A model of state secularism

Abstract

This paper posits a minimalist state interested in taxation and religion and explores the conditions conducive to the separation of state and religion. It shows that a ruler is secular and does not favour his religion as the state religion if he is absolutely tolerant, he faces a homogeneous, co-religionist society, and/or punishing violations of the state’s religious policy is prohibitively costly. Secular rulers are accordingly classified into three types: innately, coincidentally, and instrumentally secular. In the short run, individuals are equally well-off under different secular regimes. But among rulers, instrumentally secular rulers are relatively worse-off.

Appendices

Appendix 1: Proof of Proposition 1 (a)–1 (d)

The choice problem facing individuals belonging to religion $k$ can be expressed as:

$$\begin{aligned} \mathrm{{max}}_{r_k } \varPi _k =\pi _k \left( {r_k } \right) -0.5\delta _0 \cdot \left( {r_k -r_k^*} \right) ^{2};\quad \varPi _k^{\prime } \left( {r_k^*} \right) =\pi _k^{\prime } \left( {r_k^*} \right) =0 \end{aligned}$$

(8)

Under a ruler, the individual choice problem transforms to:

$$\begin{aligned} \mathrm{{max}}_{r_k } \varPi _k =\pi _k \left( {r_k } \right) -\pi _s \left( {r_s^*} \right) \cdot t_k -0.5\delta _0 \cdot \left( {r_k -r_k^*} \right) ^{2} \end{aligned}$$

(4)

The first and second order conditions follow, where $r_{ks}$ is denoted by $r_s$ to simplify expressions:

$$\begin{aligned}&\partial \varPi _k /\partial r_k =\pi _k^{\prime } -\pi _s \left( {r_s^*} \right) \cdot t_1 \cdot \left( {r_k -r_s } \right) -\delta _0 \cdot \left( {r_k -r_k^*} \right) =0\end{aligned}$$

(9)

$$\begin{aligned}&\partial ^{2}\varPi _k /\partial r_k ^{2}=\pi _k^{{\prime }{\prime }} -\pi _s \left( {r_s^*} \right) \cdot t_1 -\delta _0 <0 \end{aligned}$$

(10)

The second order condition is satisfied because all the terms are negatively signed. Individual reaction curve and its slope can be expressed as follows:

$$\begin{aligned}&r_k =r_s +\frac{\pi _k^{\prime } -\delta _0 \cdot \left( {r_k -r_k^*} \right) }{t_1 \pi _s \left( {r_s^*} \right) }=r_s +f\left( {r_k } \right) \end{aligned}$$

(11)

$$\begin{aligned}&\left. {\partial r_s /\partial r_k } \right| _{individual} =1-\frac{\pi _k^{{\prime }{\prime }} -\delta _0 }{t_1 \pi _s \left( {r_s^*} \right) }>1 \end{aligned}$$

(12)

Both $sgn\left( {\pi _k^{\prime } } \right)$ and $sgn\left( {-\delta _0 \cdot \left( {r_k -r_k^*} \right) } \right)$ are negative (positive) for $r_k >\left( < \right) r_k^*$. The individual reaction curve passes through $\left( {r_k^*,r_k^*} \right)$. Its slope is always positive and, in fact, greater than one because the numerator of the second term in Eq. (12) is always negative, while the denominator is a fixed positive quantity. So, individual reaction curves are monotonically increasing.

Now let us turn to the ruler, who faces as many choice problems as there are communities.¹⁰ The first and second order conditions for the ruler are as follows:

$$\begin{aligned}&\partial \varPi _s /\partial r_s =\pi _s \left( {r_s^*} \right) \cdot \left( {-t_1 \cdot \left( {r_k -r_s } \right) } \right) -\theta _1 \cdot \left( {r_s -r_k^*} \right) -\lambda _0 \cdot \left( {r_s -r_s^*} \right) =0\,\forall k\end{aligned}$$

(13)

$$\begin{aligned}&\partial ^{2}\varPi _s /\partial r_s ^{2}=\pi _s \left( {r_s^*} \right) \cdot t_1 -\theta _1 -\lambda _0 <0\,\forall k \end{aligned}$$

(14)

