Extreme Parties and Political Rents

Proof of Lemma 1:

Suppose we have a voting equilibrium where the profile of vote shares of parties is ( α 1 ,   … ,   α i ,   … ,   α n ) , 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 j ≠ i . Then, the vote shares of parties i and j would change to

and

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

and

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

Equivalently,

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

Given the definition of e i , j , this is equivalent to

if j > i , and to

if j < i .

Since voter x should not deviate for any j ≠ i , it should be that, for i ≠ 1 ,   n ,

for i = 1 ,

for i = n ,

Proof of Proposition 1:

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

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 p k < p k + 1 .

If k ∈ V , then e k , k + 1 > e k , k − 1 , since otherwise e ¯ k < e ¯ k . Then, α k = e k , k + 1 − e k , k − 1 2 for every k ∈ V , k ≠ 1 ,   K , α 1 = e 1 , 2 − ( − 1 ) 2 and α K = 1 − e K , K − 1 2 .

Assume party i does not have votes (i.e. i ∉ V ) and p k < p i < p k + 1 for some k ,   k + 1 ∈ V . Party i could have votes if and only if e i , k + 1 − e i , k > 0 . Equivalently,

Equivalently,

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

Assume now that i ∉ V and p i < p 1 . Party i could have votes if and only if e i , 1 − ( − 1 ) > 0 . Equivalently,

Equivalently,

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

Assume now that i ∉ V and p i > p K . Party i could have votes if and only if 1 − e i , K > 0 . Equivalently,

Equivalently,

Since r K > 0 and p i > p K , such r i > 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 α i = e i , i + 1 − e i , i − 1 2 for any i ≠ 1 ,   n . The vote shares of parties 1 and n are respectively α 1 = e 1,2 − ( − 1 ) 2 and α n = 1 − e n , n − 1 2 . Q.E.D.

Proof of Lemma 2:

The final policy is given by p = ∑ i = 1 n α i p i . The vote share of party i, i = 2 ,   … ,   n − 1 , is

The vote shares of parties 1 and n are respectively

and

After replacing these vote shares,

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. α i r i is quasiconcave in r i for every i. Then, standard arguments guarantee the existence of an equilibrium see e.g. Friedman 1977).

For i = 2 ,   3 ,   … ,   n − 1 ,

given that adjacent parties have also some votes. In this case, α i is linear in r i , and

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

Also in this case, α i is linear in r i , and

Hence, the vote share function has a kink at r i 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 p i + 2 > p i + 1 . The same logic applies when r i is so low that even party i + 2 gets no vote at all, and so on.

If r i is low enough so that even party n does not have any vote, then the final policy becomes p = 1 2 + p i − p 1 ( p 1 + p i + r 1 − r i 2 ) and the vote share of party i becomes

Also in this case, α i is linear in r i , and

Hence, the vote share function has a kink at r i 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 1 2 ( 2 + p i − p 1 ) > 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, α i is concave in r i for i = 2 ,   3 ,   … ,   n − 1 .

For party 1,

given that party 2 gets some votes. In this case, α 1 is linear in r 1 , and

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

Also in this case, α 1 is linear in r 1 , and

Hence, the vote share function has a kink at r 1 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 p 3 > p 2 . The same line of reasoning applies when r 1 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, α 1 is concave in r 1 .

Similarly, it can be shown that α n is concave in r n . Hence, when positive, α i is concave in r i for every i. Therefore, for every i, α i r i is concave in r i when positive, and equal to 0 when r i is too high. Hence, α i r i is quasiconcave in r i 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 r 3 as a linear function of r 1 and r 2 , calling this function as R 3 , r 3 = R 3 ( r 1 ,   r 2 ) . Similarly, r 4 = R 4 ( r 2 ,   r 3 ) . Replacing r 3 , this becomes r 4 = R 4 ( r 2 ,   R 3 ( r 1 ,   r 2 ) ) which can be written as r 4 = h 4 ( r 1 ,   r 2 ) . Given that R 3 and R 4 are linear, h 4 is linear too. Continuing this procedure, we have r n − 1 = h n − 1 ( r 1 ,   r 2 ) and r n = h n ( r 1 ,   r 2 ) , both h n − 1 and h n being linear. Replacing these values on Eqs. (3) and (4), we obtain two linear equations of two unknowns r 1 and r 2 . We have already proved that an equilibrium exists. Hence, if there is not a unique equilibrium, then the two linear equations on r 1 and r 2 should coincide and every point of this straight line should be an equilibrium. Specifically, there should be an equilibrium where r 1 = 0 or r 2 = 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 r 1 * and r 2 * . Since, r i , i = 3 ,   … ,   n , is uniquely defined in terms of r 1 and r 2 by the procedure explained above, there is a unique set of equilibrium rents ( r 1 * ,   r 2 * ,   … ,   r n * ) . Q.E.D.

