I would like to evaluate the following higher order moments of a multivariate normal distribution in the case of mean $0$ and in the case of mean $\mu$: \begin{equation} E[X_i^{2 n}] \qquad E[(X_i^2 X_{i+1}^2)^n] \end{equation} In the $0$ mean case I understand from the Wick Theorem that we should have $E[X_i^{2 n}]= \frac{(2 n -1)!}{2^{n-1}(n-1)!}E[X_i^{2}]^n$ but I cannot obtain the combinatorial factors of the other. In the non-central case I am quite lost.
2026-03-25 16:05:10.1774454710
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Multivariate normal distribution moments
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(Sorry I don't have enough reputation to comment, but) What's the significance of the index $i$ here? Are you looking at a stochastic process? Otherwise, can we simply consider the bivariate case? If so, let's denote $i$ by $1$ and $j$ by $2$.
If $X_1$ and $X_2$ are correlated, we can represent $X_2$ as a linear combination of $X_1$ and some independent $Y$. In any case, moments of products of normal variables (and their powers) can be found here, for example.
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Let $Z$ be a standard normal r.v. and set $\sigma_i\equiv\sigma_{ii}$. If $\mu_i\ne 0$, \begin{align} \mathsf{E}X_i^{2n}&=\mathsf{E}(\sigma_iZ+\mu_i)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k} \sigma_i^{k}\mu_i^{2n-k}\mathsf{E}Z^k \\ &=\sum_{k=0}^{n}\binom{2n}{2k} \sigma_i^{2k}\mu_i^{2(n-k)}(2k-1)!! \end{align} because $\mathsf{E}Z^{2k}=(2k-1)!!$. When $\mu_i=0$, $$ \mathsf{E}X_i^{2n}=\mathsf{E}(\sigma_iZ)^{2n}=\sigma_i^{2n}\mathsf{E}Z^{2n}=\sigma_i^{2n}(2n-1)!!. $$
For the expectation of cross-products let $Z_1$ and $Z_2$ be independent standard normal r.v.s. Then $(X_i,X_j)\overset{d}{=}(v_iZ_1,v_{ij}Z_1+v_j Z_2)+(\mu_i,\mu_j)$, where $$ \begin{bmatrix} v_i & 0 \\ v_{ij} & v_j \end{bmatrix}=\frac{1}{\sigma_i} \begin{bmatrix} \sigma_i^2 & 0 \\ \sigma_{ij} & \sqrt{\sigma_i^2\sigma_j^2-\sigma_{ij}^2} \end{bmatrix} $$ is the Cholesky decomposition of $\operatorname{Var}([X_i, X_j]^{\top})$. Using the multinomial theorem (when $\mu_i\ne 0$, $\mu_j\ne 0$, and $\sigma_{ij}\ne 0$), $$ \mathsf{E}[X_iX_j]^{2n}=\sum_{k_1+\cdots+k_5=2n}\binom{2n}{k_1,\ldots,k_5}\prod_{l=1}^5 \alpha_l^{k_l}\times \mathsf{E}Z_1^{k_1+2k_2+k_3}\mathsf{E}Z_2^{k_1+k_4}, $$ where $$ \begin{align} \alpha_1&=v_iv_j, \quad \alpha_2=v_iv_{ij}, \\ \alpha_3&=v_i\mu_j+v_{ij}\mu_i, \\ \alpha_4&=v_j\mu_i, \quad \alpha_5=\mu_i\mu_j. \end{align} $$ When $\mu_i=\mu_j=0$ and $\sigma_{ij}\ne 0$, $$ \mathsf{E}[X_iX_j]^{2n}=\sum_{k=0}^n \alpha_1^{2k}\alpha_2^{2(n-k)}(2(2n-k))!!\,(2k-1)!!. $$
Let us take $n \ge 2$ and $(i,j)$ such that $1 \le i \le n$ and $1 \le j \le n$ and $i \neq j$. Then we have: \begin{eqnarray} &&E\left[ X_i^{2 n} X_j^{2 n} \right] = \\ && \int_{{\mathbb R}^n} x_i^{2 n} x_j^{2 n} \cdot \frac{\exp\left[ -\frac{1}{2} (\vec{x}-\vec{\mu})^{T} \cdot {\bf C}^{-1} \cdot (\vec{x}-\vec{\mu})\right]}{\sqrt{(2\pi)^{n} \det({\bf C})}} d^n \vec{x} =\\ && \int_{{\mathbb R}^n} (x_i+\mu_i)^{2 n} (x_j+\mu_j)^{2 n} \cdot \frac{\exp\left[ -\frac{1}{2} (\vec{x})^{T} \cdot {\bf C}^{-1} \cdot (\vec{x})\right]}{\sqrt{(2\pi)^{n} \det({\bf C})}} d^n \vec{x} =\\ && \sum\limits_{p=0}^{2 n} \sum\limits_{q=0}^{2 n} \binom{2 n}{p} \binom{2 n}{q} \mu_i^{2n-p} \mu_j^{2n-q} \left. \frac{\partial ^p}{\partial t_i^p} \frac{\partial ^q}{\partial t_j^q} e^{\frac{1}{2} \vec{t}^{T} \cdot {\bf C} \cdot \vec{t}} \right|_{\vec{t}=\vec{0}}=\\ && \sum\limits_{p=0}^{n} \sum\limits_{q=0}^{n} \binom{2 n}{2p} \binom{2 n}{2q} \mu_i^{2n-2p} \mu_j^{2n-2q} \left. \frac{\partial ^{2 p}}{\partial t_i^{2 p}} \frac{\partial ^{2 q}}{\partial t_j^{2 q}} e^{\frac{1}{2} \vec{t}^{T} \cdot {\bf C} \cdot \vec{t}} \right|_{\vec{t}=\vec{0}}=\\ &&\sum\limits_{p=0}^{n} \sum\limits_{q=0}^{n} \binom{2 n}{2p} \binom{2 n}{2q} \mu_i^{2n-2p} \mu_j^{2n-2q} \cdot \frac{1}{(p+q)!