# really confused about expected value

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

• $E(Y^2)=3$
• $E((Y+1)^2)=4$
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 $(E(Y+1))^2$.
• 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 $Var(X)=1$ – chRrr Oct 12 '17 at 16:37