Step 1: Derivation of u˜S(x,x_), u˜S(x,γ) and u˜S(x,y)

We start by illustrating how u˜S(x,x_),u˜S(x,γ) and u˜S(x,y) are derived. To this end, we need to determine the updated beliefs for a bidder who lost at stage one – because he did not participate in the auction, bid r1, or he bid βn−1,r2(y). These beliefs are conditional on the information the bidder learns at stage one: his bid at stage one (which we denote with b) and the winning bid at stage one (which we denote with bw). In order to shorten the notation, we set Γ≡F(γ) and Λ≡F(λ).

Step 1.1: Updated Beliefs for a Bidder who has not Bid at Stage One, and u˜S(x,x_).

Consider a bidder with type x who has made no bid at stage one. Here we describe his beliefs upon learning bw, and his expected payoff u˜S(x,x_) from (5).

1. In case there has been no bid by any bidder, an event with probability Γn−1 from the bidder’s ex ante point of view, his beliefs are given by the c.d.f. G˜(⋅|no,no) such that

   (30)G˜(s|no,no)=Fn−1(s)Γn−1if s∈[x_,γ)1if s∈[γ,xˉ]

2. In case bw=r1, an event with probability Λn−1−Γn−1 from the bidder’s ex ante point of view, his beliefs are given by the c.d.f. G˜(⋅|no,r1) such that

   (31)G˜(s|no,r1)=(n−1)(Λ−Γ)Λn−1−Γn−1Fn−2(s)if s∈[x_,γ)Λ−ΓΛn−1−Γn−1Fn−1(s)−Γn−1F(s)−Γif s∈[γ,λ]1if s∈(λ,xˉ]

   About the derivation of G˜(s|no,r1), consider the point of view of, say, bidder 1; the following probabilities refer to the n−1 bidders different from 1. For s∈[x_,γ), G˜(s|no,r1) is obtained from the probability that one of the other bidders has value in [γ,λ] and each other bidder has value smaller than s. This probability is equal to (n−1)(Λ−Γ)Fn−2(s).

   For s∈[γ,λ], G˜(s|no,r1) is obtained from the probability that at least one of the other bidders has value in [γ,λ], none of them has value bigger than λ, and each of the non winning bidders with value x∈[γ,λ] is such that x≤s. Such a probability is given by^[20]

   (32)(n−1)(Λ−Γ)∑j=0n−2Cn−2,jj+1Γn−2−j(F(s)−Γ)j

   Specifically, Λ−Γ is the probability that a bidder (the winner) has value in [γ,λ] and we have n−1 possible ways of picking a winner. If there are j other bidders (from the remaining n−2) whose value is greater than γ, we need each of them to have value less than s, and 1j+1 is the probability that our initially selected bidder wins. Remark that

   (33)Cn−2,jj+1Γn−2−j(F(s)−Γ)j=Cn−1,j+1(n−1)(F(s)−Γ)Γn−2−j(F(s)−Γ)j+1

   for j=0,1,...,n−2. The right hand side of (33) is equal to Cn−1,h(n−1)(F(s)−Γ)Γn−1−h(F(s)−Γ)h, for h=1,2,...,n−1 (with h=j+1). Hence (32) is equal to

   (n−1)(Λ−Γ)∑h=1n−1Cn−1,h(n−1)(F(s)−Γ)Γn−1−h(F(s)−Γ)h=Λ−ΓF(s)−ΓFn−1(s)−Γn−1

3. In case bw=b˜S(1)(z) for some z∈(λ,xˉ], an event with probability 1−Λn−1 from the bidder’s ex ante point of view, his beliefs are given by the c.d.f. with value Fn−2(s)/Fn−2(z) if s∈[x_,z), with value 1 if s∈[z,xˉ]. This c.d.f. applies for each stage one bid b≤b˜S(1)(z) as long as the winning bid has been b˜S(1)(z) for some z∈(λ,xˉ]; hence we define G˜(s|b,b˜S(1)(z)) such that

   (34)G˜(s|b,b˜S(1)(z))=Fn−2(s)Fn−2(z)if s∈[x_,z)1if s∈[z,xˉ]for each b≤b˜S(1)(z)

When he decides to make no bid, the bidder’s expected beliefs are represented by the c.d.f. G˜no such that

using eqs. (30), (31) and (34). Hence the payoff of a type x from not bidding at stage one is

Step 1.2: Updated Beliefs for a Bidder who has Bid r1 at Stage one but has not Won at Stage one, and u˜S(x,γ).

For future convenience, we introduce the following function M, defined for a∈[0,1] and b∈[0,1]:

Multiplying (a−b)2 by (n−1)an−2+(n−2)an−3b+...+2abn−3+bn−2 reveals that

and therefore M is strictly increasing both with respect to a and with respect to b.

