When do Firms Break the Law in
Order to Reduce Marginal Cost? An Application to the Problem of
Environmental Inspection
Jonas Häckner and Mathias Herzing
Abstract
This study attempts to identify …rm characteristics that are important in determining whether or not a speci…c …rm has strong
incentives for non-compliance with environmental laws. In particular, we analyze how these incintives are related to the size
of the cost reductions associated with non-compliance, business
cycle conditions, the degree of product di¤erentiation, market
structure, and price versus quantity competition. When cost reductions are non-dramatic, in the sense that they do not lead to
monopoly, the following rules of thumb are suggested. 1) Inspection should be intensi…ed during booms, 2) …rms that face high
costs of compliance should be inspected more intensely and 3)
…rms that are insulated from competition by product di¤erentiation or by lack of competitors should be inspected more intensely.
Although our prime focus is environmental inspection, the theoretical …ndings readily extends to other similar applications such
as VAT fraud and violations against import restrictions. They
can also have some bearing on the monitoring of …nancial markets that are subject to regulation.
JEL Classi…cation: Q58, L13, K32
Keywords: Environmental Inspection, Market Structure, Product Di¤erentiation, Bertrand, Cournot
Correspondence: Department of Economics, Stockholm University, 106 91 Stockholm, Sweden. E-mail: [email protected]. Financial support from the Swedish
Environmental Protection Agency (Naturvårdsverket) is gratefully acknowledged.
1
1
Introduction
Firms often have incentives to try to reduce marginal costs by violating
various types of legislation. The focus of this study is on environmental
inspections and the incentives to violate environmental law, but the results extend readily to similar examples such as VAT fraud and violations
of various types of import restrictions.They can also have some bearing
on the monitoring of …nancial markets that are subject to regulation.
Environmental concerns play an increasingly important role in modern societies. Individual consumers a¤ect emissions through their consumption patterns. Firms’production choices have an important impact
on the environment both globally, e.g. through emissions of greenhouse
gases, and locally, by polluting the environment surrounding the production facilities. Since the core problem is a standard economic externality,
the market solution will fail to internalize the environmental impact of
consumption and production decisions. Hence, there is scope for government intervention. Indeed, most countries try to a¤ect the behavior of
economic agents by implementing various laws protecting the environment, and in order to make sure that …rms, in particular, live up to the
standards it is also common to introduce some kind of environmental
inspection and enforcement regime.
There is a fairly large economics literature on environmental inspection. Following Cohen (1998), it can be divided into a positive strand,
that aims at understanding the objectives of …rms and inspection agencies, and a normative strand focusing on optimal regulation.
The most important positive question is perhaps the "Harrington
paradox" which essentially asks why …rms tend to comply with environmental laws to such a great extent despite the fact that monitoring is
limited and …nes are low. The explanation o¤ered by Harrington (1988)
is that non-complying …rms face the threat of being re-categorized into a
group of …rms facing tougher regulation and higher …nes. See e.g., Heyes
& Rickman (1999), Heyes (1996) for complementary explanations. Other
important positive questions relate to the objective function applied by
the inspection agency. Is the agency budget maximizing (Lee, 1983),
compliance maximizing (Garvie & Keeler, 1994) or perhaps governed by
a median voter process (Selden & Terrones, 1993)?
Normative studies typically look at the interplay between a representative …rm and an inspection agency, given various assumptions regarding e.g. informational structures. While …rms are often modeled
quite simplistically, the inspection and enforcement technology tends to
be very carefully described. Important tradeo¤s studied include those
2
between inspection and self-reporting (See e.g., Kaplow & Shavell (1994)
and Bontems & Rotillon (2000), between ex ante and ex post monitoring (Cohen (1987)) and between inspection intensity and …nes in the
tradition of Becker (1968) (See e.g., (Cohen (1987)).
Our study has a normative perspective too, but the focus is more
or less reversed in relation to most previous studies. We focus on …rm
heterogeneity and take an agnostic position regarding the exact inspection technology. Speci…cally, we try to model carefully characteristics
that we believe are important in determining whether or not a …rm has
strong incentives for non-compliance.
The situation we have in mind is the following. There is an environmental inspection agency. Its task is to make sure that all …rms in the
region comply with the law. Although the exact technology available
to the agency is not explicitly modeled, we assume that more resources
must be devoted to …rms facing stronger incentives to break the law. A
central assumption is that compliance with the law increases marginal
production costs.1
Firms belong to markets that may, or may not, coincide with the
jurisdiction of the agency. Some …rms belong to concentrated markets,
others to competitive markets. Some …rms produce di¤erentiated products, while others do not. Some …rms may cut costs a lot by deviating
while others can comply at a low cost. Some …rms compete à la Cournot
while others compete à la Bertrand. Finally, the temptation to deviate
might be a¤ected by the business cycle. The objective of this study
is simply to try to identify …rms that are especially likely to break the
law. Evidently, those are …rms that the environmental inspection agency
should devote special attention to. Note that we assume that the agency
allocates its resources in a way that is successful in the sense that it actually prevents all …rms from breaking the law. Hence, from a technical
point of view, the experiment is basically to compare symmetric pro…ts
under compliance with the pro…t obtained by a …rm that unilaterally
deviates from compliance.2
From a theoretical point of view, it can of course be argued that
an inspection agency that has perfect information about market mechanisms should be able to identify non-compliance immediately just by
observing changes in quantities and prices. From a real life perspective,
1
It is conceivable that clean production processes are associated with increases
in both …xed and variable costs, but here we focus entirely on the latter e¤ect.
Obviously, adding …xed costs would introduce a strong link between the level of
these costs and market structure.
2
This makes our analysis di¤erent from studies on cost reducing innovations.
There the focus is on cost reductions as an equilibrium outcome. See d’Aspremont
& Jacquemin (1988) for an early reference.
3
however, changes in prices and quantities may have hundreds of explanations. Hence, it is virtually impossible for an inspection agency (or a
competing …rm) to attribute them to compliance speci…cally.
Our main …ndings are the following. Not surprisingly, the incentives
to deviate from compliance are stronger the larger the cost saving and
the higher the overall level of demand in the economy.
There is a u-shaped relationship between the degree of product differentiation and the incentives to deviate, i.e., incentives are strongest
when di¤erentiation is very small and very large. Basically, when di¤erentiation is small a deviating …rm may increase demand signi…cantly by
reducing price by a small amount. Hence, the increase in pro…ts will be
large. When di¤erentiation is large the deviator does not have to bother
much about the strategic responses of the competitors. This too results
in a large increase in pro…ts.
The relationship between deviation incentives and market structure
is the following. For small cost savings, incentives are weakened as the
number of …rms increases. If the cost reduction is large, the opposite
relation holds. When the market is competitive, prices are low initially.
Hence a deviator gaining a small cost advantage will …nd it di¢ cult to
exploit this in a way that increases pro…ts signi…cantly. When cost savings are large however, they will permit the deviator to earn signi…cant
pro…ts. Hence, given the competitive point of departure, the deviator
will experience a large increase in pro…ts.
Finally, when the cost saving is small, quantity competition implies
stronger deviation incentives than price competition. If the cost reduction is large, the opposite relation holds. The intuition is the following.
Suppose that a …rm is considering a deviation from compliance and that
…rms compete in prices. Typically, price competition is more intense
than quantity competition. Hence, price competition will lead to a situation where initial prices are low. If, in that case, a deviation leads to
a small reduction in cost, the deviator will not be in a position where
he/she can use this advantage in a way that increases pro…ts to a large
extent. Under quantity competition, initial prices are higher, which
makes it easier to bene…t from the cost advantage. If the cost reduction
is large the situation is di¤erent. Then, it is relatively easy to increase
pro…ts by capturing a large fraction of the competitors’demand. However, attracting a large fraction of industry demand is less costly under
Bertrand than under Cournot. The reason is that a deviator competing in quantities will have to expand output a lot in order to achieve
this, thereby putting a signi…cant downward pressure on price. Under
Bertrand competition, a relatively small reduction in price is generally
su¢ cient to shift demand, and especially so when products are close
4
substitutes.
The paper is organized as follows. In section 2 the basic model is
presented and equilibria are characterized. Section 3 presents the main
results. The paper concludes with some …nal remarks in section 4.
2
The Model
The basic model is taken from Häckner (2000) where the utility function
is of the following type:
U (q; I) =
n
X
i=1
qi
i
n
n X
n 1
X
1 X 2
(
qi qj ) + I:
qi + 2
2 i=1
i=1 j6=i
(1)
Thus, utility is quadratic in the consumption of q-goods and linear
in the consumption of other goods I. The parameter 2 [0; 1] measures the substitutability between the products. If = 0, each …rm
has monopolistic market power, while if = 1, the products are perfect
substitutes.3 Finally, i is a …rm speci…c demand shifting parameter.
However, we will assume that i = for all i. Hence, can be can be
treated as a parameter re‡ecting the business P
cycle. Consumers maximize utility subject to the budget constraint
pi qi + I
m, where
m denotes income and the price of the composite good is normalized to
one. The …rst-order condition determining the inverse demand of good
k is
X
@U
=
qk
qj pk = 0:
(2)
@qk
j6=k
Now, let ck be the constant marginal cost of a …rm violating environmental legislation by using a low-cost, high-pollution technology. Also,
assume that all other …rms comply with the law and that they face a
common marginal cost c such that ck < c. To simplify notation, let us
de…ne the parameters zk
ck and z
c.
There are basically two ways to proceed when calculating the level of
pro…ts for a deviating …rm. Either, one can assume that competing …rms
observe true cost levels and react strategically to changes in these, or it
can be assumed that changes in cost come as a surprise, in which case
there is no strategic response. In the latter scenario, competitors’will of
course realize that they are not maximizing pro…ts and hence they will
want to adjust their behavior accordingly. Considering that real world
…rms learn over time, we believe that the former path is more realistic.
3
A negative implies that the goods are complementary, a scenario that will not
be further analyzed in this study.
