On optimal enforcement in international crime setting

Abstract

National and international criminal courts often choose to focus prosecutions on the heads of organizations that commit international crimes. In this article we consider a game between a law enforcement authority and a head of a criminal organization who decides on his level of personal exposure to crime and the number of individual criminals he recruits. Our results highlight that, depending on the level of social harm and detection costs, optimal enforcement does not always imply concentrating enforcement resources on the head of the organization and may involve investing resources in detecting and sanctioning individual criminals who execute the crime for the head.

Appendix A

The point $\displaystyle \left( n^{BR}, e^{BR} \right) = \left( \frac{2(g-pf)^2}{9(1+qs)}, \frac{g - pf}{3(1+qs)}\right)$ is candidate and the Hessian matrix is :

$$\begin{aligned}&H_S=\begin{pmatrix} \left[ S_{nn}(n_0,e_0)\right] &{} \left[ S_{ne} (n_0,e_0)\right] \\ \left[ S_{en}(n_0,e_0)\right] &{} \left[ S_{ee} (n_0,e_0)\right] \end{pmatrix} = \begin{pmatrix} \left[ \frac{-1}{e}\right] &{} \left[ -1-qs+\frac{n}{e^2}\right] \\ \left[ \frac{1}{2e^2}\right] &{} \left[ \frac{-n}{e^3}\right] \end{pmatrix} \\&H_S= \begin{pmatrix} \left[ \frac{-3(1+qs)}{g-pf}\right] &{} \left[ -1-qs+\left( \frac{2(g-pf)^2}{9(1+q)}\right) \times \left( \frac{3(1+qs)}{g-pf}\right) \right] \\ \left[ \frac{1}{2}\times \left( \frac{3(1+qs)}{g-pf}\right) ^2\right] &{} \left[ \left( \frac{-2(g-pf)^2}{9(1+qs)}\right) \times \left( \frac{3(1+qs)}{g-pf)}\right) ^3 \right] \end{pmatrix} \end{aligned}$$

The determinant is:

$$\begin{aligned} d_H = \left[ S_{nn}(n_0,e_0)\right] \times \left[ S_{ee}(n_0,e_0)\right] - \left[ (S_{en}(n_0,e_0))^2\right] \end{aligned}$$

After some simplification :

$$\begin{aligned} d_H = \frac{18(1+qs)^4}{(g-pf)^3} - \frac{9(1+qs)^3}{2(g-pf)^3} \end{aligned}$$

Therefore, $\displaystyle \frac{18(1+qs)^4}{(g-pf)^3} > \frac{9(1+qs)^3}{2(g-pf)^3}$ and $d_H>0$. Since $S_{nn}(n_0,e_0)<0$, then $(n_0,e_0)$ is a maximum.