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.

  • 1
    $\begingroup$ I believe the answer is 11. What makes you say it's incorrect? $\endgroup$ Oct 22, 2019 at 1:31
  • 2
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.