Appendix. Taxation in a one-dimensional context

As an exercise, we find the equilibrium tax rate when the policy space is one-dimensional. In this situation, the party that aligns its preferences to those of the poor (w<µ) proposes a tax rate of unity in the Stackelberg equilibrium. Understanding this exercise should help the reader to maintain their bearings in the more complicated two-dimensional problem explained in the paper.

Assume that the ideological issue is not important then α=0 in equation (3). The indirect utility function of citizen w at tax rate t is:

Now suppose that the distribution of voters, that is, of citizens who go to the polls on elections day, is g _s (w), where s is a random variable (state) uniformly distributed on [0,1].

Denote the mean of g _s by µ _s Let G _s be the cumulative distribution function of g _s Assume that G _s (µ) is strictly decreasing in s.¹¹

Let t ₁ >t ₂ be two tax policies. It is obvious from (11) that the set of citizens who prefer t ₁ to t ₂ , denoted W(t₁,t₂), is:

In state s the measure of this set is G _s (µ) That is, G _s (µ) is the fraction of voters who vote for t ₁ over t ₂ in state. Now t ₁ defeats t ₂ just in case it has a majority, i. e., when

As G _s (µ) is strictly decreasing in s, (13) is true just in case s<s ^* , where s ^* is defined by:

Assuming that there is an s ^* ϵ(0,1) satisfying (14), then the probability that t ₁ defeats t ₂ is just s ^* , since s is uniformly distributed on [0,1].

Now assume the P1 aligns its preferences to those of the poor, w ₁ <µ, while P2 aligns its preferences to the ones of the rich w ₂ <µ. Then, P1 proposes t ₁ , P2 proposes t ₂ , and t ₁ >t ₂ . As P1 wins with probability s ^* and P2 wins with probability 1- s ^* , parties’ expected utilities are Π₁ (t _1, t ₂ ) = s ^* v(t ₁ ;w ₂ )+( 1- s ^* )v(t ₂ ;w ₂ ) and respectively.

We next compute the Stackelberg equilibrium. Assume that P1 is the ‘incumbent’ and P2 is the ‘challenger’, where by definition, the challenger moves first. A Stackelberg equilibrium exists because the pay-off functions are continuous on the compact set [0,1]² Let T ₂ be P2’s equilibrium policy, and assume T ₂ <1 Then P1 obviously maximizes Π₁ (t _1, T ₂ ) at t ₁ =1.

Alternatively, suppose P2 is the incumbent. Let t ₁ be any proposal; P2 maximizes Π₂ by choosing t ₂ =0. Then P1’s problem is to choose t ₁ to maximize s ^* (t ₁ ;w ₁ )+(1-s ^* )v(0;w ₁ ): the solution is t ₁ =1.

Hence, irrespective of whether P1, that is the party that aligns its preferences to the ones of the poor, is the incumbent or challenger, the equilibrium in the game of party competition involves P1 proposing a tax rate of unity. In sum:

Proposition A.1 Let w ₁ <µ, let G _s ^* (µ) be strictly decreasing in s, and let u(x)=x be the universal von Neumann-Morgenstern utility function. Suppose there exists s ^* ϵ(0,1) such that Gs*μ=12. Then, whether the party P1 is the incumbent or challenger, the unique electoral equilibrium in the game of party competition entails T ₁ =1 and T ₂ =0.

Alternatively, when P1 aligns its preferences to those of the rich w ₁ >µ, while P2 aligns its preferences to those of the poor w ₂ >µ, then, P1 proposes t ₁ and P2 proposes t ₂ and t ₁ <t ₂ . Under such a framework, the equilibrium is such that P1 always proposes a tax rate of zero, regardless of whether it is the incumbent or the challenger. We summarize the equilibrium tax rate in Table 2:

From the proposition we can conclude that when there is no ideology and the only matter of interest is the income of the voters, if a party aligns its preferences to those of the poor, it chooses a tax rate of unity, T¯=1, to benefice the poor. If the party aligns its preferences to those of the rich, it chooses the best policy for them, t¯=0. Under this situation, we can set the stage for our study. After including ideology in the preferences, will the party P _j that aligns its preferences to those of the poor w _j <µ, compromise the radical redistributive policy it advocates when only income is the issue?

Proof of important theorems and propositions

Proof of proposition 3.1 Let (T₁(α), T₂(α)) be a sequence of Stackelberg equilibria in the games Ψ _α, and let z ₁ (α) and z ₂ (α) converge to z ₁ (1) and z ₂ (1), respectively. Suppose, contrary to the claim, that (z ₁ (1), z ₂ (1)) is not a Stackelberg equilibrium in Ψ ₁. Then,z ₁ (1) must not be a best response to z ₂ (1); so it must therefore be that there exists an equilibrium pair z1~,z2~ such that z1~ is a best response to z2~ and

Π2z~1,z~2;1>Π2(z11,z21;1)

Let (t1^α,z1^α) be P1’s best response to (t2α,z2~) in Ψ _α. Then z1^(1)≡limαz1^(α) is a best response to z2~ in Ψ ₁. By A2b,  z1^1=z1~ . Hence Π2(t1^α,z1^α, t2α,z2~) approaches Π2(z1~,z2~;1) as α approaches 1. In particular, by the above inequality, for large α:

Π2t1^α,z1^α, t2α,z2~>Π2(t1α,z1α,t2α,z2α;α)

This contradicts the fact that (t1α,z1α,t2α,z2α) is a Stackelberg equilibrium in Ψ _α, which establishes the claim. It is immediate to do the proof for P1.

By the upper-hemi-continuity of the equilibrium correspondence θ at 1, any converging subsequence of the continuum (T ₁ (α), T ₂ (α)) converges to an equilibrium of Ψ ₁. The claims follow immediately from A2(a).

Proof of proposition 3.2 Let A2(a) hold. For α=1

we have:

a1<z1*<z2*<a2

a1-z1*<0<z2*-z1*<a2-z1*

We end up with

And

a1<z1*<z2*<a2

a1+z2*<z1*+z2*<2z2*<a2+z2*

a1+z2*2<z1*+z2*2<z2*<a2+z2*2

In one side:

0<z2*-a12<z¯*-a1<z2*-a1

then we have:

In the other side:

a1+z2*-2a22<z1*+z2*2-a2<z2*-a2<0

Then we have:

Proof of theorem 3.3 First, we are proving that t ₁ (α)=0 for the case ∆=t₂-t₁<0. Suppose to the contrary: that for a sequence of α‘s tending to one, there is a Stackelberg equilibrium of Ψ _α in which t ₁ (α)>0. We know ∆z(α)>0 by Proposition 3.2; hence, for large α, π(T ₁ (α), T ₂ (α)) is indeed given by (7), and hence, either π(T ₁ (α), T ₂ (α))=s ^* (T ₁ (α), T ₂ (α)), where s ^* is defined by (6), or π(T ₁ (α), T ₂ (α))ϵ{0,1}. But by , since for all equilibria game Ψ ₁,πT1α, T2α∉{0,1} ,it follows that for sufficiently large , , and therefore π(T ₁ (α), T ₂ (α))=s ^* (T ₁ (α), T ₂ (α)).

Differentiating (6) implicitly w.r.t ₁ , we may write:

as long as the denominator in (15) does not vanish, where we have omitted the argument ‘α’ on the variables z¯ , ∆t and ∆z. But assumption A1 tells us that the expression ∫W∫-∞-z¯+1-α∆tw-μα∆z∂s*∂swra∖wdadw, since this expression is just the derivative of Φ (z,s) w.r.t. s, and so the denominator of (15) does not vanish.

