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We work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F})_{t\in[0,T]},P)$. Let $X,Y$ be two làdlàg 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?

1 Answers1

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You are correct that 2 implies 1, since every constant $t$ is also a stopping time.

1 need not imply 2. In 1 there are uncountably many null sets, once for each $t$. That is: For every $t \in [0,T]$ there is an event $N_t$ such that $\mathbb P(N_t) = 0$ and $X_t(\omega) = Y_t(\omega)$ for all $\omega \in \Omega \setminus N_t$. It could happen that $\bigcup_{t \in [0,T]} N_t$ no longer has probability zero. Assuming only 1, we can only say that $X_\tau(\omega) = Y_\tau(\omega)$ for all $\omega \in \Omega \setminus \bigcup_{t \in [0,T]} N_t$. Which is not enough to prove 2.

You say "làdlàg". I guess that means they have left and right limits a.e.

Example. Take $\Omega = [0,1]$, $T=1$, and for all $t \in [0,1]$ let $\mathcal F_t = $ Lebesgue measurable sets (a trivial sort of filtration) and $\mathbb P = $ Lebesgue measure. Define

\begin{align} X_t(\omega) &= 0,\quad\forall t \in [0,1],\quad\forall \omega \in \Omega \\ Y_t(\omega) &= \begin{cases} 0,\quad t \ne \omega \\ 1,\quad t = \omega \end{cases} \end{align} These are both làdlàg, and adapted to $\mathcal F_t$ because of the trivial nature of this filtration.
For any $t$ note that $\{\omega : X_t(\omega) \ne Y_t(\omega)\} = \{t\}$ has probability zero. Thus 1 holds.

Let $\tau(\omega) = \omega$. This is a stopping time, again because of the trivial nature of this filtration. But $\{\omega : X_\tau(\omega) \ne Y_\tau(\omega)\} = \Omega$ is not a null event. Thus 2 fails.

GEdgar
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  • Is the reason for posing this as làdlàg that you know there is no such example with càdlàg ? – GEdgar Jan 23 '22 at 11:23
  • Yes, I indeed mean làdlàg in the sense you wrote above. The only reason why I posed the question with làdlàg processes is that I am working with such processes. Nevertheless, do you know if there is such an example with càdlàg processes? –  Jan 23 '22 at 15:01
  • I think with càdlàg, it would be enough to use null events $N_t$ merely for rational $t$? Two càdlàg functions which agree on the rationals, agree everywhere. – GEdgar Jan 23 '22 at 15:09
  • I am not sure if I understand you correctly. Are you trying to give an intuition why in the càdlàg case it cannot happen that $P[N_t]=0$ for each $t$, but $P[\cup_t N_t]>0$? –  Jan 23 '22 at 15:49