) = E(X^2) + 2E(XY) + E(Y^2)$$ So your initial thought was correct.

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