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Payment for Environmental Services and Environmental Tax Under Imperfect Competition

  • Anneliese Krautkraemer ORCID logo and Sonia Schwartz EMAIL logo
Published/Copyright: July 16, 2025

Abstract

This paper designs the second-best Payment for Environmental Services (PES) when it interacts with a Pigouvian tax under market imperfections. Following Lankoski and Ollikainen (2003. “Agri-Environmental Externalities: A Framework for Designing Targeted Policies.” European Review of Agricultural Economics 30 (1): 51–75), we study the optimal allocation of land between two crops with different environmental impacts and fallow land. The regulator sets a Pigouvian tax on the agricultural production generating environmental damage and a PES on uncultivated land, as fallow buffer strips promote biodiversity. We assume an economy with market power and distortionary taxation. We show that the second-best level of the Pigouvian tax is higher than the marginal damage, contrary to Barnett (1980. “The Pigouvian Tax Rule Under Monopoly.” The American Economic Review 70 (5): 1037–41) whereas the PES is lower than the marginal benefit. The Pigouvian tax increases with the marginal social cost of public funds while the PES decreases with the marginal social cost of public funds provided that demand for the environmentally damaging agricultural good is inelastic. We thus highlight a contributory component of the environmental incentive tax. This paper also identifies specific cases where the PES is ineffective in promoting biodiversity.

JEL Classification: Q57; Q58

Corresponding author: Sonia Schwartz, LEO-UCA, Université Clermont Auvergne, Université d’Orléans, LEO, 45067 Orléans, France; and Pôle Tertiaire, 26 Avenue Léon Blum, 63000 Clermont-Ferrand, France, E-mail: 

This work benefited from the financial support of the Region AuRA (AuvergneRhône Alpes), through the Contrat de Plan Etat Region and the program BIOECO.


Appendix A: Welfare Function Concavity and Existence of the Solution

If ψ = 0, we construct the Hessian matrix, I(W) from equations (10) and (11):

I ( W ) = 2 X 1 s 2 [ F ] + X 1 s 2 [ F ] + d 2 X 2 d s 2 [ G ] + d X 2 d s 2 [ G ] 2 X 1 s t [ F ] + X 1 s X 1 t [ F ] 2 X 1 s t [ F ] + X 1 s X 1 t [ F ] 2 X 1 t 2 [ F ] + X 1 t 2 [ F ]

where

F = p 1 ( X 1 ) c 1 X 1 n B y D ( X 1 ) F = p 1 ( X 1 ) 1 n c 1 X 1 n + B y y D ( X 1 ) G = p 2 ( X 2 ) c 2 X 2 n B y G = p 2 ( X 2 ) 1 n c 2 X 2 n + B y y

Following our assumptions we can simplify the above matrix to

I ( W ) = X 1 s 2 [ F ] + d X 2 d s 2 [ G ] X 1 s X 1 t [ F ] X 1 s X 1 t [ F ] X 1 t 2 [ F ]

We have F′ < 0 and G′ < 0. We calculate the determinant of I:

D e t ( I ) = X 1 s 2 [ F ] + d X 2 d s 2 [ G ] X 1 t 2 [ F ] X 1 s X 1 t [ F ] X 1 s X 1 t [ F ]

After simplification, we obtain: D e t ( I ) = ( d X 2 d s ) 2 [ G ] ( X 1 t ) 2 [ F ] > 0 .

Thus, the welfare function is concave because the determinant is positive while [ d X 2 d s ] 2 [ G ] + [ X 1 t ] 2 [ F ] < 0 .

Equation (10) sets Ω1(s, t) and equation (11) sets Ω2(s, t). We must establish that at least one pair (s, t) satisfies Ω1(s, t) = 0 and Ω2(s, t) = 0. We use Brouwer’s fixed-point theorem. X 1(t, s) and X 2(s) are defined on [s min, s max] and [t min, t max]. Ω1(s, t) and Ω2(s, t) are continuous in s and t due to the regularity of the functions P 1(X 1), P 2(X 2), c 1(X 1), c 2(X 2), B(Y) and D(X 1). The convexity and increasing nature of the functions c 1(X 1), c 2(X 2) and D(X 1) and concavity of B(Y) guarantee that the equations are well-defined and do not diverge. Thus according to Brouwer’s fixed-point theorem, the solution (s, t) exists.

Because the welfare function is concave, the solution given by equations (10) and (11) is so unique.

