Finding $E[YZ]$ when $E[Y]$ and $E[Z]$ are known

Question

From an earlier problem:

If $X_1$, $X_2$, and $X_3$ are independent and have the means 4, 9, and 3 and the variances 3, 7, and 5, find the mean and the variance of

(a) $Y = 2X_1 - 3X_2 + 4X_3$

(b) $Z = X_1 + 2X_2 - X_3$

I found that the means of Y and Z are -7 and 19, respectfully. Also, the variances are 155 and 36, respectfully. Now, I need to find the $cov(Y,Z)$.

I know that

$cov(Y,Z) = E[YZ] - E[Y]E[Z]$

But I'm stuck on how to get $E[YZ]$ just from the information I know.

Answer 1

Hint:

To find $E[YZ]$ simply substitute the values for $Y$ and $Z$

i.e. $$E[YZ]=E[(2X_1-3X_2+4X_3)(X_1+2X_2-X_3)]$$

and solve for $E(X_1^2), E(X_2^2)$ and $E(X_3^2)$ by adding the mean squared (i.e $[E(X_i)]^2$) to the respective variances.

Hope this is helpful.

Answer 2

You can write $$ E[YZ] = E[(2X_1-3X_2+4X_3)(X_1+2X_2-X_3)] \\= E(2X_1^2-6X_2^2-4X_3^2-X_1X_2+2X_1X_3+11X_2X_3) \\ = 2E(X_1^2) -6E(X_2^2) -4E(X_3^2)-E(X_1X_2)+\ldots$$ where I just got tired of writing. As a further hint, don't calculate this as it is. First subtract off $E(Y)E(Z)$ after expressing it in terms of the expectation values of the $X$'s. This will let you express everything in terms of variances and covariances of the $X'$s. You've already calculated the variances and the covariances of the $X$'s are easy (why?).

Answer 3

Hints:

1. If $X_i$ and $X_j$ are independent random variables, then $E(X_i X_j) = E(X_i)E(X_j).$

2. Write $E(X_i^2)$ in terms of $\mathrm{Var}(X_i)$ and $E(X_i)^2.$