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Announced or Surprise Inspections and Oligopoly Competition

Emmanuel Dechenaux and Andrew Samuel

Abstract

To enforce compliance, regulators often choose between announced or unannounced (surprise) inspections. We analyze the impact of these inspection regimes on firms’ compliance choices in a multiple stage oligopoly game of quantity competition with endogenous compliance, monitoring and avoidance. In equilibrium, whether unannounced inspections achieve a higher level of compliance than announced inspections depends on the number of firms, demand and the cost of compliance. Furthermore, the impact on compliance of increasing the fine, the supervisor’s wage or the probability of inspections also depends on market size and structure and may be non-monotonic. Finally we provide conditions under which a welfare maximizing regulator will prefer an unannounced to an announced regime. Thus, our results suggest that when choosing the appropriate inspection regime, regulators should account for market characteristics, especially if compliance maximization is the objective.

JEL Classification: L13; L15; K42

A Appendix

A.1 Proof of Proposition 1

Under announced inspections, avoidance xA and monitoring mA do not depend on σ. Moreover, it is clear from the expected payoff functions that if σ = 1, then the players’ best response functions are identical under the two regimes and so is the Nash equilibrium, xU=xA and mU=mA. Furthermore, xUσ>0, which implies xU<xA for σ < 1. Similarly, mUσ<0, which implies mU>mA for σ < 1. It then follows at once that gAgU, with a strict inequality if σ < 1.

The signs of the comparative statics derivatives follow from deriving gρz for z ∈ { f,r,σ} and routine calculations.

A.2 Proof of Lemma 1

The equilibrium prices, quantities and profits follow from the application of well-known Cournot equilibrium results. Specifically firm i ’s output is

qi=1γαc+jiejNeiN+1,

and profit is

πi=(Pcei)qi=1γαc+jiejNeiN+12.=γqi2.

Assuming that all firms are active in the Cournot equilibrium, in the first stage, firm i chooses ai to maximize

Πi(ai,ai)=πi12kai2.

Given an inspection regime ρ={A,U}, a straightforward calculation yields

πiai=2N(δh+gρ)γ(N+1)2(αc+jiejNei).

The FOC for an interior solution to profit maximization with respect to ai is

Πiai=02N(δh+gρ)γ(N+1)2(αc+jiejNei)=kai.

The SOC is 2N2(δh+gρ)2γ(N+1)2k<0, which is satisfied under Assumption 2. Returning to the FOC, we have

jiej=ji(1aj)(δh+gρ)=(N1)(δh+gρ)(δh+gρ)jiaj.

Therefore,

2N(δh+gρ)γ(N+1)2(αc+jiejNei)=kaiαc+(N1)(δh+gρ)gρjiajNei=γ(N+1)22N(δh+gρ)kai.

In a symmetric equilibrium, a1=a2==aN=a,

αc+(N1)gρgρjiajNei=γ(N+1)22N(δh+gρ)kai.

After some algebra we obtain

αc(δh+gρ)=kγ(N+1)22N(δh+gρ)22N(δh+gρ)a,

which implies

a=2N(αcδhgρ)(δh+gρ)kγ(N+1)22N(δh+gρ)2.

The expressions for a firm’s equilibrium output and profit follow from straightforward calculations and are standard expressions for a Cournot equilibrium.

A.3 Deriving Assumption 2

We show that Assumption 2 guarantees that the second-order condition for the firm’s compliance maximization problem is satisfied and that it is sufficient to ensure an interior equilibrium in compliance. The second-order condition for the firm’s profit maximization with respect to compliance is

2N2(δh+gρ)2γ(N+1)2k<0.

Noting that αc>δh+gρ and that N2(N+1)2<1 it follows that if

2(αc)2γk,

then the second-order condition is also satisfied. This last inequality is simply Assumption 2.

Next aρ in eq. (2) is strictly less than 1 if and only if

aρ=2N(αcδhgρ)(δh+gρ)kγ(N+1)22N(δh+gρ)2<1.

