Joint moments of Brownian motion

Solution 1:

Another formula for this joint moment can be found as Lemma 4.5 in a paper by J. Rosen and M.B. Marcus [Annals of Probability, vol. 20, no. 4 (1992) pp. 1603-1684]; the authors refer to it as "well-known". The formula is this (for even $n$): The joint moment $E[W(t_1)W(t_2)\cdots W(t_n)]$ is the sum over all pairings $\{\{a(1),b(1)\},\{a(2),b(2)\},\ldots,\{a(n/2),b(n/2)\}\}$ of $\{1,2,\ldots,n\}$ of $$ \prod_{i=1}^{n/2}E[W(t_{a(i)},t_{b(i)})]. $$ A pairing is simply a partition of $\{1,2,\ldots,n\}$ into $n/2$ doubletons.

Solution 2:

By considering $t_1=t_2=...=t_{2n}$, the sum of the coefficients is the $2n$th moment of $W(1)$, $(2n-1)!!$, which is the number of paths of length $2n$ in Young's lattice from the empty partition to itself.

I haven't yet proven the following, but I think induction should work.

Conjecture:

The indices of the terms with positive coefficients correspond to Dyck words, so the number of terms in $E[W(t_1)W(t_2)...W(t_{2n})]$ is the $n$th Catalan number. $t_1 t_2 t_4$ corresponds to the Dyck word $++-+--$ with pluses in positions $(1, 2, 4)$.

The coefficient of the term corresponding to a Dyck word is the number of paths of length $2n$ in Young's lattice from the empty partition to itself such that the partition at each step has size equal to the height of the Dyck path. The coefficient of $t_{i_1}t_{i_2}...t_{i_n}$ is $\prod_{k=1}^n (2k-i_k)$. For example, the coefficient of $t_1 t_2 ... t_n$ equals $\prod_{k=1}^n (2k-k) = n!$ and paths in Young's lattice which ascend $n$ times and then descend $n$ times correspond to pairs of standard tableaux of the same shape, which are in bijection with permutations of $\{1,2,...,n\}$.