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In response to the former question, whether an arbitrary statistical distribution -- with non-zero kurtosis and skewness and higher order cumulants -- can be represented in terms of loop diagrams, the answer is yes! After looking at Zee's "Child Problem" In "Quantum field theory in a nutshell", we see the following multidimensional integral: \begin{eqnarray} Z(J) &=&\int d^Nq \mathrm{Exp}\left(\frac{-\vec{q}\cdot \mathbf{A} \cdot \vec{q}}{2}+\vec{J}\cdot{q}-\frac{\lambda}{4!}q^4 \right) \end{eqnarray} Which can be described in terms of a Green's function series expansion in powers of J: \begin{eqnarray} Z(J) &=& \sum_{s=0}^N \frac{1}{s!}J_{i_1}J_{i_2}\cdots J_{i_s} G_{i_1i_2\cdots i_s} \\ G_{i\cdots j} &=& \int d^Nq \left(q_i \cdots q_j \right)\mathrm{Exp}\left(\frac{-\vec{q}\cdot \mathbf{A} \cdot \vec{q}}{2}-\frac{\lambda}{4!}q^4 \right) = \langle q_i \cdots q_j \rangle \end{eqnarray} Now G is a rank 's' tensor, and we interpret J physically, as an excitation of the system, but in statistics, this is just an offset to the total random vector q. The matrix A is the covariance matrix, or our multidimensional second cumulant -- rank 2 tensor -- from before. We can immediately write down the two point Green's function for this expansion to zeroth and first loop order: \begin{eqnarray} G_{ij} &=& \mathbf{A}^{-1}_{ij}-\lambda \left( \sum_n \mathbf{A}^{-1}_{in} \mathbf{A}^{-1}_{jn} \mathbf{A}^{-1}_{nn} \right) \end{eqnarray} Which, in statistics-speak reads: \begin{eqnarray} G_{ij} &=& \underline{\underline{c_2}}_{ij}-\lambda \left( \sum_n \underline{\underline{c_2}}_{in}\underline{\underline{c_2}}_{jn}\underline{\underline{c_2}}_{nn} \right) \end{eqnarray} Those underlines are messy, but we now see that we are taking into account cross correlation between two variables -- labeled by subscripts i and j -- as well as "one-loop" terms which allow for correlation with some intermediate variable "n", and even accounting for "n"'s correlation with itself. I think this is the correct interpretation, and if one were to allow even more expansion, we'd have \begin{eqnarray} G_{ij} &=& \mathbf{A}^{-1}_{ij}-\lambda \left( \sum_n \mathbf{A}^{-1}_{in} \mathbf{A}^{-1}_{jn} \mathbf{A}^{-1}_{nn} +\mathbf{A}^{-1}_{ij}\mathbf{A}^{-1}_{nn}\mathbf{A}^{-1}_{nn}\right) \end{eqnarray} for the two-point green's function to oneloop order. This third term corresponds to a "disconnected" diagram, since the point -- or random variable -- n is in no way correlated with the initial "source" points i and j. Now to write the total generating function, we would have something like \begin{eqnarray} Z(J) &=& \left(\frac{(2\pi)^N}{\vert \mathbf{A}\vert }\right)^{1/2}\left[1+\vec{J}\cdot \mathbf{A}^{-1}\cdot \vec{J} -\lambda \vec{J}_i (\sum_n \mathbf{A}^{-1}_{in} \mathbf{A}^{-1}_{jn} \mathbf{A}^{-1}_{nn} +\mathbf{A}^{-1}_{ij}\mathbf{A}^{-1}_{nn}\mathbf{A}^{-1}_{nn})\vec{J}_j \right] \end{eqnarray} It's a bit unclear where to go from here, but these green's functions certainly correspond to expectation values of a random vector. The question is, when on introduces anharmonicity into the field -- for instance a non-zero fourth cumulant -- what happens to our expectation values? And, how do we represent those new 2-point and 4-point green's functions diagrammatically?