Distortion-Free Logrolling Mechanism

Abstract

In a large population, the first mover advantage causes only a vanishingly small distortion to the unanimity bargaining outcome. However, if the bargainers form parties through which bargaining takes place—as is the case in representative democracies—the first mover distortion remains significant. The main result of the chapter is that the distortion-free allocation can be restored under a party system if the first proposer right is allocated via a logrolling mechanism (an auction where the winning bid is added to the pool of resources) prior to the actual bargaining. Thus logrolling can be justified as a mechanism that corrects the distortion inherent to representative democracy.

I thank an anonymous referee for very useful comments. I have also benefitted from conversations with Klaus Kultti and Hannu Salonen.

Appendix: Proofs

Proof of Lemma 3.

Define a function $\bar{v}$ such that

$$\begin{aligned} \bar{u}(\bar{v}(z_{u}))=\bar{u}(z_{u})\delta , \text{ for} \text{ all}\, z_{u}\in [0,1]. \end{aligned}$$

By the concavity of $\bar{u},$ $\bar{v}$ and $x-\bar{v}(x)$ are both increasing functions. Rewrite condition (9) as

$$\begin{aligned} \bar{v}_{i}^{-1}\left( z_{i}(X)\right) -z_{i}(X)=X-{\textstyle \sum _{i\in U}}z_{j}(X). \end{aligned}$$

Since $x-\bar{v}(x)$ is an increasing function, and $z_{i}$ continuous, $z_{i}$ is strictly increasing if $X-{\textstyle \sum } z_{j}(X)$ is. Since this applies to all $i$, ${\textstyle \sum } z_{j}(X)$ is strictly increasing if $X-{\textstyle \sum } z_{j}(X)$ is. But then, since ${\textstyle \sum } z_{j}(X)$ being weakly decreasing means that $X-{\textstyle \sum } z_{j}(X)$ is strictly increasing, it cannot be the case that ${\textstyle \sum } z_{j}(X)$ is not strictly increasing. Thus ${\textstyle \sum } z_{j}(X)$ is strictly increasing and hence $z_{i}$ is strictly increasing.    $\blacksquare$

Proof of Lemma 4

Only if: First we argue that there are at least two highest bids. Suppose that there is a single highest bid. Then buying the proposing right with price $p$ must be at least profitable as the opportunity cost of lowering the bid by a small $\varepsilon >0:$

$$\begin{aligned} \left[ 1+p-{\textstyle \sum \limits _{j\not =i}}z_{j}(1+p)\right] -p\ge \left[ 1+p-\varepsilon -{\textstyle \sum \limits _{j\not =i}} z_{j}(1+p-\varepsilon )\right] -(p-\varepsilon ). \end{aligned}$$

That is

$$\begin{aligned} 0\ge {\textstyle \sum _{j\not =i}}[z_{j}(1+p)-z_{j}(1+p-\varepsilon )]. \end{aligned}$$

But by Lemma 3 this cannot hold.

Thus at least two bidders bid the winning bid $p$. Then buying the proposing right under $p$ must be at least profitable as the opportunity cost of letting the other highest bidder win with price $p$:

$$\begin{aligned} \left[ 1+p-{\textstyle \sum \limits _{j\not =i}}z_{j}(1+p)\right] -p\ge z_{i}(1+p). \end{aligned}$$

(12)

Since increasing ones bid is not profitable for the losing bargainer $j$ that bids $p,$

$$\begin{aligned} \left[ 1+p+\varepsilon -{\textstyle \sum \limits _{k\not =j}} z_{k}(1+p+\varepsilon )\right] -(p+\varepsilon )\le z_{j}(1+p),\text{ for} \text{ all}\, \varepsilon > 0. \end{aligned}$$

(13)

Since $z_{k}$ is continuous and (13) holds for all $\varepsilon >0,$ it follows that

$$\begin{aligned} \left[ 1+p-{\textstyle \sum \limits _{k\not =j}} z_{k}(1+p)\right] -p\le z_{k}(1+p). \end{aligned}$$

(14)

Combining (12) and (14) gives

$$\begin{aligned} 1={\textstyle \sum _{i\in \bar{U}}}z_{i}(1+p). \end{aligned}$$

Thus by (9),

$$\begin{aligned} u_{i}(z_{i}(1+p))=u_{i}(z_{i}(1+p)+p)\delta , \text{ for} \, \text{ all}\, i=1,\ldots ,n. \end{aligned}$$

By Lemma 5, this yields $z_{i}(1+p)=z_{i}^{*}$ for all $i,$ and $p=p^{*}$.

