Strength in numbers: robust mechanisms for public goods with many agents

Abstract

This study examines the mechanism design problem for public goods provision in a large economy with n independent agents. We propose a class of dominant-strategy incentive compatible and ex-post individually rational mechanisms, which we call the adjusted mean-thresholding (AMT) mechanisms. We show that when the cost of provision grows slower than the $\sqrt{n}$-rate, the AMT mechanisms are both eventually ex-ante budget balanced and asymptotically efficient. When the cost grows faster than the $\sqrt{n}$-rate, in contrast, we show that any incentive compatible, individually rational, and eventually ex-ante budget balanced mechanism must have provision probability converging to zero and hence cannot be asymptotically efficient. The AMT mechanisms have a simple form and are more informationally robust when compared to, for example, the second-best mechanism. This is because the construction of an AMT mechanism depends only on the first moment of the valuation distribution.

Appendices

Technical proofs

Appendix A.1 presents general convergence results dervied based on the Berry-Esseen theorem. Appendix A.2 presents the proofs for theorems in the main text.

1.1 Preliminary results

This section provides some general results as useful lemmas. We denote $\phi$ and $\Phi$ respectively as the pdf and cdf of the standard normal distribution. We use C to denote a generic constant that does not depend on n, which may have different values at each appearance.

Lemma 2

Let X and Y be two random variables with ${\mathbb {E}}|{X}| < \infty$, then for any $\alpha \in {\mathbb {R}}$,

$$\begin{aligned} {\mathbb {E}} \left[ X {\mathbf {1}}_{[\alpha ,\infty )}(Y) \right] = \int _{0}^\infty {\mathbb {P}}\left( X > x,Y \ge \alpha \right) dx - \int _{-\infty }^0 {\mathbb {P}}\left( X < x,Y \ge \alpha \right) dx. \end{aligned}$$

In particular, if $X=Y$, then

$$\begin{aligned} {\mathbb {E}} \left[ X {\mathbf {1}}_{[\alpha ,\infty )}(X) \right] = \int _{0}^\infty {\mathbb {P}}\left( X >x \right) dx - \int _{\alpha }^0 {\mathbb {P}}\left( X < x\right) dx, \text { for } \alpha \le 0, \end{aligned}$$

and

$$\begin{aligned} {\mathbb {E}} \left[ X {\mathbf {1}}_{[\alpha ,\infty )}(X) \right] = \int _{\alpha }^\infty {\mathbb {P}}\left( X>x \right) dx, \text { for } \alpha > 0. \end{aligned}$$

Proof of Lemma 2

Define $X^+ = \max \{X,0\}$ and $X^- = \max \{-X,0\}$. For $X^+$, notice that

$$\begin{aligned} X^+ = \int _0^\infty {\mathbf {1}}_{[0,X^+]}(x) dx. \end{aligned}$$

Using Fubini theorem, we have

$$\begin{aligned} {\mathbb {E}} \left[ X^+ {\mathbf {1}}_{[\alpha ,\infty )}(Y) \right]&= {\mathbb {E}} \left[ \int _0^\infty {\mathbf {1}}_{[0,X^+]}(x) {\mathbf {1}}_{[\alpha ,\infty )}(Y) dx \right] \\&= \int _0^\infty {\mathbb {E}} \left[ {\mathbf {1}}_{[0,X^+]}(x) {\mathbf {1}}_{[\alpha ,\infty )}(Y) \right] dx \\&= \int _0^\infty {\mathbb {P}} \left( X^+> x, Y \ge \alpha \right) dx \\&= \int _0^\infty {\mathbb {P}} \left( X > x, Y \ge \alpha \right) dx. \end{aligned}$$

Similarly, for $X^-$, we have

$$\begin{aligned} {\mathbb {E}} \left[ X^- {\mathbf {1}}_{[\alpha ,\infty )}(Y) \right]&= \int _0^\infty {\mathbb {P}} \left( X^- > x, Y \ge \alpha \right) dx \\&= \int _{-\infty }^0 {\mathbb {P}} \left( X < x, Y \ge \alpha \right) dx. \end{aligned}$$

The result then follows from the fact that ${\mathbb {E}}[X] = {\mathbb {E}}[X^+] - {\mathbb {E}}[X^-]$. $\square$

Let $(X_i,Y_i), 1 \le i \le n$ be an independent sequence of random vectors in ${\mathbb {R}}^2$ with zero mean and ${\mathbb {E}}|{X_i}|^3,{\mathbb {E}}|{Y_i}|^3 < \infty$. We introduce the following set of notations for their marginal moments: $\sigma ^2_{X} = {\mathbb {E}}X_i^2, \sigma _{Y} = {\mathbb {E}}Y_i^2,\rho _{X} = {\mathbb {E}}|{X_i}|^3, \rho _{Y} = {\mathbb {E}}|{Y_i}|^3,$ and correlation: $\sigma _{XY} = {\mathbb {E}}[X_iY_i].$ All these moments are finite. Define $S^X_n = \sum X_i$ and $S^Y_n = \sum Y_i$. Let $(Z^X,Z^Y)$ be a joint normal random vector with zero mean and the same covariance structure as $(S^X_n,S^Y_n)$. We use $\Vert {\cdot }\Vert$ to denote both the induced 2-norm of matrices and the Euclidean norm of vectors.

We want to bound the difference between the distributions of $(S^X_n,S^Y_n)$ and $(Z^X,Z^Y)$ using the Berry-Esseen theorem. The following lemma is the univariate Berry-Esseen theorem.

Lemma 3

There is a constant $C > 0$ such that

$$\begin{aligned} \sup _{x \in {\mathbb {R}}} |{ {\mathbb {P}}\left( S_n^X \le x \right) - \Phi \left( \frac{x}{\sqrt{n} {\sigma }_{X}} \right) }| \le \frac{C}{\sqrt{n}} . \end{aligned}$$

Then we prove the multivariate case.

Lemma 4

Assume X and Y are not perfectly correlated. The following bound holds between the joint distributions of $\left( S^X_n,S^Y_n \right)$ and $\left( Z^X,Z^Y \right)$: let ${\mathcal {B}}$ be the set of all measurable convex sets in ${\mathbb {R}}^2$, then there is a constant $C>0$ such that

$$\begin{aligned}&\sup _{B \in {\mathcal {B}}} |{{\mathbb {P}}\left( \left( S^X_n,S^Y_n \right) \in B \right) - {\mathbb {P}}\left( \left( Z^X,Z^Y \right) \in B\right) }| \le \frac{C}{\sqrt{n}}. \end{aligned}$$

Proof of Lemma 4

We use $\varSigma _{XY}$ to denote the normalized (by n) covariance matrix of $(S^X_n,S^Y_n)$,

$$\begin{aligned} \varSigma _{XY} = \begin{pmatrix} {\sigma }^2_{X} &{} {\sigma }_{XY} \\ {\sigma }_{XY} &{} {\sigma }^2_{Y} \end{pmatrix}. \end{aligned}$$

The multivariate Berry-Esseen theorem (Bentkus 2005) says there exists a universal constant C, such that

$$\begin{aligned}&\sup _{B \in {\mathcal {B}}} |{{\mathbb {P}}\left( (S^X_n,S^Y_n) \in B \right) - {\mathbb {P}}\left( (Z^X,Z^Y) \in B\right) }| \\&\le C \sum {\mathbb {E}} \left[ \Vert {(n\varSigma _{XY})^{-1/2} (X_i,Y_i)'}\Vert ^3 \right] \\&\le \frac{C}{\sqrt{n}} \left( \frac{1}{n} \sum {\mathbb {E}} \left[ \Vert {\varSigma _{XY}^{-1/2}}\Vert ^3 \Vert {(X_i,Y_i)'}\Vert ^3 \right] \right) \\&= \frac{C}{\sqrt{n}} \Vert {\varSigma _{XY}^{-1/2}}\Vert ^3 {\mathbb {E}} \left[ \Vert {(X_i,Y_i)'}\Vert ^3 \right] . \end{aligned}$$

The induced 2-norm of a positive semi-definite matrix equals to its largest eigenvalue. So we have

$$\begin{aligned} \Vert {\varSigma _{XY}^{-1/2}}\Vert ^3 = \lambda _{\text {min}}^{-3/2}, \end{aligned}$$

where $\lambda _{\text {min}}$ is the smallest eigenvalue of $\varSigma _{XY}$. We next compute $\lambda _{\text {min}}$. The characteristic function of $\varSigma _{XY}$ is

$$\begin{aligned} \text {det} \begin{pmatrix} {\bar{\sigma }}^2_{X} - \lambda &{} {\sigma }_{XY} \\ {\sigma }_{XY} &{} {\sigma }^2_{Y} - \lambda \end{pmatrix} = \lambda ^2 - ({\sigma }^2_{X} + {\sigma }^2_{Y}) \lambda + {\sigma }^2_{X}{\sigma }^2_{Y} - {\sigma }_{XY}^2. \end{aligned}$$

The smaller eigenvalue is

$$\begin{aligned} \lambda _{\text {min}} =&\frac{1}{2} \left( {\sigma }^2_{X} + {\sigma }^2_{Y} - \sqrt{({\sigma }^2_{X} + {\sigma }^2_{Y})^2 - 4({\sigma }^2_{X}{\sigma }^2_{Y} - {\sigma }_{XY}^2)} \right) \\ =&\frac{1}{2} \left( {\sigma }^2_{X} + {\sigma }^2_{Y} - \sqrt{({\sigma }^2_{X} - {\sigma }^2_{Y})^2 - 4{\sigma }_{XY}^2} \right) , \end{aligned}$$

which is a positive constant when $\sigma ^2_X\sigma ^2_Y - \sigma _{XY}^2 \ne 0$.

