# $X_n/(1+r)^n$ 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