Politicians, governed versus non-governed interest groups and rent dissipation

Abstract

Government intervention often gives rise to contests and the government can influence their outcome by choosing their type. We consider a contest with two interest groups: one that is governed by a central planner and one that is not. Rent dissipation is compared under two well-known contest success functions: the generalized logit and the all-pay auction. We also consider the case in which the government can limit the size of the non-governed interest group in order to determine the scope of rent dissipation, with the goal of either increasing the rent obtained by the government or reducing the wasted resources invested in the contest.

Appendix

Proof of proposition 1

a. For $k\ge 1$ .

$$\begin{aligned} RD_\mathrm{L}^*=\frac{\alpha N^{1-\alpha }k^{\alpha }\left( {n+m} \right) }{\left( {N^{1-\alpha }k^{\alpha }+1} \right) ^{2}}<\frac{n+m}{2k}=RD_\mathrm{A}^*\end{aligned}$$

(1a)

Rewriting (1a), we obtain that (1a) holds if the following inequality holds:

$$\begin{aligned} N^{2-2\alpha }k^{2\alpha }+2N^{1-\alpha }k^{\alpha }\left( {1-\alpha k} \right) +1>0 \end{aligned}$$

(2a)

Denote $z=N^{1-\alpha }$. (1a) becomes

$$\begin{aligned} z^{2}k^{2\alpha }-2zk^{\alpha }\left( {\alpha k-1} \right) +1>0. \end{aligned}$$

(3a)

We now wish to find the $N$ that satisfies (3a).

If $(1\le )k\le \frac{2}{\alpha }$, the determinant of (3a) is not positive, and thus $z^{2}k^{2\alpha }-2zk^{\alpha }\left( {\alpha k-1} \right) +1>0$ always holds. However, if $k>\frac{2}{\alpha }$ then (3a) holds for the following values:

$$\begin{aligned} \begin{array}{l} N<\left[ {\frac{\alpha k-1-\sqrt{\alpha k\left( {\alpha k-2} \right) }}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}\\ \hbox {or }\\ N>\left[ {\frac{\alpha k-1+\sqrt{\alpha k\left( {\alpha k-2} \right) }}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}\\ \end{array} \end{aligned}$$

(4a)

We will show that for $k>\frac{2}{\alpha }$ the first inequality is not possible; therefore, the second inequality remains. $\alpha k-1-\sqrt{\alpha k\left( {\alpha k-2} \right) }<1$ holds for all values of $k>\frac{2}{\alpha }$. Since $k>\frac{2}{\alpha }>1$, then $k^{\alpha }>1$. Thus, $\left[ {\frac{\alpha k-1-\sqrt{\alpha k\left( {\alpha k-2} \right) }}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}<1<N$. We conclude that if $k>\frac{2}{\alpha }$, then for $N$ that satisfies $N>\left[ {\frac{\alpha k-1+\sqrt{\alpha k\left( {\alpha k-2} \right) }}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}$ it holds that $RD_\mathrm{L}^*<RD_\mathrm{A}^*$. If $N=\left[ {\frac{\alpha k-1+\sqrt{\alpha k\left( {\alpha k-2} \right) }}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}$, then $RD_\mathrm{L}^*=RD_\mathrm{A}^*$; otherwise $RD_\mathrm{L}^*>RD_\mathrm{A}^*$.

