Efficient lottery design

Abstract

There has been a surge of interest in stochastic assignment mechanisms that have proven to be theoretically compelling thanks to their prominent welfare properties. Contrary to stochastic mechanisms, however, lottery mechanisms are commonly used in real life for indivisible goods allocation. To help facilitate the design of practical lottery mechanisms, we provide new tools for obtaining stochastic improvements in lotteries. As applications, we propose lottery mechanisms that improve upon the widely used random serial dictatorship mechanism and a lottery representation of its competitor, the probabilistic serial mechanism. The tools we provide here can be useful in developing welfare-enhanced new lottery mechanisms for practical applications such as school choice.

Appendix

Proof of Lemma 1

Let $L=\sum _{s\in S}w_{s}\mu _{s}$ be a lottery. The lemma is obvious if S is a singleton. Thus, suppose not. Without loss of generality, let $S=\left\{ 1,\ldots ,|S|\right\}$. By (L2) and (L3), for some $n\in \mathbb {N}$, for each $s\in S$, there is $m_{s}\in \mathbb {N}$ such that $w_{s}=m_{s}/n$ and $\sum _{s\in S}m_{s}=n$. Then, $\pi (L)=\pi \left( \sum _{s\in S}\frac{m_{s}}{n}\mu _{s}\right) =\pi [\frac{1}{n}\sum _{s\in S}(\overbrace{\mu _{s}+\ldots +\mu _{s}}^{m_{s}})]$. We iteratively define a collection of sets, $\left\{ M_{s}\right\} _{s\in S}$: $M_{1}=\left\{ 1,\ldots ,m_{1}\right\}$, for $s\ge 2$, $M_{s}=\left\{ \sum _{k=1}^{s-1}m_{k}+1,\ldots ,\sum _{k=1}^{s-1}m_{k}+m_{s}\right\}$. Moreover, let $M=\cup _{s\in S}M_{s}$. Also, we define a collection of assignments, $(\nu _{m})_{m\in M}$ as follows: for each $m\in M$, since there is a unique $s\in S$ with $m\in M_{s}$, let $\nu _{m}=\mu _{s}$. Then, the lottery $\frac{1}{n}\sum _{m\in M}\nu _{m}$ is of equal weights and equivalent to L. $\square$

Proof of Claim 1

Part (1) is obvious by construction of $B^{t}(\cdot )$.

Part (2) Let $t\in \left\{ 0\right\} \cup \mathbb {N}$. Suppose $B^{t}(\left\{ a\right\} )\cap X=\emptyset$, but $B^{t}(\left\{ a\right\} )=B^{t+1}(\left\{ a\right\} )$. Let $\left\{ i_{1},\ldots ,i_{M}\right\} :=\left\{ i\in I\mathrel {|}\mu _{S}(1,i)\in B^{t}(\left\{ a\right\} )\right\}$, and for each $m\in \left\{ 1,\ldots ,M\right\}$, let $a_{m}:=\mu (1,i_{m})\in B^{t}\left\{ a\right\}$. Since $B^{t}(\left\{ a\right\} )\cap X=\emptyset$, $a_{m}\not \in X$, i.e., for each $m\in \left\{ 1,\ldots ,M\right\}$, $|\mu _{1}^{-1}(a_{m})|\ge q_{a_{m}}$. This inequality is strict for at least one m, as $\left\{ a\right\} \in B^{t}(\left\{ a\right\} )$ and $|\mu _{1}^{-1}(a)|>q_{a}$. Thus, $\sum _{a\in \left\{ a_{1},\ldots ,a_{m}\right\} }|\mu _{1}^{-1}(a)|=\sum _{m=1}^{M}|\mu _{1}^{-1}(a_{m})|>\sum _{m=1}^{M}q_{a_{m}}\ge \sum _{a\in \left\{ a_{1},\ldots ,a_{m}\right\} }q_{a}$, which contradicts the feasibility of $\mu _{S}$.

