Published by De Gruyter May 26, 2020

R. Emre Aytimur

Abstract

We study the rent-seeking behavior of political parties in a proportional representation system. In our model, the final policy choice of the parliament is a weighted average of parties’ policy positions, weights being their vote shares. An extreme party chooses a higher rent level than a moderate party in exchange for greater policy influence, except in some cases of unlikely distributions of parties. Moreover, political rents are not eliminated even with free entry, unless the entry cost is arbitrarily small.

JEL Classification: D72; D73; D78

Corresponding author: R. Emre Aytimur, Division of Economics, University of Leicester School of Business, Leicester, UK, E-mail:

Proof of Lemma 1:

Suppose we have a voting equilibrium where the profile of vote shares of parties is (α1,,αi,,αn), the final policy is p and political parties’ total rent is r. Consider that a voter x of party i deviates and votes for party ji. Then, the vote shares of parties i and j would change to

αi=αiε

and

αj=αj+ε.

In this case, the new final policy p and the new total rent r would be respectively

p=pεpi+εpj

and

r=rεri+εrj.

If the initial situation is an equilibrium, it should be that this deviation is not profitable for voter x. Hence,

(px)2r(px)2r.

Equivalently,

2x(pjpi)2p(pjpi)+(rjri)+ε(pjpi)2.

Since this should hold for any ε>0, it reduces to

2x(pjpi)2p(pjpi)+(rjri).

Given the definition of ei,j, this is equivalent to

xei,j

if j>i, and to

xei,j

if j<i.

Since voter x should not deviate for any ji, it should be that, for i1,n,

maxj<i{ei,j,1}xminj>i{ei,j,1}

for i=1,

1xminj>i{ei,j,1}

for i=n,

maxj<i{ei,j,1}x1.Q.E.D.

Proof of Proposition 1:

Assume party i has votes in equilibrium, then it can choose ri>0, since its vote share is continuous in ri.

Call V{1,,n} the subset of parties having positive vote shares in equilibrium. Call parties in V as 1,2,,k,,K such that pk<pk+1.

If kV, then ek,k+1>ek,k1, since otherwise e¯k<e¯k. Then, αk=ek,k+1ek,k12 for every kV, k1,K, α1=e1,2(1)2 and αK=1eK,K12.

Assume party i does not have votes (i.e. iV) and pk<pi<pk+1 for some k,k+1V. Party i could have votes if and only if ei,k+1ei,k>0. Equivalently,

rk+1ri2(pk+1pi)rirk2(pipk)>0.

Equivalently,

ri<(pipk)rk+1+(pk+1pi)rkpk+1pk.

Since rk>0 and rk+1>0, such ri>0 exists. Hence, party i is able to have votes and to choose a positive rent level.

Assume now that iV and pi<p1. Party i could have votes if and only if ei,1(1)>0. Equivalently,

r1ri2(p1pi)(1)>0.

Equivalently,

ri<r1+2(p1pi).

Since r1>0 and p1>pi, such ri>0 exists. Hence, party i is able to have votes and to choose a positive rent level.

Assume now that iV and pi>pK. Party i could have votes if and only if 1ei,K>0. Equivalently,

1rirK2(pipK)>0.

Equivalently,

ri<rK+2(pipK).

Since rK>0 and pi>pK, such ri>0 exists. Hence, party i is able to have votes and to choose a positive rent level.

Hence, to sum up, in equilibrium, every party get votes and choose positive rent levels.

The vote share for party i is αi=ei,i+1ei,i12 for any i1,n. The vote shares of parties 1 and n are respectively α1=e1,2(1)2 and αn=1en,n12. Q.E.D.

Proof of Lemma 2:

The final policy is given by p=i=1nαipi. The vote share of party i, i=2,,n1, is

(6)αi=12(ei,i+1ei,i1)=12(ri+1ri2(pi+1pi)riri12(pipi1)).

The vote shares of parties 1 and n are respectively

α1=12(e1,2(1))=12(p+r2r12(p2p1)+1),

and

αn=12(1en,n1)=12(1(p+rnrn12(pnpn1))).

