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I have been trying very hard to prove Isserlis' theorem for n=4 case, i.e when we have 4 random variables that are jointly Gaussian variables with zero-means.

$E[X_1X_2X_3X_4]=E[X_1X_2]E[X_3X_4]+E[X_1X_3]E[X_2X_4]+E[X_1X_4]E[X_2X_3]$

Where $X_1,X_2,X_3$ and $X_4$ are jointly Gaussian random variables. Can anyone help me in proving this. If you can provide me a material for proving this theorem for higher moments I will be really grateful.

Thanks in advance

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  • $\begingroup$ I have done fare amount of googling but couldnt get what i wanted $\endgroup$ – Aditya Oct 3 '14 at 22:31
  • $\begingroup$ Maybe look into Rosen & Marcus's paper referred to here: math.stackexchange.com/questions/27921/… $\endgroup$ – user940 Oct 3 '14 at 22:33
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Let $X=(X_i)$ and $C=\mathrm{Cov}(X,X)$, then $X$ is centered hence its Laplace transform is such that $$E(\mathrm e^{s'X})=\exp\left(\tfrac12s'Cs\right),$$ for every $s=(s_i)$. The homogenous terms of degree $4$ with respect to $s$ from both sides coincide, hence $$E((s'X)^4)=\frac{4!}{2!2^2}(s'Cs)^2=3(s'Cs)^2.$$ Note that $$(s'Cs)^2=\left(\sum_{ij}C_{ij}s_is_j\right)^2=\sum_{ijk\ell}C_{ij}C_{k\ell}s_is_js_ks_\ell.$$ The product $s_1s_2s_3s_4$ appears on the RHS for every $(i,j,k,\ell)$ such that $\{i,j,k,\ell\}=\{1,2,3,4\}$. For example, $C_{12}C_{34}s_1s_2s_3s_4$ appears once for each $(i,j,k,\ell)$ in the collection $$(1,2,3,4),\ (1,2,4,3),\ (2,1,3,4),\ (2,1,4,3),\ (3,4,1,2),\ (4,3,1,2),\ (3,4,2,1),\ (4,3,2,1),$$ that is, $8$ times. Thus, the contribution of $s_1s_2s_3s_4$ to $3(s'Cs)^2$ is $$ 24(C_{12}C_{34}+C_{13}C_{24}+C_{14}C_{23})s_1s_2s_3s_4.$$ On the other hand, $$E((s'X)^4)=\sum_{ijk\ell}E(X_iX_jX_kX_\ell)s_is_js_ks_\ell.$$ hence $E(X_1X_2X_3X_4)s_1s_2s_3s_4$ appears once for every $(i,j,k,\ell)$ such that $\{i,j,k,\ell\}=\{1,2,3,4\}$, that is, $4!=24$ times. To sum up, $$ 24E(X_1X_2X_3X_4)=24(C_{12}C_{34}+C_{13}C_{24}+C_{14}C_{23}),$$ which proves the desired result since $C_{ij}=E(X_iX_j)$ for every $(i,j)$.

More generally, by the same approach, for every $n\geqslant1$, $$\frac1{(2n)!}E((s'X)^{2n})=\frac1{n!2^n}(s'Cs)^n,$$ hence $$E\left(\prod_{k=1}^{2n}X_k\right)=\sum_\tau C^{(\tau)},$$ where the sum on the RHS is over every partition $\tau$ of $\{1,2,\ldots,2n\}$ into $n$ disjoint pairs $(\tau(2k-1),\tau(2k))$ such that $\tau(2k-1)\lt\tau(2k)$ for every $1\leqslant k\leqslant n$ and $\tau(2k-1)\lt\tau(2k+1)$ for every $1\leqslant k\leqslant n-1$, and, for every such partition, $$C^{(\tau)}=\prod_{k=1}^nC_{\tau(2k-1)\tau(2k)}. $$

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  • $\begingroup$ is there a reason why you have both $s$ and $s'$? or is that just a typo? $\endgroup$ – glS Nov 3 '17 at 14:13
  • $\begingroup$ @glS No typo, $s'$ is the transposed vector of $s$, thus, a line vector instead of a column vector. $\endgroup$ – Did Nov 3 '17 at 14:20
  • $\begingroup$ right, thanks for the clarification (wasn't familiar with that notation for the transpose), makes sense now. $\endgroup$ – glS Nov 3 '17 at 14:40

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