Part (3) If the claim is not true, we have $\left\{ a\right\} \subsetneq B^{1}(\left\{ a\right\} )\subsetneq \cdots \subsetneq B^{t}(\left\{ a\right\} )\subsetneq \ldots$, which contradicts the finiteness of A. $\square$

To prove Theorems 1 and Part (2) of Theorem 3, we need the following notion and lemma.

Definition 5

Let $\succ \in \mathbf {P}^{N}$ and $P,R\in \mathcal {S}$. A temporary list of size $\varvec{m}$ is $(a^{1},i^{1},\ldots ,a^{m},i^{m},a^{m+1})$ such that for each $\ell \in \left\{ 1,\ldots ,m\right\}$, (1) $a^{\ell +1}\mathrel {\succ _{i^{\ell }}}a^{l}$, (2) $p_{i^{\ell },a^{\ell }}<r_{i^{\ell },a^{\ell }}$, (3) $p_{i^{\ell },a^{\ell +1}}>r_{i^{\ell },a^{\ell +1}}$, and (4) $a^{1},\ldots ,a^{m}$ are distinct. An improvement cycle from R to P is a temporary list of size m, $(a^{1},i^{1},\ldots ,a^{m},i^{m},a^{m+1})$ , such that $a^{m+1}=a^{1}$.

Lemma 5

Let $\succ \in \mathbf {P}^{N}$, $i\in N$, and $P,R\in \mathcal {S}$ be non-wasteful at $\succ$. Suppose that P stochastically dominates R at $\succ$. Then there is an improvement cycle from R to P.

Proof of Claim 1

We first construct a temporary list of size 1, $(a^{1},i^{1},a^{2})$, where $a^{1}$ and $a^{2}$ are distinct. Since $P\ne R$, there is $i^{1}\in N$ such that $P_{i^{1}}\ne R_{i^{1}}$. Thus, since $P_{i^{1}}$ stochastically dominates $R_{i^{1}}$ at $\succ _{i^{1}}$, there are $a^{1},a^{2}\in A$ such that $a^{2}\mathrel {\succ _{i^{1}}}a^{1}$, $p_{i^{1},a^{2}}>r_{i^{1},a^{2}}$, and $p_{i^{1},a^{1}}<r_{i^{1},a^{1}}$. Thus $a_{1}\ne a_{2}$. Then $(a^{1},i^{1},a^{2})$ is the desired list.

Suppose we are given a temporary list of size m, $(a^{1},i^{1},\ldots ,a^{m-1},i^{m-1},a^{m})$, where $a^{1},\ldots ,a^{m}$ are distinct. Then (1) $a^{m}\mathrel {\succ _{i^{m-1}}}a^{m-1}$, (2) $p_{i^{m-1},a^{m-1}}<r_{i^{m-1},a^{m-1}}$, and (3) $p_{i^{m-1},a^{m}}>r_{i^{m-1},a^{m}}$. Then, since $r_{i^{m-1},a^{m-1}}>p_{i^{m-1},a^{m-1}}\ge 0$, by the feasibility of P and non-wastefulness of R, we have $\sum _{j\in N}p_{j,a^{m}}\le q_{a^{m}}=\sum _{j\in N}r_{j,a^{m}}$. Thus, since $p_{i^{m-1},a^{m}}>r_{i^{m-1},a^{m}}$, there is $i^{m}\in N$ such that $p_{i^{m},a^{m}}<r_{i^{m},a^{m}}$. Thus, since $P_{i^{m}}$ stochastically dominates $R_{i^{m}}$ at $\succ _{i^{m}}$, there is $a^{m+1}\in A$ such that $a^{m+1}\mathrel {\succ _{i^{m}}}a^{m}$ and $p_{i^{m},a^{m+1}}>r_{i^{m},a^{m+1}}$. Thus $(a^{1},i^{1},\ldots ,a^{m},i^{m},a^{m+1})$ is a temporary size of m. Then, if $a^{m+1}=a^{\ell }$ for some $\ell \in \{1,\ldots ,m\}$, then the list $(a^{\ell },i^{\ell },\ldots ,a^{m},i^{m},a^{m+1})$ is an improvement cycle from R to P. Otherwise we continue this process. However, since |A| is finite, we eventually obtain an improvement cycle from R to P. $\square$