5
However, as will be discussed in section 3, most results still hold if the
assumption of strategic responses by compliant …rms is relaxed.
If …rm k unilaterally deviates from compliance with environmental
legislation, the other …rms will adapt by reducing production. There are
two possible outcomes, either an interior solution with all …rms continuing to produce some quantity, or a corner solution, where it is optimal
for the other …rms to cease production.4
Given that …rms i 6= k do not cease production, the pro…t maximizing
quantity and price of …rm k under Cournot competition equal (see the
Appendix, A1.1.1.1 and A1.2.1.1, for derivation):
qkC = pC
ck =
k
1
2 + (n 1)
z+
2 + (n
2
2)
(zk
z) ;
(3)
while under Bertrand competition:
(1
)[2 + (2n 3) ]z
+[2 + 3(n 2) + (n2 5n + 5) 2 ](zk
[2 + (n 3) ][2 + (2n 3) ]
pB
k = ck +
z)
(4)
and
(1
)[2 + (2n 3) ]z
+[2 + 3(n 2) + (n2 5n + 5) 2 ](zk z)
B
qk = [1 + (n 2) ]
:
(1
)[1 + (n 1)][2 + (n 3) ][2 + (2n 3) ]
(5)
Letting M 2 fB; Cg denote the mode of competition, the symmetric
equilibrium pro…t under compliance is given by
M
M 2
=
z ;
M
where M
( ; n) > 0 (unless = 1 and n 2 when …rms compete
in prices, in which case B (1; n) = 0) captures the e¤ects of the degrees
of product di¤erentiation and market concentration on pro…ts. More
speci…cally (see the Appendix, A1.1.1.1 and A1.2.1.1, for derivation),
1
;
[2 + (n 1) ]2
(1
)[1 + (n 2) ]
B
( ; n) =
:
[1 + (n 1) ][2 + (n 3) ]2
C
( ; n) =
4
It is assumed that a …rm’s capacity for production is not eliminated instantly
and hence, …rms remain present in the market even when no production is optimal.
If, however, …rm k is able to maintain its cost advantage, it is likely that a monopoly
will eventually emerge as competitors leave the market.
6
It is easily veri…ed that
@
M
@
@
< 0 for n
2 (
M
( ; 1) = 14 ) and that
M
< 0 (unless = 1 and n 2 when …rms compete in prices). Thus,
can be interpreted as an inverse measure for the degree of market
competition. A higher number of …rms or a lower degree of product
di¤erentiation decrease the value of M and hence, pro…ts fall as the
degree of competition increases.
Under Cournot competition the pro…t maximizing price and quantity of …rms i 6= k equal (see the Appendix, A1.1.1.2 and A1.2.1.2, for
derivation):
@n
M
qiC = pC
i
c=
1
2 + (n 1)
z
2
(zk
z) ;
while under Bertrand competition:
pB
i = c+
(1
)[2 + (2n 3) ]z
[1 + (n 2) ](zk
[2 + (n 3) ][2 + (2n 3) ]
z)
and
qiB = [1 + (n
2) ]
(1
(1
)[2 + (2n 3) ]z
[1 + (n 2) ](zk z)
:
)[1 + (n 1) ][2 + (n 3) ][2 + (2n 3) ]
For notational convenience, we de…ne the variable t = zzk = cck 2
(0; 1]. Basically, t is a measure of the cost saving obtained by a noncomplying …rm. When the cost saving is small, t is close to one, but the
lower is ck , the lower is t.
There will be an interior solution, such that qiM > 0, if and only if
M
t > t , where (see the Appendix, A1.1.1.2 and A1.2.1.2, for derivation)
C
t =
B
2
and t =
[1 + (n
2 + 2(n 2)
2) ]
(n 1)
2
:
Hence, to avoid a monopoly-like outcome, the degree of cost savings
must not be too large. It is easily established that the threshold value
M
t increases in , taking on values in the interval [0; 21 ] in the Cournot
case, implying that the above condition is satis…ed for all degrees of
1
product di¤erentiation when cost savings are su¢ ciently small (t
),
2
and taking on values in the interval [0; 1] in the Bertrand case. Thus,
the above condition is not satis…ed for any degree of cost savings when
C
B
products are undi¤erentiated and …rms compete in prices. Since t < t
for > 0, a corner solution is more likely in the Bertrand case, which
is to be expected, given that the scope for eliminating competitors is
larger when …rms compete in prices, especially at low degrees of product
7
C
di¤erentiation. The threshold value t is independent of the degree
of market concentration, as can be expected when …rms compete in
quantities; if the degree if product di¤erentiation is not too low or the
degree of cost savings is not too large, there is still scope for compliant
…rms to produce some small quantity when facing a violating competitor.
B
The threshold value t , however, also depends on the degree of market
B
concentration. It is easily established that t increases in n when < 1
B
(t = 1 for = 1), taking on values in the interval [ 2 ; 2 ). When …rms
compete in prices the e¤ect on pro…ts of being exposed to a violation by
a competitor is stronger the …ercer is competition and hence, an interior
solution is less likely the larger is the number of …rms.
M
Given that the degree of cost savings is not too large (i.e. t > t ),
such that …rm k’s violation does not lead to zero production by compliant
…rms, its pro…ts will be given by
DM
k
=
M
[z + (zk
z)
M 2
];
M
where M
( ; n) > 0 captures the e¤ects of the degrees of product di¤erentiation and market concentration on the extent to which cost
savings impact on pro…ts under unilateral deviation, taking strategic
responses by competitors into account. More speci…cally (see the Appendix, A1.1.1.1 and A1.2.1.1, for derivation),
2 + (n 2)
;
2
2 + 3(n 2) + (n2 5n + 5)
B
( ; n) =
(1
)[2 + (2n 3) ]
C
( ; n) =
It is easily veri…ed that
@
M
@
@
> 0 for n
2 (
M
2
:
( ; 1) = 1) and that
M
> 0 (unless = 1 when …rms compete in prices, in which case,
B
however t = 1, such that there exist no interior solutions for t < 1).
Thus, M can be interpreted as a measure for the impact of the degree
of market competition on cost savings. A higher number of …rms or
a lower degree of product di¤erentiation increase the value of M ; a
more competitive market increases the scope for taking advantage of
M
unilaterally lowering costs by violating legislation. When t > t , the
pro…tability of violating the law is denoted by M and given by
@n
M
DM
M
= (zk
z)
M
M
[2z + (zk
z)
M
]:
M
If the degree of cost savings is su¢ ciently large (i.e. t t ), …rms
i 6= k cease to produce. In this case the optimal quantities and prices of
…rm k are given by (see the Appendix, A1.1.2 and A1.2.2, for derivation)
1
ck = zk
2
C
qC
k = pk
8
in the Cournot case, while under Bertrand
pB
k = ck +
qB
k =
(1
(1
)z + [1 + (n 2) ](zk z)
;
2[1 + (n 2) ]
)z + [1 + (n 2) ](zk z)
:
2(1
)[1 + (n 1) ]
Hence, …rm k’s pro…ts are given by
DC 1 2
k = zk
4
when …rms compete in quantities, while under price competition (see
the Appendix, A1.2.2, for derivation)
DB
=
B
[z +
B
(zk
z)]2 ;
where
1
4[1 + (n 1) ][1 + (n
1 + (n 2)
B
=
.
1
B
=
2) ]
;
The gain from breaching legislation when the degree of cost savings is
M
and given by
su¢ ciently large is denoted by
C
DC
k
C
B
DB
k
B
1 2
z
4 k
B
=
[z +
=
C 2
z ;
B
(zk
z)]2
B 2
z
in the Cournot and Bertrand case, respectively.
3
Results
We will now conduct a number of experiments in order to analyze how
various market characteristics a¤ect the incentives to violate environmental legislation. Speci…cally, we study how these incentives are related
to
1) the size of the cost reduction
2) business cycle conditions
3) the degree of product di¤erentiation
4) market structure
5) Bertrand versus Cournot competition
Our …rst result is intuitively straightforward.
9
Proposition 1 The incentive to violate the environmental law is stronger
the larger the reduction in cost.
M
Proof. First, consider the case when t > t , i.e., when zk is small: It is
M
easily veri…ed that @@zk = 2 M M [z + (zk z) M ]. Since M M > 0
M
and M > 0, it immediately follows that @@zk > 0, regardless of whether
…rms compete in prices or in quantities. Next, consider the case when
C
M
t t , i.e. when zk is large: It is easily veri…ed that @@zk = z2k > 0 and
that
M
B
@
@zk
M
=2
B
B
[z + (zk
z)
B
], which is strictly positive, because
M
> 0 and
> 0. Thus, the gain from violating legislation
increases in cost savings, regardless of whether competing …rms cease to
produce in response to the violation. See also Figure 1.
Figure 1 about here
The next proposition is also quite intuitive. The pro…tability of a
cost reduction is simply larger the higher the overall level of demand.
Proposition 2 The incentive to violate the environmental law is stronger
in booms than in recessions.
M
M
Proof. For the case when t > t , it is easily estanlished that @ @ =
M
2(zk z) M M . Since M M > 0, it immediately follows that @ @ >
0 when zk > z, regardless of whether …rms compete in prices or in
M
M
quantities. For the case when t
t it can be shown that @ @ > 0,
both under Bertrand and Cournot competition (see the Appendix, A2.1
and A2.2). See also Figure 2.
Figure 2 about here
The e¤ect of the degree of product di¤erentiation on the gains from
violating legislation is not straightforward. A higher implies a higher
degree of competition on the one hand, but more scope for taking advantage of cost savings by breaching legislation on the other hand. It
turns out that the former e¤ect dominates for high degrees of product
di¤erentiation, while the latter e¤ect dominates for low degrees of product di¤erentiation. Hence, as illustrated by Figure 3, the relationship
between and is U-shaped. This means that the incentives for noncompliance are strong for remote and for close substitutes while they are
weak for intermediate levels of product di¤erentiation.
