Non-manipulability of uniform price auctions with a large number of objects

Abstract

When agents (bidders) have multi-demand preferences, uniform price auctions are generally not immune to agents’ strategic manipulation, and they may achieve an inefficient allocation. We consider economies in which a large number of identical objects have to be allocated. Agents have quasi-linear preferences with non-increasing incremental valuations. We explore the incentives of agents in uniform price auctions. An important assumption on preferences is proposed, called “no monopoly.” It requires that preferences should be correlated in such a way that no agent’s incremental valuation for an additional object when he receives sufficiently many objects is higher than those of the other agents. We show that under no monopoly and other mild assumptions on preferences, as the number of objects goes to infinity, the payment in any uniform price auction converges to that in a Vickrey auction. We deduce that when there are sufficiently many objects, truth-telling is an approximate Bayesian Nash equilibrium in each uniform price auction.

Appendix

1.1 Appendix A: Preliminaries

Given $N'\subseteq N$ and $v \in \mathcal {V} ^{N'}$, an object allocation $(x_i)_{i\in N}\in A$ is efficient for v if $\sum _{i\in N {\setminus } N'}v_i(x_i)=\max _{(y_i)_{i\in N}\in A}\sum _{i\in N'}v_i(y_i)$.¹¹ For each $N'\subseteq N$ and each $v \in \mathcal {V} ^{N'}$, let P(v) be the set of efficient object allocations for v.

Remark 1

Let $N'\subseteq N$, $v\in \mathcal {V}^{N'}$ and $(x_i)_{i\in N}\in P(v)$. For each pair $i, j\in N'$ with $i\ne j$, if $x_j>0$, then $v_i(x_i+1)-v_i(x_i)\le v_j(x_j)-v_j(x_j-1).$

Remark 2

In a uniform price auction f, for each $v\in \mathcal {V}^N$, $x^f(v)\in P(v)$.

A uniform price auction requires agents to pay more money than a Vickrey auction if they have the same object allocation rule (Gul and Stacchetti 1999).

Fact 1

(Gul and Stacchetti 1999) Let f and g be a uniform price auction and a Vickrey auction, respectively. If $x^f=x^g$, then for each $v\in \mathcal {V} ^N$ and each $i\in N$, $t^f_i(v)\ge t^g_i(v)$.

Now we state the formal definition of strategy-proofness.

Definition 5

A mechanism f is strategy-proof if for each $v\in \mathcal {V}^N$, each $i\in N$, and each $v'_i\in \mathcal {V}$, $u(f_i(v);v_i)\ge u(f_i(v'_i,v_{-i});v_i)$.

It is known that Vickrey auctions are strategy-proof. Using Fact 1 and the strategy-proofness of Vickrey auctions, we obtain the following sufficient condition for truth-telling to be an $\epsilon$-Bayesian Nash equilibrium in a uniform price auction.

Proposition 3

Let f and g be a uniform price auction and a Vickrey auction such that $x^f=x^g$. Let $\epsilon \in \mathbb {R}_+$. Suppose that for each $i\in N$ and each $v\in \mathrm {supp}_{\Phi _i}(\mathcal {V}^N)$, $t^f_i(v)-t^g_i(v) \le \epsilon$. Then, truth-telling is an $\epsilon$-Baysian Nash equilibrium in f.

Proof

Let $i\in N$ and $v_i\in \mathcal {V}$. For each $(v_i, v_{-i})\in \mathrm {supp}_{\Phi _{i}}(\mathcal {V}^N)$, $t^f_i(v_i, v_{-i}) \le t^g_i(v_i, v_{-i})+ \epsilon$. Thus, for each $v'_i\in \mathcal {V}$,

$$\begin{aligned}&\int _{ v _{ -i}\in \mathcal {V} ^{N _{ -i}}} [v _i(x^f _i(v_i, v_{-i}))-t^f _i(v_i, v_{-i}) ]\varphi _i(v_{-i}|v_i)d v _{ -i}\\&\quad \ge \int _{ v _{ -i}\in \mathcal {V} ^{N _{ -i}}} [v _i(x^g _i(v_i, v_{-i}))-t^g_i(v_i, v_{-i})]\varphi _i(v_{-i}|v_i) d v _{ -i} - \epsilon \\&\quad \ge \int _{ v _{ -i}\in \mathcal {V} ^{N _{ -i}}}[v _i(x^g _i(v'_i, v_{-i}))-t^g _i(v'_i, v_{-i})]\varphi _i(v_{-i}|v_i)d v _{ -i}- \epsilon \\&\quad \ge \int _{ v _{ -i}\in \mathcal {V} ^{N _{ -i}}} [v _i(x^f _i(v'_i, v_{-i}))-t^f _i(v'_i, v_{-i})]\varphi _i(v_{-i}|v_i) d v _{ -i}- \epsilon , \end{aligned}$$

where the second inequality follows from the strategy-proofness of Vickrey auctions, and the last inequality follows from Fact 1. Hence, truth-telling is an $\epsilon$-Bayesian Nash equilibrium. $\square$

The following is an analogue of Proposition 3.

Proposition 4

Let f and g be a uniform price auction and a Vickrey auction such that $x^f=x^g$. Let $\epsilon \in \mathbb {R}_+$ and $v\in \mathcal {V}^N$ be such that $t^f(v)-t^g(v) \le \epsilon$ for each $i\in N$. Then, truth-telling is an $\epsilon$-Nash equilibrium in (f, v).

We omit the proof since it is similar to the proof of Proposition 3.