Proof of Theorem 1

Let L be a lottery with the support $\mu _{S}$: ($\Rightarrow$) We show the contrapositive. Suppose that the support $\mu _{S}$ of L is not Pareto efficient at $\succ _{S}$. Then there is an $|S|-$fold replica assignment $\nu _{S}$ that Pareto dominates $\mu _{S}$ at $\succ _{S}$. As in Lemma 1, there is an equal-weight lottery $L^{e}=(1/|M|)\sum _{m\in M}\mu _{m}^{\prime }$ that is equivalent to L such that for each $m\in M$ there is a unique $s(m)\in S$ with $\mu _{m}^{\prime }=\mu _{s(m)}$. Now we define an $|S|-$fold replica assignment $\nu _{M}^{\prime }$: for $m\in M$, $\nu _{m}^{\prime }=\nu _{s(m)}$. Then, $\nu _{M}^{\prime }$ Pareto dominates $\mu _{M}^{\prime }$ at $\succ _{M}$. By Lemma 3, the equal-weight lottery with the support $\nu _{M}^{\prime }$ stochastically dominates the equal-weight lottery $\mu _{M}^{\prime }$ at $\succ$. Thus L is not sd-efficient at $\succ$.

($\Leftarrow$) We show the contrapositive. Suppose that L is wasteful (and thus not sd-efficient) at $\succ$. Let $R=\pi (L)$ be the stochastic assignment induced by L. Then there is $i\in N$, $a\in A$ with $r_{i,a}>0$, and $b\in A$ with $b\mathrel {\succ _{i}}a$ such that $\sum _{j\in N}r_{j,b}<q_{b}$. As $r_{i,a}>0$, there is $s\in S$ such that $\mu _{s}(i_{s})=a$. Then, let $\nu _{s}$ be an $s-$replica assignment such that $\nu _{s}(i_{s})=b$ and for each $j\in N$, $\nu _{s}(j_{s})=\mu _{s}(j_{s})$. Then, the $|S|-$fold replica assignment $(\nu _{s},\mu _{S\setminus \left\{ s\right\} })$ Pareto dominates $\mu _{s}$ at $\succ _{S}$.

Suppose that L is non-wasteful but not sd-efficient at $\succ$. Then, there is a stochastic assignment $P\ne R$ that stochastically dominates R at $\succ$. By Lemma 5, there is an improvement cycle, denoted by $(a^{1},i^{1},\ldots ,a^{m},i^{m})$, from R to P. Then, we can find indices $s^{1},\ldots ,s^{m}\in S$ such that $\mu _{s^{1}}(i^{1})=a^{1},\ldots ,\mu _{s^{m}}(i^{m})=a^{m}$. Then, define an $|S|-$fold replica assignment $\nu _{S}$ such that $\nu _{s^{1}}(i^{1})=a^{2},\ldots ,\nu _{s^{m-1}}(i^{m-1})=a^{m},\nu _{s^{m}}(i^{m})=a^{1}$, and any other agent is assigned the same object as in $\mu$. Then, $\nu _{S}$ Pareto dominates $\mu _{S}$ at $\succ _{S}$. $\square$

