# Whether the optional stopping theorem is true for a positive continuous martingale and integrable stopping times

We know that for a uniformly integrable continuous martingale $$M$$ there holds the optional stopping theorem for all stopping times: $$E[M_T|\mathscr{F}_S]=M_S$$ for stopping times $$S\leq T$$. I'm wondering whether the optional stopping theorem is true for a positive continuous martingale and integrable stopping times.

Consider a positive continuous martingale $$M=(M_t)_{t\geq 0}$$, and two integrable stopping times $$S\leq T$$, then do we have $$E[M_T|\mathscr{F}_S]=M_S$$?

If we don't require the integrability of stoppping times, then it is false. A counterexample is given by the exponential Brownian motion $$M_t=\exp\left(B_t-\frac{t^2}2\right)$$ and the stopping time $$T=\inf\left\{t\geq0: B_t\leq \frac{t^2}2-\log 2\right\}=\inf\{t\geq0: M_t=1/2\},$$ where we have $$T<+\infty$$ a.s. and $$E[M_T|\mathscr{F}_0]=1/2\neq M_0$$. But I cannot figure how to prove that $$T\in L^1$$.

Related: There is a theorem in Le Gall's Brownian Motion, Martingales, and Stochastic Calculus (Theorem 3.25) states that for nonnegative continuous supermartingale and arbitrary stopping times $$S\leq T$$, we have $$E[M_T|\mathscr{F}_S]\leq M_S$$.

Any help would be appreciated.

## 1 Answer

Relating to your counterexample: for $$t\ge0$$ fixed, the stopping time $$T\wedge t$$ is bounded, so we can apply the stopping theorem: $$E[B_{T\wedge t}]=E[B_0]=0.$$ Since $$B_{T\wedge t}\ge\frac{(T\wedge t)^2}2-\log 2$$, this gives $$0\ge E\!\left[\frac{(T\wedge t)^2}2-\log2\right],\quad\text{i.e.,}\quad E[(T\wedge t)^2]\le2\log2.$$ Letting $$t\to\infty$$ yields $$E[T^2]<\infty$$ by Fatou's lemma (or monotone convergence). In particular, $$T$$ is integrable.

This even shows that $$L^2$$-integrability of $$T$$ is not sufficient to apply the stopping theorem.