Lobbying as a multidimensional tug of war

Abstract

We analyze lobbying as contest in which lobbyists exert effort to pull a policy outcome in a multidimensional space in their preferred directions. We prove existence and uniqueness of equilibrium and perform comparative statics on the cost of effort and policy utility of the lobbyists. As cost of effort increases, the equilibrium policy outcome and inefficiency (i.e., total effort expended) are constant. Assuming power utility, the equilibrium policy minimizes a social loss function that depends on curvature of utilities. As lobbyists become less tolerant of larger losses, the equilibrium policy outcome converges to the Rawlsian policy, which maximizes the payoff of the worst-off lobbyist, and inefficiency may become large or go to zero, depending on the configuration of ideal points. As lobbyists become less tolerant of smaller losses, the equilibrium outcome converges to the mean of the lobbyists’ ideal points, and inefficiency goes to zero.

A Proofs of propositions

Proof of Proposition 1:

We verify (A2) and (A3). For the former, it suffices to show that the second derivative of $u(\Vert x^{*}-\hat{x}^{i}\Vert )-c\alpha _{i}^{2}$ with respect to $\alpha _{i}$ is negative. For any lobbyist i, the first derivative with respect to $\alpha _{i}$ is the following:

$$\begin{aligned} \frac{r}{\sum _{j} \alpha _{j}} \left[ \sum _{m=1}^{d} \left( x^{*}_{m}-\hat{x}_{m}^{i} \right) ^{2} \right] ^{\frac{r}{2}} - 2c\alpha _{i} \end{aligned}$$

The second derivative simplifies to:

$$\begin{aligned}&-\frac{1}{\left( \sum _{j} \alpha _{j} \right) ^{2}} \left( r^{2} \left[ \sum _{m=1}^{d} \left( x^{*}_{m}- \hat{x}_{m}^{i} \right) ^{2} \right] ^{\frac{r}{2}} + r \left[ \sum _{m=1}^{d} \left( x^{*}_{m}- \hat{x}_{m}^{i} \right) ^{2} \right] ^{\frac{r}{2}} \right) -2c \\&\quad = -\frac{1}{\left( \sum _{j} \alpha _{j} \right) ^{2}} \left( r^{2} + r\right) \left[ \sum _{m=1}^{d} \left( x^{*}_{m}- \hat{x}_{m}^{i} \right) ^{2} \right] ^{\frac{r}{2}} - 2c\\&\quad < 0, \end{aligned}$$

as required. To verify (A3), write

$$\begin{aligned} 2u'(z) z + u''(z) z^{2}&= - 2rz^{r-1}z - r(r-1)z^{r-2}z^{2}\\&= -2rz^{r} - r(r-1)z^{r}\\&=- \left( r^{2} + r \right) z^{r}\\&< 0, \end{aligned}$$

as required. $\square$

Proof of Proposition 2:

Suppose toward a contradiction that there exist distinct policies $x'$ and $x''$ such that $\min _{i} f_{i}(x') = \min _{i} f_{i}(x'') = \max _{x} \min _{i} f_{i}(x)$, and let $\underline{u}$ denote this value. Set $x''' = \frac{1}{2}x' + \frac{1}{2}x''$, and let k solve $\min _{i} f(x''')$, so that $f_{k}$ is among the functions taking the lowest value at $x'''$. By construction, we have

$$\begin{aligned} \min \left\{ f_{k} \left( x' \right) ,f_{k} \left( x'' \right) \right\}\ge & {} \underline{u}. \end{aligned}$$

Then strict quasi-concavity implies $f_{k}(x''') > \underline{u}$, but this implies that $x'$ and $x''$ do not solve $\max _{x}\min _{i}f_{i}(x)$, a contradiction. We conclude that the latter problem has a unique solution, as required. $\square$

Proof of Proposition 3:

First, assume $x=x^{R}$, and suppose toward a contradiction that x is not a convex combination of ideal points of lobbyists belonging to N(x). Then x is not Pareto optimal for the lobbyists in this set, and there exists $x'$ such that for all $i \in N(x)$, we have $u(\Vert x'-\hat{x}^{i}\Vert ) > u(\Vert x-\hat{x}^{i}\Vert )$. For each $\epsilon \in (0,1)$, define $x^{\epsilon }=(1-\epsilon )x + \epsilon x'$, and note that strict quasi-concavity implies that for all such $\epsilon$ and all $i \in N(x)$, we have $u(\Vert x^{\epsilon }-\hat{x}^{i}\Vert ) > u(\Vert x-\hat{x}^{i}\Vert )$. Furthermore, by continuity of u, we can choose $\epsilon >0$ sufficiently small that for all $i \in N \setminus N(x)$, we have $u(\Vert x^{\epsilon }-\hat{x}^{i}\Vert ) > u(\Vert x-\hat{x}^{i}\Vert )$. But then we have

