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I am trying to understand how the following properties for a stochastic process

  1. $B_t$ is a Gaussian process
  2. $B_t$ has independent increments
  3. $t\to B_t(\omega)$ is continuous for almost all $\omega$

Uniquely determine that a process is a Brownian process centered at some $x\in\mathbb{R}$. This post talks about knowing all the finite distributions will tell us that the measure for the stochastic process is uniquely generated by the set of all finite distributions: Uniqueness of Brownian motion. My guess is that I will need to show that these three properties uniquely determine the finite distributions, but I have no idea how to generate finite the finite distributions with these properties.

Edit: Does Kolmogorov's Extension Theorem imply that the measure generated is unique?

Andrew Shedlock
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  • Isn't stationary increments also required? – Sarvesh Ravichandran Iyer Jul 24 '20 at 05:17
  • It may, these were just the only properties that we proved so I was wondering if they uniquely determine brownian motion – Andrew Shedlock Jul 24 '20 at 05:21
  • I think uniqueness requires stationary increments. If the increments are allowed to be merely of Gaussian nature but not the same, I think uniqueness is violated. However, you say "we proved" these properties : so what was your definition of the Brownian motion, with which you were able to prove these? – Sarvesh Ravichandran Iyer Jul 24 '20 at 05:22
  • We used Kolmogorov's Extension Theorem on the collection of all finite distributions to give the existence of a process. – Andrew Shedlock Jul 24 '20 at 05:24
  • That is ok, but you must have a (mathematical) description of the finite dimensional distributions, right? Otherwise, if you don't know the f.d.distributions, how would you get the above properties? Besides, the continuity criterion comes afterwards : there is a unique (up to modification) process which is Gaussian with stationary independent increments. It exists by specifying its f.d.ds and using Kolmogorov extension. The existence of a continuous modification is shown by the Kolmogorov-Chentsov theorem, et voila the Brownian motion. – Sarvesh Ravichandran Iyer Jul 24 '20 at 05:26
  • Finally, according to the Kolmogorov extension theorem, the measure generated is unique, because the proof proceeds via extending the f.d.ds to so called cylindrical sets, which happen to generate the sigma-algebra for the space on which the bigger measure is sitting. So, every set in that sigma-algebra is approximated by the f.d.ds, ergo two processes having the same fdds will have the same extension. – Sarvesh Ravichandran Iyer Jul 24 '20 at 05:29
  • Yes we do have the finite dimensional distributions when we construct a process that becomes Brownian motion. But I am wondering if there are properties that only Brownian motion can in the way that these characteristics https://en.wikipedia.org/wiki/Wiener_process#Characterisations_of_the_Wiener_process imply a Wiener process. – Andrew Shedlock Jul 24 '20 at 06:03
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    Yes, and those properties are the ones you've listed as 1,2,3 above, along with stationary increments. (In textbooks, stationary and independent increments are combined and written as one point) – Sarvesh Ravichandran Iyer Jul 24 '20 at 06:03

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