The second order condition is satisfied for ruler if $\theta _1 +\lambda _0 >t_1 \pi _s \left( {r_s^*} \right)$, which holds if the ruler is sufficiently intolerant $\left( {\lambda _0 } \right)$ and/or cost of extracting punitive taxes $\left( {\theta _1 } \right)$ is sufficiently high. Even when the ruler is absolutely tolerant, $\lambda _0 =0, \theta _1 +\lambda _0 >t_1 \pi _s \left( {r_s^*} \right)$ can be satisfied because $\theta _1 >0$ always holds (see Remark 3). Ruler’s reaction function and its slope are given by:

$$\begin{aligned}&r_s =\frac{-t_1 \pi _s \left( {r_s^*} \right) r_k +\left( {\theta _1 r_k^*+\lambda _0 r_s^*} \right) }{-t_1 \pi _s \left( {r_s^*} \right) +\theta _1 +\lambda _0 }=g\left( {r_k } \right) \end{aligned}$$

(15)

$$\begin{aligned}&\left. {\frac{\partial r_s }{\partial r_k }} \right| _{ruler} =\frac{-t_1 \pi _s \left( {r_s^*} \right) }{-t_1 \pi _s \left( {r_s^*} \right) +\theta _1 +\lambda _0 } \end{aligned}$$

(16)

The slope of ruler’s reaction curve is constant and strictly negative. However, if $\theta _1$ or $\lambda _0 >> t_1 \pi _s \left( {r_s^*} \right)$ then the slope is zero. The ruler’s reaction curve passes through $(\bar{r}_{s}, \bar{r}_{s})$, where $\bar{r}_{s}=(\theta _{1}r_{k}^{*}+\lambda _{0}r_{s}^{*})/(\theta _{1}+\lambda _{0})$

There exists a unique equilibrium because of the monotonicity of individual and ruler’s reaction curves. The equilibrium is stable as well if the absolute value of the slope of ruler’s reaction curve is less than that of individual’s reaction curve.

$$\begin{aligned} \left| {\left( {\left. {\frac{\partial r_s }{\partial r_k }} \right| _{ruler} } \right) } \right| <\left. {\frac{\partial r_s }{\partial r_k }} \right| _{individual} \end{aligned}$$

(17)

The above condition holds if $\theta _1 +\lambda _0 >2t_1 \pi _s \left( {r_s^*} \right)$:

$$\begin{aligned} \theta _1 +\lambda _0 >2t_1 \pi _s \left( {r_s^*} \right) \Rightarrow 1-\frac{\pi _k^{{\prime }{\prime }} -\delta _0 }{t_1 \pi _s \left( {r_s^*} \right) }>1>\frac{t_1 \pi _s \left( {r_s^*} \right) }{-t_1 \pi _s \left( {r_s^*} \right) +\theta _1 +\lambda _0 } \end{aligned}$$

(18)

We can conclude that given Remarks 1, 2 (Conversion Constraint), and 3 ($\theta _1 +\lambda _0 >2t_1 \pi _s \left( {r_s^*} \right)$) the equilibrium specified in Eq. (19) not only exists but is also stable and unique.

$$\begin{aligned} \left( {\widehat{r_k }=g\left( {\widehat{r_k }} \right) +f\left( {\widehat{r_k }} \right) ,\widehat{r_s }=g\left( {\widehat{r_k }} \right) } \right) \end{aligned}$$

(19)

Now we can check what happens as parameters change and establish the specific claims in propositions.

• (a1) $\lambda _0 \in \left( {0,\infty } \right)$: Ruler is not absolutely intolerant.

  • (a11) $\delta _0 \in \left[ {0,} \right. \left. \infty \right)$: The solution is given by the intersection of the reactions curves given by Eqs. (11) and (15).

  • (a12) $\delta _0 \rightarrow \infty : \widehat{r_k }=r_k^*$ from Eq. (11). $\widehat{r_s }$ is obtained by substituting $r_k =r_k^*$ in Eq. (15).

• (a2) $\lambda _0 \rightarrow \infty$: Ruler is absolutely intolerant. From Eq. (15), $\widehat{r_s }=r_s^*$.

  • (a21) $\delta _0 \in \left[ {0,} \right. \left. \infty \right) :\widehat{r_k}$ is obtained by substituting $r_s =r_s^*$ in Eq. (11).

  • (a22) $\delta _0 \rightarrow \infty$: Individual chooses optimal effort level $\widehat{r_k }=r_k^*$ as per Eq. (11).

• (b) $\lambda _0 =0$: Ruler is absolutely tolerant. From Eqs. (11) and (15) at equilibrium $\widehat{r_s }=\widehat{r_k }=r_k^*$.

• (c) Substituting $\theta _1 \rightarrow \infty$ in Eq. (15) gives $\widehat{r_s }=r_k^*$. Substituting $r_s =r_k^*$ in Eq. (11) gives $\widehat{r_k }=r_k^*$.

• (d) $r_s^*=r_k^*$: From Eq. (15) $\widehat{r_s }=r_k^*$, which when substituted in Eq. (11) gives $\widehat{r_k }=r_k^*$.

