Jesus and the Ratchet

Abstract

This paper offers a rational economic reading of the career of the historical Jesus of Nazareth. A simple two-period model shows that Jesus may have been trapped by the incentive to maintain and increase his following, which led him to raise his challenge to ever higher levels and confront an increasingly hostile opposition; in a kind of ratchet mechanism, at each stage stepping back to the previous one was foreclosed and exit was made more and more difficult, leading ultimately to his ominous death. A detailed review of the Gospel evidence, informed by recent historical research, provides good support to the predictions of the model. The approach outlined here is potentially applicable to other prophets and leaders of radical religious or political movements.

Appendix

Starting from the first-order conditions (4a) and (4b), some algebraic manipulation yields the following condition for $\widehat{{e_{2} }}$ > $\widehat{e}_{1}$:

$$ \frac{{1 - \delta (e_{2} )}}{{ - \delta^{{\prime }} (e_{2} )}} \le \frac{{n(e_{1} )}}{{n^{{\prime }} (e_{1} )}} + \frac{{n^{{\prime }} (e_{2} ) - n^{{\prime }} (e_{1} )}}{{ - \delta^{{\prime }} (e_{2} )n^{{\prime }} (e_{1} )}} $$

(5)

where specification of the inequality sign depends on whether c′(e) is constant or increasing. To ensure concavity of the maximization problem, we assume that n′(e) is decreasing [i.e. n(e) is strictly concave] if c′(e) is constant, while it may be constant if c′(e) is increasing [i.e. if c(e) is strictly convex]. For expositional convenience, and without loss of generality, suppose n(e) is strictly concave: then by definition $n(e_{1} )/n^{{\prime }} (e_{1} )$ > e ₁ . Also for convenience, call the two terms of the summation on the RHS of (5) A and B respectively, hence RHS = A + B. If e ₂  > e ₁ then B < 0. Therefore we have the two conditions e ₁  < A and RHS < A. Looking now at the LHS of (5), if the function 1 − δ(e ₂ ) is (at least weakly) convex, which implies that the attrition function δ(e ₂ ) is (at least weakly) concave, then by definition LHS ≤ e ₂ , and so must be the RHS. Conversely, if the attrition function is strictly convex, then both LHS and RHS of (5) must be greater than e ₂ . Using all these inequalities, the reader can verify that the two conditions stated above can be satisfied with any shape of the δ(e ₂ ) function.

If we now carry out the same exercise for the case e ₂  ≤ e ₁ , the sign of inequality (5) is reversed and B ≥ 0, so the two conditions become e ₁  < A and RHS ≥ A. It is straightforward to check that these conditions cannot be satisfied if LHS ≤ e ₂ but require LHS > e ₂ . This proves that a sufficient condition for $\widehat{e}_{2}$ > $\widehat{e}_{1}$ is that δ(e ₂ ) be (weakly) concave.

Turning now to comparative statics, to see the effect of marginal cost, re-write the first-order conditions (4a) and (4b) with linear costs, hence c′(e) = c. Differentiating these totally with respect to e ₁ , e ₂ , and c yields:

$$ \frac{{de_{1} }}{dc} = \frac{{ - \delta^{{\prime \prime }} (e_{2} )n(e_{1} ) + n^{\prime \prime } (e_{2} ) + \delta^{{\prime }} (e_{2} )n^{{\prime }} (e_{1} )}}{{\left[ {2 - \delta (e_{2} )} \right]n^{{\prime \prime }} (e_{1} )\left[ { - \delta^{\prime \prime } (e_{2} )n(e_{1} ) + n^{\prime \prime } (e_{2} )} \right] - \left[ {\delta^{{\prime }} (e_{2} )n^{{\prime }} (e_{1} )} \right]^{2} }} $$

