# "Brownian motion" without assuming continuity of path at origin of state space

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).

• 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. Apr 17, 2017 at 21:34
• @AnthonyQuas Yes, that is why I modified my question later. Thanks for you comment :) Apr 17, 2017 at 21:36
• 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. Apr 18, 2017 at 0:21
• Do K&T also have càdlàg as a standing assumption, or something like that? Apr 18, 2017 at 0:24
• @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 Apr 18, 2017 at 1:18

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.
• 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. Apr 17, 2017 at 21:31
• Ok, that is morbid enough and satisfies axiom 1,2,3 since $s_1,s_2,\cdots$ cannot be dense in the unit interval. Apr 17, 2017 at 22:51
• 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)? Apr 18, 2017 at 2:08