The Desirability of forgiveness in regulatory enforcement

Abstract

I present a model that explains two common features of regulatory enforcement: selective forgiveness of noncompliance, and the collection of information on a firm’s compliance activities and not just its compliance status. I show that forgiving noncompliance is optimal if the information on a firm’s compliance activities constitutes sufficiently strong evidence of the firm having exerted a high level of compliance effort. The key benefit of forgiving noncompliance is a reduction in the probability with which the firm needs to be monitored.

Appendix

1.1 Proof of result 2

This result is most easily established by working with the reciprocals of the optimal monitoring probabilities with and without forgiveness, and examining the difference ($1/m^{*}-1/\overline{m}^{*}$). The larger is this difference, the larger is the reduction in the monitoring probability achieved by forgiveness. Using Result 1, along with (3) and (9), the difference can be written

$$\begin{aligned} \frac{1}{m^{*}}-\frac{1}{\overline{m}^{*}}\equiv \frac{F}{C^{\prime }(a)}\left\{ [\theta _{N|N}-p(a)\alpha ]\sum _{s\in S_{0}}q_{a}(s|a)-p^{\prime }(a)\alpha \sum _{s\in S_{0}}q(s|a)\right\} . \end{aligned}$$

(24)

We can establish that the expression in braces is positive as follows. By definition, for $s\in S_{0}$, noncompliance is forgiven and the expression in braces in the first condition in (10) is positive. Summing the expression in braces in (10) for all $s\in S_{0}$ yields the expression in braces in (24), which, in turn, must be positive.

Now consider the effect on (24) of a reduction in $\alpha$ from $\alpha ^{\prime }$ to $\alpha ^{\prime \prime }$, due either to a reduction in $\theta _{N|N}$ or an increase in $\theta _{N|C}$. We know from Proposition 1 that $S_{0}$ is weakly larger the smaller is $\alpha$. Thus we can write $n(S_{0}^{\prime \prime })\ge n(S_{0}^{\prime })$ with $S_{0}^{\prime }\subset S_{0}^{\prime \prime }$. First consider the case where $S_{0}$ is unchanged: $S_{0}^{\prime }=S_{0}^{\prime \prime }$. A reduction in $\alpha$ due to an increase in $\theta _{N|C}$ unambiguously increases the difference ($1/m^{*}-1/\overline{m}^{*}$), since it increases the magnitude of the first term in braces in (24) while reducing the magnitude of the second one. The effect of a reduction in $\theta _{N|N}$ is less transparent, since it reduces the magnitude of both terms. However, we know that the first sum in braces is positive, as is the second sum. As is easily verified, the relative weight attached to the first sum, $[\theta _{N|N}-p(a)\alpha ]/p^{\prime }(a)\alpha$, is decreasing in $\theta _{N|N}$. Therefore, a reduction in $\theta _{N|N}$ increases the difference ($1/m^{*}-1/\overline{m}^{*}$).

Now consider the case where $n(S_{0}^{\prime \prime })>n(S_{0}^{\prime })$. Let $D\equiv S_{0}^{\prime \prime }\backslash S_{0}^{\prime }$ denote the set of additional signals for which noncompliance is forgiven as a result of the reduction in $\alpha$. The expression in braces in (24) will now be augmented by the term

$$\begin{aligned}{}[\theta _{N|N}-p(a)\alpha ]\sum _{s\in D}q_{a}(s|a)-p^{\prime }(a)\alpha \sum _{s\in D}q(s|a). \end{aligned}$$

(25)

This term must also have a positive sign given (10). Thus, the magnitude of the difference $(1/m^{*}-1/\overline{m}^{*})$ is further increased.

Finally, using (14), Eq. (24) can be rewritten as

$$\begin{aligned} \frac{1}{m^{*}}-\frac{1}{\overline{m}^{*}}\equiv \frac{\sigma (a)[\theta _{N|N}-p(a)\alpha ]F}{C{^{\prime }}(a)}\left\{ \frac{\sigma _{a}(a)}{\sigma (a)}-\frac{p{^{\prime }}(a)\alpha }{[\theta _{N|N}-p(a)\alpha ]}\right\} . \end{aligned}$$

(26)

The expression in braces is the difference in likelihood ratios, $\Delta ^{\sigma }$, specified in Result 2. As can be seen from (26), an increase in $\Delta ^{\sigma }$ will imply a larger value for $(1/m^{*}-1/ \overline{m}^{*})$, unless the increase in $\Delta ^{\sigma }$ is accompanied by a proportionally larger reduction in the value of $\sigma (a)[\theta _{N|N}-p(a)\alpha ].$ The possibility of such a reduction cannot be ruled out.

1.2 Signs of $\partial K/\partial \theta _{N|N}$ and $\partial K/\partial \theta _{N|C}$

Recall that $K$ represents the RHS of (15). Differentiating $K$ with respect to $\theta _{N|N}$ yields

$$\begin{aligned} \frac{\partial K}{\partial \theta _{N|N}}\equiv \frac{-q_{a}(s_{M}|a)}{ p^{\prime }(a)\alpha ^{2}}+\frac{q(s_{M}|a)[1-p(a)]}{[\theta _{N|N}-p(a)\alpha ]^{2}}. \end{aligned}$$

(27)

It can be verified that this expression is negative given the fact that $\alpha [1-p(a)]<[\theta _{N|N}-p(a)\alpha ]$ and the premise that forgiveness is optimal when $s_{M}$ is observed (so (11) holds).

Differentiating $K$ with respect to $\theta _{N|C}$ yields

$$\begin{aligned} \frac{\partial K}{\partial \theta _{N|C}}\equiv \frac{q_{a}(s_{M}|a)}{ p^{\prime }(a)\alpha ^{2}}+\frac{p(a)q(s_{M}|a)}{[\theta _{N|N}-p(a)\alpha ]^{2}}, \end{aligned}$$

(28)

which is positive since $q_{a}(s_{M}|a)>0$ when forgiveness is optimal, as is true by assumption when $s=s_{M}$.

1.3 Establishing $\mu >0$ and $\tau >0$ given self-reporting

I first rule out $\mu =0$ and then rule out $\mu <0$. If $\tau >0$, then $\mu =0$ implies that the first condition in (22) always holds as a strict inequality, which in turn implies $f_{N}(s)=0\, \forall s$. But then (16) cannot be binding, contradicting the premise that $\tau >0$. If $\tau =0$, then the following first-order condition for the monitoring probability, $m_{C}$, cannot hold when $\mu =0$:

$$\begin{aligned} p(a)[c_{m}+\theta _{N|C}c_{S}]=\tau \theta _{N|N}f_{NC}-\mu \theta _{N|C}p^{\prime }(a)\,f_{NC}. \end{aligned}$$

(29)

Now suppose $\mu <0$. Then (22) implies $f_{N}(s)>0$ only if $[1-p(a)]q_{a}(s|a)-p^{\prime }(a)q(s|a)>0$, which in turn implies that the sum of the second and third terms in (18) must have a positive sign. But then (18) cannot hold, because the fourth term is non-negative.

We can now rule out $\tau =0$. Given $\mu >0$, (29) does not hold if $\tau =0$.