The efficacy and efforts of interest groups in post elections policy formation

Abstract

This paper presents a new model of interest groups and policy formation in the legislature. In our setting, the already given party ideological predispositions and power distribution determine the expected policy outcome. Our analysis applies to the case of un-enforced or enforced party discipline as well as to two-party and multi-party (proportional representation) electoral systems. The interest groups’ objective is to influence the outcome in their favor by engaging in a contest that determines the final decision in the legislature. Our first result clarifies how the success of an interest group hinges on the dominance of its ideologically closer party and, in general, the coalition/opposition blocks of parties under un-enforced party or coalition/opposition discipline. Such dominance is defined in terms of ideological inclination weighted by power. Our second result clarifies how the success of an interest group hinges on the dominance of its ideology in the ruling coalition (party) in a majoritarian system with enforced coalition (party) discipline. We then clarify under what condition an interest group prefers to direct its lobbying efforts to two parties or the two coalition and opposition blocks of parties under un-enforced discipline rather than to the members of the ruling coalition (party) under enforced discipline. The lobbying efforts under un-enforced and enforced party discipline are also compared. Finally, we clarify the effect of ideological predispositions and power on the efforts of the interest groups.

Appendix

Proof of Theorem 1

An interest group’s participation $x$ in the contest is effective if:

$$\begin{aligned} p_x =\frac{k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }}{k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1}>d_1 \delta _{1x} +d_2 \delta _{2x} =p_x^0 \end{aligned}$$

That is,

$$\begin{aligned} k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }\left[ {1-\left( {d_1 \delta _{1x} +d_2 \delta _{2x} } \right)} \right]>d_1 \delta _{1x} +d_2 \delta _{2x} \end{aligned}$$

After substituting $A=\sum _{i=1}^2 {\left( {d_i \delta _{ix} } \right)^{\frac{1}{1-\alpha }}}$ and $B=\sum _{i=1}^2 {\left( {d_i \delta _{iy} } \right)^{\frac{1}{1-\alpha }}}$ in the above inequality and some simplifications we get:

$$\begin{aligned} k^{\frac{\alpha }{1-\alpha }}\left[ {\frac{\left( {d_1 \delta _{1x} } \right)^{\frac{1}{1-\alpha }}+\left( {d_2 \delta _{2x} } \right)^{\frac{1}{1-\alpha }}}{\left( {d_1 \delta _{1y} } \right)^{\frac{1}{1-\alpha }}+\left( {d_2 \delta _{2y} } \right)^{\frac{1}{1-\alpha }}}} \right]>\left( {\frac{d_1 \delta _{1x} +d_2 \delta _{2x} }{d_1 \delta _{1y} +d_2 \delta _{2y} }} \right)^{\frac{1}{1-\alpha }} \end{aligned}$$

Define $\beta =\frac{1}{1-\alpha }>1$. Then the condition becomes:¹⁶

$$\begin{aligned} k^{\frac{\alpha }{1-\alpha }}\left[ {\frac{\left( {d_1 \delta _{1x} } \right)^{\beta }+\left( {d_2 \delta _{2x} } \right)^{\beta }}{\left( {d_1 \delta _{1y} } \right)^{\beta }+\left( {d_2 \delta _{2y} } \right)^{\beta }}} \right]>\left( {\frac{d_1 \delta _{1x} +d_2 \delta _{2x} }{d_1 \delta _{1y} +d_2 \delta _{2y} }} \right)^{\beta } \end{aligned}$$

After multiplication of both sides of the inequality by $\frac{\left( {d_1 \delta _{1y} } \right)^{\beta }}{\left( {d_1 \delta _{1x} } \right)^{\beta }}$ and some manipulations we get:

$$\begin{aligned} k^{\frac{\alpha }{1-\alpha }}\left[ {\frac{1+\left( {\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }} \right)^{\beta }}{1+\left( {\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }} \right)^{\beta }}} \right]>\left( {\frac{1+\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }}{1+\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }}} \right)^{\beta } \end{aligned}$$

