Wiener process related counterexample - MathOverflow

Wiener process related counterexample

2010-09-25T05:22:53Z

The Wiener process is defined by the three properties: 1. $W(0) = 0$, 2. $W(t)$ is almost surely continuous, and 3. $W(t)$ has independent increments with $W(t) - W(s) \sim N(0, t-s)$ (for $0 ≤ s < t$).

What would be an example of a process which satisfies 1) and 3), but not 2) ?

I am going to teach an introductory class on Brownian motion at advanced undergrad level. Just wanted to make sure that all the conditions are mutually independent.

Answer by hawai for Wiener process related counterexample

2010-09-25T05:48:38Z

2) is necessary for the definition of topology of the process

Answer by Byron Schmuland for Wiener process related counterexample

2010-09-25T06:08:15Z

This is not hard to find such an example. Let $P$ be Wiener measure on the space $\Omega = C([0,\infty))$ of continuous functions $t\mapsto \omega(t)$. Then the process $\omega(t)$ satisfies all three conditions of a Brownian motion.

Now let's define a new process $W(t)$ that is "almost" equal to $\omega(t)$, but where we deliberately wreck the sample path continuity.

Take any random time $T:\Omega\to [0,\infty)$ that has a continuous distribution on $(\Omega, P)$, and let $W(t,\omega)=\omega(t)$ when $t\not=T(\omega)$, but $W(t,\omega)=\omega(t)+1$ otherwise. The process $W(t)$ still satisfies 1 and 3 but the sample path continuity fails at exactly at the time point $T(\omega)$ for each $\omega$.

There are many such random times $T$, for example you could use $T(\omega):=\inf [t>0: \omega(t)=1 ]$, i.e. the hitting time of 1.

Answer by Reda for Wiener process related counterexample

2010-09-25T07:46:45Z

To continue on Byron's answer, properties 1) and 3) specify the law of the process, but not the topological features of a given trajectory $\omega$. In a sense, 1) and 3) are enough to define the Wiener measure, since generating n points from a brownian path only needs those. Byron exhibited another 'version' of the process that is not continuous by changing the process on a set of measure 0 (his stopping time will never hit a specific point $t$ taken in advance because the law of T has a density).

A typical verification that needs to be done is if a process defined by its law has continuous versions, which is what entails Kolmogorov continuity theorem (link wikipedia page). Basically, the idea is that if $X_t$ and $X_s$ are close on average when $t$ and $s$ are close, you can change the process on a set of zero measure to get something continuous.

I can also recommend reading the beginning of the classical Revuz and Yor about definitions of 'undistinguishable processes' and 'versions of the same process'.

Cheers

Answer by Shai Covo for Wiener process related counterexample

2010-10-21T08:59:36Z

The author of this question might be more pleased with the following answer. Let $U$ be a uniform(0,1) random variable, independent of a Brownian motion $W$. Then, the process $W'$ defined by $W'(t) = W(t) + {\mathbf 1}(t=U)$, where ${\mathbf 1}$ denotes indicator function, is discontinuous at time $U$. However, for any choice of (fixed) times $t_i$, $i=1,...,n$, we have, almost surely, $W'(t_i) = W(t_i)$ for all $i$, and hence, trivially, $W'$ has the same distributional properties stated for $W$. Furthermore, if we define $W'$ by $W'(t) = W(t) + {\mathbf 1}(t \in UA)$, where $A$ is a dense set in $(0,\infty)$ of measure zero (and where $UA:= \{Ua: a \in A\}$ ), then $W'$ is nowhere continuous (since $UA$ is dense in $(0,\infty)$); nevertheless, as before, almost surely $W'(t_i) = W(t_i)$ for all $i=1,...,n$ (since ${\rm P}(t \in UA) = {\rm P}(t/U \in A) = 0$).

Side notes: 1) Actually, as follows from the theory of Lévy processes, the almost sure continuity in the definition of Brownian motion is equivalent to almost sure cadlaguity (right-continuity with left limits); 2) The answer can be adapted to Lévy processes in general ($W$ is a special case), showing that the almost sure cadlaguity in the definition of Lévy process is not implied by the other conditions.

Finally, the author of this question ``wanted to make sure that all the conditions are mutually independent.'' This is, however, not the case, if we split condition 3) into subconditions. See this thread: link text