Brownian motion

[amir]

that was a great work
thank you bob

 

[pruse]

Now generalize. Find (P(B_{\pi(1)} < B_{\pi(2)} < \cdots < B_{\pi(n)})), where (\pi) is a permutation of (1, ..., n).

 

[bob]

Hm, well in the 3-case the ({1,2,3}) and ({3,2,1}) cases are equivalent, each with probability (\frac{1}{2^2}). There are 4 permutations left, each of which is a single exchange of adjacent elements away from one of the strictly increasing / decreasing cases, and each of which evidently has probability (\frac{1}{2^3}).

In the 4 case, the permutations ({1,2,3,4}) and ({4,3,2,1}) each have probability (\frac{1}{2^3}). There are 6 permutations that are exactly 1 exchange of adjacent elements away from one of these. So does the pattern hold that each of these has probability (\frac{1}{2^4})?