} \cdot \frac{1}{2^{p+q}} \cdot \left.\frac{\partial^{2 p}}{\partial t_i^{2 p}} \frac{\partial^{2 q}}{\partial t_j^{2 q}} \left( \sum\limits_{\xi,\eta=1}^n {\bf C}_{\xi,\eta} t_\xi t_\eta\right)^{p+q}\right|_{\vec{t}=\vec{0}}=\\ &&\sum\limits_{p=0}^{n} \sum\limits_{q=0}^{n} \binom{2 n}{2p} \binom{2 n}{2q} \mu_i^{2n-2p} \mu_j^{2n-2q} \cdot \frac{(2p)! (2q)!}{(p+q)!2^{p+q}} \cdot \sum\limits_{\sigma \in \Pi(\underbrace{i,\cdots,i}_{2 p},\underbrace{j,\cdots j}_{2 q})} \prod\limits_{\xi=1}^{p+q-1} C_{\sigma_\xi,\sigma_{\xi+1}} \end{eqnarray} In the second line from the top we wrote out the definition of the expectation value. In the third line we changed variables $x_\xi \leftarrow x_\xi-\mu_\xi$ for $\xi=1,\cdots,n$. In the fourth line we expanded the power terms in the integrand in a series and introduced the characteristic function of a zero-mean multivariate Gaussian. In the fifth line we took into consideration that the only non-zero terms will be even moments. In the sixth line we expanded the exponential in the characteristic function in a Taylor series and took into account the fact that the only non-vanishing expansion terms is the one with $m=p+q$ and finally in the seventh line we evaluated the derivatives by introducing a sum over permutations $\Pi$ of a sequence of length $2(p+q)$. The following Mathematica code illustrates the step from the sixth to the seventh line:
Update: Let $p=0,\cdots,n$ and $i=1,\cdots,n$. Then the following identity below holds true: \begin{eqnarray} &&\left.\frac{\partial^{2 p}}{\partial t_i^{2 p}} \left( \sum\limits_{\xi,\eta=1}^n {\bf C}_{\xi,\eta} t_\xi t_\eta\right)^{p}\right|_{\vec{t}=\vec{0}} = (2p)! {\bf C}_{i,i}^p \end{eqnarray}
Now, let $p,q=0,\cdots,n$ and $i,j=1,\cdots,n$. Then the following identity below holds true: \begin{eqnarray} &&\left.\frac{\partial^{2 p}}{\partial t_i^{2 p}} \frac{\partial^{2 q}}{\partial t_j^{2 q}} \left( \sum\limits_{\xi,\eta=1}^n {\bf C}_{\xi,\eta} t_\xi t_\eta\right)^{p+q}\right|_{\vec{t}=\vec{0}} = &&(2 p)!(2 q)!{\bf C}_{i,i}^{p-q} \sum\limits_{l=0}^q \left({\bf C}_{i,j}+{\bf C}_{j,i}\right)^{2q-2l} \cdot {\bf C}_{i,i}^l {\bf C}_{j,j}^l \cdot \binom{p+q}{2q-l} \binom{2 q-l}{l} \end{eqnarray}
Likewise let $p,q,r=0,\cdots,n$ and $i,j,k=1,\cdots,n$. Then the identity below holds true: \begin{eqnarray} &&\left.\frac{\partial^{2 p}}{\partial t_i^{2 p}} \frac{\partial^{2 q}}{\partial t_j^{2 q}} \frac{\partial^{2 r}}{\partial t_k^{2 r}} \left( \sum\limits_{\xi,\eta=1}^n {\bf C}_{\xi,\eta} t_\xi t_\eta\right)^{p+q+r}\right|_{\vec{t}=\vec{0}} =\\ &&(2 p)!(2 q)!(2 r)! \sum\limits_{l=0}^r \sum\limits_{l_1=0}^{2 q} \sum\limits_{l_2=0}^{l_1/2} \binom{2r-l}{l} \binom{p+q+r}{2r-l} \binom{2r-2l}{2q-l_1} \binom{l_1-l_2}{l_2} \binom{p+q-r+l}{l_1-l_2} \cdot \\ && {\bf C}_{i,i}^{p+q-r+l-l_1+l_2} {\bf C}_{j,j}^{l_2} {\bf C}_{k,k}^{l} \left( {\bf C}_{i,j}+{\bf C}_{j,i}\right)^{l_1-2 l_2} \left( {\bf C}_{i,k}+{\bf C}_{k,i} \right)^{2 r-2 q+l_1-2 l} \left( {\bf C}_{j,k} + {\bf C}_{k,j} \right)^{2 q-l_1} \end{eqnarray}