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Let $(X_t)_{t \in [0,\infty)}$ be a stochastic process, and let $\mathcal{F}_t$ be the natural filtration generated by $X_t$; i.e. $$\mathcal{F}_t=\sigma(X_s:0\leq s \leq t).$$ Now we fix a $t \in [0,\infty)$. Let $A \in \mathcal{F}_t$ and let $\omega \in A$. Suppose that $\omega'\in \Omega$ satisfies $$X_s(\omega)=X_s(\omega')$$ for any $s \in [0,t]$. How can we show that $\omega'\in A$?

inbrevi
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    have you tried the principle of good sets? – user159517 May 23 '21 at 17:29
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    You must want $X_s(\omega) = X_s(\omega')$ for all $s\in[0,t]$, to acknowledge that possibility of a jump at time $t$. – John Dawkins May 23 '21 at 18:13
  • @JohnDawkins Definitely, this is a typo. The statement is quite intuitive. But do you know how to prove it rigorously? – inbrevi May 24 '21 at 09:15
  • @user159517 I tried that. And it works! Thanks! – inbrevi May 24 '21 at 10:05
  • @inbrevi You're welcome. If you managed to answer the question yourself, I would recommend you answer your question also here (otherwise this question might spend eternity in the "unanswered" queue). – user159517 May 24 '21 at 10:24
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    Unless mistaken this is called Galmarino's test, you can have a look here :https://math.stackexchange.com/questions/1309853/proving-galmarinos-test – TheBridge May 24 '21 at 17:22

2 Answers2

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Let $(\Omega, \mathcal{F}, P)$ be the underlying probability space. We say that $A \in \mathcal {F}$ satisfies the property $\mathcal P$ if and only if for any $\omega \in A$ and $\omega' \in \Omega$ with $X_s(\omega')=X_s(\omega) \quad \forall s \in [0,t]$, we have $\omega'\in A$. Now we use the principle of good sets as suggested by @user159517.

Define $\mathcal A=\{A \in \mathcal {F}: A \text{ satisfies property } \mathcal P\}$. We need to show two statements:

  1. $\mathcal A$ is a $\sigma$-algebra;
  2. $\bigcup_{s \in [0,t]}\sigma(X_s) \subset \mathcal A$.

It follows that $\mathcal F \subset\sigma(\mathcal A)=\mathcal A\subset \mathcal F$ and hence $\mathcal A=\mathcal F$.

Argument for 1:

Trivially $\emptyset \in \mathcal A$. Let $A \in \mathcal A$ and suppose for the sake of contradiction that $A^c \notin \mathcal A$. Then there exists $\omega \in A^c$ and $\omega' \in (A^c)^c=A$ such that $X_s(\omega)=X_s(\omega')$ for every $s\in [0,t]$. This is a contradiction to $A \in \mathcal A$. The $\sigma-$additivity is trivially true.

Argument for 2:

Let $A \in \bigcup_{s\in[0,t]}\sigma(X_s)$. Then $A \in \sigma(X_s)$ for some $s\in [0,t]$. Hence, $A=X_s^{-1}(B)$ for some Borel set $B$. Suppose $\omega \in A$, then $$\omega'\in X_s^{-1}(X_s(\omega'))=X_s^{-1}(X_s(\omega))\subset X_s^{-1}(B)=A.$$

inbrevi
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There is a result (due to Doob, I think), following from the monotone class theorem, that if $A\in\mathcal F_t$ then there is a sequence $(t_n)\subset[0,t]$ and an element $B\in\mathcal B^{\Bbb N}$ such that $$ A=\{\omega: (X_{t_1}(\omega), X_{t_2}(\omega),\ldots)\in B\}. $$ From this it is clear that if $\omega\in A$ and $X_s(\omega') =X_s(\omega)$ for all $s\in[0,t]$ (or just for each $s=t_n$, $n\in\Bbb N$) then $\omega'\in A$ as well.

John Dawkins
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  • Is it me or does this prove that $\sigma(X_{s} , \mid , s \in [0,t])$ never contains ${X , , \text{is measurable in} , , [0,t]}$ (or continuous, for that matter)? –  May 24 '21 at 16:48
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    @PeterMorfe: That depends on what $\Omega$ is. If $\Omega$ is the space of cadlag paths from $[0,1]$ to $\Bbb R$ and $X_s(\omega) =\omega(s)$, then whether $s\mapsto X_s(\omega)$ is continuous is determined by the restriction $(X_s(\omega))_{s\in\Bbb Q\cap[0,1]}$ of $X$ to the rational times. But if $\Omega=\Bbb R^{[0,1]}$ then you are correct — the set of continuous paths isn't measurable. – John Dawkins May 24 '21 at 17:48
  • Thanks! Do you know what this result is called and where can I find the proof? – inbrevi May 24 '21 at 20:27