Dynamic Regulation Design Without Payments:
The Importance of Timing
Online Appendix
Ralph Boleslavsky
Department of Economics
University of Miami
Box 248126
Coral Gables, FL 33124
[email protected]
and
David L. Kelly
Department of Economics
University of Miami
Box 248126
Coral Gables, FL 33124
[email protected]
May 20, 2014
This appendix contains the proofs of the most propositions given in Boleslavsky and
Kelly (2014), hereafter BK. All references to equations numbered from 1.1 to 8.x are in BK.
Equations numbered 9.x are in the online appendix. The proofs of propositions 1 and 2 are
in the main paper.
9
Appendix: Proofs of Remaining Propositions and Remaining Calculations
9.1
Welfare and Uncertainty
Here we show how welfare is affected by changes in uncertainty. This subsection assumes
C (q,π) = πc (q).
1. Mean preserving increase in the spread. Let:
πH (ǫ) = πH + γǫ , πL (ǫ) = πL − (1 − γ) ǫ.
(9.1)
Since the mean π̄ is unchanged and C = πc (q), (2.14) implies q̄ is unchanged as well.
Next, we can compute the change in welfare from using the mechanism as:
∆W = W (ǫ) − W0 ,
(9.2)
= max γ B (q1L ) − πL (ǫ) c (q1L ) + δ (B (q2L ) − π̄c (q2L )) + (1 − γ) B (q1H ) −
πH (ǫ) c (q1H ) + δ (B (q2H ) − π̄c (q2H )) + λ πL (ǫ) (c (q1L ) − c (q1H )) +
δπ̄ (c (q2L ) − c (q2H ))
− max {(1 + δ) (B (q̄) − π̄c (q̄))}
(9.3)
Applying the envelope theorem to both maximization problems in the above equation,
we see that:
∂∆W (ǫ)
λ
λ
− c (q1H ) 1 −
.
= γ (1 − γ) c (q1L ) 1 +
∂ǫ
γ
1−γ
(9.4)
The above derivative is positive since q1L > q1H . Equation (9.4) also implies that the
welfare of the prior policy does not change with an increase in the spread. Instead,
the welfare differential increases with the spread because welfare under the mechanism
increases with the spread.
1
2. Welfare and γ: The proof has several steps. We first show that ∆W = 0 for γ = 0,1.
For γ → 1, π̄ → πL , and so from (2.14), q̄ → qL∗ . Hence the prior policy approaches
first best:
W0 → (1 + δ) (B (qL∗ ) − πL c (qL∗ )) .
(9.5)
Next proposition 2.3 implies λ → 0. Thus, (2.7) and (2.9) and π̄ → πL imply q1L and
q2L approach first best. Since q1H and q2H are implemented with probability 0, (2.6)
implies:
W → (1 + δ) (B (qL∗ ) − πL c (qL∗ )) .
(9.6)
Hence, ∆W → 0.
For γ → 0, a symmetric argument implies W approaches first best. As γ approaches 0,
(2.7) implies λ → 0. Thus, (2.8) and (2.10) and π̄ → πH imply q1H and q2H approach
first best. Since q1L and q2L are implemented with probability 0, (2.6) implies the
mechanism approaches first best. Hence ∆W → 0.
Finally, note that since the prior policy is feasible under the mechanism, we have
∆W ≥ 0. The envelope theorem implies continuity of ∆W. These facts, and that
∆W (1) = ∆W (0) = 0, imply ∆W achieves a maximum at some 0 < γ < 1.
9.2
First Order Conditions With a Cost of Funds and Proof of Proposition 3
Let λLL , λLH , λL , and µij be the Lagrange multipliers on (3.1), i = L,H, (3.3), and (3.5),
i,j = L,H, respectively. The first order conditions for problem (3.7) are:
λL
Bq (q1L ) = Cq (q1L , πL ) 1 +
,
γ
λL Cq (q1H , πL )
,
Bq (q1H ) = Cq (q1H , πH ) 1 −
1 − γ Cq (q1H , πH )
λL λLL
+ 2 ,
Bq (q2LL ) = Cq (q2LL , πL ) 1 +
γ
γ
λL
λLL Cq (q2LH , πL )
Bq (q2LH ) = Cq (q2LH , πH ) 1 +
,
−
γ
γ (1 − γ) Cq (q2LH , πH )
λLH
λL
,
+
Bq (q2HL ) = Cq (q2HL , πL ) 1 −
1 − γ γ (1 − γ)
2
(9.7)
(9.8)
(9.9)
(9.10)
(9.11)
λLH Cq (q2HH , πL )
λL
,
−
Bq (q2HH ) = Cq (q2HH , πH ) 1 −
1 − γ (1 − γ)2 Cq (q2HH , πH )
λL λLL
µLL
=φ−
− 2 ,
2
γ
γ
γ
µLH
λL
λLL
= φ−
+
,
γ (1 − γ)
γ
γ (1 − γ)
λL
λLH
µHL
= φ+
−
,
γ (1 − γ)
1 − γ γ (1 − γ)
µHH
λLH
λL
+
,
2 = φ+
1 − γ (1 − γ)2
(1 − γ)
µij tij = 0, i,j = L,H.
