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Is it true that every constant mean ($F_t$ adapted) Markov process must be a martingale?

I have not found this statement anywhere but I feel like it must be true. Below I attempt a proof.

\begin{align} &E(X_t) = E(X_s) \\ \implies &E(X_t - X_s) = 0 \quad (**)\\ \implies &E(X_t - X_s | F_s) = 0 \quad (*)\\ \implies &E(X_t | F_s) = E(X_s|F_s) = X_s \\ \end{align}

I think the lines (*) and (**) are equal since the process is Markov.

Thanks

  • ( * ) says that giving all the information at time s (i.e. the values of all $X_u$ for u smaller than s), the next increment $X_t-X_s$ has zero expectation value. The Markov property says that the increment $X_t-X_s$ depends not on the entire trajectory but just on the value $X_s$. Neither of these conditions imply that the increment $X_t-X_s$ is independent of $X_s$, which is what you are thinking probably when passing from ( ** ) to ( * ). This is a charachteristic of brownian motion for example, but not of a generic markov process (not even stationary). – Thomas Mar 23 '20 at 09:24

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It's not true. Assume $X_0=0$ and $X_n$ is uniformly distributed on $\{-1,1\}$ conditional on $X_{n-1}$ being $0$ and $X_n=sgn(X_{n-1})(|X_{n-1}|+1)$ otherwise. Then, clearly, $X_n$ is Markov, $X_n$ is uniformly distributed on $\{-n,n\}$ and hence, has mean $0$.

However, $(X_n)_{n\in\mathbb{N}}$ is by no means a martingale, since for $n\geq 2$, we have $$\mathbb{E}(X_n-X_{n-1}|X_{n-1})=1_{(X_{n-1}>0)}-1_{(X_{n-1}<0)}\neq 0$$

WoolierThanThou
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