Optimal Border Policies for Invasive Species Under Asymmetric Information

Abstract

This paper analyzes the problem faced by a border protection agency if endogenous exporter abatement activities affect invasive species risk, allowing for unobservable differences in abatement cost. We show how the optimal inspection/penalty regime differs from the symmetric information case. Departing from previous literature, we allow for technical assistance, a policy instrument specifically permitted and commonly employed under Article 9 of the World Trade Organization Sanitary and Phytosanitary Agreement. We find the information asymmetry can make it optimal for the importing country to provide technical assistance grants for exporter risk abatement, even if it would otherwise be inefficient. Further, we show that fungibility of technical assistance with inputs in other sectors of the exporting economy affects the qualitative nature of optimal policy. If technical assistance has no outside value in the exporter’s country, optimal policy is characterized by a menu of contracts balancing higher tariffs with lower penalties for being caught with an invasive. If technical assistance can be used in other sectors of the exporter’s economy, it can introduce countervailing incentives making a uniform tariff/penalty combination optimal.

Appendix

Proof of Lemma 1

From Eq.(1), the effort maximization problem is

$$\begin{aligned} e(\theta )=\underset{e}{\arg \max }\;\left\{ P(M)-\tau (\theta )-\theta e- r(I)q(e+\phi )[P(M)+t(\theta )]\right\} . \end{aligned}$$

(26)

with first order conditions,

$$\begin{aligned} -r(I)q^{\prime }(e(\theta )+\phi )[P(M)+t(\theta )]-\theta&\le 0;\end{aligned}$$

(27)

$$\begin{aligned} e(\theta )[r(I)q^{\prime }(e(\theta )+\phi )[P(M)+t(\theta )]+\theta ]&= 0, \text{ for} \text{ all} \, \theta . \end{aligned}$$

(28)

For strictly positive levels of abatement effort, $\partial e(\theta )/\partial \phi =-1$; increasing technical assistance offsets exporter effort without reducing risk. Only if there is a corner solution with zero effort will a marginal increase in technical assistance reduce overall risk. $\square$

Proof of Proposition 1

Solving Eq. (2) using pointwise optimization, the first order conditions for $t(\theta )$ are¹⁴

$$\begin{aligned} -q^{\prime }(e(\theta )+\phi )[r(I)P(M)+[1-r(I)]\delta ]-\theta&\le 0;\end{aligned}$$

(29)

$$\begin{aligned} e(\theta )\left[q^{\prime }(e(\theta )+\phi )[r(I)P(M)+[1-r(I)]\delta ]+\theta \right]&= 0\text{,} \text{ for} \text{ all} \, \theta . \end{aligned}$$

(30)

Combined with the Eqs. (27) and (28) these imply

$$\begin{aligned} e( \theta )\left[[1-r(I)]\delta -r(I)t(\theta )\right]=0,\quad \text{ for} \text{ all} \quad \theta . \end{aligned}$$

(31)

Since the optimal penalty does not vary by type, Eqs. (27) and (28) imply that induced effort is non-increasing in type. Let $e^*_0(\theta )$ denote the optimally induced effort if there were no technical assistance. Since effort is non-increasing in $\theta$ there is a threshold type $\hat{\theta }^*(\phi )=\inf \{\theta :\phi = e^*_0(\theta )\}$ below which $e^*_0(\theta )$ exceeds technical assistance, and above which technical assistance exceeds $e^*_0(\theta )$. The assumptions on $q(\cdot )$ ensure that $\hat{\theta }^*(0)=1$. Using Eq. (29), the first order conditions for technical assistance are then

$$\begin{aligned} \int \limits _0^{\hat{\theta }^*(\phi )}\theta \mathrm{d}G(\theta )-\int \limits _{\hat{\theta }^*(\phi )}^1q^{\prime }(\phi )\left[r(I)P(M)+[1-r(I)]\delta \right]\mathrm{d}G(\theta )&\le \bar{\theta };\end{aligned}$$

(32)

$$\begin{aligned} \phi \left[\int \limits _0^{\hat{\theta }^*(\phi )}\theta \mathrm{d}G(\theta )-\int \limits _{\hat{\theta }^*(\phi )}^1q^{\prime }(\phi )\left[r(I)P(M)+[1-r(I)]\delta \right]\mathrm{d}G(\theta )-\bar{\theta }\right]&= 0. \end{aligned}$$

(33)

Since we assume $\bar{\theta }>\int _0^1\theta \mathrm{d}G(\theta )$, this inequality in condition (32) is strict and it is optimal for the regulator to set $\phi =0$. Equation (5) is simply the first order condition for the choice of $I$. $\square$

Proof of Lemma 2

By (27), the first order condition for an interior solution to (6) is

$$\begin{aligned} -r(I)q(e(\theta )+\phi )t^{\prime }(\theta )-\tau ^{\prime }(\theta )=0 \text{ for} \text{ all} \,\theta . \end{aligned}$$

(34)

