Ex-ante efficiency in assignments with seniority rights

Abstract

We study random assignment economies with expected-utility agents, each of them eventually obtaining a single object. We focus attention on assignment problems that must respect object-invariant (or uniform) weak priorities such as seniority rights in student residence assignment. We propose the sequential pseudomarket mechanism: the set of agents is partitioned into ordered priority groups that are called in turns to participate in a pseudomarket for the remaining objects. SP is characterized by the concept of consistent weak ex-ante efficiency (CWEE), that is, weak ex-ante efficiency complemented by consistency to economy reduction. Moreover, it is shown that CWEE generically implies ex-ante efficiency.

Appendix

1.1 Proof of Proposition 1

Let a price vector $P^{*}$ constitute a pseudomarket equilibrium for an economy E with associated feasible random assignment $Q^{*}$. Then, under Assumptions 1 and 2, there is no other feasible random assignment $Q\ne Q^{*}$ such that $q_{x}\cdot v_{x}=q_{x}^{*}\cdot v_{x}$ for every $x\in N.$

Proof

Take a feasible random assignment $Q\ne Q^{*}$ such that $q_{x}\cdot v_{x}=q_{x}^{*}\cdot v_{x}$ $\forall x\in N$.

Since by Assumption 1 no agent x is indifferent between any two objects, we cannot have $P^{*}\cdot q_{x}<P^{*}\cdot q_{x}^{*}$ for any $x\in N$, provided $q_{x}\cdot v_{x}=q_{x}^{*}\cdot v_{x}$. (Else $q_{x}^{*}$ would not be an optimal choice with prices $P^{*}$: being $M_{x}$ x’s certain assignment to her unique most preferred object type, and for $\alpha >0$ small enough, $\alpha M_{x}+(1-\alpha )q_{x}$ would be a better and affordable choice). Therefore $P^{*}\cdot q_{x}\ge P^{*}\cdot q_{x}^{*}$ for all $x\in N$. On the other hand, we cannot have $P^{*}\cdot q_{x}>P^{*}\cdot q_{x}^{*}$ for any $x\in N$. Otherwise we would have $\sum _{x\in N}q_{x}>\sum _{x\in N}q_{x}^{*}$ for some object type i such that $P^{i*}>0$ . Since this price is positive, it must be the case that $\sum _{x\in N}q_{x}^{*}=\eta ^{i}$. Hence Q is not feasible, a contradiction. We conclude that $P^{*}\cdot q_{x}=P^{*}\cdot q_{x}^{*}$ for all $x\in N$. That is, Q is an equilibrium assignment associated to $P^{*}$.

Let $W=\{i\in S:P^{i*}=0\}$. Unify all object types with zero price as the same object type w. Its supply is $\eta ^{w}=\sum _{i\in W}\eta ^{i}$. Each agent x’s valuation for this object is $\tilde{v}_{x}^{w}=\max _{i\in W}v_{x}^{i}$, $x\in N.$ Valuations for the remaining objects are unaltered: $\tilde{v}_{x}^{j}=v_{x}^{j}$ whenever $j\notin W$. Consider such a ${W-}$unified economy with object types $\tilde{S}=\{w\}\cup S\backslash W.$ Obviously, there is a competitive equilibrium in this economy with prices equal to $\tilde{P} ^{i*}=P^{i*},$ $i\ne w,$ and $\tilde{P}^{w*}=0$. Equilibrium assignments are $\tilde{q}_{x}^{w*}=\sum _{i\in W}q_{x}^{i*}$ and $\tilde{q}_{x}^{i*}=q_{x}^{i*}$, $i\ne w,$ for assignment $\tilde{Q} ^{*}$ analogous to $Q^{*}$. An identical transformation yields $\tilde{Q}$ from Q. Notice that $Q\ne Q^{*}$ implies $\tilde{Q}\ne \tilde{Q}^{*},$ provided Assumption 1. No differences in the assignments Q and $Q^{*}$ can only arise from differences in the assignments of the free goods, since this latter fact is only possible when some agent is indifferent between two free goods.

