A top dog tale with preference complementarities

Abstract

The emergence of a winner-take-all (top dog) outcome is generally due to political or institutional constraints or to specific technological features which favour the performance of just one individual. In this paper we provide a different explanation for the occurrence of a top-dog equilibrium in exchange economies. We show that once heterogeneous complementarities (i.e. Scarf’s preferences) are analysed with general endowment distributions, a variety of equilibria different from the well-known symmetric outcome with full utilisation of resources can emerge. Specifically, we show that stable corner equilibria with a winner-take-all (top dog) individual arise that are Pareto optima although the remaining individuals are no better off than with zero consumption and resources can be unused. Because of heterogenous complementarities, market mechanisms are weak and cannot overcome the top dog’s power. Voting mechanisms or taxation policies can reduce the top dog’s privileged position.

Appendix

Proof of Proposition 1

(Case with different prices)

• (a) Suppose that the price are unequal and such that: $p_{x}=\alpha ,$ $p_{y}=\beta$ and $(1-\alpha -\beta )=p_{z},$ with $0<1-\alpha -\beta <1,$ and $0<\alpha \ne \beta \ne \gamma <1$. The equilibrium conditions become:

  $$\begin{aligned} f_{y1}= & {} \frac{\alpha X_{1}+\beta Y_{1}+(1-\alpha -\beta )Z_{1}}{(1-\alpha ) }=1/2, \\ f_{x2}= & {} \frac{\alpha X_{2}+\beta Y_{2}+(1-\alpha -\beta )Z_{2}}{(1-\beta )} =1/2, \nonumber \end{aligned}$$

  (9)

  which imply:

  $$\begin{aligned} \alpha X_{1}+\beta Y_{1}+(1-\alpha -\beta )Z_{1}= & {} (1-\alpha )/2 \\ \alpha X_{2}+\beta Y_{2}+(1-\alpha -\beta )Z_{2}= & {} (1-\beta )/2 \end{aligned}$$

• (b) Conversely suppose the conditions (3) hold. Then we have to show that this implies that $p_{x}=\alpha$; $p_{y}=\beta .$ Again multiplying through (9), we get the linear system:

  $$\begin{aligned} p_{x}X_{1}+p_{y}Y_{1}+(1-p_{x}-p_{y})Z_{1}= & {} (1-p_{x})/2, \\ p_{x}X_{2}+p_{y}Y_{2}+(1-p_{x}-p_{y})Z_{2}= & {} (1-p_{y})/2. \end{aligned}$$

  Solving these linear equations we get:

  $$\begin{aligned} p_{y}= & {} -\frac{(-\frac{1}{2}X_{2}+Z_{1}X_{2}+\frac{1}{2}X_{1}-Z_{2}X_{1}- \frac{1}{2}Z_{1}+\frac{1}{4})}{(Y_{2}X_{1}-Y_{2}Z_{1}+\frac{1}{2} Y_{2}-Z_{2}X_{1}-\frac{1}{2}Z_{2}+\frac{1}{2}X_{1}-\frac{1}{2}Z_{1}+\frac{1}{ 4}+Y_{1}Z_{2}-Y_{1}X_{2}+Z_{1}X_{2})}, \nonumber \\ p_{x}= & {} \frac{(Y_{2}-Z_{2}+\frac{1}{4}-Z_{2}Z_{1}-\frac{1}{2} Y_{1}+Y_{1}Z_{2})}{(Y_{2}X_{1}-Y_{2}Z_{1}+\frac{1}{2}Y_{2}-Z_{2}X_{1}-\frac{1 }{2}Z_{2}+\frac{1}{2}X_{1}-\frac{1}{2}Z_{1}+\frac{1}{4} +Y_{1}Z_{2}-Y_{1}X_{2}+Z_{1}X_{2})}. \nonumber \\ \end{aligned}$$

  (10)

  This solution requires that the determinant condition

  $$\begin{aligned} \left( Y_{2}X_{1}-Y_{2}Z_{1}+\frac{1}{2}Y_{2}-Z_{2}X_{1}-\frac{1}{2}Z_{2}+\frac{1}{ 2}X_{1}-\frac{1}{2}Z_{1}+\frac{1}{4}+Y_{1}Z_{2}-Y_{1}X_{2}+Z_{1}X_{2}\right) \ne 0 \end{aligned}$$

  should hold. Recalling the general Hirota conditions (3):

  $$\begin{aligned} X_{1}=\left( \frac{-\beta Y_{1}-(1-\alpha -\beta )Z_{1}+(1-\alpha )/2}{\alpha }\right) , \end{aligned}$$
  $$\begin{aligned} X_{2}=\left( \frac{-\beta Y_{2}-(1-\alpha -\beta )Z_{2}+(1-\beta )/2}{\alpha }\right) , \end{aligned}$$

  and substituting them in (10) gives $p_{x}=\alpha ;p_{y}=\beta$.

$\square$

The sufficient and necessary conditions to decentralise the other special cases of the interior Pareto optimum with (i) $p_{y}$ costing twice $p_{x}$, or (ii) $p_{x}=p_{y}$ can be shown by simply assuming in the above proof respectively that (i) $p_{x}=\alpha ,$ $p_{y}=2\alpha$ and $\alpha (X_{1}+2Y_{1})+(1-\alpha -\beta )Z_{1}=(1-\alpha )/2,$ $\alpha (X_{2}+2Y_{2})+(1-\alpha -\beta )Z_{2}=(1-\beta )/2,$ (ii) $p_{x}=p_{y}=\alpha ,$ $\alpha (X_{1}+Y_{1})+(1-\alpha -\beta )Z_{1}=(1-\alpha )/2$ and $\alpha (X_{2}+Y_{2})+(1-\alpha -\beta )$.

The sufficient and necessary conditions are derived using the same procedure as the Proof of Proposition 1 and imposing $p_{x}=p_{y}=\alpha =\beta =1/3.$