Define $z=\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }$ and $w=\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }$. Then the last inequality becomes:

$$\begin{aligned} k^{\frac{\alpha }{1-\alpha }}\left( {\frac{1+z^{\beta }}{1+w^{\beta }}} \right)>\left( {\frac{1+z}{1+w}} \right)^{\beta } \end{aligned}$$

$$\begin{aligned} k^{\frac{\alpha }{1-\alpha }}\left[ {\frac{1+z^{\beta }}{\left( {1+z} \right)^{\beta }}} \right]>\frac{1+w^{\beta }}{\left( {1+w} \right)^{\beta }} \end{aligned}$$

To check when the last condition is satisfied, define $f(a)=\frac{1+a^{\beta }}{\left( {1+a} \right)^{\beta }}$ and let’s find the extreme values by looking at the equation:

$$\begin{aligned} \frac{df(a)}{da}=\frac{\beta a^{\beta -1}\left( {1+a} \right)^{\beta }-\beta \left( {1+a} \right)^{\beta -1}\left( {1+a^{\beta }} \right)}{\left( {1+a} \right)^{2\beta }}=0 \end{aligned}$$

The solution of this equation is $a=1$ and the second order condition reveals that at $a=1$, $f$ has a minimum point (recall that $\beta >1$):

$$\begin{aligned} \frac{d^{2}f(a)}{da^{2}}=2^{\beta -1}\beta \left( {\beta -1} \right)>0 \end{aligned}$$

This means that in the range $0<a<1$, $f(a)$ decreases and in the range $a>1$ it increases. Without loss of generality, let $k\ge 1$, and $k^{\alpha }f(z)=k^{\alpha }\left[ {\frac{1+z^{\beta }}{\left( {1+z} \right)^{\beta }}} \right]>\frac{1+w^{\beta }}{\left( {1+w} \right)^{\beta }}=f(w)$ if $f(z)>f(w)$. Therefore, if $k\ge 1$ then $f(z)>f(w)$ is a sufficient condition, whereas if $k=1$ this condition is sufficient and necessary. Since $f(a)=f\left( {\frac{1}{a}} \right)$, $f(z)>f(w)$ if and only if $z>w>\frac{1}{z}$ (which means that $z>1$) or $\frac{1}{z}>w>z$ (which means that $z<1)$. Consider the first condition $z>w>\frac{1}{z}$. Setting $z=\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }$ and $w=\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }$ in $z>w>\frac{1}{z}$, yields: $\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }>\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }>\frac{d_1 \delta _{1x} }{d_2 \delta _{2x} }$. From $\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }>\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }$ we get $\delta _{2x} >\delta _{1x}$ (therefore $\delta _{2y} <\delta _{1y} )$ and from $\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }>\frac{d_1 \delta _{1x} }{d_2 \delta _{2x} }$ we get $d_2 \sqrt{\delta _{2x} \delta _{2y} }>d_1 \sqrt{\delta _{1x} \delta _{1y} }$. The results from the second condition, $\frac{1}{z}>w>z$, are the same just the roles of the two parties are reversed.

To sum up, the interest group $x$ is effective if there exists a party $i$, such that $\delta _{ix} >\delta _{jx}$ and $d_i \sqrt{\delta _{ix} \delta _{iy} }>d_j \sqrt{\delta _{jx} \delta _{jy} }$.$\square$

Proof of Theorem 2

Substituting $d_1 =1$ and $d_2 =0$ in Eqs. (1) and (4)–(7), we get that in this case, $A=\delta _{1x} ^{\frac{1}{1-\alpha }}$ and $B=\delta _{1y} ^{\frac{1}{1-\alpha }}$, the efforts of the interest groups in the members of party 1 are $x_1 =\frac{\alpha n_x k^{\alpha }\delta }{\left( {k^{\alpha }\delta +1} \right)^{2}}$ and $y_1 =\frac{\alpha n_y k^{\alpha }\delta }{\left( {k^{\alpha }\delta +1} \right)^{2}}$ and the winning probability of interest group $x$ is $p_x =\frac{k^{\alpha }\delta }{k^{\alpha }\delta +1}$, where $\delta =\frac{\delta _{1x} }{\delta _{1y} }$ is the ideological bias of party 1 to the target policy of interest group $x$. Hence, interest group $x$ is effective if $p_x =\frac{k^{\alpha }\delta }{k^{\alpha }\delta +1}>\delta _{1x} =p_x^0$. By the proof of Theorem 1, the inequality $p_x >p_x^0$ is equivalent to $k^{\frac{\alpha }{1-\alpha }}\left( {\frac{1+z^{\beta }}{1+w^{\beta }}} \right)>\left( {\frac{1+z}{1+w}} \right)^{\beta }$, where $z=\frac{d_2 \delta _{2x} }{d_1 \delta _{1x} }$ and $w=\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }$. Since in our case, $d_1 =1$ and $d_2 =0$, $z=w=0$. Substituting in $k^{\frac{\alpha }{1-\alpha }}\left( {\frac{1+z^{\beta }}{1+w^{\beta }}} \right)>\left( {\frac{1+z}{1+w}} \right)^{\beta }$ therefore yields that $p_x >p_x^0$ iff $k>1$.$\square$