10
Figure 3 about here
Proposition 3 The relationship between the incentive to violate the environmental law and the degree of product di¤erentiation is U-shaped.
Hence, incentives are strongest when di¤erentiation is very small and
very large.
M
Proof. When t > t the total e¤ect of changes in the degree of product
di¤erentiation can be assessed as follows:
@
M
=@
=
M
@
M
=@
M
M
@
+
=@
M
@ M =@ (zk
2z + M (zk
+
z)
:
z)
Hence,
@
M
=@
M
M
>0,1+
(1
t)
M (1
2t +
t)
>
@
M =@
@
M =@
M
M
:
M
It can be established that M decreases in (see the Appendix, A3.1).
n
C
B
B
In particular, C
= 1. The left1 = n+1 , and lim !1
0 = 2,
0 =
2
hand side equals 1+t
2 (1; 2) when = 0 and increases in (unless
t = 1, in which case it equals 1 for all ). Hence, for every t there exists
M
a unique bt 2 (0; 1) such that @ @ > 0 if and only if > bt , i.e. M
is U-shaped in . Furthermore, since the left-hand side of the above
condition decreases in t, it immediately folllows that bt will be closer
to zero the higher are cost savings. The above condition can also be
rearranged in the following way:
@
M
=@
M
> 0 , 2t + 2
,t <
2(
M
M
(1
t) > [2t +
M
(2
)
1) + (2
M
(1
M
M
M)
M
t)]
M
1
=
1+
2(
(2
M
M)
1)
M
b
tM :
M
M
Since @ @ > 0 for all t t (see the Appendix, A3.2) and b
tM > t (see
the Appendix, A3.3), it can be concluded that the gain from violating
legislation increases in if the degree of cost savings is su¢ ciently high
M
M
M
M
t<b
tM ). Moreover, since
( @ @ > 0 for t t and @ @ > 0 for t
M
decreases and M increases in , it immediately follows that b
tM
n(n+2)
bB
bC
bB
increases in ; b
tC
0 = t0 = 0, t1 = n(n+2) 2 > 1 and t1 = 1. See also
Figure 4.
11
Figure 4 about here
The intuition for Proposition (3) is the following. When products
are close substitutes, even a small reduction in price will lead to a large
increase in demand. Hence, a cut in costs will translate into a large
increase in pro…ts. When products are remote substitutes, …rms are
essentially insulated from competition. Hence, a …rm that cuts prices
will not have to bother much about other …rms responding strategically
by cutting their prices too. A given reduction in cost can therefore be
exploited in a way that increases demand and pro…ts signi…cantly.
The relationship between and the market structure is also complex.
While a higher n implies …ercer competition, it also increases the scope
for taking advantage of cost savings by violating legislation. For any
given level of product di¤erentiation, there is a threshold level of t such
that
decreases in the number of …rms whenever t is larger than this
threshold, and vice versa. Hence, if the cost reduction is small enough,
the temptation to violate environmental legislation is largest when the
market is concentrated. If the cost reduction is large, the opposite relation holds. Proposition (4) summarizes the results illustrated by Figure
5.
Figure 5 about here
Proposition 4 When cost savings are small/large the incentive to violate the environmental law becomes stronger/weaker the more concentrated the market structure. The set of parameters for which there is
a positive relationship between incentives to deviate and the number of
…rms is larger the smaller the degree of product di¤erentiation.
M
M
M
M
Proof. When t > t , @@n = 0 and @@n = 0 and hence, @ @n = 0 for
M
M
all zk and n if = 0, while @@n < 0 and @@n > 0 if > 0. Analogously
to the proof of Proposition 3, we obtain the following condition:
@
M
=@n
M
M
>0,1+
2t +
(1
M (1
t)
t)
>
@
M =@n
@
M =@n
M
M
n :
M
It can be established that M
(see the Appendix, A4.1).
n decreases in
2n
B
C
In particular, C
=
=
2
for
=
0,
=
2 [1; 2) for = 1 and
n
n
n
n+1
2(n
1 2
)
2
lim !1 B
2 [ 12 ; 2). The left-hand side equals 1 for all when
n =
n2
2
t = 1. For t < 1, it increases in , taking on values in the interval [ 1+t
; 2]
2
2n
in the Bertrand case, and taking on values in the interval ( 1+t ; n+1 ] in
12
the Cournot case (for n > 1). Hence, given n, there exists a unique
M
et 2 (0; 1) for every t such that @ @n > 0 if and only if > et . Moreover,
since the left-hand side of the above condition decreases in t, et increases
in t. Rearrangement analogously to the proof of Proposition 3 renders
the following condition:
@
M
=@n
M
>0,t<
M
1
1+
M
2(
(2
M
n
M)
n
1)
M
e
tM
n :
M
Since @ @n > 0 when n > 1 and @ @n = 0 when n = 1 for all t
t
M
M
(see the Appendix, A4.2), and e
tn > t (see the Appendix, A4.3), it
can be concluded that the gain from violating legislation increases in n
M
M
if the degree of cost savings is su¢ ciently high ( @ @n > 0 for t
t
M
M
M
M
and @ @n > 0 for t
t<e
tM
n ). Moreover, since
n decreases and
eC eB
increases in , it immediately follows that e
tM
n increases in ; tn = tn = 0
n
1
eB
for = 0, e
tC
= 1.
n = 2n 1 2 ( 2 ; 1] and tn = 1 for
The intuition for Proposition (4) is the following. Suppose that the
number of …rms is large and that a …rm is considering not to comply with
the law. Obviously, the competitive pressure will be high initially which
leads to low prices. If, in that case, a deviation leads to a small reduction
in cost, the deviator will not be in a position to use this advantage in
a way that increases pro…ts to a large extent. If the cost reduction is
large the situation is di¤erent. Since the number of …rms is large, the
potential deviator will initially have a small market share and earn low
pro…ts. However, for large cost reductions, the deviator is in a position
to capture a large fraction of industry demand and pro…ts. Hence, the
increase in pro…ts will tend to be large.
It should be noted that the threshold depicted in Figure 5 is itself a
function of the degree of market concentration. However, as illustrated
by Figures 6 and 7 the threshold is surprisingly robust, and especially
so under price competition.
Figures 6 and 7 about here
The …nal proposition deals with Cournot versus Bertrand competition. The results are illustrated in Figure 8. For any given level of
product di¤erentiation, there is a threshold level of t such that quantity
competition yields stronger incentives to violate the law than price competition whenever t is larger than this threshold, and vice versa. Hence,
if the cost reduction is small enough, the temptation to deviate from
compliance is largest under Cournot competition. If the cost reduction
13
is large, the opposite relation holds. Proposition (5) summarizes these
results.
Figure 8 about here
Proposition 5 When cost savings are small/large quantity competition
leads to stronger/weaker incentives to violate the environmental law than
price competition. The set of parameters for which quantity competition
leads to stronger incentives increases in the degree of product di¤erentiation.
B
Proof. First consider the case when t 2 (t ; 1), i.e. we have interior
solutions under both Bertrand and Cournot competition. If n = 1 or
= 0 it is easily veri…ed that C = B = 14 and C = B = 1 and
hence, C = B for any t. When n > 1 and > 0, C > B if and
only if
(zk
C
,
C
z) C ] > (zk
C
2t
2t
[
+ C] >
+ B
B 1
t
1 t
z)
B
Let
C
C
B
B
C
[2z + (zk
and T
2t
.
1 t
z)
B
Thus, for n > 1 and
B
[2z + (zk
>
B
, (T +
C
)>T+
B
,T >
B
]
> 0,
B
C
z)
C
1
T :
Since T increases in (see the Appendix, A5.1), taking on values in
the interval [0; 1), it immediately follows that C > B if and only if
T
t > 2+T
t , where t increases in , taking on values in the interval
B
[0; 1). Moreover it can be shown that t
t (see the Appendix, A5.2).
C B
When t 2 (t ; t ], such that compliant …rms cease to produce under
Bertrand, but not under Cournot competition, it is the case that B
B
B
and hence, C
(see the Appendix, A5.3).
C
When t t , such that compliant …rms cease to produce under both
C
C
Bertrand and Cournot competition, it is the case that
and
C
B
hence,
<
(see the Appendix, A5.4). Thus, the gain from violating legislation is larger under Cournot than under Bertrand competition
if and only if the degree of cost savings is su¢ ciently small.
Suppose that a …rm is considering a deviation from compliance and
that …rms compete in prices. Typically, price competition is more intense than quantity competition.5 Hence, price competition will lead to
5
For a discussion of situations where this is not the case, see Häckner (2000).
14
a situation where initial prices are low. If, in that case, a deviation leads
to a small reduction in cost, the deviator will not be in a position to use
this advantage in a way that increases pro…ts to a large extent. Under
quantity competition, initial prices are higher, which makes it easier to
increase pro…ts. If the cost reduction is large the situation is di¤erent.
Then, it is relatively easy to increase pro…ts by capturing a large fraction
of the competitors’ customers. However, attracting a large fraction of
industry demand is less costly under Bertrand than under Cournot competition. The reason is that a deviator competing in quantities will have
to expand output a lot in order to achieve this, thereby putting a signi…cant downward pressure on price. Under Bertrand, a relatively small
reduction in price is generally su¢ cient to shift demand, and especially
so when products are close substitutes.
In section 2, we made the assumption that changes in cost are observable from the point of view of competing …rms. Hence, competitors act
strategically by adjusting prices or quantities. It is straightforward to go
through the same exercise as above assuming non-strategic responses.6
Fortunately, most results go through unchanged (see the Appendix, A6).
An exception is Proposition (5) which would then state that the incentives for non-compliance are always stronger under Bertrand than under
Cournot. The reason is simply that competitors no longer match price
reductions. Hence, it becomes easier than before to attract new customers. Propositions (1)-(4) go through unchanged assuming Bertrand
competition. Under Cournot competition, Propositions (1) and (2) go
through unchanged. Proposition (3), however, would indicate a positive relationship between deviation incentives and the degree of product
di¤erentiation. Likewise, Proposition (4) would indicate a positive relationship between deviation incentives and the degree of market concentration.
