A (standard, real-valued) Brownian motion $W = \{W(t): t \geq 0\}$ is commonly defined by the following properties: 1) $W(0) = 0$ a.s., 2) the process has independent increments, 3) for all $s,t \geq 0$ with $s<t$, the increment $W(t) – W(s)$ is normally distributed with mean zero and variance $t-s$, and 4) almost surely, the function $t \mapsto W(t)$ is continuous.


As is well known, the above set of conditions can be reduced to 2), 3') for all $t \geq 0$, $W(t)$ has mean zero and variance $t$, and 4). [Note that, in 3'), $W(t)$ is not assumed to be normally distributed.] But what about omitting condition 2)? Can you find an example of a process $W$ satisfying conditions 1), 3), and 4), but not 2)? [Note that such $W$ must have (the Brownian motion) covariance $E[W(s)W(t)] = s$, $0 \leq s \leq t$; hence, it cannot be a Gaussian process, for otherwise it would be a Brownian motion.]



The conditions in the definition of Brownian motion
No, it is not true that a process *W* satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows.


This construction is rather contrived, and I don't know if there's any simple examples.
Start with a standard Brownian motion *W*. The idea is to apply a small bump to its distribution while retaining the required properties. I will do this by first reducing it to the discrete-time case. So, choose a finite sequence of times 0 = *t*0 < *t*1 < ... < *t**n*. Then define a piecewise linear process *X* by *X**t**k* = *W**t**k* (*k* = 0,1,...,*n*) and such that *X* is linearly interpolated across each of the intervals [*t**k*-1,*t**k*] and constant over [*t**n*,∞).


Then, *Y* = *W* - *X* is a continuous process independent from *X*. In fact, *Y* is just a sequence of [Brownian bridges](http://en.wikipedia.org/wiki/Brownian_bridge) across the intervals [*t**k*-1,*t**k*] and is a standard Brownian motion on [*t**n*,∞). Also by linear interpolation, for any time *t* ≥ 0, *X**t* is a linear combination of at most two of the random variables *X**t*1,...,*X**t**n*. The increments of *W*,
$$
W\_t-W\_s = X\_t-X\_s + Y\_t-Y\_s,
$$
are then a linear combination of at most 4 of the random variables *X**t*1,...,*X**t**n* plus an independent term. So, choosing *n* ≥ 5, if it is possible to replace (*X**t*1,...,*X**t**n*) by any other ℝn-valued random variable without changing the joint-distribution of any 4 elements, then the distributions of the increments *W**t* - *W**s* will be left unchanged. So, properties (1), (3), (4) will still be satisfied but the new process for *W* will not be a standard Brownian motion. It is possible to change the distribution in this way:



> 
> Let *X* = (*X*1,*X*2,...,*X**n*) be an ℝn-valued random variable with a continuous and strictly positive probability density *p**X*: ℝn → ℝ. Then, there exists a random variable *Y* = (*Y*1,*Y*2,...,*Y**n*) with a different distribution than *X* but for which the projection onto any *n* - 1 elements has the same distribution as for *X*.
> 
> 
> 


That is, for any *k*1,*k*2,...,*k**n*-1 in {1,...,*n*}, (*Y**k*1,*Y**k*2,...,*Y**k**n*-1) has the same distribution as (*X**k*1,*X**k*2,...,*X**k**n*-1).


We can construct the probability density *p**Y* of *Y* by applying a bump to the probability distribution of *X*,
$$
p\_Y(x)=p\_X(x)+\epsilon f(x\_1)f(x\_2)\cdots f(x\_n).
$$
Here, ε is a fixed real number and *f*: ℝ → ℝ is a continuous function of compact support and zero integral, $\int\_{-\infty}^\infty f(x)\,dx=0$. Then, $\int\_{-\infty}^\infty p\_Y(x)\,dx\_k=\int\_{-\infty}^\infty p\_X(x)\,dx\_k$ for each *k*. So, the integral of *p**Y* over ℝ*n* is 1 and, by choosing ε small, *p**Y* will be positive. Then it is a valid probability density function. Finally, as the integral along the *k*th direction (any *k*) agrees for *p**X* and *p**Y*, the projection of *X* and *Y* onto ℝ*n*-1 along the *k*th direction give the same distribution.