Proof of Theorem 3

In Eq. (7), the term $\left( {\frac{A}{B}} \right)^{1-\alpha }$ changes to $\delta$ when we move from the case of a two-party system with un-enforced party discipline to the case of a majoritarian system with a single ruling party under enforced party discipline. That is, in the latter case $p_x =\frac{k^{\alpha }\delta }{k^{\alpha }\delta +1}$. Since $\frac{\partial p_x }{\partial \delta }>0$, interest group $x$ will prefer the two-party system with unenforced party discipline provided that $\left( {\frac{A}{B}} \right)^{1-\alpha }>\delta$ or

$$\begin{aligned} \left[ {\frac{\left( {d_1 \delta _{1x} } \right)^{\frac{1}{1-\alpha }}+\left( {d_2 \delta _{2x} } \right)^{\frac{1}{1-\alpha }}}{\left( {d_1 \delta _{1y} } \right)^{\frac{1}{1-\alpha }}+\left( {d_2 \delta _{2y} } \right)^{\frac{1}{1-\alpha }}}} \right]^{1-\alpha }>\frac{\delta _{1x} }{\delta _{1y} } \end{aligned}$$

And after simplification we get $\delta _{2x} >\delta _{1x}$.$\square$

Proof of Theorem 4

Larger efforts under un-enforced party discipline means that

$$\begin{aligned} \sum _{i=1}^2 {x_i } +\sum _{i=1}^2 {y_i } =\frac{\alpha k^{\alpha }(n_x +n_y )\left( {\frac{A}{B}} \right)^{1-\alpha }}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{2}}>\frac{\alpha k^{\alpha }(n_x +n_y )\delta }{\left( {k^{\alpha }\delta +1} \right)^{2}}=x_1 +y_1 \end{aligned}$$

or, after simplification

$$\begin{aligned} \left[ {k^{2\alpha }\delta \left( {\frac{A}{B}} \right)^{1-\alpha }-1} \right]\left[ {\delta -\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]>0 \end{aligned}$$

This inequality is satisfied when the two multiplied expressions are either positive or negative. Note that $\left( {\frac{A}{B}} \right)^{1-\alpha }<\delta$ if and only if $\delta _{2x} <\delta _{1x}$. We thus get that the efforts under unenforced party discipline are larger than those enforced discipline if:

• (i) $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }>1$, $k^{\alpha }\delta >1$ and $\delta _{2x} <\delta _{1x}$, or

• (ii) $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }<1$, $k^{\alpha }\delta <1$ and $\delta _{2y} <\delta _{1y}$.

$\square$

Proof of Theorem 5 part (a)

Since (see (5) and (6)):

$$\begin{aligned} \sum _{i=1}^2 {x_i } =\alpha n_x \frac{k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{2}}, \quad \sum _{i=1}^2 {y_i } =\alpha n_y \frac{k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{2}} \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^2 {x_i } +\sum _{i=1}^2 {y_i } =\alpha (n_x +n_y )\frac{k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{2}}, \end{aligned}$$

when $h=\frac{k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{2}}$ is maximal, $\sum _{i=1}^2 {x_i }$, $\sum _{i=1}^2 {y_i }$ and $\sum _{i=1}^2 {x_i } +\sum _{i=1}^2 {y_i }$ are maximal. A necessary condition for the maximization of $h$ is:

$$\begin{aligned} \frac{\partial h}{\partial \left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]}=\frac{1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{3}}=0 \end{aligned}$$

or

$$\begin{aligned} k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }=1 \end{aligned}$$

