Optimal mechanism for land acquisition

Abstract

Sellers with one unit of land each are located at the nodes of a graph. Two sellers are contiguous if they are connected by an edge in the graph. The buyer realizes a positive value only if he can purchase plots that constitute a path of given length. We characterize the optimal mechanism when the buyer and sellers have independently distributed private valuations.

Appendices

Proof of Theorem 1

The following result is standard in the literature and known as Revenue Equivalence. It states that for any mechanism satisfying BIC, expected utility of every agent is determined by the allocation rule up to an additive constant.

Lemma 2

For any mechanism (P, t) satisfying BIC, for all $j\in \{0,\ldots , n\}$,

$$\begin{aligned} E_{-j}U_j(v_j, v_{-j}\mid v_{j})&= E_{-j}U_j(x_j, v_{-j}\mid v_{j})+\int _{x_j}^{v_j} E_{-j}P_j(z, v_{-j}) dz \text { for all } v_{j}, \end{aligned}$$

(12)

where $x_{0}={\underline{v}}_{0}$ and $x_{i}={\bar{v}}$ for $i\in \{1,\ldots , n\}$.

Proof

See Krishna (2002). $\square$

Refer to (3).

Proof

Let (P, t) be BIC and BB. First, we will show that (P, t) satisfies (4).

First, we expand the following expression.

$$\begin{aligned}&\int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} v_0(P_0 (v))f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \\&\qquad +\sum _{i=1}^{n} \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} v_i P_i(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \\&\quad =\int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} v_0(P_0 (v))f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \\&\qquad - \sum _{i=1}^{n} \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} t_i(v)f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n\\&\qquad + \sum _{i=1}^{n} \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}t_i(v)f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \\&\qquad -\sum _{i=1}^{n} \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} v_i P_i(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \\&\quad =\int _{{\underline{v}}_0}^{{\bar{v}}_0}\left[ \int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} \left\{ v_0(P_0 (v))- \sum _{i=1}^{n}t_i(v)\right\} f(v_1)\ldots f(v_n)dv_1 \ldots dv_n \right] g(v_0)dv_0 \\&\qquad +\sum _{i=1}^{n} \int _{{\underline{v}}}^{{\bar{v}}}\left[ \int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n-1 \text { times}}\left\{ v_i P_i(v)-t_i(v)\right\} g(v_0) \frac{f(v_1)\ldots f(v_n)}{f( v_i)}\frac{dv_0 \ldots dv_n}{dv_i}\right] \\&\quad \qquad f(v_i)dv_i\\&\quad =\int _{{\underline{v}}_0}^{{\bar{v}}_0}E_{-0}U_0(v\mid v_0)g(v_0)dv_0 + \sum _{i=1}^{n}\int _{{\underline{v}}}^{{\bar{v}}} E_{-i}U_i(v\mid v_i)f(v_i)dv_i\\&\quad =E_{-0} U_0({\underline{v}}_0, v_{-0}\mid {\underline{v}}_0)+ \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}_0}^{v_0} E_{-0} P_0 ( z_0, v_{-0})dz_0 g(v_0)dv_0\\&\qquad +\sum _{i=1}^{n}\left[ E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})+\int _{{\underline{v}}}^{{\bar{v}}}\int _{v_i}^{{\bar{v}}}E_{-i}P_{i}(z_i, v_{-i})dz_i f(v_i)dv_i\right] \\&\quad = E_{-0} U_0({\underline{v}}_0, v_{-0}\mid {\underline{v}}_0)+ \sum _{i=1}^{n}E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})\\&\qquad + \sum _{i=1}^{n}\left[ \int _{{\underline{v}}}^{{\bar{v}}}E_{-i}P_{i}(v_i, v_{-i})F(v_i)dv_i\right] \\&\qquad + \int _{{\underline{v}}_0}^{{\bar{v}}_0}E_{-0} P_0 (v_0, v_{-0})(1-G(v_0))dv_0 \\&\quad = E_{-0} U_0({\underline{v}}_0, v_{-0}\mid {\underline{v}}_0)+ \sum _{i=1}^{n}E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})\\&\qquad +\sum _{i=1}^{n} \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} P_i(v)\frac{F(v_i)}{f(v_i)} f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n\\&\qquad + \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} P_0 (v)\frac{1-G(v_0)}{g(v_0)}f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n. \end{aligned}$$

