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\begin{center}
A copy of this paper, and related work, is at
www.nuff.ox.ac.uk/economics/people/klemperer.htm%
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{\bf The Generalized War of Attrition}%
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A slightly revised version is published in \ \
{\it American Economic Review, }1999, 89(1), 175-89.
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Jeremy Bulow%
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Graduate School of Business, Stanford University, USA
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and%
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Paul Klemperer%
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Nuffield College, Oxford University
Oxford OX1 1NF
England%
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Int Tel: +44 1865 278588
Int Fax: +44 1865 278557%
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email: paul.klemperer@economics.ox.ac.uk%
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November 1997
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\noindent
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\underline{Abstract:}
We model a War of Attrition with $N+K$ firms competing for $N$ prizes. If
firms must pay their full costs until the whole game ends, even after
dropping out themselves (as in a standard-setting context), each firm's exit
time is independent both of $K$ and of other players' actions. If, instead,
firms pay no costs after dropping out (as in a natural oligopoly), the field
is immediately reduced to $N+1$ firms. Furthermore, in this limit it is
always the $K-1$ lowest-value firms who drop out in zero time, even though
each firm's value is private information to itself. (100 words)
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\noindent \underline{Keywords:} War of Attrition, Auctions, Standards,
Natural Monopoly, Oligopoly, ``Twoness'', ``Strategic Independence''.%
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\noindent \underline{JEL Nos:} D43, D44, L13, O30%
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\underline{Acknowledgment}: We are extremely grateful to Faruk Gul for
extensive conversations about the war of attrition. We also received very
valuable help from other colleagues and seminar participants, especially
Alan Beggs, Meg Meyer and Markus M\"{o}bius.\quad
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This paper analyzes wars of attrition with many players. Examples of such
wars are common, and are sometimes referred to as ``industry shakeouts''.
For example, five to six firms are committed to making major investments in
wireless phone service in each United States market. While Los Angeles and
New York will be able to accomodate this number, many other markets are
probably natural oligopolies that can profitably support only three or four
firms. A similar battle is taking place in the Canadian long distance market.%
\footnote{%
The battle for long distance market share [in Canada] has turned into a war
of attrition....... Losses are piling up ....... Analysts and industry
officials predict the ranks of Canada's long distance companies will shrink
sharply.''\ See {\it The Wall Street Journal}, July 25, 1997 p. B4.}
Battles to control new technologies often resemble wars of attrition. For
example, five firms---Zenith, Thomson, AT\&T, General Instruments, and
Philips Electronics, initially worked on competing HDTV standards, and
Microsoft, Netscape, and Lotus are fighting to dominate the ``groupware''
that is used within corporate intranets. In interactive Videotex, the seven
or more competing national standards that emerged in the 1970s were reduced
to three by 1984 after long negotiations in the CCITT (the international
standards organization in telecommunications), but the battle between these
three incompatible systems is still unresolved. Remarkably, we have gone
from over 300 word processing programs ten years ago to just WordPerfect and
Word.\footnote{%
Some specialized programs like Scientific Word also exist, but the other
mass-market word processors, like Xywrite and Wordstar, have vanished.}
There are now three competing standards for digital wireless phone systems
in the United States---CDMA (code division multiple access), TDMA (time
division multiple access), and GSM (global system for mobile communications,
the European standard). Consumers who purchase one type of handset will not
be able to make or receive calls over a network that uses a different
technology. While several large manufacturers have now decided to incur the
expense of producing to three standards, others have a vested interest in
the outcome; an estimated \$29 billion annual market for equipment is at
stake. The simultaneous development of three standards has deprived the
industry of the scale and manufacturing economies of a single standard, and
reduced demand by increasing consumer uncertainty. The result has been much
lower volumes and slower growth than in Europe for the whole
industry---including firms that have no interest in which standard is
adopted.\footnote{%
We are grateful to Preston McAfee for suggesting this example.
\par
The Cellular Telephone Industry Association decided in 1989 to adopt TDMA as
the U.S. standard but was persuaded in 1993 to sanction CDMA as an
alternative standard, contributing to the current upheaval. Obviously it
will be better for the industry if the ``right'' standard wins, but many
firms would be better off with any of the three standards than with the
current confusion. See {\it Business Week, }February 24, 1997, p. 44, and
June 2, 1997 p.132.}, \footnote{%
Two recent papers, Farrell (1993) and David and Monroe (1994), have already
argued that the way firms negotiate in industry standard-setting committees
is most appropriately modeled as a war of attrition. See also Farrell and
Saloner (1988).}
Multiple-player wars of attrition are also prominent in politics. In August,
1993 the United States Congress passed the Clinton Administration's budget
by the narrowest possible margin; if a single supporter in either the House
or Senate had switched their vote, the plan would have been defeated. But
although many Democrats would have preferred to vote against this unpopular
bill, they were unwilling to see the new Democratic President defeated on a
measure of such importance so, in the words of the New York Times ``one
member after another reluctantly fell into line to provide the 218-216
victory'' in the House of Representatives.\footnote{%
This is the narrowest possible margin, since in the U.S., unlike some other
countries, representatives do not wish to abstain on a measure of this
importance.}$^{,}$\footnote{%
See {\it The New York Times}, August 6, 1993, section 1 p.7.} The last
Congresswoman to vote for the budget bill virtually ensured her defeat in
the next election by supporting the President;\footnote{%
The bill was especially unpopular in very affluent districts like Marjorie
Margolies-Mezvinsky's. The Republicans chanted ``Bye-bye Marjorie'' as she
cast the final vote in favour, and she was indeed comfortably defeated in
the 1994 election.} while she was promised a good job in the Administration
in return for sacrificing her seat in Congress, it seems clear that she was
a big loser relative to other Democrats who were then able to vote against
the bill without affecting the outcome.\footnote{%
It is not known how many additional representatives would have voted for the
President if their votes had been required. Representatives' incentives
were, of course, to deny any willingness to do this, but even so it was
reported that Thornton of Arkansas and, perhaps, Minge of Minnesota were
available to vote yes if necessary.} Similarly, the bill was approved by the
Senate, after an unusually protracted debate, by 50 votes plus the
Vice-President's casting vote to 50.\footnote{%
In our interpretation of this as a war of attrition for the prizes of being
among the non-supporters of a successful bill, individual Democrats' costs
of holding out included the private costs of enduring pressure from the
Administration, and the public costs, borne by all the Democrats, of
delaying passage of the bill. The delay increased public frustration with
the political process, delayed the bill's benefits, increased the
probability of the bill failing (perhaps through the President giving up on
it) and left the Democrats less time to work on the rest of their agenda.
