Why is Brownian motion required to be merely almost surely continuous instead of continuous?

For example, this is stated as condition 2 in this article in section 1, Characterizations of the Wiener process, where it says "The function $t \rightarrow W_t$ is almost surely everywhere continuous." What is an example of a Brownian motion where there is a point at which the motion is not continuous?

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    $\begingroup$ A very warm welcome to Quant.SE and thank you for your great question! $\endgroup$ – vonjd Feb 23 '15 at 13:16
  • $\begingroup$ Is there anything else we could do for you? Otherwise it would be great if you could accept one of the answers - Thank you :-) $\endgroup$ – vonjd Mar 9 '15 at 14:00

Exhibiting a counter-example is straight-forward enough. For example, let $B_{t}(\omega)$ be a Brownian motion and $\mathcal{T}(\omega)$ a stopping time on $(\Omega,\mathbb{P})$ with a continuous distribution.

Then with


$B'_{t}(\omega)$ satisfies (1) and (2) below, but is discontinuous precisely when $t=T(\omega)$. Therefore, $B_{t}(\omega)$ is a particular realization of Brownian motion that is not everywhere continuous.

There are lots of other ways to obtain a "bad" Brownian motion. Another example is $$B'_{t}(\omega)=B_{t}(\omega)\mathbb{1}_{\{B_{t}(\omega)\;\text{irrational}\}},$$

but this is less straight-forward to prove.

The reason for stipulating almost sure continuity has to do with the way one constructs Brownian motion, and the issue can be completely dispensed with dependent on one's approach.

The usual presentation in finance texts is the abstract one, namely given a probably space $(\Omega,\mathbb{P})$, one has a Brownian motion $B_{t}(\omega)$ on this space if

  1. For every set of times $0\leq t_{1}<t_{2}<\ldots<t_{n}$ the increments $B_{t_{1}},B_{t_{2}}-B_{t_{1}},\ldots,B_{t_{k}}-B_{t_{k-1}}$ form a mutually independent set of random variables on $(\Omega,\mathbb{P}).$
  2. The increments above are normally distributed with mean $0$ and variance $\Delta t$.
  3. For almost every $\omega\in\Omega$ the path $t\mapsto B_{t}(\omega)$ is continuous.

Most texts also include a section that sketches a concrete realization of Brownian motion as the limit of scaled random walks. If one does this rigorously, one sees that (3) upgrades to for every $\omega\in\Omega$

Indeed, if we start with $(\Omega,\mathbb{P})$ satisfying the above and let $\mathcal{P}$ denote the collection of continuous functions $[0,\infty)\to\mathbb{R}$ with $p(0)=0$, then we get from (3) the inclusion map $$\mathcal{i}:\Omega\to\mathcal{P},$$ defined on a set $\Omega'\subset\Omega$ of full measure, and the push-forward measure of $\mathbb{P}$ onto $\mathcal{P}$ under this inclusion map turns out to be equal to the Wiener measure $\mathbb{W}$ on $\mathcal{P}$, which is unique.

Conversely, one can construct $(\mathcal{P},\mathbb{W})$ directly by starting with the set $\mathcal{P}$ (where every element of this set is continuous a priori) and demonstrating that the measures $\mu_{N}$ on $\mathbb{Z}^{\infty}_{2}$ arising from the appropriately scaled random walks $S_{t}^{N}(\omega)$ ($\omega\in\mathbb{Z}^{\infty}_{2})$ induce a collection of tight measures on $\mathcal{P}$ which converge weakly to $\mathbb{W}$: $$\mu_{N}\Longrightarrow\mathbb{W}\;\text{(weakly)}$$

One then defines $$\tilde{B}_{t}(\omega):=p(t)\in\mathcal{P}$$ and readily shows that under $\mathbb{W}$, $\tilde{B}_{t}$ satisfies (1)-(3) and that therefore $$\tilde{B}_{t}(\omega)=B_{t}(\omega),$$ but that now every Brownian motion is continuous.

The equivalence of the implications above show the existence of Brownian motion is essentially tantamount to the existence of a Wiener measure on $\mathbb{W}$ arising from the sequence of measures arising naturally from the scaled random walks. If one starts from the goal of obtaining this measure, one gets continuity for every Brownian motion $p(t)=B_{t}(\omega)$.

Other constructions of Brownian motion require us stipulate almost sure continuity due to technicalities arising from measure theory on product spaces. The quickest construction of Brownian motion in this direction is by applying Kolmogorov's extension theorem on a suitable class of processes; details can be found in Durrett.