If: Let all $U$ bargainers bid $p=p^{*}$. By construction, $z_{i}(1+p^{*})=z_{i}^{*}$ for all $i\in \bar{U}.$ We show this does constitute an equilibrium. Since $n>1$ and

$$\begin{aligned} 1={\textstyle \sum _{i\in \bar{U}}} z_{i}(1+p^{*}), \end{aligned}$$

(15)

it follows that

$$\begin{aligned} \left[ 1+p^{*}-{\textstyle \sum \limits _{j\not =i}}z_{j}(1+p^{*})\right] -p^{*}=z_{i}(1+p^{*}). \end{aligned}$$

Thus decreasing one’s bid does not have payoff consequences. Increasing one’s bid by $\varepsilon >0$ is strictly profitable if

$$\begin{aligned} \left[ 1+p^{*}+\varepsilon -{\textstyle \sum \limits _{j\not =i}} z_{j}(1+p^{*}+\varepsilon )\right] -(p^{*}+\varepsilon )> z_{i}(1+p^{*}). \end{aligned}$$

That is, by (15),

$$\begin{aligned} 1-{\textstyle \sum _{j\not =i}}z_{j}(1+p^{*}+\varepsilon )>1-{\textstyle \sum _{j\not =i}} z_{j}(1+p^{*}), \end{aligned}$$

which is in conflict with Lemma 3. Thus all players bidding $p^{*}$ does constitute an equilibrium.    $\blacksquare$

Proof of Proposition 4

Since $\bar{u}\left(\bar{v}(z_{u})\right)=\bar{u}(z_{u})\delta$ and (8) imply $u(\lambda _{u}^{-1}\bar{v}(z_{u}))=u(\lambda _{u}^{-1}z_{u})\delta$ and the definition of $v$ implies $u(\lambda _{u}^{-1}z_{u})\delta =u(v(\lambda _{u}^{-1}z_{u}))$ we have $\bar{v}(z_{u})=\lambda _{u}v(\lambda _{u}^{-1}z_{u}).$ Thus (10) and (11) can be written

$$\begin{aligned} u(\lambda _{u}^{-1}z_{u}^{*})\delta&=u\left( \lambda _{u}^{-1} (z_{u}^{*}+p^{*})\right) , \text{ for} \text{ all}\ u\in U,\\ {\textstyle \sum _{u\in U}} z_{u}^{*}&=1. \end{aligned}$$

By Lemma 4, this characterizes the equilibrium. Letting $y_{u}^{*}=\lambda _{u}^{-1}z_{u}^{*}$ for all $u,$ and $d^{*}=\lambda ^{-1}p^{*},$ this transforms into

$$\begin{aligned} u(y_{u}^{*})&=u\left( y_{u}^{*}+d^{*}\right) \delta ,\\ {\textstyle \sum _{u\in U}} y_{u}^{*}\lambda _{u}&=1. \end{aligned}$$

Constructing $x^{*}$ as in (5) now gives the result.   $\blacksquare$

Note that, given $u_{i},$ there is a function $v_{i}$ that specifies the present consumption value of $x_{i}$ in date $1$ such that

$$\begin{aligned} u_{i}\left(v_{i}(x_{i})\right)=u_{i}(x_{i})\delta ,\quad \text{ for} \text{ all}\, x_{i}\in [0,1]. \end{aligned}$$

(16)

By the concavity of $u_{i}$, $v_{i}^{-1}(x_{i})-x_{i}$ is a continuous and monotonically increasing function of $x_{i}$.

Proof of Lemma 1

Recall that $c_{i}>0$ for all $i$ and $Y\ge 0.$ Thus, the function $e_{i}(\cdot )$ defined by

$$\begin{aligned} e_{i}(x_{i}):=\frac{v_{i}^{-1}(c_{i}x_{i})}{c_{i}}-x_{i},\text{ for} \text{ any}\ x_{i}\ge 0, \end{aligned}$$

(17)

is continuous and monotonically increasing.