To deal with the remaining part of the bound, we employ the Minkowski inequality.

$$\begin{aligned} {\mathbb {E}} \left[ \Vert {(X_i,Y_i)'}\Vert ^3 \right]&= {\mathbb {E}} \left[ (X_i^2 + Y_i^2)^{3/2} \right] \\&\le \left( \left( {\mathbb {E}}\left[ (X_i^2)^{3/2} \right] \right) ^{2/3} + \left( {\mathbb {E}}\left[ (Y_i^2)^{3/2} \right] \right) ^{2/3} \right) ^{3/2} \\&= \left( \left( {\mathbb {E}} |{X_i}|^{3}\right) ^{2/3} + \left( {\mathbb {E}}|{Y_i}|^{3} \right) ^{2/3} \right) ^{3/2} \\&= \left( \rho _{X}^{2/3} + \rho _{Y}^{2/3} \right) ^{3/2} \\&\le \sqrt{2}(\rho _{X} + \rho _{Y}). \end{aligned}$$

The last inequality follows from the fact that the function $x \mapsto x^{2/3}$ is concave so that for any two positive numbers a and b, it holds that

$$\begin{aligned} \frac{a^{2/3}+b^{2/3}}{2} \le \left( \frac{a+b}{2} \right) ^{2/3} \implies \left( a^{2/3}+b^{2/3} \right) ^{3/2} \le \sqrt{2}(a+b). \end{aligned}$$

$\square$

Lemma 5

The following expression of the normal truncated mean holds true:

$$\begin{aligned} {\mathbb {E}}\left[ Z^X {\mathbf {1}}\left\{ Z^Y \ge \alpha _n \right\} \right] = \sqrt{n} \frac{{\sigma }_{XY}}{{\sigma }_{Y}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{Y}} \right) . \end{aligned}$$

In particular, for $X = Y$, we have

$$\begin{aligned} {\mathbb {E}}\left[ Z^X {\mathbf {1}}\left\{ Z^X \ge \alpha _n \right\} \right] = \sqrt{n} {\sigma }_{X} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{X}} \right) . \end{aligned}$$

Proof of Lemma 5

Define $e = Z^X - \frac{{\sigma }_{XY}}{{\sigma }^2_{Y}} Z^Y$. It is straightforward to compute that ${\mathbb {E}}e=0$. Since e and $Z^Y$ are jointly normal and uncorrelated, $e \perp Z^Y$. Therefore,

$$\begin{aligned} {\mathbb {E}}\left[ Z^X {\mathbf {1}}{\left\{ Z^Y \ge \alpha _n \right\} } \right]&= {\mathbb {E}}\left[ \left( \frac{{\sigma }_{XY}}{{\sigma }^2_{Y}} Z^Y + e \right) {\mathbf {1}}\left\{ Z^Y \ge \alpha _n \right\} \right] \\&= \frac{{\sigma }_{XY}}{{\sigma }^2_{Y}} {\mathbb {E}}\left[ Z^Y {\mathbf {1}}\left\{ Z^Y \ge \alpha _n \right\} \right] \\&= \sqrt{n} \frac{{\sigma }_{XY}}{{\sigma }_{Y}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{Y}} \right) . \end{aligned}$$

$\square$

Lemma 6

For any sequence of numbers $\alpha _n$, there is a constant $C>0$ such that

$$\begin{aligned} \bigg|{{\mathbb {E}}\left[ S_n^X {\mathbf {1}}\left\{ S^X_n \ge \alpha _n \right\} \right] - \sqrt{n} {\sigma }_{X}\phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{X}} \right) }\bigg| \le C n^{1/4} . \end{aligned}$$

Proof of Lemma 6

By Lemma 2 and 5, we have

$$\begin{aligned}& \bigg|{{\mathbb {E}}\left[ S_n^X {\mathbf {1}}\left\{ S^X_n \ge \alpha _n \right\} \right] - \sqrt{n} {\sigma }_{X}\phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{X}} \right) }\bigg| \\ \le&\int _{0}^\infty \bigg|{{\mathbb {P}}\left( S^X_n> x \right) - {\mathbb {P}}\left( Z^X > x \right) }\bigg| dx + \int _{-\infty }^0 \bigg|{{\mathbb {P}}\left( S^X_n< x\right) - {\mathbb {P}}\left( Z^X < x\right) }\bigg| dx. \end{aligned}$$

Using Chebyshev’s inequality, we have

$$\begin{aligned} \bigg|{{\mathbb {P}}\left( S^X_n> x \right) - {\mathbb {P}}\left( Z^X > x \right) }\bigg| \le \frac{n {\sigma }^2_{X}}{x^2}. \end{aligned}$$

Together with Lemma 3, we have

$$\begin{aligned} \int _{0}^\infty \bigg|{{\mathbb {P}}\left( S^X_n> x \right) - {\mathbb {P}}\left( Z^X > x \right) }\bigg| dx&\le \int _{0}^{n^{3/4}} \frac{C}{\sqrt{n}} \frac{{\rho }_{X}}{{\sigma }_{X}^{3}} dx + \int _{n^{3/4}}^\infty \frac{n {\sigma }^2_{X}}{x^2} dx \\&= n^{1/4} \left( C \frac{{\rho }_{X}}{{\sigma }_{X}^{3}} + {\sigma }_{X}^2 \right) \end{aligned}$$

$\square$

Lemma 7

Suppose X and Y are not perfectly correlated. For any sequence of numbers $\alpha _n$, there is a constant $C>0$ such that

$$\begin{aligned} \bigg|{{\mathbb {E}}\left[ S_n^X {\mathbf {1}}\left\{ S^Y_n \ge \alpha _n \right\} \right] - \sqrt{n} \frac{{\sigma }_{XY}}{{\sigma }_{Y}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{Y}} \right) }\bigg|\le C n^{1/4}. \end{aligned}$$

