1
$\begingroup$

i have a problem with the following exercise: Let $X$ be an measurable $\mathbb{R}^d$ valued stochastic process on $(\Omega,\mathcal{F},\mathbb{P})$ and $T$ a finite $T : \Omega\to \big[ 0, +\infty \big)$ random time. . Show that $X_T$ is $\mathcal{F}$-measurable.

I know how to show $\{\omega\in \Omega|{X_{T(\omega)}(\omega)}\in B\} \in \mathcal{F}$, but only if T would take values in $\mathbb{N}$. I have no clue how to show it for a real valued random time.

thank you

edit:Yes $X$ is measurable on the space $(\Omega \times [ 0, +\infty),\mathcal{F}\otimes \mathcal{B}([0,\infty))$

$\endgroup$
3
  • 1
    $\begingroup$ Does measurable mean that the mapping $(\omega,t)\mapsto X_t(\omega)$ is measurable with respect to the sigma-algebra $\mathcal{F}\otimes \mathcal{B}([0,\infty))$ on $\Omega\times [0,\infty)$? $\endgroup$ Apr 22, 2014 at 17:49
  • $\begingroup$ I added the information to the Question. yes measurable was ment like you wrote $\endgroup$
    – sasas2
    Apr 22, 2014 at 18:40
  • 3
    $\begingroup$ My answer here may help: math.stackexchange.com/questions/190220/… $\endgroup$
    – user940
    Apr 22, 2014 at 18:51

1 Answer 1

1
$\begingroup$

Thank you Bryon Schmuland, the answer of "The strong Markov property with an uncountable index set" does answer my question too. I was first put off why $T$ should be finite, but since then $\{T<\infty \}=\Omega, X_T$ is well defined for every $\omega$ without having to look at the Limits.

$\endgroup$
1
  • $\begingroup$ Can you please explain a bit more about why T should be finite? $\endgroup$
    – Percy Wong
    Sep 15, 2023 at 2:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .