Majority rule and selfishly optimal nonlinear income tax schedules with discrete skill levels

Abstract

Röell (Voting over nonlinear income tax schedules, unpublished manuscript, School of International and Public Affairs, Columbia University, New York, 2012) shows that Black’s median voter theorem for majority voting with single-peaked preferences applies to voting over nonlinear income tax schedules that satisfy the constraints of a finite type version of the Mirrlees optimal income tax problem when voting takes place over the tax schedules that are selfishly optimal for some individual and preferences are quasilinear. An alternative way of establishing Röell’s median voter result is provided that offers a different perspective on her findings, drawing on insights obtained by Brett and Weymark (Games Econ Behav 101:172–188, 2017) in their analysis of a version of this problem with a continuum of types. In order to characterize a selfishly optimal schedule, it is determined how to optimally bunch different types of individuals.

Appendix

Proof of Lemma 1

Suppose by way of contradiction that (26) does not hold. Let h be the smallest type in $\{j, \ldots , m\}$ such that $c^{k*}_h>c^{k*}_{h-1}$. Because $\mu _h = 0$, summing (39) over types $h,\ldots , m$ yields

$$\begin{aligned} v^\prime (c^{k*}_i)\left[ \sum _{i=h}^m n_i \beta ^k_i\right] = \sum _{i=h}^m n_i +\mu _{m+1}. \end{aligned}$$

(A.1)

The left-hand side of (A.1) is nonpositive by (25), whereas the right-hand side is positive, a contradiction. $\square$

Proof of Lemma 2

If $j = m$, (27) is vacuous and the lemma is trivially true.

Now, suppose that $j < m$. By way of contradiction, suppose that there is some type in $\{j,\ldots ,m\}$ with $c^{k*}_i>c^{k*} _j$. Let l be the lowest such type. Then, $\mu _l =0$.

Suppose, first, that type l is bunched with type m (l need not be distinct from m). By (39),

$$\begin{aligned} v^\prime (c^{k*}_j) \sum _{i=j}^{l-1} n_i \beta ^k_i \ge \sum _{i=j}^{l-1} n_i - [\mu _j + \mu _{l-1}], \end{aligned}$$

(A.2)

which implies that

$$\begin{aligned} v^\prime (c^{k*}_j)\left[ \frac{ \sum _{i=j}^{l-1} n_i \beta ^k_i }{\sum _{i=j}^{l-1} n_i}\right] \ge 1 - \frac{[\mu _j + \mu _{l-1}]}{\sum _{i=j}^{l-1} n_i}. \end{aligned}$$

(A.3)

By assumption, the fraction on the left-hand side of (A.3) is positive. Because $v^\prime (0) = \infty$, if $c^{k*}_j = 0$, this inequality is violated. Hence, $c^{k*}_j > 0$ and both (A.2) and (A.3) hold with equality. Also by (39)

$$\begin{aligned} v^\prime (c^{k*}_l)\left[ \frac{ \sum _{i=l}^{m} n_i \beta ^k_i }{\sum _{i=l}^{m} n_i}\right] = 1 + \frac{[\mu _{l+1} + \mu _{m+1}]}{\sum _{i=l}^{m} n_i}. \end{aligned}$$

(A.4)

It follows from the equality version of (A.3), (A.8), and the concavity of v that

$$\begin{aligned} v^\prime (c^{k*}_j)\left[ \frac{ \sum _{i=j}^{l-1} n_i \beta ^k_i }{\sum _{i=j}^{l-1} n_i}\right]<v^\prime (c^{k*}_l)\left[ \frac{ \sum _{i=l}^{m} n_i \beta ^k_i }{\sum _{i=l}^{m} n_i}\right] < v^\prime (c^{k*}_j)\left[ \frac{ \sum _{i=l}^{m} n_i \beta ^k_i }{\sum _{i=l}^{m} n_i}\right] . \end{aligned}$$

(A.5)

Thus,

$$\begin{aligned} \frac{ \sum _{i=j}^{l-1} n_i \beta ^k_i }{\sum _{i=j}^{l-1} n_i} < \frac{ \sum _{i=l}^{m} n_i \beta ^k_i }{\sum _{i=l}^{m} n_i}, \end{aligned}$$

(A.6)

which contradicts (27).