(9.12)
(9.13)
(9.14)
(9.15)
(9.16)
(9.17)
∗
Imposing the solution q1L = q1LL = q1HL = qL∗ and q1H = q2LH = q2HH = qH
on (9.7)(9.12), we see that the solution satisfies the first order conditions (9.7)-(9.12) if and only if
λL = λLL = λLH = 0. Imposing this and φ = 0 on (9.13)-(9.17), we see that the solution
satisfies the first order conditions (9.13)-(9.17) if and only if µij = 0 for all i,j.
We next show a set of positive payments exists that satisfies all constraints. First, let
tHH = 0, then constraint (3.1), i = H, is satisfied for:
∗
tHL = ∆CL ≡ C (qL∗ , πL ) − C (qH
, πL ) > 0.
(9.18)
∗
∆CL and q2LL = qL∗ and q2LH = qH
into (3.1),
Next, substitute tLH = ∆Cδ L and tLL = 1+δ
δ
i = L, implies (3.1) holds with equality. Substituting the proposed solution for tLH and tLL
and the first best solutions for all q’s into (3.3) implies (3.3) holds with equality.
Finally, substituting the proposed solution for all tij and the first best solution for all q’s
into (3.2), i = L,H and (3.4), we see that all three constraints hold since C is super modular.
Therefore, since all payments are positive and all constraints and first order conditions are
satisfied, the first best level of regulation is optimal for φ = 0.
9.3
Proof of Proposition 4
We must show the solution to (2.7)-(2.11) with tij = 0 for all i,j satisfies all first order
conditions and constraints for (3.7). Comparing (2.7) and (9.7), we see that (9.7) is satisfied
if and only if λL = λ, where λ is the multiplier for problem (2.6). Condition (2.8) implies
condition (9.8) is also satisfied for λL = λ.
3
Next, imposing λL = λ and q2LL = q2L on (9.9), we see that (9.9) holds if and only if:
λ λLL
Bq (q2L )
= 1+ + 2 .
(9.19)
Cq (q2L , πL )
γ
γ
Using (2.9), this is equivalent to:
γCq (q2L , πL ) + (1 − γ) Cq (q2L , πH )
Cq (q2L , πL )
λ
λ λLL
1+
= 1+ + 2 .
γ
γ
γ
(9.20)
Using the definition of R:
λLL
λ
= γ (1 − γ) (R − 1) 1 +
.
γ
2
(9.21)
So if (9.21) holds, (9.9) is satisfied. Further, imposing q2LH = q2L on (9.10) and using (2.9)
to eliminate Bq , we see that (9.10) holds if and only if (9.21) holds.
For (9.11), we impose q2HL = q2H and λL = λ, yielding:
λ
λLH
Bq (q2H )
=1−
+
.
Cq (q2H , πL )
1 − γ γ (1 − γ)
(9.22)
Imposing (2.10) gives:
γCq (q2H , πL ) + (1 − γ) Cq (q2H , πH )
Cq (q2H , πL )
1−
λ
1−γ
=1−
λ
λLH
+
.
1 − γ γ (1 − γ)
(9.23)
Using the definition of R:
λLH
= γ (1 − γ) (R − 1) 1 −
2
λ
1−γ
.
(9.24)
So condition (9.11) is satisfied if and only if (9.24) holds. Further, an analogous argument
shows that (9.12) holds if and only if (9.24) holds.
Next, our solution requires all payments to be zero, and thus all of the multipliers µij to
be positive. From (9.13) given λL = λ, this requires:
µLL
λ λLL
=
φ
−
− 2 > 0.
