Sigma algebra generated by the stopped process.

Question

Let $(X_n)_{n \geq 0}$ be a sequence of random variables. Let $\mathcal{F}_n = \sigma (X_0, \dots, X_n)$ be a filtration and $T$ is a $(\mathcal{F}_n)_{n\geq 0}$-stopping time. I want to understand whether $$\sigma (X_{n \wedge T}, n \geq 0) =^? \sigma \left( \mathcal{F}_{n \wedge T}, n \geq 0 \right).$$ My intuition tells me that $\sigma \left( \mathcal{F}_{n \wedge T}, n \geq 0 \right)$ must be larger, but I can't find an example showing it.

Answer 1

Yes, the two $\sigma$-algebras coincide.

For any discrete stopping time $\tau: \Omega \to \mathbb{N}_0$ the random variable $X_{\tau}$ is $\mathcal{F}_{\tau}$-measurable, and this implies that

$$\sigma(X_{n \wedge T}; n \geq 0) \subseteq \sigma(\mathcal{F}_{n \wedge T}; n \geq 0).$$

Moreover, we have $\mathcal{F}_{n \wedge T} \subseteq \mathcal{F}_T$ (as $ T \wedge n \leq T$), and therefore

$$\sigma(X_{n \wedge T}; n \geq 0) \subseteq \sigma(\mathcal{F}_{n \wedge T}; n \geq 0) \subseteq \mathcal{F}_T.$$

Consequently, we are done if we can show that

$$\mathcal{F}_T \subseteq \sigma(X_{n \wedge T}; n \geq 0).\tag{1}$$

Let $A \in \mathcal{F}_T$. We show by induction that $$\forall k \geq 0: \quad A \cap \{T = k\} \in \sigma(X_{n \wedge T}; n \geq 0). $$

• $k=0$: $$A \cap \{T=0\} \in \mathcal{F}_0 = \sigma(X_0) = \sigma(X_{0 \wedge T}) \subseteq \sigma(X_{n \wedge T}; n \geq 0)$$
• $k-1 \to k$: We have $A \cap \{T =k\} \in \mathcal{F}_k$, and therefore there exists by the factorization lemma a Borel measurable function $f$ such that $$1_{A \cap \{T=k\}} = f(X_1,\ldots,X_k).$$ As $1_{\{T =k\}} = 1_{\{T=k\}} 1_{\{T \geq k\}}$ we find $$1_{A \cap \{T=k\}} = f(X_1,\ldots,X_k) 1_{\{T \geq k\}} = f(X_{1 \wedge T},\ldots,X_{k \wedge T}) 1_{\{T \geq k\}}. \tag{2}$$ Because of the induction hypothesis, we have $$\{T \geq k\} = \{T \leq k-1\}^c \in \sigma(X_{n \wedge T}; n \geq 0),$$ and therefore it follows from $(2)$ that $1_{A \cap \{T=k\}}$ is $\sigma(X_{n \wedge T}; n \geq 0)$-measurable; this is equivalent to $A \cap \{T =k\} \in \sigma(X_{n \wedge T}; n \geq 0)$.