From now on we assume that $|\tilde{S}|>2.$ If $|\tilde{S} |=1$ this would directly negate $\tilde{Q}\ne \tilde{Q}^{*}$. If $|\tilde{S}|=2$ then for each agent the optimal choice is unique: either picking the free good for sure, or combining the non-free good with the free good if necessary. No indifference between these two options is possible since indifference between them arises only if the agent is indifferent between the free object and the non-free object. Once again, this would contradict $\tilde{Q}\ne \tilde{Q}^{*}$.

Denote with A the (nonempty) set of agents x such that $\tilde{q}_{x}\ne \tilde{q}_{x}^{*}.$ For each $x\in A,$ let $S_{x}=\{i\in \tilde{S}:\tilde{q}_{x}^{i}+\tilde{q} _{x}^{i*}>0\},$ the set of objects with positive demand at either or both allocations. The binding budget constraint guarantees that $|S_{x}|\ge 3$ for each $x\in A$. (Either $\tilde{q} _{x}^{i}$ or $\tilde{q}_{x}^{i*}$ or both contain at least two object types with positive purchased probabilities. Assumption 1 ensures that only one optimally chosen bundle may consist of a sure allocation of one object type). Since $\tilde{q}_{x}\cdot \tilde{v}_{x}= \tilde{q}_{x}^{*}\cdot \tilde{v}_{x}$ and $\tilde{P}^{*}\cdot \tilde{q}_{x}=\tilde{P}^{*}\cdot \tilde{q}_{x}^{*}$ $\forall x\in A$, for each $x\in A$ there is $\alpha _{x},\beta _{x}\ge 0$ such that for any $i\in S_{x}$ we have $\tilde{v}_{x}^{i}=\alpha _{x}+\beta _{x}\tilde{P}_{t^{*}}^{i*}$. Particularly, this implies that for any triple $\{i,j,k\}\subset S_{x}$ we have $\rho _{x}(i,j,k)\equiv \frac{\tilde{v}_{x}^{j}-\tilde{v}_{x}^{k}}{\tilde{v}_{x}^{i}-\tilde{v}_{x}^{k}}= \frac{\tilde{P}^{j*}-\tilde{P}^{k*}}{\tilde{P}^{i*}-\tilde{P} ^{k*}}$. Under Assumption 1 (no pairwise indifference), this is always well-defined, since $\tilde{P}^{i*}=\tilde{P}^{k*}$ is in contradiction with both i and k being purchased with positive probability. Let $\Sigma _{x}$ denote the collection of all three-element sets in $S_{x}:$ $\Sigma _{x}=\{\sigma =\{i,j,k\}\subset S_{x}\}$.

For each $\sigma \in \Sigma \,_{x}\$ let $\delta _{\sigma }$ be the only direction in the $\tilde{S}-$simplex in which one can modify quantities of only objects in $\sigma =\{i,j,k\}$ along the budget frontier (i.e. $\delta _{\sigma }\cdot 1_{|\tilde{S}|}=0$ and $\delta _{\sigma }\cdot \tilde{P} ^{*}=0$)$:\delta _{\sigma }^{i}=\frac{\tilde{P}^{j*}-\tilde{P}^{k*}}{\tilde{P}^{i*}-\tilde{P}^{k*}},$ $\delta _{\sigma }^{j}=-1,$ $\delta _{\sigma }^{k}=1-\frac{\tilde{P} ^{j*}-\tilde{P}^{k*}}{\tilde{P}^{i*}-\tilde{P}^{k*}},$ $\delta _{\sigma }^{l}=0$ for all $l\notin \sigma$ ($\delta _{\sigma }$ is well-defined under Assumption 1: recall that $\tilde{P}^{i*}=\tilde{P}^{k*}\Longrightarrow \sigma \notin \Sigma _{x}$ for any $x\in A$) Since $\sum _{x\in A}(\tilde{q}_{x}-\tilde{q}_{x}^{*})=0$ after the preceding ${W-}$unification, the components of that sum can be ordered in a path $(\tilde{q}_{x}-\tilde{q}_{x}^{*})_{x\in A}$ that starts and ends at the origin. We must then have at least one finite set of pairs agent–object sets $(\{x_{r},\sigma _{r}\})_{\sigma _{r}\in \Sigma _{x_{r}},r=1\ldots T}$ that induce a collection of linearly dependent vectors $\Delta =\{\delta _{\sigma _{r}}\}_{\sigma _{r}\in \Sigma _{x_{r}},r=1\ldots T}.$ Should all elements in $(\delta _{\sigma })_{\sigma \in \Sigma _{x},x\in A}$ be linearly independent, there would be no path $(\tilde{q}_{x}-\tilde{q}_{x}^{*})_{x\in A}$ from the origin back to the origin with one-shot moves along different, linearly independent directions.