1.2 Appendix B: Proof of Theorem 1

Let $\epsilon \in \mathbb {R}_{++}$. By Assumptions 1 and 2, there is $x ^*(\epsilon )\in \mathbb {Z}_+$ such that for each $v _i\in \mathcal {V}$ and each $x\in \mathbb {Z}_+$ with $x\ge x^*(\epsilon )$,

$$\begin{aligned} x\cdot | v_i(x+1)-v_i(x)- v^\infty _i|< \epsilon . \end{aligned}$$

(1)

Suppose $\overline{x}\ge |N|\cdot x^*(\epsilon )$.

Let f be a uniform price auction. Let $p:\mathcal {V}^N\rightarrow \mathbb {R}$ be the price scheme associated with f. Note that there can be several uniform price auctions that have the same price scheme. However, they always assign each agent bundles that he finds indifferent. Thus, if truth-telling is an $\epsilon$-Bayesian Nash equilibrium in a uniform price auction, then the same conclusion holds for any other uniform price auction that has the same price scheme. Hence, without loss of generality, we can assume that f always assigns all the objects, that is, for each $v\in \mathcal {V}^N$, $\sum _{i\in N}x^f_i(v)=\overline{x}$.

Let g be a Vickrey auction such that $x^g=x^f$. By Proposition 3, all we need to prove Theorem 1 is that for each $i\in N$ and each $v\in \mathrm {supp}_{\Phi _i}(\mathcal {V}^N)$, $t^f_i(v)-t^g_i(v)<\epsilon$. Let $i\in N$ and $v\in \mathrm {supp}_{\Phi _i}(\mathcal {V}^N)$. Denote $v^\infty := \max _{j\in N}v^\infty _j$ and let $N ^* := \{j\in N: v^\infty _j = v^\infty \}$.

Step 1

$|N^*|\ge 2$.

Proof

Since N is finite, there is $j\in N$ such that $v^\infty _j=v^\infty$. Thus, $|N^*|\ge 1$. Suppose, by contradiction, that $|N^*|=1$. Then, for each $k\in N {\setminus } \{j\}$, $v^\infty _j>v^\infty _k$. Let $\delta =v^\infty _j- \max _{k\in N {\setminus }\{j\}} v^\infty _k$. Since N is finite, $\delta >0$.

Note that there is $x'\in \mathbb {Z}_+$ such that for each $x\in \mathbb {Z}_+$ with $x \ge x'$ and each $k\in N$, $v_k(x+1)-v_k(x) \le v^\infty _k+ \delta /2$. Then, for each $x\in \mathbb {Z}_+$ with $x\ge x'$ and each $k\in N {\setminus }\{j\}$,

$$\begin{aligned} v_j(x+1)-v_j(x)\ge v^\infty _j> \delta /2+ v^\infty _k \ge v_k(x+1)-v_k(x). \end{aligned}$$

However, since $v\in \mathrm {supp}_{\Phi _i}(\mathcal {V}^N)$, this inequality contradicts Assumption 3. $\square$

Step 2

$t^f _i(v) < x^f _i(v) \cdot v^\infty + \epsilon$.

Proof

Let $j\in \mathop {\mathrm{arg~max}}\limits _{ k\in N} x _k^f(v)$. Since $\overline{ x }\ge |N|\cdot x ^*(\epsilon )$ and $\sum _{k\in N}x^f_k(v)=\overline{x}$, $x^f _j(v)\ge x ^*(\epsilon )$. Then, by (1), $x^f _j(v)\cdot ( v_j(x^f_j(v))-v_j(x^f_j(v)-1)-v^\infty _i)< \epsilon$. Thus, by the definition of p(v),

$$\begin{aligned} p(v)\le v_j(x^f_j(v))-v_j(x^f_j(v)-1) < v^\infty _j+ \frac{\epsilon }{x^f _j(v)} \le v^\infty + \frac{\epsilon }{x^f _j(v)}. \end{aligned}$$

Hence,

$$\begin{aligned} t^f_i(v)= p (v)\cdot x _i^f(v) <x^f _i(v)\cdot v^\infty + \frac{\epsilon \cdot x^f _i(v)}{x^f _j(v) } \le x^f _i(v)\cdot v^\infty + \epsilon . \end{aligned}$$

$\square$

Step 3

There is $(x_j)_{j\in N}\in P(v_{-i})$ such that for each $j\in N_{-i}$, $x_j\ge x^f_j(v)$.

Proof

Suppose by contradiction that for each $(x_j)_{j\in N}\in P(v_{-i})$, $x_j<x^f_j(v)$ for some $j\in N_{-i}$. Let

$$\begin{aligned} P^*:=\mathop {\mathrm{arg~min}}\limits _{(x_j)_{j\in N}\in P(v_{-i})} |\{j\in N_{-i}: x_j<x^f_j(v)\}|. \end{aligned}$$

Since N is finite, $P^*\ne \emptyset$. Let $j\in N_{-i}$ be such that $x_j<x^f_j(v)$ for some $(x_j)_{j\in N}\in P^*$, and let $P^*(j):=\{(x_k)_{k\in N}\in P^*:x_j<x^f_j(v)\}$. Let

$$\begin{aligned} (x_k)_{k\in N}\in \mathop {\mathrm{arg~max}}\limits _{(y_k)_{k\in N}\in P^*(j)}y_j. \end{aligned}$$

Claim 1

There is $k\in N {\setminus } \{i,j\}$ such that $x_k>x^f_k(v)$.

