Abstract
In the private values single object auction model, we construct a satisfactory mechanism—a dominant strategy incentive compatible and budgetbalanced mechanism satisfying equal treatment of equals. Our mechanism allocates the object with positive probability to only those agents who have the highest value and satisfies expost individual rationality. This probability is at least \((1\frac{2}{n})\), where n is the number of agents. Hence, our mechanism converges to efficiency at a linear rate as the number of agents grow. Our mechanism has a simple interpretation: a fixed allocation probability is allocated using a secondprice Vickrey auction whose revenue is redistributed among all the agents in a simple way. We show that our mechanism maximizes utilitarian welfare among all satisfactory mechanisms that allocate the object only to the highestvalued agents.
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Notes
They consider a more general problem with property rights. In our problem, there are no property rights. We can assign equal property rights to all the agents and apply their result.
We are grateful to an anonymous referee for suggesting this.
This means that for \(n=5\), the difference in expected welfare is 0.05.
This follows from the toponly property of our mechanism.
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The authors are grateful to an anonymous referee whose detailed comments has improved the paper significantly. The authors also thank Bhaskar Dutta, Herve Moulin, Anish Sarkar, and seminar participants at Indian Statistical Institute for their comments.
Appendix
Appendix
1.1 Proof of Theorem 1
First, we show that \(M^*\) is a satisfactory mechanism—it is clearly a toponly mechanism. Fix a valuation profile \({\mathbf {v}}\) with \(v_1 \ge v_2 \ge \cdots \ge v_n\), and observe the following using the definition of \(p_i({\mathbf {v}})\) for each i:
This establishes that \(M^*\) is budgetbalanced. For DSIC, we invoke the characterization of Myerson (1981), which states that an arbitrary mechanism \(M \equiv (f,{\mathbf {p}})\) is DSIC if and only if

1.
Monotonicity. for all \(i \in N\), for all \(v_{i}\), and for all \(v_i,v'_i\) with \(v_i > v'_i\), we have
$$\begin{aligned} f_i(v_i,v_{i}) \ge f_i(v'_i,v_{i}). \end{aligned}$$(1) 
2.
Revenue equivalence. for all \(i \in N\), for all \(v_{i}\), and for all \(v_i\), we have
$$\begin{aligned} p_i(v_i,v_{i})=p_i(0,v_{i}) + v_if_i(v_i,v_{i})  \int _0^{v_i}f_i(x_i,v_{i})dx_i. \end{aligned}$$(2)
Monotonicity is clearly satisfied by \(f^*\) and revenue equivalence is satisfied by \({\mathbf {p}}^*\) by definition. Hence, \(M^*\) is DSIC. Finally, since \(f^*\) is symmetric, \({\mathbf {p}}^*\) is also symmetric by construction. Hence, \(M^*\) satisfies ETE. This implies that \(M^*\) is a toponly satisfactory mechanism.
For individual rationality, note that for every \(i \in N\) and for all \({\mathbf {v}}\), using revenue equivalence, we have
where the inequality follows since \(p^*_i(0,v_{i}) \le 0\) by definition.
Now, we move to the second part of the proof where we show that our toponly satisfactory mechanism maximizes utilitarian welfare in the class of all toponly satisfactory mechanisms. To do this, we define some additional properties of an allocation rule, which is satisfied by \(f^*\).
Definition 9
An allocation rule f satisfies property
 P0. :

if for every \({\mathbf {v}}\) with \({\mathbf {v}}[1] = 2\), we have \(f_i({\mathbf {v}})=0\) for all \(i \notin {\mathbf {v}}[1]\).
 P1. :

if for every \({\mathbf {v}}\) with \({\mathbf {v}}[1] > 2\), we have \(\sum _{i \in {\mathbf {v}}[1]}f_i({\mathbf {v}})=1\).
 P2. :

if for every \({\mathbf {v}}\) with \({\mathbf {v}}[1]=\{k\}\) and \({\mathbf {v}}[2] > 1\), we have \(f_k({\mathbf {v}})=1\).
Notice that \(f^*\) satisfies Properties P0, P1, and P2. Before completing the proof of the theorem, we state and prove an important proposition.
Proposition 3
Suppose \((f,{\mathbf {p}})\) is a satisfactory mechanism and f satisfies Properties P0, P1, and P2. Then, for every \({\mathbf {v}}\) with \(v_1 \ge v_2 \ge v_3 \ge \cdots \ge v_n\), we have
Proof
We start off by establishing a property of payments.
Lemma 1
Suppose \((f,{\mathbf {p}})\) is a satisfactory mechanism and f satisfies Properties P0, P1, and P2. For every \(v_{1} \equiv (v_2,v_3,\ldots ,v_n)\) with \(v_2 \ge v_3 \ge \cdots \ge v_n,\) we have
Proof
We do the proof in three steps.