(8)

or

$$\begin{aligned} k^{\frac{\alpha }{1-\alpha }}\left[ {\frac{\left( {d_1 \delta _{1x} } \right)^{\frac{1}{1-\alpha }}+\left( {d_2 \delta _{2x} } \right)^{\frac{1}{1-\alpha }}}{\left( {d_1 \delta _{1y} } \right)^{\frac{1}{1-\alpha }}+\left( {d_2 \delta _{2y} } \right)^{\frac{1}{1-\alpha }}}} \right]=1 \end{aligned}$$

and the SOC when $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }=1$ is satisfied:

$$\begin{aligned} \frac{\partial ^{2}h}{\partial \left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]^{2}}=-\frac{1}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{3}}<0 \end{aligned}$$

Notice that the expressions $\sum \nolimits _{i=1}^2 {x_i }$, $\sum \nolimits _{i=1}^2 {y_i }$ and $\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i }$ are differentiated with respect to $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }$. This means that when (8) is satisfied, a change in $\frac{A}{B}$ reduces the contestants’ efforts (separately and jointly). However, if (8) is satisfied, a change in $k$ reduces efforts, but it also involves a change in $n_x$ and $n_y$ that also affect efforts. Hence, a change in $k$ reduces $\sum \nolimits _{i=1}^2 {x_i }$, provided that $n_x$ is unaltered ($n_y$ is changed). Similarly, $\sum \nolimits _{i=1}^2 {y_i }$ is reduced, provided that $n_y$ is unaltered ($n_x$ is changed) and $\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i }$ will reduce when $(n_x +n_y )$ is unchanged.

Substituting (8) in (5), (6) and (7) gives $\sum \nolimits _{i=1}^N {x_i } =0.25\alpha n_x$, $\sum \nolimits _{i=1}^N {y_i } =0.25\alpha n_y$, $\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } =0.25\alpha (n_x +n_y )$ and $p_x =p_y =0.5$. Substituting $\delta _{1x} =\delta _{2x}$ and $\delta _{1y} =\delta _{2y}$ in Eq. (8) and combining $\delta _{1x} +\delta _{1y} =\delta _{2x} +\delta _{2y} =1$ implies that $\delta _{1x} =\delta _{2x} =\frac{1}{1+k^{\alpha }}$ and $\delta _{1y} =\delta _{2y} =\frac{k^{\alpha }}{1+k^{\alpha }}$. Therefore, in this case the efforts directed by the interest groups to the parties are maximal, $\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } =0.25\alpha (n_x +n_y )$ and $p_x =p_y =0.5$.$\square$

Proof of Theorem 5 part (b)

By the proof of Theorem 5 part (a), Substituting $\delta _{1x} =1$ and $\delta _{2x} =0$ in Eq. (8) and combining $\delta _{1x} +\delta _{1y} =\delta _{2x} +\delta _{2y} =1$ gives $d_1 =\frac{1}{1+k^{\alpha }}$. Therefore, in this case the efforts of the interest groups are maximal, $\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } =0.25\alpha (n_x +n_y )$ and $p_x =p_y =0.5$.

Proof of the Remark

When both interest groups are not effective, from the proof of Theorem 1 get that the following equality is satisfied:

$$\begin{aligned} f(z)=\frac{1+z^{\beta }}{\left( {1+z} \right)^{\beta }}=\frac{1+w^{\beta }}{\left( {1+w} \right)^{\beta }}=f(w) \end{aligned}$$

since $f(a)=f\left( {\frac{1}{a}} \right)$, $f(z)=f(w)$ if and only if $w=z$ or $w=\frac{1}{z}$:

• 1. When $w=z$ we get that $\delta _{1x} =\delta _{2x}$ (or $\delta _{2y} =\delta _{1y} )$. Since the efforts are maximal, $p_x =p_y =0.5$. Since no interest group is effective, the probability of interest group $x$’s with no contest is $d_1 \delta _{1x} +d_2 \delta _{2x} =0.5$. Substituting in the last equality $\delta _{1x} =\delta _{2x}$ we get that $\delta _{1x} =\delta _{2x} =0.5$ and therefore $\delta _{1x} =\delta _{2x} =\delta _{1y} =\delta _{2y} =0.5$. By Theorem 5 part (a), for these values the efforts directed by the interest groups to the parties are maximal.