\par
Of course, we are abstracting away from many important features of the
problem such as coalition-formation, vote-buying, and bargaining in which
the legislators can actually affect the details of the bill. For an
excellent survey of the rent-seeking literature, which encompasses some of
the collective aspects, see Nitzan (1994).}
Until now, the war of attrition literature\footnote{%
The war of attrition has also been used to describe labor strikes (see, e.g.
Kennan and Wilson (1989)), litigation, and biological competition (see, e.g.
Maynard Smith (1974), and Riley (1980)), and the process of agreement to
macroeconomic stabilizations (see, e.g. Alesina and Drazen (1991) and
Casella and Eichengreen (1990)). The classic reference on industrial
competition is Fudenberg and Tirole (1986). For a more general survey of
rent seeking see Nitzan (1994).} has focused on games with two players, or
the straightforward generalization to $N+1$ players competing for $N$
prizes. While many of the best examples do involve only two players,
multiple player games are also important. We consider a generalized war of
attrition in which $N+K$ players are competing for $N$ prizes, so that $K$
must exit for the game to end.
Our examples have highlighted an important issue in modeling the generalized
war of attrition. In a natural monopoly (or oligopoly) setting, once a firm
has conceded defeat it drops out of the game and stops paying costs. In a
battle over standards, as with PCS, even a firm that does not try to enforce
its own standard will continue to bear higher costs until the remaining
firms agree on a common standard (because of reluctance of consumers to buy
and potentially higher manufacturing costs){\bf .} With just two firms (or $%
N+1$), this problem never arises because once any firm drops out the war
automatically ends.
To make the distinction clear, consider the following example: The chairman
of the economics department calls a meeting, and says that he needs five
volunteers to serve on a committee. The meeting will not end until the
committee is chosen. In the ``natural oligopoly'' game, a faculty member is
allowed to leave the meeting as soon as he agrees to serve on the committee.
In the ``standards'' game, everyone must stay in the meeting until the whole
committee is selected. Obviously, there is much less incentive to concede
quickly in the second game.
The natural oligopoly case yields a striking result: $K-1${\it \ firms will
exit immediately,} leaving only $N+1,$ or {\it one too many} firms to battle
for the $N$ prizes.\footnote{%
One example of this game is that Avinash Dixit every year offers a \$20
prize to the student who continues clapping the longest at the end of his
game theory course. Our analysis shows that if they have understood the
material, no more than two students should continue applauding after the
time when everyone would otherwise have stopped.} To understand the result,
imagine that when $K>1$ exits are still required for the game to end, a
player is within $\varepsilon $ of his planned dropout time. Then the
player's cost of waiting as planned is of order $\varepsilon $, but his
benefit is of order $\varepsilon ^{K}$ since only when $K$ other players are
within $\varepsilon $ of giving up will he ultimately win. So for small $%
\varepsilon $ he will prefer to quit now rather then wait, but in this case
he should of course have quit $\varepsilon $ earlier, and so on. So only
when $K=1$ is delay possible. This result helps explain why so many wars of
attrition, like Kodak vs. Polaroid in instant photography and Microsoft vs.
Netscape in web browsers, quickly devolved into two-horse races.
In the standards version of the game, in which all players pay until the
game ends, even if they have already conceded, the result is perhaps equally
surprising: {\it players' strategies are independent of }$K${\it \ and of
other players' dropout behaviour.} Why does this kind of {\it strategic
independence} arise? Because, as before, when there are still $K>1$ too many
firms, a player within $\varepsilon $ of his planned exit knows he has no
chance (to first order) of winning. So since in this case quitting early
does not affect the rate at which the firm pays costs, the firm would quit
early if he thereby shortened the expected length of the whole game. Only if
the firm's exit decision has no effect on the length of the whole game will
it be willing to exit at the ``correct'' equilibrium time. So no firm can
either affect, or be affected by, any other firm's dropout behaviour.
In our general model, which encompasses both the natural oligopoly and the
standards versions as special cases, each player's value of surviving in the
market is private information to that player. However we always get perfect
sorting. Thus even in the {\it one too many} limit in which the field is
immediately sorted down to $N+1$ players, it is the $K-1$ weakest players
that leave in zero time.
Section 1 presents the model. Section 2 analyses the general case. Sections
3 and 4 discuss the {\it one too many} result and {\it strategic independence%
} respectively. Section 5 illustrates the analysis with an example inspired
by the 1993 budget battle, and Section 6 concludes. Formal proofs are
collected in the Appendix.
\section{The model}
There are $N+K$ risk-neutral firms in a market. As long as a firm continues
to compete, it pays a cost that is normalized to 1 per unit time. If it
exits, it subsequently pays a cost $c>0$ per unit time until a total of $K$
firms have quit.\footnote{%
Typically we expect $c$ to be no more than 1, but note 30 suggets a context
in which it might exceed 1. Another example in which $c$ would exceed 1 is a
``contributions'' game in which $N$ people must each pay for one stage of a
building project before it yields any benefits, and discounting means
earlier contributions cost more than later ones.} If a firm $i$ is one of
the $N$ firms which survives, then it wins a prize of $v^{i}$ which is
private information to firm $i$ at the beginning of the game.\footnote{%
If flow costs differ among firms, we can simply reinterpret $v^{i}$ as the
ratio for player $i$ of the prize value to the flow cost, since this ratio
is all that matters to any firm. So all our results will still go through.
(Units for measuring costs are of course then different for different firms
so that the flow rate of costs is measured as 1 per unit time for each.)}
The values $v^{i}$ are drawn independently from the distribution $F(v),$
with $F(\underline{V})=0,F(\overline{V})=1,\underline{V}>0$ and $\overline{V}%
<\infty .$ We assume $F(\cdot )$ has a strictly positive finite derivative
everywhere. It will be convenient to write $h(v)$ for the ``hazard rate'' $%
\frac{f(v)}{1-F(v)}$. We also write $v_{j}$ for the $j^{th}$ highest of the $%
N+K$ firms' values, and $E(v_{j})$ for the expectation of this value. We
restrict attention to symmetric equilibria.
