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\begin{document}

\title{{\LARGE First versus Second-Mover Advantage with Information
Asymmetry about the Profitability of New Markets}}
\author{Eric Rasmusen\thanks{%
Dan R. and Catherine M. Dalton Professor, Department of Business Economics
and Public Policy, Kelley School of Business, Indiana University,
Bloomington, IN 47405, Tel: 812-855-3356, E-mail: Erasmuse@indiana.edu. Web: %
\url{http://www.rasmusen.org}.\newline
} \ \ \ \ \ \ \ \ \ \ \ Young-Ro Yoon\thanks{%
Department of Economics, Wayne State University, Detroit, MI 48202, Tel:
313-577-0603, E-mail: yoonyr@wayne.edu. }}
\date{Feb 21, 2010}
\maketitle

\begin{abstract}
Is it better to move first, or second--- to innovate, or to imitate? We show
that if one player's information about the profitability of new markets is
only modestly superior, the possibility of foreclosing the market can lead
to a first-mover advantage. On the other hand, more extreme information
superiority can reverse this, leading to a second-mover advantage. Knowing
more surely what is the best choice, the better-informed player wants to
delay to keep his information private and the less-informed player wants to
delay to learn. Because of this, more accurate information can actually lead
to inefficiency by increasing the incentive to delay, and exogenous costs of
delay can aid efficiency by neutralizing that strategic incentive. In fact,
in some circumstances a player may purposely coarsen his information to
deter imitation.

\vspace{0.5in}
\end{abstract}

\title{JEL Codes: D81, D82, L13\\
\vspace{0.2in}\\
Keywords: Market Entry, First- and Second-Mover Advantage, Payoff
Externalities, Informational Externalities, Endogenous Timing}

\bigskip \pagebreak

\section{Introduction}

Some companies are better than others at introducing new products or
entering new geographic markets. One kind of advantage is technological---
some companies can serve customers at lower cost. Another is
informational--- some companies are better at predicting which new products
or markets will be profitable. If this advantage is known to rivals,
however, it brings with it the peril of attracting imitation. Sony has
generally played the role of leader in new markets, while Matsushita has
been a follower (Boulding \& Christen [2003]). Lu (2002), shows that later
entrants in Japan tend to follow the entry mode of innovators. Baum \&
Haveman (1997) show that in the U.S. new hotels locate close to established
hotels. Esty \& Ghemawat (2002) study the particular case of superjumbo jet
design in the 1990's, where both pre-emption and imitation were central
concerns for Airbus and Boeing. At that time, it was uncertain whether
"hub-and-spoke" or "point-to-point" would be more common in the future.
"Hub-and-spoke" uses bigger and slower planes; "point-to-point" uses smaller
but faster planes. Wrong choices in that industry take years to reverse, and
each firm had to choose both which kind of plane to design and whether to
first wait and see what the other was doing.\footnote{%
After an initial false start by Boeing, Airbus came up with its new plane,
the big and slow A380, in 2005. The smaller and faster Boeing 787 took its
first flight in 2009 (see Kohli \& Venkatraman [2006] and %
\url{http://boeing.mediaroom.com/index.php?s=43&item=997}).}

A firm with information that a new market will be profitable must choose its
entry time to trade off the advantage of preemption against the disadvantage
of disclosing its private information. Whether it is better to move first or
second is an old question, the subject of an extensive literature that we
will later discuss. Usually, it is framed as a choice between committing by
moving first or outbidding by moving second--- of whether actions are
strategic substitutes or strategic complements.

Uncertainty is an important reason to delay. Uncertainty has two dimensions:
when it is resolved, and whether it is asymmetric. Even when information is
symmetric there is a second-mover advantage if the leader's choice causes
uncertainty to be resolved--- for example, through profits observed after
entry. If, on the other hand, information is asymmetric, although the
uncertainty is not resolved there can be a second mover advantage. The
less-informed player may want to delay so as to observe and imitate the
better-informed player's move. If imitation reduces profit, the
better-informed player will want to delay to prevent it. The intuition is
simple if the better-informed player has perfect information. If his
information is superior but still imperfect, the situation is less clear, as
we will see below.

There is a large literature on first and second mover advantages, but we
will defer explaining its intricacies until after we have laid out our own
model. Analyzing the effects of information quality on the strategic
decision on delay when both information spillover and payoff externalities
matter is our present  topic. We will use a setting in which two firms
decide which of two markets to enter. One firm has better but not perfect
information about which market is better. The firms must decide whether to
enter immediately or wait, but entry in the first period does not resolve
the underlying uncertainty. We describe the situation in terms of entering
new geographic markets, but the model is equally suited to new products or
any of the other of the varieties of innovation.

Whether it is best to move first or to move second depends on the quality of
information. If the informed player's information is inaccurate, there is a
first mover advantage for both players, the advantage of being able to
foreclose the market. Both players know that the informed player's
information is weak, so their main concern is to avoid competing in the same
market. Choosing the same market does not necessarily put them both in the
best market, however, since even the informed player's information is
imperfect. Instead, they might both end up in the less profitable market,
the worst possible outcome.