• 2. When $w=\frac{1}{z}$ we get $\frac{d_2 \delta _{2y} }{d_1 \delta _{1y} }=\frac{d_1 \delta _{1x} }{d_2 \delta _{2x} }$ or $\frac{d_2 }{d_1 }=\left( {\frac{\delta _{1x} \delta _{1y} }{\delta _{2x} \delta _{2y} }} \right)^{0.5}$. Since the efforts are maxima, $p_x =p_y =0.5$. Since no interest group is effective, the probabilities of both interest groups with no contest are identical:

$$\begin{aligned} d_1 \delta _{1x} +d_2 \delta _{2x} =d_1 \delta _{1y} +d_2 \delta _{2y} (=0.5) \end{aligned}$$

$$\begin{aligned} \frac{d_2 }{d_1 }=\frac{\delta _{1x} -\delta _{1y} }{\delta _{2y} -\delta _{2x} } \end{aligned}$$

Substituting in the last equality the additional condition for no effectiveness of the interest groups, $\frac{d_2 }{d_1 }=\left( {\frac{\delta _{1x} \delta _{1y} }{\delta _{2x} \delta _{2y} }} \right)^{0.5}$, we get the following equivalent conditions: $\left( {\frac{\delta _{1x} \delta _{1y} }{\delta _{2x} \delta _{2y} }} \right)^{0.5}=\frac{\delta _{1x} -\delta _{1y} }{\delta _{2y} -\delta _{2x} }$ or after some simplifications we get:

$$\begin{aligned} \left( {\delta _{1x} -\delta _{2x} } \right)\left( {\delta _{1y} -\delta _{2x} } \right)=0 \end{aligned}$$

This equality is satisfied if $\delta _{1x} =\delta _{2x}$ ($\delta _{1y} =\delta _{2y} )$ or $\delta _{1y} =\delta _{2x}$ ($\delta _{1x} =\delta _{2y} )$. Let’s check both cases:

• 2.1 $\delta _{1x} =\delta _{2x}$—since no interest group is effective, $d_1 \delta _{1x} +d_2 \delta _{2x} =0.5$. Substituting in the last equality $\delta _{1x} =\delta _{2x}$ we get $\delta _{1x} =\delta _{2x} =0.5$ and therefore $\delta _{1x} =\delta _{2x} =\delta _{1y} =\delta _{2y} =0.5$. By Theorem 5 part (a), for these values the efforts of the interest groups are maximal.

• 2.2 $\delta _{1y} =\delta _{2x}$ ($\delta _{1x} =\delta _{2y} )$—Substituting in the equality $\frac{d_2 }{d_1 }=\left( {\frac{\delta _{1x} \delta _{1y} }{\delta _{2x} \delta _{2y} }} \right)^{0.5}$ we get $d_1 =d_2 =0.5$. For these values the efforts directed by the interest groups to the parties are maximal.

Proof of Theorem 6

$$\begin{aligned}&\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial \delta _{jx} }\\&\quad =\frac{\alpha k^{\alpha }(n_x +n_y )d_j ^{\frac{1}{1-\alpha }}\left[ {\delta _{jx} ^{\frac{\alpha }{1-\alpha }}+\left( {1-\delta _{jx} } \right)^{\frac{\alpha }{1-\alpha }}\frac{A}{B}} \right]\left[ {1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{3}\left( {\frac{A}{B}} \right)^{\alpha }B} \end{aligned}$$

Therefore,$\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } } \right)}{\partial \delta _{jx} }>0$, $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial \delta _{jx} }>0$ and $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial \delta _{jx} }>0$, if and only if $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }<1$.