At any point of the game let $N+k$ be the remaining number of firms (so $k$
more firms must exit before the game finishes), let $\underline{v}$ be the
lowest possible remaining type conditional on all other firms having thus
far followed (symmetric) equilibrium strategies, write $T(v;\underline{v},k)$
for the additional amount of time a still-surviving firm of type $v$ will
wait before exiting if none of the other remaining $N+k$ firms exits
beforehand, and $P(v;\underline{v},k)$ for the firm's probability of being
among the $N$ ultimate survivors.%
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\section{The general solution}
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{\bf Lemma 1}: {\it In} {\it any equilibrium, for all }$\underline{v}${\it \
and all }$k,$ $T(v;\underline{v},k)$ {\it is strictly increasing in} $v$
{\it and }$P(v;\underline{v},k)$ {\it equals the probability that }$v${\it \
is one of the }$N${\it \ highest values conditional on }$N+k-1${\it \ other
firms' values exceeding }$\underline{v}.\footnote{%
So $P(v;\underline{v},k)=\sum_{j=k}^{N+k-1}\frac{(N+k-1)!}{(N+k-1-j)!j!}%
\left( \frac{F(v)-F(\underline{v})}{1-F(\underline{v})}\right) ^{j}\left(
\frac{1-F(v)}{1-F(\underline{v})}\right) ^{N+k-1-j}$}$%
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Lemma 1 follows because higher-valued firms exit later in a symmetric
equilibrium.%
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{\bf Lemma 2}: {\it There is at most one equilibrium of the game.}%
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The reason for Lemma 2 is that the difference between the expected surpluses
of any two types is uniquely determined by standard incentive compatibility
arguments.\footnote{%
The absolute level of a player's surplus cannot be determined prior to
determining the actual equilibrium because, in contrast to many problems in
which the bottom type's surplus is fixed at zero, in our problem the bottom
type receives negative surplus for $c>0.$} But, since any type's probability
of winning a prize is fixed by Lemma 1, the difference between the two
types' waiting costs is therefore also uniquely determined. However, if
there were two different equilibria specifying different quitting times $T(v;%
\underline{v},k),$ these two equilibria would yield different differences
between types' waiting costs, for at least one pair of types.\footnote{%
This is easy to see if $k=1$ (i.e. when the game ends after one more quit).
So the $k=1$ subgame is unique. But then if $k=2$, waiting costs are fixed
after one more quit, so two different functions $T(v;\underline{v},2)$ would
yield different differences in total waiting costs for some pair of types,
so the $k=2$ subgame is also unique. And so on.}%
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{\bf Lemma 3:}\footnote{%
This result can also be found in Bliss and Nalebuff (1984), Nalebuff (1982),
and elsewhere.} {\it The unique symmetric perfect Bayesian equilibrium}%
\footnote{%
For this $(k=1)$ case this is also the unique symmetric Nash equilibrium,
since each firm knows that its decision to exit is only relevant in the case
in which no other firm has previously exited, so the game is strategically
equivalent to a static game in which firms simultaneously choose exit times.}
{\it of the subgame in which just one more exit is required to end the game
is defined by}
\begin{eqnarray}
T(v;\underline{v},1) &=&\int_{\underline{v}}^{v}Nxh(x)dx \label{max6} \\
&& \nonumber
\end{eqnarray}
The intuition is straightforward: at each moment the marginal firm with type
$v$ faces the prospect of paying an extra $T^{\prime }(v;\underline{v},1)dv$
to outlast any firms with types between $v$ and $v+dv,$ and equates these
costs to the value of being a winner, $v$, times the probability, $%
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\frac{Nf(v)dv}{1-F(v)}=Nh(v)dv,$ that one of the other $N$ remaining firms
will in fact be revealed to have a type below $v+dv$. So $T^{\prime }(v;%
\underline{v},1)=Nvh(v).$ Furthermore $T(\underline{v};\underline{v},1)=0,$
since a player of type $\underline{v}$ will never win and so exits
immediately. So $T(v;\underline{v},1)=0+\int_{\underline{v}}^{v}T^{\prime
}(x;\underline{v},1)dx=\int_{\underline{v}}^{v}Nxh(x)dx.$
We can now state our main result.%
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{\bf Proposition: }{\it The unique symmetric perfect Bayesian equilibrium of
the Generalized War of Attrition is defined by}
\begin{eqnarray}
T(v;\underline{v},k) &=&c^{k-1}\int_{\underline{v}}^{v}Nxh(x)dx \label{max7}
\\
&& \nonumber
\end{eqnarray}
The intuition is that when $k>1,$ quitting $\varepsilon $ early or late
would not, to first order, affect type $v$'s probability of winning (since
only when $k$ other firms are within $\varepsilon $ of quitting can $v$ win
in the next $\varepsilon $). But quitting does slow down the rate at which $%
v $ pays costs to fraction $c$ of the previous rate, so for $v$ to be
indifferent about quitting, it must also slow down other players' rates of
quitting by the same fraction $c$.\footnote{%
Of course, this argument is not complete since it only shows other players
slow down to fraction $c$ on average.} That is, $T^{\prime }(v;\underline{v}%
,k)=cT^{\prime }(v;\underline{v},k-1),$ hence also $T^{\prime }(v;\underline{%
v},k)=c^{k-1}T^{\prime }(v;\underline{v},1)=c^{k-1}Nvh(v).$
So, for example, if $N=1,K=3$ and $c=\frac{1}{2},$ the equilibrium goes
through types four times as fast as in the two firm game $(N=K=1)$ until one
firm drops out, then twice as fast as the two firm game until a second firm
drops out, and then finally at the speed of the two firm game until the
final exit.