On the other hand, if the informed player's information is relatively
accurate, the second-mover advantage dominates. Both players know that the
informed player has a good chance of picking the big market, and this
outweighs the disadvantage of competing in the same market. The uninformed
player wants to imitate, and the informed player wants to evade imitation.
There are both offensive and defensive reasons to delay. If duopoly
competition is not severe, greater precision of information can lead to
inefficiency because it increases the informed player's incentive to conceal
his valuable information through delay. In this case, a cost of delay can
raise welfare by countering that effect because the less-informed player is
more patient to a cost of delay.

We also consider the case where a player can choose how informed to be. Does
a player always wants to have the most precise information? Oddly enough, a
player might prefer to be less informed even if acquiring better information
is costless. The reason is that very good information attracts imitation,
whereas slightly superior information does not.

\section{The Model}

We will use the terminology of geographical markets rather than product
markets in the model. An informed player (I) and an uninformed player (U)
each will enter either the North (N) or South (S) market. One of which is
bigger than the other, but which market is a bigger one is not known. There
are two periods during which each player can enter. In the first period they
choose simultaneously to enter North, enter South, or wait. If one player
waits and the other does not, the waiting player can observe the other
player's first-period choice before his own choice in the second period. The
second mover cannot observe profits, however, which are received only at the
end of the second period. Player $i$'s action set is thus $A=\{a_{i},t_{i}\}$%
, where $i\in \{U,I\}$ denotes the player, $a_{i}\in L=\{N,S\}$ denotes the
market entered, and $t=\{t_{1},t_{2}\}$ denotes the period of entry.

Table 1 shows the ex-post payoffs, with $x<\alpha y$ for $0<\alpha <1$. A
monopolist earns $x>0$ or $y>x$ depending on whether its market is small or
large. Each of two duopolists would earn $\alpha $ as much as a monopolist.%
\footnote{%
If being in a duopoly instead of a monopoly hurts a player's profits, as it
would unless the two players' products were complements, then $0<\alpha <1$.
If a duopoly industry earns less than a monopoly, as in the Cournot model
with identical products, then $0<\alpha <0.5$. If consumers sufficiently
value differentiated products, then $\alpha >0.5$ and the industry earns
more as a duopoly, though each firm would still prefer to be a monopoly. We
allow for both cases. The parameter $\alpha $ increases with: (1) the degree
of product differentiation, (2) the degree to which the two goods are
complements, and (3) the ability of the two players to collude when they are
a duopoly. If the products are identical, then $\alpha \leq 0.5$, with
perfect collusion having $\alpha =0.5$, Bertrand competition having $\alpha
=0$, and Cournot competition having $0<\alpha <0.5$. If there is perfect
collusion, then $0.5\leq \alpha <1$, depending on the degree of product
differentiation and product complementarity.}

\begin{center}
\begin{tabular}{llccc}
&  & \multicolumn{3}{c}{\textbf{Uninformed player }} \\ 
&  & Big Market &  & Small Market \\ 
& Big Market & $\alpha y,\alpha y$ &  & ${y,x}$ \\ 
\textbf{Informed player } &  &  &  &  \\ 
& Small Market & ${x},{y}$ &  & $\alpha x,\alpha x$ \\ 
&  &  &  & 
\end{tabular}

\textbf{Table 1: Ex-Post Payoffs for Big and Small Markets \\[0pt]
(for $y>x>0$ and $0<\alpha <1$) }
\end{center}

We will assume that $x<\alpha y$; that is, the single-firm duopoly profit in
a big market is greater than the monopoly profit in a small market. Thus,
the follower would be willing to crowd into a market despite the leader's
presence if he were sure the market was big.

The common prior is that both markets are equally likely to be the big
market. Before the first period, the informed player observes the private
signal $\theta \in \Theta =\{N,S\}$ which correctly identifies the big
market with probability $p\in \left( \frac{1}{2},1\right) $. As the
precision, $p$, approaches $\frac{1}{2},$ the signal becomes useless; as it
approaches $1,$ it becomes perfect. The uninformed player does not observe
the informed player's signal, but he does know $p$.

The informed player's pure strategy is 
\begin{equation}
s_{I}=(t_{I}(\theta ),a_{I}(\theta |t_{I}=t_{1}),a_{I}(\theta
|t_{I}=t_{U}=t_{2}),a_{I}(\theta |t_{U}=t_{1},t_{I}=t_{2}))  \label{1}
\end{equation}%
For given $\theta $, the informed player decides when to enter and whether
to follow his signal or not. If $a_{I}=\theta $, we will say that he
\textquotedblleft \textit{uses the signal}\textquotedblright . The
uninformed player's strategy is 
\begin{equation}
s_{U}=(t_{U},a_{U}|(t_{U}=t_{1}),a_{U}|(t_{I}=t_{U}=t_{2}),a_{U}(a_{I}|t_{I}=t_{1},t_{U}=t_{2}))
\label{2}
\end{equation}%
since he observes no his own private signal. We will allow mixed-strategies
for both players.