We can find a “multi-agent” $\Delta$ in the sense that $\bigcup \nolimits _{r=1\ldots T}\{x_{r}\}$ is not a singleton. We show this by contradiction. Let $X\subset A$ be the set of agents such that for each $x\in X,$ the collection of elements in $(\delta _{\sigma })_{\sigma \in \Sigma _{x}}$ is linearly dependent, but none of its elements is linearly independent from $(\delta _{\sigma })_{\sigma \in \Sigma _{y},y\in A\backslash \{x\}}$. By way of contradiction, $X\ne \varnothing ,$ and the collection $(\delta _{\sigma })_{\sigma \in \Sigma _{x},y\in A\backslash X}$ would contain linearly independent vectors, among themselves and also with respect to $(\delta _{\sigma })_{\sigma \in \Sigma _{y},y\in A\backslash X}.$ But then, since the path $(\tilde{q}_{x}-\tilde{q}_{x}^{*})_{x\in A}$ starts and ends at the origin, we must conclude that $A=X.$ But, again, since for each $x\in X,$ the collection of elements in $(\delta _{\sigma })_{\sigma \in \Sigma _{x}}$ is linearly independent from $(\delta _{\sigma })_{\sigma \in \Sigma _{y},y\in X\backslash \{x\}}$, each agent in X will have her own isolated path starting and ending at the origin, that is, $\tilde{q}_{x}- \tilde{q}_{x}^{*}=0$ for all $x\in X$. This contradicts the definition of A.

In the set D of all such “multi-agent” $\Delta$’s, we focus on some $\Delta ^{*}\in \arg \min _{\Delta \in D}|\Delta |$ with the minimum number of vectors, a number we denote with $\tilde{n}.$ Let $S^{\prime }=\bigcup \nolimits _{\{\delta _{\sigma _{r}}\}\in \Delta ^{*}}\sigma _{r}.$ Notice that $m\notin S^{\prime }$ implies that the corresponding coordinate for m is zero for every vector in $\Delta ^{*}$. Then, provided the two constraints $\delta _{\sigma }\cdot 1_{|\tilde{S}|}=0$ and $\delta _{\sigma }\cdot \tilde{P}^{*}=0$, there are at most $|S^{\prime }|-2$ independent vectors in $\Delta ^{*}$. Actually, there are exactly $|S^{\prime }|-2$ independent vectors, since $\tilde{n}$ is minimal. This implies $\tilde{n}=|S^{\prime }|-2+1$, or $|S^{\prime }|=\tilde{n}+1$.

Now notice that for a collection of vectors $\{d_{\sigma }\in \mathbb {R} ^{|S^{\prime }|},\sigma =(i,j,k)\}_{\delta _{\sigma }\in \Delta }$ meeting: $d_{\sigma }^{l}=0$ for $l\notin \sigma ,$ $d_{\sigma }^{i}=\frac{\tilde{P}^{j*}-\tilde{P}^{k*}}{\tilde{P} ^{i*}-\tilde{P}^{k*}},$ $d_{\sigma }^{j}=-1,$ $d_{\sigma }^{k}=1-\frac{\tilde{P}^{j*}-\tilde{P}^{k*}}{\tilde{P} ^{i*}-\tilde{P}^{k*}}$, this collection is linearly dependent (in comparison to $\Delta ^{*},$ we have only erased coordinates $m\notin S^{\prime }$.) Finally, notice that $\frac{\tilde{P}^{j*}-\tilde{P}^{k*}}{\tilde{P}^{i*}-\tilde{P}^{k*}} =\rho _{x}(i,j,k)$ for every $x\in A$ such that $\{i,j,k\}\in \Sigma _{x}$. This concludes the proof, since we are contradicting Assumption 2. $\square$

1.2 Illustrations of Assumption 2

Assumption 2 (Regularity) We assume that there is no $W-$unification economy $\tilde{E}$ with a subset of object types $S^{\prime }\subset \tilde{S},$ and a set of $\tilde{n}=|S^{\prime }|-1$ agent–object set pairs $\{x_{r},\{i_{r},j_{r},k_{r}\}\}_{r=1,\ldots ,\tilde{n}}$ such that $(d_{x_{r}}(i_{r},j_{r},k_{r}))_{r=1,\ldots ,\tilde{n}}$ are linearly dependent.

