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Investment in Green Technology and Entry Deterrence

  • John C. Strandholm and Ana Espínola-Arredondo EMAIL logo

Abstract

This paper analyzes an entry-deterrence model in which the incumbent decides whether to invest in research and development (R&D) that promotes clean technology. We consider the case in which the entrant could benefit from a technology spillover and analyze the conditions that facilitate the incumbent’s entry-deterrence behavior. We show that higher levels of the spillover make entry more attractive compared to a standard entry game. In addition, regulator and incumbent prefer entry when the spillover from clean technology is sufficiently high and the cost of investing in R&D is relatively low. However, preferences are misaligned when the spillover and cost of R&D are low.

JEL Classification: H23; L12; O32; Q58

Acknowledgements

We would like to especially thank Michael Delgado, Félix Muñoz-García, and Tristan Skolrud for their insightful comments and suggestions. We are also grateful to the participants at the Association of Environmental and Resource Economists session at the 91st annual Western Economic Association International conference, the 12th Conference of the Spanish Association for Energy Economics, and the First Catalan Economic Society Conference.

Appendix

A

A.1 Proof of Proposition 1

In the third stage, the incumbent acts as a monopolist and solves

maxqiπine=(aqi)qicqit(qiz),

with first-order condition

πineqi=act2qi=0.

Solving for qi gives qi(t)=act2. Hence, qi(t) > 0 if t > a – c. The regulator’s problem in in the second stage is

maxtSW=0qi(t)(acx)dx12dqi(t)z212γz2,

where the first-order condition is

SWt=14(a(d1)d(c+t+2z)+ct)=0.

Solving for t yields t=(d1)(ac)2dzd+1, which guarantees qi(t) > 0 since t < a – c for all parameter values. In the first stage, the monopolist maximization problem is

maxzΠine=((aqi)qicqit(qiz))12γz2,

with first-order condition

Πinez=d(d+2)(ac2z)+(ca)+γ(d+1)2z(d+1)2=0.

Solving for z gives us the no entry equilibrium level of investment zne=(ac)(d(d+2)1)γ(d+1)2+2d(d+2). Plugging this into the results from the second and third stages gives us the remainder of proposition 1:

(12)tne=(ac)γd212dγ(d+1)2+2d(d+2),
(13)qne=(ac)(γ(d+1)+d(d+3))γ(d+1)2+2d(d+2).
zne and qne are positive since a > c, β ∈ [0, 1], and d > 1. However, the sign of tne depends on the sign of γ(d2 – 1) – 2d, which gives the condition that tne > 0 if γ>2dd21. Finally, we need to guarantee that tne < a – c, which is the condition that supports qne > 0. Therefore, we need that
(ac)(γd212d)γ(d+1)2+2d(d+2)<acγ[(d21)(d+1)2]<2d(d+2)+2d2γ[d+1]<2d(d+3),

which holds for all parameter values.

A.2 Proof of Proposition 2

In the third stage, the incumbent and entrant solve,

maxqiπient=(aqiqe)qicqit(qiz),andmaxqeπeent=(aqeqi)qecqet(qeβz)F.

Taking first-order conditions and solving for the output level, we obtain the symmetric solution of a standard Cournot model, qi=qe=q=act3. In the second stage, the regulator’s problem is

maxtSW=0Q(t)(acx)dx12dQ(t)(1+β)z212γz2+F.

with first-order condition

SWt=19(2)(2ad+a+d(2c+2t+3(β+1)z)c+2t)=0.

Solving for t gives t(z)=(ac)(2d1)3(β+1)dz2(d+1). In the first stage, the incumbent solves

max z Π i e n t = ( ( a q i q e ) q i c q i t ( q i z ) ) 1 2 γ z 2 ,

with first-order condition

Πientz=(ac)(d(β+2d+2)1)+(β+1)dz((β5)d6)2γ(d+1)2z2(d+1)2=0.

