# Difference of two stopping times

Let $$(X_{n})_{n \geq 0}$$ be a sequence of random variables and $$\tau,t$$ stopping times with respect to the sequence $$(X_{n})_{n \geq 0}$$

\begin{align*}\{\tau+t =n\} = \{\tau+t = n\} \cap \{t \leq n\} &= \bigcup_{k=0}^n \{\tau+t = n\} \cap \{t = k\} \\ &= \bigcup_{k=0}^n \{\tau=n-k\} \cap \{t=k\}. \end{align*}

As $$\{\tau =n-k\} \in \mathcal{F}_{n-k} \subseteq \mathcal{F}_n$$ and $$\{t = k\} \in \mathcal{F}_k \subseteq \mathcal{F}_n$$ for any $$k \leq n$$, this implies that $$\{\tau+t=n\} \in \mathcal{F}_n$$, and so $$\tau+t$$ is a stopping time.

Now, my question is, let assume that I am considering $$\tau-t$$. I know that in general, $$\tau-t$$ is not a stopping time. However, if I were to consider my birthday this year (a stopping time), which is a deterministic stopping time. At any time, I know exactly when my birthday occurs. Also, I know two days before my birthday i.e, $$\tau-2$$. What kind of a formulated counterexample will show that $$\tau-2$$ is indeed a stopping time in this setting.

• I'm not sure I understand what you're asking. Are you asking whether or not $\tau-2$ is a stopping time when $\tau$ is deterministic? Commented Nov 12, 2021 at 3:28
• @user6247850 Yes. I am asking whether or not τ−2 is a stopping time in this case, and if so, what is your reasoning? Commented Nov 12, 2021 at 3:49
• To be clear, by "in this case" you mean "$\tau$ is deterministic"? Commented Nov 12, 2021 at 4:02
• @user6247850 yes. $\tau$ is deterministic. We need to stop two days before my birthday. Commented Nov 12, 2021 at 4:31

If $$\tau$$ is deterministic, $$\tau-2$$ is deterministic. As long as $$\tau-2 \ge 0$$, it is a stopping time because $$\{\tau-2 = n\} = \emptyset$$ or $$\{\tau-2 = n\} = \Omega$$ for all $$n$$.
I think the key is that your $$\tau$$ is deterministic, so $$\tau-2$$ is deterministic. But if $$\tau$$ is not deterministic, i.e. $$\tau$$ is not a constant random variable, we want to know if $$\{\tau-2\leq n\}$$ happened or not, we need the information in $$\mathscr{F}_{n+2}$$, since $$\{\tau-2\leq n\}\equiv\{\tau\leq n+2\}$$.