Stopped process

Remarks

The notation $t \wedge \tau = \min(t, \tau)$ is standard, so $X^\tau_t = X_{t \wedge \tau}$. On the event $\set{\tau = \infty}$, the stopped process equals $X$.

In discrete time, if $X$ is adapted, then so is $X^\tau$, because $$X^\tau_n = X_0 + \sum_{k=1}^{n} \mathbf{1}_{\set{\tau \ge k}} (X_k - X_{k-1})$$ for real-valued $X$, and $\set{\tau \ge k} = \Omega \setminus \set{\tau \le k - 1} \in \mathcal{F}_{k-1}$. The same formula shows that a stopped martingale is a martingale. In continuous time, the stopped process of a progressively measurable process is progressively measurable.

Example: let $S$ be a simple symmetric random walk started at $0$ and let $\tau$ be the first time with $\abs{S_n} = 5$. Then $\abs{S^\tau_n} \le 5$ for all $n$, so $S^\tau$ is a bounded martingale.

Non-example: if $\tau$ is not a stopping time, the formula still defines a process, but it need not be adapted. Take the walk above and let $\sigma$ be the last visit to $0$ before time $10$. The value $S^\sigma_1$ equals $S_1$ if the walk returns to $0$ between times $1$ and $10$, and $0$ otherwise. This is not determined by $\mathcal{F}_1$, so $S^\sigma$ is not adapted.