Solving for z gives the equilibrium level of investment under entry zent=(ac)(d(β+2d+2)1)d(β+1)((5β)d+6)+2γ(d+1)2. Plugging this into the results from the second and third stages gives us the equilibrium values of the emission fee and quantity:

(14)qent=(ac)(d(β+1)(2d+5)+2γ(d+1))4γ(d+1)2+2d(β+1)((5β)d+6).
(15)qent=(ac)(d(β+1)(2d+5)+2γ(d+1))4γ(d+1)2+2d(β+1)((5β)d+6).
zent and qent are positive since a > c, β ∈ [0, 1], and d > 1. However, tent depends on the sign of the numerator, determined by 2γ2d2+d1d(β+1)(2d(β2)+3). The numerator, and thus tent, is positive if one of the following conditions hold:
  1. If 1<d<32 and 0<β<4d32d,

    The numerator is positive if d(β+1)(2d(β2)+3)<0. Simplifying yields 2d(β2)+3<0β2<32dβ<4d32d. This condition, 4d32d<1, and binding on β, if d<32.

  2. if 1<d<32, 4d32d<β<1, and γ>(β+1)d(2(β2)d+3)22d2+d1.

    If d<32 and 4d32dβ1, then the entire numerator needs to be positive, which happens when γ is high.

  3. or if d>32,

    from case (1), we are looking for 2d(β – 2) + 3 < 0. Rearranging gives 32<d(2β), which always holds if d>32.

Otherwise, tent < 0. We also need to guarantee that tent < (a - c), so that qent > 0. Therefore, we need that

(ac)2γ2d2+d1d(β+1)(2(β2)d+3)4γ(d+1)2+2d(β+1)((5β)d+6)<(ac)2γ2d2+d1d(β+1)(2(β2)d+3)<4γ(d+1)2+2d(β+1)((5β)d+6)2γ[(2d2+d1)2(d+1)2]<d[2(β+1)((5β)d+6)+(β+1)(2d(β2)+3)]6γ(d+1)<3d(2d+5)(1+β),

which holds for all parameter values.

A.3 Proof of Corollary 1

First, we can show that zent<zne for all parameter values:

zent<zne(ac)(d(β+2d+2)1)d(β+1)((5β)d+6)+2γ(d+1)2<(ac)(d(d+2)1)γ(d+1)2+2d(d+2),

Which simplifies to,

d β + d 4 β + β d ( d ( 4 β ) + 12 2 β ) β + d ( d + 4 ) + 1 6 β 2 ( d + 1 ) 2 ( ( β 2 ) d + 1 ) < γ

which holds for all parameter values since the numerator of the left-hand side is positive and the denominator is negative given that β ∈ [0, 1].

A.4 Proof of Corollary 2

Let us next examine if the output level under no entry exceeds that under entry.

qne>qent(ac)(γ(d+1)+d(d+3))γ(d+1)2+2d(d+2)>(ac)(d(β+1)(2d+5)+2γ(d+1))4γ(d+1)2+2d(β+1)((5β)d+6)γ(d+1)+d(d+3)γ(d+1)2+2d(d+2)(β+1)d(2d+5)+2γ(d+1)4γ(d+1)22(β+1)d((β5)d6)>0,

which we can simplify to

γd(d+1)7β+2d2(1β)+d(3β)(2β+5)+11+2(β+1)d2(8+3d(d+4)βd(d+3))>0

which holds for all allowable parameter values since 3d(d + 4) > βd(d + 3). Total quantity produced when there is entry is

Q=(ac)((β+1)d(2d+5)+2γ(d+1))2γ(d+1)2(β+1)d((β5)d6).

The comparison of total quantity when entry ensues to no entry, Q=2qent<qne, is

(ac)(β+1)2d2+2γ(1+d)2(β+1)2d2+4γ(1+d)2<(ac)(γ(d+1)+d(d+3))γ(d+1)2+2d(d+2),d{γ(d+1)[1β+d(β(β+2d+3)2)]+d(1+β)((1β)d(d+3)2)}<0,γ<d(1+β)((1β)d(d+3)2)(d+1)(1β+d(β(β+2d+3)2)).

