Optimal Marginal Deterrence and Incentives for Precaution

A.1 Appendix A

The Lagrangian is

where λj,j=L,H (respectively μj,j=L,H) are Kuhn and Tucker’s multipliers associated with eq. [5] (respectively eq. [6]).

Necessary conditions are

1. It is straightforward that we cannot have simultaneously μj>0 and λj>0j=L,H.

2. Consider μj>0,j=L,H, so fj=0,j=L,H (see eq. [31]). Using eq. [30], this implies λj=0,j=L,H. From eqs [28] and [29], we must have pfs−hs>0. This is impossible since fj=0,j=L,H.

3. It follows that we have either λj>0 and μj=0 or λj=μj=0, j=L,H. Using eqs [28] and [29], the first case, implying fL=fH=F, arises if hs>pfs=pF or h>hs−1(pF). The second case requires hs=pfs.

Using iii, we have

1. hs−pfs>(=)0 if hs>(≤)pF,

2. fs=F if hs≥pF.

A.2 Appendix B

First, notice that the constraints fL≤F and fH≥0 are never binding. Using eq. [3], the fine fL is necessarily strictly lower than fH. Hence, if fH=F, then fL<F or if fL=0, then fH>0. So, the constraints above can be ignored.

The Lagrangian is

where μ (respectively λH and μL) is Kuhn and Tucker’s multiplier associated with eq. [3] (respectively eq. [5] for H only and eq. [6] for L only).

Necessary conditions are

Fines

1. Under assumption 1, the case where μL>0 and λH>0 is not possible. Consider the contrary. Using eqs [36] and [37], this implies fL=0 and fH=F, so that eq. [7] is strictly satisfied. Thus, from eq. [35], we have necessarily μ=0. Combining eqs [33] and [34], we get μL=−λHαs1−αs<0, a contradiction since μL and λH are assumed to be positive.

2. Consider λH>0 (thus, μL=0 from i). Adding eqs [33] and [34] we obtain

Thus, using eq. [33], λH>0 implies μ>0. So, we have fH=F and pfL=pF−γΔα. Using eq. [39], this gives

1. Consider μL=0 (thus, λH=0 from i). Similar development to ii implies μLq=(pfs−hs)g(γ+pfs,h)p, μ>0, fL=0 and pfH​=γΔα. So, this case arises if hs<γ1−αsΔα. But, using assumption 2, this inequality leads to the contradiction h+(1−αs)hc(h)−γΔα<0. So, we have necessarily μL=0.

2. Consider both μ=λH=0. Hence, from eqs. [33] and [34], we have pfs=hs.

From iii, we have μL=0. So, using eq. [33], we get μq=(hs−pfs)g(γ+pfs)αsΔα. From ii, we know that fH−fL=γpΔα and fs=F−αsγpΔα. Plugging these expressions into eq. [38], then simplifying, we get eq. [12] given that hs−1pF−γαsΔα comes from the threshold hs=pF−γαsΔα determined in ii.

A.3 Appendix C

First, notice that

So, using eqs [11] and [40], and adding then subtracting ∫hs−1pF−γαsΔαhs−1(pF)∫hs+γ∞Vsgdbdh in eq. [13], we find

But, for h∈hs−1(pF−γαsΔα),hs−1(pF), there is underdeterrence, so pF−γαsΔα<hs. Hence, we obtain

Second, notice that

and

Moreover, adding then subtracting ∫hs−1pF−γαsΔαhs−1(pF)∫hs+γ∞Vngdbdh and ∫hs−1(pF)∞∫pF+γ∞Vngdbdh in [14], we obtain after simplifications,

Since Vn=Vs−ΔV, we obtain

Comparing eqs [41] and [42] leads to the proposition.

A.4 Appendix D

The Lagrangian is, for j=L,H

where μ (respectively λj and μj) is Kuhn and Tucker’s multiplier associated with eq. [19] (respectively [5] and [6]).

Necessary conditions are, since xj≥0,j=L,H

1. Consider μj>0,j=L,H. By eq. (51), this implies fL=fH=0. Thus, by eq. [50], λL=λH=0. By combining eq. [43] with [45] and eq. [44] with eq. [47], we have

Using eqs [46] and [48], it implies xL=xH=0. So, since fL=fH=0, eq. [3] is not satisfied.

1. Consider μL>0 and μH=0. Following similar developments to i, μL>0 implies fL=λL=xL=0. Two subcases arise. If λH=0 (thus fH<F), by eqs [44], [47] and [48], we have xH=0. Using eqs [3] and [7], μ is necessarily strictly positive. This subcase is not possible since it boils down to iii in Appendix B. If λH>0, fH=F and constraint [3] is free using eq. [7]. As xH cannot be negative, this subcase boils down to i in Appendix B and is impossible.

2. Consider μH>0 and μL=0, and follow similar developments. We get, μH>0, which implies fH=λH=xH=0. Therefore, eq. [3] reduces to −pfL+axLΔα−γ≥0, which cannot be satisfied with fL≥0 and xL≥0.

3. Consider μj=0,j=L,H. Combining eq. [43] with eq. [45] and eq. [44] with eq. [47], we have

Thus, consider λH=λL=0. Using eqs [46] and [48], we get

Using eqs [43] and [44], this implies pfs=hs.

Otherwise, λH and λL are simultaneously positive. Then fH=fL=F, so pfs=pF. Besides, eqs [45] and [47] are satisfied at equality. Then, adding eqs [45] and [47], xs must verify

given that eq. [19] is binding. This implies, since fH−fL=0,

This case arises if hs>pF or h>hs−1(pF).

A.5 Appendix E

Ignoring the constraint [6] for the sake of simplicity, the Lagrangian is

where μ (respectively λj,j=L,H) the Kuhn and Tucker’s multiplier associated with eq. [21] (respectively [5]).

Necessary conditions are

From eqs [55] and [56], we have λj≥0,j=L,H. Using eqs [52] and [53], it follows that

From eq. [59], we have

By contrast, λL>0 can happen if μ≥0. Consider that eq. [21] is slack, so μ=0 from eq. [54]. Thus, λj>0,j=L,H, eqs [59] and [60] require hs>psfs with psfs=F(αspL+(1−αs)pH), since fj=F,j=L,H. This is equivalent to

But eq. [21] is slack. After rearrangements, this requires that the solutions for pL and pH given by eqs [57] and [58] given eq. [62] (equivalently by eqs [24] and [25]) satisfy eq. [26].

If eq. [26] is not satisfied, μ becomes positive. Thus, from eq. (61) λL>0 and we have necessarily fL=F. Using eqs [57], [58] and cj′′>0,j=L,H, it follows that pL must be decreased and pH increased compared to their values given by eqs [24] and [25].