Oates’ decentralization theorem with imperfect household mobility

Abstract

This paper studies how Oates’ trade-off between centralized and decentralized public good provision is affected by changes in households’ mobility. We show that an increase in household mobility favors centralization. This results from two effects. First, mobility increases competition between jurisdictions in the decentralized régime, resulting in lower levels of public good provision. Second, while tyranny of the majority creates a gap between social welfare in different jurisdictions in the centralized régime, mobility allows agents to move to the majority jurisdiction, raising average social welfare. Our main result is obtained in a baseline model where jurisdictions first choose taxes, and households move in response to tax levels. We show that the result is robust to changes in the objective function and the strategic variable of local governments.

Appendix

Proof of Proposition 1: We first prove that $(\tau ^*, \tau ^*)$ is a pure strategy Nash equilibrium of the taxation game. Suppose that jurisdiction $2$ chooses $\tau ^*$. Using Eq. (8), we compute the marginal effect of an increase in $\tau _1$ on $n_1$:

$$\begin{aligned} \frac{\partial n_1}{\partial \tau _1} = \frac{n_1 (U_g(G,1-\tau _1)-(U_g(G, 1-\tau ^*)) -U_e(G,1-\tau _1)}{(U_g(G,1-\tau _1)-U_g(G,1-\tau ^*))(\tau ^*- \tau _1) + 2 \lambda }. \end{aligned}$$

Notice that the denominator is always positive as $(U_g(G,1-\tau _1)-U_g(G,1-\tau ^*))(\tau ^*-\tau _1) >0$. Next, we compute the derivative of the resident’s utility with respect to an increase in taxes:

$$\begin{aligned} \frac{\partial U_1}{\partial \tau _1} = n_1 U_g(G,1-\tau _1) - U_e(G,1-\tau _1) +(\tau _1-\tau ^*)U_g(G,1-\tau _1) \frac{\partial n_1}{\partial \tau _1}. \end{aligned}$$

Developing, we find that the sign of $\frac{\partial U_1}{\partial \tau _1}$ is the same as the sign of

$$\begin{aligned} 2 \lambda (n_1 U_g(G,1-\tau _1) - U_e(G,1-\tau _1)) + n_1 U_g(G,1- \tau ^*) U_g(G,1- \tau _1) (\tau ^*-\tau _1). \end{aligned}$$

If $\tau _1 < \tau ^*$, this expression is positive and $\frac{ \partial U_1}{\partial \tau _1} > 0$. If $\tau _1 > \tau ^*$, the expression is negative and $\frac{\partial U_1}{\partial \tau _1 }< 0$, showing that $\tau _1 = \tau ^*$ is a best response to $\tau _2 = \tau ^*$.

To show that there cannot be any other symmetric equilibrium, we compute $\frac{\partial U_1}{\partial \tau _1}$ along the diagonal when $\tau _1=\tau _2 = \tau$:

$$\begin{aligned} \frac{\partial U_1}{\partial \tau _1}|_{\tau _1=\tau _2=\tau } = U_g(G,1-\tau ) - U_e(G,1-\tau ). \end{aligned}$$

The only point at which $\frac{\partial U_1}{\partial \tau _1}=0$ is the point $\tau _1 = \tau _2 = \tau ^*$.

Proof of Proposition 2: We first verify that $(\tau ^{**}, \tau ^{**})$ is a pure strategy Nash equilibrium of the taxation game. As $U(\tau ^{**}, 1-\tau ^{**}) > U(\tau , 1-\tau )$ for any $\tau \ne \tau ^*$, if the other jurisdiction charges $\tau ^*$, any deviation to another tax rate $\tau$ induces a migration out of the jurisdiction, resulting in a utility

$$\begin{aligned} U(n_1 \tau , 1-\tau ) < U(\tau , 1-\tau ) < U(\tau ^{**}, 1-\tau ^{**}). \end{aligned}$$

Hence, when the other jurisdiction chooses tax rate $\tau ^{**}$, any deviation to $\tau \ne \tau ^{**}$ results in a loss of utility.

We now verify that $(\tau ^{**}, \tau ^{**})$ is the unique symmetric Nash equilibrium. To this end, compute first:

$$\begin{aligned} \frac{\partial U_1}{\partial \tau _1} = \frac{n_1 U_g(n_1 \tau _1, -\tau _1)- U_e(n_1 \tau _1,1-\tau _1)}{-\tau _1 U_g(n_1 \tau _1, 1-\tau _1)-\tau _2 U_g((2-n_1)\tau _2,1-\tau _2)+2 \lambda }, \end{aligned}$$

showing that the only tax level at which $\frac{\partial U_1}{ \partial \tau _1}$ is equal to zero along the diagonal is $\tau _1=\tau _2 = \tau ^{**}$.

