Equicorrelation matrix

The equicorrelation matrix is the $\mathbb{R}^{n \times n}$ matrix where the entries on the diagonal are all equal to $1$ and all off-diagonal entries are equal to some parameter $\rho$ which lies in $[-1, 1]$ . If we were to write out the matrix, it would look something like this:

${\bf \Sigma} = \begin{pmatrix} 1 & \rho & \rho & \dots & \rho \\ \rho & 1 & \rho & \dots & \rho \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \rho & \rho & \rho &\dots & 1 \end{pmatrix}$

Alternatively, we can write it as ${\bf \Sigma} = \rho {\bf 11}^T + (1-\rho){\bf I}$ , where ${\bf 1} \in \mathbb{R}^n$ is the column vector with all entries being 1 and $\bf I$ is the identity matrix.

Here are some useful properties of the equicorrelation matrix:

It is a Toeplitz matrix, and hence has the properties that all Toeplitz matrices has (see e.g. this link).
It has two eigenvalues. The first eigenvalue is $1 + (n-1)\rho$ , with associated eigenvector $v_1 = {\bf 1}$ . The second eigenvalue is $1 - \rho$ , with $n-1$ associated eigenvectors $v_2, \dots, v_n$ , where the entries of $v_k$ are $\begin{aligned} (v_k)_j = \begin{cases} 1 &\text{if } j = 1, \\ -1 &\text{if } j = k, \\ 0 &\text{otherwise.} \end{cases} \end{aligned}$ This can be verified directly by doing the matrix multiplication.
$\text{det} ({\bf \Sigma}) = (1-\rho)^{n-1}[1 + (n-1)\rho]$ . This is because the determinant of a square matrix is equal to the product of its eigenvalues.
$\bf \Sigma$ is positive definite if and only if $-\frac{1}{n-1} < \rho < 1$ . A sketch of the proof can be found in the answer here. It boils down to proving some properties of the determinant expression in the previous point.
${\bf \Sigma}^{-1} = \dfrac{1}{1-\rho} \left( {\bf I} - \dfrac{\rho}{1 + (n-1)\rho} {\bf 11}^T \right)$ . This can be verified directly by matrix multiplication. It can also be derived using the Sherman-Morrison formula.

4 thoughts on “Equicorrelation matrix”

szcfweiya on February 22, 2020 at 10:05 am said:

Hi, there should be a typo in the inverse of Sigma, the `{\bf 11}^T` should be in the numerator, not in the denominator.

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- kjytay on February 22, 2020 at 3:27 pm said:
  
  Thanks for spotting that! I’ve amended the formula.
  
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Mattheu on September 15, 2020 at 8:54 pm said:

What about the covariance matrix version of this correlation matrix, i.e. with diagonal element \sigma^2_i and off-diagonal value \rho\sigma_i\sigma_j ? Can we deduce some of its properties? Thanks!!

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- kjytay on September 15, 2020 at 9:14 pm said:
  
  If E is the equicorrelation matrix and S is the diagonal matrix with the \sigma_i’s on the diagonal, then the covariance matrix is simply SES. With this formula we should be able to derive analogs of the properties above.
  
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Equicorrelation matrix

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