Define $\bar{e}_{i}\in (0,\infty ]$ by

$$\begin{aligned} \sup _{x_{i}\ge 0}e_{i}(x_{i}):=\bar{e}_{i}. \end{aligned}$$

Since $e_{i}(\cdot )$ is continuous and monotonically increasing, also its inverse

$$\begin{aligned} x_{i}(e):=e_{i}^{-1}(e), \text{ for} \text{ all}\ e\in [0,\bar{e}_{i}], \end{aligned}$$

is continuous and monotonically increasing in its domain $[0,\bar{e}_{i}]$. Condition (17) can now be stated in the form

$$\begin{aligned} x_{i}(e)=\frac{v_{i}(c_{i}(x_{i}(e)+e))}{c_{i}}, \text{ for} \text{ all}\ e\in [0,\bar{e}_{i}]. \end{aligned}$$

(18)

Moreover, since $0=x_{i}(0)$ and $\infty =x_{i}(\bar{e}_{i}),$ there is, by the Intermediate Value Theorem, a unique $d>0$ such that

$$\begin{aligned} {\textstyle \sum _{i=1}^{n}}x_{i}(d)=Y. \end{aligned}$$

   $\blacksquare$

Proof of Proposition 2

Only if: In a stationary SPE the game ends in finite time. Assume that it never ends. Then each player receives zero. This means that in all subgames each player must get zero. Otherwise there would be a subgame where some offer $y=\left( y_{1},\ldots ,y_{n}\right)$ is accepted. Because of stationarity this offer is accepted in every subgame. In particular, player $1$ can deviate in the first period and offer $y=\left( y_{1},\ldots ,y_{n}\right)$. This is a profitable deviation and constitutes a contradiction with the assumption that there is a stationary SPE where the game never ends.

Assume next that there is a stationary SPE where an offer $x(i)$ by some player $i\in \left\{ 1,2,\ldots ,n\right\}$, is not accepted immediately. Denote by $z(i)$ the equilibrium outcome in a subgame that starts with an offer $x(i)$ of player $i$. But now player $i$ could offer $z(i)$ instead of $x(i)$; everyone else would accept the offer as in the stationary equilibrium acceptance depends only on the offer.

Thus, in any equilibrium, $i(t)$’s offer $x(i(t))=(x_{j}(i(t)))_{j\in N}$ is accepted at stage $t\in \{0,1,2,..\}$. In stationary equilibrium the time index $t$ can be relaxed from $x(i(t)).$ An offer $x$ by $i$ is accepted by all $j\not =i$ if

$$\begin{aligned} x_{j}(i)\ge v_{j}(x_{j}(j)), \text{ for} \text{ all}\ j\not =i. \end{aligned}$$

(19)

Player $i$’s equilibrium offer $x(i)$ maximizes his payoff with respect to constraint (19) and the resource constraint. By A3, all constraints in (19) and the resource constraint must bind. That is,

$$\begin{aligned} v_{j}(x_{j}(i))=v_{j}(x_{j}(j)), \text{ for} \text{ all}\ j\not =i, \end{aligned}$$

(20)

and

$$\begin{aligned} {\textstyle \sum _{i=1}^{n}}x_{i}(j)=X,\,\text{ for} \text{ all}\, j. \end{aligned}$$

(21)

Since player $i$’s acceptance decision is not dependent on the name of the proposer, there is $x_{i}>0$ such that $x_{i}(j)=x_{i}$ for all $j\not =i.$ By (20), $x_{j}(i)<x_{j}(j)$ for all $j.$ Hence there is $d>0$ such that

$$\begin{aligned} {\textstyle \sum _{i=1}^{n}}x_{i}=X-d. \end{aligned}$$

(22)

By (20) and (22), $x$ and $d$ do meet (1) and (2). Since $1$ is the first proposer, the resulting outcome is $x(1)=(x_{1} +d,x_{2},\ldots ,x_{n})$.

