# $X_n/(1+r)^n$ martingale implies $X_n$ not Markov?

If $X_n/(1+r)^n$ is a martingale, I can conclude that $X_n$ is not a martingale. But can I also conclude that $X_n$ is not Markov?

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 aaa: No you cannot. – Did Nov 6 '12 at 10:49 Why not? How can I prove it? – aaa Nov 6 '12 at 10:52 For a trivial counterexample, try every Xn deterministic. – Did Nov 6 '12 at 10:55 how can I prove that the def. of markov is satisfied then? E[ g(X_(n+1)) |info up to n ]= h(X_n) – aaa Nov 6 '12 at 11:02 This is not the definition of a Markov chain. Do you know the definition of a Markov chain? – Did Nov 6 '12 at 11:03