The above condition simplifies to,

2N(N+1)2(αc)(δh+gρ)γ<k.

Since the LHS of the above expression is strictly decreasing in N, N 1, and once again using the fact that gρ+δh<αc it follows that the above condition is always weaker than Assumption 2.

A.4 Proof of Proposition 2

From Proposition 1, we have gUgA. Using this fact, aUaA0 holds if and only if[23]

(6)kγ(N+1)2(αc2δhgUgA)+2N(δh+gU)(δh+gA)(αc)0.

Define αc+2δh+gU+gA. If αα, then (6) is satisfied because all terms are non-negative. Therefore, if αα, unannounced inspections lead to higher compliance than announced inspections.

Now suppose α<α, then (6) is satisfied if and only if

(7)kγ(N+1)2((αc2δh)+gU+gA)+2N(δh+gU)(δh+gA)(αc)0.

Isolating α shows that (7) holds if and only if

αc+(2δh+gU+gA)γk(N+1)22N(δh+gU)(δh+gA)+γk(N+1)2A(N).

It is straightforward to verify that

A(N)>0.

It follows that if (7) is not satisfied at N = 2, then it is not satisfied at any N. Hence, if α < A(2), announced inspections yield higher strictly compliance regardless of the number of firms. Moreover, for finite N, we have A(N)<α and limNA(N)=α. Since A(N) is continuous and strictly increasing, if α[A(2),α), then there exists a unique N 2 such that if NN, (7) holds and thus compliance is higher with unannounced inspections. Instead, if N>N, (7) is violated and compliance is higher with announced inspections.

A.5 Proof of Proposition 3

The equilibrium level of compliance aρ depends on f, r and σ solely through the level of enforcement gρ. Therefore, for any τ ∈ { f,r,σ},

aρτ=aρggρτ.

Viewing the enforcement burden as a variable g, where g=gρ under regime ρ, and differentiating aρ with respect to g, we obtain

(8)aρg02N(δh+g)2γ(N+1)22(δh+g)αc1k.

If αc+2(δh+gρ)=αρ, then eq. (8) is satisfied and aρg0. Therefore, aρτ is either equal to zero or it has the same sign as gρτ. Proposition 1 characterizes the derivatives gρτ and point (i) above follows from a direct application of the result in Proposition 1.

Now suppose α<c+2(δh+gρ). Then there are two cases. The first case is when eq. (8) is satisfied at N = 2 or

4(δh+gρ)29γ2(δh+gρ)αc1kαc+18γk(δh+gρ)9γk+4(δh+gρ)2=α_ρ,

where α_ρ<αρ follows from gρ>0. In this case, aρg0 at N = 2. However, in (8), the left-hand side is strictly decreasing in N, whereas the right-hand side does not depend on N. Furthermore, limN2N(δh+g)2γ(N+1)2=0, which is less than the right-hand side of (8) due to the fact that α<αρ. Therefore there exists a Nˆρ2 such that for N>Nˆρ, aρg<0. Hence, for α(α_ρ,αρ), if NNˆρ, then for any τ ∈ { f,r,σ}, the sign of aρτ is given by the sign of gρτ as in point (i). Otherwise, if N>Nˆρ, the sign of aρτ is given by the sign of gρτ so that the effect on compliance of each policy variable is the reverse of that in (i).

The second case is when (8) is not satisfied at N = 2 or αα_ρ. In this case, aρg<0 for every N. It thus follows that for any τ ∈ { f,r,σ}, the sign of aρτ is given by the sign of gρτ. Thus the effect of each policy variable is the reverse of the effect characterized in point (i).

A.6 Proof of Lemma 2

In the proof, we omit the ρ subscript in M(gρ), aρ and qρ. First, consider the case of direct harm. When δ = 1, the market surplus is

M(g)=NN+22γq2N12ka2.

