Undominated mechanisms and the provision of a pure public good in two agent economies.

Abstract

We revisit the problem of providing a pure public good in economies involving two agents, under the assumption that preferences are quasilinear. We characterize the class of strategy-proof, anonymous and feasible mechanisms that are not dominated by another strategy-proof, anonymous and feasible mechanism. Mechanisms that decide, at some profiles of preferences, against the alternative that is preferred by all agents are members of this class. On the contrary, Groves mechanisms, with the sole exception of the Pivotal mechanism, are not. Finally, the only budget-balanced mechanism that belongs to this class is the Unanimity mechanism.

Appendix

1.1 Proof of Proposition 1

Proof

The proof of the if part of our statement is straightforward and thus omitted. The proof of only if part involves three preliminary steps.

Step 1. If a mechanism $\varphi =(f,t)$ satisfies Strategy-proofness then for each economy $(\theta _{1},\theta _{2})\in \mathbb {R}^{2},$

$$\begin{aligned}&f(\theta _{1},\theta _{2})=1 \implies \forall i\in N,\; \forall \theta _{i}^{\prime }> \theta _{i},\; \varphi _{i}\left( \theta _{i}^{\prime },\theta _{-i}\right) =\varphi \left( \theta _{1},\theta _{2}\right) ,\\&f(\theta _{1},\theta _{2})=0 \implies \forall i\in N,\; \forall \theta _{i}^{\prime }< \theta _{i},\; \varphi _{i}\left( \theta _{i}^{\prime },\theta _{-i}\right) =\varphi \left( \theta _{1},\theta _{2}\right) . \end{aligned}$$

For each bundle $(\lambda ,z) \in \{0,1\}\times \mathbb {R}$ and each $\theta \in \mathbb {R}$ let

$$\begin{aligned} LC\big ((\lambda ,z);\theta \big )= & {} \big \{(\lambda ',z') \in \{0,1\}\times \mathbb {R}:u((\lambda ,z);\theta )\ge u((\lambda ',z');\theta )\big \},\\ UC\big ((\lambda ,z);\theta \big )= & {} \big \{(\lambda ',z') \in \{0,1\}\times \mathbb {R}:u((\lambda ,z);\theta )\le u((\lambda ',z');\theta )\big \}, \end{aligned}$$

denote respectively the lower and upper contour sets at the bundle $(\lambda ,z)$ from the point of view of preference parameter $\theta$. Consider some arbitrary $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $i\in N$ and $\theta ^{\prime }_{i}\in \mathbb {R}$, with $\theta ^{\prime }_{i}\ne \theta _{i}$. By Strategy-proofness,

$$\begin{aligned} \varphi _{i}(\theta _{i}^{\prime },\theta _{-i})\in LC\big (\varphi _{i}(\theta _{1},\theta _{2});\theta _{i}\big )\;\text {and}\; \varphi _{i}(\theta _{i}^{\prime },\theta _{-i})\in UC\big (\varphi _{i}(\theta _{1},\theta _{2});\theta _{i}^{\prime }\big ) \end{aligned}$$

(6.1)

If $f(\theta _{1},\theta _{2})=1$ and $\theta _{i}^{\prime }> \theta _{i}$ or if $f(\theta _{1},\theta _{2})=0$ and $\theta _{i}^{\prime }< \theta _{i}$ one has that $LC\big (\varphi _{i}(\theta _{1},\theta _{2});\theta _{i}\big )\subseteq LC\big (\varphi _{i}(\theta _{1},\theta _{2});\theta _{i}^{\prime }\big )$. Hence,

$$\begin{aligned} LC\big (\varphi _{i}(\theta _{1},\theta _{2});\theta _{i}\big )\cap UC\big (\varphi _{i}(\theta _{1},\theta _{2});\theta _{i}^{\prime }\big )=\{\varphi _{i}(\theta _{1},\theta _{2})\}. \end{aligned}$$

Hence, by 6.1, $\varphi _{i}(\theta _{1},\theta _{2})=\varphi _{i}(\theta _{i}^{\prime },\theta _{-i})$.

Step 2. If a mechanism $\varphi =(f,t)$ satisfies Strategy-proofness then for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$,

$$\begin{aligned}&f(\theta _{1},\theta _{2})=1 \implies \forall \left( \theta ^{\prime }_{1},\theta ^{\prime }_{2}\right) \ge (\theta _{1},\theta _{2}),\; f\left( \theta ^{\prime }_{1},\theta ^{\prime }_{2}\right) =1,\\&f(\theta _{1},\theta _{2})=0 \implies \forall \left( \theta ^{\prime }_{1},\theta ^{\prime }_{2}\right) \le (\theta _{1},\theta _{2}),\; f\left( \theta ^{\prime }_{1},\theta ^{\prime }_{2}\right) =0. \end{aligned}$$

Suppose that for some pair $(\theta ^{\prime }_{1},\theta ^{\prime }_{2}),(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and some $j \in N$, $\theta ^{\prime }_{j} > \theta _{j}$ and $\theta ^{\prime }_{-j}=\theta _{-j}$. By Step 1,

$$\begin{aligned} f(\theta _1,\theta _2)=1 \Rightarrow f\left( \theta ^{\prime }_j,\theta _{-j}\right) =1 \Rightarrow f\left( \theta '_1,\theta '_2\right) =1. \end{aligned}$$

A similar argument proves the remaining case.

Step 3. If $\varphi =(f,t)$ satisfies Strategy-proofness, then for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, each $i\in N$ and each $x,y\in \mathbb {R}$, with $x<y$,

• if $f(x,\theta _{-i})=f(y,\theta _{-i})$, then $t_{i}(y,\theta _{-i})=t_{i}(x,\theta _{-i})$,

• if $f(x,\theta _{-i})<f(y,\theta _{-i})$, then $t_{i}(y,\theta _{-i})=t_{i}(x,\theta _{-i})-\inf \big \{\theta _{i}\in \mathbb {R}\; :\; f(\theta _{i},\theta _{-i})=1\big \}$.

Consider some arbitrary profile $(\theta '_1,\theta '_2) \in \mathbb {R}^{2}$, some $i\in N$ and some pair $x,y\in \mathbb {R}$, with $x<y$.

Case 1. $f(x,\theta ^{\prime }_{-i})=f(y,\theta ^{\prime }_{-i})$. By Step 1, $t_{i}(y,\theta ^{\prime }_{-i})=t_{i}(x,\theta ^{\prime }_{-i})$.

Case 2. $f(x,\theta ^{\prime }_{-i})<f(y,\theta ^{\prime }_{-i})$. Let $p^{\prime }=\inf \big \{\theta \in \mathbb {R} : f(\theta ,\theta ^{\prime }_{-i})=1\big \}$. By assumption, $p^{\prime }\in [x,y]$. Suppose first that for some $\mu >0$, $t_{i}(y,\theta ^{\prime }_{-i})=t_{i}(x,\theta ^{\prime }_{i})-p^{\prime }+\mu$. Note that, by Step 1, for each $\theta _{i}>p^{\prime }$, $f(\theta _{i},\theta ^{\prime }_{-i})=1$. Moreover, again by Step 1,

$$\begin{aligned} \text {for each } \theta >p^{\prime },\;t_{i}\big (\theta ,\theta ^{\prime }_{-i}\big )=t_{i}\big (y,\theta ^{\prime }_{-i}\big ). \end{aligned}$$

(6.2)

Hence, for some $\epsilon \in (0,\mu )$, we obtain

$$\begin{aligned} \begin{array}{lcll} u\Big (\varphi _{i}\big (p^{\prime }+\epsilon ,\theta ^{\prime }_{-i}\big );p^{\prime }-\epsilon \Big ) &{} = &{} p^{\prime }-\epsilon +t_{i}\big (p^{\prime }+\epsilon ,\theta ^{\prime }_{-i}\big ) &{} \text {[by construction],} \\ &{} = &{} p^{\prime }-\epsilon +t_{i}\big (y,\theta ^{\prime }_{-i}\big )&{} \text {[from 6.2],}\\ &{} = &{} p^{\prime }-\epsilon +t_{i}\left( x,\theta ^{\prime }_{-i}\right) -p^{\prime }+\mu , &{} \text {[by assumption],}\\ &{}=&{}u\big (\varphi _{i}(p^{\prime }-\epsilon ,\theta ^{\prime }_{-i});p^{\prime }-\epsilon \big )+(\mu -\epsilon ). \end{array} \end{aligned}$$

By assumption $\epsilon <\mu$, therefore Strategy-proofness is violated, a contradiction. If for some $\mu >0$, $t_{i}(y,\theta ^{\prime }_{-i})=t_{i}(x,\theta ^{\prime }_{-i})-p^{\prime }-\mu$ we would arrive at the same contradiction by a similar argument.

Thus, $t_{i}(y,\theta ^{\prime }_{-i})=t_{i}(x,\theta ^{\prime }_{-i})-p^{\prime }$.

We now turn to the definition of $g:\mathbb {R}\rightarrow \mathbb {R}$. By Step 3, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, each $i\in N$ and some constant $c(\theta _{-i})\in \mathbb {R}$,

$$\begin{aligned}&f(\mathbb {R},\theta _{-i})=\{0,1\}\implies \text {for each } y\in \{x\in \mathbb {R}\; :\; f(x,\theta _{-i})=0\},\; t_{i}(y,\theta _{-i})=c(\theta _{-i}),\\&\text {and }f(\mathbb {R},\theta _{-i})=k \in \{0,1\}\implies \text {for each } x\in \mathbb {R},\; t_{i}(x,\theta _{-i})=c(\theta _{-i}). \end{aligned}$$

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i\in N$, let $g(\theta _{-i})=c(\theta _{-i})$. Hence, by Step 3, we recover the structure of transfers as they appear in the statement.

1.2 Proof of Theorem 1

1.2.1 Necessity

Lemma 1

If the mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), satisfies Feasibility, Strategy-proofness and Anonymity then $\varphi$ is Undominated only if there exist $y,z\in \mathbb {R}$, with $y<z$, such that $f(y,y)<f(z,z)$.

Proof

Suppose, by way of contradiction, that for each $x\in \mathbb {R}$, $f(x,x)=k\in \{0,1\}$. Let $k=0$. A similar argument applies if $k=1$. Let $f^{\prime }\in \mathcal {F}$ be such that $f^{\prime }(\theta _{1},\theta _{2})=1$ if and only if $(\theta _{1},\theta _{2})\ge (0,0)$. By construction, $f^{\prime } \in \mathcal {F}$. Hence, by Proposition 1, the mechanism $\varphi ^{\prime }=(f^{\prime },t^{\prime })$, associated with payment scheme $(p^{\prime },g)$, is strategy-proof and anonymous. Moreover, by construction, for each $\theta \in \mathbb {R}$, $p(\theta )=p^{\prime }(\theta )=0$. Hence, by construction, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i\in N$, $t_{i}(\theta _{1},\theta _{2})=t^{\prime }_{i}(\theta _{1},\theta _{2})$. Hence, since $\varphi$ is feasible so is $\varphi ^{\prime }$. It can be readily verified that $\varphi ^{\prime }$ dominates $\varphi$ which yields the contradiction.

Lemma 2

If the mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), satisfies Strategy-proofness and Anonymity then $\varphi$ is Feasible only if, for each $\theta \in \mathbb {R}$,

$$\begin{aligned} \begin{array}{cll} \theta \le \beta ^{f} \text { and } f\left( \beta ^{f},\beta ^{f}\right) =0 &{} \implies &{} g(\theta )\le 0 \\ \theta \ge \beta ^{f} \text { and } f\left( \beta ^{f},\beta ^{f}\right) =1&{} \implies &{} g(\theta )\le p(\theta ). \end{array} \end{aligned}$$

Proof

Suppose first that $\theta '\le \beta ^{f}$ and $f(\beta ^{f},\beta ^{f})=0$. If $\theta '<\beta ^{f}$, by the definition of $\beta ^{f}$, $f(\theta ',\theta ')=0$. Hence, $f(\theta ',\theta ')=0$. By Proposition 1, $\sum _{i\in N}t_{i}(\theta ',\theta ')=2g(\theta ')$. By Feasibility, $g(\theta ')\le 0$.

Suppose next that $\theta '\ge \beta ^{f}$ and $f(\beta ^{f},\beta ^{f})=1$. If $\theta '>\beta ^{f}$, by the definition of $\beta ^{f}$, $f(\theta ',\theta ')=1$. Hence, $f(\theta ',\theta ')=1$. By Proposition 1, $\sum _{i\in N}t_{i}(\theta ',\theta ')=2\big (g(\theta ')-p(\theta ')\big )$. By Feasibility, $g(\theta ')\le p(\theta ')$.

Lemma 3

If $f,f'\in \mathcal {F}$ are such that $\beta ^{f}=\beta ^{f'}=\beta$, $f(\beta ,\beta )>f'(\beta ,\beta )$ and for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\setminus \{(\beta ,\beta )\}$, $f(\theta _{1},\theta _{2})=f'(\theta _{1},\theta _{2})$, then the mechanism $\varphi =(f,t)$, associated with payment scheme (p, g), is welfare equivalent with the mechanism $\varphi '=(f',t')$, associated with payment scheme $(p',g)$.