Notice that a feature of the equilibrium is that the $K-1$ lowest-valued
firms are actually indifferent about staying past their equilibrium dropout
points; each would be willing to delay until $K-1$ others have quit
(assuming each thought the others were following the equilibrium
strategies). Of course, if any one of these firms were to delay its
departure until $K-1$ others had left, that would speed the game and benefit
everyone else.\footnote{%
Of course, if $c>1$ each firm that leaves speeds up the game, so leaving
late hurts others.}
Note also that, by contrast, the highest-valued losing firm (the only loser
in the standard $N+1$ firm model) would hurt everyone else by delaying its
exit, so the equilibrium length of the game is non-monotonic in players'
valuations. For example, a game with two ``tough'' players and one ``weak''
player competing for one prize takes longer than either a game with three
tough players or a game with one tough player and two weak players.%
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{\it Expected Time Between Exits}
The expected time between successive departures increases in later stages
for three separate reasons. First, there are fewer players who might leave ($%
N+k$ falls). Second, the remaining players are stronger ($E(v_{N+k}$) rises
as $k$ falls). And third, each exit must slow the game ($c^{k-1}$ rises as $%
k $ falls) in order to make the next firm which drops out indifferent
between paying the full costs of remaining in the game a little longer or
paying the lower costs per period of being out. The Corollary to our
Proposition demonstrates exactly these features:
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{\bf Corollary}: {\it The expected time taken to reduce from} $N+k$ {\it %
firms to} $N+k-1$ {\it firms is}
\begin{eqnarray}
&&Nc^{k-1}\frac{E(v_{N+k})}{N+k} \label{max44} \\
&& \nonumber
\end{eqnarray}
The Appendix offers a purely algebraic proof of the Corollary. However, an
approach that is more economic (and economical) is to consider a game in
which, after all but $j$ players have been revealed as having values above $%
v_{j+1},$ as in our problem, the remaining players fight a standard
one-stage war of attrition for $j-1$ prizes. Since this game requires just
one more exit, lemma 3 tells us that the time until the lowest of the $j$
remaining firms (with value $v_{j})$ quits is $%
\int_{v_{j+1}}^{v_{j}}(j-1)xh(x)dx$. But by the Revenue Equivalence Theorem
the expected costs per player must be the same as in an English auction, in
which $(j-1)$ players win at price $v_{j}$, that is, $E\{(j-1)v_{j}/j\}$.%
\footnote{%
We can use the Revenue Equivalence Theorem because in both a one-stage war
of attrition and an English auction a player who quits immediately receives
an expected surplus of zero.} So, $E\left\{
\int_{v_{j+1}}^{v_{j}}xh(x)dx\right\} =\frac{E(v_{j})}{j}.$ Now in our
problem the expected time between the exits of the $(j+1)^{st}$ and $j^{th}$
highest-value firms (who have actual values $v_{j+1}$ and $v_{j})$ is, from (%
\ref{max7}), $E\left\{ c^{j-(N+1)}\int_{v_{j+1}}^{v_{j}}Nxh(x)dx\right\} ,$
so substitution yields the corollary.%
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{\it Expected Length of the Game}
Simple summation of (\ref{max44}) yields
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{\bf Corollary: }{\it The expected length of the Generalized War of
Attrition is}
\begin{eqnarray}
&&N\sum_{j=N+1}^{N+K}c^{j-(N+1)}\frac{E(v_{j})}{j} \label{max55} \\
&& \nonumber \\
&& \nonumber
\end{eqnarray}
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{\it Firms' Costs Varying with }$k$
It is easy to extend the model to allow firms' costs to be a function of $k$%
. (For example, in an oligopoly context losses are probably increasing in $k$%
.) If costs are $\ell _{k}$ times as great as in our model when $k$ more
firms are required to exit, then equilibrium requires that types leave $\ell
_{k}$ times as fast at any point of time. Thus the total costs firms incur
in the war of attrition are independent of $\ell _{k}$.%
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{\it Discounting}
It is also easy to see that discounting would have no effect on how firms
play the game at any moment of time, since discounting is just equivalent to
there being some exogenous flow of probability that the game will end and
firms will stop accruing further costs or benefits. So our results and our
formulae for $T(v;\underline{v},k)$ are unchanged by discounting, but
discounting makes the costs of the war of attrition even greater relative to
the discounted value of the prizes.%
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{\it Deadlines}
In many contexts there is a deadline at which the game ends and no-one wins
a prize.\footnote{%
For example, the government imposes a standard that is no-one's preference,
or firms go their separate ways and choose different standards, or the
industry dies. Of course there are other possible models of deadlines e.g.
winners are chosen randomly (e.g. by whips in a voting context), or everyone
pays a cost at the deadline (e.g. the bill fails to pass which is everyone's
worst outcome).} Since this corresponds to infinitely heavy discounting
taking place at the moment of the deadline, this too has no effect on how
the game is played prior to the deadline.
\section{The special case $c=0$: `` One Too Many''}
In the limit as $c$ approaches 0, firms drop out arbitrarily fast until only
$N+1$ remain. That is, if $N$ firms can be profitable in a market and
dropouts pay no costs after exiting, then competition in the symmetric
equilibrium will immediately shake out to just {\it one too many} firms to
be profitable.\footnote{%
In an {\it open-loop} equilibrium (i.e., when firms choose an exit time at
the start of the game that cannot be revised based on when other firms quit)
the {\it one too many} result does not arise; see Krishna and Morgan (1997)
who analyse an open-loop $c=0$ model. Hillman and Riley (1989) analyse an
all-pay auction without private information but with asymmetric bidders and
show that just two bidders make positive bids. However, this ``twoness''
result does not survive if the bidders are symmetric (as ours are); in this
case there is a symmetric equilibrium in which all bidders make positive
bids (see also Baye et al (1996)). Nor does Hillman and Riley's ``twoness''
result survive in a ``second-price all-pay'' auction, i.e. an all-pay
auction except that the winner pays only the second bid (as in our war of
attrition), since this is equivalent to the open-loop {\it c=0} model
analysed by Krishna and Morgan.}
For example, when there is just one winner, competition effectively reveals
the third-highest value, that is, $v_{3}$, immediately, and then yields a
conventional two firm game beginning with $\underline{v}=v_{3}.$
An alternative way of deriving this result that should appeal to auction
theorists is to consider the expected total costs paid by the remaining two
firms after the buyer with the third-highest value drops out. The Revenue
Equivalence Theorem tells us that these costs must be the same as the
expected costs in a second-price auction between these firms, namely the
expectation of the second-highest value, $v_{2}$.\footnote{%
See Myerson (1981) and Riley and Samuelson (1981) for the earliest
statements of the Revenue Equivalence Theorem. Our 1994 paper, Bulow and
Klemperer (1994), states and applies the theorem for a setting which, like
this one, has multiple objects that are sold in a dynamic game.} Compare
this with the expected total costs paid by all the firms in the initial game
with $K$ eventual losers. Again by Revenue Equivalence with the second-price
auction, total expected costs must be the expectation of $v_{2}$ in the
initial game. So the expected costs paid to get from the initial game to the
subgame with two firms remaining must be zero.\footnote{%
The reason this logic only holds when $c\rightarrow 0$ is that this assures
that the expected surplus of a firm with type $\underline{v}$ is zero, since
the firm can exit immediately at no further cost. This in turn guarantees
that the expected surplus of a firm with type $v$ equals $\int_{\underline{v}%
}^{v}P(x;\underline{v},k)dx$ (see equation (\ref{max4}) in the appendix)$,$
and hence that the expected total costs paid by all firms (which must equal
the sum of the expected gross income to the survivors less the sum of firms'
expected surpluses) are the same regardless of whether there is a
second-price auction or a symmetric war of attrition in any subgame. If $%
c>0, $ then the expected surplus of a firm with the lowest possible type is
negative, so the war of attrition will be more costly to the firms, in
expectation, than a second-price auction.}
To understand the result observe that if there were positive delay while $%
K>1 $ exits were still required, then a firm that quit $\varepsilon $
earlier than it had originally planned would save $\varepsilon $ in waiting
costs but would reduce its probability of winning by an amount of order $%
\varepsilon ^{K}.$ So all firms would drop out at least a little earlier
than planned, so firms must in fact quit without delay until only a single
firm remains in excess of the number who can ultimately survive.