If ψ > 0, (17) sets a function h(t). We obtain:

h ( t ) = d 2 W d t 2 = d 2 X 1 d t 2 p 1 ( X 1 ) p 2 ( T X 1 ) c 1 X 1 n + c 2 T X 1 n D ( X 1 ) + d X 1 d t 2 p 1 ( X 1 ) + p 2 ( T X 1 ) 1 n c 1 X 1 n 1 n c 2 T X 1 n D ( X 1 )

Under our assumptions, we have:

h ( t ) = 2 W t 2 = d X 1 d t 2 p 1 ( X 1 ) + p 2 ( T X 1 ) 1 n c 1 X 1 n 1 n c 2 T X 1 n D ( X 1 ) < 0

The limit h(t) when t → 0 is positive and the limit h(t) when t → +∞ is −∞ because X ̄ 1 n u ( t ) t < 0 and X ̄ 2 n u ( t ) t > 0 . According to the intermediate value theorem, it exists t nu such as h(t nu ) = 0. As h′(t) < 0 the welfare function is concave when ψ > 0. The solution given by equation (17) is so unique.

Appendix B: Welfare Function Concavity and Existence of the Solution with MCF

If ψ = 0, we use (20) and (21) to create the Hessian matrix:

H = a b c d

where

a = 2 X 1 s 2 [ A + ϵ ( t + s ) ] + X 1 s 2 [ A ] + 2 ϵ X 1 s + d 2 X 2 d s 2 [ B + ϵ s ] + d X 2 d s 2 [ B ] + 2 ϵ d X 2 d s b = 2 X 1 s t [ A + ϵ ( t + s ) ] + X 1 t X 1 s [ A ] + ϵ X 1 s + X 1 t c = 2 X 1 t s [ A + ϵ ( t + s ) ] + X 1 t X 1 s [ A ] + ϵ X 1 s + X 1 t d = 2 X 1 t 2 [ A + ϵ ( t + s ) ] + X 1 t 2 [ A ] + 2 ϵ X 1 t

and

A = p 1 ( X 1 ) c 1 X 1 n B y D ( X 1 ) B = p 2 ( X 2 ) c 2 X 2 n B y A = p 1 ( X 1 ) 1 n c 1 X 1 n + B y y D ( X 1 ) B = p 2 ( X 2 ) 1 n c 2 X 2 n + B y y

Thanks to our assumptions, we can simplify the Hessian to:

H = X 1 s 2 [ A ] + d X 2 d s 2 [ B ] + 2 ϵ X 1 s + d X 2 d s X 1 t 2 [ A ] + 2 ϵ X 1 t X 1 t 2 [ A ] + 2 ϵ X 1 t X 1 t 2 [ A ] + 2 ϵ X 1 t

So the determinant is:

D e t = X 1 s 2 [ A ] + d X 2 d s 2 [ B ] + 2 ϵ X 1 s + d X 2 d s X 1 t 2 [ A ] + 2 ϵ X 1 t X 1 t 2 [ A ] + 2 ϵ X 1 t 2

Simplifying, we find:

D e t = d X 2 d s 2 X 1 t 2 [ A ] [ B ] + 2 ϵ X 1 t d X 2 d s d X 2 d s [ B ] + X 1 t [ A ] + 4 ϵ 2 d X 2 d s X 1 t > 0

With A′ < 0 and B′ < 0, we find a positive determinant. And, because ( X 1 s ) 2 [ A ] + [ d X 2 d s ] 2 [ B ] + 2 ϵ ( X 1 s + d X 2 d s ) < 0 , we have a concave function.

Equation (20) sets ϒ1(s, t) and equation (21) sets ϒ2(s, t). We must establish that at least one pair (s, t) satisfies ϒ1(s, t) = 0 and ϒ2(s, t) = 0. We use Brouwer’s fixed-point theorem. X 1(t, s) and X 2(s) are defined on [s min, s max] and [t min, t max]. ϒ1(s, t) and ϒ2(s, t) are continuous in s and t due to the regularity of the functions P 1(X 1), P 2(X 2), c 1(X 1), c 2(X 2), B(Y) and D(X 1). The convexity and increasing nature of the functions c 1(X 1), c 2(X 2) and D(X 1) and concavity of B(Y) guarantee that the equations are well-defined and do not diverge. Thus according to Brouwer’s fixed-point theorem, the solution (s, t) exists.

As the welfare function is concave, the solution given by equations (20) and (21) is unique.