In order for the right-hand side to be positive, we need that θ<β<d2+3d2d2+3d, where θ2d23d+12d+124d4+12d3+13d210d+1d2.

If β < θ, then the inequality condition on γ switches signs (since we multiply through by a negative), and the condition always holds:

γ>0>d(1+β)((1β)d(d+3)2)(d+1)(1β+d(β(β+2d+3)2)),

and, therefore, 2qent<qne.

A.5 Proof of Corollary 3

To find the level of investment that deters entry, zˆ, we evaluate the entrant’s profit as a function of the incumbent’s investment z, where t(z) and q(t) are the equations from states 2 and 3 of the no entry equilibrium, respectively, that are used to obtain proposition 2. Therefore, the entrant’s zero-profit condition becomes,

πeent=(aqe(t(z))qi(t(z)))qe(t(z))cqe(t(z))t(z)(qe(t(z))βz)F=0,

where,

t ( z ) = ( a c ) ( 2 d 1 ) 3 ( β + 1 ) d z 2 ( d + 1 ) , a n d q ( t ) = 1 3 ( a c t )

Plugging in and solving for z gives the entry-deterring level of investment

zˆ=ωα(d+1)2β24d2+1ω24d(5d+6)F+4βdω22(2d+3)F+4d2F(d+1)2(β+1)d(6β+(5β1)d).

where ω = a – c and α=(β(2d(d+1)1)+d). This high level of entry-deterring investment is

zˆhigh=ωα+(d+1)2β24d2+1ω24d(5d+6)F+4βdω22(2d+3)F+4d2F(d+1)2(β+1)d(6β+(5β1)d).

Under entry deterrence, using zˆ and the emission fee and quantity used for proposition 1 (tne(zˆ) and q=(actne(zˆ))/2 from stages 2 and 3 of the no entry equilibrium, respectively), the incumbent’s profit is

ΠiED(zˆ)=(aczˆ)(ac+d(d+2)zˆ)(d+1)212γzˆ2.

We compare this to the incumbent’s profit when it accommodates entry, which comes from the equilibrium values in proposition 2 (zent, tent, and qent), which is

Πient=(ac)2(2γ+4d(β+d)+1)4(β+1)d((5β)d+6)+8γ(d+1)2.

If ΠiED(zˆ)>Πient, then it is more profitable for the incumbent to deter entry than to accommodate entry.

The analytic comparison between the incumbent’s profit under entry deterrence and entry is intractable. Hence, we next provide a numerical comparison using the parameter values for Figure 1: a=10,c=1,d=2.5,β=0.25,γ=1.5, and F = 3. If the incumbent accommodates entry, the incumbent’s profit evaluated at zent = 1.66, tent = 2.91, and qent = 2.03 (from proposition 2) is Πient=6.89. If the incumbent engages in entry deterrence, the incumbent will lower its investment to zˆ=0.57 (from corollary 3), where the accompanying emission fee is set at tED = 3.04, and output is qED = 2.98 (the emission fee and quantity used for proposition 1, evaluated at zˆ, as previously explained). The incumbent’s profit when it deters entry is ΠiED=10.37.

A.6 Proof of Proposition 3

The entrant’s profit function in equilibrium is

πeent=(ac)2(β+1)dΓ+4γ2(d+1)2+4γ(d+1)η4(β+1)d((β5)d6)2γ(d+1)22F,

where Γ=6β+4(β(2β5)1)d3+4β36β5d2+(β5)(2β+5)d, and η=β+dβ2+β+4βd2+2(β+1)2d+5. We obtain the level of fixed cost that blocks entry by setting Πeent=0 and solving for F which is:

Fˉ=(ac)2(β+1)dΓ+4γ2(d+1)2+4γ(d+1)η4(β+1)d((β5)d6)2γ(d+1)22.

The value F_ is the limit of the region where Πient>ΠIne(zˆ), where zˆ is a function of F. Solving Πient=ΠIne(zˆ) for F defines F_. An analytic solution to this cannot be found, so we turn our attention to the numerical analysis of F_ found in Figure 2, Figure 4, and Figure 5.

