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.(adsbygoogle = window.adsbygoogle || []).push({});

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 R^{2}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)

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# Confusion over Brownian sample paths

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