If ψ > 0, equation (23) sets g(t):

g ( t ) = d 2 W d t 2 = d 2 X 1 d t 2 [ E + ϵ t ] + d X 1 d t 2 [ E ] + 2 ϵ d X 1 d t

where

E = p 1 ( X 1 ) p 2 ( T X 1 ) c 1 X 1 n + c 2 T X 1 n D ( X 1 ) E = p 1 ( X 1 ) + p 2 ( T X 1 ) 1 n c 1 X 1 n 1 n c 2 T X 1 n D ( X 1 ) < 0

With our assumptions we can simplify this to:

g ( t ) = d 2 W d t 2 = d X 1 d t 2 [ E ] + 2 ϵ d X 1 d t < 0

The limit g(t) when t → 0 is positive and the limit g(t) when t → +∞ is −∞ because X ̄ 1 n u ( t ) t < 0 and X ̄ 2 n u ( t ) t > 0 . According to the intermediate value theorem, it exists t n u mcf such as g t n u mcf = 0 . As g′(t) < 0, the welfare function is concave when ψ > 0. The solution given by equation (23) is so unique.

Appendix C: Tax and PES Changes with MCF if ψ > 0

According to (22), t and s depend on ϵ. Moreover t and s must satisfy conditions (20) and (21). We set:

q = p 1 ( X 1 ( t ( ϵ ) , s ( ϵ ) ) ) 1 n c 1 X 1 ( t ( ϵ ) , s ( ϵ ) ) n D ( X 1 ( t ( ϵ ) , s ( ϵ ) ) ) < 0 z = p 2 ( X 2 ( s ( ϵ ) ) ) 1 n c 2 X 2 ( s ( ϵ ) ) n < 0

Additionally, we know: X 1 t = X 1 s < 0 .

We differentiate (20) and (21) with respect to ϵ and rearrange the equations into the following matrix form:

K d s d ϵ d t d ϵ = X 1 s [ t + s ] X 2 s s + ( T X 1 X 2 ) X 1 t ( t + s ) X 1

where K = i j k l , with:

i = X 1 s ( q + B y y ) X 1 s + 2 ϵ + B y y X 2 s + X 2 s ( z + B y y ) X 2 s + B y y X 1 s + 2 ϵ j = X 1 s ( q + B y y ) X 1 t + 2 ϵ + B y y X 2 s k = X 1 t ( q + B y y ) X 1 s + B y y X 2 s + 2 ϵ l = X 1 t ( q + B y y ) X 1 t + 2 ϵ

We multiply each side of the equation by K −1 to isolate d s d ϵ and d t d ϵ :

(A1) d s d ϵ d t d ϵ = K 1 X 1 s [ t + s ] X 2 s s + ( T X 1 X 2 ) X 1 t ( t + s ) X 1

where

K 1 = 1 det K l j k i

We calculate det K:

D e t = X 1 t ( q + B y y ) X 1 t + 2 ϵ X 1 s ( q + B y y ) X 1 s + 2 ϵ + B y y X 2 s + X 2 s ( z + B y y ) X 2 s + B y y X 1 s + 2 ϵ X 1 s ( q + B y y ) X 1 t + 2 ϵ + B y y X 2 s 2 D e t = X 1 t 2 X 2 s 2 q z + X 1 t 2 X 2 s 2 q B y y + X 1 t 2 X 2 s 2 z B y y + 2 X 1 t 2 X 2 s q ϵ + 2 X 1 t X 2 s 2 z ϵ + 2 X 1 t 2 x 2 s B y y ϵ + 2 X 1 t X 2 s 2 B y y ϵ + 4 X 1 t X 2 s ϵ 2 > 0

because q < 0 and z < 0, X 1 t = X 1 s < 0 and X 2 s < 0 .

We calculate d s d ϵ , using (A1):

s ϵ = 1 det X 1 t ( q + B y y ) X 1 t + 2 ϵ X 1 s [ t + s ] X 2 s s + ( T X 1 X 2 ) + 1 det X 1 s ( q + B y y ) X 1 t + 2 ϵ + B y y X 2 s X 1 t ( t + s ) X 1 s ϵ = 1 det X 1 t 2 X 2 s q s + X 1 t 2 q ( T X 2 ) + X 1 t 2 X 2 s t B y y > 0 + X 1 t 2 B y y ( T X 2 ) + X 1 t X 2 s X 1 B y y 2 X 1 t X 2 s s ϵ + 2 X 1 t ϵ ( T X 2 )

So  s ϵ < 0  if  X 1 t 2 X 2 s t B y y + X 1 t X 2 s X 1 B y y < 0 X 1 t x 2 s B y y X 1 t t + X 1 < 0

i.e. if X 1 t t + X 1 > 0 X 1 t t X 1 + 1 > 0 X 1 t t X 1 e X 1 / t > 1 .