We claim that this assumption can embed the assumption that bans multiplicity of equilibria in linear utility economies. Indeed, there is no difference between our model and a linear utility model when there is only one object type that is affordable for every agent, which we call w . This object has zero price in equilibrium.

Lemma 2

For an economy E,  let a Pseudomarket equilibrium price vector $P^{*}$ have an associated random assignment $Q^{*}$ such that $P^{i*}>P^{*}\cdot q_{x}^{*}>P^{w*}=0,$ $\forall i\in S\backslash \{w\},$ $\forall x\in N.$ Then there is no other feasible assignment $Q\ne Q^{*}$ such that $q_{x}\cdot v_{x}=q_{x}^{*}\cdot v_{x}$ for every $x\in N$ if there is no cycle of agents and object types $(\{x_{r},i_{r}\})_{r=1,\ldots ,\tilde{n}}$ (with not all agents nor all objects identical) such that

$$\begin{aligned} \rho _{x_{1}}(i_{1},i_{2},w)\cdot \rho _{x_{2}}(i_{2},i_{3},w)\cdot \cdots \cdot \rho _{x_{\tilde{n}-1}}(i_{\tilde{n}-1},i_{\tilde{n}},w)\cdot \rho _{x_{\tilde{n}}}(i_{\tilde{n}},i_{1},w)=1 \end{aligned}$$

Proof

We ignore the agents whose favorite object type is w. They obtain sure assignment of this object in both allocations. The rest of agents have to choose among the different combinations of some object $i\ne w$ and w. Hence is it without loss of generality in this setup that we focus on $d_{x}(i,j,k)$ such that $k=w$ . Under Assumption 2, there is no $\tilde{n}\times (\tilde{n}+1)$ matrix (where at least one agent is different)

$$\begin{aligned} \begin{bmatrix} 1-\rho _{x_{1}}(i_{1},i_{2},w)&\rho _{x_{1}}(i_{1},i_{2},w)&-1&0&\cdots&0 \\ 1-\rho _{x_{2}}(i_{2},i_{3},w)&0&\rho _{x_{2}}(i_{2},i_{3},w)&-1&\ddots&\vdots \\ \vdots&\vdots&\ddots&\ddots&\ddots&0 \\ 1-\rho _{x_{\tilde{n}-1}}(i_{\tilde{n}-1},i_{\tilde{n}},w)&0&0&\ddots&\rho _{x_{\tilde{n}-1}}(i_{\tilde{n}-1},i_{\tilde{n}},w)&-1 \\ 1-\rho _{x_{\tilde{n}}}(i_{\tilde{n}},i_{1},w)&-1&0&\cdots&0&\rho _{x_{\tilde{n}}}(i_{\tilde{n}},i_{1},w) \end{bmatrix} \end{aligned}$$

with rank lower than $\tilde{n}.$ It means that the determinant of this matrix after we eliminate the first column is zero. And this determinant is precisely

$$\begin{aligned} \rho _{x_{1}}(i_{1},i_{2},w)\cdot \rho _{x_{2}}(i_{2},i_{3},w)\cdot \cdots \cdot \rho _{x_{\tilde{n}-1}}(i_{\tilde{n}-1},i_{\tilde{n}},w)\cdot \rho _{x_{\tilde{n}}}(i_{\tilde{n}},i_{1},w)-1 \end{aligned}$$

proving the desired result. $\square$

Notice that, if we normalize valuations by subtracting $v_{x}^{w}$ from all valuations of agent x,  and we do it for all $x\in N,$ we obtain the same condition as in Lemma 4.1 in Bonnisseau et al. (2001).