Fudenberg and Kreps (1987) and Haigh and Cannings (1989) have already
considered the case $c=0$ in the special case in which all firms' values are
equal (i.e., every firm $i$ has value $v^{i}=\underline{V}=\overline{V})$,
so there is no private information.\footnote{%
Haigh and Cannings consider exactly this game. Fudenberg and Kreps' model is
more complex, but the situation their weak entrants face has essentially
these features (see their section 5).} Then the symmetric equilibrium (again
in continuous time) is in mixed strategies, and all firms mix across all
possible dropout times.\footnote{%
See Fudenberg and Tirole (1991, p230-232) for discussion of how the mixed
strategy equilibrium of the two-player war of attrition with complete
information corresponds to the equilibria of wars of attrition with
incomplete information in which every type plays a (different) pure strategy.%
} In this case, if a firm survives to be one of the final $N+1$ firms in the
market, its expected future payoff is zero (since it is indifferent to
dropping out immediately). But it can also earn zero by dropping out at the
beginning of the entire game. Therefore, firms will only be willing to wait
to become one of the final $N+1$ firms if the cost of waiting is zero---that
is, the time that it takes to reduce the field to $N+1$ must be zero
regardless of $K$.
Note that, strictly speaking, the game has no symmetric equilibrium actually
at the limit $c=0$. Our argument has made clear that there cannot be an
equilibrium in which the types separate with all except the lowest type
waiting a strictly positive length of time. But nor can there be positive
probability of any firm quitting in zero time, since in symmetric
equilibrium this would imply positive probability of {\it all} firms
quitting in zero time, so every firm would do better to wait a little (see
Lemma 1).\footnote{%
Haigh and Cannings get around the problem of the absence of a symmetric
equilibrium in their model by imposing a series of instantaneous
randomizations to eliminate $k-1$ bidders in zero time (and Fudenberg and
Kreps take a similar approach). This seems very natural in their special
case in which all players have identical values. To the best of our
knowledge our paper is the first to allow $c>0$, and so to interpret the $%
c=0 $ case as the limit as c$\rightarrow 0.$}
\section{The special case $c=1$: `` Strategic Independence''}
Now consider the special case in which all firms pay full costs until the
game is resolved. This is the polar opposite of the previous case, and can
be thought of as the standards case, where all firms lose until a standard
is established, with losses independent of whether a firm is one of the
remaining competitors for establishing the standard.
When $c=1$, the Proposition yields that types leave at the rate
\[
\frac{1}{T^{\prime }(v;\underline{v},k)}=\frac{1}{Nvh(v)}
\]
when the marginal remaining type is $v$, so we have ``strategic
independence''. That is, each firm chooses the same dropout time as if it
were in a game with just $N+1$ firms and $N$ prizes, independent of the
actual number of remaining firms. Having chosen its dropout time at the
beginning of the game, the firm then sticks with it.
The intuition is that because a firm's flow costs are unaffected by dropping
out before the end of the game, and its probability of winning is also
unaffected (to first order) by small changes in its exit time when $K>1$
exits are still required to end the game, the firm's exit decision cannot
affect the length of the game. So other firms' decisions are unaffected by
this firm's actions.
As for the ``one too many'' case, it may help in understanding this result
to consider the mixed strategy equilibrium of the limiting case of our game
in which all firms' values are known to be equal. In this case firms must be
indifferent to dropping out at any time prior to the end of the game. Thus a
firm must be indifferent between dropping out now, when more than $N+1$
firms remain, or waiting until exactly $N+1$ remain and then dropping out
immediately, or dropping out at any time in between. Since in any of these
cases the firm does not win and pays costs proportional to the length of the
game, the length of the game must be independent of the firm's choice. So
the dropout decisions of the first $K-1$ firms to exit do not affect the
decisions of the remaining firms.
The length of the ``strategic independence'' game is strictly a function of
the $N+1^{st}$ highest value, but the larger $K$ is the longer the game will
take in expectation, because the expected value of the $N+1^{st}$ highest
value rises as $N+K$ rises.
Both the ``one too many'' and the ``strategic independence'' games take
equally long to reduce from $N+1$ firms down to $N$. The difference is that
in the ``one too many'' game we get down to $N+1$ firms immediately, and so
only have to incur costs running through the types between the $N+2^{nd}$
highest value and $N+1^{st}$ highest value. However, with ``strategic
independence'' all types must be run through in real time, and the amount of
time required for the industry to shake down from $N+K$ firms to $N+1$ may
far exceed the time needed to get from $N+1$ to $N$.
\section{An example}
We illustrate our model with an example inspired by the 1993 budget battle.
Assume 51 senators would each like to see a bill passed but would prefer not
to have to vote for it. Each senator has a value independently drawn from a
uniform distribution on [0,$\overline{V}$] of being the one person who need
not vote for the bill. So $N=1$ and $K=50.$ We normalise units of time so
that the costs of holding out are 1 per unit time for those who have not yet
pledged support.
In the ``strategic independence'' case all 51 senators suffer political
costs equally until the impasse is resolved, whether or not they have
themselves given in. In this case $c=1,$ so using (\ref{max55}) the expected
delay before passage of the bill is $\approx $2.55 $\overline{V}.$\footnote{%
For the uniform distribution $E(v_{j})=\left( 1-\frac{j}{N+K+1}\right)
\overline{V},$ so (\ref{max55}) implies the expected delay is $N\overline{V}%
\sum_{N+1}^{N+K}\left( \frac{1}{j}-\frac{1}{N+K+1}\right) $which with $N=1$
and $K=50$ yields $\overline{V}\left( -\frac{50}{52}+\sum_{j=2}^{51}\frac{1}{%
j}\right) \approx \overline{V}\left( -2+\sum_{j=1}^{53}\frac{1}{j}\right)
\approx \overline{V}(-2+\log 53+\gamma )$in which $\gamma $ $(\approx .58)$
is the Euler number.} So even the ``winner'' suffers costs that are on
average more than two and one half times as great as his prize!\footnote{%
Of course, this does not mean that a player should refuse to play. It is
common knowledge that each player anticipates negative surplus relative to
the bill passing immediately with his vote, but legislators may still obtain
positive surplus relative to the bill not passing at all.