We assume that P1 is the incumbent and P2 is the challenger (i.e., P2 moves first). Since s ^* is differentiable for large α, so is Π ₁ (T ₁ ,T ₂ ;α) differentiable at (T ₁ ,T ₂ )= (T ₁ (α),T ₂ (α)) , for large α. Since T ₁ (α) is a best response to T ₂ (α), it thereforeδΠ1δz1T1α,T2α,α=0, since z ₁ (α) is an interior solution (as the domain of possible z ₁ ’s is the real line). This first-order condition can be solved to yield:

Π1τ1,τ2;α=s*vt1,z1;w1,a1+(1-s*)v(t2,z2;w1,a1)

Π1τ1,τ2;α=s*[1-αw1+t1μ-w1-α2(z1-a1)2]+1-s*[1-αw1+t2μ-w1-α2(z2-a1)2]

The F. O. C. subject to is given by:

ðΠ1ðz1=s*-αz1-a1+∂s*ðz11-αw1+t1μ-w1-α2(z1-a1)2-∂s*ðz11-αw1+t2μ-w1-α2z2-a12=0

∂s*∂z11-αw1+t1μ-w1-α2z1-a12-1-αw1+t2μ-w1+α2(z2-a1)2=s*α(z1-a1)

∂s*∂z11-αw1-μt2-t1+α2(z2-a1)2-(z1-a1)2=s*α(z1-a1)

∂s*∂z1=s*α(z1-a1)1-αw1-μt2-t1+α2(z2-a1+z1-a1)(z2-a1-z1+a1)

Similarly, it follows that ∂Π1(τ1,τ2;α)∂t1≥0, since by hypothesis t ₁ (α)>0 for all (finite) α.

The just stated inequality can be solved to yield:

ðΠ1ðt1=s*(1-α)(μ-w1)+∂s*∂t11-αw1+t1μ-w1-α2(z1-a1)2-∂s*∂t11-αw1+t1μ-w1-α2(z2-a1)2≥0

∂s*∂t11-αw1+t1μ-w1-α2z1-a12-1-αw1+t2μ-w1+α2z2-a12≥s*(1-α)(w1-μ)

∂s*∂t11-αt1μ-w1-1-αt2μ-w1+α2(z2-a1)2-(z1-a1)2≥s*(1-α)(w1-μ)

an expression whose derivation uses the fact that the denominator of (17) is positive, which follows from Proposition 3.2.

Next, differentiating (6) w.r.t. z ₁ yields:

Let the (common) denominator in the fractions on the r.h.s. of (18) and (15) be denoted ‘D’. Using (18) and (16), we can solve for D, eliminating ∂s*∂z1;

-∫wgs*wr(z¯+1-α∆tw-μα∆z2)dwD=s*α(z1-a1)1-α∆tw1-μ+α(z¯-a1)∆z

D=-11-α∆tw1-μ+αz¯-a1∆z∫wgs*wr(z¯+1-α∆tw-μα∆z|w)12+(1-α)∆t(w-μ)α(∆z)2dws*α(z1-a1)

substituting the expression for D into (15) yields:

In turn, (19) and (17) imply:

s*α(z1-a1)∫wgs*wr(z¯+1-α∆tw-μα∆z|w)1-αw-μα∆zdw1-α∆tw1-μ+αz¯-a1∆z∫wgs*wr(z¯+1-α∆tw-μα∆z|w)12+(1-α)∆t(w-μ)α(∆z)2dw≥s*1-αw1-μ1-α∆tw1-μ+αz¯-a1∆z ,

(z1-a1)∫wgs*wr(z¯+1-α∆tw-μα∆z|w)μ-w∆zdw∫wgs*wr(z¯+1-α∆tw-μα∆z|w)12+(1-α)∆t(w-μ)α(∆z)2dw≥(w1-μ)

1∆z(z1-a1)∫wgs*wr(z¯+1-α∆tw-μα∆z|w)μ-w∆zdw1α(∆z)2∫wgs*wr(z¯+1-α∆tw-μα∆z|w)12+(1-α)∆t(w-μ)α(∆z)2dw≤(μ-w1)

