Proof of Lemma 1. Suppose that n out of m collaborating pairs separate, and index these pairs by i: (Ui; Di), with i = 1; : : : ; n. For pair i to separate, we must have Vii = Ui + Di, so their joint payo¤ in the product market must not exceed the joint value of their outside options. Summing over igivesn Pni=1( Ui+ Di).
The total expected payo¤, gross of investment costs, for all …rms that separate at s= 1 cannot exceed the total value that …rms from the n collaborating pairs can generate as of s = 2. This implies Pni=1( Ui + Di) n . Combining with n
Pn
i=1( Ui+ Di) gives n n , which rules outn 1. If insteadn= 0, then the same reasoning shows that each collaborating pairj will go to the product market, because > Uj+ Dj. Hence, in equilibrium, no collaborating pair will separate.
Now suppose …rms (Uj; Dk) are matched at some stage s 1 but that neither …rm was part
of a collaborating pair. In this case, Vjk = 0. All …rms can guarantee themselves zero payo¤ by
separating, so Uj+ Di 0. It follows that Vjk Uj+ Di, so the …rms will separate, given our tie-breaking rule that …rms separate when indi¤erent.
Proof of Lemma 2. Lemma 1 shows that on the equilibrium path, the rematching market consists only of …rms that did not collaborate. Consider a one-stage deviation where collaborating pair i
in fact separates ats= 1. Then Ui and Di rematch into new pairs(Ui; Dj) and (Uj; Di), that can
earn and (1 ) respectively in the product market. If instead (Ui; Dj)or(Uj; Di)separate,
then Lemma 1 shows that neitherDj norUj would ever go to the product market, implying outside
options Dj = Uj = 0for all s 2.
Consider the outside option ofUi, whenm 1downstream …rms remain in the game with which
it has not yet matched. We show by induction that this outside option is equal to
Ui(m 1) = " 1 1 2 m 1# : (7)
First suppose thatUi and Dj are matched whenm 1 = 0. If(Ui; Dj) separates, then no further
rematching is possible, so Ui faces outside option Ui(m 1 = 0) = 0. This outside option corresponds to (7) evaluated at m 1 = 0. Hence,Ui and Dj will go to the product market when
m 1 = 0, byVij = >0 = Ui(0) + Dj.
Suppose instead that Ui and Dj are matched when m 1 1. The induction hypothesis is as
the game with whichUi has not yet matched (with m0 1 m 2),Ui will immediately go to the
product market with its partner, and will face outside option (7) evaluated atm=m0.
If (Ui; Dj) separates, then Ui will rematch with some Dk in the next stage, where m 2
downstream …rms remain with whichUihas not yet matched. By the induction hypothesis,(Ui; Dk)
will then go to the product market, where (1) implies Ui = ( + Ui(m 2) Dk)=2. Substituting for Ui(m 2) from (7) along with Dk = 0yields Ui = (1
1 2
m 1
). This payo¤ represents
Ui’s outside option when matched with Dj, so that Ui(m 1) = (1
1 2
m 1
), as required by (7). If s 2, then this means that (Ui; Dj) will indeed go to the product market: Vij = >
(1 12 m 1) = Ui(m 1) + Dj.
The proof for Di is entirely analogous, but with replaced by1 . Thus, the outside option
ofDi, whenm 1 upstream …rms remain in the game with which it has not yet matched, is equal
to Di(m 1) = " 1 1 2 m 1# (1 ) : (8)
To complete the inductive argument, it remains to con…rm that collaborating pair (Ui; Di) will
go to the product market at s = 1. This is indeed the case: Vii = > (1 12 m 1
) =
Ui(m 1) + Di(m 1), as required, using (7) and (8). From (1) and (2), the gross payo¤s to
Ui and Di from going to the product market are Ui = ( + Ui(m 1) Di(m 1))=2 and
Di = ( Ui(m 1) + Di(m 1))=2, where again using (7) and (8) gives the payo¤s speci…ed in the lemma.
Proof of Proposition 1. We …rst rule out any candidate equilibrium where m 1 pairs collaborate, with 1 m 1 N 1. Lemma 1 shows that on the equilibrium path, each collaborating pair
i immediately goes to the product market at s = 1; all other …rms separate and remain in the rematching market for all later stages. Firms Ui and Di in collaborating pair i earn Ui I=2 and Di I=2 respectively, with Ui and Di given by Lemma 2. Firms Uj and Dj in a non-collaborating pairj both earn zero.