γ2
γ
γ
(9.25)
Given (9.21), the above inequality holds if and only if:
φ > (1 − γ) (R − 1) +
λ
(γ + (1 − γ) R) .
γ
4
(9.26)
So for φ sufficiently large, tLL = 0 as required, with µLL defined by substituting (9.21) and
λL = λ into (9.13). Recall λ is the multiplier from problem (2.6), and thus is independent
of φ, so a φ sufficiently large always exists. For tLH , (9.14) requires:
λ
λLL
µLH
= φ− +
> 0,
γ (1 − γ)
γ γ (1 − γ)
(9.27)
Using (9.21) and simplifying gives:
λ
λ
.
φ > − γ (R − 1) 1 +
γ
γ
(9.28)
So for φ large, tLH = 0, with µLH defined by substituting (9.21) and λL = λ into (9.14).
Condition (9.26) is more restrictive than condition (9.28) if and only if:
λ
λ
λ
(1 − γ) (R − 1) + (γ + (1 − γ) R) > − γ (R − 1) 1 +
,
(9.29)
γ
γ
γ
λ
(R − 1) 1 +
γ
> 0,
(9.30)
which holds since R > 1. For tHL , (9.15) requires:
µHL
λ
λLH
= φ+
−
> 0.
γ (1 − γ)
1 − γ γ (1 − γ)
(9.31)
Substituting in (9.24) and simplifying gives:
λ
λ
−
φ > (1 − γ) (R − 1) 1 −
.
1−γ
1−γ
(9.32)
Given (9.32), tHL = 0 and µHL is defined by (9.24) and (9.15). Clearly (9.26) implies (9.32).
Finally, note that µHH > 0 and so tHH = 0, with the µHH defined by (9.16) and (9.24).
Therefore, given (9.26), the proposed solution satisfies the first order conditions and
the period one incentive constraints, which are identical to the incentive constraints from
the problem without payments. The period two incentive constraints hold since tij = 0,
q2LH = q2LL = q2L , and q2HL = q2HH = q2H .
Finally, since λ < 1 − γ, a sufficient condition for (9.26) is:
φ > (1 − γ) (R − 1) +
1−γ
(γ + (1 − γ) R) .
γ
5
(9.33)
The above equation simplifies to the desired result:
φ>
9.4
(1 − γ) R
.
γ
(9.34)
Proof of Proposition 5
Let λLL , λLH , λL , and µij be the Lagrange multipliers on (3.1), i = L,H, (3.3), and (3.5),
i,j = L,H, respectively. The first order conditions for problem (3.7) are:
λL
,
Bq (q1L ) = Cq (q1L , πL ) 1 +
γ
λL Cq (q1H , πL )
Bq (q1H ) = Cq (q1H , πH ) 1 −
,
1 − γ Cq (q1H , πH )
λL λLL
+ 2 ,
Bq (q2LL ) = Cq (q2LL , πL ) 1 +
γ
γ
λL
λLL Cq (q2LH , πL )
Bq (q2LH ) = Cq (q2LH , πH ) 1 +
,
−
γ
γ (1 − γ) Cq (q2LH , πH )
λLH
λL
,
+
Bq (q2HL ) = Cq (q2HL , πL ) 1 −
1 − γ γ (1 − γ)
λLH Cq (q2HH , πL )
λL
,
−
Bq (q2HH ) = Cq (q2HH , πH ) 1 −
1 − γ (1 − γ)2 Cq (q2HH , πH )
λL λLL
µLL
=φ−
− 2 ,
2
γ
γ
γ
µLH
λL
λLL
= φ−
+
,
γ (1 − γ)
γ
γ (1 − γ)
λL
λLH
µHL
= φ+
−
,
γ (1 − γ)
1 − γ γ (1 − γ)
µHH
λL
λLH
,
+
2 = φ+
1 − γ (1 − γ)2
(1 − γ)
µij tij = 0, i,j = L,H.
(9.35)
(9.36)
(9.37)
(9.38)
(9.39)
(9.40)
(9.41)
(9.42)
(9.43)
(9.44)
(9.45)
The incentive constraints are:
− C (q2HL , πL ) + tHL ≥ −C (q2HH , πL ) + tHH ,
(9.46)
− C (q2LL , πL ) + tLL ≥ −C (q2LH , πL ) + tLH ,
(9.47)
−C (q1L , πL ) − δγC (q2LL , πL ) − δ (1 − γ) C (q2LH , πH ) + δγtLL + δ (1 − γ) tLH ≥
6
−C (q1H , πL ) − δγC (q2HL , πL ) − δ (1 − γ) C (q2HH , πH ) +
δγtHL + δ (1 − γ) tHH
(9.48)
We wish to show that q2LL ≤ q2HL and q2LH ≤ q2HH , with at least one inequality being
strict. We show these conditions hold using four possible cases.
