Please forgive me if this is a stupid question -- literally all I've done rigorously in stochastic processes so far is to look at the definitions ... but this Gaussian condition has me confused.

Here's what I understand as the definition of a Wiener process:

Definition: (Wiener Process)

Let $(\Omega,\mathcal{A},\mathbb{P})$ be a measure space. Let $T=\mathbb{R}_{\ge 0}$ . A Wiener process is a random variable $X:T\times\Omega\to\mathbb{R}^d$ that satisfies the following conditions:

  1. (Independent increments) For any times $s_1\le t_1\le s_2\le t_2\le\cdots\le s_n\le t_n$ in $T$, the variables $\{X(t_i,\cdot)-X(s_i,\cdot)\}_{i=1}^n$ are independent.
  2. (Stationary increments) For any $s<t$ the variables $X(t,\cdot)-X(s,\cdot)$ are equal in distribution to $X({t-s},\cdot)-X(0,\cdot)$.
  3. (Gaussian increments) For every $t>s\in T$, $X(t,\cdot)-X(s,\cdot)$ is multivariate normally distributed.
  4. (Continuity) For almost every $\omega\in\Omega$, the functions $t\mapsto X(t,\omega)$ are continuous.

Specifically I'm wondering,

Question a) If we also assume that $X(0,\cdot)$ is constant almost everywhere, then is this definition equivalent to the definition obtained by replacing condition 3 with 3':

3': $\mathrm{E}[(X(t,\cdot)-X(s,\cdot))^2]<\infty$ for every $t>s\in T$?

The intuition is that if the $X(t,\cdot)$ are like the limits of sums of i.i.d. variables, then by the Central Limit Theorem, finite variance of the "microscopic" variables would seem to imply Gaussian $X(t,\cdot)$ for $t>0$.

Question b) Is condition 3 superfluous given the other conditions? In the Wikipedia page for Levy Processes it is stated

Aside from Brownian motion with drift, all other proper Lévy processes have discontinuous paths.[citation needed]

The page defines a Levy process as starting at 0 ($X(0,\omega)=0$ a.s.), having properties 1 and 2 above, and the condition

4' (for Levy process): paths are continuous in probability ($\forall \epsilon>0, \forall t\ge0$, $\displaystyle\lim_{h\to 0^+}\mathbb{P}(|X_{t+h}-X_t|)>\epsilon$).

My reasoning is that I believe condition 4 of the Wiener definition is stronger than (i.e. implies) condition 4'. Thus any process satisfying 1,2, and 4 of the Wiener definition is a Levy process with (almost surely) continuous paths, so that by the assertion it is Brownian motion, which by definition satisfies property 3.

If the condition is not superfluous are there any easy examples of processes satisfying 1,2, and 4 but not 3?

  • 1
    $\begingroup$ You may want to check this almostsure.wordpress.com/2010/06/16/… $\endgroup$
    – m_gnacik
    Jun 8, 2017 at 19:04
  • $\begingroup$ @m_gnacik this is just the type of reference I was looking for! Seems that the proof that 1,2,4 imply 3 will be too long for an answer on this site. $\endgroup$
    – cantorhead
    Jun 9, 2017 at 13:25

1 Answer 1


It seems like you are right; Brownian motion possesses Gaussian increments as a consequence of the other properties. However, I doubt anyone would define Brownian motion without mentioning the Gaussian increments as really this is the fundamental property of Brownian motion. Moreover, the fact that any sample path continuous Levy process is a Wiener process is not trivial.

You can think of it like the holomorphic/analytic nomenclature - defining analytic functions (i.e. those possessing convergent power series) is redundant, since all holomorphic functions are analytic, right? Well, not really. This is a fairly remarkable result in complex analysis, and restricting ourselves to only one definition can be limiting in other scenarios, such as if we want to consider real-analytic functions.

Another way to think of it is: Brownian motion is any stochastic process which satifies these conditions. Whether or not any such process actually exists is another question. Theorem: condition $3$ is redundant; any process which satisfies $1$,$2$ and $4$ must be a Brownian motion. There are quite a few constructions of Brownian motion, and all of the ones I have seen use some nice properties of Gaussian random variables as a jumping off point, usually proving continuity as the last step, and certainly not using the above theorem.

In fact, many authors do not specify that Brownian motion has almost surely continuous sample paths, but rather prove that given any Brownian motion $B$, there exists a "continuous version", i.e. there exists a process $X$ with continuous sample paths such that $\mathbb P(B(t)=X(t))=1$ for all $t$. Of course, once we have a continuous version we use it exclusively, so many authors naturally decide that they may as well include continuity in the definition. If you are uncomfortable with including redundancies in the definition, this might be a good workaround.

  • $\begingroup$ I like the analogy to holomorphic functions. $\endgroup$
    – cantorhead
    Jun 9, 2017 at 13:09
  • $\begingroup$ I'm just waiting to see the reception of this answer before I accept. Sounds reasonable, but I'm obviously in no position to make that judgement. $\endgroup$
    – cantorhead
    Jun 9, 2017 at 13:27

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