Proof

Since for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\setminus \{(\beta ,\beta )\}$, $f(\theta _{1},\theta _{2})=f'(\theta _{1},\theta _{2})$, for each $i\in N$,

$$\begin{aligned} \text {for each } \theta _{-i}\in \mathbb {R}\setminus \{\beta \}, \; f(\mathbb {R},\theta _{-i})=f'(\mathbb {R},\theta _{-i}) \text { and }f(\beta ,\theta _{-i})=f'(\beta ,\theta _{-i}). \end{aligned}$$

(6.3)

Hence,

$$\begin{aligned} \text {for each }\theta \in \mathbb {R},\; p(\theta )=p'(\theta ). \end{aligned}$$

(6.4)

Hence, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\setminus \{(\beta ,\beta )\}$ and each $i\in N$,

$$\begin{aligned} u\big (\varphi ^{\prime }_{i}(\theta _1,\theta _{2});\theta _{i}\big )=u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big ). \end{aligned}$$

At $(\beta ,\beta )$, by assumption, $f(\beta ,\beta )>f'(\beta ,\beta )$. Hence, for each $i\in N$,

$$\begin{aligned}&u\big (\varphi _{i}(\beta ,\beta ) ; \beta \big )=\beta -p(\beta )+g(\beta ),\text { and}\\&u\big (\varphi '_{i}(\beta ,\beta ) ; \beta \big )=g(\beta ). \end{aligned}$$

Moreover, by 6.3, the assumption that $f(\beta ,\beta )>f'(\beta ,\beta )$ and the fact that $f,f^{\prime }\in \mathcal {F}$ are weakly monotonic, we obtain

$$\begin{aligned} f(\beta ,\beta )=1\implies p(\beta )\le \beta \text { and }f^{\prime }(\beta ,\beta )=0\implies p'(\beta )\ge \beta . \end{aligned}$$

By 6.4, $p(\beta )=p'(\beta )$. Hence, $p(\beta )=\beta$. Hence, for each $i\in N$,

$$\begin{aligned} u\big (\varphi _{i}(\beta ,\beta ) ; \beta \big )=u\big (\varphi '_{i}(\beta ,\beta ) ; \beta \big ). \end{aligned}$$

Lemma 4

If the mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), satisfies Feasibility, Strategy-proofness, Anonymity and $\beta ^{f}<0$, then $\varphi$ is Undominated only if $f \in \mathcal {F}$ is such that for each $\theta >\beta$, $p(\theta )\ne \beta$ and for each $\theta \in \mathbb {R}$,

$$\begin{aligned} g^{f}(\theta )= \left\{ \begin{array}{ll} 0 &{} \text {if } \theta <\beta ^{f},\\ 0 &{} \text {if } \theta =\beta ^{f} \text { and }f(\beta ^{f},\beta ^{f})=0, \\ p(\theta ) &{} \text {if } \theta =\beta \text { and }f(\beta ^{f},\beta ^{f})=1,\\ p(\theta )+\beta ^{f} &{} \text {if } \theta > \beta ^{f}. \end{array} \right. \end{aligned}$$

Proof

The proof involves several steps.

Step 1. If $\beta ^{f}<0$ then, for each $\theta \in \mathbb {R}, \; g(\theta ) \le 0$.

This follows by Lemma 2 and the fact that f is weakly monotonic.

Step 2. For each $\theta <\beta ^{f}$, $p(\theta ) < 0\implies g(\theta ) \ge p(\theta ).$

Let there exist $\theta ' <\beta ^{f}$ such that $p(\theta ') < 0$ and $g(\theta ') < p(\theta ')$. We will show that $\varphi$ is dominated by the mechanism $\varphi ':=(f,t')$ associated with the payment scheme $(p,g')$, with

$$\begin{aligned} g^{\prime }(\theta )= \left\{ \begin{array}{ll} p(\theta ') &{} \text {if } \theta = \theta ', \\ g(\theta ) &{} \text {otherwise}.\\ \end{array} \right. \end{aligned}$$

By construction, $\varphi '$ is strategy-proof and anonymous. Moreover, again by construction, for each $\theta \in \mathbb {R}$, $g'(\theta ) \ge g(\theta )$ and $g'(\theta ') > g(\theta ')$. Hence, for each $i\in N$ and each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $t_{i}(\theta _{1},\theta _{2})\le t_{i}^{\prime }(\theta _{1},\theta _{2})$ and $t_{i}(\theta ^{\prime },\theta ^{\prime })< t_{i}^{\prime }(\theta ^{\prime },\theta ^{\prime })$. There only remains to prove that $\varphi '$ is feasible. By construction, for each $(\theta _{1},\theta _{2}) \in \mathbb {R}^2$, if for each $i \in N$, $\theta _i \ne \theta '$, then $\sum _{i\in N}t_i'(\theta _1,\theta _{2})=\sum _{i\in N}t_i(\theta _1,\theta _{2})\le 0$. The latter inequality holds since, by assumption, $\varphi$ is feasible. In light of Step 1 and Lemma 2 one can easily prove that if for some $i \in N$, $\theta _i=\theta '$, then also $\sum _{i\in N}t_i'(\theta _{1},\theta _{2})\le 0$. Hence, $\varphi '$ is feasible. Hence $\varphi ^{\prime }$ dominates $\varphi$, a contradiction.

Step 3. For each $\theta <\beta ^{f}$, $g(\theta ) = 0$.

Let there exist some $\theta '<\beta ^{f}$ such that $g(\theta ') \ne 0$. By Lemma 2, $g(\theta ')<0$. To prove our statement we will consider two separate cases.

Case 1. $p(\theta ')\ge 0$. Consider the mechanism $\varphi ':=(f,t')$, associated with the payment scheme $(p,g')$, with

$$\begin{aligned} g^{\prime }(\theta )= \left\{ \begin{array}{ll} 0 &{} \text {if } \theta = \theta ', \\ g(\theta ) &{} \text {otherwise}.\\ \end{array} \right. \end{aligned}$$

By employing the same reasoning as in the previous step, it can be readily verified that $\varphi ^{\prime }$ is feasible and, moreover, dominates $\varphi$.

Case 2 : $p(\theta ')<0$. Let

$$\begin{aligned} \delta ^*=-\sup \{g(\theta )-p(\theta )|\theta <\beta ^{f}\}. \end{aligned}$$

(6.5)

By construction and by Lemma 2, $\beta ^{f}\le \delta ^{*}\le 0$. Let $\hat{f}:\mathbb {R}^{2}\rightarrow \{0,1\}$ be such that for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $\hat{f}(\theta _{1},\theta _{2})=0$ if and only if either $f(\theta _{1},\theta _{2})=0$ or $(\theta _{1},\theta _{2})\ll (\delta ^*,\delta ^*)$. By construction, $\beta ^{\hat{f}}=\delta ^*$ and

$$\begin{aligned} \hat{p}(\theta )= \left\{ \begin{array}{ll} \max \big \{\delta ^*, p(\theta )\big \} &{} \text {if } \theta < \delta ^*\\ p(\theta ) &{} \text {if } \theta \ge \delta ^*. \end{array} \right. \end{aligned}$$

Let

$$\begin{aligned} \hat{g}(\theta )= \left\{ \begin{array}{ll} 0 &{} \text {if } \theta < \delta ^* \\ g(\theta ) &{} \text {if} \ \theta \ge \delta ^*. \end{array} \right. \end{aligned}$$

Consider $\hat{\varphi }=(\hat{f},\hat{t})$, associated with the payment scheme $(\hat{p},\hat{g})$. By construction, $\hat{f}\in \mathcal {F}$. Hence, by Proposition 1, $\hat{\varphi }$ is strategy-proof and anonymous. Let us show that $\hat{\varphi }$ is feasible.

By construction, if $(\theta _{1},\theta _{2}) \ge (\delta ^*,\delta ^*)$, then $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\sum _{i\in N}t_{i}(\theta _{1},\theta _{2})$. By assumption, $\varphi$ is feasible. Hence, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$. Hence, without loss of generality, let $\theta _{1}\in (-\infty ,\delta ^*)$. We distinguish between two cases.

(i) : $\hat{f}(\theta _1,\theta _2)=0$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\hat{g}(\theta _{1})+\hat{g}(\theta _{2})$. By construction, $\hat{g}(\theta _1)=0$. Moreover, by construction and Step 1, $\hat{g}(\theta _2) \le 0$. Hence, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$.

(ii) : $\hat{f}(\theta _1,\theta _2)=1$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\hat{g}(\theta _{1})-\hat{p}(\theta _{1})+\hat{g}(\theta _{2})-\hat{p}(\theta _{2})$. By construction, $\hat{g}(\theta _{1})=0$. Moreover, since $\hat{f}(\theta _1,\theta _2)=1$ and $\theta _{1}<\delta ^{*}$, $\theta _{2}\ge \delta ^*$. Hence, by construction, $\hat{g}(\theta _{2})=g(\theta _{2})$ and $\hat{p}(\theta _{2})=p(\theta _{2})$. Hence,

$$\begin{aligned} \sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=-\hat{p}(\theta _{1})-p(\theta _{2})+g(\theta _{2})\le -\delta ^*-p(\theta _{2})+g(\theta _{2}). \end{aligned}$$

The latter inequality holds since $\theta _{1}<\delta ^*\implies \hat{p}(\theta _{1})\ge \delta ^*$. We claim that $g(\theta _{2})-p(\theta _{2})\le \delta ^*$. If $\delta ^{*}=0$, since $\theta _{2}\ge \delta ^{*}\ge \beta ^{f}$, this follows from Lemma 2. By way of contradiction, let $\delta ^{*}<0$ and suppose that $g(\theta _{2})-p(\theta _{2})>\delta ^{*}=-\sup \{g(\theta )-p(\theta )|\theta <\beta ^{f}\}$. By definition, for each $\theta \in \mathbb {R}$, both $p(\theta )$ and $g(\theta )$ are finite, so $-\delta ^{*}$ is finite. Moreover, by the definition of $\delta ^{*}$, there exists some $\theta ' \le \beta ^{f}$ such that $g(\theta ^{'})-p(\theta ^{'})\simeq -\delta ^{*}.$ Hence,

$$\begin{aligned} \sum _{i\in N} t_{i}(\theta ',\theta _{2})\simeq -\delta ^*-p(\theta _{2})+g(\theta _{2}) > 0, \end{aligned}$$

This contradicts the fact that $\varphi$ is feasible. Therefore, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$.

Finally, in what follows we show that $\hat{\varphi }$ dominates $\varphi$. Let

$$\theta ^*= \inf \{\theta \le \beta ^{f} \; :\; p(\theta )=\delta ^*\}.$$

By construction $\theta ^*$ exists and $\theta ^* < \delta ^*$.

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, if $f(\theta _{1},\theta _{2})=\hat{f}(\theta _{1},\theta _{2})=1$, then for each $i \in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&=\hat{t}_{i}\big (\theta _{1},\theta _{2})- t_{i}\big (\theta _{1},\theta _{2}\big ) \\ \nonumber&=-\hat{p}(\theta _{-i})+\hat{g}(\theta _{-i})+p(\theta _{-i})-g(\theta _{-i}). \end{aligned}$$

If $\theta _{-i}\in (-\infty ,\theta ^{*})$, $-\hat{p}(\theta _{-i})+\hat{g}(\theta _{-i})+p(\theta _{-i})-g(\theta _{-i})=-g(\theta _{-i})\ge 0$, by Step 1. If $\theta _{-i} \in [\theta ^{*},\delta ^{*})$, then $-\hat{p}(\theta _{-i})+\hat{g}(\theta _{-i})+p(\theta _{-i})-g(\theta _{-i})=p(\theta _{-i})-\delta ^*-g(\theta _{-i})$. Then if $\theta _{-i} < \beta ^{f}$, by the definition of $\delta ^{*}$, $p(\theta _{-i})-\delta ^*-g(\theta _{-i})\ge 0$. If $\theta _{-i} \ge \beta ^{f}$, then since $\delta ^*\le 0$, $p(\theta _{-i})-\delta ^*-g(\theta _{-i})\ge p(\theta _{-i})-g(\theta _{-i})$. Moreover, by Lemma 2, $p(\theta _{-i})-g(\theta _{-i})\ge 0$. Finally, if $\theta \in [\delta ^{*},+\infty )$, $-\hat{p}(\theta _{-i})+\hat{g}(\theta _{-i})+p(\theta _{-i})-g(\theta _{-i})=-g(\theta _{-i}) = 0$.

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, if $f(\theta _{1},\theta _{2})=\hat{f}(\theta _{1},\theta _{2})=0$, then for each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&= \hat{t}_{i}\big (\theta _{1},\theta _{2})- t_{i}\big (\theta _{1},\theta _{2}\big ) \\ \nonumber&= \hat{g}(\theta _{-i})+-g(\theta _{-i})\ge 0. \end{aligned}$$

The latter inequality holds because of Step 2. By construction, there does not exist $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ such that $f(\theta _{1},\theta _{2})<\hat{f}(\theta _{1},\theta _{2})$. Hence, there remains one final case:

$$\begin{aligned} (\theta _{1},\theta _{2})<(\delta ^*,\delta ^*)\text { and } f(\theta _{1},\theta _{2})=1. \end{aligned}$$

We obtain, for each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&=\hat{t}_{i}\big (\theta _{1},\theta _{2}) - \theta _i - t_{i}\big (\theta _{1},\theta _{2}\big ) \\ \nonumber&= \hat{g}(\theta _{-i})-\theta _{i}+p(\theta _{-i})-g(\theta _{-i}).\end{aligned}$$

Suppose, first, that for some $i\in N$, $\theta _i <\beta ^{f}$. Therefore, since $f(\theta _1,\theta _2)=1$, we obtain $\theta _{-i}\ge \beta ^{f}$. Hence, by Lemma 2, $p(\theta _{-i})\ge g(\theta _{-i})$. Moreover, since $\theta _i <\beta ^{f}$, by the definition of $\delta ^*$, $p(\theta _i)-g(\theta _i) \ge \delta ^*$. Hence,

$$\begin{aligned}&u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )\ge \hat{g}(\theta _{-i})-\theta _{i},\;\text {and}\\&u\big (\hat{\varphi }_{-i}(\theta _{1},\theta _{2});\theta _{-i}\big )- u\big (\varphi _{-i}(\theta _{1},\theta _{2}) ; \theta _{-i}\big )\ge \hat{g}(\theta _{i})-\theta _{-i}+\delta ^*. \end{aligned}$$

By assumption, $\theta _{-i}<\delta ^*$ and $\theta _{i}<\delta ^*$. Hence, by construction, $\hat{g}(\theta _{-i})=\hat{g}(\theta _{i})=0$. By assumption, $\theta _{i}<\beta ^{f}<0$. Hence, $-\theta _{i}>0$. Hence,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )>0 \text { and } u\big (\hat{\varphi }_{-i}(\theta _{1},\theta _{2});\theta _{-i}\big )- u\big (\varphi _{-i}(\theta _{1},\theta _{2}) ; \theta _{-i}\big )>0. \end{aligned}$$

Suppose finally that $(\beta ^{f},\beta ^{f}) \le (\theta _1,\theta _2) < (\delta ^*,\delta ^*)$. Therefore, by Lemma 2, for each $i\in N$, $p(\theta _{i})-g(\theta _{i})\ge 0$. Moreover, as demonstrated above, $\delta ^{*}\le 0$. Hence, for each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )>0. \end{aligned}$$

Hence, for each $(\theta _{1},\theta _{2})<(\delta ^*,\delta ^*)$ such that $f(\theta _{1},\theta _{2})=1$ and each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )>0. \end{aligned}$$

Step 4. For each $\theta '>\beta ^{f}$, $p(\theta ') \ne \beta ^{f}$.