Suppose that Di deviates by not investing, so that no collaboration takes place between pair i.
Then by Lemma 1,Di will enter the rematching market and remain in the game for all later stages,
earning a payo¤ of zero. To rule out this deviation, we require Di I=2 0, where Lemma 2 implies 2 1 2 " 1 1 2 m 1# I 2 0 (9)
which is equivalent to (3). Now suppose instead thatUj andDj jointly deviate by investing, so the
number of non-collaborating pairs is m 2. Uj and Dj then earn Uj I=2 and Dj I=2, with
Uj and Dj given by Lemma 2, but with m 1replaced bym 2. To rule out this deviation, we must show that it is not pro…table for at least one …rm. Lemma 2 showed that Uj Dj, so the condition for ruling out the deviation is Dj I=2 0, or equivalently
2 1 2 " 1 1 2 m 2# I 2 0: (10)
It follows immediately from >1=2and >0 that (9) and (10) cannot both hold, so there is no equilibrium with1 m 1 N 1.
Now consider a candidate equilibrium where all …rms invest. Then Lemma 2 evaluated at
m 1 = 0 implies Ui = Di = ( I)=2. As argued above, if either Ui or Di deviate by not investing, then they will remain in the game for all later stages and earn a payo¤ of zero. By
I >0, this deviation is not pro…table, so an equilibrium exists where all …rms invest.
Finally consider a candidate equilibrium where no …rm invests, so where all …rms earn zero. Suppose thatUj andDj jointly deviate by investing. By the same argument as above, the condition
to rule out this deviation is (10) evaluated at m 2 = N 1. Given (3) and De…nition 1, this condition is equivalent toN N . Hence, an equilibrium exists where no …rm invests if and only ifN N .
Proof of Proposition 2. We …rst adapt the argument of Lemma 1 to show that all collaborating pairs immediately go to the product market ats= 1, after any history. Suppose instead thatm pairs in partnerships andnintegrated pairs collaborate and then separate. If pairiin a partnership instead went to the product market, then (1) and (2) imply that Ui would earn ( + Ui Di)=2 I=2 andDi would earn ( + Di Ui)=2 I=2. The fact that they separate implies either( + Ui
Di)=2 I=2 Ui I=2or( + Di Ui)=2 I=2 Di I=2, where both inequalities reduce to Ui+ Di.
If integrated pair jthat separates instead went to the product market, they would earn a joint payo¤ of I L. The fact that they separate implies I L Uj + Dj I L, or equivalently Uj + Dj. Summing over alli and j therefore gives (m+n)
Pm
i=1( Ui+
Di) +
Pn
j=1( Uj+ Dj).
The total expected payo¤, gross of costs, for all …rms that separate at s= 1 cannot exceed the total value that …rms from the m+n collaborating pairs that separate can generate as of s= 2.
This implies Pmi=1( Ui+ Di) +
Pn
j=1( Uj + Dj) (m+n) . It then follows from > that
m+n= 0. Indeed, whenm=n= 0, …rms in each pair will prefer to immediately go to the product market and earn ( I)=2>0, rather than separate and earn zero.
We now show that an integrated pair j will always collaborate, regardless of how many pairs form partnerships. Collaborating gives Uj and Dj a joint payo¤ of I L >0. If the …rms did
not invest and then went to the product market, they would earn a joint payo¤ of L <0, equal to the cost of integration. If Uj and Dj did not invest and then separated, then each …rm would
remain in the game for all later stages and earn L=2<0. Either way,Uj and Dj maximize their
joint payo¤ by collaborating.
This in turn implies that on the equilibrium path, each pairithat forms a partnership must also collaborate. If pair idid not collaborate, then Ui and Di would separate, remain in the game for
all later stages, and earn zero. The …rms could therefore increase their joint payo¤ by integrating and both investing, to earn I L >0. Thus, for pairiin a partnership, Ui andDi both earn
an equilibrium payo¤ of( I)=2, given Lemma 2 evaluated at m 1 = 0. For integrated pair j, the joint payo¤ to Uj and Dj is I L.
