# Condition on the covariance matrix of a gaussian process needed to have the Markov property

Let suppose to have a realization $$\mathbf{X}=(\mathbf{X}_1,\dots, \mathbf{X}_n)$$, where $$\mathbf{X}_i \in \mathcal{R}^d$$, from a $$d-$$variate Gaussian process.

Let also suppose that $$E(\mathbf{X}_i)= \mathbf{0}_d$$ and $$Cov(\mathbf{X}_i)= \boldsymbol{\Sigma}$$.

If I indicate with $$C_{ij}$$ the portion of the covariance matrix of $$\mathbf{X}$$ that rules the dependence between $$\mathbf{X}_i$$ and $$\mathbf{X}_j$$, and i assume that this must depend on the distance $$|i-j|$$, which are the necessary conditions needed to have the following? $$f(\mathbf{X}_i|\mathbf{X}_{i-1},\mathbf{X}_{1-2},\dots,\mathbf{X}_1) = f(\mathbf{X}_i|\mathbf{X}_{i-1})$$ where $$f()$$ is the normal density, i.e. the process is Markovian.

A reference with a proof is highly appreciated.

Thanks

Since the process is Gaussian and zero-mean, formulas for conditional means and variances for the multivariate normal distribution imply that you can write each $$\mathbf{X}_t$$ as a linear combinations of $$\mathbf{X}_{t-1}$$, $$\mathbf{X}_{t-2}$$, ... plus an independent Gaussian error term. But since the process is also Markovian, only the first term will have a non-zero matrix of coeffients $$\boldsymbol{\phi}_1$$. Thus, we can write $$\mathbf{X}_t = \boldsymbol{\phi}_1\mathbf{X}_{t-1}+\mathbf{w}_t. \tag{1}$$ The Markov property also implies that the error terms $$\mathbf{w}_1, \mathbf{w}_2, \dots$$ must all be independent and covariance stationary implies that $$\boldsymbol{\phi}_1$$ and the variance matrix of $$\mathbf{w}_t$$ remains constant across time. Hence, the process you describe is a vector AR(1) process. Its autocovariance matrix function is $$\boldsymbol\Gamma(k)=\mbox{Cov}(\mathbf{X}_{t-k},\mathbf{X}_{t})=\boldsymbol\Gamma(0) (\boldsymbol{\phi}_1')^k$$ for $$k\ge 1$$, see Wei, 2007, section 16.3.1. A further restriction is that $$\boldsymbol\phi_1$$ must have eigenvalues with modulus smaller than 1, otherwise the process is not stationary. Note also that $$\boldsymbol\Gamma(-k)=\boldsymbol\Gamma(k)'$$ by definition of the auto covariance matrix function (so the autocovariance matrix function can not depend only on $$|i-j|$$).
A sufficient and necessary condition for a Gaussian process to be also Markovian is to have a triangular covariance function, i.e., $$\mathrm{Cov}[X_{t_{1}}, X_{t_2}] = r_1(\min(t_1, t_2))r_2(\max(t_1, t_2)),$$ for unique (up to multiplicative constants) functions $$r_1$$ and $$r_2$$. The first full proof of this equivalence was provided in this paper by I. S. Borisov.