Suppose that there exists $\theta '>\beta ^{f}$ such that $p(\theta ')=\beta ^{f}$. Hence, $p(\beta ^{f})\ge \beta ^{f}$. Suppose first that $p(\beta ^{f})>\beta ^{f}$ (the case $p(\beta ^{f})=\beta ^{f}$ is based on a similar construction). Hence, there exists $\Delta \in \Big (\beta ^{f}, \min \{0,\theta '\}\Big )$ with $\Delta \le p(\beta ^{f})$. Let $\hat{f}:\mathbb {R}^{2}\rightarrow \{0,1\}$ be such that for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $\hat{f}(\theta _{1},\theta _{2})=0$ if and only if either $f(\theta _{1},\theta _{2})=0$ or, for each $i\in N$, $\theta _{i}\in (\beta ^{f},\Delta )$. By construction, $\beta ^{\hat{f}}=\Delta$, and

$$\begin{aligned} \hat{p}(\theta )= \left\{ \begin{array}{ll} \Delta &{} \text {if } \theta \in (\beta ^{f},\Delta )\\ p(\theta ) &{} \text {otherwise. } \end{array} \right. \end{aligned}$$

Let

$$\begin{aligned} \hat{g}(\theta )= \left\{ \begin{array}{ll} g(\theta )+\Delta -\beta ^{f} &{} \text {if } \theta \in (\beta ,^{f}\Delta )\\ g(\theta ) &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

Consider $\hat{\varphi }=(\hat{f},\hat{t})$, associated with the payment scheme $(\hat{p},\hat{g})$. By assumption, $f\in \mathcal {F}$. Hence, by construction, $\hat{f}\in \mathcal {F}$. Hence, by Proposition 1, $\hat{\varphi }$ is strategy-proof and anonymous. Next we show that $\hat{\varphi }$ is feasible.

By construction, if $(\theta _{1},\theta _{2}) \in \mathbb {R}^{2}$ is such that for each $i\in N$, $\theta _{i}\notin (\beta ^{f},\Delta )$, then $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\sum _{i\in N}t_{i}(\theta _{1},\theta _{2})$. By assumption, $\varphi$ is feasible. Hence, $\hat{\varphi }$ is feasible. Hence, without loss of generality, let $\theta _{1}\in (\beta ^{f},\Delta )$. By assumption, for each $\theta \in (\beta ^{f},\Delta )$, $f(\theta ,\theta )=1$. Hence, by Proposition 1 and the assumption that $\varphi$ is feasible, we obtain for each $\theta \in (\beta ^{f},\Delta )$, $g(\theta )\le p(\theta )$. Moreover, by assumption, for each $\theta \in (\beta ^{f},\Delta )$, $p(\theta )=\beta ^{f}$. Hence, for each $\theta \in (\beta ^{f},\Delta )$, $g(\theta )\le \beta ^{f}$. Hence, by construction, for each $\theta \in (\beta ^{f},\Delta )$, $\hat{g}(\theta )\le \Delta$. By construction, $\Delta <0$. Therefore,

$$\begin{aligned} \theta _{1}\in (\beta ^{f},\Delta ) \implies g(\theta _{1})\le \beta ^{f} \text { and }\hat{g}(\theta _{1})<0. \end{aligned}$$

(6.6)

We distinguish between two cases.

(i) $\hat{f}(\theta _1,\theta _2)=0$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\hat{g}(\theta _{1})+\hat{g}(\theta _{2})$. By 6.6 and Step 1, for each $\theta \in \mathbb {R}$, $\hat{g}(\theta )\le 0$. Hence, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$.

(ii) $\hat{f}(\theta _1,\theta _2)=1$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\hat{g}(\theta _{1})-\hat{p}(\theta _{1})+\hat{g}(\theta _{2})-\hat{p}(\theta _{2})$. By construction, $\hat{g}(\theta _{1})=g(\theta _{1})+\Delta -\beta ^{f}$ and $\hat{p}(\theta _{1})=\Delta$. Hence, by 6.6, $\hat{g}(\theta _{1})-\hat{p}(\theta _{1})\le 0$. Moreover, by construction, $\hat{f}(\theta _1,\theta _2)=1$ and $\theta _{1}\in (\beta ^{f},\Delta )$ imply that $\theta _{2}\ge \Delta$. Hence, by construction, $\hat{g}(\theta _{2})-\hat{p}(\theta _{2})=g(\theta _{2})-p(\theta _{2})$. Moreover, by construction, $\Delta >\beta ^{f}$. Hence, $\theta _{2}>\beta ^{f}$. Hence, by construction and Lemma 2, $\hat{g}(\theta _{2})-\hat{p}(\theta _{2})=g(\theta _{2})-p(\theta _{2})\le 0$. Thus, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$.

Finally, in what follows we will prove that $\hat{\varphi }$ dominates $\varphi$. Let

$$\begin{aligned} \Gamma (\theta )\equiv \hat{g}(\theta )-g(\theta )= \left\{ \begin{array}{ll} \Delta -\beta ^{f} &{} \text {if } \theta \in (\beta ^{f},\Delta ), \\ 0 &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

By Step 1 and by construction, for each $\theta \in \mathbb {R}$, $\Gamma (\theta )\ge 0$. Let

$$\begin{aligned} \Pi (\theta )\equiv p(\theta )-\hat{p}(\theta )= \left\{ \begin{array}{ll} \beta ^{f}-\Delta &{} \text {if } \theta \in (\beta ^{f}, \Delta ), \\ 0 &{} \text {otherwise. } \end{array} \right. \end{aligned}$$

Hence, for each $\theta \in \mathbb {R}$, $\Gamma (\theta )+\Pi (\theta )=0$.

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, if $f(\theta _{1},\theta _{2})=\hat{f}(\theta _{1},\theta _{2})=1$, then for each $i \in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&=\hat{t}_{i}\big (\theta _{1},\theta _{2})- t_{i}\big (\theta _{1},\theta _{2}\big )\\ \nonumber&= -\hat{p}(\theta _{-i})+\hat{g}(\theta _{-i})+p(\theta _{-i})-g(\theta _{-i})\\ \nonumber&= \Gamma (\theta _{-i})+\Pi (\theta _{-i}) = 0. \end{aligned}$$

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, if $f(\theta _{1},\theta _{2})=\hat{f}(\theta _{1},\theta _{2})=0$, then for each $i \in N$,

$$\begin{aligned} \nonumber u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&=\hat{t}_{i}\big (\theta _{1},\theta _{2})- t_{i}\big (\theta _{1},\theta _{2}\big )\\ \nonumber&=\hat{g}(\theta _{-i})-g(\theta _{-i})\\ \nonumber&=\Gamma (\theta _{-i})\ge 0.\end{aligned}$$

By construction, there does not exist $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ such that $f(\theta _{1},\theta _{2})<\hat{f}(\theta _{1},\theta _{2})$. Hence, there remains one final case: for each $i\in N$, $\theta _{i}<\Delta$ and $f(\theta _{1},\theta _{2})=1$. We obtain, for each $i \in N$,

$$\begin{aligned} \nonumber u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&=\theta _i + \hat{t}_{i}\big (\theta _{1},\theta _{2})- t_{i}\big (\theta _{1},\theta _{2}\big )\\ \nonumber&=\hat{g}(\theta _{-i})-\theta _{i}+p(\theta _{-i})-g(\theta _{-i}).\end{aligned}$$

Since $\theta _{-i}\in (\beta ^{f},\Delta )$, by construction, $\hat{g}(\theta _{-i})=g(\theta _{-i})+\Delta -\beta ^{f}$. Moreover, by assumption, $p(\theta _{-i})=\beta ^{f}$. Hence,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )=\Delta -\theta _{i}>0. \end{aligned}$$

Step 5. The mechanism $\varphi =(f,t)$ associated with the payment scheme $(p,g^{f})$ satisfies Feasibility, Strategy-proofness and Anonymity.

The proof is omitted for the sake of brevity (it is available on request).

We can now prove the statement. Consider the mechanism $\varphi =(f,t)$ associated with the payment scheme (p, g). Let $\varphi$ be an undominated mechanism. Hence, by Step 4 one has that $p:\mathbb {R}\rightarrow \mathbb {R}$ is such that, for each $\theta >\beta ^{f}$, $p(\theta )\ne \beta ^{f}$. We need to prove that $g(\theta )=g^{f}(\theta )$. By Step 3, for each $\theta < \beta ^{f}$, $g(\theta )=0$. Hence, by Lemma 2,

$$\begin{aligned} g(\theta )\le \left\{ \begin{array}{ll} 0 &{} \text {if } f(\beta ^{f},\beta ^{f})=0,\\ p(\beta ^{f}) &{} \text {if } f(\beta ^{f},\beta ^{f})=1. \end{array} \right. \end{aligned}$$

Finally, we claim that for each $\theta >\beta ^{f}$, $g(\theta )\le p(\theta )+\beta ^{f}$. One should first note that, by Step 4 and since f is symmetric, for some $\epsilon >0$ arbitrarily small one has $p(\beta ^{f}-\epsilon ) \simeq \beta ^{f}.$⁸ Assume, by way of contradiction, that for some $\theta ' >\beta ^{f}$, $g(\theta ')>p(\theta ')+\beta ^{f}$. Suppose first that $f(\theta ',\beta ^{f}-\epsilon )=1$. Hence, by Proposition 1, $\sum _{i\in N}t_{i}(\beta ^{f}-\epsilon ,\theta ')=g(\beta ^{f}-\epsilon )-p(\beta ^{f}-\epsilon )+g(\theta ')-p(\theta ')$. By Step 3, $g(\beta ^{f}-\epsilon )=0$. By construction, $p(\beta ^{f}-\epsilon ) \simeq \beta ^{f}$. By assumption, $g(\theta ')>p(\theta ')+\beta ^{f}$. Thus, $\sum _{i\in N}t_{i}(\beta ^{f}-\epsilon ,\theta ')>0$, which contradicts Feasibility. Suppose then that $f(\theta ',\beta ^{f}-\epsilon )=0$. Since $\epsilon >0$ is arbitrarily small, we obtain for each $\theta <\beta ^{f}$, $f(\theta ',\theta )=0$. Moreover, since by assumption, $\theta '>\beta ^{f}$, for each $\theta >\beta ^{f}$, $f(\theta ',\theta )=1$. Thus, $p(\theta ')=\beta ^{f}$ and, by Step 4 we have arrived at a contradiction.

Therefore, for each $\theta \in \mathbb {R}$, $g'(\theta )\ge g(\theta )$. By Step 5, the mechanism $\varphi '=(f,t')$, associated with the payment scheme $(p,g')$ is strategy-proof, anonymous and feasible. Moreover, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i \in N$,

$$\begin{aligned} u\big (\varphi '_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )= g'(\theta )-g(\theta )\ge 0 \end{aligned}$$

Moreover, if for some $\theta ''\in \mathbb {R}$, $g'(\theta '')>g(\theta '')$, this inequality becomes strict and $\varphi '$ dominates $\varphi$. Hence, for each $\theta \in \mathbb {R}$, $g(\theta )=g'(\theta )$.

Lemma 5

If the mechanism $\varphi =(f,t)$,associated with the payment scheme (p, g), satisfies Feasibility, Strategy-proofness, Anonymity and $\beta ^{f}>0$, then $\varphi$ is Undominated only if $f \in \mathcal {F}$ is such that for each $\theta <\beta ^{f}$, $p(\theta )\ne \beta ^{f}$ and for each $\theta \in \mathbb {R}$,

$$\begin{aligned} g^{f}(\theta )= \left\{ \begin{array}{ll} -\beta ^{f} &{} \text {if } \theta <\beta ^{f},\\ 0 &{} \text {if } \theta =\beta ^{f} \text { and }f(\beta ^{f},\beta ^{f})=0, \\ p(\theta ) &{} \text {if } \theta =\beta ^{f} \text { and }f(\beta ^{f},\beta ^{f})=1,\\ p(\theta ) &{} \text {if } \theta > \beta ^{f}. \end{array} \right. \end{aligned}$$

Proof

The proof involves several steps.

