How to find covariance of vectors?

Question

Given $X=(X_1,X_2,X_3)^T$ distributed as $N_3(\mu,\Sigma)$ with $\mu=(1,-1,2)$ and $\Sigma= \begin{pmatrix} 4&0&-1 \\ 0&5&0\\-1&0&2\\\end{pmatrix}$

Find whether $(X_1,X3)$ is independent of $X_2$

I know the definition of covariance is $cov(X,Y)=E[(X-\mu_X)(Y-\mu_Y)^T]$

But this doesnt seem to work in this case since I have $X=(X_1,X_3)$ and $Y=X_2$,

I used the definition of covariance and found that:

$cov((X_1,X_3),X_2)=E[(X_1-\mu_{X_1},X_3-\mu_{X_3})(X_2-\mu_{X_2})]=(cov(X_1,X_2),cov(X_3,X_2))=(0,0)$

I believe this means they are independent, however I don't really understand what it means to have a covariance be a vector. Whenever I've computed covariances I get real numbers because the vectors being compared are of the same length. I'm not sure how to interpret this.

Answer 1

There is a generalization of scalar covariance, namely the cross-covariance matrix, which may be of arbitrary dimensions.

Since you have found $(X_1,X_3)$ and $X_2$ are uncorrelated, they are independent, since jointly normal and uncorrelated implies independent.