# Transformation of a covariance matrix to one with more or less “extreme” diagonals

Assume $\Sigma$ is a valid correlation matrix in $p$ dimensions. (It could be a covariance matrix, but I'd rather keep ones on the diagonals if possible to keep things simple).

Is there a procedure or function that takes $\Sigma$ and operates on it to form a valid covariance matrix with off-diagonal elements that are always either further from zero, or that have moved in the direction of zero?

For example, say that $$\Sigma = \left[\begin{array}[c c] c 1 & .5\\ .5 & 1 \\ \end{array}\right]$$

an example of the values becoming more "extreme" would be

$$\Sigma_E = \left[\begin{array}[c c] c 1 & .8\\ .8 & 1 \\ \end{array}\right]$$

An example of them becoming less "extreme" would be

$$\Sigma_e = \left[\begin{array}[c c] c 1 & -.5\\ -.5 & 1 \\ \end{array}\right]$$ or $$\Sigma_e = \left[\begin{array}[c c] c 1 & .1\\ .1 & 1 \\ \end{array}\right]$$

In multiple dimensions, the criterion would be that each off-diagonal $\rho$ would be multiplied by some factor greater than 1 in the "more extreme" case, or less than 1 in the "less extreme" case. Of course, the problem is that an arbitrarily chosen set of factors can lead to negative definite matrices.

What is a procedure that is guaranteed to return a valid covariance matrix as per the above?

## 1 Answer

I assume that the only requirement for a "valid covariance matrix" is that the matrix should have $1$s on the diagonal and be positive definite.

With that being said: for any covariance matrix $\Sigma$ and any real number $t$ with $0 \leq t \leq 1$, the matrix $$(1-t) \Sigma + tI$$ is necessarily a valid covariance matrix ($I$ here is the identity matrix). In fact, this gives you a sliding scale which brings the off diagonal elements to $0$ as $t \to 1$.

Here's a nice way to go in the other direction: if you can compute the lowest eigenvalue $\lambda > 0$ (so $\Sigma$ would need to be strictly positive definite, not just semidefinite), we could compute $$\frac{\Sigma - tI}{1-t}$$ where $0 \leq t \leq \lambda$.

I'm not sure how to make things "more extreme" when $\Sigma$ is positive semidefinite (at least, how to do so while guaranteeing that the diagonal entries are still $1$). However, it should be possible to do so whenever $\Sigma$ has rank at least $2$.

• Thanks, but unless I'm missing something, this can't make the off-diagonals bigger, only smaller. – generic_user Dec 7 '16 at 19:26
• That's right. There's no way to make the off-diagonals bigger (and stay positive definite) without finding some kind of eigenvalue/eigenvector first. – Omnomnomnom Dec 7 '16 at 19:29
• Nothing stopping me finding eigenvalues... – generic_user Dec 7 '16 at 19:30
• All right, I'll think about that then – Omnomnomnom Dec 7 '16 at 19:33
• See my latest edit. – Omnomnomnom Dec 7 '16 at 19:38