After replacing these vote shares,

p=12(e1+1)p1+k=2n112(ri+1ri2(pi+1pi)riri12(pipi1))pi+12(1en1)pn.

The result follows after simplifications. Q.E.D.

Proof of Proposition 2:

To prove the existence of an equilibrium, it is sufficient to show that the objective function of party i, i.e. αiri is quasiconcave in ri for every i. Then, standard arguments guarantee the existence of an equilibrium see e.g. Friedman 1977).

For i=2,3,,n1,

αi=12(ri+1ri2(pi+1pi)riri12(pipi1))

given that adjacent parties have also some votes. In this case, αi is linear in ri, and

αiri=12(12(pi+1pi)12(pipi1)).

If ri is so low that an adjacent party gets no vote at all, say party i+1, then the vote share of party i becomes

αi=12(ri+2ri2(pi+2pi)riri12(pipi1)).

Also in this case, αi is linear in ri, and

αiri=12(12(pi+2pi)12(pipi1)).

Hence, the vote share function has a kink at ri just low enough so that an adjacent party gets no vote at all. However, note that the vote share function has now lower slope in absolute value since pi+2>pi+1. The same logic applies when ri is so low that even party i+2 gets no vote at all, and so on.

If ri is low enough so that even party n does not have any vote, then the final policy becomes p=12+pip1(p1+pi+r1ri2) and the vote share of party i becomes

αi=12(1(12+pip1(p1+pi+r1ri2)+riri12(pipi1))).

Also in this case, αi is linear in ri, and

αiri=12(12(2+pip1)12(pipi1))

Hence, the vote share function has a kink at ri just low enough so that even party n gets no vote at all. However, note that the vote share function has now lower slope in absolute value since 12(2+pip1)>0.

To sum up, the demand function between the kinks is linear and the slope of a segment between two kinks is higher than the slope of the next segment. The same argument applies when some parties to the left of party i do not have any vote or when some parties both to the left and to the right of party i do not have any vote. Hence, αi is concave in ri for i=2,3,,n1.

For party 1,

α1=12(e1(1))=12(12+pnp1(p1+pn+r1rn2)+r2r12(p2p1)+1)

given that party 2 gets some votes. In this case, α1 is linear in r1, and

α1r1=12(12(2+pnp1)12(p2p1)).

If r1 is so low that party 2 gets no vote at all, then the vote share of party 1 becomes

α1=12(12+pnp1(p1+pn+r1rn2)+r3r12(p3p1)+1).

Also in this case, α1 is linear in r1, and

α1r1=12(12(2+pnp1)12(p3p1)).

Hence, the vote share function has a kink at r1 just low enough that party 2 gets no vote at all. However, note that the vote share function has now lower slope in absolute value since p3>p2. The same line of reasoning applies when r1 is so low that even party 3 gets no vote at all, and so on.

To sum up, the demand function between the kinks is linear and the slope of a segment between two kinks is higher than the slope of the next segment. Hence, α1 is concave in r1.

Similarly, it can be shown that αn is concave in rn. Hence, when positive, αi is concave in ri for every i. Therefore, for every i, αiri is concave in ri when positive, and equal to 0 when ri is too high. Hence, αiri is quasiconcave in ri for every party, which implies the existence of an equilibrium by standard arguments.

Next, we show that the equilibrium is unique. Using equation (2), we can express r3 as a linear function of r1 and r2, calling this function as R3, r3=R3(r1,r2). Similarly, r4=R4(r2,r3). Replacing r3, this becomes r4=R4(r2,R3(r1,r2)) which can be written as r4=h4(r1,r2). Given that R3 and R4 are linear, h4 is linear too. Continuing this procedure, we have rn1=hn1(r1,r2) and rn=hn(r1,r2), both hn1 and hn being linear. Replacing these values on Eqs. (3) and (4), we obtain two linear equations of two unknowns r1 and r2. We have already proved that an equilibrium exists. Hence, if there is not a unique equilibrium, then the two linear equations on r1 and r2 should coincide and every point of this straight line should be an equilibrium. Specifically, there should be an equilibrium where r1=0 or r2=0, as any given line must intersect one of the axes.[11] However, this contradicts with the result of the last section proving that in any equilibrium, all parties choose positive rents. Then, there exists a unique pair of equilibrium rents r1* and r2*. Since, ri, i=3,,n, is uniquely defined in terms of r1 and r2 by the procedure explained above, there is a unique set of equilibrium rents (r1*,r2*,,rn*). Q.E.D.

