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Equicorrelation Matrix

Posted on (Update: )
Tags: Equicorrelation, Toeplitz, Woodbury Formula

kjytay’s blog summarizes some properties of equicorrelation matix, which has the following form,

\[\Sigma = \begin{bmatrix} 1 & \rho & \rho & \cdots & \rho\\ \rho & 1 & \rho & \cdots & \rho\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \rho & \rho & \rho & \cdots & 1 \end{bmatrix}\,,\]

where the off-diagonals $\rho \in[-1,1]$.

Rewrite in matrix form, we have

\[\Sigma = \rho \1\1^T+(1-\rho)\I\,.\]

Toeplitz matrix

First of all, $\Sigma$ is the so-called Toeplitz matrix, in which each descending from left to right is constant.

Formally, any $n\times n$ matrix $A$ of the form,

\[A = \begin{bmatrix} a_0 & a_{-1} & a_{-2} & \cdots & \cdots & a_{-(n-1)}\\ a_1 & a_0 & a_{-1} & \ddots & & \vdots\\ a_2 & a_1 & \ddots & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & a_{-1} & a_{-2}\\ \vdots & & \ddots & a_1 & a_0 & a_{-1}\\ a_{n-1} & \cdots & \cdots & a_2 & a_1 & a_0 \end{bmatrix}\]

is a Toeplitz matrix, and the elements satisfy

\[A_{ij} = A_{i+1,j+1} = a_{i-j}\,.\]

Note that a Toeplitz is not necessarily symmetric, but it is persymmetric. In R, toeplitz() returns a symmetric Toeplitz matrix with the given first row.

More properties can be found in Robert M. Gray. Toeplitz and Circulant Matrices: A review


Firstly consider the eigenvalues of $\1\1^T$, since it is a rank-1 matrix, which implies that it has $n-1$ eigenvalues which are zero. Denote the eigenvectors as $\xi_i, i=1,\ldots,n$. Suppose the nonzero eigenvalue is $\lambda_1$ with its associated eigenvector $\xi_1$, then

\[\1\1^T\xi_1 = \lambda_1\xi_1\,,\]

left multiplying $\1^T$ get

\[n\1^T\xi_1 = \lambda_1 \1^T\xi_1\,,\]

which implies that $\lambda_1 = n$.

Note that

\[\Sigma\xi_i = (\rho \1\1^T+(1-\rho)\I)\xi_i = \rho\lambda_i\xi_i + (1-\rho)\xi_i = \begin{cases} [(n-1)\rho + 1]\xi_1 & i=1\\ (1-\rho)\xi_i & i=2,3,\ldots \end{cases}\,,\]

which means that the eigenvalues of $\Sigma$ is $(n-1)\rho+1$ and $n-1$ eigenvalues with value $1-\rho$, and the same eigenvectors as $\1\1^T$.

In the words of a handout, the rank-1 modification $\rho\1\1^T$ to the matrix $(1-\rho)\I$ only changes one of the eigenvalues, i.e., updating one eigenvalue $1-\rho$ to $(n-1)\rho+1$.

Based on the eigenvalues, we can easily get

  • $\det(\Sigma) = (1-\rho)^{n-1}[1+(n-1)\rho]$
  • $\Sigma$ is positive definite iff $-\frac{1}{n-1} < \rho < 1$.


By the Woodbury matrix identity, or say Sherman-Morrison formula, we can get

\[\Sigma^{-1} = \frac{1}{1-\rho}\left(\I - \frac{\rho \1\1^T}{1+(n-1)\rho}\right)\,.\]

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