We will now show that the condition $N>\left[ {\frac{\alpha k-1+\sqrt{\alpha k\left( {\alpha k-2} \right) }}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}$ has substance. Denote $f(k)=\left[ {\frac{\alpha k-1+\sqrt{\alpha k(\alpha k-2)}}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}$. We will show that for all $0<\alpha <1$ and $k>\frac{2}{\alpha }$, $f(k)$ increases with $k$ and that when $k\rightarrow \frac{2}{\alpha }$ it holds that $f<1$ (namely $N>f$ for all $N)$ and when $k\rightarrow \infty$, then $f\rightarrow \infty$ (thus, there exist values for which $N>f)$. First, $\frac{\partial f}{\partial k}=\frac{1}{\left( {1-\alpha } \right) k^{2\alpha }}\left\langle {\begin{array}{l} \left\{ {\alpha +0.5\left[ {\alpha k(\alpha k-2)} \right] ^{-0.5}\left( {2\alpha ^{2}k-2\alpha } \right) } \right\} k^{\alpha } \\ -\alpha k^{\alpha -1}\left\{ {\alpha k-1+\left[ {\alpha k(\alpha k-2)} \right] ^{0.5}} \right\} \\ \end{array}} \right\rangle ^{\frac{\alpha }{1-\alpha }}$ which is identical to $\frac{\partial f}{\partial k}=\frac{\alpha }{\left( {1-\alpha } \right) k^{\alpha }}\left[ {1-\alpha +\frac{1}{k}+\frac{\left( {\alpha k-1} \right) \left( {1-\alpha } \right) +\alpha }{\left[ {\alpha k(\alpha k-2)} \right] ^{0.5}}} \right] ^{\frac{\alpha }{1-\alpha }}>0$. Since when $k=\frac{2}{\alpha }(>2)$ we obtain $f=\left[ {\frac{1}{k^{\alpha }}} \right] ^{\frac{1}{1-\alpha }}=k^{\frac{\alpha }{\alpha -1}}=k^{\frac{2}{2-k}}<1$ and $f=\left[ {\frac{\alpha -\frac{1}{k}+\sqrt{\alpha \left( {\alpha -\frac{2}{k}} \right) }}{k^{\alpha -1}}} \right] ^{\frac{1}{1-\alpha }}$therefore when $k\rightarrow \infty$ we obtain $f\rightarrow \left( {2\alpha } \right) ^{\frac{1}{1-\alpha }}k\rightarrow \infty$.

b. For $k<1$. (1a) becomes:

$$\begin{aligned} RD_\mathrm{L}^*=\frac{\alpha N^{1-\alpha }k^{\alpha }\left( {n+m} \right) }{\left( {N^{1-\alpha }k^{\alpha }+1} \right) ^{2}}<\frac{k\left( {n+m} \right) }{2}=RD_\mathrm{A}^*. \end{aligned}$$

This inequality is identical to $N^{2-2\alpha }k^{1+\alpha }+2N^{1-\alpha }\left( {k-\alpha } \right) +k^{1-\alpha }>0$ and its determinant is $4\alpha k^{2\alpha }\left( {\alpha -2k} \right)$. If the determinant is negative, which means that $\frac{\alpha }{2}<k(<1)$, then $RD_\mathrm{L}^*<RD_\mathrm{A}^*$. If the determinant is positive, which means that $k<\frac{\alpha }{2}$, then $RD_\mathrm{L}^*<RD_\mathrm{A}^*$ if:

$$\begin{aligned}&N^{1-\alpha }<\frac{2k^{\alpha }\left( {\alpha -k} \right) -\sqrt{4\alpha k^{2\alpha }\left( {\alpha -2k} \right) }}{2k^{2\alpha +1}} \hbox { or }\\&N^{1-\alpha }>\frac{2k^{\alpha }\left( {\alpha -k} \right) +\sqrt{4\alpha k^{2\alpha }\left( {\alpha -2k} \right) }}{2k^{2\alpha +1}} \end{aligned}$$

or

$$\begin{aligned}&N^{1-\alpha }<\frac{2k^{\alpha }\left( {\alpha -k} \right) -2k^{\alpha }\sqrt{\alpha \left( {\alpha -2k} \right) }}{2k^{2\alpha +1}} \hbox { or }\\&N^{1-\alpha }>\frac{2k^{\alpha }\left( {\alpha -k} \right) +2k^{\alpha }\sqrt{\alpha \left( {\alpha -2k} \right) }}{2k^{2\alpha +1}} \end{aligned}$$

or

$$\begin{aligned} N<\left[ {\frac{\alpha -k-\sqrt{\alpha \left( {\alpha -2k} \right) }}{k^{1+\alpha }}} \right] ^{\frac{1}{1-\alpha }}\hbox { or } N>\left[ {\frac{\alpha -k+\sqrt{\alpha \left( {\alpha -2k} \right) }}{k^{1+\alpha }}} \right] ^{\frac{1}{1-\alpha }} \end{aligned}$$

(5a)

From $a$ and $b$ we obtain Proposition 1. $\square$

Proof of Proposition 2

The first-order condition for maximization is:

$$\begin{aligned} \frac{\partial RD_\mathrm{L}^*}{\partial N}=\alpha k^{\alpha }\left( {n+m} \right) \left( {1-\alpha } \right) N^{-\alpha }\frac{\left( {1-k^{\alpha }N^{1-\alpha }} \right) }{\left( {N^{1-\alpha }k^{\alpha }+1} \right) ^{3}}=0. \end{aligned}$$

Solving the first-order condition yields $N=k^{\frac{\alpha }{\alpha -1}}$ and it can be verified that the second-order conditions for maximization hold.