(6)

$$ \frac{{de_{2} }}{dc} = \frac{{\left[ {2 - \delta (e_{2} )} \right]n^{\prime \prime } (e_{1} ) + \delta^{{\prime }} (e_{2} )n^{{\prime }} (e_{1} )}}{{\left[ {2 - \delta (e_{2} )} \right]n^{{\prime \prime }} (e_{1} )\left[ { - \delta^{\prime \prime } (e_{2} )n(e_{1} ) + n^{\prime \prime } (e_{2} )} \right] - \left[ {\delta^{{\prime }} (e_{2} )n^{{\prime }} (e_{1} )} \right]^{2} }} $$

(7)

The denominator of both equations is the Hessian determinant of the second-order conditions (not reported here), which is required to be positive for a maximum. Given our assumptions about the n(e) and δ(e ₂ ) functions, the numerator of (7) is certainly negative, while the sum of the first two terms in the numerator of (6) is required to be negative by the second-order conditions (which implies that δ″, if negative, must be sufficiently small in absolute value). Hence both total derivatives are negative. It is then straightforward to compute $d\widehat{K}_{1} /dc$ and $d\widehat{K}_{2} /dc$ from Eqs. (2) and (3) in the text and, using these results and the first-order conditions, find that the second is negative while the sign of the former is ambiguous.

Specifying $n(e) = \sqrt e$ and hence $\widehat{K}_{1} = \sqrt {e_{1} } - ce_{1}$, the derivatives de ₁ /dc and $d\widehat{K}_{1} /dc$ can be recalculated. Substituting into the latter $\left[ {2 - \delta (e_{2} )} \right]$ = $2c\sqrt {e_{1} }$ from the counterpart to FOC (4a), we find that a sufficient condition for $d\widehat{K}_{1} /dc$ > 0 is $(1 - c\sqrt {e_{1} } )$ ≤ 0, which implies $\widehat{K}_{1}$ ≤ 0. In turn, using again the above equality from (4a) to substitute into the above expression for $\widehat{K}_{1}$, we find that a necessary and sufficient condition for $\widehat{K}_{1}$ ≤ 0 is $\delta (e_{2} )$ ≤ 0 at the optimal solution.

To see the effect of the attrition rate, re-write the FOCs (4a) and (4b) using the linear function $\delta (e_{2} ) = 1 - \alpha e_{2}$, with 0 ≤ α ≤ 1. Differentiating these totally with respect to e ₁ , e ₂ , and α yields:

$$ \frac{{de_{1} }}{d\alpha } = \frac{{n^{{\prime }} (e_{1} )\left\{ {\alpha n(e_{1} ) - e_{2} [n^{\prime \prime } (e_{2} ) - c^{\prime \prime } (e_{2} )]} \right\}}}{{[(1 + \alpha e_{2} )n^{\prime \prime } (e_{1} ) - c^{\prime \prime } (e_{1} )][n^{\prime \prime } (e_{2} ) - c^{\prime \prime } (e_{2} )] - [\alpha n^{{\prime }} (e_{1} )]^{2} }} $$

(8)

$$ \frac{{de_{2} }}{d\alpha } = \frac{{ - n(e_{1} )[\left( {1 + \alpha e_{2} } \right)n^{\prime \prime } (e_{1} ) - c^{\prime \prime } (e_{1} )] + \alpha e_{2} [n^{'} (e_{1} )]^{2} }}{{[(1 + \alpha e_{2} )n^{\prime \prime } (e_{1} ) - c^{\prime \prime } (e_{1} )][n^{\prime \prime } (e_{2} ) - c^{\prime \prime } (e_{2} )] - [\alpha n^{'} (e_{1} )]^{2} }} $$

(9)

The denominator of both equations is the Hessian determinant of the second-order conditions (recalculated for this specification), which is required to be positive for a maximum. The bracketed expression in the numerator of (8) is required to be negative by the second-order conditions, as is the expression in the first brackets of the numerator of (9). So both total derivatives are positive. Finally, it is straightforward to compute $d\widehat{K}_{1} /d\alpha$ and $d\widehat{K}_{2} /d\alpha$ from Eqs. (2) and (3) in the text and, using de ₁ /dα and the FOCS, find that the former is negative and the latter positive.