A.7 Comparative Statics with Respect to the Spillover

Let us now consider an increase in the spillover in the case of entry:

tentβ=3d(ac)(β+1)2d2(2d+5)+2γ(d+1)(d(2β+2d+3)1)2(β+1)d((β5)d6)2γ(d+1)22<0qentβ=(ac)(β+1)2(2d+5)d3+2γ(d+1)d(d(2β+2d+3)1)2(β+1)d((β5)d6)2γ(d+1)22>0zentβ=d(ac)(d(d(β(β+4)+4(β2)d15)2(β+1))+6)+2γ(d+1)2(β+1)d((β5)d6)2γ(d+1)22>0.

The sign of zentβ is determined by the sign of (d(d(β(β+4)+4(β2)d15)2(β+1))+6)+2γ(d+1)2). If γ > ϕ, where ϕd(2(β+1)d(β(β+4)+4(β2)d15))62(d+1)2, then zentβ>0. If γ < ϕ, then zentβ<0.

A.8 Social Welfare

Social welfare under the cases of no entry and entry are

SWne=(ac)2(γ1)γ+(γ+3)d4+γ2+6γ+13d3+3γ2+12γ+13d2+3γ2+10γ1d2γ+(γ+2)d2+2(γ+2)d2,SWent=(ac)2(β+1)2dd2β+4d2(4β)+(5β)(β+9)d+291γ2(β+1)d((β5)d6)2γ(d+1)22γd22β+dβ2+44β+8βd(d+4)+4d(d+7)+48+244γ2(d+1)32(β+1)d((β5)d6)2γ(d+1)22F.

Solving SWneSWent(F)=0 for F, we can obtain F˜:

F˜=(ac)2γ+(β+1)2dd2β+4(β4)d2+(β5)(β+9)d29+12(β+1)d((β5)d6)2γ(d+1)22γd22β+dβ2+44β+8βd(d+4)+4d(d+7)+48+244γ2(d+1)32(β+1)d((β5)d6)2γ(d+1)22+γ2(d+1)3+γ(d(d(d(d+6)+12)+10)1)+d(d(d(3d+13)+13)1)2(γ+(γ+2)d(d+2))2.

Using parameter values from Figure 2 (a = 10, c = 1, d = 2.5, and γ = 1.5), we can simplify F˜ as a function of β that can be plotted in Figure 6:

F˜=β(β(β(15.40136.6407β)+814.382)+1213.04)160.77(β(β6.4)13.28)2.

A.9 Robustness Check

We next present a robustness check for Figure 3, Figure 7, and Figure 6. First, note that Figure 3 and Figure 7 coincide with the top-left figure inside Figure 8 where β = 0 and γ = 1.5. We observe that if γ increases but β remains at zero, cutoff F˜ shifts upward. However, it does not overlap with either F_ or Fˉ. This is also true for the case of β = 0.25. Second, as the spillover increases the area for which the regulator and firm preferences are aligned expands. Finally, in the bottom-left figure, where the cost of investment is low but the spillover is high, the cutoff for blockaded entry lies below the cutoff where the incumbent prefers to deter entry. In this case, the incumbent will not deter entry as it either prefers to accommodate at a low fixed entry cost, or entry is blockaded if the fixed cost is high enough such that the incumbent would otherwise prefer to deter entry.

Figure 8: 
              Regions where social welfare under entry and no entry coincide when β and γ increases.
Figure 8:

Regions where social welfare under entry and no entry coincide when β and γ increases.

Figure 9 shows the robustness check for Figure 6. Note that the middle-left figure evaluated at d = 2.5 and γ = 1.5 represents Figure 6. If environmental damages increase from d = 2.5 to d = 5, a higher spillover induces a higher social welfare under entry than under no entry. The top-left figure represents a situation where a subsidy is in place.

Figure 9: 
              Regions where social welfare under entry and no entry coincide when d and γ increases.
Figure 9:

Regions where social welfare under entry and no entry coincide when d and γ increases.

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Published Online: 2020-02-01

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