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

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When X and Y are independent with E(X) = 5 and E(Y) = 6, the expected value E(Z) for Z = (2+X)(3X + 4Y) can be partially computed as E(Z) = 6E(X) + 8E(Y) + 3E(X^2) + 4E(X)E(Y). However, to find E(X^2), additional information such as the variance of X is required, as E(X^2) cannot be determined solely from E(X). The independence of X and Y allows for the calculation of E(XY) as E(X)E(Y), but this does not extend to E(X^2) without further data. Thus, while some components of E(Z) can be calculated, the complete determination of E(Z) is not possible with the given information. Understanding the relationship between expected values and independence is crucial in this context.
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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