Let $\lambda $ be the uninformed player's belief as to the probability that
the informed player uses the signal in choosing a market. The strategy
profile $s=\{s_{U},s_{I}\}$ and $\lambda $ is a perfect Bayesian equilibrium
if $E\pi _{I}(s_{I},s_{U})$ and $E\pi _{U}(s_{I},s_{U})$ are maximized for
given $\lambda $ and $s=\{s_{U},s_{I}\}$ and $\lambda $ is consistent with $%
s_{I}$ in terms of Bayesian updating.

Throughout this paper, one particular value of $p$ is critical for
determining the equilibrium, so let us define: 
\begin{equation}
\overline{p}\equiv \frac{y-x\alpha }{\left( y-x\right) \left( \alpha
+1\right) }  \label{3}
\end{equation}%
It will turn out that for $p<\overline{p}$ there is a first-mover advantage
and for $p>\overline{p}$ there is a second-mover advantage.

\section{Exogenous Timing of Entry}

We will start by assuming that the sequence of entry is exogenous, a
necessary prelude to the analysis of endogenous entry. This analysis is in
order to derive both firms' best responses for each available outcome used
in calculating the players' expected payoffs.\footnote{%
When firms decide their timing of choice endogenously, they should consider
their expected payoffs for all available outcomes which depend on both
players' best responses.} The possible exogenous-timing games are (1) the
players move simultaneously, (2) the uninformed player moves first, and (3)
the informed player moves first.

If the uninformed player has no chance to observe I's choice, he has only
prior belief about which is a big market. In this case, the informed
player's best response is contingent on his belief as to the uninformed
player's choice.

\bigskip

\noindent \textbf{Lemma 1}\newline
\noindent \textit{Suppose that U has only prior belief about which is a big
market. Let us denote by }$q$\textit{\ I's belief that U selects "North".
Then any }$q\in \left[ 0,1\right] $\textit{\ is a consistent belief. }

\noindent \textbf{Proof: }\textit{In the appendix. }

\bigskip

The key reasoning is as follows. As the uninformed player should make a
decision on timing of entry before round 1 starts, he has no chance to
observe the informed player's signal. This implies that what the uninformed
player should expect is not whether the informed player will select to
"North" or "South", but whether he will follow the private signal or not.
The uninformed player's expectation, based on the belief about both the true
state and private signal, makes him indifferent between selecting "North"
and "South" regardless of the informed player's best response contingent on
his belief. Different equilibria will result depending on the informed
player's belief.\footnote{%
In all equilibria, it is same that the uninformed player is indifferent
between his choices. Just the informed player's best responses contingent on
his belief are different.} For the simple analysis, throughout this paper,
we assume that $q=\frac{1}{2}$.\footnote{%
Lemma A.1 (Page 19) describes the critical value of informed player's belief 
$q^{\ast }$ which yields his different best response. The assumption that $q=%
\frac{1}{2}$ corresponds to the case where $q<q^{\ast }$ because $\frac{1}{2}%
<q^{\ast }$.} As we assume no communication between players before making a
choice, this would be a reasonable assumption.

\bigskip

\noindent \textbf{Assumption 1}\newline
\textit{When U makes a choice without observing I's choice, I believes that }%
$\Pr (a_{U}=N)=0.5$.

\bigskip

Then both players' equilibrium behaviors in each exogenous ordering case can
be described as follows.\footnote{%
If the informed player is indifferent between using and deviating from his
signal for given $p$, we assume that he uses a signal. Also if the
uninformed player is indifferent, for given $p$, between imitating and
deviating from the informed player's choice, we assume that he imitates the
informed player's choice.
\par
\bigskip}

\bigskip

\noindent \textbf{Proposition 1}.\textit{\ }\newline
\textit{1)\ Suppose that the choice of location is simultaneous. Then, the
informed player uses his signal and the informed player chooses randomly
between North and South. \newline
2)\ Suppose that the choice of location is sequential. \newline
2-1)\ If the uninformed player chooses first, the informed player uses the
signal if it is accurate enough (if }$\overline{p}\leq p<1$\textit{), and
otherwise (if }$\frac{1}{2}<p<\overline{p}$\textit{) chooses the opposite of
the uninformed player.\newline
2-2)\ If the informed player chooses first, he uses the signal. The
uninformed player imitates him if the signal is accurate enough (if }$%
\overline{p}\leq p<1$\textit{), but otherwise (if }$\frac{1}{2}<p<\overline{p%
}$\textit{) he chooses the opposite. }

\noindent \textbf{Proof}: \textit{In the appendix}.

\bigskip

Proposition 1 says that when the informed player is the leader, he should
use his signal rather than try to conceal it by randomization. The
equilibrium is separating, so the uninformed player can infer the signal $%
\theta $ perfectly.

When the informed player is the follower, he has even less reason to
randomize. He knows he is better informed, so it is natural for him to use
the signal, but he must also consider the competition that arises when both
players are in the same market. Hence, his information quality affects
whether he uses his signal or not. If it is low, i.e., $p< \overline{p} $,
he has more reason to worry that the signal is wrong. If the uninformed
player already accidentally selected the location signalled by $\theta $,
the informed player will be reluctant to do the same because both players
might end up in the small market.

