Covariance of partitioned linear combination

[showzen]

Homework Statement

Given random vector $X'=[X_1,X_2,X_3,X_4]$ with mean vector $\mu '_X=[4,3,2,1]$ and covariance matrix
$$\Sigma_X=\begin{bmatrix}
3&0&2&2\\
0&1&1&0\\
2&1&9&-2\\
2&0&-2&4
\end{bmatrix}.$$
Partition $X$ as
$$X=\begin{bmatrix}
X_1\\X_2\\\hline X_3\\X_4\end{bmatrix}
=\begin{bmatrix}
X^{(1)}\\\hline X^{(2)}\end{bmatrix}.$$
Let $A=[1,2]$ and $B=\begin{bmatrix}1&-2\\2&-1\end{bmatrix}$. Find Cov$(AX^{(1)},BX^{(2)})$.

Homework Equations

Cov$(CX)=C\Sigma_X C'$
Cov$(X^{(1)},X^{(2)})=\Sigma_{12}$

The Attempt at a Solution

Cov($AX^{(1)},BX^{(2)})=A\Sigma_{12}B'=[1,2]\begin{bmatrix}2&2\\1&0\end{bmatrix}\begin{bmatrix}1&2\\-2&-1\end{bmatrix}=[0,6]$

Although I have arrived at an answer, I do not know how to interpret it. We have scalar $AX^{(1)}$ and vector $BX^{(2)}$, and we arrive at row vector covariance?

 

[StoneTemplePython]

in general, nice use of latex here. One nitpick: $\text{Cov}(C \mathbf X)=C\Sigma_X C^T$ is wrong. It should read $\text{Cov}(C \mathbf X, C \mathbf X)=C\Sigma_X C^T$ -- the idea of covariance with one argument only doesn't make much sense. The idea more generally here is that $\text{Cov}(C \mathbf X, D \mathbf X)=C\Sigma_X D^T$. In general a non-symmetric covariance matrix can irk me a bit so I get your question on how it can be a (row) vector result.

The idea here is suppose you have a scalar random variable $Y_1$ and a vector of

$\mathbf Y = \begin{bmatrix} Y_2\\ Y_3 \end{bmatrix}$

so the covariance of $\text{cov}(Y_1, \mathbf Y)$ is just covariance of $Y_1, Y_2$ and also covariance of $Y_1, Y_3$. Collect these results in a vector -- that's it.

- - - -

Another way to approach this problem, would be to use

$A := \left[\begin{matrix}1 & 2 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right]$

and

$B := \left[\begin{matrix}0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 1 & -2\\0 & 0 & 2 & -1\end{matrix}\right]$

now apply

$\text{Cov}(A \mathbf X, B \mathbf X) = A \text{Cov}( \mathbf X, \mathbf X) B^T = A \Sigma_X B^T= \left[\begin{matrix}0 & 0 & 0 & 6\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right]$

and you can just read off the result. E.g. the covariance of the scalar $\big(A \mathbf X\big)_1$ with scalar $\big(B \mathbf X\big)_4$ is given in the top right corner of the resulting matrix. At the end of the day you're interested in that and $\big( A\mathbf X\big)_1$ with scalar $\big(B \mathbf X\big)_3$ which is given in row 1, column 3, and of course is zero.

 

[tnich]

I would interpret it this way: $U = AX^{(1)}$ is a linear combination of the elements of $X^{(1)}$, which results in a scalar. $V = BX^{(2)}$ is a a pair of linear combinations of the elements of $X^{(2)}$, which results in a column vector, $V = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$. Then $$Cov(AX^{(1)}, BX^{(21)}) = Cov(U, V) =
\begin{bmatrix}
Cov(U, V_1) &
Cov(U, V_2)
\end{bmatrix}
$$.

 

[showzen]

Thanks everyone for your help. I see now that the answer is a matrix with elements that are the covariance between $AX^{(1)}$ and the elements of $BX^{(2)}$.