Variance of the sum of random variables

Question

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?

Answer 1

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 $B$.

Answer 2

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

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