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This question is inspired partly by this question Any reference on Brownian Motion continuity. In this post, the author asked if the following three axioms can define a Brownian motion without assuming the continuity axiom

"4.$W(t)$ is continuous with probability one. i.e. $\lim _{h\rightarrow 0}P(|W(t+h)-W(t)|>\epsilon )=0,\forall \epsilon>0, t\in S$" By assuming this, Brownian motion is a special case of Levy process.

  1. $W(0) = 0$.
  2. For all $0 \le t_1 \le t_2 \le t_3 \le t_4$, $W(t_2) - W(t_1)$ and $W(t_4) - W(t_3)$ are independent random variables.
  3. For all $0 \le t_1 \le t_2$ , $W(t_2) - W(t_1)$ is normally distributed with mean 0 and variance $\sigma^2\,(t_2 - t_1)$.
OP

In fact, [Karlin&Taylor] defined Brownian motion to be a stochastic process satisfying 1,2,3 axioms with an additional stipulation

"4*.$W(t)$ is continuous at $t=0$"

And they derived continuity of Brownian path as a result using Karhunen–Loève representation Theorem at Sec 7.4. A possible relevant clue is that we always require the characteristic function $E(e^{Xt})$ to be continuous around origin in order to determine a random variables in distribution via characteristic functions. So I guess axiom 4* is a guarantee that some transform exists?

My question is that: If we only assume axiom 1,2,3 on a stochastic process as above, can we construct a stochastic process $W(t)$ that is not a Brownian motion (which is defined as a stochastic process with axiom 1,2,3,4 satisfied OR axiom 1,2,3,4* satisfied in [Karlin&Taylor])? OR Alternatively, is the continuity axiom redundant? (I do not think so but it does not seem very clear how I can construct a counter example to illustrate the point.)

After looking at @Bjørn Kjos-Hanssen's answer, I felt a more appropriate question to ask is that if there is a stochastic process that is not càdlàg and satisfies axioms 1,2,3.

[Karlin&Taylor]Karlin, S., and H. M. Taylor. "A first course in stochastic processes" Academic Press. New York (1975).

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  • $\begingroup$ From @BjornKjosHanssen's answer, you see there is a unique version of the process restricted to the rationals; and there is a set of realizations of full measure that is uniformly continuous on this subset. A càdlàg process agreeing with a process like this is precisely Brownian motion. $\endgroup$ – Anthony Quas Apr 17 '17 at 21:34
  • $\begingroup$ @AnthonyQuas Yes, that is why I modified my question later. Thanks for you comment :) $\endgroup$ – Henry.L Apr 17 '17 at 21:36
  • $\begingroup$ I'm a bit skeptical that the result in K&T is really what you've quoted, since as in Bjørn's answer, you can construct a process satisfying 1, 2, 3, 4*, yet which almost surely has a discontinuity, and it would be very weird for anyone to use the word "Brownian motion" for such a process. Even 4 is the wrong axiom, since it's also preserved by modifications, so Bjørn's example will also satisfy 4. $\endgroup$ – Nate Eldredge Apr 18 '17 at 0:21
  • $\begingroup$ Do K&T also have càdlàg as a standing assumption, or something like that? $\endgroup$ – Nate Eldredge Apr 18 '17 at 0:24
  • $\begingroup$ @NateEldredge 4 is copied directly from the linked post as the OP described; To make sure, I looked up K&T, on p.343 Def 2.1 they defined Brownian motion exactly as a sto. process satisfying 1,2,3,4*. Would you write down your example in mind so that we can discuss? Thanks for your interest $\endgroup$ – Henry.L Apr 18 '17 at 1:18
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Yes, let $W$ be Brownian motion and let $V$ be the following modification: $V_t=W_t$ except that we pick a number $s\in [0,1]$ according to the uniform distribution, independently of $W$, and let $V_s=0$.

Then 1,2,3 are satisfied but the sample path of $V$ is almost surely discontinuous (at $s$).

To get almost sure discontinuity at 0, use $s_1,s_2,\dots$ in the unit interval, all $s_i$ independent of each other and of $W$, with say $$V_{s_i}=1\ne 0$$ for all $i$. Note that $S=\{s_i:i\ge 1\}$ is almost surely dense in the unit interval, but $S$ is random relative to $W$ so it will be disjoint from any countable set of $t$'s considered "in advance".


Background: $W$ will be uniformly continuous on the rationals by axioms 1, 2, 3. But without axiom 4, the question whether the paths are continuous almost doesn't make sense -- the set $$\{f:f \text { is continuous}\} $$ is not measurable. So one solves this by redefining $W$ to be the unique continuous extension of the values of $W$ on the rationals.

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  • $\begingroup$ Fantastic, thank you. So it is safe to say removing continuity at origin $X(0)=0$ will cause the same gap due to the increment arguement right? It suffices to define $Y_t=X_{t-s}$ after $s$ is drawn. $\endgroup$ – Henry.L Apr 17 '17 at 21:31
  • $\begingroup$ Ok, that is morbid enough and satisfies axiom 1,2,3 since $s_1,s_2,\cdots$ cannot be dense in the unit interval. $\endgroup$ – Henry.L Apr 17 '17 at 22:51
  • $\begingroup$ Nate cast some doubts in the comments, and by $\{f:f \text { is continuous}\}$, do you mean the collection of all continuous paths OR the set of continuous points of a given path(I think is the latter)? $\endgroup$ – Henry.L Apr 18 '17 at 2:08

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