Regulatory Contestability and Cost Pass-Through

Dennis L. Weisman

doi:10.1515/bejeap-2024-0379

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Regulatory Contestability and Cost Pass-Through

Dennis L. Weisman

Veröffentlicht/Copyright: 5. Februar 2025

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Aus der Zeitschrift The B.E. Journal of Economic Analysis & Policy Band 25 Heft 2

Abstract

Regulatory contestability posits that the threat of economic regulation disciplines pricing even in the absence of explicit regulatory mechanisms (e.g., cost-of-service regulation or price cap regulation). Formal modeling and numerical simulations reveal how regulatory contestability interacts with cost-reducing effort to limit the pass-through of costs. This may suggest that under certain conditions regulatory contestability can substitute for explicit regulatory mechanisms and potentially reduce the cost of regulation.

Keywords: regulatory contestability; competition; cost pass-through; cost-reducing effort

JEL Classification: L51; L96; L98

Corresponding author: Dennis L. Weisman, Department of Economics, Kansas State University, Manhattan KS, 66506-4001, USA, E-mail: weisman@ksu.edu

Acknowledgments

I am grateful to an anonymous referee and the editor, Till Requate, for thoughtful suggestions for revision that improved the original submission.

Appendix

Proof of Proposition 1

The result follows upon rearranging the terms in (1) and appealing to the definition of ɛ. ■

Proof of Proposition 2

(A1) sgn d p * d M = sgn τ ′ p d 2 π / d p 2 = + − < 0 .

(A2) sgn d p * d c = sgn D ′ p d 2 π / d p 2 = − − > 0 .

(A3) sgn ∂ ∂ M d p ∗ d c = sgn τ ′ ′ p D ′ p d 2 π / d p 2 2 = − + < 0 .

■

Proof of Proposition 3

Let the subscripts denote partial derivatives. Differentiating (4) and (5) with respect to p and e yields

(A4) π p p = 2 D ′ p + p − c D ′ ′ p − τ ′ ′ p M < 0 ;

(A5) π p e = π e p = − c ′ e D ′ p < 0 ;

and

(A6) π e e = − c ′ ′ e D p − ψ ′ ′ e < 0 .

Sufficient second-order conditions for a maximum require that π _pp < 0, π _ee < 0 and π _pp π _ee − π _pe π _ep > 0. Employing Cramer’s rule,

(A7) d p ∗ d M = τ ′ p π p e 0 π e e π p p π e e − π p e π e p = τ ′ p π e e π p p π e e − π p e π e p < 0 ;

(A8) d e ∗ d M = π p p τ ′ p π e p 0 π p p π e e − π p e π e p = − τ ′ p π e p π p p π e e − π p e π e p > 0 .

(A9) d p ∗ d c 0 = 1 − δ e D ′ p π p e − δ ′ e D p π e e π p p π e e − π p e π e p = 1 − δ e D ′ p π e e ︷ + + π p e δ ′ e D p ︷ − π p p π e e − π p e π e p .

Setting the numerator on the right-hand side of (A9) to be less (greater) than or equal to zero and solving for δ ′ e yields Proposition 3(iii) since π _pp π _ee − π _pe π _ep > 0.

(A10) d e ∗ d c 0 = π p p 1 − δ e D ′ p π e p − δ ′ e D p π p p π e e − π p e π e p = − π p p δ ′ e D p ︷ + − 1 − δ e D ′ p π e p ︷ + π p p π e e − π p e π e p .

Setting the numerator on the right-hand side of (A10) to be greater (less) than or equal to zero and solving for δ ′ e yields Proposition 3(iv) since π _pp π _ee − π _pe π _ep > 0. ■

Proof of Proposition 4

It follows from (A4) and (A9) that

(A11) ∂ ∂ M d p ∗ d c 0 = τ ′ ′ p π e e 1 − δ e D ′ p π e e + π p e δ ′ e D p π p p π e e − π p e π e p 2 < 0

since τ ′ ′ p > 0 , π _ee < 0 and 1 − δ e D ′ p π e e + π p e δ ′ e D p > 0 when d p ∗ d c 0 > 0 _.

It follows from (A4) and (A10) that

(A12) ∂ ∂ M d e ∗ d c 0 = − τ ′ ′ p π p p π e e δ ′ e D ( p ) + π p e π e p δ ′ e D p − π p p π e e δ ′ e D p + 1 − δ ( e ) D ′ ( p ) π e e π e p π p p π e e − π p e π e p 2 .