Proof

Suppose by contradiction that for each $k\in N {\setminus } \{i,j\}$, $x_k\le x^f_k(v)$. Then, by $x_j<x^f_j(v)$, $\sum _{k\in N_{-i}}x_k<\sum _{k\in N_{-i}}x^f_k(v)\le \overline{x}$. Let $(y_k)_{k\in N}\in A$ be such that $y_i=0$, $y_j=x_j+1$, and for each $k\in N {\setminus } \{i, j\}$, $y_k=x_k$. Note that $(y_k)_{k\in N}$ is feasible because $\sum _{k\in N}y_k=1+\sum _{k\in N_{-i}}x_k\le \overline{x}$. Moreover, by $v_j(y_j)=v_j(x_j+1)\ge v_j(x_j)$, $\sum _{k\in N_{-i}}v_k(y_k)\ge \sum _{k\in N_{-i}}v_k(x _k )$. Thus, $(x_k)_{k\in N}\in P(v_{-i})$ implies $(y_k)_{k\in N}\in P(v_{-i})$.

By $x_j<x^f_j(v)$, we have either $y_j=x^f_j(v)$ or $y_j<x^f_j(v)$. If $y_j=x^f_j(v)$, then

$$\begin{aligned} |\{k\in N_{-i}:y_k<x^f_k(v)\}|<|\{k\in N_{-i}:x_k<x^f_k(v)\}|, \end{aligned}$$

which contradicts $(x_k)_{k\in N}\in P^*$. Thus, $y_j<x^f_j(v)$. This implies $(y_k)_{k\in N}\in P^*(j)$. By $y_j>x_j$, however, this also contradicts the definition of $(x_k)_{k\in N}$. $\square$

Claim 2

$v_j(x_j+1)-v_j(x_j)<v_k(x_k)-v_k(x_k-1)$.

Proof

Suppose by contradiction that $v_j(x_j+1)-v_j(x_j)\ge v_k(x_k)-v_k(x_k-1)$. Let $(y_\ell )_{\ell \in N}\in A$ be such that $y_i=0$, $y_j=x_j+1$, $y_k=x_k-1$, and for each $\ell \in N {\setminus } \{i,j,k\}$, $y_{\ell }=x_{\ell }$. Note that $(y_\ell )_{\ell \in N}$ is feasible because $\sum _{\ell \in N}y_{\ell }=\sum _{\ell \in N_{-i}}x_{\ell } \le \overline{x}$. Moreover,

$$\begin{aligned} \sum _{\ell \in N_{-i}}v_{\ell }(y_{\ell })=v_j(x_j+1)+v_k(x_k-1)+\sum _{\ell \in N {\setminus }\{i,j,k\}}v_{\ell }(x_{\ell })\ge \sum _{\ell \in N_{-i}}v_{\ell }(x_{\ell }). \end{aligned}$$

By $(x_\ell )_{\ell \in N}\in P(v_{-i})$, we have $(y_\ell )_{\ell \in N}\in P(v_{-i})$.

By $x_j<x^f_j(v)$, we have either $y_j=x^f_j(v)$ or $y_j<x^f_j(v)$. If $y_j=x^f_j(v)$, then

$$\begin{aligned} |\{\ell \in N_{-i}:y_{\ell }<x^f_{\ell }(v)\}|<|\{\ell \in N_{-i}:x_{\ell }<x^f_{\ell }(v)\}|, \end{aligned}$$

which contradicts $(x_\ell )_{\ell \in N}\in P^*$. Thus, $y_j<x^f_j(v)$. This implies $(y_k)_{k\in N}\in P^*(j)$. By $y_j>x_j$, however, this also contradicts the definition of $(x_\ell )_{\ell \in N}$. $\square$

By $x_j<x^f_j(v)$, non-increasing incremental valuations of $v_j$, Claims 1 and ,

$$\begin{aligned} v_j(x^f_j(v))-v_j(x^f_j(v)-1)\le & {} v_j(x_j+1)-v_j(x_j)\\< & {} v_k(x_k)-v_k(x_k-1)\\\le & {} v_k(x^f_k(v)+1)-v_k(x^f_k(v)). \end{aligned}$$

This contradicts Remark 1. $\square$

Let

$$\begin{aligned} P^*(v_{-i})&:=\{(x_j)_{j\in N}\in P(v_{-i}):\text { for each }j\in N_{-i}, x_j\ge x^f_j(v)\}. \end{aligned}$$

By Step 3, $P^*(v_{-i})\ne \emptyset$.

Step 4

Let $(x_j)_{j\in N}\in P^*(v_{-i})$ and $j\in N _{ -i}$. If $x _j> x^f _j(v)$, $v _j(x _j)-v _j(x _j-1) \ge v^\infty$.

Proof

Suppose, by contradiction, that $x _j> x^f _j(v)$ and $v _j(x _j)-v _j(x _j-1) < v^\infty$. By Step , $N ^* {\setminus }\{i\}\ne \emptyset$. Let $k\in N ^* {\setminus }\{i\}$. Then, $v _k(x _k+1)-v _k(x _k) \ge v^\infty$, and thus, $k\ne j$. Thus, $v _k(x _k+ 1)-v _k(x _k) \ge v^\infty > v _j(x _j)-v _j(x _j-1)$. By $(x_\ell )_{\ell \in N}\in P^*(v_{-i})$, this inequality contradicts Remark 1. $\square$

Step 5

Completing the proof.