We have thus shown that there must be a maximal type $q<m$ that is bunched with l. Note that it may be the case that $q = l$. By (40),

$$\begin{aligned} v^\prime (c^{k*}_l)\left[ \frac{ \sum _{i=l}^{q} n_i \beta ^k_i }{\sum _{i=l}^{q} n_i}\right] = 1. \end{aligned}$$

(A.7)

Using (39) for types $q+1$ to m implies

$$\begin{aligned} v^\prime (c^{k*}_l)\left[ \frac{ \sum _{i=q+1}^{m} n_i \beta ^k_i }{\sum _{i=q+1}^{m} n_i}\right] = 1 + \frac{[\mu _{q+2} + \mu _{m+1}]}{\sum _{i=q+1}^{m} n_i}. \end{aligned}$$

(A.8)

The argument leading to (A.6) may be repeated to conclude that

$$\begin{aligned} \frac{ \sum _{i=l}^{l-1} n_i \beta ^k_i }{\sum _{i=j}^{l-1} n_i} < \frac{ \sum _{i=l}^{m} n_i \beta ^k_i }{\sum _{i=l}^{m} n_i}, \end{aligned}$$

(A.9)

which contradicts (27).

Having shown that both possibilities for type l result in a contradiction, no such type is possible and, hence, (28) holds. $\square$

Proof of Theorem 2

Because ${\bar{f}}_{\sigma ^k}$ is a convex function and $v^{\prime }$ is decreasing, the consumption vector ${\mathbf {c}}^{k*}$ defined in (36) satisfies the constraints in proposer $k$’s reduced-form problem. Because ${\bar{\beta }}^k$ is nondecreasing in type, the set of all types that share a common value of the adjusted virtual wage is an interval of types. Let $\{j, \ldots , m\}$ be such an interval. Because ${\bar{f}}_{\sigma ^k}$ is the highest convex function that lies nowhere above $f_{\sigma ^k}$, (25) holds if ${\bar{\beta }}^k_j \le 0$ and (27) holds if ${\bar{\beta }}^k_j > 0$. Hence, by Lemmas 1 and 2, respectively, all types in $\{j, \ldots , m\}$ are bunched together. Furthermore, because the types for which the adjusted virtual wages are nonpositive are the first l types for some $l \in \{0, 1, \ldots , n\}$, Lemma 1 also implies that $c^{k*}_i$ is zero for any type $i \le l$.

For any type i for which ${\bar{\beta }}^k_i > 0$, let $c^{k \circ }_i$ be the common value of the consumption of the types $\{j, \ldots , m\}$ that are bunched with type i. The optimal value of $c^{k \circ }_i$ is the solution to

$$\begin{aligned} \max \ \sum _{h=j}^m n_h {\bar{\beta }}^k_i v(c^{k \circ }_i) - \sum _{h=j}^m n_h c^{k \circ }_i \ \ \text {subject to } \ \ c^{k \circ }_i \ge 0. \end{aligned}$$

(A.10)

Because ${\bar{\beta }}^k_i > 0$, v is strictly concave, and $v^{\prime }(0) = \infty$, this problem has a unique solution and it is positive. This solution is given by the first equality in (36). $\square$

Proof of Lemma 3

Suppose, by way of contradiction, that there is a complete bunching interval $\{j,\ldots ,m\}$ with $j < m$ that does not include type k. We first consider the case in which $m < k$. Because $\mu _j = \mu _{m+1} = 0$, by (39) and (41),

$$\begin{aligned} n_j \beta ^n_j v^\prime (c^{k*}_j) \ge n_j+\mu _{j+1} \quad \text {and} \quad n_m \beta ^n_m v^\prime (c^{k*}_j) \ge n_m-\mu _{m} \end{aligned}$$

(A.11)

or, equivalently,

$$\begin{aligned} \beta ^n_j \ge \frac{1}{v^\prime (c^{k*}_j)}\left[ 1+\frac{\mu _{j+1}}{n_j}\right] \quad \text {and} \quad \beta ^n_m \ge \frac{1}{v^\prime (c^{k*}_j)}\left[ 1-\frac{\mu _m}{n_m}\right] . \end{aligned}$$

(A.12)

The inequalities in (A.11) and (A.12) bind if $c^{k*}_j > 0$. In this case, (A.12) implies that $\beta ^n_j \ge \beta ^n_m$. If $c^{k*}_j = 0$, then $j = 1$ and (A.12) implies that $\beta ^n_j \ge 0$. Because there are no types above m with zero consumption, by Theorem 2 and the way that the adjusted virtual wages are determined, it must be the case that $\beta ^n_m \le 0$. Thus, in this case as well, we have $\beta ^n_j \ge \beta ^n_m$. However, because type $n$’s optimal consumption schedule does not exhibit bunching, by Theorem 1, $\beta ^n_j < \beta ^n_m$, so we have a contradiction.