• Case 1: tLL > 0, tHL > 0. In this case (9.41), (9.43), and (9.45) imply:
µLL
λL λLL
=φ−
− 2 = 0,
2
γ
γ
γ
(9.49)
µHL
λL
λLH
=0=φ+
−
.
γ (1 − γ)
1 − γ γ (1 − γ)
(9.50)
Solving these two equations for λLL and λLH and substituting into (9.37)-(9.40) gives:
G (q2LL ) = 1 + φ,
G (q2LH ) = R −
(9.51)
(1 − γ) R + γ
γ
φ+
λL ,
1−γ
γ (1 − γ)
G (q2HL ) = 1 + φ,
G (q2HH ) = R −
G (q) ≡
(9.52)
(9.53)
(1 − γ) R + γ
γ
φ−
λL ,
1−γ
(1 − γ)2
Bq (q)
, Gq < 0.
Cq (q, πL )
(9.54)
(9.55)
First, q2LL = q2HL , is immediate from (9.51) and (9.53). Second, q2LH < q2HH , holds
if and only if G (q2LH ) > G (q2HH ), which holds by (9.52) and (9.54).
• Case 2: tLL = tHL = 0. Note that from (9.41) and (9.42) we have µLH /γ (1 − γ) ≥
7
µLL /γ 2 > 0, which implies from (9.45) that tLH = 0. Since tHH = 0 from (9.44) and
(9.45), we have all payments equal to zero. Thus, this case is equivalent to theorem
4, which shows the original timing mechanism is optimal. Therefore, we have: q2LL =
q2LH = q2L < q2HL = q2HH = q2H , and so both conditions hold with strict inequality.
• Case 3: tLL > 0 and tHL = 0. First, µHL ≥ 0 and (9.43) implies:
φ≥
λL
λLH
−
.
γ (1 − γ) 1 − γ
(9.56)
Further, case 1 proves that tLL > 0 implies (9.51) and (9.52). Therefore q2LL ≤ q2HL
if and only if G (q2LL ) ≥ G (q2HL ), which by (9.51) and (9.39) holds since (9.56) holds.
Next, tLL > 0, tLH = 0, and (9.47) implies q2LL < q2LH . Thus G (q2LL ) < G (q2LH ),
which along with (9.51) and (9.52) implies:
1+φ<R−
γ
(1 − γ) R + γ
φ+
λL ,
1−γ
γ (1 − γ)
φ < (1 − γ) (R − 1) +
(1 − γ) R + γ
λL .
γ
(9.57)
(9.58)
Therefore (9.56) and (9.58) are the lower and upper bounds for φ for this case.
Next, tHL = tHH = 0 and (9.46) implies q2HH ≥ q2HL . Thus, G (q2HH ) ≤ G (q2HL ),
which along with (9.39) and (9.40) implies:
λLH
λL
≥ γ (1 − γ) (R − 1) 1 −
1−γ
2
.
(9.59)
Since λLH > 0, (9.46) binds and therefore q2HH = q2HL . Repeating the above argument
with equality gives:
λLH
λL
= γ (1 − γ) (R − 1) 1 −
.
1−γ
2
(9.60)
Finally, we must show q2LH < q2HH , which holds if and only if G (q2LH ) > G (q2HH ).
8
Using (9.40) and (9.52), we must show:
γ
R+γ
λLH
.
φ<
λL +
1−γ
γ (1 − γ)
(1 − γ)2
(9.61)
Substituting in (9.60) and simplifying gives:
1+γ
φ < (1 − γ) (R − 1) +
γ
R (1 − γ) + γ
γ
λL ,
(9.62)
which holds from (9.58).
• Case 4: tLL = 0 and tHL > 0. We will prove that this case cannot be a solution. Since
tLL = 0, (9.41), and (9.45) imply:
λL λLL
µLL
=
φ
−
− 2 > 0.
γ2
γ
γ
(9.63)
We will derive an equation which contradicts (9.63), which establishes that case 4
cannot be a solution.
