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Hi I'm currently working through past exam questions and am stuck with the following question:

Random variables $X_1$, $X_2$ and $X_3$ are identically distributed, with common mean $\mu$, common variance $\sigma^2$, and common pairwise correlation $\rho$ (i.e. $corr(X_i,X_j) =$ $\rho$ for all i $\neq$ j). Let Y = $X_1$ - $2X_2$ + $X_3$.

a) Fin the expectation and variance of Y.

b)Find the correlation between Y and $X_1$

Am I wrong in assuming that the expectation of Y would simply be $E(X_1) - E(2X_2) + E(X_3)$ given that E(X+Y)=E(X)+E(Y). Am I completely off? Any hints would be greatly appreciated.

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  • $\begingroup$ Yes, you are on the right track with the expectation. You should also know that $E(aX) = aE(X)$. For the variance of Y, remember the formula $Var(aX + bY) = a^2Var(X) + b^2Var(Y) + 2abCov(X, Y)$. $\endgroup$ – James Jun 11 '15 at 11:03
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You're right. Note though that since those variables are identically distributed, $E(X_1) = E(X_2) = E(X_3) = \mu$ so in the end $E(Y) = 0$.

For the variance you have a different formula (which you have to prove:)

$$Var(X_1 + X_2) = Var(X_1) + Var(X_2) + 2\text{Cov}(X_1, X_2)$$

Use this formula to find the variance of $Y$.

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  • $\begingroup$ Thanks for a swift response! Am I correct in thinking that the: Var($X_1$ - $2X_2$ + $X_3$) = Var($X_1$) + 4*Var($X_2$) + Var($X_3$) - 4Cov($X_1$,$X_2$) + 2Cov($X_1$,$X_3$) - 4Cov($X_2$,$X_3$) ? $\endgroup$ – orwellian Jun 11 '15 at 11:57
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You are correct about the expectation of $Y$ so you find easily that $\mathbb EY=0$.

Advice on the rest:

First solve it for special case $\mu=0$ and $\sigma=1$.

Then realize that in the general case (co)variances are multiplied with $\sigma^2$ and that correlations are not affected.

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