Let $(x_j)_{j\in N}\in P^{*}(v_{-i})$. Without loss of generality, we assume that $\sum _{j\in N_{-i}}x_j=\overline{x}$. Note that $t^g_i(v)=\sum _{j\in N_{-i}}v_j(x_j)-\sum _{j\in N_{-i}}v_j(x^g_j(v))=\sum _{j\in N_{-i}}v_j(x_j)-\sum _{j\in N_{-i}}v_j(x^f_j(v))$. Thus,

$$\begin{aligned} t^f _i(v)-t^g _i(v)&< x^f_i(v)\cdot v^\infty + \epsilon -\Biggl (\sum _{j\in N_{-i}}v_j(x_j)-\sum _{j\in N_{-i}}v_j(x^f_j(v))\Biggr )\\&= x^f_i(v)\cdot v^\infty + \epsilon \\&\quad -\sum _{j\in N_{-i},\ x_j>x^f_j(v)}(v_j(x_j)-v_j(x_j-1)+v_j(x_j-1)-v_k(x_j-2)\\&\quad +\cdots +v_j(x^f_j(v)+1)-v_j(x^f_j(v)))\\&\le x^f_i(v)\cdot v^\infty + \epsilon -\sum _{j\in N_{-i},\ x_j>x^f_j(v)}(v_j(x_j)-v_j(x_j-1)) \cdot (x_j-x^f_j(v))\\&\le x^f_i(v)\cdot v^\infty + \epsilon -\sum _{j\in N _{-i},\ x_j>x^f_j(v)}v^\infty \cdot (x_j-x^f_j(v)), \end{aligned}$$

where the first inequality follows from Step 2, the second inequality from non-increasing incremental valuations, and the last inequality from Step 4. Note that

$$\begin{aligned} \sum _{j\in N_{-i},\ x_j>x^f_j(v)}(x_j-x^f_j(v))&=\sum _{j\in N_{-i},\ x_j>x^f_j(v)}(x_j-x^f_j(v)) \\&\quad +\sum _{j\in N_{-i},\ x_j=x^f_j(v)}(x_j-x^f_j(v))\\&=\overline{x}-\sum _{j\in N_{-i}}x^f_j(v)\\&=x^f_i(v). \end{aligned}$$

Hence, $t^f _i(v)-t^g _i(v) < \epsilon$. $\square$

1.3 Appendix C: Proof of Corollary 1

Let $\epsilon \in \mathbb {R}_{++}$. Let $r\in \mathbb {R}_+$ and f be a uniform price auction with reserve price r. Let $p: \mathcal {V}^N\rightarrow \mathbb {R}$ be the price scheme associated with f. Note that there can be several uniform price auctions with reserve price r that have the same price scheme. However, they always assign each agent bundles that he finds indifferent. Thus, if truth-telling is an $\epsilon$-Bayesian Nash equilibrium in a uniform price auction with reserve price r, then the same conclusion holds for any other uniform price auction with reserve price r that has the same price scheme. Hence, without loss of generality, we can assume that f always assigns as many objects as possible, that is, for each $v\in \mathcal {V}^N$,

$$\begin{aligned} \sum _{i\in N}x^f_i(v)=\min \left\{ \overline{x}, \sum _{i\in N}|\{x\in X{\setminus } \{\overline{x}\}: v_i(x+1)-v_i(x)\ge p(v)\}|\right\} . \end{aligned}$$

(2)

Now, we construct a new model where there are two additional agents. Let $N':=\{1.\dots , n, n+1, n+2\}$, and $v^*\in \mathcal {V}^*$ be such that for each $x\in \mathbb {Z}_+$, $v^*(x)=x\cdot r$. Let $\mathcal {V}':=\mathcal {V}\cup \{v^*\}$. For each $i\in N'$, we denote i’s prior belief in the new setting by $\Phi '_i$ and its probability density function by $\varphi '_i$. For each $\mathcal {V}''\subseteq \mathcal {V}^N$, let $(\mathcal {V}'', v^*):=\{v\in (\mathcal {V}')^{N'}: v_N\in (\mathcal {V}'')^N \text { and }v_{n+1}=v_{n+2}=v^*\}$. For each $i\in N$ and each $\mathcal {V}''\subseteq \mathcal {V}^N$, we assume $\Phi '_i (\mathcal {V}'', v^*)=\Phi _i (\mathcal {V}'')$. Note that for each $i\in N$ and each $v\in (\mathcal {V}')^{N'}$, $v\in \mathrm {supp}_{ \Phi ' _i}((\mathcal {V}') ^{N'})$ if and only if $v_{n+1}=v_{n+2}=v^*$ and $v_N\in \mathrm {supp}_{ \Phi _i}(\mathcal {V} ^N)$. For each $i\in \{n+1, n+2\}$, we assume $\Phi '_i=\Phi '_1$. Note that in the new model, Assumptions 1, 2, and 3 are satisfied.

We define a mechanism g in the new model as follows. Let $v\in (\mathcal {V}')^{N'}$. If $v_{n+1}=v_{n+2}=v^*$, then

$$\begin{aligned}&\text {for each }i\in N,\ x^g_i(v)=x^f_i(v_N),\ x^g_{n+1}(v)=\overline{x}-\sum _{i\in N}x^g_i(v),\text { and }x^g_{n+2}(v)=0, \text { and}\\&\text {for each }i\in N', t^g_i(v)=x^g_i(v)\cdot p(v_N). \end{aligned}$$

Otherwise, g(v) is an allocation given by a uniform price auction (without a reserve price).

Claim 3

g is a uniform price auction in the new model.

Proof

Let $v\in (\mathcal {V}')^{N'}$. If $v_{n+1}\ne v^*$ or $v_{n+2}\ne v^*$, then by the definition of g, g(v) is an allocation given by a uniform price auction. Thus, suppose $v_{n+1}=v_{n+2}=v^*$. Denote $p:= p(v_N)$. We show that $p\in [V_{\overline{x}+1}(v), V_{\overline{x}}(v)]$ and for each $i\in N'$,

$$\begin{aligned}&|\{x \in X{\setminus }\{\overline{x}\}: v_i(x+1)-v_i(x)> p\}|\le x^g_i(v)\\&\quad \le |\{x \in X{\setminus }\{\overline{x}\}: v_i(x+1)-v_i(x)\ge p\}|. \end{aligned}$$

Note that the latter claim is done for each $i\in N$ because $x^g_i(v)=x^f_i(v_N)$. We have two cases.