Now consider a candidate equilibrium where K pairs integrate and N K N pairs form partnerships. Since all integrated pairs immediately go to the product market, the incentives for collaborating pairs are identical to those considered in Proposition 1, but with N replaced by
N K. Recall that in any subgame, …rms play the equilibrium with the lowest level of aggregate investment. Thus, Proposition 1 combined withN K N implies that no …rm in a partnership will invest. This contradicts the fact that all pairs collaborate on the equilibrium path, and implies thatN K < N must hold in equilibrium.
Consider a candidate equilibrium whereN K N < N . SupposeK 1, and that integrated pair j deviates to a partnership. The number of partnerships then becomes N K + 1, and Proposition 1 implies these pairs will all collaborate, byN K+ 1 N < N . It follows thatUj
and Dj earn a joint payo¤ of I from the deviation, which exceeds their joint payo¤ I L
in the candidate equilibrium. Thus, any candidate equilibrium with N K N < N must have K = 0. All pairs then form partnerships, and all collaborate by N < N . The joint payo¤ to each pair is therefore I, which exceeds their joint payo¤ from deviating to integration,
I L. Hence, whenN < N , there is a unique equilibrium where all pairs form partnerships and collaborate.
to a partnership, then the number of partnerships becomes N K+ 1. Proposition 1 implies that all of these pairs will collaborate if N K+ 1 < N , in which case the deviation increases pair
j’s joint payo¤ from I L to I. Thus, any candidate equilibrium with N N must have N K+ 1 N , where combining with N K < N pins down the number of integrated pairs: N N < K N N + 1. We now con…rm that there is no incentive to deviate from this candidate equilibrium. If integrated pair j deviates to a partnership, then Proposition 1 implies they will not collaborate, by N K+ 1 N , so the deviation reduces pairj’s joint payo¤ from
I L to zero. For pair iin a partnership, deviating to integration will reduce its joint payo¤ from I to I L. If either Ui and Di instead deviate by not investing, then their payo¤
would drop from ( I)=2 to zero. Thus, when N N , there is a unique equilibrium where K
pairs integrate, with N N < K N N + 1, and where all pairs collaborate.
Proof of Proposition 3. As in the proof of Proposition 2, we show that all collaborating pairs immediately go to the product market at s = 1, after any history. Suppose that m pairs in unstructured partnerships, n integrated pairs, and l pairs in structured partnerships collaborate and then separate. The total expected payo¤, gross of costs, for …rms that separate at s = 1
cannot exceed the total value that …rms from these m+n+l collaborating pairs can generate as of s= 2. This implies Pmi=1( Ui + Di) +
Pn
j=1( Uj+ Dj) +
Pl
k=1( Uk+ Dk) (m+n+l) . Rematching is independent of initial organizational form, hence Ui = Uj = Uk U and
Di = Dj = Dk D, which implies (m+n+l)( U + D) (m+n+l) .
For any one of the l pairs, indexed by k, the structured partnership cannot be implementable, since otherwise neitherUknorDkwould choose to separate. It follows thatUk andDkbargain over
with weights equal to one half, just like pairs in unstructured partnerships. Thus, (1) and (2) imply thatUkand Dk would earn( + U D)=2 I=2 C=2and ( + D U)=2 I=2 C=2
respectively from going to the product market. The fact that they separate implies either ( +
U D)=2 I=2 C=2 U I=2 C=2 or( + D U)=2 I=2 C=2 D I=2 C=2,
which is equivalent to U + D. Combined with (m+n+l)( U + D) (m+n+l) , it
follows from > that all collaborating pairs in structured partnerships go immediately to the product market, l = 0. An identical argument to that in the proof of Proposition 2 then shows thatm=n= 0, so all other collaborating pairs do the same.
As a result, the proof of Proposition 2 immediately shows that an integrated pair will always collaborate, and that all pairs in unstructured partnerships must collaborate on the equilibrium path. For pair k in a structured partnership, not investing will give both Uk and Dk a payo¤ of
C=2, since they can never rematch with a …rm that collaborated. IfUkandDkinstead integrated
and both invested, then their joint payo¤ would increase to I L >0. It follows that on the equilibrium path, all …rms will invest.
Consider pair k, with structured partnership 1 (I +C)=2 , that collaborates. If is implementable, then pair kimmediately goes to the product market, whereUk earns I=2
C=2 = I C >0and Dk earns (1 ) I=2 C=2 = 0. Deviating to no investment yields
C <0, which is not pro…table. The same argument shows that if neitherUk norDk invest, then
both …rms can strictly increase their payo¤s by jointly deviating to investment. It follows that pair
kwill collaborate if is implementable.
