### On law enforcement

**Matthew Martin**8/19/2014 05:28:00 PM

The crisis in Ferguson has prompted a national dialogue about law enforcement tactics and the unfair targeting of innocents through "tough-on-crime" policies like racial profiling, mandatory minimums, and criminal procedures that make it easier to convict. However, economics tells us that "tough-on-crime" tactics do not always maximize law and order. The unfair targeting of innocent people for criminal investigation through racial profiling and stop-and-frisk style tactics actually increases the incidence of criminality at the margins. Here's how.

### A Model in two parts

### 1 Modeling the incidence of criminality

We consider a model with a large number of households with preferences over consumption [$] C [$] and jail [$] J [$] according to [$] U=E_B\left[u\left(C\right)-J\right] [$] where [$] u [$] is strictly concave increasing in [$] C. [$] The household is endowed with lawful income of [$] y [$] and has the option of committing a burglary to steal [$] B [$] units of consumption. If the household is convicted, he serves jail time that yields [$] J [$] units of disutility (we can think of disutility from jail as being a function [$] j\left(s\left(B\right)\right) [$] where [$] s\left(B\right) [$] is a policy function prescribing sentences based on the magnitude [$] B [$] of the crime, and [$] j\left(\cdot\right) [$] as the utility function of jail time. However, we consider the extensive margin where [$] B [$] is fixed.)The judicial authority investigates a share [$] q [$] of the population and the investigation leads to a conviction rate of [$] r_1\lt 1 [$] among those who are investigated and committed the crime (that's a sensitivity of [$]r_1[$]), and [$] r_2\lt r_1 [$] among individuals who are investigated but innocent (that's a specificity of [$]1-r_2[$]). Therefore, the probability of being convicted given that the household commits the crime is [$] p_1=q_1r_1 [$] and the probability of conviction given hat the household does not commit the crime is [$] p_2=q_2r_2, [$] where [$]q_1,q_2[$] are the probabilities of investigating an innocent and guilty person respectively, which are functions of the total share of the population [$]q[$] that is investigated, as will be defined in section 2 below.

*[update: this paragraph has been edited to correct some of the notation*]

The household's budget constraint is [$] C\leq y+B [$] if he commits the crime, and [$] C\leq y [$] otherwise. The household choice has two regimes, one where he commits crimes with probability 1, and one where he commits crime with probability 0, where utility from the former is [$] u\left(y+B\right)-p_1 J [$] and the utility from the latter is [$] u\left(y\right)-p_2 J. [$]

Borrowing from the indivisible labor literature

^{1}and assuming a functional form for the utility function, we can rewrite the above in terms of a representative agent that chooses a probability of committing crime [$] \alpha [$] according to

\begin{align*}

\max_{\alpha}~ ln\left(C\right)-\alpha p_1J-\left(1-\alpha\right)p_2J&\\

subject~to~C\leq y+\alpha B&

\end{align*}

Solving yields [$$] \alpha^*=\frac{1}{J\left(p_1-p_2\right)}-\frac{y}{B} [$$] This model captures our intuitions about criminal justice. For example, it is plainly apparent from the solution that increasing penalties [$] J, [$] all else equal, reduces crime rates. It is often assumed that stepping up investigations--that is, targeting a larger share of the population for investigations--will result in a reduction in crime rates. To examine this we take the derivative of [$] \alpha [$] with respect to the investigation rate [$] q: [$] [$$] \frac{\partial \alpha}{\partial q}=\frac{J}{\left(J\left(p_1-p_2\right)\right)^2}\left(\frac{\partial p_2}{\partial q}-\frac{\partial p_1}{\partial q}\right) [$$] which is less than zero if and only if [$$] \frac{\partial p_1}{\partial q}>\frac{\partial p_2}{\partial q}. [$$]

### 2 Modeling the profiling decision

To understand that last derivative, we need a model of police profiling. Mixing models of heterogeneity with a representative agent framework can be problematic, but lets assume that utilities are such that this is valid. Assume that individuals are heterogeneous in such a way that their probability [$] \tilde{\alpha} [$] of committing a crime--as measured by a hypothetical social planner--is distributed according to a continuously differentiable distribution function [$] F\left(\tilde{\alpha}\right) [$] with support [$] \left[0,1\right]. [$] The judicial authority prioritizes investigations of individuals so that individuals with the highest [$] \tilde{\alpha} [$] probabilities are investigated first, followed by progressively lower probability types until they've exhausted their investigative resources--that is, until the share of the population being investigated equals the policy parameter [$] q. [$] Thus we can write [$$] q\equiv 1-F\left(\bar{\alpha}\right) [$$] where [$] \bar{\alpha} [$] is the lowest probability type to be investigated. Therefore, we have that\begin{align*}

p_1&=\underbrace{\left(1-F\left(\bar{\alpha}\right)\right)}_{q}\underbrace{\int^{1}_{\bar{\alpha}}\tilde{\alpha}f\left(\tilde{\alpha}\right)d\tilde{\alpha}}_{\alpha}r_1\\

p_1&=\underbrace{\left(1-F\left(\bar{\alpha}\right)\right)}_{q}\underbrace{\left(1-\int^{1}_{\bar{\alpha}}\tilde{\alpha}f\left(\tilde{\alpha}\right)d\tilde{\alpha}\right)}_{1-\alpha}r_2

