# Difference between $X_t = Y_t$ a.s. and $X_{\tau} = Y_{\tau}$ a.s. ($\tau$ a stopping time) for càdlàg processes $X,Y$?

We work on a filtered probability space $$(\Omega,\mathcal{F},(\mathcal{F})_{t\in[0,T]},P)$$. Let $$X,Y$$ be two càdlàg adapted stochastic processes. What is the difference between the following two conditions:

1. $$X_t= Y_t \enspace P$$-a.s. $$\forall t\in [0,T]$$
2. $$X_{\tau}= Y_{\tau} \enspace P$$-a.s. $$\forall [0,T]$$-valued stopping times $$\tau$$

Question: Which of the two conditions implies the other one, and are they even equivalent?

Partial answer: I think that 2. $$\Rightarrow$$ 1. holds. Indeed, if 2. holds, then for $$t\in [0,T]$$ fixed, defining $$\tau(\omega)$$ to be equal to $$t$$ for all $$\omega$$, yields that $$\tau$$ is a stopping time, and thus 1. holds.

What about the other direction, i.e. does 1. $$\Rightarrow$$ 2. hold?

Note: The same question under the assumption that $$X,Y$$ are làdlàg is discussed in this post. In this case 1. $$\Rightarrow$$ 2. does not hold. There is a counterexample in the accepted answer.

Indeed, 2. implies 1. by the argument you have presented.

Now note the following:

Theorem ([1;Lemma 21.5]). Fix a probability space as you have done in your question and let $$I\subset\mathbb R$$. We will assume that the probability space is complete, i.e. that all subsets of null sets are measurable (though the reference given formulates this Theorem without this assumption).
If $$X=(X_t)_{t\in I},Y=(Y_t)_{t\in I}$$ are two stochastic processes taking values in some metric space $$(E,d)$$ such that $$\mathsf P(X_t=Y_t)=1$$ for all $$t\in I$$, then, if either

1. $$I$$ is countable; OR
2. $$I$$ is a connected set and $$X,Y$$ are almost surely right-continuous,

we have that $$\mathsf P(X_t\neq Y_t \text{ for some }t\in I)=0.$$

Let $$I=[0,T]$$ for some $$T>0$$. Then the Theorem is a non-trivial result because the event "$$X_t\neq Y_t \text{ for some }t\in[0,T]$$" is an uncountable union of the events "$$X_t\neq Y_t$$", and in general the uncountable union of null sets need not be a null set.

In any case, this shows that 1. also implies 2., because for any function $$\tau:\Omega\to[0,T]$$ (whether a stopping time or not, $$\tau$$ doesn't even need to be measurable), we have that $$\{\omega\in\Omega: X_{\tau(\omega)}(\omega)\neq Y_{\tau(\omega)}(\omega)\}\subset\{\omega\in\Omega:X_t(\omega)\neq Y_t(\omega)\text{ for some }t\in[0,T]\},$$ so that by completeness of the probability space, $$\mathsf P(X_\tau\neq Y_\tau)=0.$$ If the probability space is not complete, you may run into trouble with non-measurable sets appearing.

# Literature

[1] Achim Klenke, Wahrscheinlichkeitstheorie. 3. Auflage. Springer-Verlag Berlin/Heidelberg (2013).

• Actually, I haven’t looked too much into the measurability issues. But I believe you can get useful results also for non-complete spaces. Jan 27, 2022 at 11:05
• Thank you! In my case the probability space is complete.
– user711386
Jan 27, 2022 at 11:12
• I have a follow-up question: If $X,Y$ are cadlag, adapted, then $P[X_t=Y_t]=1$ for each $t$ fixed implies that $P[X_t=Y_t, \forall t]=1$. Further, your answer shows that for a fixed stopping time $\tau$, $P[X_{\tau}=Y_{\tau}]=1$. But what can we say about $P[X_{\tau}=Y_{\tau}, \forall [0,T]\text{-valued stopping times } \tau]$?
– user711386
Feb 10, 2022 at 10:38
• It is better to submit a new post, because the conditions are changed many times Feb 11, 2022 at 3:30
• @hannah It's still $1$ because if $\omega\in\Omega$ is such that $X_t(\omega)=Y_t(\omega)$ for all $t\in[0,T]$, then also $\forall \tau:\Omega\to[0,T], X_{\tau(\omega)}(\omega)=Y_{\tau(\omega)}(\omega)$. Feb 11, 2022 at 10:03