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Let $X$ be a stochastic process, and let $\tau$ be a discrete $\{\mathcal{F}_t^X\}-$stopping time. Show that

$$\mathcal{F}_\tau^X = \sigma(X(t\wedge \tau):t\ge 0).$$

I am struggling to find a way to show this identity. I know that $\mathcal{F}_{\tau}^X=\{A\in \mathcal{F}_\infty: A \cap \{\tau\le t\} \in \mathcal{F}_t\}$. But I don't know how to use this fact to prove the above identity. I would greatly appreciate any help.

saz
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nomadicmathematician
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  • Well, a first step would be to show that $X(\tau \wedge t)$ is $\mathcal{F}_{\tau}^X$-measurable... – saz Jun 12 '17 at 17:32
  • @saz I know that result holds when $X$ is continuous or progressive, but how can I show that given no assumptions on these? – nomadicmathematician Jun 12 '17 at 18:20
  • You have to use that the stopping time is discrete... – saz Jun 12 '17 at 18:31
  • @saz Ok, so using discreteness, we need to show that for any $s\ge 0$, and $a\in \mathbb{R}$, $n\in \mathbb{N}$, ${X(\tau \wedge s)\le a}\cap {\tau = n} \in \mathcal{F}_n$. But then separating two cases, $s <n$ and $s\ge n$, we get ${X(s)\le a} \cap {\tau=n}$ and ${X(n)\le a} \cap {\tau=n}$. Both of which are in $\mathcal{F}_n$ using filtration property and measurability of each $X_t$. So I get one side of inclusion, can you help me with the other direction? – nomadicmathematician Jun 12 '17 at 18:41
  • Please use $\mathcal{F}^X_\tau=\bigcup_{k=1}^\infty (\mathcal{F}^X_k1_{(\tau=k)})$ and $\mathcal{F}^X_k1_{(\tau=k)}=\sigma(X_{j\wedge \tau},j\ge 1)1_{(\tau=k)}$. – JGWang Jun 16 '17 at 09:35
  • @JGWang But then you still have to show that $\tau$ is $\sigma(X(t \wedge \tau); t \geq 0)$-measurable... or am I missing something? – saz Jul 11 '17 at 17:44
  • @saz Thank you for your reply. You are right, $\tau$ is $\sigma(X(t\wedge\tau); t\ge 0)$-measurable, since $1_{(\tau=k)}=f(X_1,\cdots,X_k)1_{(\tau=k)}=f(X_{1\wedge\tau},\cdots,X_{k\wedge\tau})1_{(\tau=k)}$. – JGWang Jul 12 '17 at 03:42
  • @JGWang What is $f$......? – saz Jul 12 '17 at 03:48
  • @saz Since $\tau$ is a ${\mathcal{F}^X_t}$-stopping time, then $(\tau=k)\in\mathcal{F}k^X=\sigma(X_1,\cdots,X_k)\vee\mathcal{N}$ and there exists a Borel function $f$ such that $1{(\tau=k)}=f(X_1,\cdots,X_k)$ a.s.. – JGWang Jul 12 '17 at 08:57
  • @JGWang Thanks; I still don't see how you deduce from $$1_{{\tau=k}} = f(X_{\tau \wedge 1},\ldots,X_{\tau \wedge k}) 1_{{\tau=k}}$$ that ${\tau=k} \in \sigma(X_{\tau \wedge n}; n \geq 1)$. Clearly $f(\dots)$ is $\sigma(X(\tau \wedge n); n \geq 1)$, but how do you get rid of the indicator function on the right-hand side? – saz Jul 12 '17 at 13:44
  • @saz, Thank you for your help to point the problems in my above statements, also thank you to give a correct proof of the question. – JGWang Jul 13 '17 at 01:32
  • Does this work for non-discrete stopping times? – Niebla Jul 04 '24 at 10:57

1 Answers1

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Let $(X_n)_{n \in \mathbb{N}}$ be a stochastic process and $\tau: \Omega \to \mathbb{N}$ an $\mathcal{F}^X$-stopping time. You have already shown that $X(\tau \wedge n)$ is $\mathcal{F}_{\tau}^X$-measurable for all $n \geq 1$, and therefore it just remains to show that

$$\mathcal{F}_{\tau}^X \subseteq \sigma(X(\tau \wedge n); n \geq 1) =: \mathcal{H}. \tag{1} $$

Let us first recall the factorization lemma:

Let $Y:\Omega \to \mathbb{R}^d$ be a random variable. Then a mapping $T: \Omega \to \mathbb{R}$ is $\sigma(Y)$-measurable if, and only if, there exists a Borel-measurable mapping $f:\mathbb{R}^d \to \mathbb{R}$ such that $T = f \circ Y$.

Proof of $(1)$: Since $\tau$ is a discrete stopping time, we have

$$\begin{align*} \mathcal{F}_{\tau}^X &\stackrel{\text{def}}{=} \{A; \forall k \geq 1: \, \, A \cap \{\tau \leq k\} \in \mathcal{F}_k^X\} \\ &= \{A; \forall k \geq 1: \, \, A \cap \{\tau = k\} \in \mathcal{F}_k^X\}. \tag{2} \end{align*}$$

Let $A \in \mathcal{F}_{\tau}^X$. We show by induction that $A \cap \{\tau = k\} \in \mathcal{H}$ for all $k \geq 1$.

  • $k=1$: $$A \cap \{\tau = 1\} \in \mathcal{F}_1^X = \sigma(X_1) = \sigma(X_{\tau \wedge 1}) \subseteq \mathcal{H}.$$

  • $k-1 \to k$: By $(2)$ we have $A \cap \{\tau =k\} \in \mathcal{F}_k = \sigma(X_1,\ldots,X_k)$. Applying the above theorem shows that there exists a mapping $f: \mathbb{R}^k \to \mathbb{R}$ such that $$1_{A \cap \{\tau = k\}} = f(X_1,\ldots,X_k). $$ Using that $1_{\{\tau=k\}} = 1_{\{\tau=k\}} 1_{\{\tau \geq k\}}$ we find $$\begin{align*} 1_{A \cap \{\tau = k\}} = f(X_1,\ldots,X_k) 1_{\{\tau \geq k\}} &= f(X_{\tau \wedge 1},\ldots,X_{\tau \wedge k}) 1_{\{\tau \geq k\}}. \end{align*}$$ By the induction hypothesis, we have $\{\tau \geq k\} = \{\tau \leq k-1\}^c \in \mathcal{H}$, and therefore the right-hand side is $\mathcal{H}$-measurable. This, however, is equivalent to saying that $A \cap \{\tau=k\} \in \mathcal{H}$.

Finally, we conclude that

$$A = \bigcup_{k \geq 1}(A \cap \{\tau=k\}) \in \mathcal{H}.$$

Remarks:

  • It is also possible to show that $$\sigma(X_{n \wedge \tau}; n \geq 1) = \sigma(\mathcal{F}_{\tau \wedge n}; n \geq 1),$$ see this question.
  • See this question for the analogous result for time-continuous processes.
saz
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  • Could you please explain why the following statement is true? Thank you. > By the induction hypothesis, we have ${\tau \geq k} = {\tau \leq k-1}^c \in \mathcal{H}$. – Petra Axolotl Mar 27 '23 at 17:36