Can E(Z) be determined from E(X) and E(Y) when X and Y are independent?

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SUMMARY

The discussion centers on determining the expected value E(Z) for the expression Z = (2+X)(3X + 4Y) given that X and Y are independent random variables with E(X) = 5 and E(Y) = 6. The correct approach involves calculating E(Z) as E(Z) = 6E(X) + 8E(Y) + 3E(X^2) + 4E(X)E(Y). However, to compute E(X^2), knowledge of the variance Var(X) is necessary, as it relates to E(X^2) through the identity Var(X) = E(X^2) - E(X)^2. Thus, without additional information about the variance, E(Z) cannot be fully determined.

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oyth94
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Hi I know this may be a silly question but i am doubting myself on how i did this question:

Suppose X and Y are independent, with E(X) = 5 and E(Y) = 6. For each of the following variables Z, either compute E(Z) or explain why we cannot determine
E(Z) from the available information:
Z = (2+X)(3X + 4Y)= 6X + 8Y + 3X^2 + 4XY

So I did E(Z= 6X + 8Y+ 3X^2 + 4XY) = 6E(X) + 8E(Y) + 3E(X^2) + 4E(X)(Y)
Im not sure if i am on the right track so far (ie i just have to plug in E(X) and E(Y) to find E(Z)
but doesn't this require some integrals since it is in the absolute continuous case? Otherwise if we plug it in isn't it not independent because the answer doesn't equal to E(XY)?
 
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Re: expected value independence

The problem is that You know $E\{X\}$ and $E\{Y\}$, that X and Y are independent and nothing else. That permits You to find $E\{X Y\}$ but for $E\{X^{2}\}$ You have to know $\text{Var} \{X\}$ and apply the identity $\text{Var} \{X\} = E \{X^{2}\} - E^{2} \{X\}$... Kind regards $\chi$ $\sigma$
 

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