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?
Best Answer
Neat question! It is true, if I'm not mistaken.
Let $n$ be arbitrary and set $A = \{E[X_{n+1} \mid F_n] > X_n\} \in F_n$. Set $\tau = (n+1) 1_A + n 1_{A^c}$; you may verify that $\tau$ is a bounded stopping time. Then $X_\tau = X_{n+1} 1_A + X_n 1_{A^c}$.
Let $Y = E[X_\tau \mid F_n] = E[X_{n+1} \mid F_n] 1_{A} + X_n 1_{A^c}$ (by properties of conditional expectation). Note that $Y \ge X_n$, and $Y > X_n$ on $A$. But $E[Y] = E[X_\tau] = E[X_0] = E[X_n]$ since $n$ is also a bounded stopping time. So we conclude that $P(A) = 0$, which is to say $E[X_{n+1} \mid F_n] \le X_n$ almost surely. The opposite inequality can be shown in the same way, so we get $X_n = E[X_{n+1} \mid F_n]$ almost surely, and $n$ was arbitrary. Thus $X_n$ is a martingale.
Intuitively, if you think of $X_n$ as a stock price, then $A$ is the event that "information available by time $n$ suggests that, on average, the price is going to rise tomorrow". So $X_\tau$ corresponds to the strategy "if on day $n$, it looks like the price is going to rise, then hold until tomorrow; otherwise sell today". If $A$ had positive probability then this investment strategy would be profitable on average, which is supposed to be impossible if the stock price is a martingale.