For a bidder bidding r1, the probability to win at stage one is

Let p˜ℓ denote the probability that another bidder wins at stage one with a bid of r1. The probability that another bidder wins at stage one with a bid b˜S(1)(z) for some z∈(λ,xˉ] is 1−Λn−1. Since p˜(γ)+p˜ℓ+1−Λn−1=1, it follows that p˜ℓ=Λn−1−p˜(γ), that is

Now consider a bidder who has bid r1 at stage one but has not won. Then either bw=r1, or bw=b˜S(1)(z) for some z∈(λ,xˉ].

1. In case bw=r1 and another bidder has won, an event with probability p˜ℓ from the bidder’s ex ante point of view, his beliefs are given by G˜(⋅|r1,r1) such that

   (40)G˜(s|r1,r1)=(n−1)(Λ−Γ)2p˜ℓFn−2(s)if s∈[x_,γ)Λ−Γnp˜ℓM(F(s),Γ)if s∈[γ,λ]1if s∈(λ,xˉ]

   Considering the point of view of bidder 1, the derivation of G˜(s|r1,r1) for s∈[x_,γ) is similar to the derivation of G˜(s|no,r1) for s∈[x_,γ), taking into account that bidder 1 has bid r1 rather than abstaining from bidding.

   For s∈[γ,λ], G˜(s|r1,r1) is obtained from the probability that none of the other bidders has value greater than λ, at least one of them has value in [γ,λ] and wins, and each losing bidder with value x∈[γ,λ] is such that x≤s. This probability is equal to

   (41)(n−1)(Λ−Γ)∑j=0n−2Cn−2,jj+2Γn−2−j(F(s)−Γ)j

   From (39) we see that ∑j=0n−2Cn−2,jj+2Γn−2−j(Λ−Γ)j=M(Λ,Γ)n(n−1). Hence (41) is equal to

   (n−1)(Λ−Γ)∑j=0n−2Cn−2,jj+2Γn−2−j(F(s)−Γ)j=Λ−ΓnM(F(s),Γ)

2. In case bw=b˜S(1)(z) for some z∈(λ,xˉ], an event with probability 1−Λn−1 from the bidder’s ex ante point of view, his beliefs are given by G˜(⋅|r1,b˜S(1)(z)) in (34).

When he decides to bid r1 at stage one, the bidder expects to lose with probability 1−p˜(γ)=p˜ℓ+1−Λn−1, hence his expected beliefs are represented by the c.d.f. G˜r1 such that

using (40) and (34). Hence the payoff of a type x from bidding r1 at stage one is

Step 1.3: Updated Beliefs for a Bidder who has Bid b˜S(1)(y), with y∈(λ,xˉ], at Stage one but has not Won at Stage one, and u˜S(x,y).

If a bidder has bid b˜S(1)(y) at stage one and has not won, then bw=b˜S(1)(z) for some z≥y, and his beliefs are given by G˜(⋅|b˜S(1)(y),b˜S(1)(z)) in (34). Hence, the payoff of a type x from bidding b˜S(1)(y) at stage one, for y∈(λ,xˉ], is

and notice that the second term in the right hand side in (43) is equal to

Step 2: Derivation of γ and λ: Existence of a Unique Solution for eqs. (16)-(17), and Definition of r˜1

Using (35), (42), and (43) we find

Hence eqs. (16) and (17) reduce, after some rearranging, respectively to

with

Step 2.1: Definition of λ∗.

Define τ(λ)=Fn−1(λ)λ−r1−vn(λ), a strictly increasing function such that τ(r1)<0 and τ(xˉ)=rˉ1−r1>0. Hence there exists λ in the interval (r1,xˉ), which we denote λ∗, such that τ(λ)<0 for λ∈(r1,λ∗), τ(λ∗)=0, τ(λ)>0 for λ∈(λ∗,xˉ].

Step 2.2: If λ∈(r1,λ∗), then There Exists no γ∈(r1,λ] such that A(γ,λ)=0; if λ∈[λ∗,xˉ], then There Exists a Unique γ∈(r1,λ] such that A(γ,λ)=0, Denoted γA(λ).

Given a function h of two variables, here and in the remainder of the Appendix we write hi to denote the partial derivative of h with respect to its i-th variable, i=1,2.

First we prove that the function A is strictly increasing with respect to γ:

From the definition of p˜(γ) in (38), we see that

and, moreover, ∂p˜(γ)∂γ(γ−r1)>0 and n−12f(γ)vn−1(γ)>0.

Now we examine the sign of A(r1,λ) and of A(λ,λ). We have that

and A(λ,λ)=τ(λ). Therefore, if λ∈(r1,λ∗) then A(λ,λ)<0 and there is no solution to A(γ,λ)=0 in the interval (r1,λ]; if λ∈[λ∗,xˉ], then there exists (A is continuous in λ) a unique solution to A(γ,λ)=0 in the interval (r1,λ], which we denote γA(λ).

