Cross-border shopping and the Atkinson–Stiglitz theorem

Abstract

We introduce cross-border shopping and indirect tax competition into a model of optimal taxation. The Atkinson–Stiglitz result that indirect taxation cannot improve the efficiency of information-constrained tax-transfer policies, and that indirect taxes should not be differentiated across goods, is shown to hold in this case even if countries are asymmetric. However, if the tax system must contain indirect taxation, differentiated indirect tax rates arise in the equilibrium and restricting differentiated indirect taxation can be welfare-increasing.

Appendix: Derivation of key expression (5)

Consider the problem of maximizing (2) s.t. (3) and (4). The Lagrangian is

where $v_{h}^{i}=v_{h}^{i} ( f ( q_{z}^{i},q_{z}^{-i},c_{h}^{i} ) ,\frac{y_{h}^{i}}{w_{h}^{i}} )$, and $\hat{v}_{h}^{i}=v_{h}^{i} ( f ( q_{z}^{i},q_{z}^{-i},c_{l}^{i} ) ,\frac{y_{l}^{i}}{w_{h}^{i}} )$ denotes the utility of a mimicker. The government budget constraint can be simplified using

$$\sum\limits_{j=h,l}{A}^{-i} {\lambda}_{j}^{-i}{t}_{z}^{i}{z}_{j}^{-i,i}- \sum\limits_{j=h,l}{\lambda}_{j}^{i}{A}^{i}{t}_{z}^{i}{z}_{j}^{i,-i}=\frac{t_{z}^{i} ( t_{z}^{-i}-t_{z}^{i} ) }{a}. $$

We leave out the first-order conditions with respect to $y_{l}^{i}$, $y_{h}^{i}$, μ and γ and focus on those with respect to $c_{l}^{i}$, $c_{h}^{i}$ and $q_{z}^{i}$:

(6)

(7)

(8)

These conditions implicitly define the best response and must be, together with the counterparts for country −i, fulfilled in equilibrium. We multiply (6) by $z_{l}^{i,i}$ and (7) by $z_{h}^{i,i}$ and add them to (8) to find

By Roy’s Identity the first two terms in the first line add to zero. The last three terms in the second line also equal zero. Given identical transport costs high and low productivity individuals buy the same quantity abroad, and low productivity individuals and mimickers consume the same quantities of good z such that the last line equals zero. Using the Slutsky equation for z and x, this leads to (5).

Proof of Proposition 1

First, with $t_{x}^{i}=0$ the second term of F ⁱ in (5) disappears. Next, $t_{z}^{i}=t_{z}^{-i}=0$ is compatible with (5) and the corresponding expression F ⁻ⁱ. We now show that no other combination of tax rates can be an equilibrium. If $t_{z}^{i}<0$ in (5), then F ⁱ>0, contradicting (5). The same follows from F ⁻ⁱ for $t_{z}^{-i}<0$. Consider now $t_{z}^{i}>t_{z}^{-i}$, such that $z_{j}^{-i,i}=0$. This implies

(9)

which is only fulfilled for $t_{z}^{i}=0$ and therefore $t_{z}^{-i}<0$. Since neither country subsidizes z, an equilibrium with $t_{z}^{i}>t_{z}^{-i}$ can be ruled out. The combination with $t_{z}^{i}$ and $t_{z}^{-i}$ interchanged is analogous, such that an equilibrium with $t_{z}^{i}<t_{z}^{-i}$ cannot exist. Finally, $t_{z}^{i}=t_{z}^{-i}>0$ can be ruled out, since it implies (9), which requires $t_{z}^{i}=0$. Thus, only $t_{z}^{i}=t_{z}^{-i}=0$ is compatible with an equilibrium. □

Proof of Proposition 2

(i) Consider the case $t_{z}^{-i}\leq t_{z}^{i}$. Expression (5) for country i reduces to

(10)

where $t_{k}^{i}\equiv\tau_{k}^{i}q_{k}^{i}$. Consider now the Hicksian demands and their properties. We know that $\sum_{j=h,l}\lambda_{j}^{i}q_{z}^{i}\frac{\partial\tilde{z}_{j}^{i}}{\partial q_{z}^{i}}+\sum_{j=h,l}{}\lambda_{j}^{i}q_{x}^{i}\frac {\partial \tilde{x}_{j}^{i}}{\partial q_{z}^{i}}=0$. Multiplying this expression by $\tau_{z}^{i}$, using (10), and rearranging yields

(11)