On the other hand, if his information quality is high, i.e., $\overline{p}%
\leq p$, he has relatively strong confidence in the correctness of the
signal. Even if the other player already chose the location signalled by $%
\theta $, it is better to join him there in what is very likely the best
market, because $\alpha y>x$. Hence, regardless of the uninformed player's
choice of location, the informed player uses the signal.

Similar reasoning applies to the uninformed player's strategy. If and only
if the informed player's information quality is low, the uninformed player
thinks mainly of avoiding competition and diverges in his choice of market.

From (3), it can be checked that 
\begin{equation}
\frac{\partial \overline{p}}{\partial y}=\frac{\left( \alpha -1\right) x}{%
\left( \alpha +1\right) \left( y-x\right) ^{2}}<0\text{, }\frac{\partial 
\overline{p}}{\partial x}=\frac{\left( 1-\alpha \right) y}{\left( \alpha
+1\right) \left( y-x\right) ^{2}}>0\text{, }\frac{\partial \left( \overline{p%
}\right) }{\partial \alpha }=-\frac{x+y}{\left( \alpha +1\right) ^{2}\left(
y-x\right) }<0  \label{4}
\end{equation}

As the large market size $y$ increases, the parameter set for which $p\in %
\left[ \overline{p},1\right) $ increases. As $y$ increases, each firm's
payoff from being a duopolist in the large market increases, whereas the
payoff $x$ from being a small-market monopolist stays the same. Hence, the
informed player puts more emphasis on being in the big market and has more
reason to follow his signal. The uninformed player knows this, so he too
relies more on the signal-- which means that as $y$ increases he becomes
more eager to imitate the informed player. On the other hand, as $x$, the
profit when a player operates as a monopolist in a small market, increases,
the parameter set for which $p\in \left[ \overline{p},1\right) $ falls. The
loss from being in a small market decreases and each player's incentive to
avoid competition in one market grows relative to the incentive to be in a
big market. As $x$ rises, whichever player is the follower becomes more
likely to diverge from the leader's choice. Finally, as $\alpha $ increases
(as competition becomes less severe), each firm's payoff from being a
duopolist in the same market increases. Hence, each firm's incentive to
avoid being in the same market decreases. Hence, the parameter set of $p$
for which the informed player sticks to his signal and and the uninformed
player wants to imitate the informed player's choice increases.

\subsection{The Expected Payoffs}

We wish to make the timing of entry over the two periods endogenous, and
this requires setting out the possible payoffs from different timings. The
informed player's expected payoff is one of two expressions, (5) or (6),
depending on whether the uninformed player will have a chance to observe the
informed player's choice or not. If the informed player goes first and the
uninformed player second, the uninformed player can deduce the signal $%
\theta $ perfectly, so his action $a_{U}$ is perfectly predictable. Hence,
the informed player's expected payoff is: 
\begin{equation}
\pi _{I}(t_{U},t_{I})=\sum_{w\in \{N,S\}}\Pr (\left. w\right\vert \theta
)\pi _{I}\left( a_{I},a_{U},w\right)  \label{5}
\end{equation}%
If, however, the uninformed player has no chance to infer the informed
player's signal, he chooses $a_{U}$ only using his prior. From Assumption 1,
the informed player's expected payoff is: 
\begin{equation}
\pi _{I}(t_{U},t_{I})=\left( 0.5\right) \left( \sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta )\pi _{I}\left( a_{I},a_{U}=N,w\right)
+\sum_{w\in \{N,S\}}\Pr (\left. w\right\vert \theta )\pi _{I}\left(
a_{I},a_{U}=S,w\right) \right)  \label{6}
\end{equation}%
These two equations cover the four possible combinations of timing. Payoff
(5) is for $(t_{U},t_{I})=(t_{2},t_{1})$ and payoff (6) is for $%
(t_{U},t_{I})=(t_{1},t_{2})$, $(t_{2},t_{2})$ and $(t_{1},t_{1})$. On the
other hand, the uninformed player's expected payoff is: 
\begin{equation}
\pi _{U}(t_{U},t_{I})=\sum_{\theta \in \{N,S\}}\sum_{w\in \{N,S\}}\Pr
(w,\theta )\pi _{U}\left( a_{I},a_{U},w\right)  \label{7}
\end{equation}%
Here, his posterior belief should be about the true state $w$ and I's
signal, $\Pr (w,\theta )$, because $I$ has no chance to observe $a_{I}$ and
therefore no chance to infer $\theta $ before he makes a decision.
Straightforward calculations yield the payoffs in Tables 1 and 2.

\begin{center}
\begin{tabular}{lllll}
&  &  & $t_{I}$ & \multicolumn{1}{l}{} \\ 
&  & $t_{1}$ &  & $t_{2}$ \\ 
& $t_{1}$ & $\frac{y+x\alpha -p\left( y-x\right) \left( 1-\alpha \right) }{2}%
,\allowbreak \frac{\left( x-px+py\right) \left( \alpha +1\right) }{2}$ &  & $%
\allowbreak \frac{\left( x+y\right) }{2},\allowbreak \frac{\left( x+y\right) 
}{2}$ \\ 
$t_{U}$ &  &  &  &  \\ 
& $t_{2}$ & $\left( y+px-py\right) ,\allowbreak \left( x-px+py\right) $ &  & 
$\frac{y+x\alpha -p\left( y-x\right) \left( 1-\alpha \right) }{2}%
,\allowbreak \frac{\left( x-px+py\right) \left( \alpha +1\right) }{2}$%
\end{tabular}