Step 2.3: There Exists r˜1∈(r2,rˉ1) such that the Equation B(γA(λ),λ)=0 has a Unique Solution in (λ∗,xˉ) if r1∈(r2,r˜1); the Equation B(γA(λ),λ)=0 has no Solution in (λ∗,xˉ) if r1≥r˜1.

First we prove that B(γA(λ),λ) is strictly decreasing in λ. Notice that Γ below is actually equal to F(γA(λ)). We have that

and we prove that dB(γA(λ),λ)dλ<0. From the previous step we have that

From the proof of Step 2.2 we know that the denominator in the right hand side of (47) is positive.

Therefore dBdλ has the same sign as

with K=p˜−n−12(Λ−Γ)Γn−2−Γn−1>0. Moreover,

hence (48) is smaller than

which is equal to −f(γA(λ))f(λ) times

We now prove that (49) is negative by showing that the terms inside the square brackets are positive.

1. The inequality γA(λ)>r1 implies (λ−r1)M1(Λ,Γ)+(λ−r1)M2(Λ,Γ)−∫γA(λ)λ(M2(F(s),Γ)ds>(λ−r1)M1(Λ,Γ)+∫γA(λ)λM2(Λ,Γ)−M2(F(s),Γ)ds, and the right hand side is positive since M1>0 and M2(a,b)=∑j=0n−3(n−2−j)(j+1)an−3−jbj is strictly increasing in a.

2. The term M(Λ,Γ)∫γA(λ)λM2(F(s),Γ)ds is positive since M2>0.

3. The term M1(Λ,Γ)M(Γ,Λ)−M(Λ,Γ)M2(Λ,Γ) is positive. In fact, from (36) we have M1(a,b)=n−1n−2an−2bn+nn−1an−2b2−2n−2nan−1b(a−b)3, and M2(a,b)=(n−2)an−bn+nab(bn−2−an−2)(a−b)3. Therefore,

   M1(Λ,Γ)M(Γ,Λ)−M(Λ,Γ)M2(Λ,Γ)= nΛ2n−2ΓΛ−Γ41−(n−1)2kn−2+2n(n−2)kn−1−(n−1)2kn+k2n−2 

   with k=ΓΛ∈(0,1). We now define

   (50)μ(k)=k2n−2−n−12kn+2nn−2kn−1−n−12kn−2+1

   and show it is positive for each k∈(0,1). Remark that μ(1)=0. We now prove that μ(k)>0 for each k∈(0,1). We find that μ′(k)=kn−3ν(k), with ν(k)=2(n−1)kn−nn−12k2+2n(n−1)n−2k−n−12(n−2) and ν(1)=0. In addition, we have that ν′(k)=2n(n−1)kn−1−n−1k+n−2, with ν′(1)=0, and ν′′(k)=−2n(n−1)2(1−kn−2)<0 for each k∈(0,1). Hence, ν′ is strictly decreasing and since ν′(1)=0 we can conclude that ν′(k)>0 for each k∈(0,1). This, in turn, implies that ν is strictly increasing, and since ν(1)=0, we obtain that ν(k)<0 for each k∈(0,1). Therefore μ′(k)<0 for each k∈(0,1), and since μ(1)=0 we can conclude that μ(k)>0 for each k∈(0,1).

Now we prove that B(γA(λ∗),λ∗)>0 and then examine the sign of B(γA(xˉ),xˉ). Given that B(γA(λ),λ) is a continuous function of λ, if B(γA(xˉ),xˉ)<0 then there exists a unique λ˜∈(λ∗,xˉ) such that B(γA(λ˜),λ˜)=0. Since A(γA(λ˜),λ˜)=0, it follows that γA(λ˜),λ˜ is a solution to (44), i.e. to eqs. (16)-(17). We prove that there exists r˜1∈(r2,rˉ1) such that B(γA(xˉ),xˉ)<0 if and only if r1∈(r2,r˜1).

Regarding B(γA(λ∗),λ∗), since A(γA(λ∗),λ∗)=0=τ(λ∗)=A(λ∗,λ∗), we have that γA(λ∗)=λ∗; hence (37) and (46) imply that

where the last equality follows from the definition of λ∗.

Regarding B(γA(xˉ),xˉ), we have that

We now take into account that γA(xˉ) is an increasing function of r1 (dγA(xˉ)dr1>0 since ∂A∂γ>0 and ∂A∂r1<0), and we view B(γA(xˉ),xˉ) as a function ℓ(r1) of r1 that is defined for r1∈(r2,rˉ1). As r1↑rˉ1, we have that γA(xˉ)→xˉ and F(γA(xˉ))→1, hence