First, note we again have tLH = 0 as in case 2. Since tLH = tLL = 0, we have from
(9.47) that q2LH ≥ q2LL . Therefore since G is decreasing G (q2LH ) ≤ G (q2LL )). Hence
from (9.37) and (9.38):
λL
λLL
λL λLL
R 1+
−
≤1+
+ 2 ,
γ
γ (1 − γ)
γ
γ
λLL
λL
≥ γ (1 − γ) (R − 1) 1 +
γ
2
.
(9.64)
(9.65)
Since λLL > 0, constraint (9.47) binds and thus q2LH = q2LL . Repeating the above
argument with equality proves that:
λLL
λL
= γ (1 − γ) (R − 1) 1 +
.
γ
2
9
(9.66)
Next, note that since tHL > 0, (9.45) implies µHL = 0 which implies from (9.43):
λLH = γ (1 − γ) φ + γλL .
(9.67)
Therefore, (9.46) binds, which, along with tHL > 0 and tHH = 0 implies q2HL > q2HH .
Therefore, G (q2HL ) < G (q2HH ). Now, from case one, we know that tHL > 0 implies
G (q2HL ) = 1 + φ. Therefore, using (9.40):
λL
λLH
,
1+φ<R 1−
−
1−γ
(1 − γ)2
φ<R−1−R
(9.68)
λL
λLH
−
.
1 − γ (1 − γ)2
(9.69)
Now using (9.67):
φ < (1 − γ) (R − 1) − λL R +
γ
1−γ
.
(9.70)
But the above equation contradicts (9.63), so case 3 cannot be a solution.
Now for the first period regulations, from (9.35) and (9.36), we see that q1L ≥ q1H if and
only if λL ≤ γ (1 − γ) (R − 1). Therefore, q1L ≥ q1H is immediate if λL = 0. If λL > 0 the
constraint (9.48) binds. Rewriting the constraint gives:
C (q1H , πL ) − C (q1L , πL ) = δE [C (q2 (πL , π) , π)] − δE [C (q2 (πL , π) , π)] −
δγ (tHL − tLL ) − δ (1 − γ) tLH .
(9.71)
For cases 1 and 3, what we have shown is sufficient for the first two terms on the right hand
side to be negative. Therefore, for case 3 we have q1L ≥ q1H since tHL < tLL and tLH = 0.
For case 1, if λLL > 0, we have:
µLH
µLL
> 2 = 0.
γ (1 − γ)
γ
(9.72)
Therefore tLH = 0. Next, (9.46) and (9.47) imply that tHL ≤ tLL if and only if:
C (q2HL , πL ) − C (q2HH , πL ) ≤ C (q2LL , πL ) − C (q2LH , πL ) ,
10
(9.73)
which holds since we have shown q2LL = q2HL and q2HH ≥ q2LH .
This argument requires (9.47) to bind. If not, then λLL = 0 and from (9.37) and (9.38),
we have q1LL = q1LH . Therefore, from (9.47), we have tLL > tLH . Now reduce tLL and tLH to
zero, holding all other decisions constant. All constraints and first order conditions continue
to hold, but welfare increases by φ (tLH + tLL ), contradicting that (9.47) not binding is a
maximum.
9.5
Proof of Proposition 6
The first order conditions for the relaxed problem are:
λ
Bq (q1L ) = πL 1 +
Cq (q1L ) ,
γ
λ πL
Cq (q1H ) ,
Bq (q1H ) = πH 1 −
1 − γ πH
λ
Bq (q2L ) = π̄L 1 +
Cq (q2L ) ,
γ
λ π̄L
Cq (q2H ) ,
Bq (q2H ) = π̄H 1 −
1 − γ π̄H
(9.74)
(9.75)
(9.76)
(9.77)
Define λL = γ (1 − γ) π̄π̄HL − 1 and λR = γ (1 − γ) ππHL − 1 . Both of λL and λR are
positive under our assumptions.
Consider first the equality case of (4.15). Consider the solution λ = λL = λR . Plugging
in the solution into (9.74)-(9.77), implies q1L = q1H = q̄1 and q2L = q2H = q̄2 . Given these
decisions, (4.13) and (4.14) hold. Therefore, the no-information regulation is optimal in both
periods.
For the non-equality case of (4.15), observe from the first order conditions that:
λ < λR ⇔ q1L > q̄1 > q1H ,
(9.78)
λ > λL ⇔ q2H > q̄2 > q2L ,
(9.79)
πH
π̄H
<
⇔ λL < λR .