Step 1. $\beta ^{f}\le \sup _{\theta \ge \beta ^{f}}g(\theta )$. Suppose that $\sup _{\theta \ge \beta ^{f}}g(\theta )<\beta ^{f}$. Let $\alpha \in \big (\sup _{\theta \ge \beta ^{f}}g(\theta ),\beta ^{f}\big )$ be such that $\alpha >0$. Since $\beta ^{f}$ is positive, such an $\alpha$ exists. Let

$$\begin{aligned} A=\{(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\; :\; \theta _{1}\ge \alpha \text { and }\theta _{2}\ge \alpha \},\text { and} \\ B=\{(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\; :\; f(\theta _{1},\theta _{2})=1\}. \end{aligned}$$

Assume first that there exists $\alpha \in \big (\sup _{\theta \ge \beta ^{f}}g(\theta ),\beta ^{f}\big )$ such that $\alpha >0$ and $\big \{\mathbb {R}^{2}\setminus A\big \}\cap B\ne \emptyset$. Let $\hat{f}\in \mathcal {F}$ be such that for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $\hat{f}(\theta _{1},\theta _{2})=1$ if and only if either $f(\theta _{1},\theta _{2})=1$ or $(\theta _{1},\theta _{2})\ge (\alpha ,\alpha )$ or both. By construction $\beta ^{\hat{f}}=\alpha$ and

$$\begin{aligned} \hat{p}(\theta )= \left\{ \begin{array}{ll} p(\theta ) &{} \text {if } \theta <\alpha ,\\ \min \{\alpha ,p(\theta )\} &{} \text {if } \theta \ge \alpha . \end{array} \right. \end{aligned}$$

Let $\hat{\varphi }=(\hat{f},\hat{t})$ be associated with the payment scheme $(\hat{p},g)$. By assumption, $f\in \mathcal {F}$. Hence, by construction, $\hat{f}\in \mathcal {F}$. By Proposition 1, $\hat{\varphi }$ is strategy-proof and anonymous. Let us show that $\hat{\varphi }$ is feasible. We need to consider two cases:

Case 1: $\hat{f}(\theta _{1},\theta _{2})=1$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=g(\theta _{1})-\hat{p}(\theta _{1})+g(\theta _{2})-\hat{p}(\theta _{2})$. We will prove that for each $\theta \in \mathbb {R}$, $g(\theta )-\hat{p}(\theta )\le 0$ so that $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$. Suppose first that $\theta \ge \beta ^{f}$ and $\alpha \le p(\theta )$. Hence, since $\theta \ge \beta ^{f}$, by construction, $g(\theta )\le \sup _{\theta \ge \beta ^{f}}g(\theta )<\alpha$. Moreover, since $\alpha <\beta ^{f}\le \theta$, $\hat{p}(\theta )=\min \{\alpha ,p(\theta )\}$. By assumption, $\alpha \le p(\theta )$. Hence, $\hat{p}(\theta )=\alpha$. Hence, $g(\theta )-\hat{p}(\theta )<\alpha -\alpha = 0$. Suppose then that $\theta \ge \beta ^{f}$ and $\alpha > p(\theta )$. Hence, $g(\theta )-\hat{p}(\theta )=g(\theta )-p(\theta )$. Consider, first, the case $\theta =\beta ^{f}$. Hence, $p(\beta ^{f})<\alpha <\beta ^{f}$. Hence, either $f(\beta ^{f},\beta ^{f})=1$ or for each $\theta \in \mathbb {R}$, $f(\beta ^{f},\theta )=0$. The latter case contradicts the assumption that $\big \{\mathbb {R}^{2}\setminus A\big \}\cap B\ne \emptyset$. Hence, $f(\beta ^{f},\beta ^{f})=1$ and by Lemma 2, since $\varphi$ is feasible, $g(\beta ^{f})-p(\beta ^{f})\le 0$. Suppose, then, that $\theta >\beta ^{f}$. Hence, by Lemma 2, since $\varphi$ is feasible, $g(\theta )-p(\theta )\le 0$. Suppose, finally, that $\theta <\beta ^{f}$. Hence, by construction, $g(\theta )-\hat{p}(\theta )=g(\theta )-\min \{\alpha ,p(\theta )\}$. By Lemma 2, $g(\theta )\le 0$. Moreover, since f is monotonic, $\theta <\beta ^{f}$ and $\beta ^{f}\ge 0$ imply that $p(\theta )\ge 0$. By construction, $\alpha >0$. Hence, for each $\theta <\beta ^{f}$, $\min \{\alpha ,p(\theta )\}\ge 0$. Hence, $g(\theta )-p(\theta )\le 0$.

Case 2: $\hat{f}(\theta _{1},\theta _{2})=0$. By construction, this implies that $f(\theta _{1},\theta _{2})=0$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=g(\theta _{1})+g(\theta _{2})=\sum _{i\in N}t_{i}(\theta _{1},\theta _{2})$. By assumption, $\varphi$ is feasible. Hence, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$.

Next, we show that $\hat{\varphi }$ dominates $\varphi$. By construction, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $\hat{f}(\theta _{1},\theta _{2})\ge f(\theta _{1},\theta _{2})$. Hence, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )= \left\{ \begin{array}{ll} 0 &{} \text {if } \hat{f}(\theta _{1},\theta _{2})=f(\theta _{1},\theta _{2})=0,\\ \theta _{i}-\hat{p}(\theta _{-i}) &{} \text {if } \hat{f}(\theta _{1},\theta _{2})>f(\theta _{1},\theta _{2}),\\ p(\theta _{-i})-\hat{p}(\theta _{-i}) &{} \text {if } \hat{f}(\theta _{1},\theta _{2})=f(\theta _{1},\theta _{2})=1. \end{array} \right. \end{aligned}$$

If $\hat{f}(\theta _{1},\theta _{2})>f(\theta _{1},\theta _{2})$ then,

$$\begin{aligned} (\theta _{1},\theta _{2})\in \{(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\; :\; (\theta _{1},\theta _{2})\ge (\alpha ,\alpha )\}\cap \{(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\; :\; f(\theta _{1},\theta _{2})=0\}. \end{aligned}$$

Hence, for each $i\in N$, $\theta _{i}\ge \hat{p}(\theta _{i})=\alpha$. Moreover, by construction, for each $\theta \in \mathbb {R}$, $p(\theta )\ge \hat{p}(\theta )$. Hence, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )\ge 0. \end{aligned}$$

There remains to show that there exists $(\theta '_{1},\theta '_{2})\in \mathbb {R}^{2}$ and $i\in N$ such that,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta '_{1},\theta '_{2});\theta '_{i}\big )- u\big (\varphi _{i}(\theta '_{1},\theta '_{2}) ; \theta '_{i}\big )> 0. \end{aligned}$$

By construction, $\alpha <\beta ^{f}$. Hence, there exists $\theta ''\in (\alpha ,\beta ^{f})$. By construction, $\hat{f}(\theta '',\theta '')>f(\theta '',\theta '')$. Hence, as demonstrated previously, for each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta '',\theta '');\theta ''\big )- u\big (\varphi _{i}(\theta '',\theta '') ; \theta ''\big )=\theta ''-\alpha > 0. \end{aligned}$$

Hence, if there exists $\alpha \in \big (\sup _{\theta \ge \beta ^{f}}g(\theta ),\beta ^{f}\big )$ such that $\alpha >0$ and $\big \{\mathbb {R}^{2}\setminus A\big \}\cap B\ne \emptyset$, then $\hat{\varphi }$ dominates $\varphi$, which yields the desired contradiction.

Suppose that for each $\alpha \in \big (\sup _{\theta \ge \beta ^{f}}g(\theta ),\beta ^{f}\big )$ such that $\alpha >0$, $\big \{\mathbb {R}^{2}\setminus A\big \}\cap B=\emptyset$. By construction, $\alpha <\beta ^{f}$. Hence, $B \subset A$ and

$$\begin{aligned} B=\{(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\; :\; \theta _{1}>\alpha \text { and }\theta _{2}>\alpha \}. \end{aligned}$$

Let

$$\begin{aligned} \gamma =\max \big \{0,\sup _{\theta \ge \beta ^{f}}g(\theta )\big \}\text { and }\gamma ^-=\min \big \{0,-\sup _{\theta \ge \beta ^{f}}g(\theta )\big \}. \end{aligned}$$

Let $f'\in \mathcal {F}$ be such that for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $f'(\theta _{1},\theta _{2})=1$ if and only if $\theta _{1}\ge \gamma$ and $\theta _{2}\ge \gamma$, so that

$$\begin{aligned} p'(\theta )= \left\{ \begin{array}{ll} 0 &{} \text {if } \theta <\gamma ,\\ \gamma &{} \text {if } \theta \ge \gamma , \end{array} \right. \end{aligned}$$

and $\beta ^{f'}=\gamma$. Let

$$\begin{aligned} g'(\theta )= \left\{ \begin{array}{ll} \gamma ^- &{} \text {if } \theta <\gamma ,\\ \gamma &{} \text {if } \theta \ge \gamma . \end{array} \right. \end{aligned}$$

Let $\varphi '=(f',t')$ be associated with the payment scheme $(p',g')$. Hence, by construction, $f'\in \mathcal {F}$. By Proposition 1, $\varphi '$ is strategy-proof and anonymous. One can easily check that $\varphi '$ is feasible. We omit the proof.

Let us show that $\varphi '$ dominates $\varphi$. By construction, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $f'(\theta _{1},\theta _{2})\ge f(\theta _{1},\theta _{2})$. Hence, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i\in N$,

$$\begin{aligned}&u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )=\\&\quad = \left\{ \begin{array}{ll} g'(\theta _{-i})-g(\theta _{-i}) &{} \text {if } f'(\theta _{1},\theta _{2})=f(\theta _{1},\theta _{2})=0,\\ \theta _{i}-p'(\theta _{-i})+g'(\theta _{-i})-g(\theta _{-i}) &{} \text {if } f'(\theta _{1},\theta _{2})>f(\theta _{1},\theta _{2}),\\ p(\theta _{-i})-p'(\theta _{-i})+g'(\theta _{-i})-g(\theta _{-i}) &{} \text {if } f'(\theta _{1},\theta _{2})=f(\theta _{1},\theta _{2})=1. \end{array} \right. \end{aligned}$$

By assumption, there exists $\theta ^*\ge \beta ^{f}$ such that $g(\theta ^*)\simeq \sup _{\theta \ge \beta ^{f}}g(\theta ).$ By assumption, for each $\theta <\beta ^{f}$, $f(\theta ,\theta ^{*})=0$. Hence, by Proposition 1 and Feasibility, for each $\theta <\beta ^{f}$, $g(\theta )\le -\sup _{\theta \ge \beta ^{f}}g(\theta )$. Moreover, by Lemma 2, for each $\theta <\beta ^{f}$, $g(\theta )\le 0$. Hence, for each $\theta <\beta ^{f}$, $g(\theta )\le \gamma ^-$. Moreover, by assumption, $\beta ^{f}>0$ and $\beta ^{f}>\sup _{\theta \ge \beta ^{f}}g(\theta )$. Hence, $\beta ^{f}>\gamma$. Therefore,

$$\begin{aligned} g'(\theta _{-i})-g(\theta _{-i})\ge \left\{ \begin{array}{ll} 0 &{} \text {if } \theta _{-i}<\gamma ,\\ \gamma -\gamma ^{-}=0 &{} \text {if } \theta _{-i}\in [\gamma ,\beta ^{f}),\\ \gamma -\sup _{\theta \ge \beta ^{f}}g(\theta )\ge 0 &{} \text {if } \theta _{-i}\ge \beta ^{f}. \end{array} \right. \end{aligned}$$

Hence, $g'(\theta )-g(\theta )\ge 0$. Moreover, by construction, $f'(\theta _{1},\theta _{2})=f(\theta _{1},\theta _{2})=1$ implies that for each $i\in N$, $\theta _{i}\ge \beta ^{f}>\gamma$. Hence, by construction, $p(\theta _{-i})-p'(\theta _{-i})=\beta ^{f}-\gamma >0$. Finally, $f'(\theta _{1},\theta _{2})>f(\theta _{1},\theta _{2})$ implies that for each $i\in N$, $\theta _{i}\ge \gamma$. Moreover, by construction $p'(\theta _{-i})=\gamma$. Hence, $\theta _{i}-p'(\theta _{-i})=\theta _{i}-\gamma \ge 0$. Thus, $\varphi '$ dominates $\varphi$, which yields the desired contradiction.

Step 2. For each $\theta <\beta ^{f}$, $p(\theta ) \ne \beta ^{f}$.

Suppose that there exists $\theta '<\beta ^{f}$ such that $p(\theta ')=\beta ^{f}$. Hence, $p(\beta ^{f})\le \beta ^{f}$. Moreover, for each $\theta >\beta ^{f}$, $f(\theta ,\theta ')=1$. Hence, for each $\theta >\beta ^{f}$, $p(\theta )\le \theta '<\beta ^{f}$. Hence, by Lemma 2, for each $\theta >\beta ^{f}$, $g(\theta )\le \theta '<\beta ^{f}$. Therefore, by Step 1 and Lemma 2, it must be $f(\beta ^{f},\beta ^{f})=1$ and $p(\beta ^{f})=g(\beta ^{f})=\beta ^{f}$. Hence, for each $\theta <\beta ^{f}$, $f(\theta ,\beta ^{f})=0$. Hence, by Proposition 1 and Feasibility, for each $\theta <\beta ^{f}$, $g(\theta )\le -\beta ^{f}$. In short,

$$\begin{aligned} \left\{ \begin{array}{ll} g(\theta )\le -\beta ^{f} &{} \text {if } \theta<\beta ^{f},\\ g(\theta )\le \theta '<\beta ^{f} &{} \text {if } \theta >\beta ^{f}, \text { and}\\ g(\beta ^{f})=p(\beta ^{f})=\beta ^{f}. &{} \end{array} \right. \end{aligned}$$

(6.7)

Since $p(\beta ^{f})=\beta ^{f}$ and $\beta ^{f}>0$, there exists $\Delta \in (\theta ', \beta ^{f})$, with $\Delta >0$ such that if for each $i\in N$, $\theta _{i}\in [\Delta ,\beta ^{f})$, then $f(\theta _{1},\theta _{2})=0$. Let $\hat{f}:\mathbb {R}^{2}\rightarrow \{0,1\}$ be such that for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, $\hat{f}(\theta _{1},\theta _{2})=1$ if and only if either $f(\theta _{1},\theta _{2})=1$ or for each $i\in N$, $\theta _{i}\ge \Delta$. By construction, $\beta ^{\hat{f}}=\Delta$, and

$$\begin{aligned} \hat{p}(\theta )= \left\{ \begin{array}{ll} \Delta &{} \text {if } \theta \in (\Delta ,\beta ^{f}],\\ p(\theta ) &{} \text {otherwise. } \end{array} \right. \end{aligned}$$

Let

$$\begin{aligned} \hat{g}(\theta )= \left\{ \begin{array}{ll} \Delta &{} \text {if } \theta \in (\Delta ,\beta ^{f}],\\ g(\theta ) &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

Consider $\hat{\varphi }=(\hat{f},\hat{t})$, associated with the payment scheme $(\hat{p},\hat{g})$. By assumption, $f\in \mathcal {F}$. Hence, by construction, $\hat{f}\in \mathcal {F}$. Hence, by Proposition 1, $\hat{\varphi }$ is strategy-proof and anonymous. Let us show that $\hat{\varphi }$ is feasible.