Proof of Lemma 7

By Lemma 2 and 5, we know that the absolute difference we want to bound is

$$\begin{aligned}&\bigg|{{\mathbb {E}}\left[ S_n^X {\mathbf {1}}\left\{ S^Y_n \ge \alpha _n \right\} \right] - {\mathbb {E}}\left[ Z^X {\mathbf {1}}\left\{ Z^Y \ge \alpha _n \right\} \right] }\bigg| \\ \le&\underbrace{\bigg|{ \int _{0}^\infty {\mathbb {P}}\left( S^X_n> x,S_n^Y \ge \alpha _n \right) - {\mathbb {P}}\left( Z^X > x,Z^Y \ge \alpha _n \right) dx}\bigg|}_{I_1} \\ +&\underbrace{\bigg|{\int _{-\infty }^0 {\mathbb {P}}\left( S^X_n< x,S^Y_n \ge \alpha _n \right) - {\mathbb {P}}\left( Z^X < x,Z^Y \ge \alpha _n \right) dx }\bigg|}_{I_2}. \end{aligned}$$

We first deal with $I_1$. Notice that

$$\begin{aligned} {\mathbb {P}}\left( S^X_n> x,S_n^Y \ge \alpha _n \right) \le {\mathbb {P}}\left( S^X_n > x\right) \le \frac{n{\sigma }^2_{X}}{x^2}, \end{aligned}$$

where the second inequality follows from the Chebyshev inequality. Similarly, we have

$$\begin{aligned} {\mathbb {P}}\left( Z^X> x,Z^Y \ge \alpha _n \right) \le {\mathbb {P}}\left( Z^X > x\right) \le \frac{n{\sigma }^2_{X}}{x^2}. \end{aligned}$$

So putting these two tail bounds together, we have

$$\begin{aligned} \bigg|{{\mathbb {P}}\left( S^X_n> x,S_n^Y \ge \alpha _n \right) - {\mathbb {P}}\left( Z^X> x,Z^Y \ge \alpha _n \right) }\bigg| \le \frac{n{\sigma }^2_{X}}{x^2}, \text { for all } x>0. \end{aligned}$$

Then, based on Lemma 4, we have

$$\begin{aligned} I_1&\le \int _{0}^{n^{3/4}} |{ {\mathbb {P}}\left( S^X_n> x,S_n^Y \ge \alpha _n \right) - {\mathbb {P}}\left( Z^X> x,Z^Y \ge \alpha _n \right) }| dx \\&+ \int _{n^{3/4}}^\infty |{ {\mathbb {P}}\left( S^X_n> x,S_n^Y \ge \alpha _n \right) - {\mathbb {P}}\left( Z^X > x,Z^Y \ge \alpha _n \right) }| dx \\&\le n^{3/4} \frac{C}{\sqrt{n}} + \int _{n^{3/4}}^\infty \frac{n{\sigma }^2_{X}}{x^2} dx \\&= n^{1/4} \left( C + {\sigma }^2_{X} \right) . \end{aligned}$$

Following the same steps we can derive a same bound for $I_2$. Then the result follows. $\square$

1.2 Proofs of results in the main text

We introduce some notations that are useful in the proofs. The moments of the valuation distribution are denoted by

$$\begin{aligned} \mu \equiv {\mathbb {E}}[V_i], \sigma ^2 \equiv {\mathbb {E}}\vert V_i - \mu \vert ^2, \rho \equiv {\mathbb {E}}\vert V_i - \mu \vert ^3, \sigma _{\psi }^2 \equiv {\mathbb {E}}\vert \psi (V_i) \vert ^2, \rho _{\psi } \equiv {\mathbb {E}}\vert \psi (V_i) \vert ^3. \end{aligned}$$

They are all finite since we assume that f has a bounded support and is bounded away from zero on the support. To keep the asymptotic analysis concise, we use the notations in the following table.³⁶

Notation	Definition	Short explanation
$d_n = o(e_n)$	$d_n/e_n \rightarrow 0$	$|d_n |$ dominated by $e_n$
$d_n = O(e_n)$	$\limsup |{d_n}|/e_n < \infty$	$|d_n |$ bounded above by $e_n$
$d_n = \omega (e_n)$	$|d_n |/e_n \rightarrow \infty$	$|d_n |$ dominates $e_n$
$d_n = \Omega (e_n)$	$\liminf |d_n |/e_n > 0$	$|d_n |$ bounded below by $e_n$
$d_n = \Theta (e_n)$	$|d_n |= O(e_n),|d_n |= \Omega (e_n)$	$|d_n |$ bounded below and above by $e_n$
$d_n = O_\varepsilon (e_n)$	$\exists \varepsilon >0, d_n = O(n^{-\varepsilon }e_n)$	$|d_n |$ nearly bounded above by $e_n$
$d_n = \omega _{\varepsilon }(e_n)$	$\forall \varepsilon >0, d_n = \omega (n^{-\varepsilon }e_n)$	$|d_n |$ nearly dominates $e_n$

