# Continuous a.s. process

In Ross's Stochastic processes:

1. A stochastic process $\{X(t), t \geq 0\}$ is said to be a Brownian motion process if $X(0) = 0$, $\{X(t), t \geq 0\}$ has stationary independent increments, and for every t > 0, $X(t)$ is normally distributed with mean 0 and variance $c^2t$.
2. Brownian motion could also be defined as a Gaussian process having $E(X(t)) = 0$ and, for $s < t, \text{Cov}(X(s), X(t))= s$.
3. The Brownian Bridge can be denned as a Gaussian process with mean value 0 and covariance function $s(l - t), s \leq t$.

I was wondering

1. if Brownian motion and Brownian Bridge are both continuous a.s.?
2. If the above three definitions for Brownian motion and Brownian Bridge already implicitly imply that the processes such defined are continuous a.s.? Or do these definitions miss the continuity a.s. requirement?
3. if a Gaussian process is always continuous a.s.?

Thanks and regards!

• Caveat: the accepted answer relies (at least partly) on the conviction that the law of an arbitrary stochastic process Z={Zt:t∈I} is determined by its finite-dimensional distributions. This is not so.
– Did
Dec 18, 2013 at 10:39

1. They are commonly defined to be continuous a.s.; 2. No; 3. No (since any non-random function corresponds to a Gaussian process).

Elaborating.

Let $X = \{X_t:t \geq 0\}$ be a continuous Brownian motion (the same idea will be true for Brownian bridge) and $U$ an independent uniform$(0,1)$ variable. Define a process $\tilde X$ by $\tilde X_t = X_t$ if $t \neq U$, and $\tilde X_t = X_t + 1$ if $t=U$. Then $\tilde X$ is discontinuous at $t=U$; nevertheless it satisfies the conditions in Definitions 1 and 2 above. So, $\tilde X$ is a Brownian motion in law (it has the finite dimensional distributions of a Brownian motion), but is not a Brownian motion, which is commonly defined to be continuous a.s.

As for the last question, let $f:I \to \mathbb{R}$ be an arbitrary non-random function, and define a stochastic process $Y$ by $Y_t=f(t)$, $t \in I$. Then, by definition, $Y$ is a Gaussian process, since the joint distribution of $(Y_{t_1},\ldots,Y_{t_n})$ is Gaussian for any $t_1,\ldots,t_n \in I$. So if $f$ is discontinuous, so is $Y$.

EDIT (more details). The law of an arbitrary stochastic process $Z=\{Z_t:t \in I\}$ is determined by its finite-dimensional distributions, that is, the distributions of $(Z_{t_1},\ldots,Z_{t_n})$ for all $n \geq 1$ and $t_1,\ldots,t_n \in I$. If $Z$ is a Gaussian process (say on an interval $I$), then its law is completely determined by its mean function $m(t)={\rm E}(Z_t)$ and covariance function $c(s,t)={\rm Cov}(Z_s,Z_t)$ (for all $s,t \in I$). Hence definitions 2 and 3 above characterize Brownian motion in law and Brownian bridge in law, respectively. In the example of the process $\tilde X$, for any $n \geq 1$ and fixed times $t_1,\ldots,t_n \geq 0$, the random vectors $(X_{t_1},\ldots,X_{t_n})$ and $(\tilde X_{t_1},\ldots,\tilde X_{t_n})$ are identically distributed (indeed, they are equal with probability $1$), and hence, in particular, the conditions in definitions 1 and 2 above are satisfied for the process $\tilde X$, which is a discontinuous Brownian motion in law. (Remark: a Brownian motion in law can moreover have sample paths which are nowhere continuous.)

EDIT. In view of definitions 1 and 2, it is interesting to note that a Brownian motion can be defined without requiring that the (one-dimensional) marginal distributions be normal (with variance proportional to $t$). Indeed, the following statement holds: A stochastic process $X = \{X_t:t \geq 0\}$ is a Brownian motion if $X_0 = 0$, $X$ has stationary independent increments, $X$ is a.s. continuous, and for every $t > 0$, $X_t$ has mean $0$.

• @Shai: Thanks! I was wondering: (1) what parts in the three definitions respectively imply continuity a.s.? (2) how "any constant function corresponds to a Gaussian process" can be a counterexample, isn't constant function continuous a.s.? (3) in Wikipedia en.wikipedia.org/wiki/…, the definition of Wiener process explicitly requires continuity a.s. besides those said in the first definition quoted in my post. So is it redundant in Wikipedia's definition?
– Tim
May 6, 2011 at 1:07
• (4) Does "2.No" mean you agree with that the three quoted definitions not having continuity a.s. are correct?
– Tim
May 6, 2011 at 1:14
• @Did and Shai: The cause of the controversy is nontrivial. By the definition the law of a stochastic process en.wikipedia.org/wiki/Law_(stochastic_processes), Shai Covo's statement "The law of an arbitrary stochastic process Z={Zt:t∈I} is determined by its finite-dimensional distributions, that is, the distributions of (Zt1,…,Ztn) for all n≥1 and t1,…,tn∈I." as such is wrong, as shown by Did's counter example. However, it becomes correct if the process under consideration is continuous.
– Hans
Dec 16, 2013 at 2:31
• @Did: I ran into this question just now because I was learning the law and continuity of a stochastic process. I am not so sure "everybody" is convinced reading the previous comments. It is, as I said before, not a trivial problem. Some theorems are needed to equate the finite dimensional distribution characterization to the law via the continuity of the process. Also, it helps to clarify things for posterity. Indeed, I would suggest you write an answer explicitly to correct and clarify the issue.
– Hans
Dec 16, 2013 at 15:48
• @Hansen I added an explicit comment to the question, "for posterity"... :-) // If you have a math question, ask a question on the site, the whole MSE thing is designed for that and you will benefit from the expertise of several users instead of one.
– Did
Dec 18, 2013 at 10:42