After differentiating this expression, the second order condition can be expressed as $-\frac{\mathrm{d}t(\theta )}{\mathrm{d}\theta }\frac{ \mathrm{d}q(e(\theta )+\phi )}{\mathrm{d}\theta }/\frac{\mathrm{d}^2q(e(\theta )+\phi )}{\mathrm{d}\theta ^2}\le 0.$ Incentive compatibility thus requires $\mathrm{d}t(\theta )/\mathrm{d}\theta \le 0$. Differentiating (1) and using (34) yields Eq. (8). $\square$

Proof of Proposition 2

The first-order conditions for $t(\theta )$ are

$$\begin{aligned}&\displaystyle -q^{\prime }(e(\theta )+\phi )[r(I)P(M)+[1-r(I)]\delta ]-\theta \le \frac{G(\theta )}{g(\theta )};\end{aligned}$$

(35)

$$\begin{aligned}&\displaystyle e(\theta )\left[-q^{\prime }(e(\theta )+\phi )[r(I)P(M)+[1-r(I)]\delta ]-\theta -\frac{G(\theta )}{g(\theta )}\right]=0. \end{aligned}$$

(36)

Using Eqs. (27) and (28), Eq. (36) simplifies to

$$\begin{aligned} e(\theta )\left[-q^{\prime }(e(\theta )+\phi )[[1-r(I)]\delta -r(I)t(\theta )]-\frac{G(\theta )}{g(\theta )}\right]&= 0. \end{aligned}$$

(37)

Equations (27) and (28) imply $\mathrm{d}e(\theta )/\mathrm{d}\theta \le 0$. Since effort is non-increasing in $\theta$ there is a threshold type $\hat{\theta }(\phi )\equiv \inf \{\theta :\phi = e_0(\theta )\}$. Similarly to the case without private information, the assumptions on $q(\cdot )$ ensure that $\hat{\theta }(0)=1$. Using Eq. (35), the first order conditions for technical assistance are

$$\begin{aligned}&\int \limits _{0}^{\hat{\theta }(\phi )}\left\{ \theta +\frac{G(\theta )}{g(\theta )}\right\} \mathrm{d}G(\theta )-\int \limits _{\hat{\theta }(\phi )}^1q^{\prime }(\phi )\left[r(I)P(M)+[1-r(I)]\delta \right]\mathrm{d}G(\theta )\le \bar{\theta };\end{aligned}$$

(38)

$$\begin{aligned}&\phi \left[\int \limits _{0}^{\hat{\theta }(\phi )}\left\{ \theta +\frac{G(\theta )}{g(\theta )}\right\} \mathrm{d}G(\theta )-\int \limits _{\hat{\theta }(\phi )}^1q^{\prime }(\phi )\left[r(I)P(M)+[1-r(I)]\delta \right]\mathrm{d}G(\theta )-\bar{\theta }\right]=0.\nonumber \\ \end{aligned}$$

(39)

In contrast to the benchmark, if there is no technical assistance then with $\epsilon$ sufficiently small the left hand side of (38) is strictly greater than $\bar{\theta }\equiv \int _0^1\theta \mathrm{d}G(\theta )+\epsilon$. This condition can then only be satisfied if the regulator provides a strictly positive amount of technical assistance. Using Eq. (35), the expression in Eq. (14) follows from first order condition for $I$. $\square$

Proof of Lemma 3

The lemma follows from the first-order conditions for (16),

$$\begin{aligned} -q^{\prime }(e(\theta ))r(I)[P(M)+t(\theta )]&\le \theta ;\end{aligned}$$

(40)

$$\begin{aligned} e(\theta )\left[-q^{\prime }(e(\theta ))r(I)[P(M)+t(\theta )]-\theta \right]&= 0, \text{ for} \text{ all} \, \theta . \end{aligned}$$

(41)

since the optimal level of effort is independent of $\phi$. $\square$

Proof of Lemma 4

Let

$$\begin{aligned} \Lambda (\theta _a,\theta _b)=r(I)q(e(\theta _a,\theta _b))[P(M)+t(\theta _a)]+\theta _b e(\theta _a,\theta _b). \end{aligned}$$

(42)

From (6), incentive compatibility requires

$$\begin{aligned} -\Lambda (\theta _a,\theta _a)-\tau (\theta _a)&\ge -\Lambda (\theta _b,\theta _a)-\tau (\theta _b);\end{aligned}$$

(43)

$$\begin{aligned} -\Lambda (\theta _b,\theta _b)-\tau (\theta _b)&\ge -\Lambda (\theta _a,\theta _b)-\tau (\theta _a). \end{aligned}$$

(44)

Summing these two equations yields

$$\begin{aligned} \Lambda (\theta _a,\theta _b)-\Lambda (\theta _a,\theta _a)\ge \Lambda (\theta _b,\theta _b)-\Lambda (\theta _b,\theta _a). \end{aligned}$$

(45)

