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, ∂xU∂σ>0, which implies xU<xA for σ < 1. Similarly, ∂mU∂σ<0, which implies mU>mA for σ < 1. It then follows at once that gA≤gU, with a strict inequality if σ < 1.

The signs of the comparative statics derivatives follow from deriving ∂gρ∂z 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

and profit is

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

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

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

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

Therefore,

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

After some algebra we obtain

which implies

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

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

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

The above condition simplifies to,

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

A.4 Proof of Proposition 2

From Proposition 1, we have gU≥gA. Using this fact, aU∗−aA∗≥0 holds if and only if^[23]

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

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

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

It is straightforward to verify that

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 limN→∞A(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 N≤N‾, (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},

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

If α≥c+2(δh+gρ)=α‾ρ, then eq. (8) is satisfied and ∂aρ∗∂g≥0. Therefore, ∂aρ∗∂τ is either equal to zero or it has the same sign as ∂gρ∂τ. Proposition 1 characterizes the derivatives ∂gρ∂τ 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

where α_ρ<α‾ρ follows from gρ>0. In this case, ∂aρ∗∂g≥0 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, limN→∞2N(δh+g)2γ(N+1)2=0, which is less than the right-hand side of (8) due to the fact that α<α‾ρ. Therefore there exists a Nˆρ≥2 such that for N>Nˆρ, ∂aρ∗∂g<0. Hence, for α∈(α_ρ,α‾ρ), if N≤Nˆρ, then for any τ ∈ { f,r,σ}, the sign of ∂aρ∗∂τ is given by the sign of ∂gρ∂τ as in point (i). Otherwise, if N>Nˆρ, the sign of ∂aρ∗∂τ is given by the sign of −∂gρ∂τ 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 α≤α_ρ. In this case, ∂aρ∗∂g<0 for every N. It thus follows that for any τ ∈ { f,r,σ}, the sign of ∂aρ∗∂τ is given by the sign of −∂gρ∂τ. 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

Differentiating with respect to g yields

Substituting the expressions for ∂q∗∂a and q∗ yields

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

which always holds because

and N≥ 2. It follows that if ∂a∗∂g≥0, then M^′(g) < 0. Thus, if M^′(g) ≥ 0, it must be the case that ∂a∗∂g<0, i.e. enforcement reduces compliance.

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

Then we have

Setting g = 0 and simplifying yields

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 gU≥gA with an equality if σ = 1. Moreover, gU is increasing and continuous in r and limr↓0gU=0. Therefore there exists a unique y ∈ (0,r^′] such that

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

Since NqA∗(1−aA∗)>0, we have

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

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

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

For this probability of detection, we obtain

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

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