So s ϵ < 0 if e X 1 / t > 1 .

We calculate d t d ϵ , using (A1):

t ϵ = 1 det X 1 t ( q + B y y ) X 1 s + B y y X 2 s + 2 ϵ × X 1 s ( t + s ) X 2 s s + ( T X 1 X 2 ) + X 1 s ( q + B y y ) X 1 s + 2 ϵ + B y y X 2 s + X 2 s ( z + B y y ) X 2 s + B y y X 1 s + 2 ϵ × X 1 t ( t + s ) X 1 t ϵ = 1 det X 1 t 2 X 2 s q s X 1 t X 2 s 2 z s X 1 t X 2 s 2 z t X 1 t 2 q T X 2 s 2 z x 1 + X 1 t 2 q x 2 X 1 t 2 X 2 s t B y y X 1 t X 2 s 2 t B y y X 1 t 2 T B y y X 1 t X 2 s T B y y X 1 t X 2 s X 1 B y y X 2 s 2 X 1 B y y + X 1 t 2 X 2 B y y + X 1 t X 2 s X 2 B y y 2 X 1 t X 2 s t ϵ 2 X 1 t T ϵ 2 X 2 s X 1 ϵ + 2 X 1 t X 2 ϵ t ϵ = 1 det X 1 t 2 X 2 s q s X 1 t 2 q ( T X 2 ) X 1 t 2 B y y ( T X 2 ) 2 X 1 t ϵ ( T X 2 ) X 1 t 2 X 2 s t B y y X 1 t X 2 s X 1 B y y X 1 t X 2 s 2 z s X 1 t X 2 s 2 z t X 1 t X 2 s B y y ( T X 2 ) X 2 s 2 X 1 B y y 2 X 2 s X 1 ϵ X 2 s 2 z X 1 2 X 1 t X 2 s t ϵ X 1 t X 2 s 2 t B y y

We know that

X 1 t 2 X 2 s t B y y X 1 t X 2 s X 1 B y y > 0 X 1 t X 2 s B y y [ X 1 t t X 1 ] > 0 X 1 t t X 1 > 1

2 X 1 t X 2 s t ϵ 2 X 2 s X 1 ϵ > 0 2 X 2 s ϵ [ X 1 t t + X 1 ] > 0 X 1 t t X 1 > 1

X 1 t X 2 s 2 z t X 2 s 2 z X 1 > 0 if X 2 s 2 z [ X 1 t t + X 1 ] > 0 X 1 t t X 1 > 1

X 2 s 2 X 1 B y y X 1 t X 2 s 2 t B y y > 0 if X 2 s 2 B y y [ X 1 t t + X 1 ] > 0 X 1 t t X 1 > 1

X 1 t X 2 s 2 z s + X 1 t 2 X 2 s q s > 0 X 1 t X 2 s s [ X 1 t q X 2 s z ] > 0 [ X 1 t q X 2 s z ] > 0 X 1 t / X 2 s > z / q ϖ .

t ϵ > 0 if e X 1 / t > 1 and X 1 t / X 2 s > ϖ ( ϵ ) .

Appendix D: Tax and PES Changes with MCF if ψ = 0

We use (23) and we set: J ( t , ϵ ) = d X 2 d t p 1 ( T X 2 ) + p 2 ( X 2 ) + c 1 ( T X 2 n ) c 2 ( X 2 n ) + D ( T X 2 ) ϵ t + ϵ ( T X 2 ) . Applying the implicit function theorem we find:

d t d ϵ = J ϵ J t = d X 1 d t t + X 1 d X 1 d t p 1 ( X 1 ) + p 2 ( T X 1 ) 1 n c 1 X 1 n 1 n c 2 T X 1 n D ( X 1 ) + 2 d X 1 d t ϵ

We know that the denominator of the above expression is negative. So we obtain d t d ϵ > 0 if e X 1 / t > 1 .

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Received: 2024-04-04
Accepted: 2025-03-10
Published Online: 2025-07-16

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