1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ $E(X_1+X_2+X_3)=E(X_1)+E(X_2)+E(X_3)$ 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
2
$\begingroup$

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 $E(W^2)=0$.

$\endgroup$
  • $\begingroup$ Ah I see, I think I knew that too. Thanks for the reminder. $\endgroup$ – TheHopefulActuary Dec 2 '12 at 23:24
-4
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Linearity of expectation has nothing to do with independence... it's just linear, period. $\endgroup$ – snar Feb 8 '14 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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