# Is every Markov Process a Martingale Process?

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!

• 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 9:54
• @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" - i.imgur.com/f3vQzI7.png – Math1000 Aug 18 '16 at 8:35
• 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 8:48

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.

• 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 7:22
• $$g(X_s) = f(X_{s+t-s})= f(X_t)$$ – Math1000 Aug 18 '16 at 8:40
• 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
• 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
• 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

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.