Proof of Theorem 7

$$\begin{aligned}&\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial d_1 }\\&\quad =\frac{\alpha k^{\alpha }(n_x +n_y )\left( {d_1 d_2 } \right)^{\frac{\alpha }{1-\alpha }}\left[ {1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]\left[ {\left( {\delta _{1x} \delta _{2y} } \right)^{\frac{1}{1-\alpha }}-\left( {\delta _{1y} \delta _{2x} } \right)^{\frac{1}{1-\alpha }}} \right]}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{3}\left( {\frac{A}{B}} \right)^{\alpha }B^{2}} \end{aligned}$$

$\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } } \right)}{\partial d_1 }>0, \quad \frac{\partial \left( {\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial d_1 }>0$ and $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial d_1 }>0$, if and only if:

$$\begin{aligned} \left[ {1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]\left[ {\left( {\delta _{1x} \delta _{2y} } \right)^{\frac{1}{1-\alpha }}-\left( {\delta _{1y} \delta _{2x} } \right)^{\frac{1}{1-\alpha }}} \right]>0 \end{aligned}$$

Moreover:

$$\begin{aligned} sign\left[ {\left( {\delta _{1x} \delta _{2y} } \right)^{\frac{1}{1-\alpha }}-\left( {\delta _{1y} \delta _{2x} } \right)^{\frac{1}{1-\alpha }}} \right]=sign\left( {\delta _{2y} -\delta _{1y} } \right) \end{aligned}$$

therefore $\left[ {1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]\left[ {\left( {\delta _{1x} \delta _{2y} } \right)^{\frac{1}{1-\alpha }}-\left( {\delta _{1y} \delta _{2x} } \right)^{\frac{1}{1-\alpha }}} \right]>0$ if and only if:

$$\begin{aligned} \left[ {1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]\left( {\delta _{2y} -\delta _{1y} } \right)>0. \end{aligned}$$

Proof of Theorem 8

 

• (a) Since,

  $$\begin{aligned}&\frac{\partial x_j }{\partial n_x }=\frac{\alpha k^{\alpha }\left( {d_j \delta _{jx} } \right)^{\frac{1}{1-\alpha }}\left( {\frac{A}{B}} \right)^{1-\alpha }\left[ {(1-\alpha )k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1+\alpha } \right]}{\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+1} \right]^{3}A}>0, \\&\quad \frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } } \right)}{\partial n_x }>0. \end{aligned}$$

  Also:

  $$\begin{aligned} \frac{\partial x_j }{\partial n_y }=\frac{\alpha ^{2}\left( {d_j \delta _{jx} } \right)^{\frac{1}{1-\alpha }}n_x ^{1+\alpha }n_y ^{2\alpha -1}\left[ {k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }-1} \right]}{\left[ {n_x ^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+n_y ^{\alpha }} \right]^{3}\left( {\frac{A}{B}} \right)^{\alpha }B} \end{aligned}$$

  and therefore $\frac{\partial x_j }{\partial n_y }>0$ and $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } } \right)}{\partial n_y }>0$, if and only if $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }>1$. Furthermore,

  $$\begin{aligned} \frac{\partial y_j }{\partial n_x }=\frac{\alpha ^{2}\left( {d_j \delta _{jy} } \right)^{\frac{1}{1-\alpha }}\left( {\frac{A}{B}} \right)^{1-\alpha }n_x ^{\alpha -1}n_y ^{1+2\alpha }\left[ {1-k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }} \right]}{\left[ {n_x ^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }+n_y ^{\alpha }} \right]^{3}B} \end{aligned}$$

  Hence, $\frac{\partial y_j }{\partial n_x }>0$ and $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial n_x }>0$, if and only if $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }<1$. Moreover, since it is always true that $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } } \right)}{\partial n_x }>0$, the last condition, $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }<1$, is a sufficient one for $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial n_x }>0$. By symmetry $\frac{\partial \left( {\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i } } \right)}{\partial n_y }>0$, if $k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }>1$.

• (b) According the proof of Theorem 2 part (a), if (8) is satisfied ($k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }=1$), a change in $k$ reduces $\sum \nolimits _{i=1}^2 {x_i } +\sum \nolimits _{i=1}^2 {y_i }$ when $(n_x +n_y )$ is unchanged.

• (c) According the proof of Theorem 5 part (a), if (8) is satisfied ($k^{\alpha }\left( {\frac{A}{B}} \right)^{1-\alpha }=1$), the maximal contestants’ total effort is $\sum \nolimits _{i=1}^2 {x_i } +\sum _{i=1}^2 {y_i } =0.25\alpha (n_x +n_y )$. Therefore, if the interest groups have equal incentives, then increasing (decreasing) both $n_x$ and $n_y$ such that $k$ is unchanged, increases (reduces) the contestants’ efforts.