# Stopped continuous-time supermartingales

I am trying to prove that if I have a right-continuous supermartingale $$(S_t,\mathcal{F}_t)_{t\geq0}$$ and $$\tau <\infty$$ a stopping time, that $$(S_{\tau \wedge t},\mathcal{F}_{\tau \wedge t})_{t\geq0}$$ is also supermartingale.

I proved it for the discrete time case but I don't know how I can take the limit and prove it for the continuous time case.

I tried to take $$\tau_n \searrow \tau$$ and show uniform integrabillity of $$S_{\tau_n \wedge t}$$ but I am stuck.

Thank you

• Usually, the idea is to approximate $\tau$ by a sequence of discrete stopping times $\tau_n$ satisfying $\tau_n \downarrow \tau$.
– saz
Jun 9, 2015 at 18:42
• It follows from a general result called "optional sampling". Since you're trying to show uniform integrability, you might have a similar theorem. What does it say? Jun 9, 2015 at 23:20
• I can assume the discrete version of the sampling theorem
– Lin
Jun 10, 2015 at 11:22

Optional sampling theorem : If $$X = (X_t,\mathcal{F}_t)$$ is a supermartingale and $$T$$ is an arbitrary stopping time, then the stopped process $$X^T=(X_{T\wedge t},\mathcal{F}_t)$$ is also a supermartingale.
Proof : Here we give a sketch of the proof. Assume $$X$$ is a supermartingale. First, for a fixed $$n\ge 1$$, let $$D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots$$ be the set of non-negative dyadic rationals of order no greater than $$n$$. It follows that $$X = (X_t,\mathcal{F}_t; t\in D_n)$$ is a super martingale (discrete time).
Second, we construct a stopping time $$T_n$$ such that $$T_n\ge T$$ and $$T_n$$ only take values in $$D_n$$. Indeed, let $$T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}.$$ Then $$T_n\ge T_{n+1} \ge \cdots$$ and $$T_n$$ is a stopping time. Fix $$0\le s\le t$$, we wish to show that $$\mathbb{E}\left[X_{t\wedge T}|\mathcal{F_s}\right]\le X_s$$ almost surely. Similarly define $$t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots$$ and $$s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots$$ It follows from the discrete time optional sampling theorem that $$\mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le X_{s_m \wedge T_n}$$ for any integers $$m\ge n$$. Letting $$m\to\infty$$, we have $$s_m \to s$$ decreasing and $$\mathcal{F}_{s_m}\to\mathcal{F}_s$$. By Lévy's Downward theorem and the continuity of process $$X$$, we have $$\mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le X_{s_m \wedge T_n}$$ Observe that $$(X_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$$ is a backward martingale, whence it is uniformly integrable. Letting $$n\to\infty$$, we arrive at $$\mathbb{E}\left[X_{t\wedge T}|\mathcal{F}_{s}\right] \le X_{s\wedge T}.$$ This completes the proof.
• How is this a backward martingale? This makes no sense since we assumed $X_t$ is a SUPERMARTINGALE, not a martingale. Mar 26, 2020 at 2:42