1.2.1 Assumption 2 in practice: a more complex example

We complete the “Appendix” with an elaborate example that illustrates how Assumption 2 generally translates into a more complex relation among agents’ preferences. Consider the following matrix:

$$\begin{aligned} \begin{bmatrix} 1-\rho _{1}(b,c,a)&\quad \rho _{1}(b,c,a)&\quad -1&\quad 0&\quad 0&\quad 0 \\ -1&\quad 0&\quad 0&\quad \rho _{2}(d,a,e)&\quad 1-\rho _{2}(d,a,e)&\quad 0 \\ 0&\quad -1&\quad 0&\quad \rho _{3}(d,b,f)&\quad 0&\quad 1-\rho _{3}(d,b,f) \\ 0&\quad 0&\quad -1&\quad 0&\quad 1-\rho _{4}(f,c,e)&\quad \rho _{4}(f,c,e) \\ 0&\quad 0&\quad 0&\quad \rho _{5}(d,e,f)&\quad -1&\quad 1-\rho _{5}(d,e,f) \end{bmatrix} \end{aligned}$$

After erasing the last column, its determinant is

$$\begin{aligned}&\rho _{1}(b,c,a)\rho _{2}(d,a,e)-\rho _{2}(d,a,e)-\rho _{1}(b,c,a)\rho _{3}(d,b,f)+\rho _{1}(b,c,a)\rho _{5}(d,e,f) \\&\quad +\,\rho _{2}(d,a,e)\rho _{5}(d,e,f)-\rho _{4}(f,c,e)\rho _{5}(d,e,f)-\rho _{1}(b,c,a)\rho _{2}(d,a,e)\rho _{5}(d,e,f) \end{aligned}$$

Since colinearity implies that this expression is zero, we can solve for $\rho _{1}(b,c,a)$ as

$$\begin{aligned} \rho _{1}(b,c,a)= & {} \frac{-\rho _{4}(f,c,e)\frac{\rho _{5}(d,e,f)}{1-\rho _{5}(d,e,f)}-\rho _{2}(d,a,e)}{\frac{\rho _{3}(d,b,f)-\rho _{5}(d,e,f)}{1-\rho _{5}(d,e,f)}-\rho _{2}(d,a,e)} \\= & {} \frac{\rho _{4}(f,c,e)\rho _{5}(d,f,e)-\rho _{2}(d,a,e)}{\frac{\rho _{3}(d,b,f)-\rho _{5}(d,e,f)}{1-\rho _{5}(d,e,f)}-\rho _{2}(d,a,e)} \end{aligned}$$

For the second equality, notice that $1-\rho _{x}(i,j,k)=\rho _{x}(k,j,i),$ and $\rho _{x}(i,j,k)/\rho _{x}(k,j,i)=-\rho _{x}(i,k,j).$ This example illustrates that Assumption 2 can be expressed as a condition on a chain of multiplications of marginal rates of substitutions only in very limited cases. In general, a violation of Assumption 2 implies that one marginal rate of substitution (with a third alternative included) can be expressed as a chain of operators of the same type. In fact, for our example, one could create an imaginary agent y with preferences such that $\rho _{y}(b,c,a)=\rho _{1}(b,c,a),$ $\rho _{y}(d,a,e)=\rho _{2}(d,a,e),$ $\rho _{y}(d,b,f)=\rho _{3}(d,b,f),$ $\rho _{y}(f,c,e)=\rho _{4}(f,c,e),$ and $\rho _{y}(d,e,f)=\rho _{5}(d,e,f).$ One could rapidly check that

$$\begin{aligned} \rho _{y}(b,c,a)= & {} \frac{\rho _{y}(f,c,e)\rho _{y}(d,f,e)-\rho _{y}(d,a,e)}{\frac{\rho _{y}(d,b,f)-\rho _{y}(d,e,f)}{1-\rho _{y}(d,e,f)}-\rho _{y}(d,a,e) } \\= & {} \rho _{y}[\rho _{y}[\rho _{y}(b,b,f),\rho _{y}(d,b,f),\rho _{y}(d,e,f)],\\&\quad \rho _{y}[\rho _{y}(f,d,e),\rho _{y}(f,c,e),\rho _{y}(f,e,e)],\rho _{y}(d,a,e)] \end{aligned}$$

For the second equality we use the tricks $\rho _{y}(i,i,j)=1,$ $\rho _{y}(i,j,j)=0$ and $\rho _{y}(i,j,k)=1/\rho _{y}(j,i,k).$ Notice that the $\rho _{y}$ operator appears inside another $\rho _{y}$ operator, and so on.