4
Concluding remarks
We may conclude that there are few general …ndings that the environmental inspection agency can use without having access to detailed …rm
level data. However, given that the agency has some idea about the
…rms’ costs of compliance and business cycle conditions, the following
rules of thumb apply.
1) Inspection should be intensi…ed during booms.
6
Note that there will be no corner solutions when compliant …rms do not react
strategically to violations. It is therefore possible that compliant …rms incur losses if
one …rm violates legislation, in contrast to the case of strategic responses to violations,
where it is possible for …rms to cease production.
15
2) Firms that face high costs of compliance should be inspected more
intensely.
If restricting attention to non-extreme levels of cost savings, further
guidelines can be established. Speci…cally, when non-complience does
not lead to a monopoly situation it is clear from Figure 5 that the set
of parameters that give rise to a positive relationship between deviation
incentives and the number of …rms is very small. Indeed, as illustrated
in Figures 9 and 10, this area is always smaller than 10 percent of the
permissable parameter space. Hence, a reasonable policy would be to
intensify inspection in concentrated markets.
Figures 9 an 10 about here
Although Proposition (3) indicates that the incentives for non-compliance
are strongest for very di¤erentiated and very homogenous products, one
might suspect that the problem is particularly severe in the former case,
where market pro…ts are high. Combining these conjectures we arrive
at a third rule of thumb.
3) Firms that are insulated from competition by product di¤erentiation
or by lack of competitors should be inspected more intensely.
It should be noted that these results are independent of whether
…rms compete in prices or quantities. The choice of strategic variable
obviously plays a role of its own (Proposition (5)), but it does not a¤ect
other mechanisms from a qualitative point of view. This is reassuring
since information on the mode of competition is likely to be di¢ cult for
the inspection agency to obtain. For the same reason, we refrain from
using the results in Proposition (5) to construct a fourth rule of thumb.
16
5
References
d’Aspremont, C., and A. Jacquemin, 1988, ”Cooperative and Noncooperative R&D in Duopoly with Spillovers”, American Economic Review
78, 1133-1137.
Becker, G., 1968, ”Crime and Punishment: An Economic Approach”,
Journal of Political Economy 76, 169-217.
Bontems, P., and G. Rotillon, 2000, ”Honesty in Environmental Compliance Games”, European Journal of Law and Economics 10, 31-41.
Cohen, M., 1998, ”Monitoring and Enforcement of Environmental
Policy”, in Tietenberg, T., and H. Folmer (eds), International Yearbook
of Environmental and Resource Economics, volume III, Edward Elgar
Publishers.
Cohen, M., 1987, ”Optimal Enforcement Strategy to Prevent Oil
Spills: An Application of a Principal-Agent Model with ’Moral Hazard’”,
Journal of Law and Economics 30, 23-51.
Garvie, D., and A. Keeler, 1994, ”Incomplete Inforcement with Endogenous Regulatory Choice”, Journal of Public Economics 55, 141-162.
Häckner, J., 2000, ”A Note on Price and Quantity Competition in
Di¤erentiated Oligopolies”, Journal of Economic Theory 93, 233-239.
Harrington, W., 1988, ”Enforcement Leverage when Penalties are
Restricted”, Journal of Public Economics 37, 29-53.
Heyes, A., 1996, ”Cutting Environmental Penalties to Protect the
Environment”, Journal of Public Economics 60, 251-265.
Heyes, A., and N. Rickman, 1999, ”Regulatory Dealing - Revisiting
the Harrington Paradox”, Journal of Public Economics 72, 361-378.
Kaplow, L., and S. Shavell, 1994, ”Optimal Law Enforcement with
Self-Reporting Behavior”, Journal of Political Economy 102, 583-606.
Lee, D.R., 1983, ”Monitoring and Budget Maximization in the Control of Pollution”, Economic Enquiry 21, 565-575.
Selden, T.M., and M.E. Terrones, 1993, ”Environmental Legislation
and Enforcement: A Voting Model under Asymmetric Information”,
Journal of Environmental Economics and Management 24, 212-228.
17
A Appendix
A1 Model solutions
A1.1 Cournot competition
A1.1.1 Interior solutions
A1.1.1.1 Firm k Given that all …rms i 6= k comply, the reaction
functions of …rm k and …rms i 6= k are given by (see Häckner, 2000)
X
1
qk (q k ) = [zk
qj ];
(6)
2
j6=k
X
1
qj
qk ]:
(7)
qi (q i ) = [z
2
j6=k;j6=i
Hence,
X
X i
1h
(n 1)z + zk (n 1)
qj
qj =
2
X
(n 1)z + zk
,
qj =
:
2 + (n 1)
(8)
Combining (6) and (8) yields
qkC =
=
1
2
zk
1
2 + (n 1)
[(n 1)z + zk ]
2 + (n 1)
2 + (n 2)
(zk
z+
2
Plugging (6) into (2) we obtain
X
pC
qkC
qj =
k =
qkC + 2qkC
z) :
(9)
zk = ck + qkC :
(10)
j6=k
Hence, …rm k’s pro…t is given by
C
k
1
2 + (n
=
z+
2
[2 + (n 1) ]
2
2)
2
(zk
z)
:
De…ning
C
( ; n)
C
( ; n)
1
;
[2 + (n 1) ]2
2 + (n 2)
;
2
pro…ts under compliance and non-compliance are thus given by
C 2
z and DC
= C [z + C (zk z)]2 .
k
18
C
k
=
A1.1.1.2 Firms i 6= k
given by (7):
The quantities of the other …rms (i 6= k) are
X
1
qiC = [z
2
X
1
qkC ] = [z
2
qjC
j6=k;j6=i
qjC + qiC ]:
By plugging (8) into this expression and rearranging terms we obtain
(2
)qiC = z
(n 1)z + zk
2z
zk
(2
)z
(zk
=
=
2 + (n 1)
2 + (n 1)
2 + (n 1)
z)
:
Hence,
qiC = pC
i
c=
1
2 + (n 1)
Plugging (7) into (2) yields
X
C
qj =
pC
=
q
i
i
z
(zk
2
qiC + 2qiC
z) :
z = c + qiC :
j6=i
Thus, to obtain an interior solution, it has to be that
z
2
(zk
z) > 0 , t >
C
2
t :
A1.1.2 Corner solutions
C
When t t , …rms i 6= k cease to produce. Using (6) and (2), …rm k’s
quantity and price are then given by
C
qC
k = pk
1
ck = zk :
2
Hence, its pro…t is given by
DC 1 2
k = zk :
4
A1.2 Bertrand competition
A1.2.1 Interior solutions
A1.2.1.1 Firm k Given that all …rms i 6= k comply, the reaction
functions of …rm k and …rms i 6= k are given by (see Häckner, 2000)
P
ck
(1
)
i6=k pi
pk (p k ) = +
+
;
(11)
2
2[1 + (n 2) ] 2[1 + (n 2) ]
P
c
(1
)
j6=i pj
pi (p i ) = +
+
(i 6= k): (12)
2 2[1 + (n 2) ] 2[1 + (n 2) ]
19
Hence,
X
pj =
, [2 + (n
X
1)c + ck
(1
)n
(n 1)
+
+
pj
2
2[1 + (n 2) ] 2[1 + (n 2) ]
X
3) ]
pj = [1 + (n 2) ][(n 1)c + ck ] + (1
)n :
(n
(13)
Plugging this expression into (11) yields
n
[1+(n
ck
(1
)
pk =
+
+
2
2[1 + (n 2) ]
2) ][(n 1)c+ck ]+(1
[2+(n 3) ]
2[1 + (n
)n
pk
2) ]
o
Rearranging terms we obtain
[1 + (n 2) ][2 + (n 3) ]ck + (1
)[2 + (n 3) ]
+ f[1 + (n 2) ][(n 1)c + ck ] + (1
)n g
pB
k =
[2 + (n 3) ][2 + (2n 3) ]
(1
)[2 + (2n 3) ]z + [2 + 3(n 2) + (n2 5n + 5) 2 ](zk
= ck +
[2 + (n 3) ][2 + (2n 3) ]
P
Plugging the expression for
i6=k pi , obtained by rearrangement of (11),
into the expression for demand, derived in Häckner (2000), we obtain
P
B
(1
)
[1 + (n 2) ]pB
k +
i6=k pi
B
qk =
(1
)[1 + (n 1) ]
1 + (n 2)
=
(pB ck ):
(1
)[1 + (n 1) ] k
z)
:
Hence, pro…ts are given by
B
k
(1
)[1 + (n 2) ]
2 + 3(n
=
z+
2
[1 + (n 1)][2 + (n 3) ]
(1
2) + (n2 5n + 5)
)[2 + (2n 3) ]
De…ning
B
( ; n)
B
( ; n)
(1
)[1 + (n 2)]
;
[1 + (n 1)][2 + (n 3)]2
2 + 3 (n 2) + 2 (n2 5n + 5)
;
(1
)[2 + (2n 3)]
pro…ts under compliance and non-compliance are thus given by
B 2
z and DB
= B [z + B (zk z)]2 .