We now show that is always implementable when N < N . Suppose that m 1 pairs do not collaborate. Then if pair k collaborates and then separates, Uk will earn Uk(m 1) given by (7), and Dk will earn Dk(m 1) given by (8). Hence, is always implementable if = I=2 C=2 > Uk(m 1) and (1 ) = I=2 + C=2 > Dk(m 1) for all
m 1 N 1. By De…nition 2, (5) holds with equality when evaluated atm 1 =N 1, where (7) then implies Uk(N 1) = I=2 C=2. Combining with Uk(m 1) + Dk(m 1) then yields Dk(N 1) +I=2 +C=2 < I=2 +C=2. Both Uk(m 1) and Dk(m 1) are increasing in m 1, so Uk(m 1)< I=2 C=2 and Dk(m 1)< I=2 +C=2 hold for all
m 1 N 1< N 1, as required for implementation.
Suppose thatN < N and consider a candidate equilibrium where some pair integrates. If the pair instead formed joint partnership , they would collaborate and go to the product market, increasing their joint payo¤ to I C > I L. It follows that no pair will integrate in equilibrium if N < N .
Suppose furthermore that N < N . Proposition 2 showed that there was a unique equilibrium where all …rms form unstructured partnerships, invest, and earn ( I)=2>0. To show that this still constitutes an equilibrium, we rule out any deviation from pair ito a structured partnership. If pairidoes not collaborate following the deviation, then bothUi andDi earn C=2<0, which is
not pro…table. If pairicollaborate, then they will immediately go to the product market and earn joint payo¤ I C. There exists no value of satisfying both I=2 C=2 ( I)=2
and (1 ) I=2 C=2 ( I)=2, so this deviation cannot be pro…table to both Ui and Di.
We now rule out any candidate equilibrium with some pair k in a structured partnership. By deviating to an unstructured partnership and collaborating, Uk and Dk would earn Uk I=2 and Dk I=2 respectively, given Lemma 2. The number of non-collaborating pairs satis…es
m 1 N 1< N 1. De…nition 1 then implies Dk I=2>0, which combined with Uk > Dk implies Uk I=2 >0. Thus, pair k will indeed collaborate following the deviation, earning joint payo¤ I, which strictly exceeds their joint payo¤ in the candidate equilibrium, I C. There is no value of satisfying both I=2 C=2 Uk I=2and(1 ) I=2 C Dk I=2, so eitherUk orDk strictly bene…ts from the deviation.
Now suppose instead that N N < N , and let M denote the number of pairs in structured partnerships. We show there is no equilibrium withM N N . Recall that all …rms must invest on the equilibrium path, and that in any subgame, …rms play the equilibrium with the lowest aggregate investment. Thus, forM N N to constitute an equilibrium, it must be thatmpairs not collaborating isinconsistent with equilibrium pay in the ensuing subgame, for all1 m N. First consider m=N, so no pair collaborates. Then Di in an unstructured partnership earns
zero, and would instead earn Di I=2 if pair i jointly deviated to investment, given Lemma 2 evaluated at m 1 = N 1. From De…nition 1, (9) does not hold strictly at m 1 = N 1, so it cannot hold strictly at m 1 = N 1 N 1. It follows that Di does not pro…t from
this deviation: Di I=2 0. The same applies to any Dj in a non-implementable structured partnership, since pair j would bargain as under an unstructured partnership. Hence, for m=N
to be inconsistent with equilibrium play, there must be some pair with structured partnership N
that (i) is implementable: N > UN(N 1)and (1 N) > DN(N 1); and (ii) ensures UN and DN both pro…t from jointly deviating to investment: N > I=2 and (1 N) > I=2. In
particular, this argument implies M 1. Moreover (7) and (8) show that both Ui(m 1) and
Di(m 1)are increasing inm, which means that (i) and (ii) also hold for allm N 1. Hence, pair N will also collaborate whenever m N 1.
Now consider m =N 1, so only pairN collaborates. If M = 1, then none of the remaining pairs i collaborating is consistent with equilibrium play in this subgame. To see why, Di in an
unstructured partnership earns zero, and jointly deviating withUi to investment yields Di I=2,