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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!

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    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.2013-12-18

1 Answers 1

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  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 $0$.

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    @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 http://en.wikipedia.org/wiki/Wiener_process#Characterizations_of_the_Wiener_process, 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?2011-05-06
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    (4) Does "2.No" mean you agree with that the three quoted definitions not having continuity a.s. are correct?2011-05-06
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    "2. No" means that those definitions miss the continuity a.s. requirement.2011-05-06
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    @Shai *The law of an arbitrary stochastic process $Z=\{Z_t:t \in I\}$ is determined by its finite-dimensional distributions*: Absolutely not (ironically you provide a counterexample).2011-05-06
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    @Didier: What exactly do you mean?2011-05-06
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    @Shai: Thanks! Why is distribution of increment being normal not required in the definition in your last edit? How is it implied from that definition? Any reference about that definition?2011-05-06
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    @Tim: This is a well-known result, and a simple corollary from the theory of L\'evy processes.2011-05-06
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    @Shai: Assume $[\forall t\in J,Z_t=0]$ has probability $1$, for every finite $J$. Take $I=[0,1]$. Then $[\forall t\in I, Z_t=0]$ may have any probability you like. This means you do not know the law of $Z$ although I gave you all its finite-dimensional marginals.2011-05-06
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    @Didier: I still don't understand your point. Is the problem with the parameter set $I$?2011-05-06
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    @Shai: I fail to see how I could make myself clearer. You write that the finite-dimensional marginals of a process determine the distribution of the process. I provide an example of some processes with the same finite-dimensional marginals but with different distributions. Ergo?2011-05-06
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    @Didier: Do you agree that "The probability law of a stochastic process is usually specified by giving all the finite-dimensional distributions"?2011-05-06
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    @Shai: (1) Whom do you cite? (2) Why "usually"? (3) What is the point of this question of yours? (4) Did you simply *read* my comments? (5) What do you think should be the reaction of any mathematician having declared *Every odd number greater than $3$ is prime* when confronted to *Let us consider $9$*? (If the problem is that *my* example is not in *your* books (which I doubt, by the way), we have very different approaches to mathematics.)2011-05-06
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    @Didier: I cite http://www.columbia.edu/~ww2040/4701Sum07/hwk9sols.pdf (page 1).2011-05-06
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    @Didier: And from a book: Two stochastic processes $\{X_t\}$ and $\{Y_t\}$ (not necessarily defined on a common probability space) are identical in law if the systems of their finite-dimensional distributions are identical.2011-05-06
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    @Shai: You just flunked my point (5). This saddens me. (Re your link: Kolmogorov provides existence, uniqueness is another matter.)2011-05-06
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    @Shai: Flunked twice.2011-05-06
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    @Shai: To complete my example and for the interested readers, take $Z_t=1$ if $t=U$ and $Z_t=0$ otherwise, where $U$ is uniform on $(0,1)$. Every $J$-marginal of $Z$ with $J$ finite (or infinite with zero Lebesgue measure) is the same as the $J$-marginal of the zero process since $[U\in J]$ has probability zero. However, $[\forall t\in I, Z_t=0]$ has probability $0$.2011-05-06
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    @Didier: We may call the process $Z$ a zero process in law. That is $Z$ is identical in law to the zero process. This is standard terminology.2011-05-06
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    @Shai: Paul Lévy must be turning in his grave hearing you. But nevermind: since you fell back on vague and unsubstantiated claims about terminology I guess you finally got the (elementary) point I was making. Hurrah.2011-05-06
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    @Didier, @Shai: I hate to interrupt your duel, but is it possible you don't have in mind the same definition of "law of a stochastic process"? Perhaps it would help if you were both to cite what you have in mind.2011-05-06
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    @Nate: Indeed. To start with the obvious, see http://en.wikipedia.org/wiki/Stochastic_process, especially the comments in section 2.2. Your remark is the reason why I provided a concrete example (the only one provided here, so far). (Thanks for your sense of humour.)2011-05-06
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    @Nate: Thanks. Consider, for example, a sentence like "the law of a Gaussian process is completely characterized by its mean and covariance functions".2011-05-06
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    @Did and Shai: The cause of the controversy is nontrivial. By the definition the law of a stochastic process http://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.2013-12-16
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    @Hansen I thought that was trivially clear to everybody from some comments on. Hence, why do you see fit to awaken a dead case? Because a partly wrong answer was accepted by a user who was given all the evidence? You must learn to live with this sort of things on MSE, I am afraid.2013-12-16
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    @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.2013-12-16
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    @Did: Is there a way to communicate with you privately? I would like to ask you a question and to check out a problem if you would be so kind to oblige.2013-12-18
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    @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.2013-12-18
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    @Did: Thank you. Actually, I have resolved my question.2013-12-18