# Stopping times and the left limit

So assuming $$X$$ is càdlàg martingale and $$\tau\leq T$$ is a stopping time. The stopping theorem gives us something like $$E[X_T| \mathcal{F}_\tau]= X_\tau.$$

But if the usual conditions hold, shouldn't there be something like $$E[X_T| \mathcal{F}_{\tau-} ]= X_{\tau-}$$ for $$X_{t-}:= \lim\limits_{ t_n\uparrow \tau, t_n\not=t} X_{t_n}$$?

I think this is correct: $$\mathcal{F}_{\tau-}= \bigcup_{ n\in \mathbb{N}} \left\{ A\in \mathcal{F}: A \cap\{ \tau\leq t-1/n\} \right\}$$ And because the filtrations is right-continious and must not be left-continious that could differ from $$\mathcal{F}_\tau$$.

So is there something like an extended stopping theorem?

I am aware of the following result, for a slightly different definition of $$\mathcal F_{\tau-}$$. I'll start with a few definitions. For two $$\sigma$$-algebras $$\mathcal F$$ and $$\mathcal G$$, we denote the smallest $$\sigma$$-algebra containing both $$\mathcal F$$ and $$\mathcal G$$ by $$\mathcal F \vee \mathcal G$$.

Definition. The $$\sigma$$-algebra $$\mathcal F_{\tau-}$$ of events strictly prior to a stopping time $$\tau$$ is given by $$\mathcal F_{\tau-} = \mathcal F_0 \vee \sigma\left(\left\{A \cap \{ t< \tau\} : A\in\mathcal F_t , t\in [0,\infty)\right\}\right).$$

In other words, $$\mathcal F_{\tau-}$$ is the $$\sigma$$-algebra generated by $$\mathcal F_0$$ and all the sets of the form $$A \cap \{ t <\tau \}$$, where $$t \in [0,\infty)$$ and $$A \in \mathcal F_t$$. I believe the spirit of this definition should be the same as the one you state. However, I think your definition of $$\mathcal F_{\tau-}$$ is, in general, not a $$\sigma$$-algebra.

I will need one more definition before I can state the result.

Definition. A stopping time $$\tau$$ is said to be predictable if there is a sequence $$\{ \tau_n \}_{n\in\mathbb N}$$ of stopping times such that:

1. $$\tau_n \uparrow \tau$$ almost surely as $$n \to \infty$$; and,
2. on the set $$\{ \tau >0 \}$$, $$\tau_n < \tau$$ a.s. for all $$n$$.

The first point above implicitly requires that the sequence $$\{ \tau_n \}_{n\in\mathbb N}$$ be almost-surely non-decreasing. When $$\tau$$ is a predictable stopping time, we call any sequence satisfying the requirements in the above definition an announcing sequence for $$\tau$$.

I can now state the result you allude to.

Theorem. If $$\tau$$ is a predictable stopping time with announcing sequence $$\{ \tau_n \}_{n\in\mathbb N}$$ and $$M$$ is a càdlàg uniformly integrable martingale, then $$M_{\tau-} = \lim_n M_{\tau_n} = E[M_{\tau} \mid \mathcal F_{\tau-}].$$

To obtain the equality you state, we can use the optional stopping theorem to conclude that $$M_{\tau} = E[M_T \mid \mathcal F_{\tau}]$$. Using the tower property of conditional expectations, we have, under the conditions of the stated theorem, that $$M_{\tau-} = E[M_T \mid \mathcal F_{\tau-}].$$

As with the optional stopping theorem, I believe that the requirement of uniform integrability can be relaxed when we assume that our stopping time is bounded, but I'll need to have a careful look at the proof of this result before I can say that we can do the same here.

Why do we need predictability of the stopping time?

In general, if $$X$$ is a random variable, and we have that $$M_t = E[X \mid \mathcal F_t]$$ up to indistinguishability, it does not follow that $$M_{\tau-} = E[X \mid \mathcal F_{\tau-}]$$. To see this, suppose $$X$$ is a Poisson process with parameter $$\lambda$$, and $$M_t = E[X_1 \mid \mathcal F_t] = X_{t \wedge 1} + \lambda (1-t)^+$$. If $$\tau$$ is the first jump time of $$X$$, then we have that $$M_{\tau-} \neq E[X_1 \mid \mathcal F_{\tau -}]$$.

Note that the jump times of a Poisson process are not predictable.

Being able to justify all this is going to make this post quite lengthy, so I simply refer you to Chapter 6 of the reference given below, which contains everything I've written here.

Reference: Cohen, S. N., & Elliott, R. J. (2015). Stochastic calculus and applications (Vol. 2). New York: Birkhäuser.

• I think that‘s what I need! Thank you! Oct 21, 2019 at 20:25
• I thought a little bit about the bounded stopping time: If we know that $X$ is càdlàg and $\tau$ is bounded by $T\in\mathbb{R}$ we could look at $(X_t)_{t\in [0,T]}$ and there we will always have uniform integrability, if I am not wrong. Oct 22, 2019 at 5:54
• @cptflint You're welcome; I'm glad the answer was helpful. As per your observation re uniform integrability, that sounds right, but I'd want to think about it some more. However, if, in addition, you know that $X_T \in L^2$, then $\{X_t\}_{t\in [0,T]}$ is certainly uniformly integrable, this guarantees that $\{X_t\}_{t\in [0,T]}$ is square-integrable and thus uniformly integrable. Oct 22, 2019 at 6:01
• @cptflint Actually, yes, you are right. This follows from the fact that, for $t\le T$, $X_t =E[X_T \mid \mathcal F_t]$, and so $\{X_t\}_{t\in[0,T]} = \{E[X_T \mid \mathcal F_t]\}_{t\in[0,T]}$, and we know the latter is always uniformly integrable. Oct 22, 2019 at 6:13
• Thank you for this clarification! I am playing around with that and a stopping time from my other question (math.stackexchange.com/q/3399633/615367?sem=2). Just assume a càdlàg martingale $(M_t)_{t\in [0,T]}$. With this concept it should be easy too proof that $P(M_T>0)=1$ implies $P( inf M_t=0)=0$, right? But we do not have $\tau_i\leq \tau$ for $\tau_i:=inf\{ t: M_t< 1/n \}$? Because $M$ can jump to zero or under zero... Do we have to use another sequence? Oct 22, 2019 at 10:33