π̄L
πL
(9.80)
Consider λ = λL . Then q2H = q2L and (4.13) is negative if and only if q1L > q1H , which by
(9.78) holds if and only if λ = λL < λR , which in turn holds if and only if (4.15) holds with
11
strict inequality, by (9.80).
Conversely, consider λ = λR . Since q1H = q1L , (4.13) is positive if and only if q2H > q2L ,
which by (9.79) holds if and only if λ = λR > λL , which in turn holds if and only if (4.15)
holds strictly, by (9.80).
Therefore, by the intermediate value theorem, if (4.15) holds strictly (does not hold), a
λ∗ ∈ (λL ,λR ) (λ∗ ∈ (λR ,λL )) exists for which (4.13) holds with equality.
Next, since (4.13) holds with equality:
πL (C (q1L ) − C (q1H )) = δπ̄L (C (q2H ) − C (q2L )) .
(9.81)
Using (9.81) to eliminate C (q2H ) − C (q2L ) from (4.14) we see that (4.14) holds if and only
if:
π̄H
πH
(C (q1L ) − C (q1H )) ≥
(C (q1L ) − C (q1H )) .
πL
π̄L
(9.82)
Condition (9.82) is satisfied since we have shown that q1L > q1H if and only if (4.15) holds.
The properties of the solution were derived in (9.78)-(9.80).
9.6
Proof of Proposition 7
The Lagrangian and first order conditions for the regulator’s problem are as follows:
Ln =
n
X
P r (m) B (mq1L,m + (n − m) q1H,m ) − mC (q1L,m , πL ) − (n − m) C (q1H,m , πH )
m=0
+δB (mq2L,m + (n − m) q2H,m ) − mδE [C (q2L,m , π)] − (n − m) δE [C (q2H,m , π)]
+nλ
n
X
m=1
P r (m|L) wf (q1L,m , q2L,m , πL ) − wf (q1H,m−1 , q2H,m−1 , πL )
(9.83)
All low cost firms have the same incentive constraints and thus λ does not vary by firm.
From the properties of the binomial distribution:
"
#
n−1
n−m
m
P r (m) , P r (m + 1|L) =
P r (m) .
P r (m|L) =
γ m−1 (1 − γ)n−m =
γn
n (1 − γ)
m−1
Thus, the first order conditions reduce to (5.5)-(5.8).
12
7.2. Starting with the second period policies, we combine (5.7) and (5.8), so that:
λ
λ
E [Cq (q2L,m , π)] 1 +
= E [Cq (q2H,m , π)] 1 −
.
γ
1−γ
(9.84)
Since Cq is an increasing function, it is immediate that q2L,m < q2H,m .
Next, q2L,m < q̄n if and only if:
Bq (nq2L,m )
Bq (nq̄n )
>
.
E [Cq (q2L,m , π)]
E [Cq (q̄n , π)]
(9.85)
Equation (5.7) and the definition of q̄n implies:
λ
1+
γ
Bq (nq2L,m )
> 1.
Bq (mq2L,m + (n − m) q2H,m )
(9.86)
Therefore, it is sufficient to show Bq (nq2L,m ) > Bq (mq2L,m + (n − m) q2H,m ), which
holds if and only if q2H,m > q2L,m , which holds as shown above. Thus, q2L,m < q̄n . A
analogous argument using (5.8) shows that q2L,m < q2H,m implies q2L,m < q̄n .
7.1. For the period one policies, we begin by showing q1H,m < q1L,m . The first step is to
show that the sign of q1L,m − q1H,m does not vary with m. Combining (5.5) and (5.6):
λ
1+
γ
Cq (q1L,m , πL ) = 1 −
λ 1
1−γR
Cq (q1H,m , πH ) ,
λ
R − 1−γ
Cq (q1L,m , πL )
=
.
Cq (q1H,m , πL )
1 + γλ
(9.87)
(9.88)
Here (9.88) follows from the definition of R. Since C is convex, q1L,m − q1H,m > 0 if
and only if the right hand side of (9.88) is greater than one or if and only if:
λ < γ (1 − γ) (R − 1) .
(9.89)
Since (9.89) is independent of m, the sign of q1L,m − q1H,m is independent of m.