If: Let $x$ and $d$ meet (1) and (2). Construct the following stationary strategy: Player $i$ always offers $x_{-i}$ and does not accept less than $x_{i}.$ Player $i$’s offer $y$ is accepted by all $j\not =i$ only if

$$\begin{aligned} y_{j}\ge v_{j}(X-{\textstyle \sum \nolimits _{k\not =j}} x_{k})=v_{j}(x_{j}+e_{1}(x_{1})), \text{ for} \text{ all}\ j\not =i. \end{aligned}$$

(23)

Since $v_{j}$ is increasing, and since

$$\begin{aligned} x_{j}=v_{j}(x_{j}+d), \text{ for} \text{ all}\ j\not =i, \end{aligned}$$

$i$’s payoff maximizing offer to each $j$ is $x_{j}.$   $\blacksquare$

Proof of Proposition 2:

Lemma 5

For any $n,$ there are unique $y(n)\in \mathbb{R }^{n}$ and $d(n)>0$ such that

$$\begin{aligned} y_{i}(n)&=v_{i}(y_{i}(n)+d(n)), \quad for \, all\ i=1,\ldots ,n,\end{aligned}$$

(24)

$$\begin{aligned} {\textstyle \sum _{u=1}^{n}}y_{i}(n)&=n-d(n). \end{aligned}$$

(25)

Proof

By Lemma 1.    $\blacksquare$

By Lemma 2, the set of allocations the planner can implement under $n$ agents is

$$\begin{aligned} \left\{ x:\frac{1}{n}{\textstyle \sum _{i=1}^{n}}x_{i}\le 1,\ \text{ and}\ x_{i}\ge y_{i}(n), \text{ for} \text{ all}\ i=1,\ldots ,n\right\} . \end{aligned}$$

Lemma 6

Let $y(n)$ and $d(n)$ be defined as in Lemma 5. Then there is $y^{*}\in \mathbb{R } ^{U}$ and $d^{*}>0$ such that $y_{i}(n)\rightarrow _{n}y_{u}^{*},$ for all $u_{i}=u$ and $u\in U,$ and $d(n)\rightarrow _{n}d^{*},$ where

$$\begin{aligned} y_{u}^{*}&=v(y_{u}^{*}+d^{*}), \text{ for} \text{ all}\ u\in U,\end{aligned}$$

(26)

$$\begin{aligned} {\textstyle \sum _{u\in U}} \lambda _{u}y_{u}^{*}&=1. \end{aligned}$$

(27)

Proof

By Lemma 5, for any $n=1,2,\ldots ,$

$$\begin{aligned} y_{i}(n)&=v_{i}(y_{i}(n)+d(n)), \text{ for} \text{ all}\ i=1,\ldots ,n,\end{aligned}$$

(28)

$$\begin{aligned} {\textstyle \sum _{i=1}^{n}}y_{i}(n)&=n-d(n). \end{aligned}$$

(29)

Dividing both sides of (29) by $n,$

$$\begin{aligned} \frac{1}{n}{\textstyle \sum _{i=1}^{n}}y_{i}(n)=1-\frac{d(n)}{n}. \end{aligned}$$

(30)

Define a function $i:U\rightarrow \{1,2,\ldots \}$ such that $u_{i(u)}=u,$ for all $u\in U.$ By stationarity, $y_{i(u)}(n)=y_{j}(n)$ if $u=u_{j}.$ The left hand side of (30) can now be written

$$\begin{aligned} \frac{1}{n}{\textstyle \sum _{i=1}^{n}} y_{i}(n)=\frac{1}{n}{\textstyle \sum _{u\in U}} y_{i(u)}(n){\textstyle \sum _{i=1}^{n}} 1_{(u_{i}=u)}. \end{aligned}$$

By the law of large numbers,

$$\begin{aligned} \lim _{n}\frac{1}{n} {\textstyle \sum _{i=1}^{n}} 1_{(u_{i}=u)}=\lambda _{u}. \end{aligned}$$

(31)