Cancelling terms and simplifying yields

(A13) ∂ ∂ M d e ∗ d c 0 = − τ ′ ′ p π e p 1 − δ e D ′ p π e e + π p e δ ′ e D ( p ) π p p π e e − π p e π e p 2 > 0

since τ ′ ′ p > 0 , π _ep < 0 and 1 − δ e D ′ ( p ) π e e + π p e δ ′ e D ( p ) > 0 when d p ∗ d c 0 > 0 from (A9). ■

Proof of Proposition 5

The necessary first-order conditions for a maximum in [M-L-P] are given by:

(A14) p : − 2 b − γ M p − b c 0 θ e + a + b c 0 = 0 ;

and

(A15) e : a − b p θ c 0 − 2 e = 0 .

These first-order conditions can be expressed in matrix form by

(A16) − 2 b − γ M − b c 0 θ − b c 0 θ − 2 p e = − a − b c 0 − a c 0 θ

Applying Cramer’s rule to (A16) yields

(A17) p * = − a − b c o − b c o θ − a c o θ − 2 − 2 b − γ M − b c o θ − b c o θ − 2 = 2 a + b c 0 − a b c 0 θ 2 4 b + 2 γ M − b c 0 θ 2 ⇒

(A18) D p * = a − b p * = 2 a b + γ M − b 2 c 0 4 b + 2 γ M − b c 0 θ 2 ;

and

(A19) e * = − 2 b − γ M − a − b c o − b c o θ − a c o θ − 2 b − γ M − b c o θ − b c o θ − 2 = c 0 θ b a − b c 0 + a γ M 4 b + 2 γ M − b c 0 θ 2 .

■

Proof of Proposition 6

Totally differentiating (A14) and (A15) with respect to M yields:

(A20) − 2 b − γ M − b c 0 θ − b c 0 θ − 2 d p * d M d e * d M = γ p * 0 .

Applying Cramer’s rule to (A20) yields

(A21) d p * d M = γ p * − b c o θ 0 − 2 − 2 b − γ M − b c o θ − b c o θ − 2 = − 2 γ p * 4 b + 2 γ M − b c 0 θ 2 = − 2 γ 2 a + b c 0 − a b c 0 θ 2 4 b + 2 γ M − b c 0 θ 2 2 < 0

since p * > 0 ⇒ 2 a + b c 0 − a b c 0 θ 2 > 0 from (A17). This proves part 6(i). Similarly,

(A22) d e * d M = − 2 b − γ M γ p * − b c 0 θ 0 − 2 b − γ M − b c o θ − b c o θ − 2 = γ p * b c 0 θ 4 b + 2 γ M − b c 0 θ 2 = γ b c 0 θ 2 a + b c 0 − a b c 0 θ 2 4 b + 2 γ M − b c 0 θ 2 2 > 0

since p * > 0 ⇒ 2 a + b c 0 − a b c 0 θ 2 > 0 from (A17). This proves part 6(ii). Totally differentiating (A14) and (A15) with respect to c ₀ yields:

(A23) − 2 b − γ M − b c 0 θ − b c 0 θ − 2 d p * d c 0 d e * d c 0 = b θ e * − 1 − a − b p * θ .

Applying Cramer’s rule to (A23) yields

(A24) d p * d c 0 = b θ e * − 1 − b c o θ − a − b p * θ − 2 − 2 b − γ M − b c o θ − b c o θ − 2 = 2 b 1 − θ e * − b a − b p * c 0 θ 2 4 b + 2 γ M − b c 0 θ 2 = 4 b 2 b + γ M − 2 θ 2 b c 0 2 a b − b 2 c 0 + 2 a γ M 4 b + 2 γ M − b c 0 θ 2 2

upon substituting p* and e* from (A17) and (A19), respectively, and simplifying. Setting the expression in (A24) equal to zero and solving for θ yields θ ̂ . This proves part 6(iii).

(A25) d e * d c 0 = − 2 b − γ M b θ e * − 1 − b c 0 θ − a − b p * θ − 2 b − γ M − b c o θ − b c o θ − 2 = 2 b + γ M a − b p * θ + b 2 c 0 θ θ e * − 1 4 b + 2 γ M − b c 0 θ 2 ⇒

(A26) d e * d c 0 = θ a b + a γ M b 2 c 0 2 θ 2 + 4 b + 2 γ M − 2 b 2 c 0 4 b + 2 γ M 4 b + 2 γ M − b c 0 θ 2 2

upon substituting p* and e* from (A17) and (A19), respectively, and simplifying. Setting the expression in (A26) equal to zero and solving for θ yields θ ̃ , which is well-defined for a 2 c 0 ≤ b 2 b + γ M ≤ b . This proves part 6(iv). ■

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Received: 2024-10-30

Accepted: 2025-01-15

Published Online: 2025-02-05

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https://doi.org/10.1515/bejeap-2024-0379

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regulatory contestability; competition; cost pass-through; cost-reducing effort