Assume that we have a hidden Markov model, where we have a sequence of hidden variables $Z_1, \dots, Z_m$ which form a Markov chain. Now, at each "time point" $i$, an observation $Y_i$ is drawn from the conditional distribution given $Z_i$, i.e., $Y_i\sim F(\cdot | Z_i)$.

This leads to a sequence $(Z_1, Y_1), \dots, (Z_m, Y_m)$ which is a Markov chain. Now assume that the generation of the hidden variables $Z_1, \dots, Z_m$ ensures that $Z_1, \dots, Z_m$ is a stationary and ergodic sequence. Further, we even know that $(Z_1, Y_1), \dots, (Z_m, Y_m)$ then is an ergodic Markov chain.

My question is whether also the sequence of observations $Y_1, \dots, Y_m$ is ergodic given that both the sequence of hidden variables, as well as the joint sequence are ergodic. Papers such as this one seem to address the opposed question whether the sequence of hidden variables is ergodic.


Theorem 1 of this paper gives an answer for finite state spaces, and has been generalized later.

  • $\begingroup$ So, what is the answer for finite state spaces? $\endgroup$ – nbro Nov 26 '17 at 13:40

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.