An Alternating Algorithm for Finding Linear Arrow-Debreu Market Equilibria

Abstract

Motivated by the convergence result of mirror-descent algorithms to market equilibria in linear Fisher markets, it is natural for one to consider designing dynamics (specifically, iterative algorithms) for agents to arrive at linear Arrow-Debreu market equilibria. Jain (SIAM J. Comput. 37(1), 303–318, 2007) reduced equilibrium computation in linear Arrow-Debreu markets to the equilibrium computation in bijective markets, where everyone is a seller of only one good and a buyer for a bundle of goods. In this paper, we design an algorithm for computing linear bijective market equilibrium, based on solving the rational convex program formulated by Devanur et al. The algorithm repeatedly alternates between a step of gradient-descent-like updates and a distributed step of optimization exploiting the property of such convex program. Convergence can be ensured by a new analysis different from the analysis for linear Fisher market equilibria.

Appendices

Appendix A: Argument of Condition (*) for Theorem 1

We restate the argument of [12] here. Assume that {k} is a singleton strongly connected component without a loop. Let T denote the set of nodes different from {k} that can be reached on a directed path in E from {k}. In an equilibrium allocation, the agents in T ∪{k} spend all their money on the goods of the agents in T, which implies that p_k = 0, contrary to our assumption that p_j > 0 for every j ∈ A.

Appendix B: Proof of Proposition 1

Let b ∈ S_β. Consider any i,k ∈ A, j,l ∈ G. First, we have

$$ \frac{\partial \phi_{\mathbf{\beta}}(\mathbf{b})}{\partial b_{ij}} = -\log u_{ij} + \log \frac{p_{j}}{\beta_{j}} +1= 1-\left( \log u_{ij} - \log \frac{p_{j}}{\beta_{j}}\right) = 1-\log \frac{u_{ij}\beta_{j}}{p_{j}}. $$

Let $p_{j}={\sum }_{i} b_{ij}$. Thus, we have

$$ \frac{\partial \phi_{\mathbf{\beta}}(\mathbf{b})}{\partial b_{ij}} = 1-\log \frac{u_{ij}\beta_{j}}{{\sum}_{i} b_{ij}}. $$

Next, note that except for i = k, we have

$$ \frac{\partial^{2} \phi_{\mathbf{\beta}}(\mathbf{b})}{\partial b_{ij} \partial b_{kl}}=0 , $$

and if i = k, we have

$$ \frac{\partial^{2} \phi_{\mathbf{\beta}}(\mathbf{b})}{\partial b_{ij} \partial a_{kl}} = \frac{1}{{\sum}_{i} b_{ij}}=\frac{1}{p_{j}} . $$

This means that each entry of the Hessian matrix ∇²ϕ_β(b) is at most n. Then for any $\mathbf {z} \in \mathbb {R}^{n\times n}$, we have

$$ \begin{array}{@{}rcl@{}} \mathbf{z}^{\top}\cdot (\nabla^{2} \phi_{\mathbf{\beta}}(\mathbf{b}))\cdot \mathbf{z} &= &\left( \sum\limits_{i} z_{ij} \frac{1}{p_{j}}\right)_{ij} \cdot \mathbf{z} \\ &=& \sum\limits_{i,j} z_{ij} \left( \sum\limits_{i} z_{ij} \frac{1}{p_{j}}\right) \\ &=& \sum\limits_{j} \frac{1}{p_{j}} \left( \sum\limits_{i} z_{ij}\right)^{2} \\ & \leq& \sum\limits_{j} \left( z_{1j}+z_{2j}+\ldots+z_{nj}\right)^{2} \\ &\leq& \sum\limits_{j} \left( |z_{1j}|+|z_{2j}|+\ldots+|z_{nj}|\right)^{2} \\ &\leq &\sum\limits_{j} \left( 1^{2}+1^{2}+\ldots+1^{2}\right)\left( z_{1j}^{2}+z_{2j}^{2}+\ldots+z_{nj}^{2}\right) \\ &=& n \left( \sum\limits_{i,j} z^{2}_{ij}\right). \end{array} $$

The first inequality is by p_i ≥ 1. The third inequality is by Cauchy-Schwarz inequality. This implies that ∇²ϕ_β(b) ≼ αI with α = n.