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Currently, I am using a Markov Chain to build a predictive model. I have done some research on the Internet, and found that a Markov Chain has a stationary distribution followed by ergodic condition.

My question is: what if I have a Markov Chain that is not ergodic, therefore, it may not have a unique stationary distribution. Despite the fact that I would not have a unique stationary distribution, can I still call my chain a Markov Chain and can I use it for a predictive model? or do I need to have an ergodic chain for a Markov Chain model to do prediction or simulation?

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  • $\begingroup$ Prediction is a continuation of the invariant from the past. You need to identify what in the past doesn’t change $\endgroup$
    – Aksakal
    Commented Jun 4, 2020 at 14:45
  • $\begingroup$ If you can train your chain on data and can find out the transition matrix, you can predict. I don't see why prediction has to be related with ergodicity. $\endgroup$
    – hbadger19042
    Commented Jun 4, 2020 at 14:57
  • $\begingroup$ Thank you for the comment.The reason why I mentioned egordicity in my question is that all the materials that I have gone through so far explain Markov Chain with egordicity, so I thought I have to have egordicity in my Markov Chain to do prediction with Markov Chain. $\endgroup$
    – xabzakabecd
    Commented Jun 4, 2020 at 15:05

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My answer will not be canonical: the distinguishing feature of Markov chains is memory loss, i.e. the next random value in the chain is independent from all the previous values when the current value is known. If your process has this property then you may call it Markov chain.

Here's the first reference to the term chain (цепь in Russian), I know of: enter image description here It's from Markov's work Rasprostranenie zakona bol’shih chisel na velichiny, zavisyaschie drug ot druga ("РАСПРОСТРАНЕНИЕ ЗАКОНА БОЛЬШИХ ЧИСЕЛ НА ВЕЛИЧИНЫ, ЗАВИСЯЩИЕ ДРУГ ОТ ДРУГА" in Russian), Izvestiya Fiziko-matematicheskogo obschestva pri Kazanskom universitete, 2-ya seriya, tom 15, 9 4, 1906, 135-156. Also, see Markov's bio here.

In the excerpt Markov describes a chain where $x_{k+1}$ is independent of $x_1,\dots,x_{k-1}$ when $x_k$ is known.

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