Case 1:$p>r$. By $p\le \max \{r, V_{\overline{x}}(v_N)\}$ and $p>r$, $p\le V_{\overline{x}}(v_N)$. By $N\subseteq N'$, $V_{\overline{x}}(v)\ge V_{\overline{x}}(v_N)$. Thus, $p\le V_{\overline{x}}(v)$. By $v_{n+1}=v_{n+2}=v^*$ and the definition of $v^*$, if $V_{\overline{x}+1}(v_N)>r$, then $V_{\overline{x}+1}(v)=V_{\overline{x}+1}(v_N)\le p$. If $V_{\overline{x}+1}(v_N)\le r$, then $V_{\overline{x}+1}(v)=r<p$. Hence, $p\in [V_{\overline{x}+1}(v), V_{\overline{x}}(v)]$.

By $p\le V_{\overline{x}}(v_N)$, $\sum _{i\in N}|\{x\in X{\setminus } \{\overline{x}\}: v_i(x+1)-v_i(x)\ge p\}|\ge \overline{x}$. Thus, by (2), $\sum _{i\in N}x^f_i(v_N)=\overline{x}$. This implies that $x^g_{n+1}(v)=0$. By the definition of $v^*$ and $p>r$, $\{x\in X{\setminus }\{\overline{x}\}: v^*(x+1)-v^*(x)>p\}=\{x\in X {\setminus }\{\overline{x}\}: v^*(x+1)-v^*(x)\ge p\}=\emptyset$. Therefore, for each $i\in \{n+1, n+2\}$, by $v_i=v^*$ and $x^g_i(v)=0$, $|\{x\in X {\setminus }\{\overline{x}\}: v_i(x+1)-v_i(x)>p\}|\le x^g_i(v)\le |\{x\in X {\setminus }\{\overline{x}\}: v_i(x+1)-v_i(x)\ge p\}|$.

Case 2.$p=r$. By $v_{n+1}=v^*$ and the definition of $v^*$, $V_{\overline{x}}(v)\ge r=p$. By $r=p\ge V_{\overline{x}+1}(v_N)$ and $v_{n+1}=v^*$, $V_{\overline{x}+1}(v)=r$. Hence, $p\in [V_{\overline{x}+1}(v), V_{\overline{x}}(v)]$.

By the definition of $v^*$, $\{x\in X {\setminus }\{\overline{x}\}: v^*(x+1)-v^*(x)>r\}=\emptyset$ and $\{x\in X {\setminus }\{\overline{x}\}: v^*(x+1)-v^*(x)\ge r\}=X {\setminus } \{\overline{x}\}$. Hence, for each $i\in \{n+1, n+2\}$, by $p=r$ and $v_i=v^*$, $|\{x\in X {\setminus }\{\overline{x}\}: v_i(x+1)-v_i(x)>p\}|\le x^g_i(v)\le |\{x\in X {\setminus }\{\overline{x}\}: v_i(x+1)-v_i(x)\ge p\}|$. $\square$

By Claim 3 and Theorem 1, there is $\hat{x}\in \mathbb {Z}_+$ such that if $\overline{x}\ge \hat{x}$, truth-telling is an $\epsilon$-Bayesian Nash equilibrium in g.

Suppose $\overline{x}\ge \hat{x}$. Let $i\in N$ and $v_i, v'_i\in \mathcal {V}$. By the definition of $\Phi '_i$ and g,

$$\begin{aligned}&\int _{v_{-i}\in \mathcal {V}^{N {\setminus }\{i\}}}[v_i(x^f_i(v_i, v_{-i}))-t^f_i(v_i, v_{-i})]\varphi _i (v_{-i}|v_i)dv_{-i}\\&\quad =\int _{v'_{-i}\in (\mathcal {V}')^{N' {\setminus }\{i\}}}[v_i(x^g_i(v_i, v'_{-i}))-t^g_i(v_i, v'_{-i})]\varphi '_i (v'_{-i}|v_i)dv'_{-i}\\&\quad \ge \int _{v'_{-i}\in (\mathcal {V}')^{N' {\setminus }\{i\}}}[v_i(x^g_i(v'_i, v'_{-i}))-t^g_i(v'_i, v'_{-i})]\varphi '_i (v'_{-i}|v_i)dv'_{-i} - \epsilon \\&\quad =\int _{v_{-i}\in \mathcal {V}^{N {\setminus }\{i\}}}[v_i(x^f_i(v'_i, v_{-i}))-t^f_i(v'_i, v_{-i})]\varphi _i (v_{-i}|v_i)dv_{-i}-\epsilon . \end{aligned}$$

Hence, truth-telling is an $\epsilon$-Bayesian Nash equilibrium in f. $\square$

1.4 Appendix D: Proofs of Propositions 1 and 2

Proof of Proposition 1

By Assumption 2 and the assumption that each valuation function takes discrete values, there is $x ^*\in \mathbb {Z}_+$ such that for each $i\in N$, each $v \in \mathrm {supp}_{\Phi _i}(\mathcal {V} ^N)$, each $j\in N$, and each $x\in \mathbb {Z}_+$ with $x>x ^*$,

$$\begin{aligned} v_j(x+1)-v_j(x)= v^\infty _j. \end{aligned}$$

Let $\overline{x}>|N|\cdot x^*$. Let f be a uniform price auction. As we explained in the proof of Theorem 1, without loss of generality, we assume that for each $v\in \mathcal {V}^N$, $\sum _{i\in N}x^f_i(v)=\overline{x}$. Let g be a Vickrey auction such that $x^g=x^f$. Then, by following the proof of Theorem 1, we can show that for each $i\in N$, each $v\in \mathrm {supp}_{\Phi _i}(\mathcal {V} ^N)$, and each $j\in N$, $t^f _j(v)= t^g_j(v)$. Thus, by Proposition 3, we obtain the desired result. $\square$

