1
$\begingroup$

I'm studying stochastic process and Markov Chain.

I was wondering if a Gaussian Process has the Markov Property (that is the conditional probability distribution (given the present states) of future states is independent of the past states). Personally, following my intuition, I would say that a Gaussian Process has the Markov Property only when the covariance of the Gaussian is a diagonal matrix.

My reasoning is the following: A Gaussian Process with a diagonal covariance matrix is a process of independently distributed random variables, and so by definition has the Markov Property.

Is my reasoning correct? Is the assumption of diagonal covariance necessary?

$\endgroup$
  • 1
    $\begingroup$ An Ornstein–Uhlenbeck process is described as a non-trivial stationary Gauss–Markov process $\endgroup$ – Henry Feb 6 at 16:01
  • $\begingroup$ Thank you for the resource, but it's really hard to understand if my reasoning is correct from that Wikipedia page $\endgroup$ – Tommaso Bendinelli Feb 6 at 22:05
  • $\begingroup$ I believe that a section of the Wikipedia article points to the covariance matrix being non-diagonal, even if you adjust the calculation to $\text{cov}(x_s,x_t-x_s)$ with $t \gt s$, something which would be $0$ with a Wiener process $\endgroup$ – Henry Feb 6 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.