k
20
B
k
=
2
2
(zk
z)
:
A1.2.1.2 Firms i 6= k
given by (12):
The prices of the compliant …rms (i 6= k) are
P
B
c
(1
)
j6=i pj
B
pi = +
+
2 2[1 + (n 2) ] 2[1 + (n 2) ]
P B
pj
(1
)
c
+
= +
2 2[1 + (n 2) ] 2[1 + (n 2) ]
pB
i
:
2[1 + (n 2) ]
Using (13) and rearrangement of terms yields
3) ]pB
i
[2 + (2n
= [1 + (n
2) ]c + (1
) +
[1 + (n
2) ][(n 1)c + ck ] + (1
2 + (n 3)
)n
= [2 + (2n 3) ]c [1 + (n 1) ]c
[1 + (n 2) ][(n 1)c + ck ] + (1
)[2 + (2n 3) ]
+
2 + (n 3)
[ 2 2(n 2) + (n 1) 2 ]c
+ [1 + (n 2) ]ck + (1
)[2 + (2n 3) ]
= [2 + (2n 3) ]c +
:
2 + (n 3)
Thus,
[2+(n 3) ][2+(2n 3) ](pB
c) = (1
i
)[2+(2n 3) ]z
[1+(n 2) ](zk z):
Hence,
)[2 + (2n 3) ]z
[1 + (n 2) ](zk z)
[2 + (n 3) ][2 + (2n 3) ]
P
Plugging the expression for
j6=i pj , obtained by rearrangement of
(12), into the expression for demand, derived in Häckner (2000), we
obtain
P
B
B
(1
)
[1
+
(n
2)
]p
+
i
j6=i pj
B
qi =
(1
)[1 + (n 1) ]
1 + (n 2)
=
(pB c):
(1
)[1 + (n 1) ] i
pB
i = c+
(1
Thus, to obtain an interior solution, it has to be that
(1
, (1
)[2 + (2n 3) ]z
[1 + (n 2) ](zk z) > 0
)[2 + (2n 3) ]t > [1 + (n 2) ](1 t)
[1 + (n 2) ]
B
,t >
t :
2 + 2(n 2)
(n 1) 2
21
A1.2.2 Corner solutions
B
When t t , …rms i 6= k cease to produce. Using (11), …rm k’s price
and quantity are then given by
ck
(1
)
(n 1) c
+
+
2
2[1 + (n 2) ] 2[1 + (n 2) ]
(1
)z + [1 + (n 2) ](zk z)
;
= ck +
2[1 + (n 2) ]
[1 + (n 2) ]
qB
(pB ck ):
k =
(1
)[1 + (n 1) ] k
pB
k =
De…ning
1
4[1 + (n 1) ][1 + (n
1 + (n 2)
B
=
,
1
B
=
the pro…t when violating legislation is given by
2) ]
DB
;
B
=
[z+
B
(zk z)]2 .
A2 Proof proposition 2
A2.1
C
increases in
It is easily veri…ed that
@
@
Since
A2.2
1
4
C
B
C
=
zk
2
2
C
z:
@
@
and zk > z, it immediately follows that
C
> 0.
increases in
It is easily established that
@
@
B
=2
B
[z + (zk
= 2zk f
When
follows that
B
B
= 1,
B
@
@
B
B
= 0 and
> 0. When
B
]
2
B
B
[
B
B
=
1
;
4n
+
B
z
B
B
]tg:
since t < 1, it immediately
< 1, it can be concluded that, because
B
and t
@
@
B
=
B
z)
t ,
B
2zk f
B
B
[
22
B
B
+
B
B
B
]t g:
Rearrangement of the term inside the bracket yields
B
B
1
=
4[1 + (n
=
2 + 2(n
B
B
[
B
+
B
B
]t
[1 + (n
2 + 2(n 2)
1) ]
2)
2(n
2
(
(1 )[1+(n 2) ]
[1+(n 1) ][2+(n 3) ]2
1
+ 4[1+(n1 1) ]
4[1+(n 1) ][1+(n 2) ]
4 (1 )[1+(n 2) ]2
[2+(n 3) ]2
(n 1) 2 ]
2
1)
2) ]
(n 1)
4[1 + (n
1) ][2 + 2(n 2)
[2 + 2(n 1) ][2 + (n 3) ]2 4 [1 + (n 2) ]2
)
4[1 + (n 1) ][2 + (n 3) ]2 [2 + 2(n 2)
(n 1) 2 ]
8 + (16n 36) + (10n2 52n + 58) 2 + (2n3 18n2 + 46n 34)
)
4[1 + (n 1) ][2 + (n 3) ]2 [2 + 2(n 2)
(n 1) 2 ]
(1
)f4(1
)[2 + (4n 5) ] + 10(n 2)2 g + 2(n 1)2 (n 2)
)
:
4[1 + (n 1) ][2 + (n 3) ]2 [2 + 2(n 2)
(n 1) 2 ]
= (1
= (1
= (1
This term is strictly positive for any n and hence,
@
@
B
> 0 if
< 1.
A3 Proof proposition 3
A3.1
M
decreases in
A3.1.1 Cournot
C
For the case when t > t it is easily established that
@
C
=@
2(n 1)
;
2 + (n 1)
@ C =@
2(n 1)
=
:
C
(2
)[2 + (n 2) ]
C
=
Hence,
C
=
@
M =@
@
M =@
M
=
(2
M
C
It is easily veri…ed that
)[2 + (n 2) ]
:
2 + (n 1)
decreases in
for all n.
A3.1.2 Bertrand
B
For the case when t > t it can be established that
@
B
=@
B
@
B
=@
B
= (n
= (n
2 + 2(2n 5) + 2(n2 5n + 7) 2 (n2 5n + 6) 3
;
(1
)[1 + (n 1) ][1 + (n 2) ][2 + (n 3) ]
2 + 4(n 2) + (2n2 7n + 7) 2
1)
:
(1
)[2 + (2n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
1)
23
3
)
Hence,
B
@
B =@
@
B =@
B
=
B
[2 + (2n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
[2 + 2(2n 5) + 2(n2 5n + 7) 2 (n2 5n + 6) 3 ]
=
:
[1 + (n 1) ][1 + (n 2) ][2 + (n 3) ]
[2 + 4(n 2) + (2n2 7n + 7) 2 ]
It is easily veri…ed that
n 3, the impact of on
@
B
=@
=
B
B
decreases in for n = 1 and n = 2. For
is assessed as follows:
B
3(n 2) + 2(n2 5n + 5)
2n 3
+
2 + (2n 3)
2 + 3(n 2) + (n2 5n + 5)
2
2(2n 5) + 4(n2 5n + 7)
3(n2 5n + 6) 2
n 1
+
2
2
2
3
2 + 2(2n 5) + 2(n
5n + 7)
(n
5n + 6)
1 + (n 1)
2
n 3
4(n 2) + 2(2n
7n + 7)
n 2
2
1 + (n 2)
2 + (n 3)
2 + 4(n 2) + (2n
7n + 7) 2
1
[1 + (n 1) ][2 + (2n 3) ]
2 4
(n2 2n 1) 2
+
[1 + (n 2) ][2 + (n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
2(2n 5) + 4(n2 5n + 7)
3(n2 5n + 6) 2
+
2 + 2(2n 5) + 2(n2 5n + 7) 2 (n2 5n + 6) 3
4(n 2) + 2(2n2 7n + 7)
2 + 4(n 2) + (2n2 7n + 7) 2
4(n 2) + (11n2 37n + 36) 2 + (13n3 48n2 + 84n 54)
+(3n4 19n3 + 47n2 51n + 27) 4
=
[1 + (n 1) ][1 + (n 2) ][2 + (n 3) ]
[2 + (2n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
=
3
4 + 4(3n 7) + (14n2 68n + 78) 2
+8(n 2)2 (n 3) 3 + (n 2)(n 3)(2n2 7n + 7) 4
:
[2 + 4(n 2) + (2n2 7n + 7) 2 ]
[2 + 2(2n 5) + 2(n2 5n + 7) 2 (n2 5n + 6) 3 ]
Since the …rst term is negative for n
for n 3, it can be concluded that
A3.2
M
2 and the second term is negative
decreases in for all n.
B
increases in
A3.2.1 Cournot
For the case when t
n > 1 and
@
@
C
C
t it is easily veri…ed that
= 0 for n = 1.
24
@
@
C
=
@
@
C
z 2 > 0 for
A3.2.2 Bertrand
B
For the case when t
follows:
@
@
t
B
B
= [z +
B
= [z +
the impact of a change in
B
2@
(zk
z)]
@
z)]2
(zk
is assessed as
B
+ 2(zk
"
B
@
=@
B
z)[z +
+
1
B
(zk
2( 1t 1)
B 1
+
(t
z)]
@
1) @
B
B@
B
@
#
z2
z2
@
@
@
@
B
B
:
It can be established that
B
@
=@
B
@
@
= (n
1)
n
(1
1
:
)2
B
=
2 + 2(n 2)
(n 2)
(1
)[1 + (n 1) ][1 + (n
B
Since @@ = 0 for n = 1, it immediately follows that
When n > 1,
2( 1t
1+
1)
@
1) @
B 1
(t
B
2(
1+
=
1+
= (n
1
B
t
B
1)
@
1) @
1)
B
2) ]
= 0 for n = 1.
2
n
(1
2 + (2n 3)
)[1 + (n 1) ][1 + (n
(1
;
B
1
B
t
5) (2n 3) 2
2 2+(2n[1+(n
2) ]
1+(n 2) 2+(2n 5) (2n 3)
1
[1+(n 2) ]
(
@
@
2
1
)2
2) ]
;
and hence,
@
B
=@
B
+
2( 1t
1+
B 1
(t
1)
@
1) @
B
=
Since
@
@
B
(n
1)
2 + 2(n 2)
(n 2)
(1
)[1 + (n 1) ][1 + (n
+(n
1)
2 + (2n 3)
)[1 + (n 1) ][1 + (n
(1
(n 1)
)[1 + (n
(1
1) ]2
< 0 for n > 1, it immediately follows that
25
@
@
;
B
> 0 for n > 1.
2
2) ]
2) ]
M
A3.3 b
tM > t
It is easily established that
M
b
tM
t
=
M
(2
2(
)
M
M
M
(1 t )
1) + (2
2(
M
M
M)
1)t
M
:
A3.3.1 Cournot
In the Cournot case the numerator is given by
C
(2
C
)
(1
C
t )
C
2(
1)t
4 + (n 2) 2 + (n 2)
(1
2 + (n 1)
2
4 + (4n 2) + (n2 2n) 2
=
2[2 + (n 1) ]
=
2
C
)
2
2 + (n 5)
2 + (n
(n
1)
2)
2
2
0:
C
Hence, b
tC
t for all n.
