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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!

Math1000
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Sagar Kolte
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    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. – Did Aug 17 '16 at 09:54
  • @Did Reproduction complete. – Math1000 Aug 17 '16 at 10:21
  • 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. – Did Aug 17 '16 at 10:41
  • @Did I said "reproduction" - https://i.imgur.com/f3vQzI7.png – Math1000 Aug 18 '16 at 08:35
  • @Math1000 Yes, and the point is? (Unrelatedly, please correct your answer.) – Did Aug 18 '16 at 08:41
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    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." – Math1000 Aug 18 '16 at 08:48
  • if f is a bounded harmonic function, then f(Markov process) is a martingale – Marouen Jun 05 '20 at 17:15

2 Answers2

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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.

Math1000
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    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 $g:x\mapsto f(x+t-s)$. – Did Aug 18 '16 at 07:22
  • $$g(X_s) = f(X_{s+t-s})= f(X_t) $$ – Math1000 Aug 18 '16 at 08:40
  • What? $ $ $ $ $ $ – Did Aug 18 '16 at 08:43
  • Correction: $$g(X_s) = f(X_s + t - s) = f(s + t - s) = f(t) = f(X_t). $$ – Math1000 Aug 18 '16 at 08:51
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    Trying hard to stay on confusing formulations instead of adopting the crystal clear one in my comment, are we? – Did Aug 18 '16 at 10:05
  • It is clear from $f(X_t)=f(t)$ that $f(X_t)$ is a (Borel)-measurable function of $G(s)=s$, and finding the explicit map $x\mapsto f(x+t-s)$ is a trivial matter that can be left to the reader. – Math1000 Aug 18 '16 at 10:18
  • See previous comment. (Doggedly "defending" one's answers, just because one posted them, is not the way to make progress, I am sorry to say.) – Did Aug 18 '16 at 10:20
  • Consider the equivalent definition $$\mathbb E[f(X_t)\mid\mathcal F_s] = \mathbb E[f(X_t)\mid \sigma(X_s)] $$ for all $0\leqslant s\leqslant t$ and $f$ (bounded) measurable. In this example, $X_t$ and $X_s$ are independent for $s\ne t$ so my answer was mathematically correct. – Math1000 Aug 18 '16 at 10:46
  • Sure, expressing a conditional expectation conditionally on $\mathcal F_s$ as a function of $X_t$ instead of a function of $X_s$ is the most congenial and pedagogically satisfying way of explaining the solution. Suuure. – Did Aug 18 '16 at 10:49
  • I don't disagree. But this was after all a "simple" counterexample and turned out not to be such a great pedagogical one. – Math1000 Aug 18 '16 at 11:03
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    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. – Did Aug 18 '16 at 12:54
  • The process is deterministic so trivially $X_t$ is independent of $\mathcal F_s$, leading to the (potentially misleading) statement $$\mathbb E[f(X_t)\mid \mathcal F_s]; (=\mathbb E[f(X_t)]); = f(X_t) $$ in the first place. – Math1000 Aug 18 '16 at 13:05
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    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... – Did Aug 18 '16 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.

Nate Eldredge
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