The first equality is obtained by first subtracting and then adding the second term on the right hand side; the second equality uses the independence of valuations. The third equality uses the definitions of the appropriate expectations; the fourth one uses Lemma 2; the fifth equality uses integration by parts to remove the second integral; the sixth equality is obtained by using the expression for the expectations. The final expression is obtained by transposing and rearranging terms on both sides. It follows that,

$$\begin{aligned}&E_{-0} U_0({\underline{v}}_0, v_{-0}\mid {\underline{v}}_0)+ \sum _{i=1}^{n}E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})\nonumber \\&\quad = \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} P_0 (v)\left[ v_0-\frac{1-G(v_0)}{g(v_0)}\right] f(v_1)\ldots f(v_n)g(v_0)dv_0\ldots dv_n \nonumber \\&\qquad +\sum _{i=1}^{n}\int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} P_i(v)\left[ v_i+\frac{F(v_i)}{f(v_i)}\right] f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \end{aligned}$$

(13)

which is the expression in (4) above.

Now we show the necessity condition in the second part of the Theorem. If a mechanism is BIC, then for any two valuations $v_0$ and ${\hat{v}}_0$ of the buyer,

$$\begin{aligned} \begin{array}{lll}E_{-0}U_0({v}_0, v_{-0}\mid {v}_0)&{}=&{} v_0 E_{-0}P_0(v_0, v_{-0})-E_{-0}t_0(v_0,v_{-0})\\ \quad &{} \ge &{} v_0 E_{-0}P_0({\hat{v}}_0, v_{-0})-E_{-0}t_0({\hat{v}}_0, v_{0}) \end{array} \end{aligned}$$

and

$$\begin{aligned}&E_{-0}U_0({\hat{v}}_0, v_{-0}\mid {\hat{v}}_0)\\&\quad = {\hat{v}}_0 E_{-0}P_0 ({\hat{v}}_0, v_{-0})-E_{-0}t_0({\hat{v}}_0, v_{-0}) \\&\quad \ge {\hat{v}}_0 E_{-0}P_0 (v_0, v_{-0})-E_{-0}t_0(v_0, v_{-0}). \end{aligned}$$

Therefore,

$$\begin{aligned}&E_{-0}U_0({\hat{v}}_0, v_{-0}\mid {\hat{v}}_0) - E_{-0}U_0({v}_0, v_{-0}\mid {v}_0)\\&\quad \ge {\hat{v}}_0 E_{-0}P_0 (v_0, v_{-0})-E_{-0}t_0(v_0, v_{-0}) \\&\qquad -v_0 E_{-0}P_0(v_0, v_{-0})+E_{-0}t_0(v_0,v_{-0})\\&\quad = ({\hat{v}}_0 - v_0)E_{-0}P_0(v_0, v_{-0}) \end{aligned}$$

and,

$$\begin{aligned}&E_{-0}U_0({\hat{v}}_0, v_{-0}\mid {\hat{v}}_0) - E_{-0}U_0({v}_0, v_{-0}\mid {v}_0) \\&\quad \le {\hat{v}}_0 E_{-0}P_0 ({\hat{v}}_0, v_{-0})-E_{-0}t_0({\hat{v}}_0, v_{-0})\\&\qquad -v_0 E_{-0}P_0({\hat{v}}_0, v_{-0})+E_{-0}t_0({\hat{v}}_0,v_{0}) \\&\quad = ({\hat{v}}_0 - v_0)E_{-0}P_0 ({\hat{v}}_0, v_{-0}). \end{aligned}$$

This gives us the following inequality:

$$\begin{aligned} ({\hat{v}}_0 - v_0)E_{-0}P_0(v_0, v_{-0}) \le EU_0({\hat{v}}_0) - EU_0(v_0)\le ({\hat{v}}_0 - v_0)E_{-0}P_0 ({\hat{v}}_0, v_{-0})\nonumber \\ \end{aligned}$$

(14)

