A Proposition 1 (formally)

For any actionable history h˜i∈H˜i where there has not been a public price signal that revealed a deviation, let mh˜i be the last element of the actionable history (which recall is the signal or message the player just received). Then define s∗ as:

Lemma 2 and Lemma 3 state that at any actionable history, firm i knows firm j’s private history and the current reserve price exactly. Assuming no public signal has revealed a deviation, the strategy s∗ and lemmas 1–3 imply that

1. Firm i’s continuation value at any history only depends on the current value of rt and not on the entire history of rt .

2. At any actionable history, firm i knows rt via the message it receives.

3. Firm i knows what observation j received at any actionable history as well as firm j’s current price.

Consequently, when firm j plays s∗ with consistent beliefs β∗ as defined in Lemma 2 and Lemma 3, it is possible to adopt a finite automaton representation of firm i’s strategies where i’s automaton states are defined by its observation at an actionable history and whether or not a public signal has revealed a deviation.^[9] The continuation values under s∗ are expressed in equation set 4. The notation Vk(m) indicates the continuation value for firm i at automaton state defined by the observation m. Vk(∅) represents a state in the CTMC where firms do not take an action and is only expressed for clarity.

The operator EJi represents the expectation taken over the amount of time the system is in the state indicated by Vi . The parenthetical comments briefly describe the event that has just occurred that gives the continuation value indicated by the equation.

By applying the Laplace-Stieltjes transform, the above equations can be rewritten as^[10]:

Holding firm j’s strategy fixed, the one-stage deviation principal says that s∗ is optimal for firm i if there is not a one-time deviation for firm i at V1 , V3 or any other state of the CTMC reachable from V1 or V3 that increases firm i’s continuation value.^[11] However, the only automaton state reachable from V1 and V3 not along the equilibrium path is the one that includes a public revelation of prices where the firms charged different prices. In this case it is (weakly) dominant for firms to set their price to 0. Therefore, the only deviations that must be evaluated are the various one-shot deviations at V1 and V3 . All possible deviations can be condensed into 1 of 5 possible continuation values, which are summarized in box 1. We now analyze all possible one-shot deviations at actionable histories in turn.

Box 1: Categorization of all possible deviations by continuation value

Case 1: Undercutting ρL slightly (deviating at V1 ) The continuation values when i deviates at V1 by undercutting slightly are given by:

Note the new action state V6 , which is the actionable state when a public price revelation indicates that i has deviated. In that case it is weakly dominant for firm i to charge 0. By again applying the Laplace-Stieltjes transform and solving the system of equations for V1 and V1′ , we get the necessary condition for i not to undercut upon learning the reserve price ρL :

or equivalently

where

and Λ=λ1+λ2 and Cα,α≠0 are given in the online supplementary materials.

Case 2: Undercutting ρH slightly (Deviation at V3 ) The continuation values when i undercuts the market slightly upon learning the reserve price is ρH are given by:

The necessary condition for i not to deviate is then given by:

or equivalently

where

and Dα,α≠0 are given in the online supplementary materials.

Case 3: Undercutting ρH by pricing at ρL (deviating at V3 ) The continuation values when i undercuts the market by charging ρL upon learning the price is ρH and tells j the price is ρH are given by:

The necessary condition for i not to deviate is then given by

or equivalently

where

and Eα,α≠0 are given in the online supplementary materials.

Case 4: Firm i charges ρL , tells firm j that the reserve price is ρL upon learning the reserve price is ρH .) The continuation values to firm i by making a one-shot deviation in which it tells firm j that the reserve price is ρL and charges ρL although firm i knows the reserve price is ρH are given by:

Applying the Laplace-Stieltjes transform gives the necessary condition for i not to have an incentive to deviate, which is given by:

or equivalently

Applying the Laplace-Stieltjes transform gives the necessary condition for i not to have an incentive to deviate, which is given by:

or equivalently

As δ goes to 0, the left hand side of equations 5, 6 and 8 go to C0=D0=E0>0 , eq. 11 goes to γλ1Λ2γ+ΛγρH−2ρL+ΛρH−ρL and eq. 13 goes to γλ1Λ2γΛ−γρH+2γ+ΛρL . Since the left hand side of eqs. 5, 6, 8, 11 and 13 are all continuous in δ, there exists a δ_ such that for all δ<δ_ the conditions in eqs. 5, 6, 8, 11 and 13 are met as long as 2γ+Λγ+Λ<ρHρL<2γ+Λγ . It is not necessary to check if firm i should deviate upon a public revelation of prices since the continuation value of such a deviation would be a convex combination of continuation values of deviating upon learning the reserve price exactly. This implies that for a low enough δ, s∗ and β∗ as described in Lemma 2 and Lemma 3 is a Perfect Bayesian Equilibrium (i.e. s∗ is a best response to s∗ and beliefs along the equilibrium path are derived through Bayes rule).

To establish that s∗ is also a sequential equilibrium, it is necessary to show that s∗ is sequentially rational off of the equilibrium path. Define Σ,(s1,β1),(s2,β2)... indexed by ε to be a sequence of strategy profiles and associated consistent beliefs (for both firms) such that at any actionable history:

1. Firms play the action prescribed by s∗ with probability 1−1ϵ .

2. Firms randomize uniformly over all actions (prices and messages) with probability 1ϵ

As ϵ→∞ , sϵ→s∗ and along the equilibrium path βiϵ converges to placing probability 1 on values of rt and h˜jt (since the probability of firm j doing anything other than what is prescribed by s∗ at or befer time t goes to 0 as ϵ→∞ ). Therefore, sϵ converges to s∗ and along the equilibrium path, βiϵ converges to β∗ .

