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According to the definition (2.3.6) of a Markov Process in Shreve's book titled Stochastic Calculus for Finance II:

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $T$ be a fixed positive number, and let $\mathcal F(t)$, $0\leqslant t\leqslant T$, be a filtration of sub-$\sigma$-algebras of $\mathcal F$. Consider an adapted stochastic process $X(t)$, $0\leqslant t\leqslant T$. Assume that for all $0\leqslant s\leqslant t\leqslant T$ and for every nonnegative, Borel-measurable function $f$, there is another Borel-measurable function $g$ such that $$\mathbb E\left[f(X(t))\mid\mathcal F(s)\right] = g(X(s)). $$ Then we say that the $X$ is a Markov process.

it seems obvious to me that every Markov Process is a Martingale Process (Definition 2.3.5):

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $T$ be a fixed positive number, and let $\mathcal F(t)$, $0\leqslant t\leqslant T$, be a filtration of sub-$\sigma$-algebras of $\mathcal F$. Consider an adapted stochastic process $M(t)$, $0\leqslant t\leqslant T$. If $$\mathbb E\left[M(t)\mid\mathcal F(s)\right] = M(s)\quad\textrm{for all } 0\leqslant s\leqslant t\leqslant T,$$ we say this process is a martingale.

Can someone please tell me if this is correct?

Thanks!

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    $\begingroup$ The answer is No. To understand why the book led you to believe otherwise, we would need the relevant definitions to be reproduced in your question. $\endgroup$
    – Did
    Aug 17, 2016 at 9:54
  • $\begingroup$ @Did Reproduction complete. $\endgroup$
    – Math1000
    Aug 17, 2016 at 10:21
  • $\begingroup$ Did you miss "for every nonnegative, Borel-measurable function f, there is another Borel-measurable function g such that..." by any chance? Yes definitions 2.3.6 (although phrased in a slightly unusual way) and 2.3.5 (although it omits an integrability condition) are (basically) correct. $\endgroup$
    – Did
    Aug 17, 2016 at 10:41
  • $\begingroup$ @Did I said "reproduction" - i.imgur.com/f3vQzI7.png $\endgroup$
    – Math1000
    Aug 18, 2016 at 8:35
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    $\begingroup$ This text is is not known for its mathematical rigour. In particular, a footnote on page 103 states "We shall not dwell on subtle differences among types of convergence of random variables." $\endgroup$
    – Math1000
    Aug 18, 2016 at 8:48

2 Answers 2

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For a simple counterexample, let $X_t=t$ and $\mathcal F_t$ be the natural filtration. Then for $s<t$ and nonnegative measurable $f$, $$\mathbb E[f(X_t)\mid\mathcal F_s] = f(X_s+t-s)=:g(X_s) $$ so that $X_t$ is Markov, but $$\mathbb E[X_t\mid \mathcal F_s] = t\ne X_s, $$ so that $X_t$ is not a martingale.

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    $\begingroup$ The reason why the process $(X_t)$ is Markov is rather that, for each $s<t$, $\mathbb E[f(X_t)\mid\mathcal F_s] = g(X_s) $ with $g:x\mapsto f(x+t-s)$. $\endgroup$
    – Did
    Aug 18, 2016 at 7:22
  • $\begingroup$ $$g(X_s) = f(X_{s+t-s})= f(X_t) $$ $\endgroup$
    – Math1000
    Aug 18, 2016 at 8:40
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    $\begingroup$ Trying hard to stay on confusing formulations instead of adopting the crystal clear one in my comment, are we? $\endgroup$
    – Did
    Aug 18, 2016 at 10:05
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    $\begingroup$ Why do you think it did not (turned out to be such a great pedagogical one)? Not that this is the subject of this thread, but I am curious. $\endgroup$
    – Did
    Aug 18, 2016 at 12:54
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    $\begingroup$ Yeah -- which seems an excellent reason to have modified your answer along the line I suggested. Glad to see that you saw the light, in the end... $\endgroup$
    – Did
    Aug 18, 2016 at 13:08
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Just to point out the error in your logic more directly:

Suppose $X(t)$ is a Markov process. Then if we take $f(x) = x$, it is true that for all $s < t$ there exists a function $g$ such that $E[X(t) \mid \mathcal{F}(s)] = g(X(s))$. For instance if $X(t)$ is an Ornstein-Uhlenbeck process, that function $g$ is something like $g(x) = e^{-(t-s)} x$.

However, in order for $X(t)$ to be a martingale, we would need specifically to end up with $g(x) = x$. In general this need not happen.

The converse isn't true either. If we know $X(t)$ is a martingale and we try to see whether it is a Markov process, we know that if we take $f(x) = x$ there is a function $g$ that works (namely $g(x)=x$), but if we take $f$ to be some other function, the definition of martingale does not guarantee that we can find a corresponding $g$ at all.

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