It follows that $E_{-0}P_0(v_0, v_{-0})$ is increasing in $v_0$. A similar argument shows that $E_{-i}P_i(v_i, v_{-i})$ is decreasing in $v_i$. Since,

$$\begin{aligned} E_{-0} U_0({\underline{v}}_0, v_{-0}\mid {\underline{v}}_0)&= \min _{v_0\in [{\underline{v}}_0,{\bar{v}} _0]}E_{-0} U_0({v}_0, v_{-0}\mid {v}_0), \end{aligned}$$

and

$$\begin{aligned} E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})&=\min _{v_i\in [{\underline{v}},{\bar{v}}]}E_{-i}U_i({v}_{i}, v_{-i}\mid {v}_{i}), \end{aligned}$$

by (13), (P, t) satisfies IIR if the expression in (4) is nonnegative.

To show sufficiency, suppose conditions (i)–(iii) hold; we show the existence of payment functions $t_j(v)$s such that (P, t) is BIC, IIR and BB. We may consider the following payment function among many other possibilities:

$$\begin{aligned} t_i(v)&= \frac{1}{n} \int _{z_0={\underline{v}}_0}^{v_0} z_0 d(E_{-0}P_0 (z_0, v_{-0})) - \int _{z_i={\underline{v}}}^{v_i} z_i d(-E_{-i}P_i(z_i, v_{-i})) \nonumber \\&\quad +{\bar{v}}E_{-i}(P_i({\bar{v}}, v_{-i}))-\frac{1}{n} E_{-i} \left( \int _{z_0={\underline{v}}_0}^{v_0} z_0 d(E_{-0}P_0 (z_0, v_{- 0}))\right) \nonumber \\&\quad + \int _{z_i={\underline{v}}}^{{\bar{v}}} z_i d(-E_{-i}P_i(z_i, v_{-i})). \end{aligned}$$

(15)

Since $E_{-0}P_0 (v)$ is increasing in $v_0$ and $E_{-i}P_i(v)$s are decreasing in respective $v_i$s, the first two terms on the right hand side of (15) are positive; the first term is a function of $v_0$ alone and the second term is a function of $v_i$ alone; the remaining part are constant terms added make $E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})=0$ for all $i=1, \ldots , n$. To check that this payment function leads to BIC, we have for the buyer,

$$\begin{aligned}&E_{-0} U_0(v\mid {\underline{v}}_0)- E_{-0} U_0({\hat{v}}_0, v_{-0}\mid {\underline{v}}_0)\\&\quad = v_0 E_{-0}P_0 (v)-v_0 E_{-0}P_0 ({\hat{v}}_0, v_{-0})-E_{-0}\left( \sum _{i=1}^{n}t_i(v)-\sum _{i=1}^{n}t_i({\hat{v}}_0, v_{-0})\right) \\&\quad = v_0 E_{-0}P_0 (v_0, v_{-0})-v_0 E_{-0}P_0 ({\hat{v}}_0, v_{-0})\\&\qquad -E_{-0}\left[ \int _{z_0={\underline{v}}_0}^{v_0} z_0 d(E_{-0}P_0 (z_0, v_{-0})) - \sum _{i=1}^{n}\int _{z_i={\underline{v}}}^{v_i} z_i d(-E_{-i}P_i(z_i, v_{-i}))\right. \\&\qquad \left. -\int _{z_0={\underline{v}}_0}^{{\hat{v}}_0} z_0 d(E_{-0}P_0 (z_0, v_{-0})) + \sum _{i=1}^{n}\int _{z_i={\underline{v}}}^{v_i} z_i d(-E_{-i}P_i(z_i, v_{-i}))\right] \\&\quad = v_0 \int _{z_0={\hat{v}}_0}^{v_0}d(E_{-0}P_0 (z_0, v_{-0}))- \int _{z_0={\hat{v}}_0}^{v_0}z_0 d(E_{-0}P_0 ( z_0, v_{- 0})) \\&\quad = \int _{z_0={\hat{v}}_0}^{v_0}(v_0 -z_0)d(E_{-0}P_0 (z_0, v_{-0})) \\&\quad \ge 0. \end{aligned}$$