\par
In fact, the legislators would be in an even worse game than this if $c>1.$
For example, if pledging one's vote increases the lobbying pressure and
hostility one faces from those opposed to the bill, then costs are largest
for those who have already pledged.} Obviously players should do everything
they possibly can to change such games.\footnote{%
This may help explain the institution of whips, whose job it is to determine
the allocation of prizes (that is, who should be permitted not to vote for a
particular measure) without recourse to a war of attrition. Whips will not
necessarily select the highest valuers of the prizes but even uninformed
randomisation is highly desirable relative to these games. Whips seem to be
more effective in resolving smaller issues than larger ones, such as the
budget bill, for which it is harder to persuade losers that they will be
compensated on future issues.}
At the other extreme, consider the ``one too many'' case in which a
senator's only costs are his personal costs of withstanding administration
pressure, and these costs stop as soon as he knuckles under. In this case,
as we have seen, 49 senators give in to the administration immediately,
while the two with the highest values hold out in a standard two-player war
of attrition. Here $c=0$, so using (\ref{max55}) the total expected time is
just half of the expected value of the lower of the holdouts, $\frac{1}{2}%
E(v_{2}),$ that is, since $F(\cdot )$ is uniform, $\frac{25}{52}\overline{V}%
. $\footnote{%
We can also obtain this result directly without need of (\ref{max55}) by
using the Revenue Equivalence Theorem which says that the expected total
costs incurred (that is, twice the expected time) must be the same as in an
English auction, that is, $E(v_{2})=\frac{50}{52}\overline{V}$. (The Revenue
Equivalence Theorem applies here because a player who quits immediately gets
zero surplus in this case.)},\footnote{%
With $c=0$, the total resources used are $E(v_{2})$ while the expected value
of the prize is $E(v_{1})$, so in the limit in which all players' values are
equal there is complete rent dissipation independent of the number of
competitors, just as in rent-seeking models such as Hillman and Samet (1987)
who analyse an all-pay auction for a prize that is worth the same to all. Of
course for larger values of $c$, and $k>1,$ expected rent dissipation in our
model typically exceeds the expected value of the prize, as in the $c=1$
example above. Expected rent dissipation, as a fraction of the value of the
prize, is typically increasing in $k$ and is always increasing in $c$.}
In the ``one too many'' case, of course, all the time is spent waiting for
the last vote, but observe that even in the ``strategic independence'' case,
the first few votes come in relatively quickly (the first 10 votes take less
than 1\% of the total time on average) while the last few votes take much
longer (the last 4 votes take almost half the total time in expectation).
The typical case will lie between these extremes. Thus our model both
explains why political decision making can sometimes take so long, and why,
even when agreements seem close to complete, the hunt for the last few votes
can often seem so excruciatingly slow.\footnote{%
Multilateral international treaties (e.g. GATT) often seem to illustrate
this.}
\section{Conclusion}
We study wars of attrition in which two or more players must exit. Except in
the final stage, a player's departure will not end the game, and a player
may continue to incur at least some costs even after he has conceded.
Therefore, except in the final stage, by the time a player exits he knows a
small delay in conceding will not allow him to win, because it is so
unlikely that two or more others will exit before him. So the player becomes
solely interested in minimizing his costs, and in equilibrium a small change
in exit strategy must have no effect on his expected costs.
If costs are as high for those who have already conceded as for those who
continue to fight, then there is no incentive to drop out early. For a
player to be satisfied with his equilibrium exit time, his departure must
neither speed nor slow the ultimate resolution of the game. So each player's
exit behavior is unaffected by the number of other competitors and their
actions. We call this ``strategic independence.'' Examples where this may
occur are battles over standard setting and building voting majorities.
If a player does not pay any costs after he has conceded, as in a natural
monopoly game, the market will immediately and efficiently sort down to the
final stage. That is, in equilibrium we should never observe more than ``one
too many'' firms competing for the prizes.
Even with strategic independence, departures occur at a faster rate in the
early stages, because at the beginning of the game there are more, and
weaker, players who might concede. If costs are lower for those who have
exited than for those still fighting, so there is an incentive to depart
early, this effect becomes even more pronounced. Each exit must slow the
game sufficiently to make the next dropout indifferent between paying the
full costs of remaining in the game a little longer or paying the lower
costs per period of being out. Of course in the limit when players pay no
costs after conceding, all but the last departure is instantaneous. So the
model explains why rounding up most of the necessary votes for a bill might
take very little time, but gathering the last few votes may be
time-consuming and costly.
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{\bf Appendix}%
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We write $C(v;\underline{v},k)$ for the firm's expected future delay costs
over the whole of the rest of the game (net of delay costs thus far
incurred), and $S(v;\underline{v},k)$ for the firm's expected future surplus
over the whole of the rest of the game (so $S(v;\cdot ,\cdot )\equiv
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{\bf Proof of Lemma 1}: A higher-value type of a firm cannot exit before a
lower-value type of the same firm would exit. (If a low type gets the same
expected surplus from strategies with two different probabilities of being
an ultimate survivor, the high type strictly prefers the high-probability
strategy, so the high type cannot choose a strategy with a lower probability
of survival than the low type.) Also, at no moment of time does any firm
exit with strictly positive probability. (By symmetry, all firms would have
strictly positive probability of exit, but then any firm would strictly
prefer exiting just after this time to exiting at this time). So $T(\cdot
;\cdot ,\cdot )$ is strictly increasing in $v$ for all $\underline{v}$ and $%
k,$ and a firm ultimately survives if and only if $k$ or more of the
remaining $N+k-1$ other current survivors have lower values than it. So
\begin{eqnarray}
P(v;\underline{v},k) &=&\sum_{j=k}^{N+k-1}\frac{(N+k-1)!}{(N+k-1-j)!j!}%
\left( \frac{F(v)-F(\underline{v})}{1-F(\underline{v})}\right) ^{j}\left(
\frac{1-F(v)}{1-F(\underline{v})}\right) ^{N+k-1-j} \label{paul1} \\
&& \nonumber
\end{eqnarray}
$\;\square $%
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{\bf Proof of Lemma 2}:\footnote{%
This is the most elegant proof we know. An alternative, but here rather
cumbersome, approach is to first show the monotonicity and continuity of $%
T(\cdot ;\cdot ,\cdot )$ in $v$ and using these show the differentiability
of $T(v)$, so that the first-order conditions characterize the equilibrium
uniquely. (As in our proof, an inductive argument is required; for arguments
along these alternative lines see the Appendix of Gul and Lundholm (1995) or
our own working paper (joint with Huang), Bulow et al (1996).)} The proof is
by induction. We assume that there is at most one equilibrium of any subgame
in which there are $N+k-1$ firms left, and show that this implies at most
one equilibrium with $N+k$ firms remaining.