Proof of Proposition 3: We first show that $(\hat{\tau }, \hat{\tau })$ is a pure strategy Nash equilibrium of the taxation game. Suppose that jurisdiction $2$ chooses $\tau _2 = \hat{\tau }$.

Consider first a strategy $\tau _1 \le \underline{\tau }$, namely a choice $\tau _1$ so low that $U_1 < U_2$ and $n_1 < 1$. We show that this choice is dominated by choosing $\tau _1 = \hat{\tau }$. Different cases have to be distinguished. First suppose that $\alpha \hat{\tau } < \tau _1$. Then

$$\begin{aligned} U_1&= \tau _1 + \alpha \hat{\tau } + (1-n_1) (\alpha \hat{\tau } - \tau _1) + v(1- \tau _1) \\&< \tau _1 + \alpha \hat{\tau } + v(1- \tau _1) \\&< (1+\alpha )\hat{\tau } + v(1- \hat{\tau }) \end{aligned}$$

where the last inequality is obtained because $\tau _1 < \hat{\tau } < \tau ^*$, so any increase in the tax rate increases $\tau + v(1- \tau )$.

Next, suppose that $\tau _1 \le \alpha \hat{\tau }$. Notice that, as $U_1 < U_2$, $\phi (\tau _1) < \phi (\hat{\tau })$ so that

$$\begin{aligned} U_1&= \tau _1(1- \alpha ) + v(1- \tau _1) + \alpha (\tau _1 + \hat{\tau }) + (1-n_1) (\alpha \hat{\tau } - \tau _1) \\&< \hat{\tau } (1- \alpha ) + v(1- \hat{\tau }) + \alpha (\tau _1 + \hat{\tau } ) + (1-n_1) (\alpha \hat{\tau } - \tau _1) \\&= (1+\alpha ) \hat{\tau } + v(1-\hat{\tau }) - n_1 (\alpha \hat{\tau } - \tau _1) - (1-\alpha ) \tau _1. \\&< (1+\alpha ) \hat{\tau } + v(1-\hat{\tau }), \end{aligned}$$

proving that choosing $\tau _1$ is dominated by choosing $\hat{\tau }$.

Next consider values of $\tau _1 > \underline{\tau }$. Compute

$$\begin{aligned} \frac{\partial n_1}{\partial \tau _1} = \frac{-v^{\prime }(1-\tau _1)+n_1(1-\alpha )}{2 \lambda - (\hat{\tau } + \tau _1)(1-\alpha )} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial U_1}{\partial \tau _1} = n_1 - v^{\prime }(1-\tau _1) + (\tau _1 - \alpha \hat{\tau }) \frac{\partial n_1}{\partial \tau _1}, \end{aligned}$$

so that $\frac{\partial U_1}{\partial \tau _1}$ is of the same sign as:

$$\begin{aligned} A&= [n_1 - v^{\prime }(1-\tau _1)][2 \lambda -(\hat{\tau } + \tau _1)(1-\alpha )] + (\tau _1 - \alpha \hat{\tau })[n_1(1-\alpha ) - v^{\prime }(1-\tau _1)] \\&= n_1 [2 \lambda -(1-\alpha )^2 \hat{\tau }] - v^{\prime }(1-\tau _1)[2 \lambda - (1-\alpha ) \hat{\tau } + \alpha (\tau _1 - \hat{\tau })]. \end{aligned}$$

If $\underline{\tau } < \tau _1 < \hat{\tau }$, $n_1 > 1$, $v^{\prime }(1-\tau _1) < v^{\prime }(1-\hat{\tau })$ and $[2 \lambda - (1-\alpha ) \hat{ \tau } + \alpha (\tau _1 - \hat{\tau })]< [2 - \lambda - (1-\alpha ) \hat{\tau }]$ , so that $A >0$ and $\frac{\partial U_1}{\partial \tau _1} > 0$. On the other hand, if $\tau _1 > \hat{\tau }$, $n_1 < 1$, $v^{\prime }(1-\tau _1) > v^{\prime }(1- \hat{\tau })$ and $[2 \lambda - (1-\alpha ) \hat{\tau } + \alpha (\tau _1 - \hat{\tau })]> [2 - \lambda - (1-\alpha ) \hat{\tau }]$, so that $A <0$ and $\frac{\partial U_1}{\partial \tau _1} <0$. Hence, $U_1$ attains its maximum at $\tau _1 = \hat{\tau }$.

In order to prove that there is no other symmetric equilibrium in the game, we compute

$$\begin{aligned} \frac{\partial U_1}{\partial \tau _1}|_{\tau _1=\tau _2 =\tau } = 1- v^{\prime }(1-\tau )+ (1-\alpha ) \tau \frac{(1-\alpha ) - v^{\prime }(1-\tau )}{2 \lambda - 2 \tau (1-\alpha )}. \end{aligned}$$

Hence, along the diagonal, $\frac{\partial U_1}{\partial \tau _1}=0$ if and only if $\tau = \hat{\tau }$.