Formulae to about Moment and Cross-moments of Stratanovitch Iterated Integrals

The title is a bit long but quite explicit, I am looking for a reference where the moments and cross moment Stratanovitch Iterated Integrals defined as :

$E[J_n(1).J_p(1)]$ with $p\not=n$

With : $J_n(1)_{0,1}=\int_0^1(\int_0^{t_n}...(\int_0^{t_2}o~dW_{t_1})....o~dW_{t_{n-1}}).o~dW_{t_n}$

An extension to multidimensional case would also be appreciated aswell as an extension to the case where some of the integrators are replaced by a $dt$ term. For example, with multi-index notations :

$$J({(1,1,0,1)})_{0,1}=\int_0^1(\int_0^{t_4}(\int_0^{t_3}(\int_0^{t_2}o dW_{t_1})o dW_{t_2})dt_3)o dW_{t_4}$$

Best regards.

PS 1: By the way I have such general formulas for the Iterated Itô Iterated Integrals (Chapter 5.7 of the Kloeden and Platen's Book Numerical Solution of Stochastic Integrals), but the generalization to the Stratanovitch case is not treated there and I was wondering if anyone has bothered calculating the Stratanovitch case which looks computationaly straithforward but seems an incredibly exhausting task to undertake.

PS 2: I googled for some reference but could find any.

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@ Myself : I wonder from the fact that Stratanovitch calculus is the same as ordinary differentail calculus, if $J_n(1)=\frac{1}{n!}W_1^n$, and from this and moments and cross moments of gaussian laws if I could get the fromulae I am looking for in a simpler way than trying to adapt those from Kloede and Platen book. – TheBridge Oct 15 '12 at 20:17
+1, although I don't have an idea. – Tim Oct 16 '12 at 21:27

Well,

Using my comment and this article in Arxiv I think I can answer the first part of my question. So first remark that :
$J_n(1)_{(0,1)}=\frac{1}{n!}(W_1)^n$ as the Stratanovitch-Stochastic Integral follows regular differential calculus rules and so by recurrence supposing the result true at order n-1, we have at order $n$:

$\int_0^1(\int_0^{t_n}...(\int_0^{t_2}o~dW_{t_1})....o~dW_{t_{n-1}}).o~dW_{t_n}=\frac{1}{(n-1)!}\int_0^1 (W_{t_n})^{(n-1)}dW_{t_n}=\frac{1}{(n-1)!}[\frac{1}{n}W_{t_n})^n]_0^1=\frac{1}{n!}(W_1)^n$

And as the result is trivial at $n=1$, conclusion follows so we have at last :

$E[J_n(1).J_p(1)]=\frac{1}{n!.p!}E[W_1^(n+p)]$ where $W_1$ is a standard gaussian variable.

So now from formulae (15), (5) in the reference article we have :

$E[J_n(1).J_p(1)]=\frac{(p+n-1)!!}{n!.p!}$ if ($n+p$ is even) $E[J_n(1).J_p(1)]=0$ if ($n+p$ is odd) Where $(p+n-1)!!=(p+n-1).(p+n-3)....3.1$ if ($n+p$ is even) from formula (5) in the reference.

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