# Relation between Vector Auto-regressive models and correlation matrix

I am generating a multivariate time series using Vector Autoregressive Models- $$X(t) = AX(t-1) + \epsilon$$ where $$X \in R^{n \times 1}$$, $$A \in R^{n \times n}$$ and $$\epsilon \in R^{n \times 1}$$ is a constant. Is there a relation between $$A$$ matrix and correlation matrix storing the correlation information between different variables of the multivariate time series?

Correlation matrix can be defined as $$\rho _{i,j} = \text{Corr}(x_i,x_j)$$ where $$\text{Corr}(x_i,x_j)$$ is the correlation between $$i^{th}$$ and $$j^{th}$$ variable of the time series and $$\rho _{i,j}$$ is the $$i,j$$ element in the correlation matrix. I have asked the same question here.

• There is a relation. If you express this VAR(1) model as a VMA($\infty$) model, you can relate the correlation of $X$ with the $A$ matrix and the covariance matrix $\Sigma$ of the errors $\epsilon$. The components of $X$ are just linear combinations of $\epsilon$ with weights that are functions of $A$. Commented Nov 14, 2018 at 19:31
• $\epsilon \in R^{n \times 1}$ is a constant? Commented Nov 14, 2018 at 20:10

You may express this VAR(1) model as a VMA($$\infty$$) model: \begin{aligned} X_t &= A X_{t-1}+\epsilon_t \\ (I-AL) X_t &= \epsilon_t \\ X_t &= (I-AL)^{-1} \epsilon_t. \\ \end{aligned} Here, $$L$$ stands for the lag operator and $$I$$ represents an identity matrix of appropriate dimension.
You can further express this as $$X_t = \sum_{q=0}^\infty \Phi_q \epsilon_{t-q}$$ with coefficient matrices $$\Phi_q$$ where $$\Phi_0=I$$. Thus the components of $$X_t$$ are linear combinations of $$\epsilon_t$$ and its infinite lags with weights that are functions of $$A$$.
Given that $$\epsilon$$ is an i.i.d. vector with a contemporary covariance matrix $$\Sigma_{\epsilon}$$, we get that the covariance matrix of $$X$$, $$\Sigma_X$$ (with a typical element $$\sigma_{X,ij}$$), is $$\Sigma_X = \text{Cov}(X_t) = \text{Cov} \left( \sum_{q=0}^\infty \Phi_q \epsilon_{t-q} \right) = \sum_{q=0}^\infty \Phi_q \Sigma \Phi_q^{\top}$$ where $$A$$ is hiding in the matrices $$\Phi_q$$. Then the correlation between elements $$X_i$$ and $$X_j$$ can be obtained as $$\rho_{X,ij} = \frac{ \sigma_{X,ij} }{ \sqrt{\sigma_{X,ii}\sigma_{X,jj}} }.$$