Question about the continuity of Brownian motion

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I've encountered slight differences when defining Brownian motion. The differences are mainly about the continuity part.

In Wiki, it states that Brownian motion$B_t$ is almost surely continuous. Also in wiki, it states that Brownian motion is sample continuous.

Durrett's book states that with probability one, $t \longrightarrow B_t$ is continuous.

And I see different modules of continuity here. https://en.wikipedia.org/wiki/Continuous_stochastic_process#Continuity_with_probability_one

The definitions of 'Continuity with probability one', 'Continuity in probability','Sample continuity' are incredibly confusing.

My question is which one is the scenario we suppose a Brownian motion should have?

This site proves 'Continuity in probability' without assumptions about the 'continuous ' part. How can we show that the sample paths of standard Brownian motion are continuous?

Thanks for the comments, I know the assumption is 'Continuity with probability one'. And from the site above I know how to prove 'Continuity in probability', but how about the 'sample continuity' part?

Also from Wiki and I quote

Continuity with probability one at time t means that $P(A_t) = 0$, where the event $A_t$ is given by

$A_{t} = \left\{ \omega \in \Omega \left| \lim_{s \to t} \big| X_{s} (\omega) - X_{t} (\omega) \big| \neq 0 \right. \right\}$, and it is perfectly feasible to check whether or not this holds for each t ∈ T.

Sample continuity, on the other hand, requires that $P(A) = 0$, where

$A = \bigcup_{t \in T} A_{t}$. $A$ is an uncountable union of events, so it may not actually be an event itself, so $P(A)$ may be undefined! Even worse, even if A is an event, $P(A)$ can be strictly positive even if $P(A_t) = 0$ for every $t ∈ T$. This is the case, for example, with the telegraph process.