  • $\begingroup$ The name is Wiener, not Weiner. $\endgroup$ – vonjd Feb 23 '15 at 17:08
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    $\begingroup$ Considering your example it feels a little bit like cheating. The OP asked why a Wiener process has as a characteristic almost sure continuity instead of sure continuity and in my understanding wanted an example which didn't completely change the rules of the game (= the construction of the stochastic path). Of course you can define a completely different process which is not continuous but why not saying: f(x) = 0 for x <= 0 and f(x) = 1 for x > 0 which is also not continuous - but also has nothing to do with the classical construction of a Wiener process. Do you see where I am getting at? $\endgroup$ – vonjd Feb 23 '15 at 17:37
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    $\begingroup$ Perhaps the problem with the question is whether the OP wants to know if almost sure continuity follows from the construction of the Wiener process or whether it is a necessary prerequisite?!? $\endgroup$ – vonjd Feb 23 '15 at 17:40
  • $\begingroup$ @vonjd I addressed all of the issues raised in your comment in my subsequent edits - you might've been typing your response as I was making the changes. And I took it as the OP wondering whether almost sure continuity was somehow fundamental to Brownian motion, which is a natural question to ask given the axiomatic definition sometimes used. But as I tried to demonstrate in my answer, it is not an the stipulation is based on one's approach. $\endgroup$ – Sargera Feb 23 '15 at 17:45

If you say that in probability theory an event happens almost surely it happens with probability one.

From Wikipedia:

The difference between an event being almost sure and sure is the same as the subtle difference between something happening with probability 1 and happening always.

If an event is sure, then it will always happen, and no outcome not in this event can possibly occur. If an event is almost sure, then outcomes not in this event are theoretically possible; however, the probability of such an outcome occurring is smaller than any fixed positive probability, and therefore must be 0. Thus, one cannot definitively say that these outcomes will never occur, but can for most purposes assume this to be true.

Intuitively the idea is a little bit like that of a limit in calculus: Oversimplified it says that you will only get there after "infinitely many" steps, if you stop after a finite number of steps you will not reach the limit. While we got used to this idea in calculus, although almost surely is similar, it is still even less intuitive because we are additionally dealing with randomness here.

Another example: When you think of spinning a smooth Wheel of Fortune (so without clicks and with an "infinitely small" hand) every single position where it could stop has probability zero, which means that it will stop at any given point almost never or it will not stop at any given point almost surely! Yet it has to stop somewhere...

To answer your question: For all intents and purposes the path of a Brownian motion (as obtained from the limit of scaled random walks) is indeed continuous everywhere. Think of almost surely as the technical way of saying surely, so there is no example of a Brownian motion where there is a point at which the path is not continuous.

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    $\begingroup$ Your last assertion is false - there are many examples. I think a better way of saying this is that for all intents and purposes, if you think of Brownian motion as obtained from the limit of scaled random walks, then yes, all Brownian motion paths are continuous. $\endgroup$ – Sargera Feb 23 '15 at 17:51
  • $\begingroup$ @TaylorMartin: Agreed, thank you, I edited the answer accordingly. $\endgroup$ – vonjd Feb 23 '15 at 18:00

I think, that the answer by Taylor Martin contains the explanation, but should be much shorter, so I'll put it here separately.

Essentially, Brownian motion is a measure on the space of continuos functions (trajectories), say on an interval on the real line . How does one describe this measure in probabilistic terms?

One sets the probability space $(\Omega, \mathbb{P})$ and a measurable function from it to the target space: $$F: \Omega\to\mathcal{C}[0,1]$$

so that the measure of a subset of trajectories $A\subset\mathcal{C}[0,1]$ is $\mathbb{P}(\omega\in\Omega | F(\omega)\in A).$ Now if you change values of $F$ on a subset of of $\Omega$ of $\mathbb{P}$-measure zero, this would have no effect on the measure on target space $\mathcal{C}[0,1].$ So $F$ is actually defined up to measure zero, and pointwise statements, like "for every $\omega$" simply do not make sense.


You shall distinguish between the random variables and their distributions. That is, when we talk about Brownian motion, we often define it in terms of finite-dimensional distributions (that is, conditional distribution is normal with the current value as the mean and the variance of $1$). By Kolmogorov's Extension Theorem (KET), there exists a unique distribution with such finite-dimensional distributions. Now, this distribution by itself does not tell you whether the resulting process is continuous or not, because you need to specify on which space this distribution is constructed. KET only guarantees that you can construct this distribution on a boring space $\Bbb R^{[0,\infty)}$, boring due to the fact that its $\sigma$-algebra does not contain "continuous-time" events. However, by Kolmogorov's Continuity Theorem, this distribution can also be constructed on a more interesting space $\mathcal C([0,\infty))$ - and this is the classical construction of Brownian motion.

So what does the latter fact mean. It means that a Brownian motion or classical Wiener process is a random variable $B:\Omega\to\mathcal C([0,\infty))$, which trivially implies that $B(\omega)\in\mathcal C([0,\infty))$ for every $\omega$, that is every realization of classically constructed Brownian motion is continuous.


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