Step 1. Pick \(v_{1}\) such that \(v_2 = v_3 = \theta \ge v_4 \ge \cdots \ge v_n\). Pick a type profile \({\mathbf {v}} \equiv (v_1,v_{1})\) such that \(v_1=\theta \). If \(\theta =0\) this is the zero type profile, and by ETE and budgetbalance, the claim is true. Hence, suppose that \(\theta > 0\). Let \(K:=(0,v_{1})[1]\). Since \(K \ge 2\), we have \({\mathbf {v}}[1] > 2\), and Property P1 implies that \(\sum _{i \in {\mathbf {v}}[1]}f_i({\mathbf {v}})=1\). Further, consider a type profile \((x_1,v_{1})\), where \(x_1 < \theta \). Such a type profile also satisfies \((x_1,v_{1})[1] > 1\), and Property P0 and P1 imply that \(f_1(x_1,v_{1})=0\).
We now do the proof using induction on K. Using the observations in the previous paragraph along with ETE and revenue equivalence formula, we get for all \(i \in {\mathbf {v}}[1]\),
If \(K=n1\), then \({\mathbf {v}}[1]=N\), and adding the above inequalities and using ETE and BB, we get
Else, we assume that for all \(K' > K\), the claim is true. Then, we have for all \(i \notin {\mathbf {v}}[1]\), \((0,v_{i})[1]={\mathbf {v}}[1] = K+1\), and induction hypothesis implies that
Adding Eqs. (3) and (4), and using BB and ETE, we get
Simplifying, we get,
This shows that if \((0,v_{1})[1] > 1\), then the claim is true.
Step 2. Let \({\mathbf {v}}\) be a type profile such that for all \(k > 2\) and for all \(i \in {\mathbf {v}}[k]\), we have \(v_i=0\), and \({\mathbf {v}}[1]=1\) and \({\mathbf {v}}[2] > 1\). In this step, we show that if \(\theta =v_i > 0\) for every \(i \in {\mathbf {v}}[2]\), then
Suppose \({\mathbf {v}}[1]=\{1\}\). By step 1,
Further, by Property P2, \(f_1({\mathbf {v}})=1\). Further, for all \(x_1 \in (\theta ,v_1)\), we have \(f_1(x_1,v_{1})=1\) and for all \(x_1 < \theta \), we have \(f_1(x_1,v_{1})=0\)—the latter observation follows from the fact that \((x_1,v_{1})[1] > 1\) and Properties P0 and P1. Hence, using Eqs. (2) and (5), we get
Suppose \({\mathbf {v}}[2]=K\). By Property P2, \(f_i({\mathbf {v}})=0\) for all \(i \in {\mathbf {v}}[2]\). Hence, for each \(i \in {\mathbf {v}}[2]\), Eq. (2) implies that
If \(K=n1\), by adding Eqs. (6) and (7), and using BB and ETE, we get for every \(i \in {\mathbf {v}}[2]\),
This simplifies to \(p_i(0,v_{i})=\frac{\theta }{n}\).
Now, we use induction on K. Suppose the claim is true for all \(K' > K\) and \(K < n1\). By construction, for all \(j > 2\) and for all \(i \in {\mathbf {v}}[j]\), \(v_i=0\). We can construct another type profile \({\mathbf {v}}'\) such that \(v'_i=\theta \) and \(v'_j=v_j\) for all \(j \ne i\). Note that \({\mathbf {v}}'[2]=K+1\). Hence, induction hypothesis implies that
Adding Eqs. (6)–(8), and using BB and ETE we get for every \(i \in {\mathbf {v}}[2]\),
This simplifies to \(p_i(0,v_{i})=\frac{\theta }{n}\), as desired.
Step 3 Now, we complete the proof. Pick a \({\mathbf {v}}\) with \({\mathbf {v}}[1]=\{1\}\) and \({\mathbf {v}}[2] > 1\). Suppose \(v_k=\theta > 0\) for all \(k \in {\mathbf {v}}[2]\). Note that by Step 1, the claim is proved if we show that for all \(i \notin {\mathbf {v}}[1]\), we have \(p_i(0,v_{i})=\frac{\theta }{n}\)—in this case \((0,v_{i})\) is a type profile such that \((0,v_{i})[1] = 1\).