Proof of Proposition 3:

(i) Assume n is odd and the parties’ ideal policy points are symmetric around pn+12=0. Let’s define δ0 as δ0=pn+12pn12=pn+32pn+12, δ1 as δ1=pn+52pn+32=pn12pn32, and so on. Then, from equation (2),

rn+12=12(δ0rn+32+δ0rn122δ0).

Given the symmetric structure of the game and the uniqueness of equilibrium, the equilibrium will be such that r1=rn, r2=rn1, etc. (Formally,ri=rni+1 for every i.)

Hence, rn+32=rn12. Then the above equation becomes simply

(7)rn+32=rn12=2rn+12.

Assume n is even and the parties’ ideal policy points are symmetric around 0. Let’s define δ0 as δ0=pn2+1pn2 and δ1 as δ1=pn2+2pn2+1=pn2pn21, and so on. Then, from equation (2),

rn2+1=12(δ0rn2+2+δ1rn2δ0+δ1).

Given the symmetric structure of the game and the uniqueness of equilibrium, the equilibrium will be such that r1=rn, r2=rn1, etc. (Formally, ri=rni+1 for every iI.)

Hence, rn2=rn2+1. Then the above equation becomes simply

rn2+2=2δ0+δ1δ0rn2+1.

Hence, we conclude that

(8)rn2+2>2rn2+1.

Now, we suppose that ri2ri1 and we will show that ri+12ri. Then, the first result of the proposition follows from equation (7) for n odd, and from equation (8) for n even.

We write equation (2) for i. This gives

ri=12(δi1ri+1+δiri1δi1+δi).

This is equivalent to

δi1ri+1=2δi1ri+2δiriδiri1.

Since ri2ri1, 2δiri4δiri1. Hence, δi1ri+12δi1ri+3δiri1. Then, we conclude that ri+12ri.

Hence, we have shown that a party chooses a rent level at least twice as high as its more moderate adjacent party.

• (ii) Now, we will show that a party’s vote share is higher than its more moderate adjacent party for n4.

From equations (2) and (6), we find the following relation between party i’s optimal rent level and vote share:

(9)αi=pi+1pi14(pi+1pi)(pipi1)ri.

Combining equations (2) and (9), we obtain the following relation between the vote shares of three adjacent parties:

(10)αi=12(pi+2pi+1pi+2piαi+1+pi1pi2pipi2αi1).

Assume n is odd. Writing equation (10) for i=n+12 for n5[12],

αn+12=12(pn+52pn+32pn+52pn+12αn+32+pn12pn32pn+12pn32αn12).

It is equivalent to

αn+12=12(δ1δ0+δ1αn+32+δ1δ0+δ1αn12).

By symmetry, αn+32=αn12. Hence,

αn+12=δ1δ0+δ1αn+32.

Hence,

αn+32>αn+12.

Assume n is even. Writing equation (10) for i=n2+1 for n6[13],

αn2+1=12(δ2δ1+δ2αn2+2+δ1δ0+δ1αn2).

This gives

δ2δ1+δ2αn2+2=2αn2+1δ1δ0+δ1αn2>αn2+1.

By symmetry, αn2+1=αn2. Then, the last inequality follows from αn2+1=αn2 and δ1δ0+δ1<1.

Since δ2δ1+δ2<1, it follows from the inequality that

αn2+2>αn2+1.

Now, we suppose that αi>αi1 and we will show that αi+1>αi.

Writing equation (10) for i such that 3in2[14],

αi=12(δi+1δi+1+δiαi+1+δi2δi2+δi1αi1).

Using αi>αi1, this gives

(2δi2δi2+δi1)αi<δi+1δi+1+δiαi+1.

Since the coefficient of αi is at least as high as the coefficient of αi+1, we have αi+1>αi.