The first equality follows by the definition of expected utilities and the second by substitution of (15); the second and the fourth term under the square braces on the right hand side of the second equality cancel out; the remaining part is a function of $v_0$, and hence taking expectation over valuations other than 0 does not affect it. The inequality follows from the fact that if $v_0 > {\hat{v}}_0$ then $v_0 -z_0$ is positive throughout the range of the integral, and the sign of the differential is positive by assumption; if on the other hand, $v_0 < {\hat{v}}_0$ , $v_0 -z_0$ is negative, so we can interchange the limits of the integral and replace $v_0 -z_0$ with $z_0 -v_0$ to obtain a positive sign. For the sellers, we have,

$$\begin{aligned}&E_{-i}U_i(v\mid v_i)- U_i({\hat{v}}_i, v_{-i}\mid v_{i}) \\&\quad = E_{-i}t_i(v)- v_i E_{-i}P_i (v)- E_{-i}t_i ({\hat{v}}_i, v_{-i})+ v_i E_{-i}P_i ({\hat{v}}_i, v_{-i})\\&\quad = E_{-i} \left( \int _{z_i={\underline{v}}}^{{\hat{v}}_i} z_i d(-E_{-i}P_i(z_i, v_{-i}))-\int _{z_i={\underline{v}}}^{v_i} z_i d(-E_{-i}P_i(z_i, v_{- i}))\right) \\&\qquad -v_i E_{-i} P_i (v) + v_i E_{-i}(P_i({\hat{v}}_i, v_{-i}))\\&\quad = E_{-i} \left( \int _{z_i = v_i}^{{\hat{v}}_i} z_i d(-E_{-i}(P_i(z_i, v_{-i}))) -v_i \int _{z_i = v_i}^{{\hat{v}}_i} d(-E_{-i}P_i(z_i, v_{-i}))\right) \\&\quad = \int _{z_i = v_i}^{{\hat{v}}_i}(z_i -v_i)d(-E_{-i}P_i(z_i, v_{-i})) \\&\quad \ge 0, \end{aligned}$$

by arguments similar to the case of the buyer. Since (P, t) is BIC, (4) is satisfied. Since $E_{-i}U_i({\bar{v}}, v_{-i}\mid {\bar{v}})=0$ for all $i=1, \ldots , n$ and the expression in (4) is pre-assumed to be positive, we must have $E_{-0}U_0({\underline{v}}_0, v_{-0}\mid {\underline{v}}_0)\ge 0$. Since $E_{-0}U_0(v\mid v_{0})$ is increasing in $v_{0}$ and $E_{-i}U_i(v\mid v_{i})$’s are decreasing in $v_{i}$ for all $i=1, \ldots , n$, (P, t) must be IR. By construction, (P, t) is BB. $\square$

Since $U_{j}^{(P,t)}(v\mid v_{j})=v_{j}P_{j}(v)-t_{j}(v)$, and (P, t) is BB, (3) can be re-written as

$$\begin{aligned} \max _{P,t} \int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n\text { times}}\sum _{j=0}^{n} v_j P_j(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \end{aligned}$$

(16)

subject to the expression (4) being nonnegative.

Then, using (1), our problem can be written as

$$\begin{aligned}&\max _{P} \int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n\text { times}}\sum _{j=0}^{n} v_j P_j(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n\\ \text {such that }&\int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n\text { times}}\sum _{j=0}^{n} c_{j}(v_j, 1) P_j(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \ge 0. \end{aligned}$$

We construct the Lagrangian objective function as follows.

$$\begin{aligned} \Lambda&=\int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n\text { times}}\sum _{j=0}^{n}\left[ v_j +\lambda c_{j}(v_j, 1)\right] P_j(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \nonumber \\&= (1+\lambda )\int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n\text { times}}\sum _{j=0}^{n}c_{j}\left( v_j, \frac{\lambda }{1+\lambda }\right) P_j(v) f(v_1)\ldots \nonumber \\&\quad f(v_n)g(v_0)dv_0 \ldots dv_n \end{aligned}$$

(17)

We observe that ${\tilde{P}}^\alpha$ maximizes $\Lambda$ for $\alpha =\frac{\lambda }{1+\lambda }$. Now we need to show that ${\tilde{P}}^\alpha$ meets the conditions of Theorem 1, and there exists an $\alpha$ for which the expression in (4) is zero.

