Constant mean Markov process must be a martingale?

Question

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

Answer 1

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 $$\mathbb{E}(X_n-X_{n-1}|X_{n-1})=1_{(X_{n-1}>0)}-1_{(X_{n-1}<0)}\neq 0$$