Proof of Theorem 3

We first show that TRSD$^{K}$ satisfies the equal treatment of equals. Let $i,j\in N$ with $i\ne j$ and $\succ$ a problem with $\succ _{i}=\succ _{j}$. For each priority f we define another priority $f^{i\leftrightarrow j}$ to be the priority where only the positions of i and j under f are switched and the other agents have the same positions as in f. Note that the size of the support is $|\mathcal {F}(K)|\times K\times K$. Consider the lottery of the TRSD$^{K}$ after $F(K)=\left\{ f_{1},\ldots ,f_{K}\right\} \in \mathcal {F}(K)$ is selected. Then agents face lottery $\frac{1}{K}\sum _{g\in F(K)}TTC_{F(K)}(\succ _{F(K)},SD_{F(K)}(\succ );g)$. Consider $F^{i\leftrightarrow j}(K):=\left\{ f_{1}^{i\leftrightarrow j},\ldots ,f_{K}^{i\leftrightarrow j}\right\}$ and the lottery $\frac{1}{K}\sum _{g\in F^{i\leftrightarrow j}(K)}TTC_{F^{i\leftrightarrow j}(K)}(\succ _{F^{i\leftrightarrow j}(K)},SD_{F^{i\leftrightarrow j}(K)}(\succ );g)$. Since the positions of agent i and j are just reversed, the resulting lotteries are the same except that agent i and j’s stochastic assignments are switched. That is, we have

$$\begin{aligned}&\frac{1}{K}\sum _{g\in F(K)}TTC_{F(K)}(\succ _{F(K)},SD_{F(K)}(\succ );g)(i)\\&\quad =\frac{1}{K}\sum _{g\in F^{i\leftrightarrow j}(K)}TTC_{F^{i\leftrightarrow j}(K)}(\succ _{F^{i\leftrightarrow j}(K)},SD_{F^{i\leftrightarrow j}(K)}(\succ );g)(j). \end{aligned}$$

Now, there exist nonempty and disjoint sets $\mathcal {H}$ and $\mathcal {H}^{\prime }$ such that $\mathcal {H}\cup \mathcal {H}^{\prime }=\mathcal {F}(K)$ and for each $F(K)\in \mathcal {H}$, $F^{i\leftrightarrow j}(K)\in \mathcal {H}^{\prime }$. Then, using the above equation and letting $\varphi (F(K),g)=\frac{1}{K}TTC[\succ _{F(K)},SD_{F(K)}(\succ );g]$,

$$\begin{aligned} TRSD^{K}(\succ )(i)\equiv & {} \frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {F}(K)}\sum _{g\in F(K)}\varphi (F(K),g)(i)\\= & {} \frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {F}(K)}\sum _{g\in F^{i\leftrightarrow j}(K)}\varphi (F^{i\leftrightarrow j}(K),g)(j)\\= & {} \frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {H}}\sum _{g\in F^{i\leftrightarrow j}(K)}\varphi (F^{i\leftrightarrow j}(K),g)(j)\\&\quad +\frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {H}^{\prime }}\sum _{g\in F^{i\leftrightarrow j}(K)}\varphi (F^{i\leftrightarrow j}(K),g)(j)\\= & {} \frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {H}^{\prime }}\sum _{g\in F(K)}\varphi (F(K),g)(j)\\&\quad +\frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {H}}\sum _{g\in F(K)}\varphi (F(K),g)(j)\\= & {} \frac{1}{\genfrac(){0.0pt}1{n!}{K}}\sum _{F(K)\in \mathcal {F}(K)}\sum _{g\in F(K)}\varphi (F(K),g)(j)=TRSD^{K}(\succ )(k). \end{aligned}$$

The equality of the first term in the second and third line comes from the following: [$F(K)\in \mathcal {H}\text { and }g\in F^{i\leftrightarrow j}(K)$] $\Leftrightarrow$ [$F^{i\leftrightarrow j}(K)\in \mathcal {H}^{\prime }$ and $g\in F^{i\leftrightarrow j}(K)$] $\Leftrightarrow$ [$F^{\prime }(K)\in \mathcal {H}^{\prime }$ and $h\in \mathcal {H}^{\prime }$]. Similarly, the equality of the second term in the second and third line comes from the following: [$F(K)\in \mathcal {H}^{\prime }$ and $g\in F^{i\leftrightarrow j}(K)$] $\Leftrightarrow$ [$F^{i\leftrightarrow j}(K)\in \mathcal {H}$ and $g\in F^{i\leftrightarrow j}(K)$] $\Leftrightarrow$ [$F^{\prime }(K)\in \mathcal {H}$ and $h\in F^{\prime }(K)$]. Hence, the TRSD$^{K}$ satisfies the equal treatment of equals. $\square$

