1
$\begingroup$

Let $B=(B_t)_{t\in[0,\infty)}$ a Brownian motion (BM) and $(\Omega,\mathcal{F},P)$ be the probability on which $B$ is defined. Some define BM as a.s. continuous, e.g., Brownian motion is almost surely continuous or Continuous a.s. process, that is, for any $\omega\in\Omega_{0}\subset\Omega$ with $P(\Omega_0)=1$, we have $B_t(\omega)$ is a continuous function. Others (e.g., Billingsley's Probability and Measure) define as "$B_t(\omega)$ is continuous for each $\omega$".

Question 1 is, is this my understanding correct?: ``Brownian motion needs to be defined as 'continuous for every $\omega$' if we want the $B:[0,\infty)\times\Omega\to\mathbb{R}$ to be $\mathcal{B}([0,\infty))\otimes\mathcal{F}$-measurable.''

I think this is correct because of Example 1.11 in Medvegyev, Stochastic Integration theory p. 8, "An almost surely continuous process is not necessarily product measurable."

In Billingsley's Probability and Measure, 3rd ed, Theorem 37.2 it is proved that BM is $\mathcal{B}([0,\infty))\otimes\mathcal{F}$-measurable, and indeed his definition is consistent with Medvegyev's example (for every $\omega$ the path is continuous).

Question 2 If my understanding ``...'' above is correct why do people define BM as a.s. continuous? Why is it ok to have potentially non jointly measurable functions? Is it because people put filtration generated by BM anyway and the original $\mathcal{F}$ does not matter eventually?

$\endgroup$
  • $\begingroup$ Can't you just redefine $B$ as a map from $[0,\infty)\times \Omega_0 \to \mathbb R$? Thus there is effectively no difference between saying it is a.s. continuous and always continuous. $\endgroup$ – Stephen Montgomery-Smith Jul 21 '15 at 5:21
  • 1
    $\begingroup$ Or to put it another way, $[0,\infty)\times(\Omega\setminus \Omega_0)$ is a set of measure zero. So when you take the completion of the measure, all of its subsets become automatically measureable. $\endgroup$ – Stephen Montgomery-Smith Jul 21 '15 at 5:23
  • 1
    $\begingroup$ @StephenMontgomery-Smith Thank you especially for the second comment. So, is saying: "we do need to define that it is always continuous if we want the jointly measurability, but two definitions are equivalent, and thus a.s. continuous will save us from one extra step of the standard discussion of measure theory. Then why not define as a.s. continuous?" correct? $\endgroup$ – shall.i.am Jul 21 '15 at 5:58
  • $\begingroup$ Right, that is what I am saying. $\endgroup$ – Stephen Montgomery-Smith Jul 21 '15 at 12:37
  • $\begingroup$ @StephenMontgomery-Smith Cool, thank you! $\endgroup$ – shall.i.am Jul 22 '15 at 1:53

Your Answer

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

Browse other questions tagged or ask your own question.