\bigskip

\textbf{Table 2: Ex-Ante expected payoffs when the signal is imprecise: }$%
\frac{1}{2}<p<\overline{p}$ \\[0pt]

\bigskip

\begin{tabular}{lllll}
&  &  & $t_{I}$ &  \\ 
&  & $t_{1}$ &  & $t_{2}$ \\ 
& $t_{1}$ & $\frac{y+x\alpha -p\left( y-x\right) \left( 1-\alpha \right) }{2}%
,\frac{\left( x-px+py\right) \left( \alpha +1\right) }{2}$ &  & $\frac{%
y+x\alpha -p\left( y-x\right) \left( 1-\alpha \right) }{2},\allowbreak
\allowbreak \frac{\left( x-px+py\right) \left( \alpha +1\right) }{2}$ \\ 
$t_{U}$ &  &  &  &  \\ 
& $t_{2}$ & $\alpha \left( x-px+py\right) ,\alpha \left( x-px+py\right) $ & 
& $\frac{y+x\alpha -p\left( y-x\right) \left( 1-\alpha \right) }{2},\frac{%
\left( x-px+py\right) \left( \alpha +1\right) }{2}$%
\end{tabular}

\bigskip

\textbf{Table 3: Ex-Ante expected payoffs when the signal is precise: }$%
\overline{p}\leq p<1$
\end{center}

In following section, we will use Tables 2 and 3 to find the equilibrium
when firms choose their timings of entry endogenously.

\section{Endogenous Timing of Entry}

In following, denote $i$'s ex-ante expected payoff when he enters as the
leader, follower, or simultaneously by $\pi _{i}^{L}$, $\pi _{i}^{F}$, and $%
\pi _{i}^{S}$.

\subsection{First-Mover and Second-Mover Advantage}

When the signal is imprecise, $p<\overline{p}$\textbf{, } Table 2 tells us
that: 
\begin{equation}
\pi _{U}^{S}<\pi _{U}^{F}<\pi _{U}^{L}\text{ and }\pi _{I}^{S}<\pi
_{I}^{F}<\pi _{I}^{L}  \label{8}
\end{equation}

Each player's best response as follower is to choose a different location
from the leader, even though we have assumed that it is better to be a
duopolist in the big market than a monopolist in the small market ($\alpha
y>x$). As the follower, he can operate as a monopolist in one market by
diverging from the leader. This follower behavior is why equation (8) says
that a player's expected profits are highest if he is the leader.

If the informed player is the leader, he uses his signal. Since the
uninformed player diverges, the informed player more likely than not ends up
as a monopolist in the big market. If he enters as the follower, however, he
should diverge from the uninformed player even if that puts him in the
market which he believes is likely to be small. Though if he chooses the
signalled market the more likely outcome is duopoly in the big market, it
might also be duopoly in the small market, the worst possible outcome.

The uninformed player's reasoning is similar. If he enters as the follower,
he should choose a location opposite to the leader, even though he knows
that the leader has chosen what is probably the big market. He would prefer
to be the leader, since then he has probability .5 of ending up as a
monopolist in the big market, compared to a probability of $1-p$, which is
less than $.5$, as the follower.

In this weak-information case, the expected payoff when sharing the market
is possible is lower than that from being the follower. For either player, $%
\pi _{U}^{S}<\pi _{U}^{F}$ and $\pi _{I}^{S}<\pi _{I}^{F}$. If $p$ is low, a
player's confidence in the signal (his or the other player's) is so low that
he puts more emphasis on avoiding competition. If both players choose
locations simultaneously, they might both end up as monopolists, but they
might not. Acting as the follower is better even if it reduces the chance of
being in the big market because it at least prevents the possibility of
sharing a small market. The best situation is to be the leader, the
next-best is to be the follower, and the worst case is to enter
simultaneously.

When information is precise, and $p\geq \overline{p}$, Table 3 tells us
that: 
\begin{equation}
\pi _{U}^{S}=\pi _{U}^{L}<\pi _{U}^{F}\ \ \text{and}\ \pi _{I}^{L}<\pi
_{I}^{S}=\pi _{I}^{F}  \label{9}
\end{equation}

Expression (9) says that if information precision is high, then each player
does best as the follower. The informed player knows that if the uninformed
player has a chance to observe his choice, he will choose the same location.
For relatively high $p$, however, he has strong confidence in his signal and
the payoff from being either a monopolist or a duopolist in the signalled
market is high. Hence, he enters late to prevent his choice from being
revealed and imitated. In fact, his payoff from simultaneous entry is just
as high as from being the follower: $\pi _{I}^{S}=\pi _{I}^{F}$. If both
players enter simultaneously, the uninformed player still has no chance to
observe the informed player's choice, and the probability the informed
player will end up sharing that market is .5.

