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For a vector (X,Y,Z) with a multivariate normal distribution, $\mu$ and $\Sigma$ are given as the following:

$ \mu = \begin{pmatrix} 76\\ 42\\ 22 \end{pmatrix} $ , and $ \Sigma = \begin{pmatrix} 16 & 5 & -16 \\ 5 & 36 & 0 \\ -16 & 0 & 25 \\ \end{pmatrix} $

However, two new variables, A and B, have been defined:

$ A=X-Z, \ \ B=X-Y $

If X, Y, and Z are independent and random, the mean and variance for the new variables can be found:

$ E[A] = E[X-Z] = E[X] - E[Z] \\ Var(A) = Var(X-Z) = Var(X) - Var(Z) $

However, we cannot assume independence. How can you find the mean and variance of A and B?

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  • $\begingroup$ Why do you say X, Y and Z are independent? $\endgroup$ – LinAlg Oct 18 '16 at 18:51
  • $\begingroup$ How did you get that formula for the variance, variance is always non negative. $\operatorname{var} (X-Z) = \operatorname{var} X + \operatorname{var} Z$ if $X,Z$ are independent. $\endgroup$ – copper.hat Oct 18 '16 at 18:51
  • $\begingroup$ @LinAlg good question—I made that assumption mistakenly. I'm correcting the question now. $\endgroup$ – mattpolicastro Oct 18 '16 at 18:52
  • $\begingroup$ Your title is inconsistent with the last statement. $\endgroup$ – copper.hat Oct 18 '16 at 18:55
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    $\begingroup$ If they are independent, variances do not subtract. They add. $\endgroup$ – Paul Oct 18 '16 at 19:02
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It all follows from linearity.

Let me use $W$ to represent the three dimensional rv. and $L$ is a linear map.

Then $E[LW] = L E[W] = L \mu$.

For variance: \begin{eqnarray} E[(LW -E[LW])((LW -E[LW])^T ] &=& E[(L(W-\mu))(L(W-\mu))^T ] \\ &=& L E[(W-\mu)(W-\mu)^T ] L^T \\ &=& L \Sigma L^T \end{eqnarray}

Then $A= L_1 W$, with $L_1 = \begin{bmatrix} 1 & 0 & -1\end{bmatrix} $ and $B = L_2 W$ with $L_2 = \begin{bmatrix} 1 & -1 & 0\end{bmatrix} $.

Grinding through the computations gives

Mean of $54$ and variance of $73$ for $A$, Mean of $34$ and variance of $42$ for $B$.

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Means are linear, so $E[A]=E[X]-E[Z]$ whether or not they independent.

Variance is more complicated. Instead, you get $$Var(A)=Var(X)+Var(Z)-2Cov(X,Z)$$.

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  • $\begingroup$ So because A is defined as X-Z, we subtract twice the covariance rather than add? And the summation of each correlated variables' variance will always be positive? $\endgroup$ – mattpolicastro Oct 18 '16 at 19:12
  • $\begingroup$ Correct on both counts. $\endgroup$ – Paul Oct 18 '16 at 19:13
  • $\begingroup$ To make sure I'm understanding correctly in this case, would $Var(A) = 16 + 25 -2 (-16) = 73$ or $Var(A) = 16 + 25 + 2 (-16) = 9$? $\endgroup$ – mattpolicastro Oct 18 '16 at 19:36

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