\end{align*}

where [$] f\left(\tilde{\alpha}\right) [$] denotes the density function of [$] \tilde{\alpha} [$] and therefore the first derivative of [$] F\left(\tilde{\alpha}\right). [$] Hereafter we will write [$] E_\bar{\alpha} [$] to denote [$] \int^{1}_{\bar{\alpha}}\tilde{\alpha}f\left(\tilde{\alpha}\right)d\tilde{\alpha}, [$] which is the expected criminality of the population being investigated.

Thanks to the Leibniz rule, we can differentiate this to get

\begin{align*}

\frac{\partial p_1}{\partial q}&=E_\bar{\alpha}r_1+\bar{\alpha}\left(1-F\left(\bar{\alpha}\right)\right)r_1\\

\frac{\partial p_2}{\partial q}&=\left(1-E_\bar{\alpha}\right)r_2-\bar{\alpha}\left(1-F\left(\bar{\alpha}\right)\right)r_2

\end{align*}

Therefore, using the result derived in the first section, increasing enforcement decreases crime only if

\begin{equation}

E_\bar{\alpha}+\bar{\alpha}\left(1-F\left(\bar{\alpha}\right)\right)\gt \frac{r_2}{r_1+r_2} \label{conditions}

\end{equation}

We can now state two propositions.

## Proposition 1.

It is optimal to investigate everyone if and only if [$] E_0\geq \frac{r_2}{r_1+r_2}. [$]

Proof: Sufficiency follows immediately from \eqref{conditions} with [$] \bar{\alpha}=0. [$] Necessity follows from Proposition 2.

## Proposition 2.

If [$] E_0\lt\frac{r_2}{r_1+r_2}, [$] then there exists [$] q^*\gt 0 [$] such that [$] \frac{\partial \alpha}{\partial q}\geq 0 [$] for all [$] q\gt q^*, [$] with strict inequality whenever [$] f\left(\tilde{\alpha}\right)\gt 0. [$] That is, there exists a point beyond which further increasing enforcement actually increases crime.

Proof: Denote [$] G\left(\bar{\alpha}\right)\equiv E_\bar{\alpha}+\bar{\alpha}\left(1-F\left(\bar{\alpha}\right)\right). [$] Then [$] G\left(1\right)=1\gt\frac{r_2}{r_1+r_2} [$] because [$] E_1=1 [$] and [$] F\left(1\right)=1. [$] Moreover, we postulated that [$] G\left(0\right)=E_0\lt \frac{r_2}{r_1+r_2}. [$] Since [$] F\left(\bar{\alpha}\right) [$] is continuously differentiable, [$] G\left(\bar{\alpha}\right) [$] is continuous and by the intermediate value theorem there exists a nonempty set [$] A [$] such that for [$] \hat{\alpha}\in A [$] we have [$] G\left(\hat{\alpha}\right)=\frac{r_2}{r_1+r_2}. [$] Let [$] \bar{\alpha}=\min \hat{\alpha}\in A, [$] and [$] q^*=1-F\left(\bar{\alpha}\right), [$] then for all [$] \tilde{\alpha}\in \left[0,\bar{\alpha}\right) [$] we have [$] G\left(\tilde{\alpha}\right)\leq \frac{r_2}{r_1+r_2} [$] with strict inequality if [$] f\left(\bar{\alpha}\right)\gt 0. [$] Furthermore, note that [$] q\equiv 1-F\left(\tilde{\alpha}\right) [$] is monotonically decreasing in [$] \tilde{\alpha}, [$] and is one-to-one when [$] f\left(\tilde{\alpha}\right)\gt 0, [$] which implies that for [$] q\in \left[0,q^*\right) [$] we must have [$] G\left(\tilde{\alpha}\right)\leq \frac{r_2}{r_1+r_2} [$] with strict inequality when [$] f\left(\bar{\alpha}\right). [$] This concludes the proof.

So what do these propositions say about stop-and-frisk? We are compelled to draw conclusions contrary to the beliefs of the New York City police commissioner: economics tells us that we can actually reduce crime by

**investigating those individuals who are least likely to commit crimes, because this will reduce the wrongful conviction rate and increase the incentive to avoid committing crime. Stop-and-frisk policies do precisely the opposite: they target investigations indiscriminately at the public, innocent and guilty alike, which will increase the wrongful convictions and obviate the disincentive our justice system aims to place on criminal acts.**

*not*So there you have it. Economics tells us that stop-and-frisk causes crime.

1. I'm probably not the first one to have applied the indivisible labor literature to criminality in this way, though I did not do a search. If you know of any papers that I have incidentally duplicated, let me know so I can give credit here.