As for the uninformed player, if the signal is relatively precise then his
ideal is to observe the informed player's choice and imitate it. If he
cannot, he is indifferent between being the leader or choosing
simultaneously: $\pi _{U}^{S}=\pi _{U}^{L}$. In either case, he has no
chance to observe the informed player's choice and to infer the signal.
Hence, his expected profits are the same in both cases.

\bigskip

\noindent \textbf{Remark 1}\newline
\textit{If information precision is relatively low (}$p<\bar{p})$\textit{,
there is a first mover advantage. On the other hand, if information
precision is relatively high (}$p\geq \bar{p}$), \textit{there is a second
mover advantage.}

\bigskip

If $p<\overline{p}$, a player would prefer to go first, but he will delay
entry if he thinks the other player will enter first, so as to avoid ending
up in the same market. If $p\geq \bar{p}$, the uninformed player would like
to delay entry in order to observe the informed player's choice, but if he
does delay, the informed player will also delay to prevent that observation.
Both end up delaying because of the conflict between two types of second
mover advantage: one from learning and the other from preventing learning.

\subsection{The Equilibrium}

Using the payoffs from Tables 2 and 3 we can characterize the endogenous
timing of entry.

\bigskip

\noindent \textbf{Proposition 2.} \textit{Suppose that entry timing is
endogenous. }\newline
\textit{1)\ If information is precise enough, i.e., }$p\geq \bar{p}$ \textit{%
, the informed player enters in the second period, and the uninformed player
is indifferent about when he enters. }\newline
\textit{2)\ If information is not precise enough, i.e., } $p<\bar{p}$\textit{%
, there are two pure-strategy equilibria, one for each of two players
entering first, and a mixed strategy equilibrium in which the uninformed
player enters without delay with probability }$z$\textit{\ and the informed
player enters without delay with probability }$w$\textit{:}%
\begin{equation*}
\left( z,w\right) =\left( \frac{\left( x-px+py\right) \left( \alpha
-1\right) }{\left( 2x\alpha -y-x-2px\alpha +2py\alpha \right) },\frac{\left(
x-px+py\right) \left( \alpha -1\right) }{\left( 2x\alpha -y-x-2px\alpha
+2py\alpha \right) }\right)
\end{equation*}

\noindent \textbf{Proof}: \textit{In the appendix.}

\bigskip

The signal's precision is high if $p\geq \overline{p}$. The informed player
delays entry and enters in period 2 to conceal his information. The
uninformed player can only use his prior belief of .5, and his expected
payoff is the same whenever and wherever he enters. The signal's precision
is low if $p<\overline{p}$. There are two pure-strategy equilibria, in which
the players enter sequentially into separate markets. Because the
information quality is low the players hesitate to rely on it and are most
concerned with avoiding competition. One player, at least, has incentive to
delay, but his benefit is not from imitating the leader, but from diverging.

If $p<\overline{p}$ there is also the mixed strategy equilibrium, in which a
player has no safe choice. Since the other player is mixing too, if he
enters early he might end up competing in the same market, but the same
thing could happen if he enters late. Entering early does have the advantage
that if the other player enters late, the leader can preempt the signalled
market (if he is the informed player) or have a .5 chance of preempting the
signalled market (if he is the uninformed player). This must be balanced by
a higher probability of ending up competing in the same market. Hence there
is some probability of early entry greater than .5 for each player which
leaves each of them indifferent about when to enter, and that is the
equilibrium mixing probability.

The comparative statics on the mixing probabilities of choosing early entry
yield that: 
\begin{align}
\frac{\partial z}{\partial p}& =\text{ }\frac{\partial w}{\partial p}=\frac{%
\left( 1-\alpha \right) \left( x+y\right) \allowbreak \left( y-x\right) }{%
\left( 2x\alpha -y-x-2px\alpha +2py\alpha \right) ^{2}}>0\text{ }  \label{10}
\\
\frac{\partial z}{\partial x}& =\frac{\partial w}{\partial x}=\frac{\left(
\alpha -1\right) \left( 2p-1\right) \allowbreak y}{\left( x+y-2x\alpha
+2px\alpha -2py\alpha \right) ^{2}}<0\text{ }  \label{11} \\
\frac{\partial z}{\partial y}& =\frac{\partial w}{\partial y}=\frac{\left(
1-\alpha \right) \left( 2p-1\right) \allowbreak x}{\left( 2x\alpha
-y-x-2px\alpha +2py\alpha \right) ^{2}}>0  \label{12}
\end{align}

Inequality (10) says that when information precision increases, both players
choose early entry with higher probability. The increase in information
precision $p$ makes choosing the signalled market more attractive. For both
players to still be willing to choose the unsignalled market, it must be
that choosing the signalled market results in greater probability of
undesirable competition. The way this probability increases is for both of
them to increase their probability of early entry until the likelihood of
competition has risen enough for them to again be indifferent about their
times of entry.