$$\begin{aligned} \min _{j}u \left( \left\| x^{\epsilon }-\hat{x}^{j} \right\| \right)> & {} \min _{j}u \left( \left\| x^{R}-\hat{x}^{j} \right\| \right) , \end{aligned}$$

a contradiction. We conclude that x is a convex combination of ideal points of lobbyists in N(x).

For the converse, assume that x is a convex combination of ideal points of lobbyists in N(x), and suppose toward a contradiction that $x \ne x^{R}$. Then we have

$$\begin{aligned} \min _{j}u\left( \left\| x^{R}-\hat{x}^{j} \right\| \right)> & {} \min _{j}u \left( \left\| x-\hat{x}^{j} \right\| \right) . \end{aligned}$$

Since $x^{R} \ne x$ and $x \in \text{ conv }\{\hat{x}^{i} \mid i \in N(x)\}$, it follows that if $\Vert \hat{x}^{i}-x^{R}\Vert < \Vert \hat{x}^{i}-x\Vert$ for some $i \in N(x)$, then there is some $k \in N(x)$ such that $\Vert \hat{x}^{k}-x\Vert < \Vert \hat{x}^{k}-x^{R}\Vert$. But this would imply

$$\begin{aligned} \min _{j}u \left( \left\| x-\hat{x}^{j} \right\| \right) \,\, = \,\, u \left( \left\| x-\hat{x}^{k} \right\| \right)> & {} u \left( \left\| x^{R}-\hat{x}^{k} \right\| \right) \,\, \ge \,\, \min _{j}u \left( \left\| x^{R}-\hat{x}^{j} \right\| \right) , \end{aligned}$$

a contradiction. Thus, we must have $u(\Vert x-\hat{x}^{i}\Vert ) = u(\Vert x^{R}-\hat{x}^{i}\Vert )$ for all $i \in N(x)$. For each $\epsilon \in (0,1)$, define $x^{\epsilon }=(1-\epsilon )x+\epsilon x^{R}$, and note that strict quasi-concavity implies that for all such $\epsilon$ and all $i \in N(x)$, we have $u(\Vert x^{\epsilon }-\hat{x}^{i}\Vert ) > u(\Vert x^{R}-\hat{x}^{i}\Vert )$. Furthermore, by continuity of u, we can choose $\epsilon >0$ sufficiently small that for all $i \in N \setminus N(x)$, we have $u(\Vert x^{\epsilon }-\hat{x}^{i}\Vert ) > u(\Vert x^{R}-\hat{x}^{i}\Vert )$, which leads to a contradiction, as above. We conclude that $x=x^{R}$, as required. $\square$

Proof of Proposition 4:

Suppose otherwise, toward a contradiction. By Theorem 1, there is a unique equilibrium outcome $x^{*}$, and this is characterized as the unique solution to (3). By exact radial symmetry, there is a bijective mapping $\phi :N \rightarrow N$ such that for each lobbyist i, we have $\hat{x}^{i} - x = x- \hat{x}^{\phi (i)}$, i.e., the ideal point of lobbyist i coincides with the reflection of the ideal point of lobbyist $\phi (i)$ through x. Since $\phi$ is bijective, we can rewrite (3), by a change of variables and multiplying by negative one, as

$$\begin{aligned} \sum _{i}u'\left( \left\| \hat{x}^{\phi (i)}-x^{*} \right\| \right) \left\| \hat{x}^{\phi (i)}-x^{*}\right\| \left( \hat{x}^{\phi (i)}-x^{*} \right) = 0. \end{aligned}$$

Using $\hat{x}^{\phi (i)}=2x-\hat{x}^{i}$, we write this as

$$\begin{aligned} \sum _{i}u'\left( \left\| \hat{x}^{i}-(2x-x^{*}) \right\| \right) \left\| \hat{x}^{i}-\left( 2x-x^{*} \right) \right\| \left( 2x-x^{*}-\hat{x}^{i} \right) = 0. \end{aligned}$$

We conclude that the policy $2x-x^{*} \ne x^{*}$ satisfies (3), as does $x^{*}$, contradicting the fact that the system has a unique solution. $\square$