Proof of Theorem 1

• (i) We first obtain an asymptotic lower bound for the ex-ante budget:

  $$\begin{aligned} {\mathbb {E}}[b^{n}(\varvec{V})]&= {\mathbb {E}} \left[ \sum \psi (V_i) {\mathbf {1}}\big \{ \sum \psi (V_i) \ge \alpha _n \big \} \right] - c_n {\mathbb {P}}\left( \sum \psi (V_i) \ge \alpha _n \right) \\&\ge {\mathbb {E}} \left[ \sum \psi (V_i) {\mathbf {1}}\big \{ \sum \psi (V_i) \ge \alpha _n \big \} \right] - c_n \\&\ge \sqrt{n} {\sigma }_{\psi } \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) - Cn^{1/4} - c_n, \end{aligned}$$

  where the last inequality follows from Lemma 6. Then we show that the first term on the RHS is the leading term. First, by the assumption $\alpha _n = o\left( \sqrt{ n \log n } \right)$, we have

  $$\begin{aligned} \log \left( \sqrt{n} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) \Big / n^{1/4} \right)&= \frac{1}{4}\log n -\left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n}} \right) ^2 + C \\&= \log n \left( \frac{1}{4} - \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n \log n}} \right) ^2 + o(1)\right) \\&= \log n (1/4 + o(1))\rightarrow \infty . \end{aligned}$$

  Next, by the assumption that $c_n = O_{\varepsilon }\left( \sqrt{n}\right)$, there exist $C,\varepsilon >0$ such that $c_n \le C n^{1/2 - \varepsilon }$ for large n. Then $\log c_n / \log n \le 1/2 - \varepsilon /2$ for large n. Therefore,

  $$\begin{aligned}&\log \left( \sqrt{n} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) \Big / c_n \right) \\ =&\frac{1}{2} \log n - \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n}} \right) ^2 - \log c_n + C \\ =&\log n \left( \frac{1}{2} - \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n \log n}} \right) ^2 - \frac{\log c_n}{\log n} + o(1) \right) \\ \ge&\log n (\varepsilon /2 + o(1)) \rightarrow \infty . \end{aligned}$$

  The above asymptotic results show that the leading term in the lower bound is $\sqrt{n} {\sigma }_{\psi } \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right)$. We then derive an upper bound for the ex-ante budget:

  $$\begin{aligned} {\mathbb {E}}[b^{n}(\varvec{V})]&\le {\mathbb {E}} \left[ \sum \psi (V_i) {\mathbf {1}}\big \{ \sum \psi (V_i) \ge \alpha _n \big \} \right] \\&\le \sqrt{n} {\sigma }_{\psi } \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) + Cn^{1/4} , \end{aligned}$$

  where the last inequality again follows from Lemma 6. By the previous analysis, the leading term in the upper bound is also $\sqrt{n}{\sigma }_{\psi } \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right)$. Therefore, we only need to derive the growth rate of that term. Take any $\varepsilon >0$, we have

  $$\begin{aligned} \log \left( \sqrt{n} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) \Big / n^{1/2-\varepsilon } \right)&= \varepsilon \log n - \frac{1}{2} \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n}} \right) ^2 +C \\&= \log n \left( \varepsilon - \frac{1}{2} \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n \log n}} \right) ^2 + o(1) \right) \\&= \log n \left( \varepsilon + o(1) \right) \rightarrow \infty . \end{aligned}$$

• (ii) First notice that ${\mathbb {E}}[w^{n}(\varvec{V})] \le {\mathbb {E}}[w^*(\varvec{V})]$. We break the ex-ante welfare into two parts

  $$\begin{aligned} {\mathbb {E}}[w^{n}(\varvec{V})]&= {\mathbb {E}} \left[ \sum V_i q^{n}(\varvec{V}) - \sum t_i^{n}(\varvec{V}) \right] \\&= {\mathbb {E}} \left[ \left( \sum V_i - c_n \right) q^{n}(\varvec{V})\right] - {\mathbb {E}}[b^{n}(\varvec{V})]. \end{aligned}$$

  We obtain a lower bound for the first term on the RHS:

  $$\begin{aligned}&{\mathbb {E}} \left[ \left( \sum V_i - c_n \right) q^{n}(\varvec{V})\right] \\ =&{\mathbb {E}} \left[ {\mathbf {1}}\big \{ \sum \psi (V_i) \ge \alpha _n \big \} \sum V_i \right] - c_n {\mathbb {P}}\left( \sum \psi (V_i) \ge c_n \right) \\ \ge&{\mathbb {E}} \left[ {\mathbf {1}}\big \{ \sum \psi (V_i) \ge \alpha _n \big \} \right] {\mathbb {E}}\left[ \sum V_i \right] - c_n {\mathbb {P}}\left( \sum \psi (V_i) \ge c_n \right) \\ =&(n {\mu }- c_n) {\mathbb {P}}\left( \sum \psi (V_i) \ge \alpha _n \right) \\ \ge&(n {\mu }- c_n) \left( 1-\Phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) - \frac{C}{\sqrt{n}} \right) \\ \ge&n {\mu }- c_n - n {\mu }\left( \Phi \left( \frac{\alpha _n}{\sqrt{n}{\sigma }_{\psi }} \right) + \frac{C}{\sqrt{n}}\right) , \end{aligned}$$