Part (1) Note that to show that $TRSD^{K}$ stochastically dominates RSD, we need to show that for some problem $\succ$, $TRSD^{K}\ne RSD$ due to the weakly stochastic dominance just proved above. If $|N|\le 3$, then RSD is sd-efficient (Bogomolnaia and Moulin 2001), and thus $TRSD^{K}=RSD$. Suppose $|N|\ge 4$. Example 3 shows that for $|N|=4$, $TRSD^{K}\ne RSD$. The extension to the case of $|N|\ge 5$ is straightforward.

Part (2) Suppose RSD is not sd-efficient for some N, A, and q. Then there is a problem $\succ$ such that $RSD(\succ )$ is not sd-efficient at $\succ$. Let $R:=RSD(\succ )$. We first show

Claim 2

there exist $\bar{K}\le n$ and $F(\bar{K}):=\{f_{1},\ldots ,f_{\bar{K}}\}$ such that $SD_{F(\bar{K})}(\succ )$ is not Pareto efficient in the |K|-fold replica problem. First consider the case where R is wasteful at $\succ$. Then there is $i\in N$, $a\in A$ with $r_{i,a}>0$, and $b\in A$ with $b\mathrel {\succ _{i}}a$ such that $\sum _{j\in N}r_{j,b}<q_{b}$. Then there is $f_{1},f_{2}\in F$ such that $SD_{f_{1}}(\succ )(i)=a$ and $\sum _{j\in N}|SD_{f_{2}}(\succ )(j)|<q_{b}$. Take $\bar{K}=2$ and $F(\bar{K}):=\{f_{1},f_{2}\}$. Then $SD_{F(\bar{K})}(\succ )$ is not Pareto efficient at $\succ$. Consider another case where R is non-wasteful but not sd-efficient at $\succ$. Then there is a stochastic assignment P such that P stochastically dominates R at $\succ$. Then, by Lemma 5, there is an improvement cycle $(a^{1},i^{1},a^{2},i^{2},\ldots ,a^{m},i^{m},a^{m+1})$ from R to P. Let $\bar{K}:=m$. Then, since $a^{1},\ldots ,a^{m}$ are distinct, we have $\bar{K}\le |A|$. Moreover, since $r_{i^{\ell },a^{\ell }}>0$ for each $\ell \in \{1,\ldots ,m\}$, there is $F(\bar{K}):=\{f_{1},\ldots ,f_{\bar{K}}\}$, where $F(\bar{K})$ allows for duplicate elements, such that for each $\ell \in \{1,\ldots ,m\}$, $SD_{f_{k}}(\succ )(i^{\ell })=a^{\ell }$. Then an assignment $\nu$ where for each $i\in \{1,\ldots ,m\}$, $\nu (i^{\ell })=a^{\ell +1}$, Pareto dominates $SD_{F(\bar{K})}(\succ )$. Hence $SD_{F(\bar{K})}$ is not Pareto efficient at $\succ$. Thus the proof of Claim 2 is completed.

Let $K\ge \bar{K}$. Then there is $F(K)\subseteq F$ such that $F(\bar{K})\subseteq F(K)$. Then, by Claim 2, $SD_{F(K)}(\succ )$ is not Pareto efficient. Thus, since $TRSD_{F(K)}(\succ _{F(K)},SD_{F(K)};g)$ for some $g\in F$ is Pareto efficient, we have $TRSD_{F(K)}(\succ _{F(K)},SD_{F(K)};g)\ne SD_{F(K)}(\succ )$. Therefore $TRSD^{K}\ne SD$.