Reduced-form budget allocation with multiple public alternatives

Abstract

We consider a reduced-form implementation problem where two players bargain over how to allocate some resources among a finite set of social alternatives. Each player’s payoff depends on how much of the resources is allocated towards each alternative. Distributional constraints on some targeted groups of alternatives are allowed. Using a network flow approach, we provide a characterization of the reduced-form implementability condition. We show that our results can be useful to study public good problems with budget constraints where a joint implementation of private and public alternatives is possible.

Appendix

Proof of Corollary 2

We show that if Q satisfies condition (16), Q is implementable by a feasible s-symmetric mechanism. By Theorem 1, condition (16) is sufficient for the existence of a feasible (possibly asymmetric) mechanism q. We can define another feasible mechanism $q'$ by interchanging the labels of agents and define $\hat{q}=1/2 q+1/2q'$. By convexity, $\hat{q}$ is feasible. Moreover, $\hat{q}$ is s-symmetric. Let $\hat{Q}$ be the reduced form of $\hat{q}$. Since Q is s-symmetric, we have $Q=Q'=\hat{Q}$. Hence Q is implemented by $\hat{q}$.

Proof of Proposition 4

(1) The monotonicity condition in (a) is familiar from the literature on incentive compatibility. For the implementability, note that the constraint structure is a two-tier laminar, by Theorem 1 and r-symmetry, we have

• (i) for all $j\in J$,

  $$\begin{aligned} M_j+N_j-Q_j=0, \end{aligned}$$

  (27)
• (ii)
  $$\begin{aligned} \sum _j M_j \lambda _j= \sum _j N_j \lambda _j, \end{aligned}$$

  (28)

• (iii) for all $A,B\subseteq J$,

  $$\begin{aligned} \sum _{j\in A}M_j \lambda _j - \sum _{j\in B} N_{j} \lambda _{j}\le \lambda (A)\lambda (B^c). \end{aligned}$$

  (29)

We can use condition (27) to project away variables $N_j=Q_j-M_j$ from the system. That is, substitute $N_j$ into conditions (28) and (29), we obtain conditions (25) and (26).

(2) We show that with incentive compatibility we can obtain a reduction of inequalities for condition (26). Note that by incentive compatibility, $M_j$ is non-decreasing in j. Also $M_j-Q_j$ is non-increasing in j, since condition (a) holds and $j/c\le 1$ for all j. By a similar argument as Che et al. (2013), to find the tightest inequalities, we can choose A and B to maximize the set function f(A, B) defined by the left hand side of (26), and the optimal solutions are of the form $A=\{j\in J: M_j\ge \lambda (B^c)\}$ and $B=\{j\in J: M_j-Q_j\ge -\lambda (A)\}$.$\square$

Proof of Proposition 5

For each $Q\in \mathcal {Q}$, define $\mathcal {F}(Q)=\{M\in \mathbb {R}^J: (Q,M)\in \mathcal {F}\}$. By construction, $\mathcal {F}(Q)$ is non-empty, i.e., there exists M such that (Q, M) is feasible. Then $(Q,M)\in \mathcal {F}$ if and only if $Q \in \mathcal {Q}$ and $M\in \mathcal {F}(Q)$. Consider a linear welfare objective $f(U)=w\cdot U$ with the interim utility weights $w=(w_j)\in \mathbb {R}_{++}^J$. By quasilinearity, we have $f(U)= f^1(Q)+f^2(M)$ and the problem becomes $\max _{(Q,M)\in \mathcal {F}}f^1(Q)+f^2(M)$. We can first find for each $Q\in \mathcal {Q}$, the solutions to the subproblem $\max _{M\in \mathcal {F}(Q)}f^2(M)$. It is well-known that for linear programming with the right hand side parameters (vector b in $Ax\le b$), the value function depends on b continuously and is piecewise linear. It implies that value function $F^2(Q)=\max _{M\in \mathcal {F}(Q)}f^2(M)$ depends on Q continuously. We then choose $Q\in \mathcal {Q}$ to maximize $f^1(Q)+F^2(Q)$. Because $F^2(Q)$ is in general nonlinear, the solution can arise at a (possibly non-extreme) boundary point of $\mathcal {Q}$.$\square$

Lemma 2

Let $J=\{a,b\}$ with $a<b\le c$. A reduced-form mechanism $(Q,M)\in \mathcal {F}$ if and only if (a) $Q_a \le Q_b$ and $(a/c) (Q_b- Q_a)\le M_b-M_a\le (b/c) (Q_b- Q_a)$, and (b)

$$\begin{aligned}&2 (M_a \lambda _a+M_b \lambda _b) = Q_a \lambda _a+ Q_b \lambda _b, \end{aligned}$$

(30)

$$\begin{aligned}&2M_a \lambda _a+M_b\lambda _b- Q_a \lambda _a\le \lambda _b, \end{aligned}$$

(31)

$$\begin{aligned}&M_a \lambda _a+M_b\lambda _b- Q_a \lambda _a\le \lambda _b^2, \end{aligned}$$

(32)

$$\begin{aligned}&M_a-Q_a \le 0, \end{aligned}$$

(33)

$$\begin{aligned}&0\le M_a, M_b \le 1. \end{aligned}$$

(34)

Proof of Lemma 2

First notice that in (26), take $B=\{\emptyset \}$ we obtain $M_a,M_b\le 1$. Take $B=\{a,b\}$ and substitute (25) into (26) we have $M_a,M_b\ge 0$. By Proposition 5 (2), we only need to consider $A= \{a,b\}, \{b\},\{\emptyset \}$ and $B=\{\emptyset \}, \{a\},\{a,b\}$. Take all pairs of such (A, B) we obtain:

$$\begin{aligned}&M_a \lambda _a+M_b \lambda _b \le 1, \end{aligned}$$

(35)

$$\begin{aligned}&2M_a \lambda _a+M_b\lambda _b- Q_a \lambda _a\le \lambda _b, \end{aligned}$$

(36)

$$\begin{aligned}&2M_a \lambda _a+2M_b\lambda _b- Q_a \lambda _a- Q_b \lambda _b\le 0, \end{aligned}$$

(37)

$$\begin{aligned}&M_b\lambda _b \le \lambda _b, \end{aligned}$$

(38)

$$\begin{aligned}&M_a \lambda _a+M_b\lambda _b- Q_a \lambda _a\le \lambda _b^2, \end{aligned}$$

(39)

$$\begin{aligned}&M_a \lambda _a+2M_b\lambda _b- Q_a \lambda _a- Q_b \lambda _b\le 0, \end{aligned}$$

(40)

$$\begin{aligned}&M_a \lambda _a- Q_a \lambda _a \le 0,\end{aligned}$$

(41)

$$\begin{aligned}&M_a \lambda _a+M_b\lambda _b- Q_a \lambda _a-Q_b \lambda _b\le 0. \end{aligned}$$

(42)

Note that the inequalities (35), (37), (40), (42) are implied by $0\le M_a,M_b\le 1$ and (30), and thus redundant. We then obtain the conditions in the Lemma.$\square$