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Malice Aforethought

  • Philip A. Curry EMAIL logo
Published/Copyright: July 12, 2016
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Abstract

This paper examines why criminal intent matters in sentencing. In particular, it considers two types of crimes, opportunistic and premeditated. Opportunistic crimes are ones that present themselves to a criminal and can be deterred if the victim makes it too costly for the criminal through private property protection. Premeditated crimes are ones sought out by the criminal, and the effect of private property protection is simply to displace crime. This difference between deterrence and displacement leads to the result that it is optimal to punish premeditated crimes more. The extent to which this is true, however, lies in the response by potential victims. If victims protect themselves from premeditated crimes in ways that also protect them from opportunistic ones, then the difference in penalties is relatively less.

JEL Classification: K14; D60

Acknowledgements

I would like to thank Martin Bouchard, Steve Easton, Steeve Mongrain, and David Scoones for comments on this paper. I also thank Marty Uhl for excellent research assistance.

Appendix

Proof to Lemma 1

Since variables aO and sO only appear in eq. (1), and aI and sI only appear in eq. (2), it is clear that aˆO is independent of sI and aˆI is independent of sO. Further, aˆOsO and aˆIsI can be found using the Implicit Function Theorem.

aˆOsO=pπaOgpsOπaaO1GpsO<0
aˆIsI=qaMIpgpsIqaa<0

Proof to Proposition 1

Consider the derivative of total costs with respect to sO, which can be rewritten as

πO[phg(psO)+c()[p[1G(psO)]pg(psO)psO]]+c()[πaO[1G(psO)]psOa^OsO]

When this expression is evaluated at sˆI, the first term is equal to πOa^IsI. We thus have

πOaˆIsI+cπaO1GpsOpsOaˆOsO>0

Thus the cost minimizing level of aI is greater than the cost minimizing level of aO. ■

Proof to Lemma 2

Since the equilibrium is symmetric, each agent’s avoidance expenditure is characterized by the equation

[πO(a)a[1G(psO)]+q(a;ai,MI)a]h+1=0

Applying the Implicit Function Theorem yields

asO=pπaOgpsOπaaO1GpsO+qaa1GpsI<0
asI=pqaMIgpsIπaaO1GpsO+qaa1GpsI<0

Proof to Proposition 2

As in the proof to Proposition 1, the derivative of total costs with respect to sO can be rewritten as

πOphgpsO+cp1GpsOpgpsOpsO+cπaO1GpsOpsOasO

When this expression is evaluated at s˜I, the first term is equal to πOasI. We thus have

πOasI+cπaO1GpsOpsOasO>0

Thus the cost minimizing level of aI is greater than the cost minimizing level of aO. ■

Proof to Lemma 3

From Lemma 1 and 2 above, we have that

a^OsO=pπaOg(psO)πaaO[1G(psO)]
a^IsI=qaMIpg(psI)qaa
asO=pπaOgpsOπaaO1GpsO+qaa1GpsI
asI=pqaMIg(psI)πaaO[1G(psO)]+qaa[1G(psI)]

First consider the effect of sI on aˆI and a. Holding sI the same, we can see that aˆIsI<asI if and only if

qaMIpg(psI)qaa<pqaMIg(psI)πaaO[1G(psO)]+qaa

At a given sI, the numerator is the same, however the denominator is greater in the term on the right (and positive). Since both terms are negative, it must be that a^IsI<asI.

Now consider the effect of sO on aˆO and a. Holding sO the same, we can see that a^OsO<asO if and only if

pπaOg(psO)πaaO[1G(psO)]<pπaOg(psO)πaaO[1G(psO)]+qaa[1G(psI)]

At a given sO, the numerator is the same, however the denominator is greater in the term on the right (and positive). Since both terms are negative, it must be that a^OsO<asO. ■

Proof to Theorem 1

Consider the first order conditions for the cost minimization problem:

phπOgpsO+cπa1GpsOpsOasOpgpsOπOpsO+pπO1GpsO=0
phgpsI+asI+c1GpsIppgpsIpsI=0

where asO and asI refer to either aˆjsj or asj, j=O,I. Let us first consider the first order condition pertaining to sO. Suppose that avoidance activities are distinct and consider evaluating it s˜O. This yields

phπOgps˜O+cπa1Gps˜Ops˜OaˆOsOpgpsOπOps˜O+pπO1Gps˜O

Since a^OsO<asO, it therefore must be the case that this expression is positive, meaning that s˜O>sˆO.

Now consider the first order condition with respect to sI. Evaluating it at s˜I when avoidance activities are distinct yields

phgps˜I+aˆIsI+c1Gps˜Ippgps˜Ips˜I

In this case, however, since a^IsI<asI, it therefore must be the case that this expression is negative, meaning that s˜I<sˆI. Since s˜O<s˜I, we therefore have that sˆO<s˜O<s˜I<sˆI. ■

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Published Online: 2016-7-12
Published in Print: 2017-3-1

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