Take any subsequence $\{n^{\prime }\}$ under which $\lim _{n^{\prime }}y_{i(u)}(n^{\prime })$ for all $u$ and $\lim _{n^{\prime }}d(n^{\prime })$ exist (the limit can be either finite or infinite)$.$ Then (28) can be written

$$\begin{aligned} \lim _{n^{\prime }} {\textstyle \sum _{u\in U}}\lambda _{u}y_{i(u)}(n^{\prime })=1-\lim _{n^{\prime }\rightarrow \infty } \frac{d(n^{\prime })}{n^{\prime }} \end{aligned}$$

(32)

By (28) $\lim _{n^{\prime }}d(n^{\prime })=\infty$ if and only if $\lim _{n^{\prime }}y_{i(u)}(n^{\prime })=\infty$ for all $u$. Thus, by (32), it must be the case that $\lim _{n^{\prime }}y_{i(u)}(n^{\prime })=y_{u}^{*}$ and $d(n)=d^{*},$ for some $(y^{*},d^{*})\in \mathbb{R } _{++}^{\left|U\right|}\times \mathbb{R } _{++}$. By (28), (32) becomes

$$\begin{aligned} \lim _{n^{\prime }}{\textstyle \sum _{u\in U}}\lambda _{u}y_{i(u)}(n^{\prime })= {\textstyle \sum _{u\in U}}\lambda _{u}y_{u}^{*}=1. \end{aligned}$$

(33)

By Lemma 1 and (28), $y_{i(u)}^{*}$ is the limit of any converging subsequence $\{y_{i(u)}(n^{\prime \prime })\}$, and $d^{*}$ is the limit of any converging subsequence $\{d(n^{\prime \prime })\}$. Thus $(y^{*},d^{*})$ is the unique limit and by (33), continuity, and (28) it meets the conditions imposed by the lemma.   $\blacksquare$

Proposition 5

As$n\rightarrow \infty ,$ allocation$x$ is implementable by the planner if and only if $x=x^{*}$.

Proof

Again, define a function $i:U\rightarrow \{1,2,\ldots \}$ such that $u_{i(u)} =u,$ for all $u\in U.$ By stationarity, $y_{i(u)}(n)=y_{j}(n)$ if $u=u_{j},$ for all $j=1,\ldots ,n.$ The set of implementable allocations can be written

$$\begin{aligned}&\left\{ x\in {\mathbb{R }}_{+}^{n}:\frac{1}{n}{\textstyle \sum _{i=1}^{n}} x_{i}\le 1,\ \text{ and}\ x_{i}\ge y_{i}(n), \text{ for} \text{ all}\ i=1,\ldots ,n\right\} \\&\\&=\left\{ x\in {\mathbb{R }}_{+}^{n}:\frac{1}{n}{\textstyle \sum _{u\in U}} x_{i(u)}{\textstyle \sum _{i=1}^{n}} 1_{(u_{i}=u)}\le 1,\ \text{ and}\ x_{j}=x_{i(u)}\ge y_{i(u)}(n)\text{ if}\ u=u_{j},\text{ for} \text{ all}\ u\in U\right\} \end{aligned}$$

Taking the limit,

$$\begin{aligned}&\lim _{n}\left\{ x\in {\mathbb{R }}_{+}^{n}:\frac{1}{n}{\textstyle \sum _{u\in U}} x_{i(u)}{\textstyle \sum _{i=1}^{n}} 1_{(u_{i}=u)}\le 1,\ \text{ and}\ x_{j}=x_{i(u)}\ge y_{i(u)}(n)\text{ if}\ u=u_{j}, \text{ for} \text{ all}\ u\in U\right\} \\&\\&=\left\{ x\in {\mathbb{R }}_{+}^{\infty }:{\textstyle \sum _{u\in U}} x_{i(u)}\lambda _{u}\le 1,\ \text{ and}\ x_{j}=x_{i(u)}\ge y_{i(u)}^{*} \text{ if}\ u=u_{j},\text{ for} \text{ all}\ u\in U\right\} . \end{aligned}$$

By (27), this reduces to

$$\begin{aligned} \left\{ x\in {\mathbb{R }} _{+}^{\infty }:x_{j}=y_{i(u)}^{*}\text{ if}\ u=u_{j},\text{ for} \text{ all}\ u\in U\right\} , \end{aligned}$$

which is a singleton $\{x^{*}\}$, as required by the proposition.    $\blacksquare$