If $z_1, z_2, ..., z_N$ is a set of $N$ Gaussian variates, then the multivariate Gaussian probability distribution has the form:
$p(z_{1},\,z_{2},\,\cdot\,\cdot\,\cdot\,\cdot\,z_{N})=\frac{1}{\pi^{N}\mathrm{det}\,\mu}\,\mathrm{e}^{-\Delta z^{\dagger}\mu^{-1}\Delta z},$
where $\Delta z$ is the column matrix with elements $\Delta z_i = z_i - \langle z_i \rangle$, and $\mu = \langle \Delta z \Delta z^\dagger \rangle$ is the covariance matrix. Assume the first moments are zero, i.e. $\langle z_i \rangle = 0$.
For such a distribution, we have the "Gaussian moment theorem", where
$\left\lt \Delta z_{i_{1}}^{\star}\Delta z_{i_{2}}^{\star}\cdot\cdot\cdot\Delta z_{i_{N}}^{\star}\Delta z_{j_{M}}\cdot\cdot\cdot\Delta z_{j_{1}}\right\gt = 0$ if $N \neq M$, and
$\left\lt \Delta z_{i_{1}}^{\star}\Delta z_{i_{2}}^{\star}\cdot\cdot\cdot\Delta z_{i_{N}}^{\star}\Delta z_{j_{M}}\cdot\cdot\cdot\Delta z_{j_{1}}\right\gt = \large \sum_{\mathrm{All \ pairings}} \langle \Delta z_{i_1}^* \Delta z_{j_1} \rangle \langle \Delta z_{i_2}^* \Delta z_{j_2} \rangle ... \langle \Delta z_{i_N}^* \Delta z_{j_N} \rangle$ if $N = M$.
If $N = M = 2$, i.e. I have the fourth order moments, and I'm given the LHS of the above equation, is there a way to calculate the second moments? The equations above would give a set of coupled quadratic equations, where the left hand side is known and a solution for the RHS needs to be determined, but can we obtain the second moments without having to solve these quadratic equations? For example, can we use the fourth moments to find the distribution, and extract the second moments from there?
First, assume that none of the $\Delta z_i$ is zero (otherwise you know its covariance with all other $\Delta z_j$, which is zero). Now you can simply use $\mathbb{E}(\Delta z_i^4) = 3 \mathbb{E}(\Delta z_i^2)^2$, so $\mathbb{E}(\Delta z_i^2) = \sqrt{\frac{1}{3}\mathbb{E}(\Delta z_i^4)}$, and after that $\mathbb{E}(\Delta z_i^3 \Delta z_j)= 3\mathbb{E}(\Delta z_i^2)\mathbb{E}(\Delta z_i \Delta z_j)$ so $\mathbb{E}(\Delta z_i \Delta z_j) = \frac{\mathbb{E}(\Delta z_i^3 \Delta z_j)}{\sqrt{3 \mathbb{E}(\Delta z_i^4)}}$.