By construction, if $(\theta _{1},\theta _{2}) \in \mathbb {R}^{2}$ is such that for each $i\in N$, $\theta _{i}\notin (\Delta ,\beta ^{f}]$, then $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\sum _{i\in N}t_{i}(\theta _{1},\theta _{2})$. By assumption $\varphi$ is feasible. Hence, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$. Therefore, without loss of generality, suppose that $\theta _{1}\in (\Delta ,\beta ^{f}]$.

(i) : $\hat{f}(\theta _1,\theta _2)=0$. Therefore, since $\theta _{1}\in (\Delta ,\beta ^{f}]$, by construction $\theta _{2}<\Delta <\beta ^{f}$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\hat{g}(\theta _{1})+\hat{g}(\theta _{2})=\Delta +g(\theta _{2})$. By 6.7 and the fact that $\theta _{2}<\beta ^{f}$, $g(\theta _{2})\le -\beta ^{f}$. Hence, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le \Delta -\beta ^{f}<0$.

(ii) : $\hat{f}(\theta _1,\theta _2)=1$. Hence, by Proposition 1, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})=\hat{g}(\theta _{1})-\hat{p}(\theta _{1})+\hat{g}(\theta _{2})-\hat{p}(\theta _{2})$. By construction, $\hat{g}(\theta _{1})=\Delta$ and $\hat{p}(\theta _{1})=\Delta$. Moreover, by construction, $\hat{f}(\theta _1,\theta _2)=1$ and $\theta _{1}\in (\Delta ,\beta ^{f})$ imply that $\theta _{2}\ge \Delta$. By 6.7 and Lemma 2,

$$\begin{aligned} g(\theta _{2})-p(\theta _{2}) \le \left\{ \begin{array}{ll} -\beta ^{f}+p(\theta _{2}) &{} \text {if } \theta _{2} \in [\Delta ,\beta ^{f}), \\ 0 &{} \text {if } \theta _2=\beta ^{f},\\ 0 &{} \text {if } \theta _{2}>\beta ^{f}. \end{array} \right. \end{aligned}$$

Finally, since $\Delta >\theta '$, $\theta _{2} \in [\Delta ,\beta ^{f})$ implies $p(\theta _{2})=\beta ^{f}$. Thus, $\sum _{i\in N}\hat{t}_{i}(\theta _{1},\theta _{2})\le 0$.

Let

$$\begin{aligned} \Gamma (\theta )\equiv \hat{g}(\theta )-g(\theta )= \left\{ \begin{array}{ll} \Delta -g(\theta ) &{} \text {if } \theta \in (\Delta ,\beta ^{f}), \\ \Delta -\beta ^{f} &{} \text {if } \theta =\beta ^{f}, \\ 0 &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

By 6.7 and by construction, for each $\theta \in \mathbb {R}\setminus \{\beta ^{f}\}$, $\Gamma (\theta )\ge 0$. Let

$$\begin{aligned} \Pi (\theta )\equiv p(\theta )-\hat{p}(\theta )= \left\{ \begin{array}{ll} \beta ^{f}-\Delta &{} \text {if } \theta \in (\Delta ,\beta ^{f}), \\ 0 &{} \text {otherwise. } \end{array} \right. \end{aligned}$$

Hence, by 6.7, for each $\theta \in \mathbb {R}$, $\Gamma (\theta )+\Pi (\theta )\ge 0$.

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, if $f(\theta _{1},\theta _{2})=\hat{f}(\theta _{1},\theta _{2})=1$. Hence, for each $i\in N$,

$$\begin{aligned} \nonumber u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&= \theta _i -\hat{p}(\theta _{-i}) +\hat{g}(\theta _{-i}) - \theta _i +p(\theta _{-i}) - g(\theta _{-i})\\ \nonumber&= \Gamma (\theta _{-i})+\Pi (\theta _{-i})\ge 0.\end{aligned}$$

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, if $f(\theta _{1},\theta _{2})=\hat{f}(\theta _{1},\theta _{2})=0$. Hence, by construction, for each $i\in N$, $\theta _{i}<\Delta <\beta ^{f}$. Hence, for each $i\in N$,

$$\begin{aligned} \nonumber u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )&= \hat{g}(\theta _{-i})-g(\theta _{-i})\\ \nonumber&= \Gamma (\theta _{-i})\ge 0.\end{aligned}$$

By construction, there does not exist $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ such that $f(\theta _{1},\theta _{2})=1$ and $\hat{f}(\theta _{1},\theta _{2})=0$. Hence, there remains one final case:

$$\begin{aligned} \text {for each } i\in N,\; \theta _{i}\ge \Delta \text { and } f(\theta _{1},\theta _{2})=0, \end{aligned}$$

or, equivalently,

$$\begin{aligned} \text {for each } i\in N,\; \Delta \le \theta _{i}<\beta ^{f}. \end{aligned}$$

We obtain, for each $i\in N$,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )=\big [ \theta _{i}-\Delta +\Delta \big ]-g(\theta _{-i}). \end{aligned}$$

Since $\theta _{-i}<\Delta$, by 6.7, $g(\theta _{-i})\le -\beta ^{f}$. Hence,

$$\begin{aligned} u\big (\hat{\varphi }_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )\ge \Delta +\beta ^{f}>0. \end{aligned}$$

Thus, $\hat{\varphi }$ dominates $\varphi$.

Step 3. The mechanism $\varphi =(f,t')$ associated with the payment scheme $(p,g')$ satisfies Feasibility, Strategy-proofness and Anonymity.

The proof is omitted for the sake of brevity (it is available on request).

We can now prove the main statement. Consider the mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g). Let $\varphi$ be an undominated mechanism. Hence, by Step 2, $p:\mathbb {R}\rightarrow \mathbb {R}$ is such that, for each $\theta <\beta ^{f}$, $p(\theta )\ne \beta ^{f}$. We will show that $g(\theta )=g^{f}(\theta )$. By weak monotonicity, the assumption that $\beta ^{f}>0$ and the definition of $\beta ^{f}$, for each $\theta >\beta ^{f}$, $p(\theta )\le \beta ^{f}$. Hence, by Lemma 2,

$$\begin{aligned} \text {for each }\theta >\beta ^{f},\; g(\theta )\le \beta ^{f}. \end{aligned}$$

By Lemma 2,

$$\begin{aligned} g(\theta )\le \left\{ \begin{array}{ll} 0 &{} \text {if } f(\beta ^{f},\beta ^{f})=0,\\ p(\beta ^{f}) &{} \text {if } f(\beta ^{f},\beta ^{f})=1. \end{array} \right. \end{aligned}$$

By Step 1, there exists $\theta ''\ge \beta ^{f}$ such that $g(\theta '')\simeq \beta ^{f}$. As demonstrated above, $p(\theta '')\le \beta ^{f}$. Hence, by Lemma 2, $p(\theta '')\simeq \beta ^{f}$. Hence, for each $\theta <\beta ^{f}$, $f(\theta ,\theta '')=0$. Hence, by Proposition 1 and Feasibility, for each $\theta <\beta ^{f}$, $g(\theta )\le -g(\theta '')$ and, since $g(\theta '')\simeq \beta ^{f}$,

$$\begin{aligned} \text {for each }\theta <\beta ^{f},\; g(\theta )\le -\beta ^{f}. \end{aligned}$$

Therefore, for each $\theta \in \mathbb {R}$, $g^{f}(\theta )\ge g(\theta )$. By Step 3, the mechanism $\varphi '=(f,t')$, associated with the payment scheme $(p,g')$ is strategy-proof, anonymous and feasible. Moreover, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i,\in N$,

$$\begin{aligned} u\big (\varphi '_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )= g^{f}(\theta )-g(\theta )\ge 0. \end{aligned}$$

Moreover, if for some $\theta ''\in \mathbb {R}$, $g^{f}(\theta '')>g(\theta '')$, this inequality becomes strict and $\varphi '$ dominates $\varphi$. Hence, for each $\theta \in \mathbb {R}$, $g(\theta )=g^{f}(\theta )$.

Lemma 6

If the mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), satisfies Feasibility, Strategy-proofness and Anonymity, and $\beta ^{f}=0$, then $\varphi$ is Undominated if and only if $f \in \mathcal {F}$ and for each $\theta \in \mathbb {R}$,

$$\begin{aligned} g^{f}(\theta )= \min \{0,p(\theta ) \}. \end{aligned}$$

Proof

Let $\varphi$ be an undominated mechanism. Let mechanism $\varphi '=(f,t')$ be associated with the payment scheme $(p,g')$. We will show that $g(\theta )=g'(\theta )$. By construction and Proposition 1, $\varphi '$ is strategy-proof and anonymous. Moreover, it can be readily verified that $\varphi '$ is feasible (the proof is omitted for the sake of brevity). We will show that for each $\theta \in \mathbb {R},$ $g^{f}(\theta )\ge g(\theta ).$

Suppose that $\theta <0$. Since $\beta ^{f}=0$, by weak monotonicity, we obtain $p(\theta )\ge 0$. Thus, $g'(\theta )=0$. By Lemma 2, $g(\theta )\le 0$. Suppose that $\theta >0$. Since $\beta ^{f}=0$, by weak monotonicity, we obtain $p(\theta )\le 0$. Thus, $g^{f}(\theta )=p(\theta )$. By Lemma 2, $g(\theta )\le p(\theta )$. Suppose that $\theta =0$ and $p(0)<0$. Hence, $g^{f}(\theta )=p(0)$. Moreover, by weak monotonicity, $p(0)<0$ implies that $f(0,0)=1$. Thus, by Lemma 2, $g(0)\le p(0)$. Finally, suppose that $\theta =0$ and $p(0)\ge 0$. Hence, $g^{f}(0)=0$. By Lemma 2, either $g(0)\le p(0)$ or $g(0)\le 0$. By assumption, $p(0)\ge 0$. Hence, in both cases, $g(0)\le g^{f}(0)$. Therefore, for each $\theta \in \mathbb {R}$, $g^{f}(\theta )\ge g(\theta )$. Moreover, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ and each $i\in N$,

$$\begin{aligned} u\big (\varphi '_{i}(\theta _{1},\theta _{2});\theta _{i}\big )- u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big )= g^{f}(\theta )-g(\theta )\ge 0. \end{aligned}$$

Moreover, if for some $\theta ''\in \mathbb {R}$, $g^{f}(\theta '')>g(\theta '')$, this inequality becomes strict and $\varphi '$ dominates $\varphi$. Hence, for each $\theta \in \mathbb {R}$, $g(\theta )=g^{f}(\theta )$.

Lemma 7

A strategy-proof and anonymous mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), is Undominated only if $f\in \mathcal {F}$ satisfies Condition W.

Proof

Suppose that $\varphi$ is Undominated. The proof proceeds by contradiction. Let $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f)$ and assume that there exists $i \in N$ such that, $f(\tilde{\theta }_i,\mathbb {R})=k \in \{0,1\}$. Let us assume that $k=0$⁹. By construction and by the weak monotonicity of f, since $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f)$,

$$\begin{aligned} \tilde{\theta }_i \le \beta ^{f} \text { and } \tilde{\theta }_{-i} \ge \beta ^{f}. \end{aligned}$$

(6.8)

Let

$${\tilde{\mathcal{D}}} = \{ (\theta _{1} ,\theta _{2} ) \in \mathbb{R}^{2} \;:\;{\text{either}}\;\theta _{1} = \tilde{\theta }_{i} \;{\text{and}}\;\theta _{2} \ge \tilde{\theta }_{{ - i}} \;{\text{or}}\;\theta _{1} \ge \tilde{\theta }_{{ - i}} \;{\text{and}}\;\theta _{2} = \tilde{\theta }_{i} \} .{\text{ }}$$

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, let

$$\begin{aligned} f^{\prime }(\theta _{1},\theta _{2})= \left\{ \begin{array}{ll} 1 &{} \text {if } (\theta _{1},\theta _{2})\in {\tilde{\mathcal{D}}},\\ f(\theta _{1},\theta _{2}) &{} \text {otherwise}. \end{array} \right. \end{aligned}$$

Step 1. $f^{\prime }$ is non-decreasing.

