On group strategyproof and optimal object allocation

Abstract

We consider a two-agent, single indivisible object allocation problem. We focus on continuous mechanisms that satisfy agent sovereignty, and investigate implications of group strategyproofness. In particular, we provide an explicit characterization of the strategyproof mechanisms and show that there are non-affine maximizer mechanisms that do not belong to the class characterized by Roberts (North-Holland, 1979). Further, we show that there are no budget-balanced strategyproof mechanisms. Also, we obtain an impossibility for existence of strong group strategyproof mechanism. We find that this impossibility goes away upon relaxing our notion of group strategyproofness, and consequently, present a class of weak group strategyproof mechanisms. Finally, we completely characterize the class of feasible strategyproof mechanisms satisfying individual rationality, and show that there are no optimal strategyproof expected revenue maximizing mechanisms under a general class of well behaved type distributions.

Appendix

1.1 Proof of Theorem 1

We first prove the following lemma that establishes two particular properties of $T^\mu _i(.)$ functions for any strategyproof mechanism $\mu \in \Gamma$.

Lemma 1

If a mechanism $\mu =(d,\tau ) \in \Gamma$ satisfies SP, then

• (1) For all $i\in N$, $T_i^\mu (.)$is a non-decreasing continuous function.

• (2) For all $x,y \ge 0$, $x=T^{\mu }_1(y) \Longleftrightarrow y=T^{\mu }_2(x)$.

Proof

To prove (1), fix $y,y'$ such that $0\le y<y'$. If $T^\mu _1(y')<T^\mu _1(y)$, then for any $x \in (T^\mu _1(y'), T^\mu _1(y))$ consider the profiles (x, y) and $(x,y')$. By Result 1 and non-decreasingness of $T_i^\mu (.)$ functions, $d_2(x,y')=0$ and $d_2(x,y)=1$, which contradict Result 1 itself. Arguing similarly for agent 2, we get that $T_i^{\mu }(.)$ is a non-decreasing function for both $i\in N$. Therefore, these functions must either be continuous or have jump discontinuities. Without loss of generality, consider the function $T_1^\mu (.)$ and suppose that there exists a $y\ge 0$ such that $T^\mu _1(y) < \lim _{z \rightarrow y+} T^\mu _1(y)$. Fix an $x \in \left( T^\mu _1(y) , \lim _{z \rightarrow y+} T^\mu _1(y) \right)$ and consider the sequence of profiles $\{(x,y^r)\}$ such that for all r, $y^r>y$ and $\{y^r\} \rightarrow y$. By Result 1, $d(x,y^r)=(0,1)$ for all r, but $d(x,y)=(1,0)$. Since, $\mu \in \Gamma$ and hence, continuous, we have $u((1, \tau _1(x,y));x) = u((0, \tau _1(x,y));x)$, which implies that $x=T^\mu _1(y)$. This contradicts our choice of x and so, $T_i^\mu (.)$ functions must be continuous. Thus, condition (1) follows.

To prove (2), fix any $x,y\ge 0$. There are two possibilities: (i) $d_1(x,y)=1$ or (ii) $d_2(x,y)=1$. If case (i) holds, then by Result 1, $x\ge T^{\mu }_1(y)$ and $y\le T^{\mu }_2(x)$. If $x>T^{\mu }_1(y)$ and $y=T^{\mu }_2(x)$, then choose $\nu >0$ such that $x>T^{\mu }_1(y+\nu )$ (by condition (1) established above, such a $\nu$ exists). By Result 1, $d_1(x,y+\nu )=d_2(x,y+\nu )=1$ and hence, a contradiction. Similarly, if $x=T^{\mu }_1(y)$ and $y<T^{\mu }_2(x)$, then choose $\nu >0$ such that $y<T^{\mu }_2(x-\nu )$. As before, Result 1 implies that $d_1(x-\nu ,y)=d_2(x-\nu ,y)=0$ and hence, a contradiction. Arguing in similar manner, we can establish a contradiction in case (ii), and so, the result follows. $\square$

Using Lemma 1 above, we present the following proofs of necessity and sufficiency:

${{\texttt {\textit{Only If:}}}}$

Fix any mechanism $\mu =(d,\tau ) \in \Gamma$ that satisfies SP. Result 1 and Lemma 1 imply the conditions (2) and (3) trivially. Further, condition (3) implies that $T^{{\mu }^{-1}}_i(.)$ functions must be well defined for all i, and so, by Lemma 1, $T_i^\mu (.)$ functions must be strictly increasing. Thus, condition (4) follows, and so, by Result 1, condition (1) follows. Finally, if $T^\mu _1(0)>0$, then by conditions (3) and (4), $T_1^\mu ( T^\mu _2(0))> T_1^\mu (0) \implies 0>T_1^\mu (0)$ which implies that there exists $\nu >0$ such that $0>T^\mu _2(\nu )$ which contradicts Result 1. Arguing similarly for $T^\mu _2(.)$, condition (5) follows.

${{\texttt {\textit{If:}}}}$

Consider any mechanism $\mu = (d,\tau )$ that satisfies the conditions (1)–(5) stated in the theorem. By Result 1, $\mu$ satisfies SP.¹⁷. Also, since $T_i^\mu (x) \in [0,\infty ), \forall \, x\ge 0, \forall \, i$, $\mu$ satisfies AS. To show continuity, consider, without loss of generality, a sequence of profiles $\{v^r\}$ converging to ${\bar{v}}$ such that the allocation decisions are $d(v^r)=(1,0)$ for all r. Therefore, by Result 1, $v^r_1 \ge T^\mu _1(v^r_2)$ and $v^r_2\le T^\mu _2(v^r_1)$ for all r. By condition (4), in limit ${\bar{v}}_1 \ge T^\mu _1({\bar{v}}_2)$ and ${\bar{v}}_2\le T^\mu _2({\bar{v}}_1)$. Now, if the allocation decisions are not preserved in limit, that is, $d({\bar{v}})=(0,1)$; then ${\bar{v}}_1=T^\mu _1({\bar{v}}_2)$ and ${\bar{v}}_2=T^\mu _2({\bar{v}}_1)$, which implies that for both $j\in N$, $u(0,\tau _j({\bar{v}});{\bar{v}}_j)=u(1,\tau _j({\bar{v}});{\bar{v}}_j) = K_j^\mu (v_i)$, where $i \ne j$. Hence, continuity of $\mu$ follows and so, we can infer that $\mu \in \Gamma$. $\square$

1.2 Proof of Theorem 2

There can be only two types of deviations by the pair $\{1,2\}$: (1) decision-preserving deviations where the allocation decision remains same before and after deviation and (2) decision-changing deviations where the allocation decision changes after deviation. Note that for all $i\in N$ and all $z \ge 0$, if $\eta =0$, then $K^\mu _i(z)=C_i$; and if $\eta =\infty$ then $K^\mu _i(z)=C_i + T^\mu _i(z)$. Now if $\eta =0$, it is easy to see that at any original type profile $v=(v_1,v_2)$, no possible misreport that any agent $j\in N$ can make, would affect the $K^\mu _i(.)$ value for the other agent $i\ne j$. So, any $\{1,2\}$ deviation $v'$ that violates WPGS must be decision changing. However, given that $K^\mu _i(.)$ functions are constant functions for both i when $\eta =0$, the winner at v would become strictly worse off by becoming loser at $v'$, which contradicts our supposed violation of WPGS. Further, if $\eta =\infty$, any decision preserving $\{1,2\}$ deviation would leave the winner’s utility unchanged, and so, any such deviation that violates WPGS must be decision changing. Now, when $\eta =\infty$, for any decision changing $\{1,2\}$ deviation from v to $v'$, the utility of w(v) at profile v (that is, $v_{w(v)} + C_{w(v)})$ is, less than her utility at $v'$ (that is, $C_{w(v)}$); which implies that no such $\{1,2\}$ deviation can violate WPGS.

