Let X be a standard two-dimensional Brownian motion. There exists a.s. t [is an element of the set] (0; 1) such that X([0; t))[intersected with] X((t; 1]) = [empty set]. It follows that X([0; 1]) is not homeomorphic to the Sierpinski carpet a.s.

Let X and Y be independent two-dimensional Brownian motions, X(0) = (0; 0); Y(0) = ([epsilon]; 0), and let p([epsilon]) = P(X[0; 1] [intersected with] Y [0; 1] = [empty set], q([epsilon]) = {Y [0; 1] does not contain a closed loop around 0}. Asymptotic estimates (when [epsilon] --> 0) of p([epsilon]); q([epsilon]), and some ...

Suppose that X is a two-dimensional Brownian motion. The trace X[0, 1] contains a self-avoiding continuous path whose Hausdorff dimension is equal to 2.