are constants the only stationary martingales?

Question

Let $(X_n,\mathcal{F}_n)_{n\in\mathbb{Z}_+}$ be a nonnegative bounded martingale. Assume that the process $X_n$ is strictly stationary. Does it imply that $X_n$ is constant almost surely?

Answer 1

Outline: It suffices to show that $X_n=X_0$ a.s. for each $n\ge 1$. By Doob's decomposition, $X_n^2 = M_n +A_n$, $n\ge 0$, where $(M_n)$ is a martingale and $(A_n)$ in predictable and increasing, with $A_0=0$. By stationarity, $E(X_n^2)=E(X_0^2)=E(M_0)$, so $E(A_n)=0$, and consequently $A_n=0$ a.s for each $n\ge 1$. This means that $(X_n^2)_{n\ge 0}$ is a martingale. Now compute $E(X_{n+1}^2|\mathcal F_n)$ in two ways to learn that $E[(X_{n+1}-X_n)^2|\mathcal F_n]=0$, and in turn $E[(X_{n+1}-X_n)^2]=0$. This means that $X_{n+1}=X_n$ a.s. for each $n\ge 0$.