Inequalities (11) and (12) say that the probability of early entry falls
with $x$, the size of the small market, and rises with $y$, the size of the
big market. When the size of the small market increases, that increases the
benefit from waiting and possibly being the only player to enter in period
2. As a result, the probability of early entry by the other player does not
have to be so high to keep the player indifferent about his time of entry.
The effect of an increase in the size of the big market is parallel: that
increases the benefit from possibly being the only early entrant and the
disadvantage of entering early and making the same choice as the other
player must increase to balance that benefit.

Thus, we have shown that whether a first- or a second-mover advantage
emerges depends on whether the preemption effect is dominant, so that the
leader does not care about information leakage, or the information effect is
dominant, so the leader's main concern is to prevent imitation. When
knowledge of which market is best is imperfect, the presence of payoff
externalities makes preemption possible even when both players would prefer
a sure big market duopoly to small market monopoly.

\subsection{Efficiency}

Usually in information models, we analyze only ex ante efficiency--- whether
equilibrium decisions maximize the sum of expected payoffs given the
information available to the players at the start of the game. Here,
however, it is also possible that equilibrium leads to ex post efficiency---
that equilibrium decisions maximize the sum of actual payoffs, as if
decisions were made by a social planner who had no uncertainty. Here, of
course, the uncertainty is over which market is big, and even the informed
player's information is imperfect. Let us use the word \textquotedblleft
efficient\textquotedblright\ to mean \textquotedblleft maximizing the sum of
each firm's ex-post payoff,\textquotedblright\ since we have not specified
demand precisely enough to discuss consumer welfare, which will depend on
product variety and the loss from a market being unserved as well as on
equilibrium prices.

In our model a player prefers to be a duopolist in the big market than a
monopolist in the small market, i.e., $\alpha y>x$. If $2\alpha y>x+y$, that
is, if $\alpha >\frac{x+y}{2y}$, it is efficient for both firms to be in the
big market, since duopoly profits are large relative to monopoly profits and
the difference between market sizes is large. Otherwise, efficiency requires
that the firms choose different markets.

Suppose the informed player's information is imprecise ($p<\overline{p}$).
From Proposition 2, the pure-strategy equilibria have sequential entry into
two different markets. If $\alpha <\frac{x+y}{2y}$, this is the efficient
outcome. Thus, if competition hurts profits enough or the markets are close
enough in size, imprecise information results in the firms maximizing
industry profits ex post, at least if the equilibrium is in pure strategies.

\bigskip

Ex ante, neither player knows with certainty which is a big market, so the
best a planner maximizing industry profit can do is to either locate both
firms at the signalled market or to put them in different markets. Locating
in the same market is ex ante efficient if and only if $2\alpha
(py+(1-p)x)\geq x+y$, which can be expressed in two ways $\alpha \geq \frac{%
x+y}{2[py+(1-p)x]}$ or $p\geq \frac{y+x-2x\alpha }{2\alpha (y-x)}$. 
\footnote{%
When both are in the same market, it is the big market with probability $p$
and the small market with probability $1-p$. Hence, $\pi _{I}=\pi
_{U}=p\alpha y+(1-p)\alpha x$ and $\pi _{I}+\pi _{U}=2\alpha \left(
py+(1-p)x\right) $. If both are in the separate market, $\pi _{I}+\pi
_{U}=x+y$. Finally, locating both firms in the same market is ex ante
efficient if and only if $2\alpha (py+(1-p)x)\geq x+y$.} If the planner has
only the players' information about which market is big, expected profits
from co-location will naturally be lower than when he knew perfectly which
was big. Thus, the degree of product differentiation or duopoly cooperation $%
\alpha $ has to be bigger than in Proposition 3. That degree will now depend
on the quality of information $p$ too. The main key results of the analysis
as to ex-ante efficiency can be summarized as follows.\footnote{%
The detailed description of the ex-ante efficiency can be found in the
appendix (Proof of Propositon 3).}

\bigskip

\noindent \textbf{Proposition 3}. \textit{\ }\newline
\textit{1) Suppose that duopoly profit is low relative to monopoly profit,
so }$\frac{x}{y}<\alpha <\frac{x+y}{2y}$\textit{. Then all pure- strategy
equilibria are efficient. The mixed equilibrium is efficient only if }$p\geq 
\bar{p}$\textit{.\newline
\noindent 2) Suppose that duopoly profit is high relative to monopoly
profit, so }$\frac{x+y}{2y}<\alpha $\textit{. Then improved information can
lead to the inefficient equilibria.\newline
}\textbf{Proof}: \textit{In the appendix.}

\begin{center}
\textit{\TEXTsymbol{<} Figure 1 here \TEXTsymbol{>}}
\end{center}

Figure 1 illustrates the possible equilibrium regions for the case where the
monopoly profits are $x=1$ in the small market and $y=5$ in the large
market. The values of $p$ lie between $.5$ and $1$, and values of the
"duopoly/monopoly profit ratio" lie between $x/y$ and $1$, given our
assumption that the duopoly profit in the big market is greater than the
monopoly profit in the small market, i.e., $\alpha y>x$. The vertical line
represents the boundary condition that $\alpha =\frac{x+y}{2y}$ and the two
sloping lines represent $p=\overline{p}$ and $p=\frac{y+x-2x\alpha }{2\alpha
(y-x)}$.