  where the second inequality follows from the fact that $\psi$ is non-decreasing and the second to last inequality follows from Lemma 3. By the result in part (i), we can bound the ex-ante budget from above by

  $$\begin{aligned} {\mathbb {E}}[b^{n}(\varvec{V})]&\le \sqrt{n}{\sigma }_{\psi } \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) + Cn^{1/4}\\&\le \sqrt{n} {\sigma }_{\psi } / \sqrt{2 \pi } + Cn^{1/4} , \end{aligned}$$

  where the last inequality follows from the fact that $\phi (\cdot ) \le 1/\sqrt{2 \pi }$. The ex-ante efficient welfare is bounded above by ${\mathbb {E}}[w^*(\varvec{V})] \le {\mathbb {E}}\left[ \sum V_i \right] = n {\mu }$. Combining these results, we get

  $$\begin{aligned} \frac{{\mathbb {E}}[w^*(\varvec{V})] - {\mathbb {E}}[w^{n}(\varvec{V})]}{{\mathbb {E}}[w^*(\varvec{V})]}&\le \frac{c_n}{n\mu } + \Phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{\psi }} \right) + \frac{C}{\sqrt{n}} \rightarrow 0. \\ \end{aligned}$$

  The remaining task is to show that the leading term in the above expression is $\Phi \left( \alpha _n / \sqrt{n} {\sigma }_{\psi } \right)$. We use the well-known lower bound on the tail of $\Phi$:

  $$\begin{aligned} \Phi (v) = 1-\Phi (|v|) \ge |{v}|\exp \left( -v^2/2 \right) /\big (\sqrt{2\pi }\left( 1+v^2 \right) \big ), v \le 0. \end{aligned}$$

  Then

  $$\begin{aligned}&\log \left( \sqrt{n} \Phi \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n}} \right) \right) \\ \ge&\frac{1}{2} \log n + \log \left( \frac{|{\alpha _n}|}{\sqrt{n}} \right) - \frac{1}{2} \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n}} \right) ^2 - \log \left( 1 + (\alpha _n/{\sigma }_{\psi }\sqrt{n})^2 \right) + C \\ =&\frac{1}{4} \log n + \log \left( \frac{|{\alpha _n}|}{\sqrt{n}} \right) - \frac{1}{2} \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n}} \right) ^2 + \log \left( \frac{n^{1/4}}{1 + (\alpha _n/{\sigma }_{\psi }\sqrt{n})^2} \right) + C \\ =&\log n \left( \frac{1}{4} - \frac{1}{2} \left( \frac{\alpha _n}{{\sigma }_{\psi }\sqrt{n \log n}} \right) ^2 \right) + \log \left( \frac{|{\alpha _n}|}{\sqrt{n}} \right) \\ +&\log \left( \frac{n^{1/4}/\log n}{1/ \log n + (\alpha _n/{\sigma }_{\psi }\sqrt{n \log n})^2} \right) + C, \end{aligned}$$

  where the last line goes to $\infty$. Therefore, $\Phi \left( \alpha _n / \sqrt{n} {\sigma }_{\psi } \right)$ is $\omega (1/\sqrt{n})$ and hence the leading term of the welfare ratio.

• (iii) The result comes directly from the previous two parts.

$\square$

Proof of Lemma 1

By the definition of $\psi$, we have

$$\begin{aligned} \sigma _{\psi h}&= {\mathbb {E}} \left[ \psi (V_i) h(V_i) \right] \\&= {\mathbb {E}} \left[ V_i h(V_i) \right] - \int _0^\infty h(v) (1-F(v)) dv. \end{aligned}$$

By the Fubini theorem, the second term on the RHS is

$$\begin{aligned} \int _0^\infty h(v) (1-F(v)) dv&= \int _0^\infty h(v) {\mathbb {E}}[{\mathbf {1}}_{[0,V_i)}(v)] dv \\&= {\mathbb {E}}\left[ \int _0^\infty h(v) {\mathbf {1}}_{[0,V_i)}(v) dv \right] \\&= {\mathbb {E}} \left[ \int _0^{V_i} h(v) dv \right] < \infty . \end{aligned}$$

The above quantity is finite since V has a bounded support and h is continuous. Then we perform integration by parts to the above Lebesgue-Stieltjes integral:

$$\begin{aligned} \int _0^{V_i} h(v) dv = vh(v)\Big |_{0}^{V_i} - \int _0^{V_i} v dh(v) = V_ih(V_i) - \int _0^{V_i} v dh(v). \end{aligned}$$

Therefore,

$$\begin{aligned} \sigma _{\psi h}&= {\mathbb {E}} \left[ V_ih(V_i) - \int _0^{V_i} h(v) dv \right] = {\mathbb {E}} \left[ \int _0^{V_i} v \text { } dh(v)\right] , \end{aligned}$$

which is positive if h is increasing. $\square$

For Theorem 2, we introduce the following notations for the moments of $h(V_i)$:

$$\begin{aligned} \mu _h \equiv {\mathbb {E}}[h(V_i)], \sigma _h^2 \equiv {\mathbb {E}}\vert h(V_i) - \mu _h \vert ^2, \rho _h \equiv {\mathbb {E}}\vert h(V_i) - \mu _h \vert ^3. \end{aligned}$$

They are finite when h is a continuous function on $[0,{\bar{v}}]$.

