1
$\begingroup$

Almost everywhere (for example on Wiki) the Wiener process is defined as a process $W(t)$ such that:

  1. $W(0)=0 $;
  2. It has independent increments;
  3. The increments in an interval $\Delta t$ are gaussian with expected value 0 and variance $D\Delta t$;
  4. The process is sample continuous;

Is this definition minimal? Or could the 4) be dropped?

My idea is trying to prove the fact that the following holds $\forall \epsilon>0$:

$$lim_{\Delta t \rightarrow0}P(|\Delta W|> \epsilon)=0$$

where $\Delta W$ is the increment of the process in $\Delta t$. This should be a sufficient condition for sample continuity (sometimes it is called Lindeberg condition) that is valid for Markov processes according to Feller "Handbook of stochastic methods". Of course the definition of the Wiener process entails that the process is Markov because it has independent increments.

$\endgroup$

1 Answer 1

0
$\begingroup$

The first $3$ conditions are not quite enough to show that $W$ has continuous sample paths, but they will ensure that there exists a modification that does have sample paths by Kolmogorov's continuity theorem. For example, let $W$ be a standard Brownian motion and define $X(t) = W(t)1_{W(t) \ne 1}$. For any fixed $t$ we have that $X(t) = W(t)$ a.s. so the first $3$ conditions will hold for $X$ ($X(0)=0$, $X(t)-X(s)$ is independent of $X(u)-X(v)$ for any $v < u < s < t$, and $X(t)-X(s) \stackrel{d}{=}W(t)-W(s) \sim N(0,t-s)$), but $X(t)$ does not have continuous sample paths.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.