Proof of Proposition 2

As we explained in the proof of Theorem 1, without loss of generality, we assume that for each $v'\in \mathcal {V}^N$, $\sum _{i\in N}x^f_i(v')=\overline{x}$. Let $i\in N$ and $v \in \mathrm {supp}_{\Phi _i}(\mathcal {V} ^N)$. Let $\epsilon \in \mathbb {R} _{ + + }$. By Assumption 1, for each $j\in N$, there is $x ^* _j(\epsilon ) \in \mathbb {Z}_+$ such that for each $x\in \mathbb {Z}_+$ with $x \ge x ^* _j(\epsilon )$,

$$\begin{aligned} x \cdot | v_j(x+1)-v_j(x)- v^\infty _j|< \epsilon . \end{aligned}$$

Let $\overline{x}> |N|\cdot \max _{j\in N} x ^* _j(\epsilon )$. Let g be a Vickrey auction such that $x^g=x^f$. Then, by following the proof of Theorem 1, we can show that for each $j\in N$, $t^f _j(v)-t^g _j(v)< \epsilon$. Therefore, by Proposition 4, we obtain the desired result. $\square$

1.5 Appendix E: Proof of Theorem  2

As we explained in the proof of Theorem 1, without loss of generality, we assume that for each $v\in \mathcal {V}^N$, $\sum _{i\in N}x^f_i(v)=\overline{x}$. Since $(N, v, \overline{ x })$ is a K-replica economy of $(N', v', \overline{ x }')$, there is $(N_i)_{i\in N'}\in (2^N)^{N'}$ such that (i) $\bigcup _{i\in N'}N_i=N$, (ii) for each pair $i,j\in N'$ with $i\ne j$, $N_i\cap N_j=\emptyset$, (iii) for each $i\in N'$, $|N_i|=K$, and (iv) for each $i\in N'$ and each $j\in N_i$, $v_j=v'_i$. Let $(x'_i)_{i\in N'}$ be an efficient object allocation in the subeconomy $(N', v', \overline{ x }')$, i.e.,

$$\begin{aligned} (x'_i)_{i\in N'} \in \mathop {\mathrm{arg~max}}\limits \left\{ \sum _{i\in N'}v'_i(x_i): (x_i)_{i\in N'}\in \{0,\dots ,\overline{x}'\}^{N'} \text { and }\sum _{i\in N'}x_i\le \overline{x}'\right\} . \end{aligned}$$

Without loss of generality, we assume $\sum _{i\in N'}x'_i=\overline{x}'$. Let $(x_i)_{i\in N}\in A$ be such that for each $i\in N$, $x_i=x'_j$ where $j\in N'$ and $i\in N_j$.

Step 1

$(x_i)_{i\in N}\in P(v)$.

Proof

Suppose by contradiction that $(x_i)_{i\in N}\notin P(v)$. Let

$$\begin{aligned} (y^*_i)_{i\in N}\in \mathop {\mathrm{arg~min}}\limits _{ (y_i)_{i\in N}\in P(v)}\sum _{ i\in N} |y _i-x _i|. \end{aligned}$$

By $(x_i)_{i\in N}\notin P(v)$ and $(y^*_i)_{i\in N}\in P(v)$, $\sum _{i\in N}v_i(y_i ^*)>\sum _{i\in N}v_i(x_i)$. Thus, there is $i\in N$ such that $v_i(y^*_i)>v_i(x_i)$. By monotonicity, $y^*_i>x_i$. By the definition of $(x_j)_{j\in N}$,

$$\begin{aligned} \sum _{j\in N}x_j=\sum _{j\in N'}K\cdot x_j'=K\cdot \overline{x}'=\overline{x}\ge \sum _{j\in N}y^*_j. \end{aligned}$$

Thus, by $y^*_i>x_i$, there is $j\in N {\setminus }\{i\}$ such that $y_j^*<x _j$. $\square$

Claim 4

$v_i(y_i^*)-v_i(y_i^*-1)=v_j(y_j^*+1)-v_j(y_j^*)$

Proof

By the definition of $(x_k)_{k\in N}$, there is $k\in N'$ such that $i\in N _k$ and $x_k'=x_i$. For the same reason, there is $\ell \in N'$ such that $j\in N_{\ell }$ and $x_{\ell }'=x_j$. Note that $x'_{\ell }=x_j>y^*_j\ge 0$. Thus, by the definition of $(x'_{i'})_{i'\in N'}$ and Remark 1, $v_k'(x_k'+1)-v_k'(x_k')\le v_{\ell }'(x_{\ell }')-v_{\ell }'(x_{\ell }'-1)$. Therefore,

$$\begin{aligned} \nonumber v_i(x_i+1)-v_i(x_i)&=v_k'(x_k'+1)-v_k'(x_k')\\ \nonumber&\le v_{\ell }'(x_{\ell }')-v_{\ell }'(x_{\ell }'-1)\\&=v_j(x_j)-v_j(x_j-1). \end{aligned}$$

(3)

By $(y ^*_{i'})_{i'\in N}\in P(v)$ and Remark 1, $v_i(y_i^*)-v_i(y_i^*-1)\ge v_j(y_j^*+1)-v_j(y_j^*)$. Thus, by $y_j^*<x_j$, (3), $y_i^*>x_i$, and non-increasing incremental valuations,

$$\begin{aligned} v_i(y_i^*)-v_i(y_i^*-1)&\ge v_j(y_j^*+1)-v_j(y_j^*)\\&\ge v_j(x_j)-v_j(x_j-1)\\&\ge v_i(x_i+1)-v_i(x_i)\\&\ge v_i(y_i^*)-v_i(y_i^*-1). \end{aligned}$$