Under Myerson regularity, for any $\alpha \in [0,1]$, $c(v_{j}, \alpha )$s are increasing in $v_{j}$. Therefore both ${\tilde{P}}_{0}^{\alpha }$ and ${\tilde{P}}_{i}^{\alpha }$ are increasing in $v_0$ and decreasing in $v_i$ and consequently $E_{-0}{\tilde{P}}_{0}^{\alpha }$ is increasing in $v_0$ and $E_{-i}{\tilde{P}}_{i}^{\alpha }$ is decreasing in $v_i$.

Let

$$\begin{aligned} L^{c}(\alpha )=\int _{{\underline{v}}_0}^{{\bar{v}}_0} \underbrace{\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}}}_{n\text { times}}\sum _{j=0}^{n}c_{j}(v_j, 1) {\tilde{P}}_{j}^{\alpha }(v) f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n \nonumber \\ \end{aligned}$$

(18)

By construction, $L^{c}(1)\ge 0$. Further, ${\tilde{P}}_{0}^{\alpha }$ is increasing in $\alpha$ and ${\tilde{P}}_{i}^{\alpha }$’s are decreasing in $\alpha$. Therefore, if $\alpha <\beta$, $L^{c}(\alpha )$ and $L^{c}(\beta )$ will be different only in profiles v such that ${\tilde{P}}^\beta (v) \equiv 0$ but ${\tilde{P}}_{0}^{\alpha }(v)=1$, ${\tilde{P}}_{i}^{\alpha }(v)=-1$ where $i\in \tilde{{\mathcal {P}}}_{[1]}(v, \alpha )$. This implies $c(v_0, \beta )\le {\tilde{S}}_{[1]}(v, \beta )$ and so $c(v_0, 1)\le {\tilde{S}}_{[1]}(v, 1)$. So, ${\tilde{P}}^{\beta }$ eliminates the virtually inefficient trades allowed by ${\tilde{P}}^{\alpha }$. The equation $c(v_0, \alpha )={\tilde{S}}_{[1]}(v, \alpha )$ has many solutions that vary continuously in $\alpha$. Therefore, $L^{c}(\alpha )$ is continuous, increasing and $L^{c}(1)\ge 0$. If a successful mechanism does not exist, $L^{c}(0)<0$. Hence there exists an $\alpha \in (0,1]$ such that $L^{c}(\alpha )=0$.

Asymptotic properties of optimal $\alpha$ when $k=1$

Suppose $k=1$, $n>k$, and all valuations are independently drawn from U[0, 1]. The expression (4) evaluated at $P^{\alpha }$ is:

$$\begin{aligned}&\int _{\frac{\alpha }{1+\alpha }}^{1}\int _{0}^{v_{0}-\frac{\alpha }{1+\alpha }}\left( 2v_{0}-2v_{1}-1\right) n(1-v_{1})^{n-1}dv_{1}dv_{0}. \end{aligned}$$

At the optimal $\alpha$, this expression must equal zero. Further, the derivative of this expression with respect to $\alpha$ can be obtained by an application of the Leibniz Rule:

$$\begin{aligned}&\int _{\frac{\alpha }{1+\alpha }}^{1}\frac{d}{d\alpha }\left[ \int _{0}^{v_{0}-\frac{\alpha }{1+\alpha }}\left( 2v_{0}-2v_{1}-1\right) n(1-v_{1})^{n-1}dv_{1}\right] dv_{0}\\&\quad = -\int _{\frac{\alpha }{1+\alpha }}^{1}\frac{1}{(1+\alpha )^{2}}\left( 2v_{0}-2\left( v_{0}-\frac{\alpha }{1+\alpha }\right) -1\right) n\left( 1-\left( v_{0}-\frac{\alpha }{1+\alpha }\right) \right) ^{n-1}dv_{0}\\&\quad = \frac{n(1-\alpha )}{(1+\alpha )^{3}}\int _{\frac{\alpha }{1+\alpha }}^{1}n\left( 1-\left( v_{0}-\frac{\alpha }{1+\alpha }\right) \right) ^{n-1}dv_{0}. \end{aligned}$$