Lastly, it is necessary to verify that s∗ is sequentially rational off of the equilibrium path given beliefs β∗ . The only actionable histories for firm i off of the equilibrium path are ones that include an undetected deviation by firm i and ones that have revealed a deviation by either firm.

Consider any actionable history h˜it where i has deviated but not been detected. By Lemma 2 and Lemma 3, β∗ places probability 1 on one value of h˜jt and rt . Therefore, firm i’s continuation value from playing s∗ after an undetected deviation can be represented by either V1 or V3 , both of which firm i would not choose to deviate from s∗ . Therefore s∗ is sequentially rational at any h˜it in which firm i has deviated but not been detected.

For any actionable history h˜it that has revealed a deviation, βi∗ specifies some distribution over past histries. However, since sϵ converges to s∗ , upon realizing a deviation, it is sequentially rational for firm i to play 0 regardless of the belief of past histories.

Lemma 4 simplifies the analysis since it says that a necessary and sufficient condition for the existence of the collusive region is that V3−V3′>0 since this condition implies that all other necessary conditions are met. In other words, V3−V3′=0 defines the collusive region. We can now prove claims 1-6 in Proposition 4 in order.

1. T is a function: Since V3−V3′ can be written as a quadratic in µ, applying the quadratic formula to V3−V3′=0 yields:

where

Since a > 0, b > 0 and c < 0, there exists exactly one positive root for all values for a fixed value of δ, γ and Λ. Therefore, T is a function since for fixed values of δ, γ and Λ, T returns only one value of µ.

2. T is a bijection in δ when holding γ and Λ fixed: We need to show that T is one-to-one and onto. To show that T is onto, we need to show that for any fixed µ, there exists a value of δ that makes V3−V3′=0 . Since all parameters are greater than 0, V3−V3′ can be written as:

For a fixed µ, as δ→∞ , V3−V3′→−∞ and as δ→0 , V3−V3′→γμΛ2γ+μ+Λ4γρL+3ΛρL>0 . Since V3−V3′ is continuous, by the intermediate value theorem there is at least one value of δ such that V3−V3′=0 for any fixed µ. This proves T is onto. To show that T is one-to-one, we need to show that for a fixed value of µ, there are not multiple values of δ that make V3−V3′=0 . To show that this is the case, we remind the reader of a well-known result:

Note that the sign of the coefficients on δ5,δ4 are unambiguously negative and the constant in equation is unambiguously positive in eq. 15. Therefore, the number of positive roots of eq. 15 is either 1 or 3. There are three roots if the signs of the coefficients of δ3,δ2 and δ1 are one of ′+−+′ , ′+−−′ ,++− or ′−+−′ (other combinations of signs yield 1 positive root). Let D3 , D2 , and D1 be the coeffecients on δ3 , δ2 and δ1 respectively. We now show that none of the three possible sign patterns are possible.

1. Assume D3>0 then it must be that μ>−Λ+32γ2+48γΛ+13Λ22 . Since D2 is increasing in µ for positive values of µ, for it to be the case that D3>0 and D2<0 it must be that at μ=−Λ+32γ2+48γΛ+13Λ22 , D2<0 . Plugging in for µ in D2 gives

   D2=ρL2γ+Λ20γ2+10Λ2+γ27Λ+32γ2+48γΛ+13Λ2>0.

   Therefore, it is impossible for D3>0 and D2<0 . This proves that it is impossible for the coefficients to have signs ′+−−′ and ′+−+′ .

2. Assume D2>0 , then it must be that μ>−γ2−2γΛ−Λ2+13γ4+54γ3Λ+61γ2Λ2+25γΛ3+3Λ43γ+2Λ . Since D1 is increasing in µ for positive values of µ, for it to be the case that D1<0 and D2>0 , it must be that at μ=−γ2−2γΛ−Λ2+13γ4+54γ3Λ+61γ2Λ2+25γΛ3+3Λ43γ+2Λ . Plugging this value in for µ in D1 gives

   (16)D1=ρL3γ+2Λ2[32γ6+180γ5Λ+4Λ6+γ4(413Λ2+16ϕ)+γ3(415Λ3+14Λϕ)+γ2(194Λ4+13Λ2ϕ)+γ(42Λ5+2Λ3ϕ)]

   where ϕ=13γ4+54γ3Λ+61γ2Λ2+25γΛ3+3Λ4 . Since all parameters are greater than 0, D1>0 and therefore it is impossible for D1<0 while D2>0 and the signs of the coefficients cannot be ′−+−′ or ′++−′ .

Since for a fixed µ there is only 1 positive δ such that V3−V3′=0 , T is one-to-one.

3. T is strictly increasing in δ: Since T is a continuous bijection it must be strictly monotone. It is straightforward to show that T is monotone increasing in δ.

4. limδ→0T=0 : Setting δ = 0 in eq. 14 and simplifying shows that limδ→0T=0 .

5. limδ→∞T=∞ : The highest order δ term under the radical in eq. 14 is δ8 and it has a positive coefficient. The highest order δ term in b is δ3 . Therefore, limδ→∞T=∞ .

6. limδ→∞∂T∂δ=1 : The expression for ∂T∂δ is given in the online supplementary material. Taking the limit as δ→∞ gives the result.   □