Suppose \(K={\mathbf {v}}[2]\). We use induction on K. If \(K=n1\), the claim follow from step 2. Suppose the claim is true for all \(K' > K\). Pick \(i \in {\mathbf {v}}[k]\), where \(k > 2\). We can construct a type profile \({\mathbf {v}}'\) with \(v'_i=\theta \) and \(v_j=v'_j\) for all \(j \ne i\). Since \({\mathbf {v}}'[2]=K+1\), induction hypothesis implies that
Now, at type profile \({\mathbf {v}}\), we know that \({\mathbf {v}}[1]=\{1\}\) and \({\mathbf {v}}[2] > 1\). By Property P2, \(f_1({\mathbf {v}})=1\) and for all \(x_1 \in (\theta ,v_1)\), we have \(f_1(x_1,v_{1})=1\). Further, by Properties P0 and P1, \(f_1(x_1,v_{1})=0\) for all \(x_1 < \theta \). Using these observations and Eq. (2), we get
where the second equality follows from step 1. Since \(f_i({\mathbf {v}})=0\) for all \(i \ne 1\), we can argue the following. For every \(i \in {\mathbf {v}}[2]\), we have
For every \(i \in {\mathbf {v}}[k]\), where \(k > 2\), using Eq. (9),
Adding Eqs. (10)–(12), and using ETE we get for every \(i \in {\mathbf {v}}[2]\),
Simplifying, we get \(p_i(0,v_{i})=\frac{\theta }{n}\), as desired. \(\square \)
Now, we complete the proof of Proposition 3. Suppose \((f,{\mathbf {p}})\) is a satisfactory mechanism and f satisfies Properties P0, P1, and P2. Using Lemma 1, we immediately get that \(p_i(0,v_{i})=\frac{v_3}{n}\) if \(i \in \{1,2\}\) and \(p_i(0,v_{i})=\frac{v_2}{n}\) if \(i \notin \{1,2\}\). Using these equations, we get \(\sum _{i \in N}p_i(0,v_{i})=\frac{1}{n} [(n2)v_2+2v_3]\). \(\square \)
Now, we complete the remaining part of Proof of Theorem 1. Assume for contradiction that mechanism \(\mathcal {\tilde{M}} \equiv ({\tilde{f}},{\tilde{\mathbf {p}}})\) is a toponly satisfactory mechanism such that for all \({\mathbf {v}}\), we have
with strict inequality holding for some \({\mathbf {v}}\).
Every toponly allocation rule satisfies Property P0. Since \(f^*\) satisfies Properties P1 and P2, Eq. (13) implies that \({\tilde{f}}\) satisfies Properties P1 and P2—this is because an implication of Eq. (13) is that \({\tilde{f}}\) is efficient at all valuation profiles where \(f^*\) is efficient, and \(f^*\) is efficient at the profiles mentioned in Properties P1 and P2.
Then, by Proposition 3, we have for all \({\mathbf {v}}\) with \(v_1 \ge v_2 \ge \cdots \ge v_n\),
Note that if \(v_2=v_3\), then Properties P1 and P2 imply that \({\tilde{f}}_1({\mathbf {v}})=f^*_1({\mathbf {v}})=1\). Now suppose \(v_2 > v_3\). If \(v_1=v_2\), then by revenue equivalence formula and using the fact that \({\tilde{f}}_1(x_1,v_{1})={\tilde{f}}_2(x_2,v_{2})=0\) for all \(x_1,x_2 < v_1 (=v_2)\), we get
Adding and using budgetbalance and ETE, we have
Using Eq. (14), we get
Hence, if \(v_1=v_2\) or \(v_2=v_3\), by toponly property \({\tilde{f}}=f^*\). Since Eq. (13) holds strictly for some \({\mathbf {v}}\), such a valuation profile must satisfy \(v_1> v_2 > v_3\). By toponly property and Eq. (13), we must have
But then,
which is a contradiction.
This completes the proof of Theorem 1.
1.2 Proof of Theorem 2
We do the proof with the help of some lemmas.
Lemma 2
Suppose f is a satisfactorily implementable and satisfies Properties P0, P1, and P2. Then, for all valuation profiles \({\mathbf {v}} \in V^n\) and for \(i \in {\mathbf {v}}[k]\) with \(k > 2,\) we have
Proof
Consider a valuation profile \({\mathbf {v}}\). Denote the valuation of agents in \({\mathbf {v}}[k]\) for any k as \(\theta _k\). Pick an agent \(i \in {\mathbf {v}}[3]\) and consider a valuation profile \({\mathbf {v}}'\) as follows: \(v'_j=v_j\) if \(j \ne i\) and \(v'_i=\theta _2\) (in other words, valuation of agent i is increased to second ranked valuation). Note that \({\mathbf {v}}'[2] > 1\) and \(i \in {\mathbf {v}}'[2]\). Hence, by Properties P0, P1, and P2, we have \(f_i({\mathbf {v}}')=0\). By monotonicity of f and ETE, we get that \(f_i({\mathbf {v}})=0\) and \(f_j({\mathbf {v}})=0\) for all \(j \in {\mathbf {v}}[3]\).
