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The expected value of X is 1 and is the same as X's SD. The expected value of Ysquare is 3, the expected value of (Y+1)square equals to 4, the expected value of (X+Y)square is 7. I got stuck at figuring out E(Y+1)square. any hint that can help me solve this? thanks

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This seems to be a homework problem. However, let me summarize, you have:

  • $E(Y^2)=3$
  • $E((Y+1)^2)=4$

You search: $$(E(Y+1))^2 =? $$ Note that $$(Y+1)^2= Y^2+2Y+1.$$ Hence using the linearity of the expectation, i.e. the rule $E(aX+b) = aE(X)+b$, you may derive $E(Y)$ from your given values and finally derive $(E(Y+1))^2$.

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  • $\begingroup$ Thanks for your answer. So I can rewrite it as E(Y^2) + 2E(Y) + 1 = 4, hence E(Y) = 0? $\endgroup$ – nervedbaby Oct 12 '17 at 16:06
  • $\begingroup$ For E(X+Y)^2 = 7, I can rewrite it as E(X^2) +2E(X)E(Y) + E(Y^2) = 7? $\endgroup$ – nervedbaby Oct 12 '17 at 16:07
  • $\begingroup$ your first comment is correct. your second comment is not correct. You have $E((X+Y)^2)= E(X^2+2XY+Y^2) = E(X^2)+ 2E(XY) +E(Y^2)$ and in general $E(XY)\neq E(X)E(Y)$. However, you may derive $E(XY)$ from $$E((X+Y)^2) = E(X^2) + E(Y^2) + 2E(XY) = 7$$ you already know $E(Y^2)$. In order to calculate $E(X^2)$ use the identity $E(X^2) =Var(X) +E(X)^2$ and note that $\sqrt{Var(X)}=1$ holds if and only if $Var(X)=1$ $\endgroup$ – chRrr Oct 12 '17 at 16:37
  • $\begingroup$ Thank you very much! Your comment really helps me a lot! $\endgroup$ – nervedbaby Oct 12 '17 at 17:01

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