By the envelope theorem, expression $\tfrac{\mathrm{d}\Lambda (\theta _a,\theta _b)}{\mathrm{d}t(\theta _a)}=\tfrac{\partial \Lambda (\theta _a,\theta _b)}{\partial t(\theta _a)}=r(I)q(e(\theta _a,\theta _b))>0$, and $\tfrac{\mathrm{d}^2\Lambda (\theta _a,\theta _b)}{\mathrm{d}t(\theta _a)\mathrm{d}\theta _{b}}=r(I)q^{\prime }(e(\theta _a,\theta _b))\tfrac{\mathrm{d}e(\theta _a,\theta _b)}{\mathrm{d}\theta _b}\ge 0$. Therefore if $\theta _a<\theta _b$ then inequality (45) requires $t(\theta _a)\ge t(\theta _b)$. Differentiating (15) and using the envelope theorem yields Eq. (20). $\square$

Proof of Proposition 3

As Lemma 5 in Lewis and Sappington (1989) provides an in-depth formal analysis of a problem with similar structure (for a more general treatment, see Maggi and Rodríguez-Clare 1995; Jullien 2000), here we only provide a sketch of the proofs required to characterize the equilibrium contract. Condition (19) requires that $t(\theta )\ge t^0$ for $\theta <\theta ^0$ and $t(\theta )\le t^0$ for $\theta >\theta ^0$. Let $\lambda (\theta )$ and $\mu (\theta )$ be Lagrange multipliers for these two constraints. Using Eq. (40), if exporters undertake a positive amount of effort the first-order conditions for $t(\theta )$ are

$$\begin{aligned}&\displaystyle \left[-q^{\prime }(e(\theta ))[[1-r(I)]\delta -r(I)t(\theta )]-\frac{G(\theta )}{g(\theta )}\right]\tfrac{\partial e(\theta )}{\partial t}+\lambda (\theta )=0\quad \text{ for}\quad \theta <\theta ^0(\phi );\quad \end{aligned}$$

(46)

$$\begin{aligned}&\displaystyle [1-r(I)]\delta -r(I)t^0=0 \text{ for}\quad \theta =\theta ^0(\phi );\end{aligned}$$

(47)

$$\begin{aligned}&\displaystyle \left[-q^{\prime }(e(\theta ))[[1-r(I)]\delta -r(I)t(\theta )]-\frac{G(\theta )-1}{g(\theta )}\right]\tfrac{\partial e(\theta )}{\partial t}-\mu (\theta )=0\quad \text{ for} \quad \theta >\theta ^0(\phi ).\nonumber \\ \end{aligned}$$

(48)

Let $t^{G}(\theta )$ denote the penalty path defined by Eq. (46), setting $\lambda (\theta )=0$:

$$\begin{aligned} t^G(\theta )=t^0+\frac{G(\theta )}{r(I)q^{\prime }(e(\theta ))g(\theta )}. \end{aligned}$$

(49)

Similarly, let $t^{G-1}(\theta )$ denote the penalty path defined by Eq. (48), with $\mu (\theta )=0$:

$$\begin{aligned} t^{G-1}(\theta )=t^0+\frac{G(\theta )-1}{r(I)q^{\prime }(e(\theta ))g(\theta )}. \end{aligned}$$

(50)

If $\lambda (\theta )=\mu (\theta )=0$, the Potential Separation assumption (see footnote 11) ensures that the paths for $t(\theta )$ as defined by Eqs. (49) and (50) are strictly decreasing as depicted in Fig. 1. Thus, any pooling interval contains $\theta ^0(\phi )$. Since $q^{\prime }(\cdot )<0$, it would be the case that $t(\theta )<t^0$ for any $\theta <\theta ^0(\phi )$ that were not in the pooling interval. Since such a result would violate monotonicity condition (45), $t(\theta )=t^0$ for all $\theta <\theta ^0(\phi )$. A similar result applies for $\theta >\theta ^0(\phi )$.

Fig. 1

If technical assistance is fungible, there is a uniform optimal penalty, $t_0$, for being caught with an invasive

The first-order condition for $\phi$ is

$$\begin{aligned} \int \limits _{0}^{\theta ^0(\phi )}G(\theta )\mathrm{d}\theta -\int \limits _{\theta ^0(\phi )}^{1}[1-G(\theta )]\mathrm{d}\theta =0. \end{aligned}$$

(51)

The first-order condition for $I$ is

$$\begin{aligned}&\int \limits _{0}^{1}r^{\prime }(I)q(e(\theta ))[\delta -P(M)]\mathrm{d}G(\theta )-k\!=\!\int \limits _{0}^{\theta ^0(\phi )}G(\theta )\tfrac{\partial e}{\partial r}r^{\prime }(I)\mathrm{d}\theta -\int \limits _{\theta ^0(\phi )}^{1}[1-G(\theta )]\tfrac{\partial e}{\partial r}r^{\prime }(I)\mathrm{d}\theta \nonumber \\&\quad =\int \limits _{0}^{\theta ^0(\phi )}\tfrac{G(\theta )q^{\prime }(e(\theta ))r^{\prime }(I)}{r(I)q^{\prime \prime }(e(\theta ))}\mathrm{d}\theta -\int \limits _{\theta ^0(\phi )}^{1}\tfrac{[1-G(\theta )]q^{\prime }(e(\theta ))r^{\prime }(I)}{r(I)q^{\prime \prime }(e(\theta ))}\mathrm{d}\theta . \end{aligned}$$

(52)

$\square$