Changing the variable of integration to $z=1-v_{0}+\frac{\alpha }{1+\alpha }$, this integral becomes

$$\begin{aligned}&\frac{n(1-\alpha )}{(1+\alpha )^{3}}\int _{\frac{\alpha }{1+\alpha }}^{1}z^{n-1}dz\\&\quad = \frac{(1-\alpha )}{(1+\alpha )^{3}}\times \left( 1-\left( \frac{\alpha }{1+\alpha }\right) ^{n}\right) \end{aligned}$$

Notice that this expression is strictly positive for all $\alpha \in (0,1)$.

Now compare the value of (4) at $n+1$ and n. Their difference is:

$$\begin{aligned}&\int _{\frac{\alpha }{1+\alpha }}^{1}\int _{0}^{v_{0}-\frac{\alpha }{1+\alpha }}\left( 2v_{0}-2v_{1}-1\right) \left[ (n+1)(1-v_{1})^{n}-n(1-v_{1})^{n-1}\right] dv_{1}dv_{0}\\&\quad =\int _{\frac{\alpha }{1+\alpha }}^{1}\int _{0}^{v_{0}-\frac{\alpha }{1+\alpha }}\left( 2v_{0}-2v_{1}-1\right) \left[ n(1-v_{1})^{n-1}\left\{ 1-v_{1}-1\right\} \right. \\&\qquad \left. +(1-v_{1})^{n}\right] dv_{1}dv_{0}\\&\quad =\frac{1}{n}\int _{\frac{\alpha }{1+\alpha }}^{1}\int _{0}^{v_{0}-\frac{\alpha }{1+\alpha }}\left( 2v_{0}-2v_{1}-1\right) n(1-v_{1})^{n-1}dv_{1}dv_{0}\\&\qquad -\int _{\frac{\alpha }{1+\alpha }}^{1}\int _{0}^{v_{0}-\frac{\alpha }{1+\alpha }}\left( 2v_{0}-2v_{1}-1\right) n(1-v_{1})^{n-1}v_{1}dv_{1}dv_{0} \end{aligned}$$

Notice that the final expression is $\frac{1}{n}$ times the expression in (4) minus another term which is the same as the expression in (4) upto the multiplicative term $v_{1}$. At the optimal $\alpha$ corresponding to n, the first term is zero while the second term is positive. Hence this difference must be negative. Since the expression in (4) is increasing in $\alpha$, the optimal $\alpha$ must decrease with n.

Proof of Proposition 2

The following Lemma is useful to prove this result.

Lemma 3

The expected sum of VCG payments satisfies

$$\begin{aligned} E\left( \sum _{j=0}^{n}t_{j}^{V}(v)\right)&=E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\right) \end{aligned}$$

Proof

The VCG mechanism is ex post efficient, BIC and as mentioned below (10), it satisfies the property $E_{-j}U_{j}^{V}(x_{j}, v_{-j})=0$, where $x_{0}={\underline{v}}_{0}$ and $x_{i}={\bar{v}}$ for $i\in \{1, \ldots , n\}$. Thus it follows from (12) that for all $j\in \{0, \ldots , n\}$,

$$\begin{aligned} E_{-j}U_{j}^{V}(v_{j})&= \int _{x_j}^{v_j} E_{-j}P_{j}^{*}(z, v_{-j}) dz \text { for all } v_{j}. \end{aligned}$$

(19)

Taking expectation with respect to $v_{j}$ on both sides, and summing over j we get

$$\begin{aligned} \sum _{j=0}^{n}EU_{j}^{V}(v)&= \int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} P_{0}^{*} (v)\frac{1-G(v_0)}{g(v_0)}f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n\\&\quad + \sum _{i=1}^{n}\int _{{\underline{v}}_0}^{{\bar{v}}_0}\int _{{\underline{v}}}^{{\bar{v}}}\ldots \int _{{\underline{v}}}^{{\bar{v}}} P_{i}^{*}(v)\frac{F(v_i)}{f(v_i)} f(v_1)\ldots f(v_n)g(v_0)dv_0 \ldots dv_n.\\&=E\left( P_{0}^{*} (v)\frac{1-G(v_0)}{g(v_0)}+\sum _{i=1}^{n}P_{i}^{*}(v)\frac{F(v_i)}{f(v_i)}\right) \end{aligned}$$

