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My approach to this SE question uses the following joint moments of Brownian motion. For $n=1,2$ they are obvious and well-known, the others are not terribly hard to work out. Is there a reference where these formulas are given, or/and is there a pattern to the coefficients?

Fix $t_1\leq t_2\leq t_3\leq\cdots \leq t_n$. For odd values of $n$ we have $\mathbb{E}[W(t_1)\ W(t_2) \cdots W(t_n)]=0$ while for even values of $n$ we get

\begin{eqnarray*} \mathbb{E}[W (t_1)\ W(t_2)]&=& t_1 \cr \mathbb{E}[W (t_1)\ W(t_2)\ W(t_3)\ W(t_4)]&=& 2t_1 t_2+t_1t_3 \cr \mathbb{E}[W (t_1)\ W(t_2)\ W(t_3)\ W(t_4)\ W(t_5)\ W(t_6)]&=& 2t_1t_2t_5+t_1 t_3 t_5 +4 t_1 t_2 t_4 +2 t_1 t_3 t_4 +6 t_1 t_2 t_3 \end{eqnarray*}

I suppose everything about Brownian motion has been worked out, but I can't find this in any of my books. It's not very important, but I'm just curious!

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By the discussion before, we can suppose that $t_1<t_2<...<t_n.$ put $\xi_1=W_{t_1}, \xi_2=W_{t_2}-W_{t_1},...,\xi_n=W_{t_n}-W_{t_{n-1}}$, which are independent Gaussian variables. Then, we have $$E[W_{t_1}W_{t_2}...W_{t_n}]=E[\xi_1(\xi_1+\xi_2)...(\xi_1+\xi_2+...+\xi_n)],$$ Which could be done. For the case three, it could be written as $$\xi_1(\xi_1+\xi_2)(\xi_1+\xi_2+\xi_3)=\sum_{i=1}^{3!}{\xi_1\xi_{[\frac{i}{3}]}‌​}\xi_{i\%3}, \xi_0=\xi_3$$. I try to treat it as a growth of tree for each term like binary code in computer science ; but I don't get the explicit formula. –  Jun Deng Mar 19 '11 at 5:26
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2 Answers 2

up vote 5 down vote accepted

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.

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Thanks very much for this. I'd learned this many years ago but had forgotten it. –  Byron Schmuland Mar 20 '11 at 0:32
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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 $\{1,2,...,n\}$.


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Thanks! This is a beautiful answer, but it will take me some time to digest it. –  Byron Schmuland Mar 19 '11 at 14:24
    
Your answer makes an unexpected and delightful connection between Brownian moments and combinatorics, but I've decided to accept John's answer. It is closer to what I was looking for, a reference and explanation in terms of probability. I'm sorry that I can't accept both answers! –  Byron Schmuland Mar 20 '11 at 19:45
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