Differentiating with respect to g yields

M(g)=NN+22γ2qg+qaagqNkaga.

Substituting the expressions for qa and q yields

M(g)=N(N+2)γqgq+N(N+2)γ(h+g)γ(N+1)agqNkaga=N(N+2)γqgq+N(N+2)(h+g)(N+1)qkaag=N(N+2)γqgq+N(N+2)(h+g)(N+1)αc(h+g)γ(N+1)+a(h+g)γ(N+1)kaag.

The first term is clearly negative. The term in the square bracket is non-positive if and only if

a(N+2)(αc(h+g))(h+g)kγ(N+1)2(N+2)(h+g)2,

which always holds because

a=2N(αc(h+g))(h+g)kγ(N+1)22N(h+g)2

and N 2. It follows that if ag0, then M(g) < 0. Thus, if M(g) 0, it must be the case that ag<0, i.e. enforcement reduces compliance.

Now turning to the case where the harm falls on third parties, the market surplus is given by

M(g)=NN+22γq2N(1a)qhN12ka2.

Then we have

M(g)=N(N+2)qgqNagq+(1a)qghNkaga.

Setting g = 0 and simplifying yields

M(0)>0N(N+2)(αc)γ(N+1)2+N1γ(N+1)+2N(αc)2kγ2(N+1)3h>0.

Clearly, the first term is negative while the second term is positive. Solving this expression for h yields the h˜ identified in Lemma 2. A straightforward computation shows that the derivative of h˜ with respect to N is negative.

A.7 Proof of Proposition 4

First we show that R(r) exists, is unique and R(r) r. Fix (f,σ) and r = r such that gA>0. From Proposition 1, we know that gUgA with an equality if σ = 1. Moreover, gU is increasing and continuous in r and limr0gU=0. Therefore there exists a unique y ∈ (0,r] such that

gA|r=r=gU|r=y.

Now setting R(r) = y proves the claim.

Second, note that all but the last term of the welfare functions depend on (f,r,σ) only through gρ. It follows that

WA(fA,rA,σA)WU(fA,R(rA),σA)=NqA(1aA)σAd(mA,xA)fAmAd(mU,xU)fAmU.

Since NqA(1aA)>0, we have

WA(fA,rA,σA)WU(fA,R(rA),σA)>0d(mA,xA)fAmAd(mU,xU)fAmU>0.

Using the expressions for gA, gU and the fact that gA=gU yields

d(mU,xU)=d(mA,xA)+σAxAxUσAfA.

After substituting the above expression for d(mU,xU) and simplifying, we obtain

d(mA,xA)fAmAd(mU,xU)fAmU>0xU+σAmU>σA(xA+mA).

It follows that announced inspections yield higher welfare only if this last condition holds. Hence if σA(xA+mA)xU+σAmU, announced inspections cannot be strictly optimal, which is what we had to prove.

We solve the model for the special case

d(m,x)={mm+xifm+x>00ifm+x=0.

For this probability of detection, we obtain

R(r)=σr.

Now set r equal to r under announced inspections and fix (f,σ). Define (mU′′,xU′′) as the equilibrium of the unannounced inspections game when r = R(r). Then, we have

σ(xA+mA)xU′′+σmU′′=σr2f1σf2+2fr+r20.

Therefore the inequality in Proposition 4 always holds and announced inspections cannot be optimal in this case.

Acknowledgements

We would like to thank an anonymous referee, Andrew Daughety, Valentina Dimitrova-Grajzl, Jennifer Reinganum, Kathy Spier, Tracy Lewis and Joel Watson for their useful suggestions, as well as participants at the Law and Economic Theory Conference (Berkeley, 2014), Southern Economic Association Meetings (Atlanta, 2014), the Public Economic Theory conference (Seattle, 2014) and in seminars at Haverford College, Oberlin College and the University of Pennsylvania for helpful comments. Ha Nguyen provided valuable research assistance.

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Published Online: 2018-12-15

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