# Introduction

Here is a draft of a method to construct precision matrices based on correlated Gaussian samples with mean zero and variance one.

# Method

Let \(y_t\) be a random vector of size \(J\) and \(t \in \left\{1, \dots, T\right\}\), where \(T\) is the number of temporal replicates. We assume that \(y_t \sim \mathcal N(0, Q^{-1})\) and that the marginal variance of the \(i\)-th element of \(y_t\), \(y_{i,t}\), is \(1\). This means that the diagonal of \(Q^{-1}\) is a vector of ones, and that

\[ E(y_t) = 0, \qquad \mathrm{cov}(y_t) = Q^{-1}. \]

Furthermore, it is assumed that Q is a sparse precision matrix. Using the properties of Gaussian conditional distributions, we have

\[ E(y_{i,t}|y_{-i,t})=-Q_{i,i}^{-1} \sum_{j\in\mathcal A_i, j\neq i} Q_{i,j}y_{j,t}, \]

\[ \mathrm{Prec}(y_{i,t}|y_{-i, t}) = Q_{i,i} = (\mathrm{var(y_{i,t}|y_{-i, t})})^{-1}=\tau_i^{-2}, \]

where \(\mathcal A_i\) is the set containing the neighbors of site \(i\), i.e. the sites that are such that \(Q_{i,j} \neq 0\) if \(j \in \mathcal A_i\).

Assume that we have realizations of \(y_1, \dots y_t\) that can be used to infer the precision matrix \(Q\). We set up a regression model to estimate the non-zero elements of \(Q\). Here, we consider \(y_{i,t}\) as a realization, i.e. as an observation. The regression model for each site, \(i\), will be

\[ y_{i,t} = \sum_{j\in\mathcal A, j\neq i} \beta_{i,j}y_{j,t} + \varepsilon_{i, t}, \quad t\in \left\{1, \dots, T\right\}. \]

At each site \(i\), we estimate the parameter vector \(\beta_i\) with

\[ \hat\beta_i = (X_i^TX_i)^{-1}X_i^Ty_i, \]

where

\[ X_i = \begin{pmatrix} y_{j_{1, i}, 1} & \dots & y_{j_{m, i}, 1} \\ \vdots & \vdots & \vdots \\ y_{j_{1, i}, T} & \dots & y_{j_{m, i}, T} \end{pmatrix}, \]

and \(y_{j_{l, i}, 1}\) is the \(l\)-th neighbor oy \(y_{i, t}\) at time \(t\). The variance of \(\varepsilon_{i, t}\) is \(\tau_i^2\) and it is estimated with

\[ \hat\tau_i^2 = T^{-1}(y_i - X_i\hat\beta_i)^T(y_i - X_i\hat\beta_i). \]

The next step is to transform \(\hat\beta_i\) and \(\hat\tau_i^2\) such that they give estimates of the elements of \(Q\), namely

\[ \hat Q_{i, j} = \begin{cases} -\hat\tau_i^2\hat\beta_{i, j}, \quad \text{if } i \neq j, \\ \hat\tau_i^2, \qquad \quad \text{ if } i = j, \end{cases} \]

where \(\hat\beta_{i, j}\) is the \(j\)-th element og \(\hat\beta_i\). Let \(\hat B\) be a matrix with \((i, j)\)-th element \(\hat\beta_{i, j}\). Note that \(\hat \beta_{i, i} = 0\), and thus \(\hat B_{i, i} = 0\). Furthermore let \(\hat K\) be a diagonal matrix such that

\[ \hat K = \mathrm{diag}\left(\hat\tau_1^{-2}, \dots, \hat\tau_J^{-2}\right). \]

An estimate of Q can now be presented as

\[ \hat Q = \hat K(I + \hat B), \]

where \(I\) is an identity matrix of size \(J\).

We have to make sure that \(\hat Q\) is symmetric. This can be achieved by setting

\[ \tilde Q{i, j} = \tilde Q_{j, i} = \frac12(\hat\tau_i^{-2}\hat\beta_{i, j} + \hat\tau_j^{-2}\hat\beta_{j, i}), \]

and defining new regression parameters \(\tilde \beta_{i, j}\) that are such that

\[ \hat\tau_i^{-2}\tilde\beta_{i,j} = \tilde Q_{ij} = \tilde Q_{j, i} = \hat \tau_j^{-2}\tilde \beta_{j, i}, \]

which gives

\[ \tilde\beta_{i, j} = \hat\tau_i^{2}\tilde Q_{i, j}, \quad \tilde\beta_{j, i} = \hat\tau_j^{2}\tilde Q_{i, j}, \]

and let \(\tilde Q\) and \(\tilde B\) be the matrices containing the \(\tilde Q_{i,j}\)’s and the \(\tilde \beta_{i, j}\)’s.

We can not be sure of \(\tilde Q\) being positive definite. One way to check whether the matrix is positive definite or not, is to compute the Cholesky decomposition of \(\tilde Q\), that is, \(\tilde Q = LL^T\), and check whether all the diagonal elements of L are positive. If the matrix \(\tilde Q\) is invertible then it is more likely that it is positive definite, while if \(\tilde Q\) is not invertible then it is not positive definite. The estimated precision matrix, \(\tilde Q\), is invertible if \((i + \tilde B)\) is invertible, where \(\tilde Q = \hat K(I + \tilde B)\). Strictly diagonally dominant matrices are invertible. In general, the \(n \times n\) A, with elements \(\left\{a_{i, j}\right\}_{i, j}\), is strictly diagonally dominant if

\[ \vert a_{i, i}\vert > \sum_{j\neq i} \vert a_{i, j}\vert, \qquad 1\leq i \leq n. \]

The matrix \((I + \tilde B)\) is strictly diagonally dominant if

\[ 1 > \sum_{j \in \mathcal A_i} \vert \tilde \beta_{i, j}|, \qquad 1 \leq i \leq J, \]

for all \(i \in \left\{1, \dots, J \right\}\). Alternatively, \(\lambda_i \in (0, 1)\) is found for each \(i\) to tune \(\tilde Q\) such taht it is strictly diagonally dominant, using

\[ \hat\tau_i^{-2} > \lambda_i \sum_{j\in\mathcal A_i, j\neq i} \vert \tilde Q_{i, j} \vert, \qquad 1\leq i \leq J, \]