Consider any pair $(\theta ^{'}_{1},\theta ^{'}_{2}),(\theta ^{''}_{1},\theta ^{''}_{2})\in \mathbb {R}^{2}$ such that $(\theta ^{'}_{1},\theta ^{'}_{2}) \le (\theta ^{''}_{1},\theta ^{''}_{2})$. Assume first that $\left( \theta ^{'}_{1},\theta ^{'}_{2}\right) \notin \tilde{D}$. Hence,

$$\begin{aligned} f\left( \theta ^{'}_{1},\theta ^{'}_{2}\right) = f\left( \theta ^{'}_{1},\theta ^{'}_{2}\right) \le f\left( \theta ^{''}_{1},\theta ^{''}_{2}\right) \le f'\left( \theta ^{''}_{1},\theta ^{''}_{2}\right) . \end{aligned}$$

The first inequality holds because f is weakly monotonic and the second holds by construction. Suppose then $(\theta ^{'}_{1},\theta ^{'}_{2}) \in \tilde{D}$. This implies that, for some $i \in N$, $\theta '_i = \tilde{\theta }_i$ and $\theta '_{-i} \ge \tilde{\theta }_{-i}$. Hence, since by assumption $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f)$ and since f is weakly monotonic,

$$\begin{aligned} \tilde{\theta }_{i} = p(\tilde{\theta }_{-i}) \Rightarrow {\theta }'_{i} \ge p(\theta '_{-i}). \end{aligned}$$

(6.9)

Since, by assumption, $(\theta ^{'}_{1},\theta ^{'}_{2}) \le (\theta ^{''}_{1},\theta ^{''}_{2})$, then either $(\theta ^{''}_{1},\theta ^{''}_{2}) \in \tilde{D}$, so that $f(\theta ^{''}_{1},\theta ^{''}_{2})=1$ or there exists $j \in N$ such that $\theta ^{''}_j > \theta '_j \ge p(\theta '_{-j})$. The latter inequality holds because of 6.9. Hence, $f(\theta ^{''}_1,\theta ^{''}_2)=1$. Hence, $f'(\theta ''_1,\theta ''_2)=1$ and $f^{\prime }(\theta ^{''}_{1},\theta ^{''}_{2})\ge f^{\prime }(\theta ^{'}_{1},\theta ^{'}_{2})$.

By construction, for each $\theta \in \mathbb {R}$,

$$\begin{aligned} p^{\prime }(\theta )= \left\{ \begin{array}{ll} \tilde{\theta }_{-i} &{} \text {if} \ \theta =\tilde{\theta _i} \\ p(\theta ) &{}\text {otherwise. } \end{array} \right. \end{aligned}$$

(6.10)

For each $\theta \in \mathbb {R}$ let $g'(\theta )=g^{f^{\prime }}(\theta )$. By Step 1 and construction, $f^{\prime }\in \mathcal {F}$. Let the mechanism $\varphi ^{\prime }=(f',t')$ be associated with the payment scheme $(p^{\prime }, g^{\prime })$. By Lemma 4, 5, 6 and by construction, $\varphi ^{\prime }$ is feasible. Moreover, since $\varphi$ is undominated, for each $\theta \in \mathbb {R}$, $g(\theta )=g^{f}(\theta )$

Step 2. $\varphi ^{\prime }$ dominates $\varphi$.

At any profile $(\theta _{1},\theta _{2}) \notin {\tilde{\mathcal{D}}}$ one has that $\varphi ^{\prime }$ and $\varphi$ are, by construction, welfare equivalent. Consider then some $(\theta _{1},\theta _{2}) \in {\tilde{\mathcal{D}}}$ such that $f^{\prime }(\theta _{1},\theta _{2})>f(\theta _{1},\theta _{2})$. Hence, for each $i\in N$,

$$\begin{aligned} u\big (\varphi ^{\prime }_{i}((\theta _{1},\theta _{2});\theta _{i}\big )-u\big (\varphi _{i}((\theta _{1},\theta _{2}) ; \theta _{i}\big )=\theta _{i}-p^{\prime }(\theta _{-i})+g^{\prime }(\theta _{-i})-g(\theta _{-i}). \end{aligned}$$

Since $(\theta _{1} ,\theta _{2} ) \in {\tilde{\mathcal{D}}}$ then, for some $j \in N$, $\theta _j = \tilde{\theta }_j$ and hence, by 6.9, $p^{\prime }(\theta _{-j})\le \theta _j$. Moreover, by appealing to 6.8 and to the definitions of $g^{f}(\theta )$ and $g^{f^{\prime }}(\theta )$, $g^{\prime }(\theta _{-j})=g(\theta _{-j})$. Hence,

$$\begin{aligned} u\big ((\varphi ^{\prime }_{j}(\theta _{1},\theta _{2});\theta _{j}\big )-u\big (\varphi _{j}((\theta _{1},\theta _{2}) ; \theta _{j}\big ) \ge 0. \end{aligned}$$

Moreover, by 6.10, $p'(\theta _{j})=\tilde{\theta }_{-j}$, and appealing again to 6.8 and the definitions of $g^{f}(\theta )$ and $g^{f^{\prime }}(\theta )$, $g^{\prime }(\theta _{j})=g(\theta _{j})$. Hence,

$$\begin{aligned} u\big (\varphi ^{\prime }_{-j}((\theta _{1},\theta _{2});\theta _{-j}\big )-u\big ((\varphi _{-j}(\theta _{1},\theta _{2}) ; \theta _{-j}\big )= \theta _{-j} - \tilde{\theta }_{-j}. \end{aligned}$$

By construction, $\theta _{-j} \ge \tilde{\theta }_{-j}$. Finally, for some $\theta _{-j} > \tilde{\theta }_{-j}$ the expression above is strictly positive. Thus, $\varphi ^{\prime }$ dominates $\varphi$, the desired contradiction.

Lemma 8

A strategy-proof and anonymous mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), is Undominated only if $f\in \mathcal {F}$ satisfies Condition F.

Proof

Suppose that $\varphi$ is Undominated. The proof proceeds by contradiction. Let $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f)$. Hence, by Condition W, for each $i\in N$, $f(\mathbb {R},\theta _i)=\{0,1\}$. Hence, a violation of Condition F ensues if¹⁰

$$\begin{aligned} \tilde{\theta }_{1}>\beta ^{f}\text { and } \tilde{\theta }_{2}>p(\tilde{\theta }_{1}). \end{aligned}$$

(6.11)

Let

$$\begin{aligned} {\tilde{\mathcal{D}}}=\Big \{(\theta _{1},\theta _{2}) \in \mathbb {R}^{2}\; :\; \text {there exists } i \in N\text { s.t. } \theta _i = \tilde{\theta }_{1} \ \text {and} \ \theta _{-i} \in \big [ p(\tilde{\theta }_{1}),\tilde{\theta }_{2}\big ] \Big \}. \end{aligned}$$

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, let

$$\begin{aligned} f^{\prime }(\theta _{1},\theta _{2})= \left\{ \begin{array}{ll} -1 &{} \text {if } (\theta _{1},\theta _{2}) \in \tilde{\mathcal {D}},\text { and}\\ f(\theta _{1},\theta _{2}) &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

By the same argument employed in Step 1 of Lemma 7, $f^{\prime }$ is non-decreasing. Moreover, by construction, $f'$ is symmetric. Hence, $f' \in \mathcal {F}$. For each $\theta \in \mathbb {R}$ let $g'(\theta )=g^{f^{\prime }}(\theta )$. Let the mechanism $\varphi ^{\prime }=(f',t')$ be associated with the payment scheme $(p^{\prime }, g^{\prime })$. By Lemma 4, 5, 6 and by construction, $\varphi ^{\prime }$ is feasible. We will now prove that $\varphi ^{\prime }$ dominates $\varphi$ which yields the desired contradiction.

By construction, for each $(\theta _{1} ,\theta _{2} ) \in \mathbb{R}^{2} \backslash \left\{ {{\tilde{\mathcal{D}}}} \right\}$ and each $i\in N$,

$$\begin{aligned} u\big (\varphi ^{\prime }_{i}(\theta _1,\theta _{2});\theta _{i}\big )=u\big (\varphi _{i}(\theta _{1},\theta _{2}) ; \theta _{i}\big ). \end{aligned}$$

Hence, without loss of generality let $\theta _{1}=\tilde{\theta }_1$. By construction, for each $\theta \in \mathbb {R}$,

$$\begin{aligned} p^{\prime }(\theta )= \left\{ \begin{array}{ll} \tilde{\theta }_2 &{} \text {if} \ \theta =\tilde{\theta _1}, \\ p(\theta ) &{}\text {otherwise. } \end{array} \right. \end{aligned}$$

(6.12)

Moreover, since $\varphi$ is undominated, for each $\theta \in \mathbb {R}$, $g(\theta )=g^{f}(\theta )$. Hence, by appealing to the definitions of $g^{f}(\theta )$ and $g^{f^{\prime }}(\theta )$, independently of the value of $\beta ^{f}\in \mathbb {R}$, we obtain,

$$\begin{aligned}&u\big (\varphi ^{\prime }_{2}(\tilde{\theta }_1,\theta _{2});\theta _{2}\big )-u\big (\varphi _{2}(\tilde{\theta }_{1},\theta _{2}) ; \theta _{2}\big )=\\&\quad = \left\{ \begin{array}{ll} \tilde{\theta }_{2}-p(\tilde{\theta }_{1}) &{} \text {if } \theta _{2}<p(\tilde{\theta }_{1}), \\ \tilde{\theta }_{2}-\theta _{2} &{} \text {if } p(\tilde{\theta }_{1}) \le \theta _{2} \le \tilde{\theta }_{2},\\ 0 &{} \text {if } \theta _{2} > \tilde{\theta }_{2}. \\ \end{array} \right. \\&u\big (\varphi ^{\prime }_{1}(\tilde{\theta }_1,\theta _{2});\tilde{\theta }_1\big )-u\big (\varphi _{1}(\tilde{\theta }_1,\theta _{2}) ; \tilde{\theta }_1\big )=0-0=0. \end{aligned}$$

By 6.11, $\tilde{\theta }_{2}>p(\tilde{\theta }_{1})$. Thus, $\varphi ^{\prime }$ dominates $\varphi$.

Lemma 9

A strategy-proof, feasible and anonymous mechanism $\varphi =(f,t)$, associated with the payment scheme (p, g), is Undominated only if $p(\beta ^{f})=\beta ^{f}$.

Proof

Suppose that $\varphi$ is Undominated and, moreover, $\beta ^{f}<0$. By Step 5 in Lemma 1, $p(\beta ^{f})\le \beta ^{f}$. Suppose, for the sake of contradiction, that $p(\beta ^{f})<\beta ^{f}$. Hence, $f(\beta ^{f},\beta ^{f})=1$. Moreover, since by assumption $\varphi$ is Undominated, then by Lemma 1 and by assumption, $g(\theta )=g^f(\theta )$ and

$$\begin{aligned} g^f(\beta ^{f})=p(\beta ^{f})< \beta ^{f}< 0. \end{aligned}$$

(6.13)

Let

$${\tilde{\mathcal{D}}} = \{ (\theta _{1} ,\theta _{2} ) \in \mathbb{R}^{2} \;:\;{\text{there exists }}i \in N{\text{ s}}{\text{.t}}{\text{. }}\theta _{i} = \beta ^{f} \;{\text{and}}\;\theta _{{ - i}} \in [p(\beta ^{f} ),\beta ^{f} )\} .$$

For each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$, let

$$\begin{aligned} f^{\prime }(\theta _{1},\theta _{2})= \left\{ \begin{array}{ll} 0 &{} \text {if } (\theta _{1},\theta _{2}) \in \tilde{\mathcal {D}},\text { and }\\ f(\theta _{1},\theta _{2}) &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

By the same argument employed in Step 1 of Lemma 7, $f^{\prime }$ is non-decreasing . Moreover, by construction, $f'$ is symmetric. Hence, $f' \in \mathcal {F}$. For each $\theta \in \mathbb {R}$, let $g'(\theta )=g^{f^{\prime }}(\theta )$. Let the mechanism $\varphi ^{\prime }=(f',t')$ be associated with the payment scheme $(p^{\prime }, g^{\prime })$. By Lemmas 4, 5, 6 and by construction, $\varphi ^{\prime }$ is feasible. We will now prove that $\varphi ^{\prime }$ dominates $\varphi$ which yields the desired contradiction.