This implies $\tau_{z}^{i}<\tau_{x}^{i}$ and therefore $t_{z}^{i}<t_{x}^{i}$, since $\sum_{j=h,l}\frac{\partial\tilde {x}_{j}^{i}}{\partial q_{z}^{i}}>0$. Since $t_{x}^{i}=t_{x}^{-i}=t_{x}>0$ and $t_{z}^{-i}\leq t_{z}^{i}$, country −i chooses a lower tax for good z as well. (ii) For symmetric countries we have A ⁱ=A ⁻ⁱ=A, $\alpha_{j}^{i}=\alpha_{j}^{-i}=\alpha_{j}$, and $\lambda_{j}^{i}=\lambda_{j}^{-i}=\lambda_{j}$. In a symmetric equilibrium, we have $z_{j}^{-i,i}=0$. From (5) we get $\sum_{j=h,l}A\lambda_{j}t_{z}\frac{\partial\tilde {z}_{j}^{i}}{\partial q_{z}}+\sum_{j=h,l}A\lambda_{j}t_{x}\frac{\partial\tilde{x}_{j}^{i}}{\partial q_{z}}-\frac{t_{z}}{a}=0$, which requires t _z >0 and corresponds to (10). Following the argumentation in part (i), we therefore have t _x >t _z . For a→∞, $\frac{t_{z}}{a}$ →0, and therefore t _z →t _x . □

Proof of Proposition 4

Consider the case $t_{x}^{b}=t_{x}^{s}=t_{x}>0$, A ^b>A ^s and $\lambda_{j}^{i}=\lambda_{j}^{-i}=\lambda _{j}$. Assume for contradiction that in equilibrium $t_{z}^{b}<t_{z}^{s}$. Expression (5) for the big country is

(12)

where again $t_{k}^{i}\equiv\tau_{k}^{i}q_{k}^{i}$. The properties of the Hicksian demands imply $\sum_{j=h,l}\lambda _{j}\*q_{z}^{b}\frac{\partial\tilde{z}_{j}^{b}}{\partial q_{z}^{b}}+\sum_{j=h,l}{}\lambda _{j}q_{x}^{b}\frac{\partial\tilde{x}_{j}^{b}}{\partial q_{z}^{b}}=0$. Multiplying this with $\tau_{z}^{b}$ and combining it with (12) yields

(13)

For the small country as the high tax country, we can use (11) and get

(14)

By (14), $\tau_{x}^{s}-\tau_{z}^{s}=\Delta>0$, where Δ is a constant. By (13), as $\frac{1}{A^{b}}\rightarrow0$, $\tau_{z}^{b}\rightarrow\tau _{x}^{b}$, since $\sum_{j=h,l}{}\lambda_{j}\frac{\partial \tilde{x}_{j}^{b}}{\partial q_{z}^{b}}>0$. For $\frac{1}{A^{b}}$ sufficiently small, this implies $\tau_{x}^{b}-\tau_{z}^{b}<\Delta$. Since $\tau_{x}^{b}=\tau_{x}^{s}$, this also implies $\tau_{z}^{b}>\tau_{z}^{s}$. But this contradicts $t_{z}^{b}<t_{z}^{s}$. □

Proof of Proposition 5

If $t_{x}^{i}>0$ and $t_{x}^{-i}=0$, (5) now differs between country i and country −i,

(15)

(16)

The following constellations are potential equilibria: (i) $t_{z}^{-i}\geq t_{z}^{i}=0$, (ii) $t_{z}^{i}=t_{z}^{-i}>\nobreak0$, (iii) $t_{z}^{i}>t_{z}^{-i}=0$, (iv) $t_{z}^{-i}>t_{z}^{i}>0$, (v) $t_{z}^{i}<0$, or $t_{z}^{-i}<0$, (vi) $t_{z}^{i}>t_{z}^{-i}>0$. Substituting these possibilities into (15) and (16) shows that only (vi) is compatible with the equilibrium. This constellation $t_{z}^{i}>t_{z}^{-i}>0$ reduces (15) to $\sum_{j=h,l}{A}^{i}{\lambda}_{j}^{i}{t}_{z}^{i}\frac{\partial\tilde{z}_{j}^{i}}{\partial q_{z}^{i}}+\sum_{j=h,l}{\lambda }_{j}^{i}{A}^{i}{t}_{x}^{i}\frac{\partial\tilde{x}_{j}^{i}}{\partial q_{z}^{i}}-\frac{t_{z}^{i}}{a}=0$, implying $t _{x}^{i}>t_{z}^{i}$. □