# Confusion over Brownian sample paths

1. May 15, 2014

This question is in this rubric because I figured it belonged to measure theory, but I am ready to move it if I am wrong.

In reading "Fractal Geometry" by Falconer, I see (Thm. 16.2 and comment following its proof) that the Hausdorff dimension of a Brownian sample path (a continuous function from the unit interval to R2 such that for all h>0 every h-interval is normally distributed with mean 0 and variance h) is 2 but its 2-dimensional Hausdorff dimension is 0. Combined with the idea (also given in Falconer) that a Hausdorff dimension of s, with r<s<t, is that point where the r-dimension is infinite and the t-dimension is zero, I end up with a shaky understanding as follows: in three dimensions, the two-dimensional Brownian sample path has H. dimension 2, but if you take the intersection of the path with any two-dimensional manifold, you get a figure with Hausdorff dimension zero. But I have my doubts about my interpretation, and would be glad to be corrected.

While I am at it, if a two-dimensional Brownian sample path has Hausdorff dimension two, then since Hausdorff and Hamel dimensions are supposed to overlap for integral values, the 2-dim. Brownian path should have also Hamel dimension 2 as well, no?

An object of dimension 2 should have a 2-dimensional area. But can you find the area of a 2-dim. Brownian sample path (when it is viewed in at least three dimensions)?

(An alternative source that seems to shed some light but which I did not fully understand is: http://stat-www.berkeley.edu/~peres/bmall.pdf)

2. Jul 2, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Jul 2, 2014

Perhaps it will help if I rephrase my question a bit better.
First let me define a Brownian sample path: strictly, a mapping from t to X(ω,t) for every point ω in the sample space, such that
(io) with probability 1, X(0) =0; also X(t) is a continuous function of t
(ii)for any t≥0 and h > 0, the increment X(t+h)-X(t) is normally distributed with mean 0 and variance h; that is, there is a probability function P such that
P(X(t+h)-X(t)) = (2πh)-1/2-∞exp(-u2/2h)du
[From these two conditions one can also deduce that if 0≤t1 ≤t2≤t3....≤t2m, the increments X(tn+1) -X(tn) are independent for 1≤n≤2m-1.]

Alternatively one can take a concrete example, where X[0,1] →R, define X(k*2-j) for 0≤k≤2j by induction on j :
set X(0) = 0 and X(1) at random from a normal distribution with mean 0 and variance 1; Next select X(1/2) from a normal distribution with mean ((X(0)+X(1)) and variance 1/2, and then at the next step X(1/4) and X(3/4) are chosen, and so forth.

Then we get the result :
With probability 1, a Brownian sample path in Rn for n≥2 has Hausdorff and box dimensions each equal to 2.

So, to restate my basic question: For integers, Hamel dimension and Hausdorff dimension are supposed to overlap. Looking at this curve, however, I have difficulty figuring out how it could have Hamel dimension 2.

Thanks in advance for any pointers.