By construction, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}$ such that $\theta _{1} \ne \beta ^{f}$ and $\theta _{2} \ne \beta ^{f}$ and each $i\in \{1,2\}$,

$$\begin{aligned} u\big (\varphi ^{\prime }_{i}(\theta _1,\theta _{2});\theta _{i}\big )=u\big (\varphi _{1}(\theta _{1},\theta _{2}) ; \theta _{i}\big ). \end{aligned}$$

Hence, without loss of generality let $\theta _{1}=\beta ^{f}$. By construction, for each $\theta \in \mathbb {R}$,

$$\begin{aligned} p^{\prime }(\theta )= \left\{ \begin{array}{ll} \beta ^{f} &{} \text {if} \ \theta =\beta ^{f}, \\ p(\theta ) &{}\text {otherwise. } \end{array} \right. \end{aligned}$$

(6.14)

By Lemma 3 we may assume, without loss of generality, that $f'(\beta ^{f},\beta ^{f})=1$. Hence, by appealing to the definition of $g^{f^{\prime }}(\theta )$,

$$\begin{aligned} g^{f^{\prime }}\left( \beta ^{f}\right) =p'\left( \beta ^{f}\right) =\beta ^{f}. \end{aligned}$$

(6.15)

Finally, in light of equations 6.13 and 6.15 we obtain,

$$\begin{aligned}&u\big (\varphi ^{\prime }_{2}(\theta _1,\theta _{2});\theta _{2}\big )-u\big (\varphi _{2}(\theta _{1},\theta _{2}) ; \theta _{2}\big )=\\&\quad = \left\{ \begin{array}{ll} \theta _{2}-p'(\beta ^{f})+g^{f^{\prime }}\left( \beta ^{f}\right) -\theta _{2}+p\left( \beta ^{f}\right) -g\left( \beta ^{f}\right) =0 &{} \text {if } \theta _{2} \ge \beta ^{f},\\ g^{f^{\prime }}\left( \beta ^{f}\right) -\theta _{2}+p\left( \beta ^{f}\right) -g\left( \beta ^{f}\right) =\beta ^{f}-\theta _{2}>0 &{} \text {if } p\left( \beta ^{f}\right) \le \theta _2< \beta ^{f},\\ g^{f^{\prime }}\left( \beta ^{f}\right) -g\left( \beta ^{f}\right) = \beta ^{f} - p\left( \beta ^{f}\right) >0 &{} \text {if } \theta _{2} < p(\beta ^{f}). \\ \end{array} \right. \end{aligned}$$

Moreover,

$$\begin{aligned}&u\big (\varphi _{1}^{\prime }(\theta _1,\theta _{2});\theta _{1}\big )-u\big (\varphi _{1}(\theta _{1},\theta _{2}) ; \theta _{1}\big )=\\&\quad = \left\{ \begin{array}{ll} 0 &{} \text {if } \theta _{2} \ge \beta ^{f}, \\ g^{f^{\prime }}(\theta _2)-\beta ^{f} + p(\theta _2)-g(\theta _2)=-\beta ^{f}-p(\theta _2) \ge 0 &{} \text {if } p\left( \beta ^{f}\right) \le \theta _2< \beta ^{f},\\ 0 &{} \text {if } \theta _{2} < p\left( \beta ^{f}\right) . \\ \end{array} \right. \end{aligned}$$

Thus, $\varphi ^{\prime }$ dominates $\varphi$, a contradiction. The proof for the cases where $\beta ^{f}=0$ and $\beta ^{f}>0$ follows very closely the proof we have just provided. For the sake of brevity they are omitted.

1.3 Sufficiency

Lemma 10

Let $f,f'\in \mathcal {F}$ and $\beta ^{f}=\beta ^{f'}$. If $\varphi =(f,t)$ is associated with payment scheme $(p,g^{f})$ and $\varphi '=(f',t')$ is associated with payment scheme $(p',g^{f'})$, then $\varphi$ does not dominate $\varphi ^\prime$ and $\varphi ^\prime$ does not dominate $\varphi$.

Proof

Suppose that $\beta ^{f}=\beta ^{f'}=\beta \ge 0$. The proof for the case $\beta ^{f}=\beta ^{f'}<0$ is similar and, thus, omitted.

Case 1: For each $\theta \in \mathbb {R}$, $p(\theta )=p'(\theta )$.

If, moreover, for each $(\theta _{1},\theta _{2})\in \mathbb {R}^{2}\setminus \{(\beta ,\beta )\}$, $f(\theta _{1},\theta _{2})=f'(\theta _{1},\theta _{2})$, then by Lemma 3, $\varphi$ and $\varphi '$ are welfare equivalent and the statement is proven. Suppose then that there exists $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in \mathbb {R}^{2}\setminus \{(\beta ,\beta )\}$ such that $f(\tilde{\theta }_{1},\tilde{\theta }_{2})\ne f'(\tilde{\theta }_{1},\tilde{\theta }_{2})$. First, we will show that $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f)$ and $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f')$. Suppose not. Without loss of generality let $(\tilde{\theta }_{1},\tilde{\theta }_{2})\notin B(f)$. If for each $i\in N$, $f(\mathbb {R},\tilde{\theta }_{i})=k \in \{0,1\}$, then for each $i\in N$,

$$\begin{aligned} f\left( \tilde{\theta }_{1},\tilde{\theta }_{2}\right) =0 \implies \tilde{\theta }_{i}\le \beta \text { and } f\left( \tilde{\theta }_{1},\tilde{\theta }_{2}\right) =1\implies \tilde{\theta }_{i}\ge \beta . \end{aligned}$$

If, moreover, for some $j\in N$, $\tilde{\theta }_{j}=\beta$, then $f(\mathbb {R},\beta )=k$. Moreover, by definition $(\beta ,\beta ) \in B(f)$. Hence, we obtain a violation of Condition W. Hence,

$$\begin{aligned}&\text {either } \big< (\tilde{\theta }_{1},\tilde{\theta }_{2})\ll (\beta ,\beta )\text { and } f'(\tilde{\theta }_{1},\tilde{\theta }_{2})=1\big>, \\&\text {or } \big < (\tilde{\theta }_{1},\tilde{\theta }_{2}) \gg (\beta ,\beta )\text { and } f'(\tilde{\theta }_{1},\tilde{\theta }_{2})=0\big >. \end{aligned}$$

Either way, this constitutes a contradiction. Hence,

$$\begin{aligned} \text {for some }j\in N,\; f(\mathbb {R},\tilde{\theta }_{-j} )=\{0,1\}. \end{aligned}$$

(6.16)

Hence, since $(\tilde{\theta }_{1},\tilde{\theta }_{2})\notin B(f)$, $p(\tilde{\theta }_{-j})\ne \theta _{j}$. Hence, since by assumption $p(\tilde{\theta }_{-j})=p'(\tilde{\theta }_{-j})$, $p'(\tilde{\theta }_{-j})\ne \theta _{j}$. Hence, if $f'(\mathbb {R},\tilde{\theta }_{-j} )=\{0,1\}$, by 6.16 and since, by assumption, $f(\tilde{\theta }_{1},\tilde{\theta }_{2})\ne f'(\tilde{\theta }_{1},\tilde{\theta }_{2})$, we obtain

$$\begin{aligned} \text {either } \big< p(\tilde{\theta }_{-j})<\tilde{\theta }_{j}< p'(\tilde{\theta }_{-j}) \big> \text { or } \big < p(\tilde{\theta }_{-j})>\tilde{\theta }_{j}> p'(\tilde{\theta }_{-j})\big >, \end{aligned}$$

which contradicts the assumption $p(\tilde{\theta }_{-j})=p'(\tilde{\theta }_{-j})$. Hence,

$$\begin{aligned} f\left( \mathbb {R},\tilde{\theta }_{-j} \right) =k\in \{0,1\}. \end{aligned}$$

(6.17)

Hence, since $p\left( \tilde{\theta }_{-j}\right) =p'(\tilde{\theta }_{-j})$, by 6.17,

$$\begin{aligned} p(\tilde{\theta }_{-j})=0.\end{aligned}$$

(6.18)

Moreover, by 6.17,

$$\begin{aligned} f'\left( \tilde{\theta }_{1},\tilde{\theta }_{2}\right) =0 \implies \tilde{\theta }_{-j}\le \beta \text { and }f'\left( \tilde{\theta }_{1},\tilde{\theta }_{2}\right) =1\implies \tilde{\theta }_{-j}\ge \beta .\end{aligned}$$

(6.19)

Hence, by 6.16, 6.18 and 6.19, we obtain $\beta =0$. Hence, if $\tilde{\theta }_{-j}=0$, then

$$\begin{aligned}&f(0,0)=0 \implies \text {for each } y>0,\; (0,y)\in B(f), \text { and }\\&f(0,0)=1 \implies \text {for each } y<0,\; (0,y)\in B(f). \end{aligned}$$

Either way a violation of Condition W ensues. Hence,

$$\begin{aligned} \tilde{\theta }_{-j}>\beta =0.\end{aligned}$$

(6.20)

Without loss of generality let $f(\tilde{\theta }_{1},\tilde{\theta }_{2})=1$. Hence, by 6.16, 6.17 and 6.18, for each $q<0$,

$$\begin{aligned} u\big (\varphi _{j}(q,\tilde{\theta }_{-j}) ; q\big )-u\big (\varphi ^{\prime }_{j}(q,\tilde{\theta }_{-j}) ; q\big )= & {}\ q, \\ \nonumber u\big (\varphi _{-j}(q,\tilde{\theta }_{-j}) ; \tilde{\theta }_{-j}\big )-u\big (\varphi ^{\prime }_{-j}(q,\tilde{\theta }_{-j}) ; \tilde{\theta }_{-j}\big )= & {}\ \tilde{\theta }_{-j}. \end{aligned}$$

By assumption, $q<0$. Hence, by 6.20, $\varphi$ does not dominate $\varphi ^{\prime }$ and $\varphi ^{\prime }$ does not dominate $\varphi$.

Therefore, for each $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in \mathbb {R}^{2}$ such that $f(\tilde{\theta }_{1},\tilde{\theta }_{2})\ne f'(\tilde{\theta }_{1},\tilde{\theta }_{2})$, $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f)$ and $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in B(f')$. Hence, by Conditions W and F, for each $(\tilde{\theta }_{1},\tilde{\theta }_{2})\in \mathbb {R}^{2}$ and each $i\in N$, $p(\tilde{\theta }_{i})=p'(\tilde{\theta }_{i})$. Hence, $\varphi$ and $\varphi '$ are welfare equivalent and the statement is proven.

Case 2. There exists $\tilde{\theta }>\beta$ such that either $p(\tilde{\theta })>p'(\tilde{\theta })$ or $p(\tilde{\theta })<p'(\tilde{\theta })$.

Without loss of generality, let $\tilde{\theta }_{1}>\beta$ and $p(\tilde{\theta }_{1})>p'(\tilde{\theta }_{1})$. Let $\tilde{\theta }_{2}\in \big (p'(\tilde{\theta }_{1}),p(\tilde{\theta }_{1})\big )$. Hence, since $\tilde{\theta }_{1}>\beta$,

$$\begin{aligned} \text {there exists } y\in \mathbb {R}\text { such that } f\left( \tilde{\theta }_{1},y\right) =f'\left( \tilde{\theta }_{1},y\right) =1. \end{aligned}$$

Hence either $f(\tilde{\theta }_{1},\tilde{\theta }_{2})<f'(\tilde{\theta }_{1},\tilde{\theta }_{2})$ or $f(\tilde{\theta }_{1},\tilde{\theta }_{2})=f'(\tilde{\theta }_{1},\tilde{\theta }_{2})=1$. Suppose first that $f(\tilde{\theta }_{1},\tilde{\theta }_{2})=0$ and $f'(\tilde{\theta }_{1},\tilde{\theta }_{2})=1$. Hence,

$$\begin{aligned} u\big (\varphi _{1}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{1}\big )-u\big (\varphi ^{\prime }_{1}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{1}\big )=&g^{f}(\tilde{\theta }_{2})-\tilde{\theta }_{1}+p'(\tilde{\theta }_{2})-g^{f'}(\tilde{\theta }_{2}),\\ \nonumber u\big (\varphi _{2}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{2}\big )-u\big (\varphi ^{\prime }_{2}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{2}\big )=&g^{f}(\tilde{\theta }_{1})-\tilde{\theta }_{2}+p'(\tilde{\theta }_{1})-g^{f'}(\tilde{\theta }_{1}). \end{aligned}$$

By assumption, $\tilde{\theta }_{1}>\beta$. Hence, by construction, $g^{f}(\tilde{\theta }_{1})=p(\tilde{\theta }_{1})$ and $g^{f'}(\tilde{\theta }_{1})=p'(\tilde{\theta }_{1})$. By assumption, $\tilde{\theta }_{2}<p(\tilde{\theta }_{1})$. Since $\tilde{\theta }_{1}>\beta \ge 0$, by weak monotonicity, $p(\tilde{\theta }_{1})\le \beta$. Hence, $\tilde{\theta }_{2}<\beta$. Hence, by construction, $g^{f}(\tilde{\theta }_{2})=g^{f'}(\tilde{\theta }_{2})=-\beta$. Therefore,

$$\begin{aligned}&u\big (\varphi _{1}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{1}\big )-u\big (\varphi ^\prime _{1}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{1}\big )= \\&\quad = \left\{ \begin{array}{ll} -\tilde{\theta }_{2}+p'(\tilde{\theta }_{2}) &{} \text {if } \tilde{\theta }_{2}<\beta , \\ -\beta -\tilde{\theta }_{2} &{} \text {if } \tilde{\theta }_{2}\ge \beta , \end{array} \right. \\&u\big (\varphi _{2}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{2}\big )-u\big (\varphi ^\prime _{2}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{2}\big )= p(\tilde{\theta }_{1})-\tilde{\theta }_{2}. \end{aligned}$$

By assumption, $p(\tilde{\theta }_{1})-\tilde{\theta }_{2}>0$. Suppose that for each $x\in \mathbb {R},\; f'(x,\tilde{\theta }_{2})=1$. Hence, $p'(\tilde{\theta }_{2})=0$. Hence, since $\tilde{\theta }_{1}>\beta \ge 0$, $-\tilde{\theta }_{1}+p'(\tilde{\theta }_{2})<0$. Suppose that $f'(\mathbb {R},\tilde{\theta }_{2})=\{0,1\}$. Hence, since $f'(\tilde{\theta }_{1},\tilde{\theta }_{2})=1$, $\tilde{\theta }_{1}\ge p'(\tilde{\theta }_{2})$. Suppose that $\tilde{\theta }_{1}= p'(\tilde{\theta }_{2})$. Hence, $p'(\tilde{\theta }_{1})= p'\big (p'(\tilde{\theta }_{2})\big )$. However, by symmetry of f one has $p'\big (p'(\tilde{\theta }_{2})\big )=\tilde{\theta _{2}}$, which yields $p'(\tilde{\theta }_{1})=\tilde{\theta _{2}}$, whereas by assumption, $\tilde{\theta }_{2}>p'(\tilde{\theta }_{1})$. Hence, we arrive at a contradiction. Hence, $\tilde{\theta }_{1}> p'(\tilde{\theta }_{2})$. Thus,

$$\begin{aligned} u\big (\varphi _{1}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{1}\big )-u\big (\varphi ^\prime _{1}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{1}\big ) <0, \\ \nonumber u\big (\varphi _{2}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{2}\big )-u\big (\varphi ^\prime _{2}(\tilde{\theta }_{1},\tilde{\theta }_{2}) ; \tilde{\theta }_{2}\big ) >0. \end{aligned}$$

Therefore, $\varphi$ does not dominate $\varphi ^\prime$ and $\varphi ^\prime$ does not dominate $\varphi$.

