TITLE: "Converse" of optional stopping theorem
QUESTION [10 upvotes]: Assume we are considering the discrete case. If $\{X_n\}_{n\in\mathbb{N}}$ is a martingale adapted to $F_n$ and $X_n\in L^1$, then for any bounded stopping time $\tau$, according to the optional stopping theorem (or optional sampling theorem), we will have that $E(X_\tau)=E(X_0)$.
I've seen some discussions on the condition such as the boundedness of the stopping time. However, now I am curious about that if we don't have the assumption that $X_n$ is a martingale does $E(X_\tau)=E(X_0)$ still holds for any bounded stopping time? In another word, does there exist any integrable random process $X_n$ adapted to $F_n$ which is not a martingale such that for any bounded stopping time $\tau$ it holds $E(X_\tau)=E(X_0)$?
Or we can rephrase this as "converse" of OST: if integrable random process $X_n$ adapted to $F_n$ and for any bounded stopping time $\tau$ it holds $E(X_\tau)=E(X_0)$, then $X_n$ is a martingale. Is this converse true or not?

REPLY [6 votes]: Nate Eldredge's answer provides some nice intuition on why the result holds true. Here's a different approach. 
The task at hand is to prove that $\forall n, E(X_{n+1}|F_n)=X_n$. Let $n$ be fixed. Going back to very definition of conditional expectation, it is sufficient to prove that for every $A\in F_n$, $E(1_A X_{n+1})=E(1_AX_n)$.
Define $\tau = n 1_A + (n+1)1_{A^c}$. $\tau$ is indeed a bounded stopping time (easy to check) and $$E(X_{\tau})=E(1_AX_n +1_{A^c}X_{n+1})$$
Since the constant $n+1$ is also a bounded stopping time, $E(X_{n+1})=E(X_0)$, thus $$E(1_AX_{n+1})+ E(1_{A^c}X_{n+1}) = E(X_{n+1})=E(X_0) = E(X_\tau)=E(1_AX_n)+ E(1_{A^c}X_{n+1})$$
hence $$E(1_AX_{n+1}) = E(1_AX_n)$$ which concludes the proof.