Since the left hand side of the above equality is

$$\begin{aligned}&E\left( \sum _{j=0}^{n}\left( v_{j}P_{j}^{*}(v)-t_{j}^{V}(v)\right) \right) , \end{aligned}$$

we get,

$$\begin{aligned} E\left( \sum _{j=0}^{n}t_{j}^{V}(v)\right)&=E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\right) -E\left( P_{0}^{*} (v)\frac{1-G(v_0)}{g(v_0)}+\sum _{i=1}^{n}P_{i}^{*}(v)\frac{F(v_i)}{f(v_i)}\right) \\&=E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\right) , \end{aligned}$$

where $c(v_{j}, 1)$s are the virtual valuations defined in (1) corresponding to $\alpha =1$. $\square$

If Part: Suppose

$$\begin{aligned}&E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\right) \rightarrow E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\right) \text { as } n\rightarrow \infty . \end{aligned}$$

Notice that trade does not take place in $P^{OPT}$ at a profile where trade does not take place in $P^{*}$, but there are profiles where trade takes place in $P^{*}$ and not in $P^{OPT}$.

The expected surplus in the optimal mechanism can be written as follows:

$$\begin{aligned}&E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\right) \nonumber \\&\quad =E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\mid P^{OPT}(v)=P^{*}(v)\right) \Pr \left( P^{OPT}(v)=P^{*}(v)\right) \nonumber \\&\qquad +E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\mid P^{OPT}(v)\ne P^{*}(v)\right) \Pr \left( P^{OPT}(v)\ne P^{*}(v)\right) \nonumber \\&\quad =E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\mid P^{OPT}(v)=P^{*}(v)\right) \Pr \left( P^{OPT}(v)=P^{*}(v)\right) \end{aligned}$$

(20)

since the second term of the first equality is zero as $\sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)=0$ at every profile v where $P^{OPT}(v)\ne P^{*}(v)$.

The expected surplus in an efficient mechanism can be written as follows:

$$\begin{aligned}&E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\right) \nonumber \\&\quad =E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\mid P^{OPT}(v)=P^{*}(v)\right) \Pr \left( P^{OPT}(v)=P^{*}(v)\right) \nonumber \\&\qquad +E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\mid P^{OPT}(v)\ne P^{*}(v)\right) \Pr \left( P^{OPT}(v)\ne P^{*}(v)\right) \end{aligned}$$

(21)

Also,

$$\begin{aligned}&E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\right) \nonumber \\&\qquad = E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\mid P^{*}(v)=P^{OPT}(v)\right) \Pr \left( P^{*}(v)=P^{OPT}(v)\right) \nonumber \\&\qquad \quad + E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\mid P^{*}(v)\ne P^{OPT}(v)\right) \Pr \left( P^{*}(v)\ne P^{OPT}(v)\right) \end{aligned}$$

(22)

As $P_{j}^{OPT}(v)=0$ for all j at every profile v where $P^{OPT}(v)\ne P^{*}(v)$, the second term on the right hand side is zero. Therefore,

$$\begin{aligned}&E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\right) \nonumber \\&\quad =E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\mid P^{*}(v)=P^{OPT}(v)\right) \Pr \left( P^{*}(v)=P^{OPT}(v)\right) . \end{aligned}$$

(23)

The expected budget surplus for VCG can be written as

$$\begin{aligned}&E\left( \sum _{j=0}^{n}t_{j}^{V}(v)\right) \nonumber \\&\quad = E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\right) \nonumber \\&\quad = E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\mid P^{*}(v)=P^{OPT}(v)\right) \Pr \left( P^{*}(v)=P^{OPT}(v)\right) \nonumber \\&\qquad + E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\mid P^{*}(v)\ne P^{OPT}(v)\right) \Pr \left( P^{*}(v)\ne P^{OPT}(v)\right) \end{aligned}$$

(24)

where the first equality follows from Lemma 3.