We now use induction. We assume that at any valuation profile \({\mathbf {v}}\) and for all \(k < K\) and \(k \ge 3\), we have \(f_j({\mathbf {v}})=0\) for all \(j \in {\mathbf {v}}[k]\). We now show that \(f_j({\mathbf {v}})=0\) for all \(j \in {\mathbf {v}}[K]\). To do so, we pick an agent \(i \in {\mathbf {v}}[K]\) and construct a valuation profile \({\mathbf {v}}'\) as follows: \(v'_j=v_j\) if \(j \ne i\) and \(v'_i=\theta _{K1}\). Since \(i \in {\mathbf {v}}'[K1]\), by the induction hypothesis, \(f_i({\mathbf {v}}')=0\). By monotonicity of f and ETE, we get that \(f_i({\mathbf {v}})=0\) and \(f_j({\mathbf {v}})=0\) for all \(j \in {\mathbf {v}}[K]\). This completes the proof. \(\square \)
The next lemma uses the following strengthening of Properties P0 and P1.
Definition 10
An allocation rule f satisfies Property C1 if for every \({\mathbf {v}}\) with \({\mathbf {v}}[1] > 1\), we have \(\sum _{i \in {\mathbf {v}}[1]}f_i({\mathbf {v}})=1\).
Note that Property C1 implies Properties P0 and P1.
Lemma 3
Suppose f is a satisfactorily implementable allocation rule satisfying Properties C1 and P2. Then, for every \(v_{2} \equiv (v_1,v_3,v_4,\ldots ,v_n)\) with \(v_1 > v_3 \ge v_4 \ge \cdots \ge v_n,\) we have
Proof
Consider \(v_{2} \equiv (v_1,v_3,v_4,\ldots ,v_n)\) with \(v_1 > v_3 \ge v_4 \ge \cdots \ge v_n\). Further, consider a type profile \({\mathbf {v}}'\) such that \(v'_2=v_1\) and \(v'_j=v_j\) for all \(j \ne 1\). Since C1 implies Properties P0 and P1, Proposition 3 gives
Observing that \({\mathbf {v}}'[1]  > 1\) and using Property C1, we get \(f_1({\mathbf {v}}')+f_2({\mathbf {v}}')=1\). By the revenue equivalence formula, \(p_i({\mathbf {v}}')=p_i(0,v'_{i})\) for all \(i \notin \{1,2\}\) and for all \(i \in \{1,2\}\),
where we used the fact that \(v'_j=v_j\) for all \(j \ne 2\) and ETE. Using budgetbalance, we get that
Using Eqs. (16) and (17), and simplifying we get
\(\square \)
We are now ready to complete the proof of Theorem 2.
Proof of Theorem 2
The fact that the MGL mechanism is satisfactory is routine to check—BB and ETE is clear, and for DSIC, one can either do a direct check of incentive constraints or verify that the revenue equivalence formula holds.
To prove that the MGL mechanism maximizes utilitarian welfare across all satisfactory mechanisms, suppose there is a satisfactory mechanism \(M \equiv (f,{\mathbf {p}})\) such that
with strict inequality satisfying for some \({\mathbf {v}}\). This implies that f is efficient at all valuation profiles where \(f^{G'}\) is efficient. Then f must satisfy Properties C1 and P2.
Choose a type profile \({\mathbf {v}}\) with \(v_1 \ge v_2 \ge \cdots \ge v_n\). Note that if \(v_1=v_2\) or \(v_2=v_3\), then Properties C1 and P2 imply that \(f^{G'}({\mathbf {v}})=f({\mathbf {v}})\). So, we consider \({\mathbf {v}}\) such that \(v_1> v_2 > v_3 \ge v_4 \ge \cdots \ge v_n\).
Now, for any \(x_2 \in (v_3,v_1)\), Lemma 2 implies that \(f_j(x_2,v_{2})=0\) for all \(j > 2\). Hence, Eq. (18) implies that
Using \(f_1(x_2,v_{2})+f_2(x_2,v_{2}) \le 1\), we simplify this to get
But \(v_1 > x_2\) implies that \(f_1(x_2,v_{2}) \ge 11/n\) and \(f_2(x_2,v_{2}) \le 1/n\). Using Lemma 3 along with monotonicity of \(f_2\), we get \(f_2(x_2,v_{2})=1/n\), and hence, \(f_1(x_2,v_{2})=11/n\) for all \(x_2 \in (v_3,v_1)\). This implies that \(f_1({\mathbf {v}})=11/n\) and \(f_2({\mathbf {v}})=1/n\) as desired.
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Mishra, D., Sharma, T. A simple budgetbalanced mechanism. Soc Choice Welf 50, 147–170 (2018). https://doi.org/10.1007/s0035501710780
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DOI: https://doi.org/10.1007/s0035501710780