# The conditions in the definition of Brownian motion

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.]

• Please use the preview. It seeems that you keep posting edits using that as a preview, and this is confusing. – Harald Hanche-Olsen Oct 21 '10 at 6:22
• A special case that might be useful to consider first: if $X$ and $Y$ are separately normally distributed and uncorrelated, and $X+Y$ is also normally distributed, must $(X,Y)$ be jointly normally distributed? – Nate Eldredge Oct 21 '10 at 15:15
• Nate - The answer to that is no. The answer to the original question is no, W need not be a BM. I have a (slightly messy) construction of a counterexample in mind, which I'll post when I have a few moments free. – George Lowther Oct 21 '10 at 18:03
• Looks like all the revision have caused this to become community wiki. So Shai won't get further rep from it. – George Lowther Oct 21 '10 at 18:08
• @George Lowther: Great, I will be interested to see it. If one can answer my question more generally with $n$ random variables, then one could probably plug the resulting joint distributions into Kolmogorov's extension theorem and produce a counterexample to the original question. – Nate Eldredge Oct 21 '10 at 19:49

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 = t0 < t1 < ... < tn. Then define a piecewise linear process X by Xtk = Wtk (k = 0,1,...,n) and such that X is linearly interpolated across each of the intervals [tk-1,tk] and constant over [tn,∞).

Then, Y = W - X is a continuous process independent from X. In fact, Y is just a sequence of Brownian bridges across the intervals [tk-1,tk] and is a standard Brownian motion on [tn,∞). Also by linear interpolation, for any time t ≥ 0, Xt is a linear combination of at most two of the random variables Xt1,...,Xtn. 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 Xt1,...,Xtn plus an independent term. So, choosing n ≥ 5, if it is possible to replace (Xt1,...,Xtn) by any other ℝn-valued random variable without changing the joint-distribution of any 4 elements, then the distributions of the increments Wt - Ws 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 = (X1,X2,...,Xn) be an ℝn-valued random variable with a continuous and strictly positive probability density pX: ℝn → ℝ. Then, there exists a random variable Y = (Y1,Y2,...,Yn) 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 k1,k2,...,kn-1 in {1,...,n}, (Yk1,Yk2,...,Ykn-1) has the same distribution as (Xk1,Xk2,...,Xkn-1).

We can construct the probability density pY 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 pY over ℝn is 1 and, by choosing ε small, pY will be positive. Then it is a valid probability density function. Finally, as the integral along the kth direction (any k) agrees for pX and pY, the projection of X and Y onto ℝn-1 along the kth direction give the same distribution.

• Bah, it's come up as community wiki! Think that's all down to the number of edits made to the original question. – George Lowther Oct 22 '10 at 0:41
• Actually, this argument shows that for any n, there is a continuous process whose distribution agrees with a standard Brownian motion at any set of n times, but is not a BM. – George Lowther Oct 22 '10 at 0:57

Hi,

No you can't, because (1-3-4) implies that $W_t$ is a continuous martingale with quadratic variation equals to $t$ so by a very well known Lévy's Theorem, it has to be a Brownian Motion.

Regards

• Why does it have to be a martingale? – George Lowther Oct 21 '10 at 7:55
• you are right, my mistake, the constancy of expactation of $W_t$ is not enough it should hold for all finite stopping time to hold which I cannot prove. – The Bridge Oct 21 '10 at 10:52
• For Lévy's theorem you only need it to be a local martingale. – Louigi Addario-Berry Oct 21 '10 at 14:11