Consider the subgame defined by $k$ and $\underline{v}$. (This is
well-defined by Lemma 1). There is no finite period of time in which there
is zero probability of exit. (If there was, then a type that was due to exit
at the end of this period would do better to exit at the beginning of this
period; because there is a unique equilibrium after the next exit, its time
cannot affect the subsequent development of the game.) So $T(v;\underline{v}%
,k)$ is continuous, and
\begin{equation}
T(\underline{v};\underline{v},k)=0 \label{max2}
\end{equation}
so also
\begin{equation}
S(\underline{v};\underline{v},k)=-C(\underline{v};\underline{v},k).
\label{max3}
\end{equation}
Now note that since in equilibrium no type of firm can gain by following any
other type's exit rule,
\[
S(v^{a};\underline{v},k)\geq S(v^{b};\underline{v},k)+P(v^{b};\underline{v}%
,k)(v^{a}-v^{b})\ \qquad \text{for all \qquad }v^{a},v^{b}\in [\underline{v},%
\overline{V}].
\]
So $S(v;\underline{v},k)$ has derivative $dS/dv=P(v;\underline{v},k)$ and
therefore
\begin{equation}
S(v;\underline{v},k)=S(\underline{v};\underline{v},k)+\int_{\underline{v}%
}^{v}P(x;\underline{v},k)dx \label{max4}
\end{equation}
and, noting (\ref{max3}),
\begin{equation}
C(v;\underline{v},k)=C(\underline{v},\underline{v},k)+vP(v;\underline{v}%
,k)-\int_{\underline{v}}^{v}P(x;\underline{v},k)dx. \label{max5}
\end{equation}
So (\ref{max5}) and (\ref{paul1}) uniquely determine $C(v;\underline{v},k)$,
since $C(\underline{v};\underline{v},k)$ equals $c$ times the expected
length of the subgame after $\underline{v}$ quits and leaves $(N+k-1)$ firms
remaining, and the equilibrium of this subgame is, by assumption, unique.
We can now show that there is at most one equilibrium $T(v;\widehat{v},k).$
Suppose instead that there are two equilibria with $\widetilde{T}(\overline{v%
};\widehat{v},k)<\stackrel{\approx }{T}(\overline{v};\widehat{v},k)$ for
some $\overline{v}.$ Then by the continuity of $\widetilde{T}(v;\widehat{v}%
,k)$ and $\stackrel{\approx }{T}(v;\widehat{v},k),$ there exists $\underline{%
v}\in [\widehat{v},\overline{v})$ such that $\widetilde{T}(\underline{v};%
\widehat{v},k)=\stackrel{\approx }{T}(\underline{v};\widehat{v},k)=\tau $
and $\widetilde{T}(v;\widehat{v},k)<\stackrel{\approx }{T}(v;\widehat{v},k)$
for all $v\in (\underline{v},\overline{v}].$ But if $\widetilde{T}(v;%
\widehat{v},k)$ and $\stackrel{\approx }{T}(v;\widehat{v},k)$ are both
equilibria, then $\widetilde{T}(v;\underline{v},k)=\widetilde{T}(v;\widehat{v%
},k)-\tau $ and $\stackrel{\approx }{T}(v;\underline{v},k)=$ $\stackrel{%
\approx }{T}(v;\widehat{v},k)-\tau $ must be equilibria of the subgame
defined by $\underline{v}$ and $k.$ But then $\widetilde{T}(v;\underline{v}%
,k)<$ $\stackrel{\approx }{T}(v;\underline{v},k)$ for all $v\in (\underline{v%
},\overline{v}]$, so any $v\in (\underline{v},\overline{v}]$ would expect
lower waiting costs under $\widetilde{T}(v;\underline{v}.k)$ than under $%
\stackrel{\approx }{T}(v;\underline{v},k)$ before the next drop out (whether
by this firm or another firm) and the same waiting costs thereafter, since
by assumption equilibrium is unique after the next dropout. But this
contradicts the fact that $C(v;\underline{v},k)$ is the same for both
equilibria $\widetilde{T}(v;\underline{v},k)$ and $\stackrel{\approx }{T}(v;%
\underline{v},k).$ This completes the inductive step, and so proves the
result, since it holds trivially when just $N$ firms remain. \quad $\Box .$%
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{\bf Proof of Lemma 3:} Given that all other firms use this exit rule, the
expected future surplus of a type $v$ who behaves as a type $v^{*}$ is
\[
U(v,v^{*})=-(1-P(v^{*};\underline{v},1))T(v^{*};\underline{v},1)+\int_{%
\underline{v}}^{v^{*}}\frac{\partial P(x;\underline{v},1)}{\partial x}(v-T(x;%
\underline{v},1))dx
\]
in which the first and second terms are $v$'s payoffs from the events that
he quits and survives, respectively, $P(v^{*};\underline{v},k)$ is defined
by (\ref{paul1}) and equals $%
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1-\left( \frac{1-F(v^{*})}{1-F(\underline{v})}\right) ^{N}$ when $k=1,$ and $%
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\frac{\partial P(x;\underline{v},1)}{\partial x}=\frac{Nf(x)}{1-F(x)}\left(
\frac{1-F(x)}{1-F(\underline{v})}\right) ^{N}$is the density with which $v$
wins and the lowest of the other $N$ firms is of type $x.$ Thus (\ref{max6})
satisfies $v$'s first-order condition
\[
\frac{\partial U}{\partial v^{*}}(v,v^{*})=0\Rightarrow
\]
\[
\begin{array}{ll}
- &
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(1-P(v^{*};\underline{v},1))T^{\prime }(v^{*};\underline{v},1)+\frac{%
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\left[ -\left( \frac{1-F(v^{*})}{1-F(\underline{v})}\right) ^{N}\left\{
T^{\prime }(v^{*};\underline{v},1)-\frac{Nvf(v^{*})}{1-F(v^{*})}\right\}
\right] =0%
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\end{array}
\]
at $v^{*}=v.$ It satisfies the second-order conditions since it implies
\[
sign\frac{\partial U}{\partial v^{*}}(v,v^{*})=sign(v-v^{*}).