Suppose next that $f(\tilde{\theta }_{1},\tilde{\theta }_{2})=f'(\tilde{\theta }_{1},\tilde{\theta }_{2})=1$. By assumption, $p(\tilde{\theta }_{1})>\tilde{\theta }_{2}$. Hence, for each $x\in \mathbb {R}$, $f(\tilde{\theta }_{1},x)=1$. Hence, $p(\tilde{\theta }_{1})=0$. By assumption, $p(\tilde{\theta }_{1})>p'(\tilde{\theta }_{1})$. Hence, $p'(\tilde{\theta }_{1})<0$. Therefore for some $\theta '_{2}<p'(\tilde{\theta }_{1})$, $f'(\tilde{\theta }_{1},\theta '_{2})<f(\tilde{\theta }_{1},\theta '_{2})$. Thus,

$$\begin{aligned} u\big (\varphi _{1}(\tilde{\theta }_{1},\theta '_{2}) ; \tilde{\theta }_{1}\big )-u\big (\varphi '_{1}(\tilde{\theta }_{1},\theta '_{2}) ; \tilde{\theta }_{1}\big )=&\tilde{\theta }_{1}-p(\theta '_{2})+g^{f}(\theta '_{2})-g^{f'}(\theta '_{2}),\\ \nonumber u\big (\varphi _{2}(\tilde{\theta }_{1},\theta '_{2}) ; \theta '_{2}\big )-u\big (\varphi ^{\prime }_{2}(\tilde{\theta }_{1},\theta '_{2}) ; \theta '_{2}\big )=&\theta '_{2}-p(\tilde{\theta }_{1})+g^{f}(\tilde{\theta }_{1})-g^{f'}(\tilde{\theta }_{1}). \end{aligned}$$

By assumption, $\tilde{\theta }_{1}>\beta$. By construction, $\theta '_{2}<\beta$. Hence,

$$\begin{aligned} u\big (\varphi _{1}(\tilde{\theta }_{1},\theta '_{2}) ; \tilde{\theta }_{1}\big )-u\big (\varphi '_{1}(\tilde{\theta }_{1},\theta '_{2}) ; \tilde{\theta }_{1}\big )=&\tilde{\theta }_{1}-p(\theta '_{2}), \\ \nonumber u\big (\varphi _{2}(\tilde{\theta }_{1},\theta '_{2}) ; \theta '_{2}\big )-u\big (\varphi '_{2}(\tilde{\theta }_{1},\theta '_{2}) ; \theta '_{2}\big )=&\theta '_{2}-p'(\tilde{\theta }_{1}). \end{aligned}$$

By construction, $\theta '_{2}-p(\tilde{\theta }_{1})<0$. By symmetry of f and the fact that for each $x\in \mathbb {R}$, $f(\tilde{\theta }_{1},x)=1$ it cannot be that for each $x\in \mathbb {R}$, $f(x,\theta '_{2})=0$. Hence, $f'(\mathbb {R},\tilde{\theta }_{2})=\{0,1\}$. Hence, by the same argument as before, $f(\tilde{\theta }_{1},\theta '_{2})=1$ implies that $\theta '_{2}-p'(\tilde{\theta }_{1})>0$. Thus,

$$\begin{aligned} u\big (\varphi _{1}\left( \tilde{\theta }_{1},\theta '_{2}\right) ; \tilde{\theta }_{1}\big )-u\big (\varphi ^{\prime }_{1}(\tilde{\theta }_{1},\theta '_{2}) ; \tilde{\theta }_{1}\big )>0,\\ \nonumber u\big (\varphi _{2}\left( \tilde{\theta }_{1},\theta '_{2}\right) ; \theta '_{2}\big )-u\big (\varphi ^{\prime }_{2}(\tilde{\theta }_{1},\theta '_{2}) ; \theta '_{2}\big )<0. \end{aligned}$$

Therefore, $\varphi$ does not dominate $\varphi ^{\prime }$ and $\varphi ^{\prime }$ does not dominate $\varphi$.

Lemma 11

Let $f,f'\in \mathcal {F}$ and $\beta ^{f}\times \beta ^{f'} \ge 0$. If $\varphi =(f,t)$ is associated with payment scheme $(p,g^{f})$ and $\varphi '=(f',t')$ is associated with payment scheme $(p',g^{f'})$, then $\varphi$ does not dominate $\varphi ^\prime$ and $\varphi ^\prime$ does not dominate $\varphi$.

Proof

Suppose that $\beta ^{f}>\beta ^{f'} \ge 0$. The proof for the case $\beta ^{f}<\beta ^{f'}\le 0$ is similar and, thus, omitted. By condition (i) in the statement of the Theorem

$$\begin{aligned} p(\beta ^{f'}) \ne \beta ^{f}. \end{aligned}$$

(6.21)

By the weak monotonicity of f, it cannot be $f(\mathbb {R},\beta ^{f'})=1$. By condition W, it cannot be $f(\mathbb {R},\beta ^{f'})=0$ and $(x,\beta ^{f'}) \in B(f)$. Hence, let $f(\mathbb {R},\beta ^{f'})=0$ and $(x,\beta ^{f'}) \notin B(f)$. Hence,

$$\begin{aligned} \text { for each } x> \beta ^{f}, p(x)>\beta ^{f}. \end{aligned}$$

(6.22)

Suppose finally that $f(\mathbb {R},\beta ^{f'})=\{0,1\}$. By 6.21 and the weak monotonicity of f one obtains $p(\beta ^{f'}) > \beta ^{f}$. Hence,

$$\begin{aligned} \text {there exists } \tilde{\theta }_{1} \in (\beta ^{f},p(\beta ^{f'})) \text { such that } f(\tilde{\theta }_{1},\beta ^{f'})=0. \end{aligned}$$

(6.23)

Moreover, by Lemma 9, $p(\beta ^{f'})=\beta ^{f'}$. Hence,

$$\begin{aligned} \text {there exists } \tilde{\theta }_{1}>\beta ^{f} \text { such that } f'(\tilde{\theta }_{1},\beta ^{f'})=1. \end{aligned}$$

(6.24)

By 6.23 and 6.24 there exists $\tilde{\theta }_{1}>\beta ^{f}$ such that $f(\tilde{\theta }_{1},\beta ^{f'})<f'(\tilde{\theta }_{1},\beta ^{f'})$. Hence,

$$\begin{aligned} u\big (\varphi _{2}(\tilde{\theta }_{1},\beta ^{f'}) ; \beta ^{f'}\big )-u\big (\varphi ^\prime _{2}(\tilde{\theta }_{1},\beta ^{f'}) ; \beta ^{f'}\big )= p(\tilde{\theta }_{1})-\beta ^{f'}. \end{aligned}$$

(6.25)

Since $f(\tilde{\theta }_{1},\beta ^{f'})=0$ then $p(\tilde{\theta }_{1})\ge \beta ^{f'}$. Suppose that for each $x \in (\beta ^{f},p(\beta ^{f'}))$ such that $f(x,\beta ^{f'})=0$, $p(x)=\beta ^{f'}$. Then $(x,\beta ^{f'}) \in B(f)$. This leads to a violation of Condition F (according to which $\beta ^{f'} < \beta ^{f} \Rightarrow x \ge p(\beta ^{f'})$). Hence, there exists $x' \in (\beta ^{f},p(\beta ^{f'}))$ such that $f(x',\beta ^{f'})=0$ and $p(x')>\beta ^{f'}$. Let $x'=\tilde{\theta }_{1}$. By 6.22 and 6.25,

$$\begin{aligned} u\big (\varphi _{2}(\tilde{\theta }_{1},\beta ^{f'}) ; \beta ^{f'}\big )-u\big (\varphi ^\prime _{2}(\tilde{\theta }_{1},\beta ^{f'}) ; \beta ^{f'}\big )>0. \end{aligned}$$

(6.26)

Moreover, for each $y<\beta ^{f'}$ and each $i \in N$, $f(y,y)=f'(y,y)=0$ and

$$\begin{aligned} u\big (\varphi _{i}(y,y) ; y\big )-u\big (\varphi ^\prime _{i}(y,y) ; y\big )=\beta ^{f'}-\beta ^{f}<0. \end{aligned}$$

(6.27)

Therefore, by 6.26 and 6.27, $\varphi$ does not dominate $\varphi ^\prime$ and $\varphi ^\prime$ does not dominate $\varphi$.

Lemma 12

Let $f,f'\in \mathcal {F}$, $\beta ^{f}\ne \beta ^{f'}$ and $\beta ^{f}\times \beta ^{f'} < 0$. If $\varphi =(f,t)$ is associated with payment scheme $(p,g^{f})$ and $\varphi '=(f',t')$ is associated with payment scheme $(p',g^{f'})$, then $\varphi$ does not dominate $\varphi ^\prime$ and $\varphi ^\prime$ does not dominate $\varphi$.

Proof

By assumption, either $\beta ^{f}<0< \beta ^{f'}$ or $\beta ^{f'}< 0<\beta ^{f}$. Suppose that the first inequality holds. The proof of the other case is similar and, thus, omitted. Hence, for each $y<\beta ^{f}$ we obtain $f(y,y)=f'(y,y)=0$. Moreover, for each $i\in N$,

$$\begin{aligned} u\big (\varphi _{i}(y,y) ; y\big )-u\big (\varphi ^\prime _{i}(y,y) ; y\big )=\beta ^{f'}>0. \end{aligned}$$

(6.28)

Also, for each $x>\beta (f')$, $f(x,x)=f'(x,x)=1$. Hence, for each $i\in N$,

$$\begin{aligned} u\big (\varphi _{i}(x,x) ; x\big )-u\big (\varphi ^\prime _{i}(x,x) ; x\big )= \beta ^{f}<0. \end{aligned}$$

(6.29)

Therefore, by 6.28 and 6.29, $\varphi$ does not dominate $\varphi ^\prime$ and $\varphi ^\prime$ does not dominate $\varphi$.

1.4 Proof of Corollary to Theorem 1

By Theorem 1, Unanimity mechanisms are undominated. It is straightforward to demonstrate that they satisfy Budget-balance. Let us turn to the only if part of the statement. Let us consider a strategy-proof, anonymous, feasible, and undominated mechanism $\varphi =(f,t)$ that satisfies Budget-balance. By Proposition 1 this mechanism can be associated with a payment scheme (p, g).

Step 1. $\beta ^{f}=0$.

Let $\epsilon >0$. Hence, by construction, $f\big (\beta ^{f}-\epsilon ,\beta ^{f}-\epsilon \big )<f\big (\beta ^{f}+\epsilon ,\beta ^{f}+\epsilon \big )$. Moreover, by Proposition 1,

$$\begin{aligned} \sum _{i\in N}t_{i}\big (\beta ^{f}-\epsilon ,\beta ^{f}-\epsilon \big )&=2g\big (\beta ^{f}-\epsilon \big ). \\ \sum _{i\in N}t_{i}\big (\beta ^{f}+\epsilon ,\beta ^{f}+\epsilon \big )&=\left\{ \begin{array}{ll} 2\beta ^{f} &{} \text {if } \beta ^{f}<0, \\ 0 &{} \text {if } \beta ^{f}\ge 0. \end{array} \right. \end{aligned}$$

By Theorem 1,

$$\begin{aligned} g\big (\beta ^{f}-\epsilon \big )= \left\{ \begin{array}{ll} 0 &{} \text {if } \beta ^{f}<0, \\ -2\beta ^{f} &{} \text {if } \beta ^{f}\ge 0. \end{array} \right. \end{aligned}$$

Hence, unless $\beta ^{f}=0$, Budget-balance is contradicted.

Let $\mathcal {U}\subset \mathcal {F}$ be the set of functions f satisfying either 1 or 2 of the following conditions:

• 1. $\forall (\theta _{1},\theta _{2})\in \mathbb {R}^{2},\; \big <\exists j\in N$ s.t. $\theta _{j}>\beta ^{f}\big > \implies f(\theta _{1},\theta _{2})=1$.

• 2. $\forall (\theta _{1},\theta _{2})\in \mathbb {R}^{2},\; \big <\exists j\in N$ s.t. $\theta _{j}<\beta ^{f}\big > \implies f(\theta _{1},\theta _{2})=0$.

Step 2. $f\in \mathcal {U}$.

By construction, if $f\notin \mathcal {U}$, then there exists $\theta _{1}^{\prime },\theta _{1}^{\prime \prime },\theta _{2}^{\prime }\in \mathbb {R}^{3}$, with $\theta _{1}^{\prime }<\theta _{1}^{\prime \prime }<\beta ^{f}<\theta _{2}^{\prime }$, such that $f(\theta _{1}^{\prime },\theta _{2}^{\prime })=0$ and $f(\theta _{1}^{\prime \prime },\theta _{2}^{\prime })=1$. By Step 1, $\beta ^{f}=0$. Hence, by Theorem 1, $\sum _{i\in N}t_{i}(\theta _{1}^{\prime },\theta _{2}^{\prime })=p(\theta _{2}^{\prime })$. Moreover, since $f(\theta _{1}^{\prime \prime },\theta _{2}^{\prime })=1$, $\theta _{1}^{\prime \prime }\ge p(\theta _{2}^{\prime })$. By assumption and Step 1, $\theta _{1}^{\prime \prime }<\beta ^{f}=0$. Hence, $\sum _{i\in N}t_{i}(\theta _{1}^{\prime },\theta _{2}^{\prime })=p(\theta _{2}^{\prime })<0$, a contradiction.

Combining Step 1 and 2 the proof is complete.