12
$\begingroup$

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.]

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Oct 21, 2010 at 15:15
  • 2
    $\begingroup$ 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. $\endgroup$ Oct 21, 2010 at 18:03
  • $\begingroup$ @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. $\endgroup$ Oct 21, 2010 at 19:49

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Oct 22, 2010 at 0:57
0
$\begingroup$

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

$\endgroup$
3
  • 6
    $\begingroup$ Why does it have to be a martingale? $\endgroup$ Oct 21, 2010 at 7:55
  • $\begingroup$ 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. $\endgroup$
    – The Bridge
    Oct 21, 2010 at 10:52
  • $\begingroup$ For Lévy's theorem you only need it to be a local martingale. $\endgroup$ Oct 21, 2010 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.