Note that since we could write equation (10) only for 3in2, we still have to show that αn>αn1. By symmetry, α1>α2 will follow.

From above, we know that

rn2rn1+3δn1δn2rn2.

Writing equation (2) for n1,

rn1=δn1rn2+δn2rn2(δn1+δn2).

Manipulating together the last two equations, we get

rn(12+3(δn1+δn2)2δn2)rn1.

Similarly as we did for i1,n, we find the following relation between party n’s optimal rent level and vote share:

(11)αn=2+pn1p14(2+pnp1)(pnpn1)rn.

Combining the last two equations, we have

αn(4δn116δn1)(12+3(δn1+δn2)2δn2)rn1.

Writing equation (9) for n1, we have

αn1=δn1+δn24δn1δn2rn1.

Comparing the last two equations, it can be shown that αn>αn1.

Hence, we have proven that a party’s vote share is higher than its more moderate adjacent party for n5.

For n=4, using equations (2), (9) and (11), it can be similarly shown that α1=α4>α2=α3. Hence, the above result is also true for n=4.

• (iii) Since a party’s payoff is αiri, the third result follows from the first two for n4. Q.E.D.

Proof of Proposition 4:

The distances between parties’ ideal policies can be made arbitrarily small for n high enough. But then, if rent levels were positive, equation (6), reproduced below for convenience, would imply that vote shares of parties would go to infinity, a contradiction.

αi=12(ei,i+1ei,i1)=12(ri+1ri2(pi+1pi)riri12(pipi1)).

Hence, the total rent, r=iαiri, goes to 0 as well. Q.E.D.

Proof of Proposition 5:

Denote pi and ri the final policy and the total rent if voter x votes for party i, fixing the voting decisions of all other voters. Voter x will prefer party i over party j if and only if

(pix)2ri(pjx)2rj.

Equivalently,

x12rirjpipj+pi+pj2=12rirjpipj+pi

if pi<pj, and the reverse inequality if pi>pj, where the last equality follows since pj approaches pi as mass ε of voter x approaches 0.

Call V{1,,n} the subset of parties having positive vote shares in equilibrium. Call parties in V as 1,2,,k,,K such that pk<pk+1.

If kV, then ek,k+1>ek,k1, since otherwise e¯k<e¯k.

Assume party i does not have votes and pk<pi<pk+1 for some k,k+1V. Party i could have votes if and only if ei,k+1ei,k>0. Equivalently,

rirkpipk<rirk+1pipk+1.

We have pipk>0 and pipk+1<0. Therefore, if ri is smaller than both rk+1 and rk+1, then rirk<0 and rirk+1<0, so the above inequality holds. Hence, party i is able to have votes and to choose a positive rent level.

Assume now that iV and pi<p1. Party i could have votes if and only if ei,1(1)>0. Equivalently,

rir1pip1+pi(1)>0.

We have pip1>0. Therefore, if ri<r1, the above inequality holds. Hence, party i is able to have votes and to choose a positive rent level.

The similar reasoning holds for iV and pi>pK.

Hence, in equilibrium, every party get votes and choose positive rent levels.

Assume rimax{ri1,ri+1}. Then,

rirkpipk<rirk+1pipk+1,

i.e. party i receive no votes. We therefore conclude that ri<max{ri1,ri+1}.

Assume n is odd and the parties’ ideal policy points are symmetric around pn+12=0. In a symmetric equilibrium, it will be such that r1=rn, r2=rn1, etc.

Hence, rn+32=rn12. Since ri<max{ri1,ri+1}, it follows that rn+12<rn+32=rn12.

Assume n is even and the parties’ ideal policy points are symmetric around 0. In a symmetric equilibrium, it will be such that r1=rn, r2=rn1, etc.

Hence, rn2=rn2+1. Since ri<max{ri1,ri+1}, it follows that rn2+2>rn2+1.

Now, we suppose that riri1. Since ri<max{ri1,ri+1}, this implies that ri+1ri. The result follows. Q.E.D.

Proof of Proposition 6:

For small enough c, the number of parties n goes to infinity by Proposition 1. Hence, Proposition 4 gives the result. Q.E.D.

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