2
$\begingroup$

We have the following settings:

Given the probability space $(\Omega,\mathscr F,P)$

$T$ is a random time, if it is a $\mathscr F$-measurable random variable with values in $[0,\infty]$

A stochastic process is called measurable if for every $A\in\mathscr B(\Bbb R^d)$, the set $\{(t,\omega);X_t(\omega)\in A\}$ belongs to the product $\sigma$-algebra $\mathscr B([0,\infty))\otimes\mathscr F$. I will be appreciate for any help.

Now I want to show that for a measurable process $X_t$(with well-defined $X_{\infty}$) and a random time $T$, the collection of all sets of the form $\{X_T\in A\}$ and $\{X_T\in A\}\cup\{T=\infty\}$,forms a sub-sigma algebra of $\mathscr F$.

I understand that we have the following mapping$$\omega\rightarrow(\omega,T(\omega))\rightarrow X(\omega,T(\omega))$$ $$\mathscr F\rightarrow\mathscr B([0,\infty])\otimes\mathscr F\rightarrow\mathscr B([0,\infty])$$.

As $X_t$ is measurable, we know that $X_T^{-1}(A)$ is somehow contained in $\mathscr B([0,\infty])\otimes\mathscr F$. As I am not sure if the preimage of $X$ can be written in the form of $C\otimes D$ and then consider $T^{-1}(D)\cap C$ in the product sigma algebra, I don't know how to proceed further to the original space $\mathscr F$

$\endgroup$
2
  • $\begingroup$ What kind of measurability assumptions do you have for $t=\infty$? Do you assume that $\{(t,\omega) \in [0,\infty] \times \Omega; X_t(\omega) \in A\}$ is in the product sigma-algebra on $[0,\infty] \times \mathcal{F}$? $\endgroup$
    – saz
    May 11 '17 at 17:17
  • $\begingroup$ @saz Sorry, you are right, it should be $\mathscr B([0,\infty])\otimes \mathscr F$ $\endgroup$ May 11 '17 at 17:20
2
$\begingroup$

By assumption, the mapping

$$([0,\infty) \times \Omega; \mathcal{B}[0,\infty] \otimes \mathcal{F}) \ni (t,\omega) \mapsto X(t,\omega) \in (\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d)) \tag{1}$$

is measurable. On the other hand, the mapping

$$(\Omega,\mathcal{F}) \ni \omega \mapsto Y(\omega):= (T(\omega),\omega) \in ([0,\infty) \times \Omega; \mathcal{B}[0,\infty] \otimes \mathcal{F}) \tag{2}$$

is also measurable. Indeed: Take $B \in \mathcal{B}[0,\infty]$ and $F \in \mathcal{F}$, then

$$\{Y^{-1}(B \times F)\} = \{\omega; \omega \in F, T(\omega) \in B) = F \cap \{T \in B\} \in \mathcal{F}.$$

Since sets of the form $B \times F$ are a generator of the product-$\sigma$-algebra, this proves that $Y$ is measurable.

Combining $(1)$ and $(2)$, we find that the mapping

$$(\Omega,\mathcal{F}) \ni \omega \mapsto X \circ Y(\omega) = X(T(\omega),\omega) \in (\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$$

is measurable as the composition of the measurable mappings $(1)$ and $(2)$. Hence, $\{X_T \in A\} \in \mathcal{F}$ for any Borel set $A$.

$\endgroup$
6
  • $\begingroup$ Can I find an explicit expression for the pre-image of $A$ under $X_T$? $\endgroup$ May 12 '17 at 12:25
  • $\begingroup$ @quallenjäger What exactly do you mean by "explicit"? $\endgroup$
    – saz
    May 12 '17 at 12:36
  • $\begingroup$ For exampe we know that $\mathscr F_{\tau}$ where $\tau$ is a stopping time can be written as $\{A\cap{\tau \leq t}|A\in \mathscr F\}$. Can we write the preimage in similiar form? Or is there any connection to it? $\endgroup$ May 12 '17 at 12:46
  • $\begingroup$ Well, in fact $X_{\tau}$ is $\mathcal{F}_{\tau}$-measurable, and therefore the $\sigma$-algebra generated by $X_{\tau}$, i.e. $$\{\{X_{\tau} \in A\}; A \in \mathcal{B}(\mathbb{R}^d)\}$$ is contained in $\mathcal{F}_{\tau}$. $\endgroup$
    – saz
    May 12 '17 at 13:37
  • $\begingroup$ Thanks, I was not quite sure if they are the same. Very clear explanation. I really appreciate your answer. $\endgroup$ May 12 '17 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.