Difference of stopping times is not a stopping time

Question

In this question Measurability of stopping times

it was explained that if $\tau $ and $\rho$ are stopping times, the difference $\tau -\rho$ is in general not a stopping time.

1. I wonder if under the additional assumption that $\tau >\rho$ that is still true.

2. In the comments it was stated that : If for for some $T$ we have $\mathcal F_{\tau}, \mathcal F_{\rho} \subseteq \mathcal F_{T}$ then, we infer that $\{τ−ρ<0\}\in \mathcal F_T$. I didn't quite get the reasons they gave. What do the subindeces in $\mathcal F_{\tau}, \mathcal F_{\rho}$ actually represent to start with? And why isn't this in contradiction with the fact that the difference of stopping times is not a stopping time?

Answer 1

Assume that $(\Omega, \mathcal{F}, P)$ is the underlying probability space.

If $\tau$ denotes a stopping time, then the sigma-field $\mathcal{F}_{\tau}$ consists of all events $A \in \mathcal{F}$ such that for all $t \in \mathbb{R}^+$ $A \cap \{T \leq t\} \in \mathcal{F}_t$

This is the meaning of the symbol you are referring to. So it is by no means obvious how to work with this object and it requires a separate proof to show that $\{\tau \leq \rho\}$ is $\mathcal{F}_{\tau}$-measurable. E.g. along the following lines (this is only one direction):

$$\{\tau < \rho, \tau \leq t\}= \cup_{r \in \mathbb{Q}, r< t} \{\tau \leq r, r< \rho\} \cup \{\tau \leq t, t<\rho\}$$

Note that for $\tau - \rho$ to be a stopping time you need to check $$\{\tau - \rho \leq t\} \in \mathcal{F}_t$$ and the above reasoning gives you $\{\tau - \rho < 0\} \in \mathcal{F}_{\tau}$. So there is no apparent relation.

In general your statement on the difference of two stopping times being a stopping time should be false, even under the additional assumptions.

For example, lets assume $\tau$ is less than the (constant) stopping time $t$. Your assertion would be that $t-\tau$ denotes a stopping time, i.e. you have to show that for any $v$: $$\{t-\tau \leq v\} \in \mathcal{F}_v$$ Pick $v=\frac{t}{4}$: $$\{t-\tau \leq v\} = \{\tau \geq \frac{3}{4}t\} \in \mathcal{F}_{\frac{t}{4}}$$ which in general can't be true because $\{\tau \geq \frac{3}{4}t\}$ is in $\mathcal{F}_{\frac{3}{4}t}$

Edit: $\mathcal{F}_{\tau}$ is not the sigma algebra generated by $\tau$. In fact it is the smallest sigma field such that $X_{\tau}$ is measurable for each stochastic process which is adapted and cadlag.$X_{\tau}$ thereby denotes the process $X$ sampled at $\tau$, i.e. $X(\omega, t=\tau(\omega))$ (assuming $\tau<\infty$)