A similar argument can be used to show that no complete bunching interval of the form $\{j,\ldots , m\}$ with $j < m$ and $j>k$ exists. In this case, $c^{k*}_j$ is necessarily positive, so the analogues to (A.11) and (A.12) for $\beta ^1_j$ and $\beta ^1_m$ hold with equality. $\square$

Proof of Theorem 4

We first suppose that $i<k$. We show that the utility that type i receives with type $k$’s selfishly optimal allocation is greater than or equal to the utility that type i receives with type $(k+1)$’s selfishly optimal allocation. Formally,

$$\begin{aligned} w_i v(c^{k}_i)-y^{k}_i \ge w_i v(c^{k+1}_i)-y^{k+1}_i. ^{15} \end{aligned}$$

(A.13)

¹⁵

Substituting (12) into the left-hand side of (A.13) yields

$$\begin{aligned} w_i v(c^{k}_i)-y^{k}_i = w_iv(c^k_i)-\left[ y^k_k + \sum _{j=i}^{k-1} w_j[v(c^k_j)-v(c^k_{j+1})] \right] . \end{aligned}$$

(A.14)

Rearranging terms in (A.14) gives

$$\begin{aligned} \begin{aligned} w_i v(c^{k}_i)-y^{k}_i = {}&w_iv(c^k_i)- y^k_k - w_i v(c^k_i) \\&{} + \sum _{j=i}^{k-2}[w_j-w_{j+1}]v(c^k_{j+1})+w_{k-1}v(c^k_k) . \end{aligned} \end{aligned}$$

(A.15)

Cancelling the terms in $w_iv(c^k_i)$ and adding and subtracting $w_k v(c^k_k)$ on the right-hand side of (A.15) yields

$$\begin{aligned} w_i v(c^{k}_i)-y^{k}_i= w_k v(c^k_k) - y^k_k + \sum _{j=i}^{k-1}[w_j-w_{j+1}]v\left( c^k_{j+1}\right) . \end{aligned}$$

(A.16)

Substituting (12) into the right-hand side of (A.13) yields

$$\begin{aligned} w_i v(c^{k+1}_i)-y^{k+1}_i= w_iv(c^{k+1}_i)-\left[ y^{k+1}_{k+1} + \sum _{j=i}^{k} w_j\left[ v\left( c^{k+1}_j\right) -v\left( c^{k+1}_{j+1}\right) \right] \right] . \nonumber \\ \end{aligned}$$

(A.17)

Rearranging terms on the right-hand side of (A.17) gives

$$\begin{aligned} w_i v(c^{k+1}_i)-y^{k+1}_i= w_k v(c^{k+1}_{k+1}) - y^{k+1}_{k+1} + \sum _{j=i}^{k-1}[w_j-w_{j+1}]v\left( c^{k+1}_{j+1}\right) . \end{aligned}$$

(A.18)

Subtracting (A.18) from (A.16), we obtain

$$\begin{aligned} \begin{aligned}&\left[ w_i v(c^{k}_i) - y^{k}_i\right] - \left[ w_i v\left( c^{k+1}_i\right) -y^{k+1}_i \right] = \left[ w_k v(c^k_k) - y^k_k\right] \\&\quad - \left[ w_k v\left( c^{k+1}_{k+1}\right) - y^{k+1}_{k+1}\right] + \sum _{j=i}^{k-1}[w_j-w_{j+1}]\left[ v\left( c^k_{j+1}\right) -v\left( c^{k+1}_{j+1}\right) \right] . \end{aligned} \end{aligned}$$

(A.19)

The first term on the right-hand side of (A.19) is the maximal utility of type k when the proposer is of this type. Because type $k$’s adjacent upward self-selection constraint binds at type $(k+1)$’s selfishly optimal bundle, the second term is the utility that type k receives when the proposer is of type $k+1$. Because someone of type k can do no better than with its own proposal, the difference between the first and second terms is nonnegative. The final term is also nonnegative because any selfishly optimal consumption schedule is nondecreasing in type. Hence, the entire right-hand side of (A.19) is nonnegative, which establishes (A.13).

If $i>k$, we can use a similar argument to show that the utility that type i receives with type $k$’s selfishly optimal allocation is greater than or equal to the utility that i receives with type $(k-1)$’s selfishly optimal allocation. This argument uses (13) instead of (12). $\square$