"Converse" of optional stopping theorem

Solution 1:

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.

Solution 2:

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.