Proof of Theorem 2

If $h(V_i)$ and $\psi (V_i)$ are perfectly correlated (i.e., h and $\psi$ are linearly dependent), then the result follows from Theorem 1. Therefore, we only need to study the case where $h(V_i)$ and $\psi (V_i)$ are not perfectly correlated.

• (i) We first obtain an asymptotic lower bound for the ex-ante budget:

  $$\begin{aligned} {\mathbb {E}}[b^{n}(\varvec{V})]&= {\mathbb {E}} \left[ \sum \psi (V_i) {\mathbf {1}}\big \{ \sum h(V_i) \ge \alpha _n \big \} \right] - c_n {\mathbb {P}}\left( \sum h(V_i) \ge \alpha _n \right) \\&\ge {\mathbb {E}} \left[ \sum \psi (V_i) {\mathbf {1}}\big \{ \sum h(V_i) \ge \alpha _n \big \} \right] - c_n \\&\ge \sqrt{n} \frac{{\sigma }_{\psi h}}{{\sigma }_{h}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right) - Cn^{1/4} -c_n, \end{aligned}$$

  where the last inequality follows from Lemma 7. The first term is strictly positive by Lemma 1. Then following the same steps as in the proof of Theorem 1(i), we can show that the leading term in the above expression is $\sqrt{n} \frac{{\sigma }_{\psi h}}{{\sigma }_{h}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right)$ under the assumptions $\alpha _n = o(\sqrt{n \log n})$ and $c_n = O_\varepsilon (\sqrt{n})$. We then derive an upper bound for the ex-ante budget:

  $$\begin{aligned} {\mathbb {E}}[b^{n}(\varvec{V})]&\le {\mathbb {E}} \left[ \sum \psi (V_i) {\mathbf {1}}\big \{ \sum h(V_i) \ge \alpha _n \big \} \right] \\&\le \sqrt{n} \frac{{\sigma }_{\psi h}}{{\sigma }_{h}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right) + Cn^{1/4} , \end{aligned}$$

  where the last inequality again follows from Lemma 7. By the previous analysis, the leading term in the upper bound is also $\sqrt{n} \frac{{\sigma }_{\psi h}}{{\sigma }_{h}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right)$, and the result follows.

• (ii) We follow the proof of Theorem 1(ii). First notice that ${\mathbb {E}}[w^{n}(\varvec{V})] \le {\mathbb {E}}[w^*(\varvec{V})]$. We break the ex-ante welfare into two parts

  $$\begin{aligned} {\mathbb {E}}[w^{n}(\varvec{V})]&= {\mathbb {E}} \left[ \sum V_i q^{n} - \sum t^{n}_i(\varvec{V}) \right] \\&= {\mathbb {E}} \left[ \left( \sum V_i - c_n \right) q^{n}(\varvec{V})\right] - {\mathbb {E}}[b^{n}(\varvec{V})]. \end{aligned}$$

  We obtain a lower bound for the first term on the RHS:

  $$\begin{aligned}{\mathbb {E}}& \left[ \left( \sum V_i - c_n \right) q^{n}(\varvec{V})\right] \\ =\,\,&{\mathbb {E}} \left[ {\mathbf {1}}\big \{ \sum h(V_i) \ge \alpha _n \big \} \sum V_i \right] - c_n {\mathbb {P}}\left( \sum h(V_i) \ge c_n \right) \\ \ge\,\,&{\mathbb {E}} \left[ {\mathbf {1}}\big \{ \sum h(V_i) \ge \alpha _n \big \} \right] {\mathbb {E}}\left[ \sum V_i \right] - c_n {\mathbb {P}}\left( \sum h(V_i) \ge c_n \right) \\ =\,\,&(n {\mu }- c_n) {\mathbb {P}}\left( \sum h(V_i) \ge \alpha _n \right) \\ \ge\,\,&(n {\mu }- c_n) \left( 1-\Phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right) - \frac{C}{\sqrt{n}} \right) \\ \ge\,\,&n {\mu }- c_n - n \mu \left( \Phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right) + \frac{C}{\sqrt{n}} \right) , \end{aligned}$$

  where the second line follows from the fact that h is non-decreasing and the second to last line follows from Lemma 3. Following the proof of part (i), we can upper bound the ex-ante budget by

  $$\begin{aligned} {\mathbb {E}}[b^{n}(\varvec{V})]&\le \sqrt{n} \frac{{\sigma }_{\psi h}}{{\sigma }_{h}} \phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right) + C n^{1/4} \\&\le \sqrt{n} \frac{{\sigma }_{\psi h}}{{\sigma }_{h} \sqrt{2 \pi }} + C n^{1/4}. \end{aligned}$$

  We use the same upper bound on the efficient ex-ante welfare ${\mathbb {E}}[w^*(\varvec{V})] \le n {\mu }$ as before. Combining these results together, we get a similar bound as in part (ii) of Theorem 1:

  $$\begin{aligned} \frac{{\mathbb {E}}[w^*(\varvec{V})] - {\mathbb {E}}[w^{n}(\varvec{V})]}{{\mathbb {E}}[w^*(\varvec{V})]}&\le \frac{c_n}{n{\mu }} + \Phi \left( \frac{\alpha _n}{\sqrt{n} {\sigma }_{h}} \right) + \frac{C}{\sqrt{n}} \rightarrow 0. \end{aligned}$$

  Similar as in the proof of Theorem 1(ii), the leading term in the above expression is $\Phi \left( \alpha _n / \sqrt{n} {\sigma }_{h} \right)$.

