So I have this problem where I need to find $E(4X+3Y-2Z^2-W^2+8)$ where $W,X,Y,Z$ are all standard normal and I'm kind of confused on how to find the expected value here. I thought to do it we just had to add the means together like this: $E(X_1+X_2+X_3)=\mu_1+\mu_2+\mu_3$ as long as they are iid of course. How come the answer is $5$?

I tried doing $$\begin{align}E(4X+3Y-2Z^2-W^2+8)\\ =E(4X)+E(3Y)-E(2Z^2)-E(W^2)+E(8)\\ =4E(X)+3E(Y)-2E(Z^2)-E(W^2)+8\\=(4*0)+(3*0)-(2*0)-0+8\\=8\end{align}$$ And quite obviously $8$ doesn't equal $5$. So I'm just not sure where I went wrong.

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    $\begingroup$ $E(X_1+X_2+X_3)=E(X_1)+E(X_2)+E(X_3)$ is true so long as the right hand side is meaningful, no matter where they are iid or not. $\endgroup$ – Henry Dec 2 '12 at 23:26

The variance of $Z$ is $E(Z^2)-(E(Z))^2$. Since the variance is $1$, and $E(Z)=0$, we have $E(Z^2)=1$. Same for $W^2$. That gets us $5$. The error was in thinking that $E(Z^2)=0$ and $E(W^2)=0$.

  • $\begingroup$ Ah I see, I think I knew that too. Thanks for the reminder. $\endgroup$ – TheHopefulActuary Dec 2 '12 at 23:24

You cant open $E(X+Y^2 ) = E(X) + E(Y^2)$. It is not linear at all .

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    $\begingroup$ Linearity of expectation has nothing to do with independence... it's just linear, period. $\endgroup$ – snar Feb 8 '14 at 0:00

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