The next step is to show q1i (m,λ), i = L,H is an increasing function of m if and only
13
if q1L,m > q1H,m . Totally differentiating (5.5) and (5.6) with respect to m and solving
′
for q1L
(m) gives:
′
q1L
(m) =
Bqq · (q1L,m − q1H,m )
λ
−Bqq · (m + (n − m) z) + Cqq (q1L,m , πL ) 1 + γ
Cqq (q1L,m , πL ) 1 + λγ
′
′
.
q1L
(m) = zq1L,m
, z≡
λ
Cqq (q1L,m , πL ) R − 1−γ
(9.90)
Here we have suppressed the function arguments for Bqq . Hence both derivatives are
positive if and only if q1L,m > q1H,m .
Finally, to prove q1L,m > q1H,m we suppose not, so that q1L,m ≤ q1H,m . If so, then step
one implies the inequality holds for all m and step two implies both derivatives are
decreasing functions of m. Rewriting the incentive constraint (5.2) gives:
n−1
X
m=1
P r (m|L) (C (q1H,m−1 , πL ) − C (q1H,m , πL )) + (C (q1H,m , πL ) − C (q1L,m , πL ))
+δE [(C (q2H,m−1 ,π) − C (q2H,m , π)) + (C (q2H,m ,π) − C (q2L,m ,π))] +
P r (n|L) (C (q1L,n−1 , πL ) − C (q1L,n , πL )) + (C (q1H,n−1 , πL ) − C (q1L,n−1 , πL )) +
δE [(C (q2L,n−1 , π) − C (q2L,n , π)) + (C (q2H,n−1 , π) − C (q2L,n−1 , π))] ≥ 0, (9.91)
with strict inequality only if λ = 0. The second term in rows one and three are nonnegative, since we have supposed q1L,m ≤ q1H,m , which holds for all m. The second
terms in rows two and four are strictly positive, since we have shown q2L,m < q2H,m .
The first term in each row is non-negative since q1L,m and q1H,m are non-increasing in
m, given q1L,m ≤ q1H,m . Thus the incentive constraint is strictly positive. Thus λ = 0.
∗
∗
Then, from (5.5) and (5.6), q1L,m = qLm
> q1H,m = qHm
results, which contradicts that
q1L,m ≤ q1H,m . Thus q1L,m > q1H,m .
We next show that q1L,m > q̄n . First, multiplying (5.5) and (5.6) by γ and 1 − γ,
respectively, and adding using the definition of R gives:
Bq (mq1L,m + (n − m) q1H,m ) = (γ + λ) Cq (q1L,m , πL ) + (R (1 − γ) − λ) Cq (q1H,m , πL ) .
14
Since q1L,m > q1H,m :
Bq (mq1L,m + (n − m) q1H,m ) < (γ + R (1 − γ)) Cq (q1L,m , πL ) = E [Cq (q1L,m , π)] .
Here the last equality follows from the definition of R. Now since q1L,m < q1H,m :
Bq (nq1L,m ) < Bq (mq1L,m + (n − m) q1H,m ) < E [Cq (q1L,m , π)] .
(9.92)
Note that (9.92) holds with equality at q̄n . Since the Bq and Cq are decreasing functions,
q1L,m > q̄n . The proof that q1H,m < q̄n follows an identical logic.
7.3. That q1L,m and q1H,m are increasing in m follows from q1L,m > q1H,m and (9.90).
10
Appendix: Endogenous Investment and Declining Costs
Frequently, firms undertake investment or R&D which reduces compliance costs over time.
In response, regulation often becomes more strict over time. Here we suppose that firms
may undertake endogenous investment which reduces compliance costs and show that our
basic result continues to hold. In particular, if regulation becomes more strict over time,
then the regulator offers one regulation which is above the expected trend in regulation in
the first period and is below the expected trend in regulation in the second period, and a
second regulation which is initially below, and subsequently above, the trend.
Consider the model of section 2.1, but assume costs C (q, π, ζ) are a function of investment
ζ in a cost saving technology. We assume investment reduces costs Cζ < 0, the cost function
is convex in [q,ζ], and investment reduces marginal costs, Cζq < 0. Investment is increasing
in regulatory stringency if and only if Cζq < 0. Finally, we assume that the cost function is
such that the second order conditions for the regulator’s problem continue to hold. Let δPζ
denote the price of investment paid in the first period and normalize ζ1 = 0.
The firm chooses a level of investment after reporting period one costs to the regulator.
