Does there exist two random vectors $\mathbf{X}$ and $\mathbf{Y}$ having the following matrix as their covariance matrix? i.e each entry $(i,j)$ of the matrix is $Cov(X_i,Y_j)$ If not, explain why. If yes, give an example.

$$\Sigma_1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \Sigma_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \Sigma_3 = \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}, \Sigma_4 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}, \Sigma_5 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix}, \Sigma_6 = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$

I don't know if there are any general properties of cross-covariance matrices like there are for regular covariance matrices? Obviously if $Cov(X,Y)$ is an $mxn$ matrix then $X$ is $mx1$ and $Y$ is $nx1$. But other than that I believe any matrix could be a cross-covariance matrix?

Not sure what examples to give of vectors other than to say the dimensions of the vectors must match the matrix (like I explained previously), and the vectors must have the property that $Cov(X_i,Y_j)$ corresponds to each entry $i,j$ in the matrix. Also, in the case of $\Sigma_1$ the vectors must be uncorrelated.


1 Answer 1


You are correct: any matrix can be a cross-covariance matrix. But it doesn't look straightforward to prove. Here is an outline of a proof and an effective algorithm for finding the random variables $X$ and $Y$.

Let $B$ be an $m\times n$ real matrix (which could be any of the $\Sigma_i$ in the question). Replacing $B$ with its transpose $B^\prime$ if necessary, we may assume without any loss of generality that $m \le n.$

First, find an $m\times m$ invertible matrix $S$ and an $n\times n$ invertible matrix $T$ for which $S\,B\,T$ is diagonal. This is the Smith normal form of $B.$ It can be found by row-reducing and then column-reducing $B.$ Because the coefficients of $B$ are in a field, we may take the first $r$ diagonal elements of $SBT$ to be unity and the remaining ones to be zero, where of course $0\le r \le m.$ In block-matrix notation this means

$$S\,B\,T = \left(\begin{array}{c|c} \mathbb{I}_r & \mathbb{O}_{r\times n-r}\\ \hline \mathbb{O}_{m-r\times r} & \mathbb{O}_{m-r\times n-r}\end{array}\right)$$

The $\mathbb{I}$ matrix is the $r\times r$ identity and all the $\mathbb{O}$ matrices are zero-matrices; subscripts denote the dimensions.

This (by inspection) could be the off-diagonal block of an $m+n\times m+n$ covariance matrix provided there are $r$ pairs of perfectly correlated variables. In light of this, consider the $m\times m$ matrix $A$ with

$$A = \left(\begin{array}{c|c} \mathbf{1}_r\,\mathbf{1}_r^\prime & \mathbb{O}_{r\times m-r}\\ \hline \mathbb{O}_{m-r\times r} & \mathbb{I}_{m-r\times m-r}\end{array}\right)$$

and the $n\times n$ matrix $\Delta$ with

$$\Delta = \left(\begin{array}{c|c} \mathbf{1}_r\,\mathbf{1}_r^\prime & \mathbb{O}_{r\times n-r}\\ \hline \mathbb{O}_{n-r\times r} & \mathbb{I}_{n-r\times n-r}\end{array}\right)$$

Here, $\mathbf{1}_r$ is the column $r$-vector of ones.

From these three ingredients form the $m+n\times m+n$ matrix

$$\Upsilon = \left(\begin{array}{c|c} A & S\,B\,T\\ \hline T^\prime B^\prime S^\prime & \Delta\end{array}\right) = \left(\begin{array}{c|c} \begin{array}{c|c} \mathbf{1}_r\,\mathbf{1}_r^\prime & \mathbb{O}_{r\times m-r}\\ \hline \mathbb{O}_{m-r\times r} & \mathbb{I}_{m-r\times m-r}\end{array} & \begin{array}{c|c} \mathbb{I}_r & \mathbb{O}_{r\times n-r}\\ \hline \mathbb{O}_{m-r\times r} & \mathbb{O}_{m-r\times n-r}\end{array}\\ \hline \begin{array}{c|c} \mathbb{I}_r & \mathbb{O}_{r\times m-r}\\ \hline \mathbb{O}_{n-r\times r} & \mathbb{O}_{n-r\times m-r}\end{array} & \begin{array}{c|c} \mathbf{1}_r\,\mathbf{1}_r^\prime & \mathbb{O}_{r\times n-r}\\ \hline \mathbb{O}_{n-r\times r} & \mathbb{I}_{n-r\times n-r}\end{array}\end{array}\right)$$