Proof of Proposition 3:

(i) Assume n is odd and the parties’ ideal policy points are symmetric around p n + 1 2 = 0 . Let’s define δ 0 as δ 0 = p n + 1 2 − p n − 1 2 = p n + 3 2 − p n + 1 2 , δ 1 as δ 1 = p n + 5 2 − p n + 3 2 = p n − 1 2 − p n − 3 2 , and so on. Then, from equation (2),

Given the symmetric structure of the game and the uniqueness of equilibrium, the equilibrium will be such that r 1 = r n , r 2 = r n − 1 , etc. (Formally,   r i = r n − i + 1 for every i.)

Hence, r n + 3 2 = r n − 1 2 . Then the above equation becomes simply

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

Given the symmetric structure of the game and the uniqueness of equilibrium, the equilibrium will be such that r 1 = r n , r 2 = r n − 1 , etc. (Formally, r i = r n − i + 1 for every i ∈ I .)

Hence, r n 2 = r n 2 + 1 . Then the above equation becomes simply

Hence, we conclude that

Now, we suppose that r i ≥ 2 r i − 1 and we will show that r i + 1 ≥ 2 r i . 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

This is equivalent to

Since r i ≥ 2 r i − 1 , 2 δ i r i ≥ 4 δ i r i − 1 . Hence, δ i − 1 r i + 1 ≥ 2 δ i − 1 r i + 3 δ i r i − 1 . Then, we conclude that r i + 1 ≥ 2 r i .

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 n ≥ 4 .

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

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

Assume n is odd. Writing equation (10) for i = n + 1 2 for n ≥ 5 ^[12],

It is equivalent to

By symmetry, α n + 3 2 = α n − 1 2 . Hence,

Hence,

Assume n is even. Writing equation (10) for i = n 2 + 1 for n ≥ 6 ^[13],

This gives

By symmetry, α n 2 + 1 = α n 2 . Then, the last inequality follows from α n 2 + 1 = α n 2 and δ 1 δ 0 + δ 1 < 1 .

Since δ 2 δ 1 + δ 2 < 1 , it follows from the inequality that

Now, we suppose that α i > α i − 1 and we will show that α i + 1 > α i .

Writing equation (10) for i such that 3 ≤ i ≤ n − 2 ^[14],

Using α i > α i − 1 , this gives

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

Note that since we could write equation (10) only for 3 ≤ i ≤ n − 2 , we still have to show that α n > α n − 1 . By symmetry, α 1 > α 2 will follow.

From above, we know that

Writing equation (2) for n − 1 ,

Manipulating together the last two equations, we get

Similarly as we did for i ≠ 1 ,   n , we find the following relation between party n’s optimal rent level and vote share:

Combining the last two equations, we have

Writing equation (9) for n − 1 , we have

Comparing the last two equations, it can be shown that α n > α n − 1 .

Hence, we have proven that a party’s vote share is higher than its more moderate adjacent party for n ≥ 5 .

For n = 4 , using equations (2), (9) and (11), it can be similarly shown that α 1 = α 4 > α 2 = α 3 . Hence, the above result is also true for n = 4 .

• (iii) Since a party’s payoff is α i r i , the third result follows from the first two for n ≥ 4 . 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.

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

Proof of Proposition 5:

Denote p ′ i and r ′ i 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

Equivalently,

if p i < p j , and the reverse inequality if p i > p j , where the last equality follows since p ′ j approaches p ′ i 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 p k < p k + 1 .

If k ∈ V , then e k , k + 1 > e k , k − 1 , since otherwise e ¯ k < e ¯ k .

Assume party i does not have votes and p k < p i < p k + 1 for some k , k + 1 ∈ V . Party i could have votes if and only if e i , k + 1 − e i , k > 0 . Equivalently,

We have p ′ i − p ′ k > 0 and p ′ i − p ′ k + 1 < 0 . Therefore, if r i is smaller than both r k + 1 and r k + 1 , then r ′ i − r ′ k < 0 and r ′ i − r ′ k + 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 i ∉ V and p i < p 1 . Party i could have votes if and only if e i , 1 − ( − 1 ) > 0 . Equivalently,

We have p ′ i − p ′ 1 > 0 . Therefore, if r i < r 1 , the above inequality holds. Hence, party i is able to have votes and to choose a positive rent level.

The similar reasoning holds for i ∉ V and p i > p K .

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

Assume r i ≥ max { r i − 1 ,   r i + 1 } . Then,

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

Assume n is odd and the parties’ ideal policy points are symmetric around p n + 1 2 = 0 . In a symmetric equilibrium, it will be such that r 1 = r n , r 2 = r n − 1 , etc.

Hence, r n + 3 2 = r n − 1 2 . Since r i < max { r i − 1 , r i + 1 } , it follows that r n + 1 2 < r n + 3 2 = r n − 1 2 .

Assume n is even and the parties’ ideal policy points are symmetric around 0. In a symmetric equilibrium, it will be such that r 1 = r n , r 2 = r n − 1 , etc.

Hence, r n 2 = r n 2 + 1 . Since r i < max { r i − 1 ,   r i + 1 } , it follows that r n 2 + 2 > r n 2 + 1 .

Now, we suppose that r i ≥ r i − 1 . Since r i < max { r i − 1 ,   r i + 1 } , this implies that r i + 1 ≥ r i . 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.