Escape from Europe: a calculus of consent model of the origins of liberal institutions in the North American colonies

Abstract

The migration out of Europe and the establishment of North American colonies presents us with a great puzzle: why did the colonists establish democratic forms of governance? Considering that early democratic colonies appeared even before philosophical works such as those of Locke and Montesquieu were written, it is difficult to make the case that ideology was the driving factor. We show that the calculus of consent model proposed by Buchanan and Tullock (The calculus of consent, Liberty Fund, Indianapolis, 1962) offers a simple but subtle solution this puzzle. Because migrants formed much more homogeneous communities, and because, thanks to the large geographical expanse, the inter-jurisdictional externalities were small, the efficient level of consensus within each colony was much greater than in Europe, and the scope of efficient centralized decision-making was much smaller. Hence, a structure of decentralized democratic communities emerged as the efficient outcome.

Appendix

The simplest case of decreasing external costs and increasing decision-making costs is when the marginal costs, $\delta E/\delta n$ and $\delta D/\delta n$, are proportional to the current levels of the costs, $E$ and $D$:

$$\left\{ {\begin{array}{*{20}l} {\delta E\left( n \right) = - \frac{1}{h}E\left( n \right) \delta n} \\ {\delta D\left( n \right) = \frac{1}{g} D\left( n \right) \delta n} \\ \end{array} } \right.$$

$h$ and $g$ are positive. This means: (a) in terms of external costs, we have diminishing returns from increasing the deciding group; and (b) in terms of decision-making costs, in line with Olson’s (1965) logic of collective action, the bigger the group, the less likely it is that one can personally affect the decision of the group, and, hence, it is increasingly more difficult to motivate people to become involved in the decision making process or to seek a convergence to complete consensus.

Under this assumption, we find the following mathematical forms for the external costs and for the decision-making costs:

$$E\left( n \right) = E_{0} \exp \left( { - \frac{n}{h}} \right)$$

$$D\left( n \right) = D_{0} \exp \left( {\frac{n}{g}} \right)$$

where $h$ and $g$ are two measures of homogeneity, and $E_{0}$ and $D_{0}$ are the costs under pure authoritarian rule (i.e. when $n \to 0$). The model is entirely agnostic with respect to why homogeneity may change, e.g. due to changes of the group size or just of the internal composition, and, hence, can be applied to a wide range of cases.

The equilibrium decision rule, minimizing total cost, $E + D$, is:

$$n^{*} = - \frac{gh}{g + h}\log \left( {\frac{{D_{0} h}}{{E_{0} g}}} \right)$$

Proof of Theorem 1

$$\frac{{dn^{*} }}{{dE\left( {n^{*} } \right)}} = \frac{{\partial n^{*} }}{{\partial E_{0} }}\frac{1}{{\left( {\frac{dE}{{dE_{0} }}} \right)_{{n^{*} }} }} + \frac{{\partial n^{*} }}{\partial h}\frac{1}{{\left( {\frac{dE}{dh}} \right)_{{n^{*} }} }} = \frac{{gh^{2} }}{{\left( {g + h} \right)n^{*} E}} > 0$$

$\square$

Proof of Theorem 2

$$\frac{{dn^{*} }}{{dD\left( {n^{*} } \right)}} = \frac{{\partial n^{*} }}{{\partial D_{0} }}\frac{1}{{\left( {\frac{dD}{{dD_{0} }}} \right)_{{n^{*} }} }} + \frac{{\partial n^{*} }}{\partial g}\frac{1}{{\left( {\frac{dD}{dg}} \right)_{{n^{*} }} }} = - \frac{{g^{2} h}}{{\left( {g + h} \right)n^{*} D}} < 0$$

$\square$

Constants $h$ and $g$ are not independent. As discussed in the main text, homogeneity manifests itself both in the realm of external costs and in the realm of decision-making costs. Let $g = \alpha h$, where $\alpha$ is a positive scaling constant accounting for the possibly different strength with which homogeneity impacts external and decision-making costs. Consequently,

$$n^{*} = - \frac{\alpha }{1 + \alpha }h\log \left( {\frac{{D_{0} }}{{\alpha E_{0} }}} \right)$$

Lemma 1

The external costs curve always starts above the decision-making curve.

Proof

The level of consensus is always positive, $n^{*} > 0$, which implies that $E_{0} > D_{0} .$$\square$

Lemma 2

The scaling parameter$\alpha$is always greater than$\frac{{D_{0} }}{{E_{0} }}$.

Proof

$n^{*} = 0 \Rightarrow \forall h, \alpha = \frac{{D_{0} }}{{E_{0} }}$ The level of consensus is always positive, $n^{*} > 0$, which implies that $\alpha > \frac{{D_{0} }}{{E_{0} }}$. This is illustrated in Fig. 5.$\square$

Fig. 5

Graphical illustration of Theorem 3

Proof of Theorem 3

$$\frac{{\partial n^{*} }}{\partial h} = - \frac{\alpha }{1 + \alpha }\log \left( {\frac{{D_{0} }}{{\alpha E_{0} }}} \right) > 0$$

$\square$

This is always positive because of the Lemmas 1 and 2. Hence, as homogeneity increases, the efficient level of consensus also increases. As shown by Fig. 5, the sensitivity of the efficient level of consensus with respect to homogeneity increases as a function of the scaling factor $\alpha$.