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Can you please help me by giving an example of a stochastic process that is Martingale but not Markov process for discrete case?

Peter
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  • The post above gives some idea, but I don't get much from it.I will greatly appreciate if someone can give me a very simple example (in discrete case), with proof. – Peter Sep 25 '13 at 02:54
  • This questions is not a duplicate of the linked question. It's much more general. The linked question is about a specific process. – tparker Aug 06 '24 at 14:10

1 Answers1

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Markov chains have a finite memory, Martingales can have an infinite one.

Pick a random value for $X_0$. Let the sequence of random variables $\{\epsilon_n,\,n>0\}$ be i.i.d. with mean$=E[\epsilon_{n}]=0$ and independent of $X_0$.

The process governed by $X_{n+1}=X_n+\epsilon_{n+1}X_0$

  • is a martingale as \begin{align} &E[X_{n+1}|X_0,\dots,X_n] =E[X_{n}|X_0,\dots,X_n]+E[\epsilon_{n+1}X_0|X_0,\dots,X_n]\\ &=X_n+E[\epsilon_{n+1}|X_0,\dots,X_n]E[X_0|X_0,\dots,X_n]\\ &= X_n+E[\epsilon_{n+1}]X_0\\ &= X_n+0X_0\\&=X_n \end{align} Here, it is key that $\epsilon_n$ is independent of the $\{X_i, 0\leq i\leq n\}$.

  • is not Markov as it is clear that $\Pr[X_n+\epsilon_{n+1}X_0|X_n]\neq \Pr[X_n+\epsilon_{n+1}X_0|X_0,\dots,X_n]$. To determine $X_{n+1}$ not only the value of $X_n$ but the entire path to it (at which value did the path start at $X_0$) is needed.

Martingale are about expectation and the Markov property about probability, which of course is also an expectation, but that's stuff for another post.

Inspitration was drawn from http://djalil.chafai.net/blog/2012/01/20/martingales-which-are-not-markov-chains/

jII
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QT_Scientist
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  • Ok, I have a question for you. That process above, I think the martingale proof is not persuasive. E[X_{n+1}] is a fixed number, while Xn is a random variable. The above actually, should have written as E[X_{n+1}|Fn] = Xn. So E[\epsilon_{n+1} | Fn] is not guaranteed to become zero. – Peter Sep 26 '13 at 03:34
  • @Peter your first remark is correct, I wrote it down in a sloppy way. corrected now. I don't get your second remark, as the $\epsilon$'s are independent of the $X_n$. – QT_Scientist Sep 26 '13 at 08:39
  • Thanks a lot QT, You point is correct. You can pull the epsilon zero out since it's independent of the process Xn. I missed that point earlier – Peter Sep 26 '13 at 15:30
  • So there is another property of a martingale, namely $\mathbb{E}[|X_n|] < \infty \forall n$. How can we see that this property is also satisfied? – Jason Apr 17 '20 at 09:35