1
$\begingroup$

Let $\mathbf{X}$ be a vector of i.i.d. random variables. Let $\mathbf{C}$ be a desired covariance matrix for which I like to determine mixing matrix $\mathbf{A}$ such that

$$ var(\mathbf{A}\mathbf{X}) = \mathbf{A}^T\mathbf{A} = \mathbf{C}. $$

Now, this could be solved by the Cholesky or eigenvalue decomposition depending on the definiteness of $\mathbf{C}$. I am, however, interested in a balanced mixing matrix, i.e.,

$$ | a_{ij} | = |a_{ji}| $$

The idea is that the resulting random variables are as mixed as possible. In contrary, if $\mathbf{A}$ is triangular (as from the Cholesky decomposition) at least one random variable is identical to the original which is undesired.

A numerical example: Let $$\mathbf{C} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ be the desired covariance matrix. Possible mixing matrices are $$\mathbf{A}_1 = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} / \sqrt{3} \textrm{ and } \mathbf{A}_2 = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ where $\mathbf{A}_1^T \mathbf{A}_1 = \mathbf{C}$ and $\mathbf{A}_2^T \mathbf{A}_2 = \mathbf{C}$ and $\mathbf{A}_1$ is more balanced than $\mathbf{A}_2$.

What is a systematic method to find a balanced mixing matrix (such as $\mathbf{A}_1$) instead of an unbalanced mixing matrix (such as $\mathbf{A}_2$)?

$\endgroup$
1
$\begingroup$

Some thoughts that might be helpful.

Let's say you can distill your balance criterion into an objective function $f(A)$. Once we find one viable matrix $A_0$, for example by Cholesky, all other matrices can be derived as $QA_0$ where $Q$ is orthogonal such that $Q^TQ=I$. Therefore your problem reduces to finding an orthogonal matrix that maximizes $f(QA_0)$. Depending on your function $f$, there might even be closed form expression for $Q$.

$\endgroup$
  • $\begingroup$ Thanks for the hint. The invariance under orthogonal transformation certainly helps. Are you aware of any terminology regarding the balance criterion which would aid literature search? $\endgroup$ – Sebastian Schlecht Jul 19 '18 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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