I have a proof of a property regarding the stopped $\sigma$-algebra, where one part I do not understand, I'll highlight what I do not get, can you please help me?

We have a probability space $(\Omega,\mathcal{F},P)$, with a filtration $\{\mathcal{F}_t\}$. We have a random variable T, which is a stopping time, i.e. $\{\omega: T(\omega)\le t\}\in \mathcal{F}_t\ \ \forall t$.

The stopped sigma-algebra is defined as $\mathcal{F}_t = \{F\in \mathcal{F}: F\cap\{\omega: T(\omega)\le t\}\in \mathcal{F}_t \ \forall t\}$.

The theorem proved is this:

Let $T$ be a finite stopping time. Then $\mathcal{F}_t$ is the smallest $\sigma$-algebra containing all càdlàg processes sampled at T. That is:

$\mathcal{F}_T = \sigma(X_T: \text{X adapted and càdlàg}).$

The proof given is this:

Let $\mathcal{G} = \sigma(X_T: \text{X adapted and càdlàg})$. Let $A \in \mathcal{F}_T$. Then $X_t=1_A(\omega)\cdot 1_{\{T(\omega)\le t\}}$ is a càdlàg process, and $X_T(\omega)=1_A(\omega)$. Hence $A \in \mathcal{G}$, and $\mathcal{F}_t \subset \mathcal{G}$.

Next let $X$ be an adapted càdlàg process. We need to show that $X_T$ is $\mathcal{F}_T$ measurable. Consider $X(s,\omega)$ as a function from $[0,\infty)\times\Omega$ into $\mathbb{R}$. Construct $\phi: \{T \le t\}\rightarrow[0,\infty)\times\Omega$ by $\phi(\omega)=(T(\omega),\omega))$. Then since X is adapted and càdlàg, we have that $X_T = X \circ \phi$ is a measurable mapping from $(\{T\le t\},\mathcal{F}_t\cap\{T \le t\})$ into $(B,\mathcal{B}(\mathbb{R})).$ Therefore

$\{\omega: X(T(\omega),\omega)\in B\}\cap \{T\le t\}$

is in $\mathcal{F}_t$, and this implies that $X_T$ is $\mathcal{F}_T$-measurable.

I have two questions regarding this proof which I do not understand.

  1. It is assumed ha T is a finite stopping time, but where in the proof is that actually used? I can't see where he uses it?

  2. This is my main question, and it is in regard to the second part which I have highlighted. Why does it follow that since X is an adapted càdlàg-process, then $X\circ \phi$ is measurable wih regards to the given spaces$ (\{T\le t\},\mathcal{F}_t\cap\{T \le t\})$ and $(B,\mathcal{B}(\mathbb{R}))$. Is this easy to see, or do we have to do some work to show this?

From what I see, in order to finish this part of the proof and show that $X_T$ is $\mathcal{F}_T$-measurable, we have to show two things. First that $X_T^{-1}(B)\in \mathcal{F}$ and for all t, $X_T^{-1}(B)\cap\{T\le t\}\in \mathcal{F}_t$for all t, and all Borel-sets. Is this showed in the proof?


1 Answer 1

  1. The assumption is needed to make sense of $X_T$. If $T(\omega)=\infty$, then $X_T(\omega) = X_{\infty}(\omega)$ is not well-defined for a stochastic process $(X_t)_{t \geq 0}$.
  2. It is not exactly obvious. Since $(X_t)_{t \geq 0}$ is adapted and has càdlàg sample paths, it holds that $$X: [0,t] \times \Omega \to \mathbb{R}$$ is $\mathcal{B}([0,t]) \otimes \mathcal{F}_t / \mathcal{B}(\mathbb{R})$-measurable. This can be shown e.g. by approximating the process with simple functions which converge pointwise to the process $X$. On the other hand, it is not difficult to see that the mapping $\phi: \{T \leq t\} \to [0,t] \times \Omega$, defined in the proof, is measurable (with respect to $\mathcal{F}_t \cap \{T \leq t\}/\mathcal{B}([0,t]) \otimes \mathcal{F}_t$). Therefore, it follows that the composition $X_T = X \circ \phi$ is $\mathcal{F}_t \cap \{T \leq t\}/\mathcal{B}(\mathbb{R})$-measurable.
  3. The proof shows that $X_T^{-1}(B) \in \mathcal{F}_t$ for all $t \geq 0$ and all Borel sets $B$. Note that this implies, in particular, $$X_T^{-1}(B) = \bigcup_{n \geq 0} (X_T^{-1}(B) \cap \{T \leq n\}) \in \mathcal{F}.$$ Hence, by the very definition of $\mathcal{F}_{\tau}$, we have $X_T^{-1}(B) \in \mathcal{F}_T$ for all Borel sets $B$.
  • $\begingroup$ @user119615 You are welcome. $\endgroup$
    – saz
    Apr 3, 2016 at 17:01

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