A Statistical Test of Walrasian Equilibrium by Means of Complex Networks Theory

Abstract

We represent an exchange economy in terms of statistical ensembles for complex networks by introducing the concept of market configuration. This is defined as a sequence of nonnegative discrete random variables $\{w_{ij}\}$ describing the flow of a given commodity from agent i to agent j. This sequence can be arranged in a nonnegative matrix W which we can regard as the representation of a weighted and directed network or digraph G. Our main result consists in showing that general equilibrium theory imposes highly restrictive conditions upon market configurations, which are in most cases not fulfilled by real markets. An explicit example with reference to the e-MID interbank credit market is provided.

Appendix

In this appendix we show how to compute the grandcanonical partition function for the network equivalents of the three ideal gas systems. Let’s start by computing $\mathcal {Q}$ in the case of a directed Fermionic (i.e. binary) network. Since $L=\sum _{i\ne j}{a}_{ij}$, the grand partition function reads (recall $\beta = \frac{1}{T}$):

$$\begin{aligned} \mathcal {Q}(z,V,T)&= \sum _{L = 0}^{n(n-1)} z^L Z_G(L,V,T) = \sum _{\{{a}_{ij}\}} e^{\beta (\mu L-H)} \end{aligned}$$

(42)

$$\begin{aligned}&= \sum _{\{{a}_{ij}\}} e^{\sum _{i\ne j}\beta (\mu - \epsilon _{ij}){a}_{ij}} \end{aligned}$$

(43)

$$\begin{aligned}&= \prod _{i \ne j} \left[ 1 + e^{\beta (\mu - \epsilon _{ij})}\right] . \end{aligned}$$

(44)

The expected occupation numbers $\langle {a}_{ij} \rangle$ are obtained in the usual way

$$\begin{aligned} \langle {a}_{ij} \rangle&= - \frac{1}{\beta } \dfrac{\partial }{\partial \epsilon _{ij}} \log \mathcal {Q}\end{aligned}$$

(45)

$$\begin{aligned}&= \dfrac{e^{\beta (\mu - \epsilon _{ij})}}{1 + e^{\beta (\mu - \epsilon _{ij})}}, \end{aligned}$$

(46)

which coincides with the results of Park & Newman for $T = 1$. In the Bosonic (weighted) case, instead of (44) we have

$$\begin{aligned} \mathcal {Q}(z,V,T)= \prod _{i \ne j} \left[ \frac{1}{1 - e^{\beta (\mu - \epsilon _{ij})}} \right] . \end{aligned}$$

(47)

The expression (47) gives

$$\begin{aligned} \langle {w}_{ij} \rangle = \frac{e^{\beta (\mu - \epsilon _{ij})}}{1 - e^{\beta (\mu - \epsilon _{ij})}}, \end{aligned}$$

(48)

that is the Bose-Einstein distribution for networks. Finally for a Boltzmann (weighted) network, taking into account that in this case $W(G) = \prod _{i \ne j} w_{ij}!$, we have [34]:

$$\begin{aligned} \mathcal {Q}(z,V,T)= \prod _{i \ne j} \exp \left[ e^{\beta (\mu - \epsilon _{ij})}\right] . \end{aligned}$$

(49)

and

$$\begin{aligned} \langle {w}_{ij} \rangle = e^{\beta (\mu - \epsilon _{ij})} \end{aligned}$$

(50)

We remark that the network analogue of the Boltzmann gas, which implies that the ${w}_{ij}$ are Poisson variables, is generally overlooked in the literature or introduced with ad hoc assumptions [38]. It’s easy to see that, with the constraints (20)–(21), in this case we arrive at the following explicit ME solution [39]:

$$\begin{aligned} \left\langle w_{ij} \right\rangle = \frac{\omega _ix_j}{L} \qquad \qquad i,j = 1, \dots , N \end{aligned}$$

(51)

From this solution we can obtain the values for the parameters $\mu , \lambda _i, \theta _i$:

$$\begin{aligned} \mu&= - T \ln L\\ \lambda _i&= - T \ln x_i \\ \theta _i&= - T \ln \omega _i \end{aligned}$$

Following [28] we can thus connect the probability distribution of these parameters with the probability distribution of the corresponding observables, namely strengths. We also observe that nodes with the same strength values are also characterized by the same parameter values. We can derive the same results for Bosonic and Fermionic systems by observing that

$$\begin{aligned} \frac{\partial \langle a_{ij} \rangle }{\partial \lambda _i}&= \frac{\partial \langle a_{ij} \rangle }{\partial \theta _j} = \beta \, \langle a_{ij} \rangle \, \left[ \langle a_{ij} \rangle - 1 \right] < 0 \end{aligned}$$