A few additional observations are in order. Actually, equilibrium exists and is stable as well if $\underline{\xi }t_1 \pi _s \left( {r_s^*} \right) \le \theta _1 +\lambda _0$, where:

$$\begin{aligned} \underline{\xi }=1+t_1 \pi _s \left( {r_s^*} \right) \left( {t_1 \pi _s \left( {r_s^*} \right) -\left( {\pi _k^{{\prime }{\prime }} -\delta _0 } \right) } \right) ^{-1}\in \left( {1,2} \right) \end{aligned}$$

(20)

$\left( {\pi _k^{{\prime }{\prime }} -\delta _0 } \right) <0\,\,\forall r_k \Rightarrow \underline{\xi }\in \left( {1,2} \right) .$ The choice of $\xi =2$ in Remark 3 $\left( \theta _1 +\lambda _0 >2t_1 \pi _s \left( {r_s^*} \right) \right)$ satisfies the condition for stability irrespective of functional forms involved.

If $\underline{\xi }t_1 \pi _s \left( {r_s^*} \right) >\theta _1 +\lambda _0 >t_1 \pi _s \left( {r_s^*} \right)$, the condition for a stable equilibrium is not satisfied. The system can move to stable equilibrium in three plausible ways: (a) ruler becomes more intolerant ($\lambda _0$ increases), (b) citizens offer more resistance making taxation expensive ($\theta _1$ increases), or (c) ruler reduces the punitive tax rate ($t_1$ decreases). $\square$

Appendix 2: Proof of Corollary 1

Using Eqs. (2) through (4) and Proposition 1, individual pay-off under a secular regime can be expressed as follows:

$$\begin{aligned}&\displaystyle \varPi _k \left( {\widehat{r_k }|\widehat{r_s }} \right) \!=\!\pi _k \left( {\widehat{r_k }} \right) \!-\!\pi _s \left( {r_s^*} \right) \cdot t_0 \!-\!0.5t_1 \cdot \pi _s \left( {r_s^*} \right) \cdot \left( {\widehat{r_k }\!-\!\widehat{r_s }} \right) ^{2}\!-\!0.5\delta _0 \!\cdot \! \left( {\widehat{r_k }\!-\!r_k^*} \right) ^{2}&\end{aligned}$$

(21)

$$\begin{aligned}&\displaystyle \varPi _k \left( {\widehat{r_k }|\widehat{r_s }} \right) =\pi _k \left( {\widehat{r_k }} \right) -\pi _s \left( {r_s^*} \right) \cdot t_0 -\Delta I=I-\Delta I&\end{aligned}$$

(22)

$\widehat{r_k }=\widehat{r_s }=r_k^*$ holds under each of the three secular regimes. $I\left( {r_k^*|r_k^*} \right) =constant>0$ and $\Delta I\left( {r_k^*|r_k^*} \right) =0$. So, in a one period model, individuals will be equally well-off under each of the three secular regimes.

Using Eqs. (3) and (5) through (7) and Proposition 1, a secular ruler’s pay-off can be expressed as:

$$\begin{aligned}&\varPi _s \left( {\widehat{r_s }|\widehat{r_k }} \right) \nonumber \\&\quad =\!t_0 \sum \nolimits _k \pi _s \left( {r_s^*} \right) \!-\!\theta _0\!+\!0.5 \sum \nolimits _k \left( {t_1 \pi _s \left( {r_s^*} \right) \cdot \left( {\widehat{r_k }\!-\!\widehat{r_s }} \right) ^{2}\!-\!\theta _1 \left( {\widehat{r_s }\!-\!r_k^*} \right) ^{2}\!-\!\lambda _0 \left( {\widehat{r_s }\!-\!r_s^*} \right) ^{2}} \right) \end{aligned}$$

(23)

$$\begin{aligned}&\varPi _s \left( {\widehat{r_s }|\widehat{r_k }} \right) =t_0 \sum \nolimits _k \pi _s \left( {r_s^*} \right) -\theta _0 +\Delta S=S+\Delta S \end{aligned}$$

(24)

$S\left( {r_k^*|r_k^*} \right) =constant>0$ and same for all regimes. But $\Delta S\left( {r_k^*|r_k^*} \right) =0$ for innately and coincidentally secular rulers, whereas $\Delta S\left( {r_k^*|r_k^*} \right) =- \sum \nolimits _k \lambda _0 \left( {r_k^*-r_s^*} \right) ^{2}<0$ for instrumentally secular rulers, which completes the proof. $\square$