Therefore, $v_i(y_i^*)-v_i(y_i^*-1)=v_j(y_j^*+1)-v_j(y_j^*)$. $\square$

Let $(y_k)_{k\in N}\in A$ be such that for each $k\in N$,

$$\begin{aligned} y_k= {\left\{ \begin{array}{ll} y_i^*-1 &{}\text {if }k=i,\\ y_j^*+1 &{}\text {if }k=j,\\ y^*_k&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

By Claim 4, $v _i(y _i ^*-1)+v _j(y _j ^*+ 1)=v _i(y _i ^*)+v_j(y _j ^*)$. Thus,

$$\begin{aligned} \sum _{k\in N}v_k(y_k)=v_i(y_i^*-1)+v_j(y_j^*+1)+\sum _{k\in N {\setminus } \{i,j\}}v_k(y_k^*) =\sum _{k\in N}v_k(y_k^*). \end{aligned}$$

Thus, by $(y^*_k)_{k\in N}\in P(v)$, we have $(y_k)_{k\in N}\in P(v)$. Moreover, by $y_i^*>x_i$ and $y_j^*<x_j$,

$$\begin{aligned} \sum _{k\in N}|y_k-x_k|&=|y_i^*-1-x_i|+|y_j^*+1-x_j|+\sum _{k\in N {\setminus } \{i,j\}}|y_k^*-x_k|\\&=-2+\sum _{k\in N}|y_k^*-x_k|\\&<\sum _{k\in N}|y_k^*-x_k|. \end{aligned}$$

This contradicts the definition of $(y^*_k)_{k\in N}$. $\square$

Let

$$\begin{aligned}&N^*:=\{i\in N: v_i(x_i+1)-v_i(x_i)=V_{\overline{x}+1}(v)\}. \end{aligned}$$

Step 2

$N^*\ne \emptyset$.

Proof

By Step 1, Remark 1, and non-increasing incremental valuations, for each $i\in N$, each $j\in N$, and each $x \in X {\setminus } \{0\}$ with $x\le x_i$,

$$\begin{aligned} v_i(x)-v_i(x-1) \ge v_i(x_i)-v_j(x_i-1)\ge v_j(x_j+1)-v_j(x_j) \end{aligned}$$

Hence, by $\sum _{i\in N}x_i=\overline{x}$ and non-increasing incremental valuations, $\max _{i\in N} \{v_i(x_i+1)-v_i(x_i)\} =V_{\overline{x}+1}(v)$, which completes the proof. $\square$

Note that for each $i\in N^*$, there are at least $K-1$ other agents who have the same valuation function and object assignment at $(x_j)_{j\in N}$ as agent i. Thus, $|N^*|\ge K$. Fix $i\in N$ and let

$$\begin{aligned}&N_1:=\{j\in N {\setminus }\{i\}:x^f_j(v)=x_j\},\\&N_2:=\{j\in N {\setminus }\{i\}:x^f_j(v)>x_j\},\text { and}\\&N_3:=\{j\in N {\setminus }\{i\}:x^f_j(v)<x_j\}. \end{aligned}$$

Let $N^*_j:=N_j\cap N^*$ for each $j= 1,2,3$.

Step 3

$x^f_i(v)\le |N^*_1|+\sum _{j\in N^*_3}(x_j-x^f_j(v)+1)+\sum _{j\in N_3 {\setminus } N^*_3}(x_j-x^f_j(v))$.

Proof

Note that $N^* {\setminus } \{i\}=N^*_1\cup N^*_2\cup N^*_3$. Thus, by $|N^*|\ge K$ and $K> \overline{ x }'\ge x_i$,

$$\begin{aligned} |N^*_1|+|N^*_2|+|N^*_3|=|N^* {\setminus } \{i\}|\ge x_i. \end{aligned}$$

(4)

By $N_2\supseteq N^*_2$ and the definition of $N_2$,

$$\begin{aligned} |N^*_2|\le |N_2|\le \sum _{j\in N_2}(x^f_j(v)-x_j). \end{aligned}$$

By $\sum _{j\in N}x^f_j(v)=\overline{x}=\sum _{j\in N}x_j$,

$$\begin{aligned}&x^f_i(v)+\sum _{j\in N_1}x^f_j(v)+\sum _{j\in N_2}x^f_j(v)+\sum _{j\in N_3}x^f_j(v) =\sum _{j\in N}x^f_j(v) =\sum _{j\in N}x_j =x_i \\&\quad +\sum _{j\in N_1}x_j+\sum _{j\in N_2}x_j+\sum _{j\in N_3}x_j. \end{aligned}$$

By $\sum _{j\in N_1}x^f_j(v)=\sum _{j\in N_1}x_j$ and $|N^*_2|\le \sum _{j\in N_2}(x^f_j(v)-x_j)$,

$$\begin{aligned} x^f_i(v)-x_i=\sum _{j\in N_3}(x_j-x^f_j(v))-\sum _{j\in N_2}(x^f_j(v)-x_j)\le \sum _{j\in N_3}(x_j-x^f_j(v))-|N^*_2|. \end{aligned}$$

(5)

Therefore, by (4) and (5),

$$\begin{aligned} x^f_i(v)&=x_i+(x^f_i(v)-x_i)\\&\le |N^*_1|+|N^*_2|+|N^*_3|+\sum _{j\in N_3}(x_j-x^f_j(v))-|N^*_2|\\&\le |N^*_1|+|N^*_3|+ \sum _{j\in N_3}(x_j-x^f_j(v))\\&\le |N^*_1|+\sum _{j\in N^*_3}(x_j-x^f_j(v)+1)+\sum _{j\in N_3 {\setminus } N^*_3}(x_j-x^f_j(v)). \end{aligned}$$