Letting α→1, (20) becomes, in the limit:

α∆z(1)(z1-a1)∫wgs*wr(z¯1|w)(w-μ)dwα(∆z(1))22∫wgs*wr(z¯1|w)dw≤(μ-w1)

Using the definitions of ρ¯,σ¯ and μ¯ provided in the text,

∫wgs*wr(z¯1|w)(w-μ)dw∫wgs*wr(z¯1|w)dw≤∆z(1)(μ-w1)2(z11-a1)

we can write the negation of (21) as

∫wgs*wr(z¯1|w)(w-μ)dw∫wgs*wr(z¯1|w)dw>∆z(1)(μ-w1)2(z11-a1)

which is precisely condition (10(a)). Hence, by A3, (21) does not hold, which contradicts the original supposition -that there is a sequence of equilibria at which t _¨1 (α)>0. The reader could verify easily that the inequality in (22) does not change for t ₁ =t ₂ . Adding the fact that should be small enough to keep 1-α∆tw1-μ+αz¯-a1∆z<0 for the case t ₁ <t ₂ , we get the same expression for (22). Then, if (10) holds and w ₁ is small enough, we have t ₁ (α)=0 for any case t ₁ <t _2, t ₁ =t _2, t ₁ >t ₂ .

Second, we prove that t ₂ (α)=0 for the case ∆t= t ₂ - t ₁ <0.

Suppose to the contrary: that for a sequence of α’s tending to one, there is a Stackelberg equilibrium of Ψ _α in which t ₂ (α)=0. We know ∆z(α)>0 by Proposition 3.2; therefore, for large α. π(T ₁ (α), T ₂ (α)) is indeed given by (7), and hence, either π(T ₁ (α), T ₂ (α))=s ^* (T ₁ (α), T ₂ (α)) where s ^* is defined by (6), or π(T ₁ (α), T ₂ (α))ϵ{0,1}. But by A2(a), since for all equilibria game Ψt,π∉{0,1} it follows that for sufficiently large π(T1(α),T2 (α))∉{0,1}, and therefore π(T ₁ (α), T ₂ (α))=s ^* (T ₁ (α), T ₂ (α)).

Differentiating (6) implicitly w.r.t. , we may write:

∂s*∂z2=-∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)1-α∆tw-μα∆zdw∫W∫-∞-z¯+1-α∆tw-μα∆z∂gs*∂s(w)ra|wdadw

as long as the denominator in (23) does not vanish, where we have omitted the argument ‘α’ on the variables z¯, ∆ _t , and ∆ _z . But assumption A1 tells us that the expression ∫W∫-∞-z¯+1-α∆tw-μα∆z∂gs*∂swra|wdadw<0 , since this expression is just the derivative of Φ(z,s)w.r.t. s, and so the denominator of (23) does not vanish.

We assume that P2 is the incumbent and P1 is the challenger (i.e., P1 moves first). Since s ^* is differentiable for large α, so is Π₂(T ₁ ,T ₂ ,α) differentiable at (T ₁ ,T ₂ )= (T ₁ (α), T ₂ (α)), for large α. Since T ₂ (α) is a best response to T ₁ (α), it therefore ∂Π2∂z2(T1α,T2α,α)=0, z ₂ (α) since is an interior solution (as the domain of possible z ₂ ’s is the real line). This first-order condition can be solved to yield:

Π2τ1,τ2;α=s*vt1,z1;w2,z2+(1-s*)v(t2,z2;w2,a2)

Π2τ1,τ2;α=s*1-αw2+t1μ-w2-α2z1-a22+1-s*1-αw2+t2μ-w2-α2z2-a22

The F. O. C. subject to is given by:

∂Π2∂z2=∂s*∂z21-αw2+t1μ-w2-α2z1-a22+1-s*-αz2-a2-∂s*∂z21-αw2+t2μ-w2-α2z2-a22=0

∂s*∂z21-αw2+t1μ-w2-α2z1-a22-1-αw2+t2μ-w2-α2z2-a22=1-s*αz2-a2

∂s*∂z21-αt2μ-w2-1-αt2μ-w2+α2z2-a22-z1-a22=1-s*αz2-a2

∂s*∂z2=1-s*αz2-a21-αw2-μt2-t1+α2(z2-a2+z1-a2)(z2-a2-z1+a2)

Similarly, it follows that ∂Π2τ1,τ2;α∂t2≥0, since by hypothesis t ₂ (α>0) for all (finite) α. The just stated inequality can be solved to yield:

∂Π2∂t2=∂s*∂t21-αw2+t1μ-w2-α2z1-a22+1-s*(1-α)μ-w2-∂s*∂t21-αw2+t2μ-w2-α2z2-a22≥0

∂s*∂t21-αw2+t1μ-w2-α2z1-a22-1-αw2+t2μ-w2-α2z2-a22≥1-s*(1-α)w2-μ

∂s*∂t21-αt1μ-w2-1-αt2μ-w2+α2z2-a22-z1-a22≥1-s*(1-α)w2-μ

an expression whose derivation uses the fact that the denominator of (25) is positive (with w ₂ small enough ), which follows from Proposition 3.2.

Next, differentiating (6) w.r.t. z ₂ yields:

Let the (common) denominator in the fractions on the r.h.s. of (26) and (23) be denoted ‘D’. Using (26) and (24), we can solve for D, eliminating ∂s*∂z2;

-∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)(12-1-α∆tw-μα(∆z)2)dwD=1-s*αz2-a21-α∆tw2-μ+α(z¯-a2)∆z

D=-1[1-α∆tw2-μ+α(z¯-a2)∆z]∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)(12-1-α∆tw-μα(∆z)2)dw1-s*αz2-a2

substituting the expression for D into (23) yields:

In turn, (25) and (27) imply:

1-s*αz2-a2∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)1-αw-μα∆zdw[1-α∆tw2-μ+α(z¯-a2)∆z]∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)(12-1-α∆tw-μα(∆z)2)dw≥1-s*αz2-a21-α∆tw2-μ+α(z¯-a2)∆z

z2-a2∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)w-μα∆zdw∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)(12-1-α∆tw-μα(∆z)2)dw≥(w2-μ)

1α∆zz2-a2∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)w-μdw1α∆z2∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)(α∆z22-1-α∆tw-μ)dw≥(w2-μ)

Letting α→1, (28) becomes, in the limit:

α∆z(1)z21-a2∫Wgs*wr(z¯+(1)|w)w-μdwα∆z(1)22∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)dw≥(w2-μ)

Using the definitions of ρ¯, σ¯ and μ¯ and provided in the text,

∫Wgs*wr(z¯+(1)|w)w-μdw∫Wgs*wr(z¯+1-α∆tw-μα∆z|w)dw≤∆z(1)(w2-μ)2z21-a2

we can write the negation of (29) as

∫Wgs*wr(z¯+(1)|w)w-μdw∫Wgs*wr(z¯+(1)|w)dw>∆z(1)(w2-μ)2z21-a2

which is precisely condition (10(b)). Hence, by A3, (29) does not hold, which contradicts the original supposition -that there is a sequence of equilibria at which t ₂ (α)>0. The reader could verify easily that the inequality in (30) does not change for the case t ₁₌ t ₂ and for the case t ₁ <t ₂ . Adding the fact that should be small enough to keep 1-α∆tw2-μ+αz¯-a2∆z>0 for the case t ₁ <t ₂ , we get the same expression for (30).

Then, if (10(b)) holds and is small enough, we have at any case t ₁ <t _2, t ₁ =t _2, t ₁ >t ₂ and the theorem is proved.