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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 $\mathscr F$

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  • $\begingroup$ 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, 2017 at 17:17
  • $\begingroup$ @saz Sorry, you are right, it should be $\mathscr B([0,\infty])\otimes \mathscr F$ $\endgroup$ May 11, 2017 at 17:20

1 Answer 1

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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 $A$.

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  • $\begingroup$ Can I find an explicit expression for the pre-image of $A$ under $X_T$? $\endgroup$ May 12, 2017 at 12:25
  • $\begingroup$ @quallenjäger What exactly do you mean by "explicit"? $\endgroup$
    – saz
    May 12, 2017 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, 2017 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, 2017 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, 2017 at 13:38

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