$\square$

By Step 3, there is $(y^*_j)_{j\in N^*_1\cup N_3}\in X^{|N^*_1\cup N_3|}$ such that $\sum _{j\in N^*_1\cup N_3}y ^*_j= x^f_i(v)$ and for each $j\in N^*_1\cup N_3$,

$$\begin{aligned} y^*_j\le {\left\{ \begin{array}{ll} 1&{}\text {if }j\in N^*_1,\\ x _j-x^f _j(v)+ 1&{}\text {if }j\in N^*_3,\\ x_j-x^f_j(v)&{}\text {if }j\in N_3 {\setminus } N^*_3. \end{array}\right. } \end{aligned}$$

Let $(y_j)_{j\in N}\in X^N$ be such that for each $j\in N$,

$$\begin{aligned} y_j= {\left\{ \begin{array}{ll} 0&{}\text {if }j=i,\\ x^f_j(v)+y^*_j&{}\text {if }j\in N^*_1\cup N_3,\\ x^f_j(v)&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Note that

$$\begin{aligned} \sum _{ j\in N} y _j= \sum _{ j\in N _1 ^*\cup N _3} y _j ^* + \sum _{ j\in N_{-i}}x^f _j(v)=x^f_i(v)+\sum _{ j\in N_{-i}}x^f _j(v)= \overline{ x }. \end{aligned}$$

Thus, $(y_j)_{j\in N}\in A$. Moreover, for each $j\in N_{-i}$, $y _j \ge x^f _j(v)$.

Step 4

Let $j\in N_{-i}$ be such that $y_j>x^f_j(v)$. For each $x\in \{x^f_j(v)+1,\dots , y_j\}$, $v_j(x)-v_j(x-1)=V_{\overline{x}+1}(v)$.

Proof

By the definition of f and non-increasing incremental valuations,

$$\begin{aligned} V_{\overline{x}+1}(v)\ge v_j(x^f_j(v)+1)-v_j(x^f_j(v)) \ge v_j(y_j)-v_j(y_j-1). \end{aligned}$$

Thus, we complete the proof if we show $V_{\overline{x}+1}(v)\le v_j(y_j)-v_j(y_j-1)$. Note that by $y_j>x^f_j(v)$, we have $j\in N^*_1\cup N_3$.

Case 1:$j\in N^*_1$. By the definition of $y^*$, we have $y_j= x^f_j(v)+y^*_j\le x_j+1$. By $y_j>x^f_j(v)=x_j$, we have $y_j=x_j+1$. Thus, by $j\in N^*$,

$$\begin{aligned} V_{\overline{x}+1}(v)=v_j(x_j+1)-v_j(x_j)= v_j(y_j)-v_j(y_j-1). \end{aligned}$$

Case 2:$j\in N^*_3$. By the definition of $y^*$, we have $y_j=x^f_j(v)+y^*_j\le x_j+1$. Thus, by $j\in N^*$ and non-increasing incremental valuations,

$$\begin{aligned} V_{\overline{x}+1}(v)=v_j(x_j+1)-v_j(x_j) \le v_j(y_j)-v_j(y_j-1). \end{aligned}$$

Case 3:$j\in N_3 {\setminus } N^*_3$. By $N^*\ne \emptyset$ and $j\in N_3 {\setminus } N^*_3$, there is $k\in N^*$ such that $k\ne j$. Note that by $j\in N_3$, $x_j>x^f_j(v)\ge 0$. Thus, by $k\in N^*$, $(x_\ell )_{\ell \in N}\in P(v)$ and Remark 1,

$$\begin{aligned} V_{\overline{x}+1}(v)=v_k(x_k+1)-v_k(x_k)\le v_j(x_j)-v_j(x_j-1). \end{aligned}$$

By the definition of $y^*$, $y_j=x^f_j(v)+y^*_j\le x_j$. Thus, by non-increasing incremental valuations,

$$\begin{aligned} V_{\overline{x}+1}(v)\le v_j(x_j)-v_j(x_j-1) \le v_j(y_j)-v_j(y_j-1). \end{aligned}$$

$\square$

Step 5

Completing the proof.

Let g be a Vickrey auction such that $x^g=x^f$. By Step 4, for each $j\in N_{-i}$ with $y_j>x^f_j(v)$,

$$\begin{aligned} v_j(y_j)-v _j(x^f_j(v))&=v_j(y_j)-v_j(y_j-1)+(v_j(y_j-1)-v_j(y_j-2)) \\&\quad +\cdots +(v_j(x^f_j(v)+1)-v_j(x^f_j(v)))\\&=V_{\overline{x}+1}(v)\cdot (y_j-x^f_j(v)). \end{aligned}$$

By the definition of y,

$$\begin{aligned} \sum _{j\in N_{-i}}(y_j-x^f_j(v))=\sum _{j\in N_1^*\cup N_3}y^*_j=x^f_i(v). \end{aligned}$$

Thus, we have

$$\begin{aligned} t^g_i(v)= & {} \max _{(x''_j)_{j\in N}\in A}\sum _{j\in N_{-i}}v_j(x''_j)-\sum _{j\in N_{-i}}v_j(x^g_j(v))\\\ge & {} \sum _{j\in N_{-i}}v_j(y_j)-\sum _{j\in N_{-i}}v_j(x^f_j(v))\\= & {} \sum _{j\in N_{-i}}(v_j(y_j)-v_j(x^f_j(v)))\\= & {} V_{\overline{x}+1}(v)\cdot \sum _{j\in N_{-i}}(y_j-x^f_j(v))\\= & {} V_{\overline{x}+1}(v)\cdot x^f _i(v)\\= & {} t^f_i(v). \end{aligned}$$

Hence, by Proposition 4, we obtain the desired result. $\square$