Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Weiner or Wiener? –  J. H. S. Sep 25 '10 at 5:33
add comment

4 Answers

up vote 12 down vote accepted

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.

share|improve this answer
    
Nice. To make it even less continuous, maybe make $W(t,\omega)=w(t)+1$ whenever $w(t)$ is at a local minimum? –  Bjørn Kjos-Hanssen Sep 25 '10 at 6:28
    
Yes there are lots of ways to do this. As Reda's answer explains, as long as you change the sample paths without changing the law of the process you get such a "bad" Brownian motion. –  Byron Schmuland Sep 25 '10 at 13:45
1  
An alternative construction that you may find more natural: Begin with a product space $C([0,\infty))\times (0,1)$ equipped with the product measure $P\times\lambda$, where $\lambda$ is Lebesgue measure. You can use the independent uniform $(0,\infty)$ random variable to play the role of $T$, which (by construction) automatically has a continuous distribution. –  Byron Schmuland Sep 25 '10 at 13:49
2  
Actually, introducing an independent random variable to wreck sample path continuity is not necessary. $W(t)=1_{\{\omega(t){\rm\ is\ irrational}\}}\omega(t)$ works fine. –  George Lowther Oct 23 '10 at 11:06
    
BTW, I just asked a very similar question at math.stackexchange.com/questions/549570/… –  Stefan Nov 3 '13 at 1:52
add comment

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

share|improve this answer
    
Very nice explanation! –  Byron Schmuland Sep 25 '10 at 13:43
1  
Hello Byron and Reda, thanks for replying. Just one thing which is confusing me a little. Reda says, "Byron exhibited another version by changing the process on a set of measure zero". Intuitively, what I understand of Byron's construction is that every path is being broken "at a different point in time". Thus, all the paths are discontinuous, but the joint distributions of the random variables W(t) remain unchanged. Is that what you mean as well ? –  Cosmonut Sep 25 '10 at 17:58
    
Indeed... In Byron's construction, every path is being broken at a random point T. However, if i choose in advance $t_0, t_1, \dots, t_n$ and look at the joint law of $(W_{t_0}, \dots, W_{t_n})$, there is no chance $T$ is one of $t_0, t_1, \dots, t_n$, so the law is the same. I can take n arbitrarily big also (if T has density, i can look at all rational times at the same time...). I just changed $W_T$, which accounts for a set of measure zero. –  Reda Sep 25 '10 at 19:36
add comment

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

share|improve this answer
    
Just curious, did you read Byron's answer? –  zhoraster Nov 9 '10 at 14:26
    
Yes, I did. I stress that: 1) As far as I can recall, I essentially created my example(s) a few years ago; 2) My second example is of a BM in law which is nowhere continuous; 3) My first example seems the most appropriate one in our context, taking into account that the author of this question might want to use it in "an introductory class on Brownian motion at advanced undergrad level" –  Shai Covo Nov 9 '10 at 15:06
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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