A3.3.1 Bertrand
In the Bertrand case the numerator is given by
(2
B
)
B
(1
B
t )
2(
B
B
1)t
[1 + (n 2) ]
2 + 3(n 2) + (n2 5n + 5) 2
2( B 1)
= (2
)
2
2 + 2(n 2)
(n 1)
2 + 2(n 2)
(n 1)
2
2
2
2
2[2 + (3n 5) + (n
4n + 3) ] [2 + (3n 4) + (n
3n + 1) ] B
=
:
2 + 2(n 2)
(n 1) 2
B
When n = 1 this term equals
the case that
2[2 + (3n
,
[2 + (3n
[2 + (2n
,
5) + (n2
2[1 + (n
5) + (n2
(4+2
4
4n + 3) 2 ]
2)
2
, which is positive. For n > 1 it
4) + (n2
3n + 1) 2 ]
1) ][1 + (n 2) ][2 + (n 3) ]
4n + 3) 2 ][2 + 4(n 2) + (2n2
7n + 7) 2 ]
[2 + (3n
B
3) ][2 + 3(n 2) + (n2 5n + 5) 2 ][2 + (3n 4) + (n2
[2 + 2(2n 5) + 2(n2 5n + 7) 2 (n2 5n + 6) 3 ]
[2 + (2n 2) ][2 + (3n 7) + (n2 5n + 6) 2 ]
[2 + (3n 5) + (n2 4n + 3) 2 ][2 + 4(n 2) + (2n2 7n + 7) 2 ]
8
9
][2 + (3n 7) + (n2 5n + 6) 2 + (1
)] =
< [2 + (2n 2)
f2 + (3n 5) + (n2 4n + 3) 2 + [1 + (n 2) ]g
:
;
f2 + 4(n 2) + (2n2 7n + 7) 2
[1 + (n 2) ][2 + (n 3) ]g
3n + 1) 2 ]
2
2) ][2 + (3n 7) + (n2 5n + 6) 2 ]
[(n 2) + (n2 3n + 3) ]g
[2 + (3n 5) + (n2 4n + 3) 2 ][2 + 4(n 2) + (2n2 7n + 7) 2 ]
[1 + (n 2) ][2 + (4n 6) + (3n2 12n + 7) 2 + (n3 6n2 + 10n 3) 3 ]g
= f[2 + (2n
26
It is easy to see that this inequality holds for all n > 1. Thus, b
tB
for all n.
A4 Proof proposition 4
M
n
A4.1
decreases in
A4.1.1 Cournot
C
For the case when t > t it is easily established that
@
C
=@n
C
@
C
=@n
C
=
=
2 + (n
2 + (n
1)
2)
;
:
Hence,
C
n
=
@
C =@n
@
C =@n
C
=
C
It is straightforward that
C
n
2[2 + (n 2) ]
:
2 + (n 1)
decreases in
for all n.
A4.1.2 Bertrand
B
For the case when t > t it can be established that
B
2 + 4(n 2) + (2n2 7n + 7) 2
;
B
[1 + (n 1) ][1 + (n 2) ][2 + (n 3) ]
2 + (4n 7) + (2n2 6n + 5) 2
@ B =@n
=
:
B
[2 + (2n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
@
=@n
=
Hence,
B
n
=
@
B =@n
@
B =@n
B
B
[2 + (2n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
[2 + 4(n 2) + (2n2 7n + 7) 2 ]
=
:
[1 + (n 1) ][1 + (n 2) ][2 + (n 3) ]
[2 + (4n 7) + (2n2 6n + 5) 2 ]
27
B
t
It can be established that B
n decreases in
B
impact of on n is assessed as follows:
B
n =@
B
n
for n = 1. For n
2, the
3(n 2) + 2(n2 5n + 5)
2n 3
+
2 + (2n 3)
2 + 3(n 2) + (n2 5n + 5) 2
n 1
4(n 2) + 2(2n2 7n + 7)
+
2
2
2 + 4(n 2) + (2n
7n + 7)
1 + (n 1)
n 2
n 3
(4n 7) + 2(2n2 6n + 5)
1 + (n 2)
2 + (n 3)
2 + (4n 7) + (2n2 6n + 5) 2
4(n 2) + (11n2 37n + 36) 2 + (13n3 48n2 + 84n 54) 3
+(3n4 19n3 + 47n2 51n + 27) 4
=
[1 + (n 1) ][1 + (n 2) ][2 + (n 3) ]
[2 + (2n 3) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
2 + 4(n 2) + (2n2 9n + 9) 2
:
[2 + 4(n 2) + (2n2 7n + 7) 2 ][2 + (4n 7) + (2n2 6n + 5) 2 ]
@
=
The …rst term is negative for n 2 and the second term is negative for
n 1. Hence, B
for all n.
n decreases in
A4.2
M
increases in n
A4.2.1 Cournot
C
For the case when t
> 0 and
C
@
@n
t it is easily veri…ed that
= 0 for
C
@
@n
@ C 2
z
@n
=
> 0 for
= 0.
A4.2.2 Bertrand
B
For the case when t
t it is easily established that
B
@
= [z +
@n
= [z +
B
B
B
2@
(zk
z)]
2
(zk
z)]
@n"
B
+ 2(zk
@
B
=@n
B
z)[z +
+
1
B
(zk
2( 1t 1)
B 1
+
(t
z)]
B
@
1) @n
It can be established that
@
B
=@n
B
=
2 + (2n 3)
[1 + (n 1) ][1 + (n
B
@
=
@n
1
:
28
B@
2) ]
;
B
#@n
z2
z
2@
@ B
@n
B
@n
:
B
Since @@n = 0 for
When > 0,
2( 1t
1+
1)
B 1
(t
B
@
@n
= 0, it immediately follows that
B
2(
@
1) @n
1+
=
1
1
B
t
B
= 0 for
= 0.
2
5) (2n 3)
B
2 2+(2n[1+(n
@
2) ]
=
2+(2n
5) (2n 3)
1+(n
2)
1+ 1
1) @n
[1+(n 2) ]
1)
1
B
t
5) (2n 3) 2
2 2+(2n[1+(n
2) ]
2+(2n 5) (2n 3)
(
+
2
2
2 + (2n 3)
[1 + (n 1) ][1 + (n
=
1
2) ]
;
and hence,
@
B
=@n
B
Since
@ B
@n
< 0 for
+
2( 1t
B
1)@
=@n
B 1
(t
1+
0:
1)
B
@
@n
> 0, it is straightforward that
> 0.
M
A4.3 e
tM
n > t
It is easily established that
M
e
tM
n
t
=
(2
M
M
(1
n )
M
2( n
1)
M
t )
+ (2
2(
M
M
n
M)
n
1)t
M
:
A4.3.1 Cournot
In the Cournot case the numerator is given by
(2
C
n)
2
=
2 + (n
Hence, e
tC
n
C
C
(1 t ) 2( C
n
2 + (n 2)
(1
1)
2
C
1)t
2
2 + (n
2
2 + (n
)
2
3)
=
1) 2
2 + (n 1)
0:
C
t for all n and all .
A4.3.1 Bertrand
In the Bertrand case the numerator is given by
B
n)
B
B
B
(1 t ) 2( B
1)t
n
2
[1 + (n 2) ]
2) + (n
5n + 5) 2
B 2 + 3(n
= (2
)
2( B
1)
n
n
2
2 + 2(n 2)
(n 1)
2 + 2(n 2)
(n 1)
2
2
2
2
2[2 + (3n 5) + (n
4n + 3) ] [2 + (3n 4) + (n
3n + 1) ] B
n
=
:
2 + 2(n 2)
(n 1) 2
(2
29
2
For this term to be positive, it has to be that
2[2 + (3n 5) + (n2 4n + 3) 2 ] [2 + (3n 4) + (n2 3n + 1) 2 ]
2[1 + (n 2) ][2 + (4n 7) + (2n2 6n + 5) 2 ]
,
[2 + (3n 5) + (n2 4n + 3) 2 ][1 + (n 1) ][2 + (n 3) ]
B
n
[2 + (2n 3) ][2 + 4(n 2) + (2n2 7n + 7) 2 ]
[2 + 3(n 2) + (n2 5n + 5) 2 ][2 + (3n 4) + (n2 3n + 1) 2 ]
, [2 + 2(n 2) ][2 + (4n 7) + (2n2 6n + 5) 2 ][2 + (3n 5) + (n2 4n + 3) 2 ]2
8
9
[1 + (n 2) ]g =
< [2 + 2(n 2) + ]f2 + (4n 7) + (2n2 6n + 5) 2
f[2 + (3n 5) + (n2 4n + 3) 2 ]
[1 + (n 2) ]g
:
;
2
2
f[2 + (3n 5) + (n
4n + 3) ] + [1 + (n 2) ]g
= f[2 + 2(n 2) ][2 + (4n 7) + (2n2 6n + 5) 2 ] + (n 1) 3 g
2
f[2 + (3n 5) + (n2 4n + 3) 2 ]2
[1 + (n 2) ]2 g
= [2 + 2(n 2) ][2 + (4n 7) + (2n2 6n + 5) 2 ][2 + (3n 5) + (n2
+(n 1) 3 [2 + (3n 5) + (n2 4n + 3) 2 ]2
2 2 [1 + (n 2) ]3 [2 + (4n 7) + (2n2 6n + 5) 2 ]:
4n + 3) 2 ]2
Since
(n 1) [2 + (3n 5) + (n2 4n + 3) 2 ]2
2[1 + (n 2) ]3 [2 + (4n 7) + (2n2 6n + 5) 2 ]
= 4(n 1) + 4(3n2 8n + 5) 2 + (13n3 59n2 + 83n 37) 3
+2(3n4 20n3 + 46n2 44n + 15) 4 + (n5 9n4 + 30n3 46n2 + 33n 9) 5
4 + 2(10n 19)
2(20n2 75n + 71) 2 2(20n3 111n2 + 207n 130) 3
2(10n4 73n3 + 201n2 248n + 116) 4 2(2n5 18n4 + 65n3 118n2 + 108n
= 4 (16n 34)
(28n2 118n + 122) 2 (27n3 163n2 + 331n 223) 3
(14n4 106n3 + 310n2 408n + 202) 4 (3n5 27n4 + 100n3 190n2 + 183n
which is strictly negative for all n, it immediately follows that the above
B
inequality holds for all n. Thus, e
tB
t for all n and all .