This matrix $\Upsilon$ is positive semi-definite (and therefore is a covariance matrix). To see this, permute the rows and columns to put the two blocks of ones in the upper left, producing the equivalent matrix

$$\Upsilon_0 = \left(\begin{array}{c|c|c} \mathbf{1}_r\,\mathbf{1}_r^\prime & \mathbb{O}_{r\times r}& \mathbb{O}_{r\times m+n-2r}\\ \hline \mathbb{O}_{r\times r} & \mathbf{1}_r\,\mathbf{1}_r^\prime & \mathbb{O}_{r\times m+n-2r}\\ \hline \mathbb{O}_{m+n-2r \times r} & \mathbb{O}_{m+n-2r\times r} & \mathbb{I}_{m+n-2r}\end{array}\right)$$

Writing an arbitrary $m+n$-row vector as $\mathbf{x}=(x_r, y_r, z_{m+n-2r})$ in terms of two $r$-vectors and an $m+n-2r$ vector, compute

$$\mathbf{x}\,\Upsilon_0\,\mathbf{x}^\prime = (x_r\mathbf{1}_r)^2 + (y_r\mathbf{1}_r)^2 + z_{m+n-2r}\,z_{m+n-2r}^\prime \ge 0,\tag{*}$$

showing $\Upsilon_0$ is positive semi-definite (this is the definition) and therefore $\Upsilon$ is positive semi-definite.

We may, however, write

$$\Upsilon = \left(\begin{array}{c|c} S & \mathbb{O}_{m\times n}\\ \hline \mathbb{O}_{n\times m} & T^\prime\end{array}\right) \ \left(\begin{array}{c|c} S^{-1}A (S^\prime)^{-1} & B\\ \hline B^\prime & (T^\prime)^{-1} \Delta T^{-1}\end{array}\right) \ \left(\begin{array}{c|c} S^\prime & \mathbb{O}_{m\times n}\\ \hline \mathbb{O}_{n\times m} & T\end{array}\right).$$

Name those three $m+n\times m+n$ matrices at the right $U,$ $\Sigma,$ and $U^\prime,$ respectively. Recalling that $S$ and $T$ are invertible it follows $U$ is invertible, whence

$$\Sigma = U^{-1}\,\Upsilon\,(U^\prime)^{-1}.$$

Now $\Sigma$ is obviously positive semidefinite, because for any $m+n$ row vector $\mathbf{x},$

$$\mathbf{x}\,\Sigma\,\mathbf{x}^\prime = (\mathbf{x} U^{-1})\, \Upsilon\, (\mathbf{x}U^{-1})^\prime \ge 0$$

by virtue of $(*).$

$\Sigma$ solves the problem: it is a covariance matrix in which $B$ is the cross-covariance between the first $m$ and last $n$ variables.

In particular, let $X$ be the $m$-variate random variable and $Y$ the $n$-variate random variable. Their variances and cross-covariance are

$$\operatorname{Var}(X) = S^{-1}A(S^\prime)^{-1};\quad \operatorname{Var}(Y) = (T^\prime)^{-1}\Delta T^{-1};\quad \operatorname{Cov}(X,Y) = B.$$


Consider $$B=\Sigma_5 = \pmatrix{0&1&0 \\ 0&1&0 \\ 0&1&0}.$$ Row-reduction produces $S$ and then column-reduction of the result produces $T$ with

$$S=\pmatrix{1&0&0 \\ -1&1&0 \\ -1&0&1},\quad T=\pmatrix{0&1&0 \\ 1&0&0 \\ 0&0&1}$$

and $r=1.$ Therefore $A = \Delta = \mathbb{I}_3$ and

$$S^{-1}A(S^\prime)^{-1} = \pmatrix{1&1&1 \\ 1&2&1 \\ 1&1&2};\quad T^{-1}\Delta(T^\prime)^{-1} = \mathbb{I}_3.$$


$$\Sigma= \pmatrix{1&1&1 &0&1&0 \\ 1&2&1 &0&1&0 \\ 1&1&2 &0&1&0 \\ 0&0&0 &1&0&0 \\ 1&1&1 &0&1&0 \\ 0&0&0 &0&0&1}.$$

You can check (by computing eigenvalues, for instance) that this matrix is positive semi-definite and you can see that $B$ is the cross-covariance of the first three and last three variables. Finally, the non-trivial upper $3\times 3$ matrix indicates there isn't any magical way to simplify the results of this analysis (as one might initially hope).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.