For a discrete stopping time $\tau$, $\mathcal{F}_\tau^X = \sigma(X(t\wedge\tau):t\ge 0).$

Question

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.

Answer 1

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.