\]
And it also satisfies the boundary condition, (\ref{max2}), $T(\underline{v};%
\underline{v},1)=0.$ \quad $\Box .$%
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{\bf Proof of Proposition: }Since Lemma 2 has shown that there is at most
one equilibrium, it suffices to show that no type, say $\widetilde{v}$, of
any player wishes to deviate from the strategy specified in (\ref{max7}),
assuming all other players follow the strategies specified in (\ref{max7}):
Assume $\widetilde{v}$ stays in beyond the time when (\ref{max7}) requires
him to quit. This makes no difference to the costs the firm incurs in
waiting for other players to quit, since quitting would reduce its cost per
unit time to a fraction $c$ of its pre-exit rate but would also multiply $%
T^{\prime }(\cdot ,\cdot ,\cdot )$ by $\frac{1}{c}$ so the density of
others' types that exit per unit time would be $c$ times as great. So type $%
\widetilde{v}$ is indifferent about waiting until $k=1,$ and if $\widetilde{v%
}$ quits before $k=1$ he will have neither gained nor lost from his
deviation assuming other players play according to the conjectured
equilibrium. But at $k=1$ type $\widetilde{v}$ will strictly wish to exit
rather than stay, since $\widetilde{v}$ will now be lower than the marginal
type, $\underline{v}$, who exits immediately in the unique equilibrium of
the remaining subgame, and staying beyond this time would lose money in
expectation (see Lemma 3).
If instead $\widetilde{v}$ quits before the time specified by (\ref{max7}),
this would also make no difference to the costs incurred in waiting for
other players to quit, since both this firm's costs and the rate at which
the other firms are exiting would be multiplied by $c.$ If by the end of the
game (when $k=0$) $\widetilde{v}$ is lower than the marginal type, $%
\underline{v}$, who just exited (assuming equilibrium behaviour) then $%
\widetilde{v}$'s total costs are unaffected by his deviation. If, however, $%
\widetilde{v}>\underline{v}$ then $\widetilde{v}$ would have lost money in
expectation by its deviation, since by Lemma 3 it would now prefer to still
be in the game (with $k=1)$ and remain until the time specified by (\ref
{max7}).
So $\widetilde{v}$ cannot gain by deviating from the strategy specified by (%
\ref{max7}), and (\ref{max7}) specifies the unique equilibrium.\quad $\Box .$%
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{\bf Algebraic Proof of First Corollary: }Using (\ref{max7}), the expected
time between the exits of the $(j+1)^{st}$ and $j^{th}$ highest-value firms
(who have actual values $v_{j+1}$ and $v_{j})$ is, $E\left\{
c^{j-(N+1)}\int_{v_{j+1}}^{v_{j}}Nxh(x)dx\right\} ,$ (in which $%
v_{N+K+1}\equiv \underline{V}).$ Now,
\[
E\int_{\underline{V}}^{v_{j}}Nxh(x)dx
\]
\begin{eqnarray*}
&=&\stackunder{\varepsilon \rightarrow 0}{\lim }\int_{\underline{V}}^{%
\overline{V}-\varepsilon }\left( \QATOP{N+K-1}{j-1}\right)
(F(v))^{N+K-j}(1-F(v))^{j-1}(N+K)f(v)\int_{\underline{V}}^{v}Nxh(x)dxdv \\
&&
\end{eqnarray*}
which, integrating by parts,
\[
=\stackunder{\varepsilon \rightarrow 0}{\lim }\left\{ \left[ \left[
\sum_{i=N+K+1-j}^{N+K}\left( \QATOP{N+K}{i}\right) (F(v))^{i}(1-F(v))^{N+K-i}%
\right] \int_{\underline{V}}^{v}Nxh(x)dx\right] \right. _{\underline{V}}^{%
\overline{V}-\varepsilon }
\]
\[
-\int_{\underline{V}}^{\overline{V}-\varepsilon }\left. \left[
\sum_{i=N+K+1-j}^{N+K}\left( \QATOP{N+K}{i}\right) (F(v))^{i}(1-F(v))^{N+K-i}%
\right] Nvh(v)dv\right\} .\footnote{%
We are being careful to take $\stackunder{\varepsilon \rightarrow 0}{\lim }%
\int^{\overline{V}-\varepsilon }$ since $\stackunder{\varepsilon \rightarrow
0}{\lim }\int^{\overline{V}-\varepsilon }Nxh(x)dx$ may be $\infty .$}
\]
So
\[
E\int_{v_{j+1}}^{v_{j}}Nxh(x)dx
\]
\[
=E\int_{\underline{V}}^{v_{j}}Nxh(x)dx-E\int_{\underline{V}%
}^{v_{j+1}}Nxh(x)dx
\]
\[
=\stackunder{\varepsilon \rightarrow 0}{\lim }\left\{ \int_{\underline{V}}^{%
\overline{V}-\varepsilon }\left( \QATOP{N+K}{N+K-j}\right)
(F(v))^{N+K-j}(1-F(v))^{j}Nvh(v)dv+\right.
\]
\[
\left. \left[ \left( \QATOP{N+K}{N+K-j}\right)
(F(v))^{N+K-j}(1-F(v))^{j}\int_{\underline{V}}^{v}Nxh(x)dx\right] _{%
\underline{V}}^{\overline{V}-\varepsilon }\right\}
\]
which after noting that the second term is zero,\footnote{%
To show the second term is zero, it suffices to show that
\[
\stackunder{\varepsilon \rightarrow 0}{\lim }\left[ 1-F(\overline{V}%
-\varepsilon )\right] \int_{\underline{V}}^{\overline{V}-\varepsilon
}Nxh(x)dx=0.
\]
A careful proof of this is as follows: define
\[
g_{\varepsilon }(x)=\left\{
\begin{array}{lll}
\lbrack 1-F(\overline{V}-\varepsilon )]Nxh(x) & \text{for} & \underline{V}0$ we can find $%
\varepsilon _{0}>0,$ such that $g_{\varepsilon }(x)<\delta $ for all $0