## 10.12 Cholesky factors of correlation matrices

A $$K \times K$$ correlation matrix $$\Omega$$ is positive definite and has a unit diagonal. Because it is positive definite, it can be Cholesky factored to a $$K \times K$$ lower-triangular matrix $$L$$ with positive diagonal elements such that $$\Omega = L\,L^{\top}$$. Because the correlation matrix has a unit diagonal,

$\Omega_{k,k} = L_k\,L_k^{\top} = 1,$

each row vector $$L_k$$ of the Cholesky factor is of unit length. The length and positivity constraint allow the diagonal elements of $$L$$ to be calculated from the off-diagonal elements, so that a Cholesky factor for a $$K \times K$$ correlation matrix requires only $$\binom{K}{2}$$ unconstrained parameters.

### Cholesky factor of correlation matrix inverse transform

It is easiest to start with the inverse transform from the $$\binom{K}{2}$$ unconstrained parameters $$y$$ to the $$K \times K$$ lower-triangular Cholesky factor $$x$$. The inverse transform is based on the hyperbolic tangent function, $$\tanh$$, which satisfies $$\tanh(x) \in (-1,1)$$. Here it will function like an inverse logit with a sign to pick out the direction of an underlying canonical partial correlation; see the section on correlation matrix transforms for more information on the relation between canonical partial correlations and the Cholesky factors of correlation matrices.

Suppose $$y$$ is a vector of $$\binom{K}{2}$$ unconstrained values. Let $$z$$ be a lower-triangular matrix with zero diagonal and below diagonal entries filled by row. For example, in the $$3 \times 3$$ case,

$z = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ \tanh y_1 & 0 & 0 \\ \tanh y_2 & \tanh y_3 & 0 \end{array} \right]$

The matrix $$z$$, with entries in the range $$(-1, 1)$$, is then transformed to the Cholesky factor $$x$$, by taking16

$x_{i,j} = \left\{ \begin{array}{lll} 0 & \mbox{ if } i < j & \mbox{ [above diagonal]} \\ \sqrt{1 - \sum_{j' < j} x_{i,j'}^2} & \mbox{ if } i = j & \mbox{ [on diagonal]} \\ z_{i,j} \ \sqrt{1 - \sum_{j' < j} x_{i,j'}^2} & \mbox{ if } i > j & \mbox{ [below diagonal]} \end{array} \right.$

In the $$3 \times 3$$ case, this yields

$x = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ z_{2,1} & \sqrt{1 - x_{2,1}^2} & 0 \\ z_{3,1} & z_{3,2} \sqrt{1 - x_{3,1}^2} & \sqrt{1 - (x_{3,1}^2 + x_{3,2}^2)} \end{array} \right],$

where the $$z_{i,j} \in (-1,1)$$ are the $$\tanh$$-transformed $$y$$.

The approach is a signed stick-breaking process on the quadratic (Euclidean length) scale. Starting from length 1 at $$j=1$$, each below-diagonal entry $$x_{i,j}$$ is determined by the (signed) fraction $$z_{i,j}$$ of the remaining length for the row that it consumes. The diagonal entries $$x_{i,i}$$ get any leftover length from earlier entries in their row. The above-diagonal entries are zero.

### Cholesky factor of correlation matrix transform

Suppose $$x$$ is a $$K \times K$$ Cholesky factor for some correlation matrix. The first step of the transform reconstructs the intermediate values $$z$$ from $$x$$,

$z_{i,j} = \frac{x_{i,j}}{\sqrt{1 - \sum_{j' < j}x_{i,j'}^2}}.$

The mapping from the resulting $$z$$ to $$y$$ inverts $$\tanh$$,

$y \ = \ \tanh^{-1} z \ = \ \frac{1}{2} \left( \log (1 + z) - \log (1 - z) \right).$

### Absolute Jacobian determinant of inverse transform

The Jacobian of the full transform is the product of the Jacobians of its component transforms.

First, for the inverse transform $$z = \tanh y$$, the derivative is

$\frac{d}{dy} \tanh y = \frac{1}{(\cosh y)^2}.$

Second, for the inverse transform of $$z$$ to $$x$$, the resulting Jacobian matrix $$J$$ is of dimension $$\binom{K}{2} \times \binom{K}{2}$$, with indexes $$(i,j)$$ for $$(i > j)$$. The Jacobian matrix is lower triangular, so that its determinant is the product of its diagonal entries, of which there is one for each $$(i,j)$$ pair,

$\left| \, \mbox{det} \, J \, \right| \ = \ \prod_{i > j} \left| \frac{d}{dz_{i,j}} x_{i,j} \right|,$

where

$\frac{d}{dz_{i,j}} x_{i,j} = \sqrt{1 - \sum_{j' < j} x^2_{i,j'}}.$

So the combined density for unconstrained $$y$$ is

$p_Y(y) = p_X(f^{-1}(y)) \ \ \prod_{n < \binom{K}{2}} \frac{1}{(\cosh y)^2} \ \ \prod_{i > j} \left( 1 - \sum_{j' < j} x_{i,j'}^2 \right)^{1/2},$

where $$x = f^{-1}(y)$$ is used for notational convenience. The log Jacobian determinant of the complete inverse transform $$x = f^{-1}(y)$$ is given by

$\log \left| \, \det J \, \right| = -2 \sum_{n \leq \binom{K}{2}} \log \cosh y \ + \ \frac{1}{2} \ \sum_{i > j} \log \left( 1 - \sum_{j' < j} x_{i,j'}^2 \right) .$

1. For convenience, a summation with no terms, such as $$\sum_{j' < 1} x_{i,j'}$$, is defined to be 0. This implies $$x_{1,1} = 1$$ and that $$x_{i,1} = z_{i,1}$$ for $$i > 1$$.↩︎