Because the regulator announces regulation for both periods in the first period, the firm
anticipates the level of regulation in period two when the investment decision is made. The
firm’s investment problem conditional on regulation q2i is then:
max −C (q1i , π, 0) − δE [C (q2i , π, ζ)] − δPζ ζ.
ζ
15
(10.1)
The firm’s first order condition is:
Pζ = −E [Cζ (q2i , π, ζ)] , i = L,H.
(10.2)
We assume a function ζi = ζ (q2i ), i = L,H, satisfying (10.2) exists.
The change in firm profits from the regulation becomes:
wf (q1 , q2 , π1 , ζ (q2 )) = −C (q1 , π1 , 0) − δE [C (q2 , π, ζ (q2 ))] − δPζ ζ (q2 ) .
(10.3)
The welfare function includes the resource costs of investment:
w (q, π, ζ (q)) = B (q) − C (q, π, ζ (q)) − Pζ · ζ (q) .
(10.4)
The incentive constraints are then:
wf (q1L , q2L , πL , ζ (q2L )) ≥ wf (q1H , q2H , πL , ζ (q2H )) ,
(10.5)
wf (q1H , q2H , πH , ζ (q2H )) ≥ wf (q1L , q2L , πH , ζ (q2L )) .
(10.6)
The regulator’s maximizes:
γ · w (q1L , πL , 0) + δE [w (q2L , π, ζ (q2L ))] + (1 − γ) · w (q1H , πH , 0) + δE [w (q2H , π, ζ (q2H ))] (10.7)
,
subject to (10.5) and (10.6). From the first order conditions, the regulator knows that after
assigning a second period level of regulation, the firm chooses the optimal level of investment
given the regulation. Because the second period regulation is suboptimal relative to the first
best level of regulation in the second period, investment is also not first best. But investment
is optimal (for both the firm and the regulator) conditional on q2 . For the properties of the
mechanism, let q̄ζ satisfy Bq (q̄ζ ) = E [Cq (q̄ζ , π, ζ (q̄ζ ))]. Then:
PROPOSITION 8 Let R be constant in q. Then the solution to problem (10.7) satisfies
q1H < q̄ < q1L and q2L < q̄ζ < q2H .
Note q̄ζ > q̄, so if the regulator has only prior information about firm costs, regulation
becomes more stringent over time since costs fall. Under our mechanism, the regulator
offers one contract that is initially above and subsequently below the trend line of regulatory
stringency in the prior information case. The other regulation option starts out below the
trend in regulatory stringency, and then is above the trend in the second period.
16
Proof of proposition 8: Starting with the second claim, using (10.2), the first order
condition of the Lagrange form of (10.7) with respect to q2L is:
G (q2L ) ≡
B (q2L )
λ
=1+ .
E [Cq (q2L , π, ζ (q2L ))]
γ
(10.8)
Differentiating the firm first order condition (10.2) gives:
ζq (q) =
E [Cζq (q, π, ζ (q))]
.
E [Cζζ (q, π, ζ (q))]
(10.9)
The assumption that C is convex in [q,ζ] and (10.9) imply that G is a decreasing function.
Therefore, q2L < q̄ζ if and only if:
G (q̄ζ ) < G (q2L ) ,
(10.10)
which holds from (10.8) and the definition of q̄ζ . An analogous argument implies q2H > q̄ζ .
For the first claim, we first show that q1L > q1H . The incentive constraint (10.5) implies:
C (q1H , πL , 0) − C (q1L , πL , 0) =
δ (E [C (q2L , π, ζL )] + Pζ ζL − E [C (q2H , π, ζH )] − Pζ ζH ) .
(10.11)
A first order Taylor expansion of E [C (q2L , π, ζ (q2L ))] + Pζ ζ (q2L ) around q2H implies the
right and side is approximately:
r.h.s ≈ δ (E [Cq (q2H , π, ζH )] + (E [Cζ (q2H , π, ζH )] + Pζ ) ζq (q2H )) (q2L − q2H ) ,
(10.12)
= δE [Cq (q2H , π, ζH )] (q2L − q2H ) < 0.
(10.13)
Here the last equality uses (10.2). Therefore, (10.13) and (10.11) imply q1L > q1H .
Given q1L > q1H , the proof that q1H < q̄ < q1L is identical to the proof of proposition
(2), which depend only on (2.7) and (2.8), which are identical to the first period first order
conditions here.
17