Answer to TYI 37: Arithmetic Progressions in 3D Brownian Motion

Consider a Brownian motion in three dimensional space. We asked (TYI 37) What is the largest number of points on the path described by the motion which form an arithmetic progression? (Namely, x_1,x_2, x_t, so that all x_{i+1}-x_i are equal.)

Here is what you voted for

TYI37 poll: Final-results

Analysis of the poll results:  Almost surely 2 is the winner with 30.14% of the 209 votes, and almost surely infinity (28.71%) comes close at second place. In the  third place is  almost surely 3 (14.83%),  and then comes positive probability for each integer (13.4%), almost surely 5 (5.26%),  almost surely 6 (2.87%), and  almost surely 4 (2.39%).

Test your political intuition: which coalition is going to be formed?

Almost surely 2 (briefly AS2) and almost surely infinity (ASI) can form a government  with no need for a larger coalition. But they represent two political extremes. Is AS3 politically closer to AS2 or to ASI? “k with probability p_k for every k>2” (briefly, COM) represent a complicated political massage. Is it closer to AS2 or to ASI? (See the old posts on which coalition will be formed.)


TYI37 poll: Partial results. It was exciting to see how the standing of the answers changed in the process of counting the votes.

And the correct answer is:

5 (FIVE)

See the paper:

Itai Benjamini and Gady Kozma: Arithmetic progressions in the trace of Brownian motion in space

Update: See Yuval Peres’ comment with an intuitive explanation.

This entry was posted in Combinatorics, Open discussion, Probability and tagged , , . Bookmark the permalink.

1 Response to Answer to TYI 37: Arithmetic Progressions in 3D Brownian Motion

  1. Yuval Peres says:

    To see why 5 is the intuitive answer, observe that for every k>1, the set of arithmetic progressions of length in R^3 is six-dimensional. The Brownian path has Hausdorff dimension 2, so insisting that a point lies on it reduces dimension by 1. Thus the dimension of length k progressions on the Brownian path in R^3 is 6-k. Critical sets for BM are empty (Just like critical Branching processes are finite), so for k=6 we get no Arithmetic progressions but for k=5 they have Hausdorff dimension 1. The same intuition leads to the correct guess that Brownian motion in R^3 has double points (Indeed a set of dimension 1 of such points) but no triple points, and BM has no double points in R^4,

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