n
A5 Proof proposition 5
A5.1 T increases in
To assess the impact of changes in
following way:
B
T =
C
1
B
=
on T , rearrange terms in the
C
C
1
30
=
C
[
B
C
C(
1)
1]:
40)
5
71) 5 ;
It is easily established that
B
Rearrangement of
C
C
=
(n
(1
)(2
1)2 3
)[2 + (2n
3) ]
:
yields
C
[1 + (n 1) ][2 + (n 2) ][2 + (n 3) ]2 [2 + (2n 3) ]
= B B =
(2
)[1 + (n 2) ][2 + (n 1) ]2 [2 + 3(n 2) + (n2 5n + 5) 2 ]
[2 + (3n 4) + (n2 3n + 2) 2 ][4 + 4(n 3) + (n2 6n + 9) 2 ][2 + (2n 3) ]
=
[2 + (2n 5)
(n 2) 2 ][4 + 4(n 1) + (n2 2n + 1) 2 ]
[2 + 3(n 2) + (n2 5n + 5) 2 ]
(14)
8 + 20(n 2) + (18n2 76n + 74) 2
[2 + (2n 3) ]
+(7n3 46n2 + 95n 60) 3 + (n4 9n3 + 29n2 39n + 18) 4
=
8 + (16n 28) + (10n2 36n + 30) 2
[2 + 3(n 2) + (n2 5n + 5) 2 ]
+(2n3 13n2 + 24n 13) 3 (n3 4n2 + 5n 2) 4
16 + (56n 104) + (76n2 292n + 268) 2 + (50n3 298n2 + 566n 342) 3
+(16n4 131n3 + 386n2 483n + 216) 4 + (2n5 21n4 + 85n3 165n2 + 153n 54)
=8
16 + (56n 104) + (76n2 292n + 268) 2 + (50n3 302n2 + 574n 346) 3
<
+(16n4 139n3 + 418n2 523n + 232) 4 + (2n5 26n4 + 117n3 237n2 + 221n 77)
:
(n5 9n4 + 30n3 47n2 + 35n 10) 6
+(5n4
(2
=1 +
= 1 + (n
1)2
= 1 + (n
1)2
(4n2 8n + 4) 3 + (8n3 32n2 + 40n 16) 4
32n3 + 72n2 68n + 23) 5 + (n5 9n4 + 30n3 47n2 + 35n 10) 6
)[1 + (n 2) ][2 + (n 1) ]2 [2 + 3(n 2) + (n2 5n + 5) 2 ]
4 + 8(n 2) + (5n2 22n + 23) 2 + (n 2)(n2 5n + 5) 3
3
(2
)[1 + (n 2) ][2 + (n 1) ]2 [2 + 3(n 2) + (n2 5n + 5) 2 ]
[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2
3
:
(2
)[1 + (n 2) ][2 + (n 1) ]2 [2 + 3(n 2) + (n2 5n + 5) 2 ]
It is easy to see that
=
B
C
C(
1)
(1
(2
)[2 + (n
> 1 for n > 1 and
5
9
=
;
(15)
> 0. Let
)[1 + (n 2) ][2 + (n 1) ]2 [2 + 3(n 2) + (n2 5n + 5) 2 ]
2) ][2 + (2n 3) ]f[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] +
31
5
2g
:
The impact of a change in
on
is assessed as follows:
n 2
2(n 1)
3(n 2) + 2(n2 5n + 5)
+
+
2
1 + (n 2)
2 + (n 1)
2 + 3(n 2) + (n2 5n + 5)
n 2
2n 3
1
+
1
2 + (n 2)
2 + (2n 3)
(n 2)[2 + 3(n 2) + (n2 5n + 5) 2 ]
+[2 + (n 2) ][3(n 2) + 2(n2 5n + 5) ] + 2
:
[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2
@ =@
=
1
+
It is easily established that
1
2
+
1
1
=
(1
1
)(2
)
> 0;
and that
n 1
n 2
2n 3
n 2
+
1 + (n 2)
2 + (n 1)
2 + (n 2)
2 + (2n 3)
n 2
2(n 2)
=
[1 + (n 2) ][2 + (n 2) ] [2 + (n 1) ][2 + (2n 3) ]
(n 2) [4 + (3n 5) ]
;
=
[1 + (n 2) ][2 + (n 1) ][2 + (n 2) ][2 + (2n 3) ]
32
2
which is positive for n
2. Moreover,
n 1
3(n 2) + 2(n2 5n + 5)
+
2 + (n 1)
2 + 3(n 2) + (n2 5n + 5) 2
(n 2)[2 + 3(n 2) + (n2 5n + 5) 2 ]
+[2 + (n 2) ][3(n 2) + 2(n2 5n + 5) ] + 2
[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2
(n 1)[2 + 3(n 2) + (n2 5n + 5) 2 ] + [2 + (n 1) ][3(n 2) + 2(n2 5n + 5) ]
=
[2 + (n 1) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
(n 2)[2 + 3(n 2) + (n2 5n + 5) 2 ]
+[2 + (n 2) ][3(n 2) + 2(n2 5n + 5) ] + 2
[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2
8
9
< 2[2 + 3(n 2) + (n2 5n + 5) 2 ]2 + (n 1) 2 [2 + 3(n 2) + (n2 5n + 5) 2 ] =
+ 2 [2 + (n 1) ][3(n 2) + 2(n2 5n + 5) ]
:
;
2 [2 + (n 1) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
=
[2 + 3(n 2) + (n2 5n + 5) 2 ]f[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2 g
2[2 + 3(n 2) + (n2 5n + 5) 2 ]2
[4 + (n 1) ][2 + 3(n 2) + (n2 5n + 5) 2 ]
+ 2 [2 + (n 1) ][3(n 2) + 2(n2 5n + 5) ]
=
[2 + 3(n 2) + (n2 5n + 5) 2 ]f[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2 g
[2 + 3(n 2) + (n2 5n + 5) 2 ][4 + (6n 16) + (2n2 11n + 11) 2 ]
+ 2 [2 + (n 1) ][3(n 2) + 2(n2 5n + 5) ]
;
=
[2 + 3(n 2) + (n2 5n + 5) 2 ]f[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2 g
which is positive for n
@ =@
3. Since
=
2(1 )+
(1 )(2
2
2)
> 0 for n = 2, it
@ =@
> 0 for n 2. Since C also increases in
can be concluded that
, it immediately follows that T increases in when n 2.
A5.2 t
t
B
B
To show that t
t , the following rearrangements can be made:
B
T
t =
2+T
, (
t ,T
B
C
)
(
Since
B
2t
B
1)(
,
B
1
t
B
2t
1
B
t
+
1
B
):
(1
)[2 + (2n 3) ]
;
2 + 2(n 2)
(n 1) 2
B
t =1
it immediately follows that
B
2t
1
B
t
=
2 [1 + (n 2) ]
:
(1
)[2 + (2n 3) ]
33
B
C
2t
1
B
t
It is easily veri…ed that
B
C
=
B
2t
B
1
B
+
t
=
(n
(1
)(2
1)2 3
)[2 + (2n
3) ]
;
4) + (n2 3n + 1)
)[2 + (2n 3) ]
2 + (3n
(1
2
:
Using expressions (14) and (15), it follows that
B
(
B
C
)
(
1)(
2t
B
)
1 t
, [1 + (n 1) ][2 + (n 2) ][2 + (n 3) ]2 [2 + (2n 3) ]
(2
)f[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] +
[2 + (3n 4) + (n2 3n + 1) 2 ]:
B
+
2
g
Since
[1+(n 1) ][2+(n 2) ] = 2+(3n 4) +(n2 3n+2)
2
2+(3n 4) +(n2 3n+1) 2 ;
and
(2
)f[2 + (n 2) ][2 + 3(n 2) + (n2 5n + 5) 2 ] + 2 g
= 8 + (16n 36) + (10n2 48n + 54) 2 + (2n3 15n2 + 36n 27)
4(n 2) 2 4(n 2)2 2 (n 2)(n2 5n + 5) 4
8 + (16n 36) + (10n2 48n + 54) 2 + (2n3 15n2 + 36n 27)
= [2 + (n 3) ]2 [2 + (2n 3) ];
B
t when n
it immediately follows that t
A5.3
B
Let
>
C
C
The impact of t on
=@t
B
=
2.
B
B
B
3
when t 2 (t ; t ]
B
@
3
2
1 + ( 1t
B
[1 + ( 1t
B [1 + ( 1
t
1)
1)
B 2
]
:
B ]2
is assessed as follows:
B
1
2
+
B 2
t 1 + ( 1t
1)
34
B
1
=
B
t2
1)
[1 + ( 1t
B
1)
B ][1
B
+ ( 1t
1)
B
2
;
2
]t
B
[1 + (
1)
B
[1 +
1)
B
1
B
t
1
(B
t
B
B
It is easily veri…ed that
. Hence,
@ B
@t
0 and
B 2
]
B ]2
)[2+(2n 3) ] 1+(n 2) 2
]
[1 + (1 [1+(n
[1 + (n 1) ][2 + (n 3) ]2
2) ]
1
=
2
4[1 + (n 1) ][1 + (n 2) ]2 [1 + (1 )[2+(2n 3) ] 2+3(n 2) +(n 5n+5) 2 ]2
[1+(n 2) ]
(1 )[2+(2n 3) ]
#2
"
[2+(2n
3)
]
1+
[1 + (n 1) ][2 + (n 3) ]2
=
4[1 + (n 1) ][1 + (n 2) ]2 1 + 2+3(n 2) +(n2 5n+5) 2
[1+(n 2) ]
2
2
=
[2 + (n 3) ]
2[1 + (n 1) ][1 + (n 2) ]
4[1 + (n 2) ]2 2 + (3n 5) + (n2 4n + 3)
B
= 1:
2
B
B
for t t . Above it was established that
Thus,
B
t . Hence, it can be concluded that
t < t and that t
B
t t .