(52)

$$\begin{aligned} \frac{\partial \langle w_{ij} \rangle }{\partial \lambda _i}&= \frac{\partial \langle w_{ij} \rangle }{\partial \theta _j} = - \beta \, \langle w_{ij} \rangle \, \left[ \langle w_{ij} \rangle + 1 \right] < 0 \end{aligned}$$

(53)

Then from

$$\begin{aligned} \sum _j \frac{e^{\beta (\mu - \lambda _{i} - \theta _{j})}}{1 \pm e^{\beta (\mu - \lambda _{i} - \theta _{j})}} = \sum _j \frac{e^{\beta (\mu - \lambda _{h} - \theta _{j})} }{1 \pm e^{\beta (\mu - \lambda _{h} - \theta _{j})}} \end{aligned}$$

(54)

we see that if $\lambda _i \ne \lambda _h$ then the equality (54) cannot be satisfied. Unfortunately, in the Bosonic and Fermionic cases we cannot write down explicitly the functional dependency between strengths and parameters. In the Boltzmann case, instead, we can derive the distribution of the parameters from the distribution of the constraints, at least in some cases. For this reason in the following and in Sect. 4 we always refer to Boltzmann systems. The authors of [28] showed that, if $\rho (w) = (\gamma - 1) \, w^{-\gamma } \, dw$ then $\phi = T \ln w$ is exponentially distributed in a Boltzmann system:

$$\begin{aligned} \rho (\phi ) = \frac{\gamma - 1}{T}\,\exp \left( -\phi \frac{\gamma - 1}{T} \right) \, d\phi \end{aligned}$$

(55)

They further observed that we can make $\rho (\phi )$ independent of T by setting $T = \gamma - 1$. In this way we obtain that the probability density of $\phi$ regains some generality, in the sense that, as long as $\rho (w)$ is in the family of Pareto distributions, $\rho (\phi )$ is unaffected by variations of T. Then we can envisage a simple procedure to estimate T from $\rho (w)$, since we can estimate $\gamma$ from the data. This argument requires that we take a continuous approximation of $\rho (w)$ in case that the strengths are integers. We can extend it to other densities, like for instance $\rho (w) = \frac{1}{w \sigma \sqrt{2 \pi }} \, \exp \left( - \frac{(\ln w)^2}{2\sigma ^2}\right) \,dw$. In this case, by operating a substitution of variables we obtain

$$\begin{aligned} \rho (\phi ) = \frac{1}{\sigma \,T\,\sqrt{2\pi }}\,\exp \left( -\frac{\phi ^2}{2\,(\sigma T)^2} \right) \,d\phi \end{aligned}$$

(56)

By setting $T = \sigma ^{-1}$ we obtain that $\phi$ is distributed like a standard normal variable. The author of [28] develop their argument in the symmetric case. In the asymmetric / directed case we are confronted with two strength distributions $\rho (\omega )$ and $\rho (x)$. Supposing that both come from the same family (Pareto) but with different parameters (respectively $\gamma _{\omega }$ and $\gamma _{x}$), we can set $\rho (\lambda ) = \exp (\lambda )$ (since $\lambda \leqslant 0$) with $T = \gamma _{x} - 1$. Then we obtain ($\theta \leqslant 0$)

$$\begin{aligned} \rho (\theta )&= \frac{\gamma _{\omega } - 1}{T} \exp \left( \theta \,\frac{\gamma _{\omega } - 1}{T} \right) \,d\theta \end{aligned}$$

(57)

$$\begin{aligned}&= \frac{\gamma _{\omega } - 1}{\gamma _{x} - 1} \exp \left( \theta \,\frac{\gamma _{\omega } - 1}{\gamma _{x} - 1} \right) \,d\theta \end{aligned}$$

(58)

In the lognormal case, supposing that $T = \sigma _{x}^{-1}$, we obtain that

$$\begin{aligned} \rho (\lambda )&= \frac{1}{\sqrt{2\pi }}\,\exp \left( \frac{-\lambda ^2}{2\,} \right) \,d\lambda \end{aligned}$$

(59)

$$\begin{aligned} \rho (\theta )&= \frac{1}{\sigma _{\omega }\sigma _{x}^{-1}\sqrt{2\pi }}\,\exp \left( \frac{-\theta ^2}{2\,\sigma _{\omega }\sigma _{x}^{-1}} \right) \,d\theta \end{aligned}$$

(60)