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Hi I am trying to calculate the expected value of $$ \mathbb{E}\big[x_i x_j...x_N\big]=\int_{-\infty}^\infty x_ix_jx_k...x_N \exp\bigg({-\sum_{i,j=1}^N\frac{1}{2}x^\top_i A_{ij}x_j}-\sum_{i=1}^Nh_i x_i\bigg)\prod_{i=1}^Ndx_i, $$ note these are higher order correlation functions for a Gaussian generating functional. Also the matrix $A_{ij}=A^\top_{ij}$ (real symmetric) and is also positive definite, thus the eigenvalues of $A_{ij}$ all satisfy $\lambda_i>0$. Note the generating functional is given by $$ \mathcal{F}(h)=\int_{-\infty}^\infty \exp\bigg({-\sum_{i,j=1}^N\frac{1}{2}x^\top_i A_{ij}x_j}-\sum_{i=1}^Nh_i x_i\bigg)\prod_{i=1}^Ndx_i=\frac{(2\pi)^{N/2}}{\sqrt{\det A_{ij}}}\exp\big( \frac{1}{2}\sum_{i,j=1}^N h_i A^{-1}_{ij}h_j\big) $$ where I used $\det(A)=\prod_{i=1}^N \lambda_i$. We calculate this by finding the minimum of the quadratic form $$ \frac{\partial}{\partial x_k}\bigg(\sum_{i,j=1}^{N} \frac{1}{2} x_i A_{ij} x_j-\sum_{i=1}^N h_i x_i \bigg)=\sum_{j=1}^{N} A_{kj}x_j- h_k=0. $$ In order to solve this we need to introduce the inverse matrix of $A$ given by $A^{-1}$. Thus we can write the solution as $$ x_i=\sum_{j=1}^{N} A^{-1}_{ij} h_j. $$ We can now make a change of variables $x_i \mapsto y_i$ to obtain $$ x_i=\sum_{j=1}^{N} K^{-1}_{ij}h_j+y_i. $$ Re-writing $\mathcal{F}(h)$ we obtain $$ \mathcal{F}(h)=\exp\bigg(\sum_{i,j=1}^{N} \frac{1}{2} h_i A^{-1}_{ij} h_j\bigg)\int_{-\infty}^\infty d^Ny \exp\bigg(-\sum_{i,j=1}^{N} \frac{1}{2} y_i A_{ij}y_j \bigg). $$ This integral is now a simple gaussian which we diagonalize by an orthogonal transformation $A=O\lambda_i\delta_{ij}O^\top$ and a linear change of variables $x=Oy$. The Jacobian of the transformation is unity since a rotation leaves the volume invariant. We write the general result as \begin{equation} \mathcal{F}(h)=\big({2\pi}\big)^{N/2} (\det A_{ij})^{-1/2} \exp\left(\sum_{i,j=1}^{N} \frac{1}{2} h_i A^{-1}_{ij} h_j\right). \end{equation} Having calculated this, I now need to calculate the expected value of the higher order moments which is what my question is.

Note in 1 dimension, the expected value I am trying to calculate is similar to $$ \int_{-\infty}^\infty x^{n} e^{-x^2/2-\alpha x}dx,\quad \Re(n)>-1, \alpha\in \mathbb{R}. $$ I have found lower order expected values given by $$ \big<x_i\big>=A^{-1}_{ij}h_j ,\quad \big<x_i x_j\big>=A^{-1}_{ik} h_k A^{-1}_{jl}h_l+A^{-1}_{ij}, $$ but am trying to generalize to higher orders.

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You can look up a result of Isserlis, which is proved by him in "On a formula for the product-moment coefficient of any order of a normal frequency distribution in any number of variables". Biometrika 12: 134–139.

Essentially, if $X_1,\ldots,X_n$ follow a zero mean multivariate gaussian distribution, $E[X_1\ldots X_n] = \sum_{\text{Partitions of $\{X_1,\ldots,X_n\}$ into pairs}} \prod_{(X_i,X_j)\text{ is a pair}}E[X_i X_j]$ when $n$ is even (and is 0 when $n$ is odd).

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  • $\begingroup$ My case is not the case of zero expected mean. I am familiar with this and Wick's theorem, I am still looking for a general proof of this method though. Also, odd moments of the random variables DO NOT vanish here because $h_i\neq 0$, thus the expected mean is shifted by h. If $h_i=0$, than yes only the even moments have non-vanishing contributions. Specifically in my example, $$ \big<x_i\big>=A^{-1}_{ij}h_j, $$ but I am trying to generalize this... $\endgroup$ – Jeff Faraci Apr 17 '14 at 22:08
  • $\begingroup$ I doubt the case for non-zero mean has a clean solution like this, but you can find the proof in the paper cited. And you can use the solution for the centered case to calculate the non-centered case in principle by linearity of expectation and a ton of algebra. $\endgroup$ – Batman Apr 17 '14 at 22:17
  • $\begingroup$ it has a very nice clean answer actually, my problem is deriving it! $\endgroup$ – Jeff Faraci Apr 17 '14 at 22:18

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