A5.4
C
C
C
[1 + ( 1t
C [1 + ( 1
t
C
@
C
=@t
C
2
=
1 + ( 1t
C
C
C
1
C
t
1
(C
t
[1 + (
[1 +
Thus,
follows that
C
B
for
for
C 2
]
1)
1)
C ]2
:
is assessed as follows:
C
2
1
+
C 2
t 1 + ( 1t
1)
It is easily veri…ed that
C
C
t
Let
C
B
C
>
>
C
when t
The impact of t on
B
1)
C
C
C
1
=
C
t2
1)
[1 + ( 1t
. Hence,
@ C
@t
[2 + (n 1) ]2
=
4
1) C ]2
[1 +
B
>
C
+ ( 1t
4
]
C
C ][1
1)
0 and
C 2
for t t . Since
C
C
>
for t t .
C
C
2
2
2+(n 2)
2
]2
= 1:
for t < t , it immediately
A6 No strategic responses
A6.1 Cournot
In this case all other …rms choose quantities as if …rm k were in compliz
ance, i.e. qj = 2+ (n
for j 6= k (see Häckner, 2000). Firm k chooses
1)
35
1)
C
2
;
2
]t
quantity qkCN according to its best response function (6):
1
qkCN = [zk
2
(n
1)
z
2 + (n
1)
]=
z
2 + (n
1)
+
zk
z
2
:
As before, the price is given by
= qkCN + ck
pCN
k
Hence, …rm k’s pro…t is given by
CN
k
1
2 + (n 1)
=
(zk
z+
2
[2 + (n 1) ]
2
2
z)
:
Naturally pro…ts under compliance are the same as before. De…ning
CN
( ; n) 2+(n2 1) , pro…ts under non-compliance can be expressed as
DCN
k
=
C
[z +
CN
(zk
z)]2 :
The gain from breaching legislation is thus given by
CN
DCN
C
= (zk
z)
C
CN
[2z + (zk
z)
CN
]:
It is easily established that CN < C and hence, DCN
< C
k
k (unless
CN
C
DCN
C
n = 1 or = 0, in which case
=
and k
= k ). Hence,
the gain from violating legislation will always be smaller compared to
the case when competitors react strategically to …rm k’s violation of
legislation. In the latter case the other …rms adapt by reducing their
produced quantities, thus making it more pro…table for …rm k to increase
its production.
It is straightforward that Propositions 1 and 2 still hold, i.e. CN increases in both zk and , although the impact of changes will be smaller
CN
now. Since CN = CN
= 2 and 1 + 2t+ CN(1(1t) t) < 2 for all , it immedin
CN
CN
ately follows that @ @ < 0 and @ @n < 0, i.e. the gain from breaching
legislation increases unambiguously both in the degree of product di¤erentiation and in the degree of market concentration. Thus, Proposition
3 no longer holds for low degrees of product di¤erentiation, and Proposition 4 no longer holds for high degrees of cost savings.
A6.2 Bertrand
Since all other …rms choose prices as if …rm k were in compliance, prices
)
will be given by pj = [1+ (n2+ 2)]c+(1
for j 6= k. Firm k chooses price
(n 3)
36
pBN
according to its best response function (11):
k
2) ]c+(1 )
(n 1) [1+(n2+(n
ck
(1
)
3)
=
+
+
2
2[1 + (n 2) ]
2[1 + (n 2) ]
[1 + (n 2) ][2 + (n 3) ]ck
+2(1
)[1 + (n 2) ] + (n 1) [1 + (n 2) ]c
= ck +
2[1 + (n 2) ][2 + (n 3) ]
2(1
)z + [2 + (n 3) ][zk z]
= ck +
:
2[2 + (n 3) ]
pBN
k
Analogously to the case where other …rms react strategically, we obtain
qkBN =
1 + (n 2)
(pBN
(1
)[1 + (n 1) ] k
ck ):
Hence, pro…ts are given by
BN
k
(1
=
[1 + (n
)[1 + (n 2) ]
2 + (n
z+
2
1) ][2 + (n 3) ]
2(1
3)
(zk
)
2
z)
:
Naturally pro…ts under compliance are the same as before. De…ning
3)
BN
( ; n) 2+(n
, pro…ts under non-compliance can be expressed as
2(1 )
DBN
k
=
B
[z +
BN
(zk
z)]2 :
The gain from breaching legislation is thus given by
BN
DBN
B
= (zk
z)
B
BN
[2z + (zk
z)
BN
]:
It is easily established that BN > B and hence, DBN
> B
k
k (unless
BN
B
DBN
B
n = 1 or = 0, in which case
=
and k
= k ). Hence,
the gain from violating legislation will now always be larger compared
to the case when competitors react strategically to …rm k’s violation of
legislation. In the latter case the other …rms adapt by reducing their
prices, thus reducing the gains from breaching legislation.
It is easy to see that Propositions 1 and 2 still hold, i.e. BN increases
in both zk and , and the impact of changes will be larger in the absence
of strategic responses.
The impact of changes in the degree of product di¤erentiation turns
out to be qualitatively similar to the case with strategic responses. Since
@ BN =@
n 1
= (1 )[2+(n
, it follows that
BN
3) ]
@
BN
=
B =@
B
@
BN =@
BN
=
2 + 2(2n
5) + 2(n2 5n + 7) 2
[1 + (n 1) ][1 + (n
37
(n 2)(n
2 ]
3)
2
:
Just as in the case of strategic responses by competitors, BN decreases
monotonously in for all n, taking on values in the interval [1; 2] (when
n = 2, BN falls below one and then increases; however, since BN
=
0
BN
2 and 1 = 1, results are not a¤ected). Hence, Proposition 3 still
holds: the incentive to deviate is U-shaped in the degree of product
di¤erentiation. However, since BN > B and BN < B , it follows
that b
tBN > b
tB , i.e. the range of degrees of cost savings, for which the
gain from violating legislation increases in , will be wider.
Changes in the degree of market concentration yield similar results
BN
as when …rms react strategically. Since @ BN=@n = 2+(n 3) it follows
that
B =@n
@
BN
n
B
=
@
BN =@n
=
BN
2 + 4(n 2) + (2n2 7n + 7)
[1 + (n 1) ][1 + (n 2) ]
2
:
decreases
Just as in the case of strategic responses by competitors, BN
n
monotonously in for all n, taking on values in the interval [2 n1 ; 2].
Hence, we obtain a result that is qualitatively similar to Proposition 4.
A6.3 Proposition 5
Analoguously to the proof of Proposition 5 the following condition is
obtained
CN
>
BN
,
C
CN
B
BN
[T +
C
]>T+
B
:
Since
C
B
CN
BN
=
[1 + (n
[1 + (n
1) ][2 + (n
2) ][2 + (n
3) ]
(n 1) 2
=1
1) ]
[1 + (n 2) ][2 + (n
which is strictly smaller than unity for n > 1 and
for n > 1 and > 0, it immediately follows that
and hence, CN < BN for n > 1 and > 0.
38
> 0, and
CN
B BN [T +
C
CN
1) ]
< BN
C
] < T+ B
;
Figure 1
delta
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
ck
The gain from deviation under Cournot (solid line) and Bertrand (dashed line) as a function of the
cost reduction, illustrating Proposition (1). α=3, c=1, γ=0.5, n=3
Figure 2
delta
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
5
alpha
The gain from deviation under Cournot (solid line) and Bertrand (dashed line) as a function of the
business cycle parameter, illustrating Proposition (2). Corresponding thresholds for interior solution
equilibria are indicated by the thin vertical lines. c=1, ck=0.5, γ=0.5, n=3
Figure 3
delta
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
gamma
The gain from deviation under Cournot (solid line) and Bertrand (dashed line) as a function of the
degree of product differentiation, illustrating Proposition (3). The threshold for interior solution
equilibria under Bertrand is indicated by the thin vertical dashed line. α=3, c=1, ck=0.5, n=3
Figure 4
t
1.0
0.9
0.8
0.7
∆
0.6
∆
0
0
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
gamma
Thresholds for the impact of product differentiation under Cournot (solid thick line) and Bertrand
(dashed thick line), illustrating Proposition (3). Corresponding thresholds for interior solution
equilibria are indicated by the thin lines. n=3
Figure 5
t
1.0
0.9
0.8
∆
0.7
0
0.6
0.5
0.4
∆
0.3
0
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
gamma
Thresholds for the impact of market concentration under Cournot (solid thick line) and Bertrand
(dashed thick line), illustrating Proposition (4). Corresponding thresholds for interior solution
equilibria are indicated by the thin lines. n=3
Figure 6
t
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
gamma
Thresholds for the impact of market concentration under Cournot for n=2 (solid line) and n=100
(dotted line).
Figure 7
t
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
gamma
Thresholds for the impact of market concentration under Bertrand for n=2 (solid line) and n=100
(dotted line).
Figure 8
t
1.0
0.9
0.8
∆
0.7
∆
0.6
∆
∆
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
gamma
The threshold determining whether Cournot or Bertrand competition leads to the strongest
incentives to pollute (dotted line), illustrating Proposition (5). Corresponding thresholds for interior
solution equilibria are indicated by the thin lines. n=3
Figure 9
Share of parameter space such that
∂Δ/∂n>0 under Cournot
0,1
0,08
0,06
0,04
0,02
0
n=2
n=3
n=4
n=5
n=6
n=7
n=8
n=9
n=10 n=100
Figure 10
Share of parameter space such that
∂Δ/∂n>0 under Bertrand
0,08
0,07
0,06
0,05
0,04
0,03
0,02
0,01
0
n=2
n=3
n=4
n=5
n=6
n=7
n=8
n=9
n=10 n=100