Now, we focus on mechanisms with a finite and positive $\eta \in (0, \infty )$, and show the sufficiency with respect to each possible kind of deviation as a separate case. First, we define for all possible $\{1,2\}$ deviations from the original valuation profile (x, y) to a misreported valuation profile $(x',y')$;

$$\begin{aligned} \Delta _1:= u_1(d_1(x,y), \tau _1(x,y);x)- u_1(d_1(x',y'),\tau _1(x',y');x), \end{aligned}$$

and

$$\begin{aligned} \Delta _2:= u_2(d_2(x,y),\tau _2(x,y);y)- u_2(d_2(x',y'),\tau _2(x',y');y). \end{aligned}$$

Case(i) Decision preserving Suppose there exists a decision preserving deviation from (x, y) to $(x',y')$ that violates WPGS, that is, $\Delta _i <0$ for both i. Consider the possibility where $d(x,y)=d(x',y')=(1,0)$, that is, $\min \{x,x'\} \ge \max \{T^\mu _1(y),T^\mu _1(y')\}$ and $\max \{y,y'\} \le \min \{T_2^\mu (x),T_2^\mu (x')\}$. Note that given the structure of $K_i^\mu (.)$ functions, $\Delta _2 <0$ is possible only if $x <x'$, and $T_2^\mu (x) < \eta$. Similarly, $\Delta _1 <0$ is possible only if $y' <y$, and $\eta <T_1^\mu (y)$. Therefore, under this possibility of violation of WPGS, we can infer that $\mathbf {(I)}\;\; 0\le y' \le y\le T^\mu _2(x)<\eta <T^\mu _1(y) \le x \le x'$. However, by condition (4) of Theorem 1 and construction of $\eta$ such that $T^\mu _i(\eta )=\eta$ for all $i\in N$; we get that

$$\begin{aligned} \left\{ T^\mu _2(x)<\eta<T^\mu _1(y)\right\} \implies x<\eta < y, \end{aligned}$$

and hence, a contradiction to (I).

Arguing in a similar manner for the possibility where $d(x,y)=d(x',y')=(0,1)$, we get that $({{\mathbf {I}}}{{\mathbf {I}}})\;\; 0\le x' \le x< T^\mu _1(y)<\eta <T^\mu _2(x) \le y \le y'$. As before, by Theorem 1 and construction of $\eta$, we get that $y<\eta <x$, which contradicts (II). Therefore, we conclude that there can be no decision preserving $\{1,2\}$-deviation that violates WPGS when $\eta \in (0,\infty )$.

Case(ii) Decision changing Suppose there exists a decision changing deviation from (x, y) to $(x',y')$ that violates WPGS, that is, $\Delta _i<0$ for all $i\in N$. Without loss of generality, suppose that $d(x,y)=(1,0)$ and $d(x', y')=1$. Now, if $\Delta _1 = x+K^\mu _1(y) - T^\mu _1(y) - K^\mu _1(y') <0$, then by Result 1, $y'>y$, $T^\mu _1(y)<\eta$ and $x<\eta$. By Theorem 1 and construction of $\eta$,

$$\begin{aligned} x< \eta \implies \left\{ T^\mu _2(x)<T^\mu _2(\eta )=\eta \right\} , \end{aligned}$$

and so, $K^\mu _2(x)= T^\mu _2(x)+C_2$.

Now, if $x' \le \eta$, then by the same logic, $K^\mu _2(x')= T^\mu _2(x')+C_2$, which implies that $\Delta _2 = K^\mu _2(x) - \{y+K^\mu _2(x') - T^\mu _2(x')\} = T_2^\mu (x) - y$. Now, by supposition $d_2(x,y)=0$, and so, $y \ge T_2^\mu (x)$ implying that $\Delta _2 \ge 0$, which contradicts our supposition. On the contrary, if $x'>\eta$, then arguing as before, $T^\mu _2(x')>\eta$ and so $K^\mu _2(x')=\eta +C_2$. Therefore, by Result 1, $\Delta _2=(T^\mu _2(x)-y)+(T^\mu _2(x') - \eta ) > 0$ and so, a contradiction to our supposition. Thus, there can be no decision changing $\{1,2\}$-deviation that can violates WPGS when $\eta \in (0,\infty )$. $\square$