Confusion over Brownian sample paths

[nomadreid]

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)

 

[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?

 

[nomadreid]

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...≤t_2m, the increments X(t_n+1) -X(t_n) 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≤2^j 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.