Part 1.a: For $k<1$, it holds that $N=k^{\frac{\alpha }{\alpha -1}}>1$, while for $N<k^{\frac{\alpha }{\alpha -1}}$, $RD_\mathrm{L}^*$ increases with $N$ and for $N>k^{\frac{\alpha }{\alpha -1}}$, $RD_\mathrm{L}^*$ decreases with $N$. Therefore, when $N_\mathrm{P} \ge k^{\frac{\alpha }{\alpha -1}}$ the maximal level of $RD_\mathrm{L}^*$ is attained by limiting the size of the non-governed interest group to $N^{*}=k^{\frac{\alpha }{\alpha -1}}$ and for $N_\mathrm{P} <k^{\frac{\alpha }{\alpha -1}}$ the maximal level of $RD_\mathrm{L}^*$ is attained at $N^{*}=N_\mathrm{P}$.

Part 1.b: For $k\ge 1$, it holds that $N=k^{\frac{\alpha }{\alpha -1}}\le 1$. Thus, for all $N>1$, $RD_\mathrm{L}^*$ decreases with $N$ and the maximal level of $RD_\mathrm{L}^*$ is attained at $N^{*}=1$.

Part 2: For $k<1$, given the proof of 1.a, it holds that $N=k^{\frac{\alpha }{\alpha -1}}>1$. For $N<k^{\frac{\alpha }{\alpha -1}}$, $RD_\mathrm{L}^*$ increases in $N$ and for $N>k^{\frac{\alpha }{\alpha -1}}$, $RD_\mathrm{L}^*$ decreases in $N$. Thus, the minimal $RD_\mathrm{L}^*$ is attained at a corner solution of either $N^{*}=1$ or $N^{*}=N_\mathrm{P}$. We now show the conditions for obtaining each of these two solutions. The legislators/regulators will be indifferent between $N=N_\mathrm{P}$ and $N=1$ when:

$$\begin{aligned} RD_\mathrm{L}^*(N_\mathrm{P} )=\frac{\alpha N_\mathrm{P}^{1-\alpha } k^{\alpha }\left( {n+m} \right) }{\left( {N_\mathrm{P}^{1-\alpha } k^{\alpha }+1} \right) ^{2}}=\frac{\alpha k^{\alpha }\left( {n+m} \right) }{\left( {k^{\alpha }+1} \right) ^{2}}=RD_\mathrm{L}^*(N=1). \end{aligned}$$

namely, if $\frac{N_\mathrm{P}^{1-\alpha } }{\left( {N_\mathrm{P}^{1-\alpha } k^{\alpha }+1} \right) ^{2}}=\frac{1}{\left( {k^{\alpha }+1} \right) ^{2}}$ which is identical to $N_\mathrm{P}^{2-2\alpha } k^{2\alpha }-N_\mathrm{P}^{1-\alpha } \left( {k^{2\alpha }+1} \right) +1=0$. Denote $w=N_\mathrm{P}^{1-\alpha }$ and the equation becomes $w^{2}k^{2\alpha }-w\left( {k^{2\alpha }+1} \right) +1=0$. Solving this equality, we obtain that $N_\mathrm{P} =\,k^{\frac{2\alpha }{\alpha -1}}$ or $N_\mathrm{P} =\,1$. At the value of $N=k^{\frac{2\alpha }{\alpha -1}}$, $RD_\mathrm{L}^*$ decreases with $N.$Therefore:

• a. If $N_\mathrm{P} >k^{\frac{2\alpha }{\alpha -1}}$, we obtain $RD_\mathrm{L}^*(N_\mathrm{P} )<RD_\mathrm{L}^*(N=1)$ and the value that minimizes $RD_\mathrm{L}^*$ is $N^{*}=N_\mathrm{P}$ such that $RD_\mathrm{L}^*=\frac{\alpha N_\mathrm{P}^{1-\alpha } k^{\alpha }\left( {n+m} \right) }{\left( {N_\mathrm{P}^{1-\alpha } k^{\alpha }+1} \right) ^{2}}$.

• b. If $(1<)N_\mathrm{P} <k^{\frac{2\alpha }{\alpha -1}}$, we obtain $RD_\mathrm{L}^*(N_\mathrm{P} )>RD_\mathrm{L}^*(N=1)$ and the value that minimizes $RD_\mathrm{L}^*$ is $N^{*}=1$ such that $RD_\mathrm{L}^*=\frac{\alpha k^{\alpha }\left( {n+m} \right) }{\left( {k^{\alpha }+1} \right) ^{2}}$.

