Intuitively, why does stopping a driftless process result in a driftless process (i.e., why is a stopped martingale still a martingale)?

Question

As the title suggests, I'm asking for an intuitive explanation/idea of why a stopped martingale is still a martingale. This result doesn't seem intuitive to me for the following reason:

If $X_{t}$ is a martingale and $\tau$ is a stopping time,then the stopped process $X_{t \wedge \tau}$ is also a martingale. The problem I have with this is that since $\tau$ is random, the paths of the stopped process $X_{t \wedge \tau}$ will become constant at different times.

So, intuitively, when one path becomes constant, its variation is no longer contributing to the drift of the process. Since $X_{t}$ was originally a martingale and equally likely to go up or down, after the first path stops, it seems like there would be a weight or drift in the opposite direction because the path that is now constant is no longer balancing the weight of the other path going in the opposite direction.

I hope the above intuitive description of my issue makes sense. What's wrong with my intuition?

Answer 1

Since by definition of a stopping time the event $\lbrace \tau \leq t \rbrace$ is $\mathcal F_t$-measurable, you don't require knowledge of the future to decide whether to stop the process.

Thinking of a martingale $X_t$ as a fair game, if you could construct a stopping time such that $ \mathbb E [X_{\tau \wedge (t + s)} \mid \mathcal F_t ] \neq X_t$, it would suggest you are able to alter the game in your favor by using knowledge of the past. This contradicts both the notion of a martingale and a stopping time.

That is the basic intuition.

It should be noted that there are exceptions to this intuition, for example that you don't in general have $\mathbb E[X_\tau \mid \mathcal F_t] = X_t$ for a martingale $X$. Conditions for when this is true is provided in the Optional Stopping Theorem.