• (iii) The result comes directly from the previous two parts.

$\square$

Proof of Theorem 3

It is well-known that the total expected payment of an incentive compatible and individually rational mechanism $(q^n,\{t_i^n\})$ is equal to

$$\begin{aligned} {\mathbb {E}} \left[ \sum t^n_i(\varvec{V}) \right] = {\mathbb {E}} \left[ \sum \psi (V_i) q^n(\varvec{V}) \right] + {\mathbb {E}} \left[ \sum t^n_i(0,\varvec{V}_{-i}) \right] . \end{aligned}$$

See, for example, Equation (3.6) in Hellwig (2003). Since the expected payment ${\mathbb {E}}[t^n_i(0,\varvec{V}_{-i})]$ needs to be non-positive for any i by the individually rational condition, the total expected payment is maximized by setting $q^n(\varvec{v}) = {\mathbf {1}}\{ \sum \psi (v_i) \ge 0 \}$ and $t^n_i(0,\varvec{v}_{-i})=0$. Such a mechanism is in fact a special case of the mechanism in Example 3 with the adjustment term $\alpha _n = 0$. From the proof of Theorem 1, we can see that the total expected payment is $O(\sqrt{n})$. Notice that even though this particular mechanism may violate incentive compatibility when the distribution is not Myerson regular, the asymptotic order $O(\sqrt{n})$ is nonetheless a valid upper bound on the growth rate of the maximum total expected payment.

Then for any sequence of incentive compatible and individually rational mechanisms $(q^n,\{t^n_i\})$ that satisfies eventually ex-ante budget balanced, it must be true that

$$\begin{aligned} 0 \le {\mathbb {E}} \left[ \sum t_i(\varvec{V}) - c_n q(\varvec{V}) \right] \le O(\sqrt{n}) - c_n {\mathbb {P}}(q^n(\varvec{V}) = 1), \text { for { n} large enough}. \end{aligned}$$

This implies that the provision probability of the public good is converging to zero:

$$\begin{aligned} {\mathbb {P}}(q^n(\varvec{V}) = 1) \le O(\sqrt{n})/c_n \rightarrow 0. \end{aligned}$$

$\square$

Discussion on Proposition 3 in Hellwig (2003)

In this section, we discuss the proof method of Proposition 3 in Hellwig (2003), hereafter H2003. We first translate the result and the proof with the terminology in our paper. Proposition 3 in H2003 shows that the second-best mechanism is asymptotically efficient when the cost of the public good does not grow with n. The proof in H2003 essentially tries to show that the AMT mechanism in Example 3 is ex-ante budget balanced and asymptotically efficient. Therefore, since the second-best mechanism must have a higher welfare by definition, it is also asymptotically efficient.

To better explain the proof, we link the notations in H2003 to the ones in our paper. The mechanism $Q^{nk}$ defined in (4.6) and (4.7) on p.597 of H2003 corresponds to the AMT mechanism with $h=\varphi$ in our Example 3. The scalar k in the mechanism $Q^{nk}$ corresponds to our adjustment term $\alpha _n$. The discussion following Inequality (4.8) on p.598 of H2003 shows that when the adjustment term k is fixed (does not vary with n), the mechanism $Q^{nk}$ is ex-ante budget balanced. The discussion following Inequality (4.10) on p.598 of H2003 shows that when the adjustment term k decreases to $-\infty$, the mechanism $Q^{nk}$ is asymptotically efficient.

We argue that this reasoning is incomplete because a discussion of the ex-ante budget when k varies with n is lacking. More specifically, the proof in H2003 requires that Inequality (4.4) on p.597 to hold for any $\varepsilon >0$ and n sufficiently large. In particular, we can take $\varepsilon = 1/n$. Then the $k(\varepsilon )$ on the second line after (4.10) depends explicitly on n and is decreasing as n increases. In this case, the discussion following Inequality (4.8) is no longer sufficient to show that the mechanism $Q^{nk}$ is ex-ante budget balanced. This is because a decreasing k would decrease the budget. The previous argument that $Q^{nk}$ is ex-ante budget balanced when k is fixed and $n \rightarrow \infty$. However, this does not address the budget when k is decreasing. For example, in the extreme case where k decreases so fast that it is equal to $-\infty$, the revenue becomes zero in the limit, leading to a budget deficit.

This is where the Berry-Esseen theorem becomes useful: we want to characterize the ex-ante budget when n increases and the adjustment term k decreases. This is possible under the Berry-Esseen theorem because it gives the convergence rate of the central limit theorem. For each n and k, we know not only that the budget can be approximated by a normal distribution but also how close this approximation is. Therefore, we can derive the rate of k under which the budget becomes balanced eventually. The details are described in Sect. 3.2 and in the proofs in Appendix A.