For $k\ge 1$, from the proof of 1.b, the maximal level of $RD_\mathrm{L}^*$ is obtained when $N=1$. This value, $RD_\mathrm{L}^*$, decreases as $N$ increases. Thus, the minimal $RD_\mathrm{L}^*$ is attained at $N^{*}=N_\mathrm{P}$. $\square$

Proof of footnote 7

Assume that $k^{\frac{\alpha }{\alpha -1}}$ is not a natural number. Denote $B$ as the highest natural number that is lower than $k^{\frac{\alpha }{\alpha -1}}$, thus it holds that $B<k^{\frac{\alpha }{\alpha -1}}<B+1$. It follows that we have to round upward $N^{*}=k^{\frac{\alpha }{\alpha -1}}$ to $B+1$, if it holds that

$$\begin{aligned} \frac{\alpha \left( {B+1} \right) ^{1-\alpha }k^{\alpha }\left( {n+m} \right) }{\left( {\left( {B+1} \right) ^{1-\alpha }k^{\alpha }+1} \right) ^{2}}>\frac{\alpha B^{1-\alpha }k^{\alpha }\left( {n+m} \right) }{\left( {B^{1-\alpha }k^{\alpha }+1} \right) ^{2}} \end{aligned}$$

or

$$\begin{aligned} \frac{\left( {B+1} \right) ^{1-\alpha }}{\left( {\left( {B+1} \right) ^{1-\alpha }k^{\alpha }+1} \right) ^{2}}>\frac{B^{1-\alpha }}{\left( {B^{1-\alpha }k^{\alpha }+1} \right) ^{2}}. \end{aligned}$$

Taking the square root of both sides we obtain

$$\begin{aligned} \frac{\left( {B+1} \right) ^{\frac{1-\alpha }{2}}}{\left( {B+1} \right) ^{1-\alpha }k^{\alpha }+1}>\frac{B^{\frac{1-\alpha }{2}}}{B^{1-\alpha }k^{\alpha }+1}. \end{aligned}$$

Rewiring this inequality we get

$$\begin{aligned}&\left( {B+1} \right) ^{0.5(1-\alpha )}\left( {B^{1-\alpha }k^{\alpha }+1} \right) >B^{0.5(1-\alpha )}\left[ {\left( {B+1} \right) ^{1-\alpha }k^{\alpha }+1} \right] \\&\left( {B+1} \right) ^{0.5(1-\alpha )}B^{1-\alpha }k^{\alpha }+\left( {B+1} \right) ^{0.5(1-\alpha )}>B^{0.5(1-\alpha )}\left( {B+1} \right) ^{1-\alpha }k^{\alpha }+B^{0.5(1-\alpha )}\\&\left( {B+1} \right) ^{0.5(1-\alpha )}-B^{0.5(1-\alpha )}>B^{0.5(1-\alpha )}\left( {B+1} \right) ^{1-\alpha }k^{\alpha }-\left( {B+1} \right) ^{0.5(1-\alpha )}B^{1-\alpha }k^{\alpha }\\&\left( {B+1} \right) ^{0.5(1-\alpha )}-B^{0.5(1-\alpha )}>B^{0.5(1-\alpha )}\left( {B+1} \right) ^{0.5(1-\alpha )}k^{\alpha }\left[ {\left( {B+1} \right) ^{0.5(1-\alpha )}-B^{0.5(1-\alpha )}} \right] \\&\left( {B+1} \right) ^{0.5(1-\alpha )}-B^{0.5(1-\alpha )} >\left[ {B\left( {B+1} \right) } \right] ^{0.5(1-\alpha )}k^{\alpha }\left[ {\left( {B+1} \right) ^{0.5(1-\alpha )}-B^{0.5(1-\alpha )}} \right] \end{aligned}$$

Since $\left( {B+1} \right) ^{0.5(1-\alpha )}-B^{0.5(1-\alpha )}>0$, we obtain that the last inequality is identical to

$$\begin{aligned} \left[ {B\left( {B+1} \right) } \right] ^{0.5(1-\alpha )}k^{\alpha }<1 \end{aligned}$$

or

$$\begin{aligned} B\left( {B+1} \right) <k^{\frac{2\alpha }{\alpha -1}} \end{aligned}$$

$\square$