If

$$\begin{aligned}&E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\right) \rightarrow E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\right) \text { as } n\rightarrow \infty , \end{aligned}$$

it follows from (20) and (21) that

$$\begin{aligned}&E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\mid P^{OPT}(v)\ne P^{*}(v)\right) \Pr \left( P^{*}(v)\ne P^{OPT}(v)\right) \rightarrow 0 \text { as } n\rightarrow \infty . \end{aligned}$$

Since $0\le \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\le {\bar{v}}_{0}-k{\underline{v}}$ for all v, the sequence of expectations $\left\{ E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\mid P^{OPT}(v)\ne P^{*}(v)\right) \right\} _n$ is bounded. Therefore,

$$\begin{aligned}&\Pr \left( P^{OPT}(v)\ne P^{*}(v)\right) \rightarrow 0 \text { as } n\rightarrow \infty . \end{aligned}$$

Now notice that at any profile v where $P^{*}(v)\ne P^{OPT}(v)$,

$$\begin{aligned} c_{0}(v_{0}, \alpha )&\le {\mathcal {S}}_{[1]}(v, \alpha ), \end{aligned}$$

where $\alpha$ characterizes the relevant optimal mechanism. Since $c_{0}(v_{0}, \alpha )$ is decreasing in $\alpha$ and $c_{i}(v_{i}, \alpha )$’s are increasing in $\alpha$, at all such profiles,

$$\begin{aligned} c_{0}(v_{0}, 1)&\le {\mathcal {S}}_{[1]}(v, 1), \end{aligned}$$

Therefore, the expectation in the second term on the right hand side of (24) is negative. Further, the sequence of such expectations is bounded below since $c_{0}(v_{0}, 1)\ge c_{0}({\underline{v}}_{0}, 1)$ and $c_{i}(v_{i}, 1)\le c_{i}({\bar{v}}, 1)$ by our regularity assumption. Therefore, the second term on the right hand side of (24) vanishes to zero as $n\rightarrow \infty$. By (23) the first term is equal to $E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\right) =0$. Hence the claim.

Only If Part: Suppose

$$\begin{aligned}&E\left( \sum _{j=0}^{n}t_{j}^{V}(v)\right) \rightarrow 0\text { as } n\rightarrow \infty . \end{aligned}$$

Since $E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\right) =0$, by (23) and (24),

$$\begin{aligned}&E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\mid P^{*}(v)=P^{OPT}(v)\right) \Pr \left( P^{*}(v)=P^{OPT}(v)\right) \\&\qquad + E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\mid P^{*}(v)\ne P^{OPT}(v)\right) \Pr \left( P^{*}(v)\ne P^{OPT}(v)\right) \\&\quad \rightarrow E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{OPT}(v)\mid P^{*}(v)=P^{OPT}(v)\right) \Pr \left( P^{*}(v)=P^{OPT}(v)\right) . \end{aligned}$$

This implies

$$\begin{aligned}&E\left( \sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\mid P^{*}(v)\ne P^{OPT}(v)\right) \Pr \left( P^{*}(v)\ne P^{OPT}(v)\right) \rightarrow 0. \end{aligned}$$

Since the sequence of expectations $\Big \{E\Big (\sum _{j=0}^{n}c_{j}(v_{j}, 1)P_{j}^{*}(v)\mid P^{*}(v)\ne P^{OPT}(v)\Big )\Big \}_n$ is bounded as argued above,

$$\begin{aligned}&\Pr \left( P^{*}(v)\ne P^{OPT}(v)\right) \rightarrow 0. \end{aligned}$$

Now, by (20) and (21),

$$\begin{aligned}&E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\right) -E\left( \sum _{j=0}^{n}v_{j}P_{j}^{OPT}(v)\right) \\&\quad =E\left( \sum _{j=0}^{n}v_{j}P_{j}^{*}(v)\mid P^{OPT}(v)\ne P^{*}(v)\right) \Pr \left( P^{OPT}(v)\ne P^{*}(v)\right) \end{aligned}$$

Since the expectation term is bounded, and the probability is vanishingly small, this expression tends to zero as $n\rightarrow \infty$. Hence the claim.