If we ignore mixed-strategy equilibria, the parameter set of $\alpha $ and $%
p $ for which the equilibria are efficient (areas A1, A2, A4 and A5)
increases as the information quality declines and duopoly relative to
monopoly profits rise. In area A3, where the duopoly competition is not
severe, ex-ante efficiency requires both firms to locate in the market
signaled as big, but the informed player delays to conceal his precise
information. The uninformed player enters randomly and they might end up in
separate markets, which is inefficient. In this sense, more accurate
information hurts efficiency.

\section{Discussion}

\subsection{Delay Cost Can Raise Welfare}

In previous sections, we have assumed that a delay in a entry does not incur
any cost. Then what would happen if there is a delay cost?\ What would be
its effects on the equilibrium and welfare?\ If information quality is weak,
such a cost has little impact because the first mover advantage dominates
and the pure strategy equilibrium is a sequential entry. Hence a delay cost
affects only on whether the sequential timing of entry can be sustained or
not.\footnote{%
If a given delay cost is too high, the follower cannot sustain a delay and
therefore the simultaneous timing of entry is derived. Also for a mixed
strategy equilibrium, only the mixed probability changes to depend on $c$.}

What is more interesting is when information quality is high, i.e., $\bar{p}%
\leq p$. It was already checked that, in this case, if no cost is imposed,
the second mover advantage dominates and both firms intend to delay a entry
if possible. Now if a delay cost is imposed, a delay decision should be
affected because a delay is allowed only if a net gain from a delay is
positive. That is there can be a case in which a firm cannot delay although
it wants to do so because of a too high cost, from which a sequential entry
by both firms can be derived. From this reasoning, in this section, we focus
on the equilibrium of the case in which information quality is high. Our
results show that the existence of delay cost can improve ex-ante efficiency.

First, let us make some observations on which player gains more from a
delay. If a delay cost is imposed, intuitively what matter would be who has
the greater gain from a delay because the one with a greater gain can be
more patient to a costly delay. \noindent Let $d_{U}$ and $d_{I}$ denote
each player's gain from being the follower. The value $d_{U}$ is the gain
from delay that allows learning (imitating the choice of the informed
player) and the value $d_{I}$ is the gain from delay that preventing
learning (preventing the uninformed player's imitation).

\bigskip

\noindent \textbf{Lemma 2}. \textit{Suppose that }$\bar{p}\leq p<1$\textit{.
There exists a value }$\alpha ^{\ast }=\frac{x+y}{2y}\in \left( \frac{x}{y}%
,1\right) $\textit{\ such that:\newline
1)\ If }$\alpha \in \left( \frac{x}{y},\alpha ^{\ast }\right) $\textit{,
then }$d_{U}<d_{I}$\textit{\ for all }$p\in \left[ \bar{p},1\right) $\textit{%
. That is, the uninformed player has less benefit from delay.\newline
2)\ If }$\alpha \in \left( \alpha ^{\ast },1\right) $\textit{, for }$p\in %
\left[ \bar{p},\frac{y+x-2x\alpha }{2\alpha (y-x)}\right) $\textit{, }$%
d_{U}<d_{I}$\textit{\ and for }$p\in \left( \frac{y+x-2x\alpha }{2\alpha
(y-x)},1\right) $\textit{, }$d_{U}>d_{I}$\textit{. For information weak
enough, the informed player has the bigger benefit from delay, but for
strong information, the uninformed player does.}

\noindent \textbf{Proof}: \textit{In the appendix.}

\bigskip

As Lemma 2 says, both firms' gains from a delay are different according to
given parameters sets. Then costly delay can yield the sequential timing of
entry because both players' cutoff values of a delay cost below which a net
positive gain is guaranteed, differ. That is, if $Min\{d_{U},d_{I}\}<c<Max%
\{d_{U},d_{I}\}$, the player with the lower delay benefit should enter in
period 1 and the other can enter in period 2.\footnote{%
If $c>Max\{d_{U},d_{I}\}$, the two players both enter in period 1. If $%
c<Min\{d_{U},d_{I}\}$, the unique equilibrium is a mixed strategy
equilibrium. To see this, consider the pure strategy combinations. Suppose $%
\left( t_{U},t_{I}\right) =(t_{1},t_{1})$. Then, the uninformed player has
an incentive to deviate to $t_{U}=t_{2}$ in order to observe the informed
player's choice. If $\left( t_{U},t_{I}\right) =(t_{2},t_{1})$, however, the
informed player has an incentive to deviate to $t_{I}=t_{2}$ to prevent the
uninformed player from observing his choice. If they both enter in the
second period, so $\left( t_{U},t_{I}\right) =(t_{2},t_{2})$, the uninformed
player has incentive to deviate to $t_{U}=t_{1}$ because delay does not
allow him to observe the informed player's move, but it does incur cost $c$.
But if $\left( t_{U},t_{I}\right) =(t_{1},t_{2}),$ the informed player might
as well deviate to enter in the first period, since in this
strong-information case he will follow his signal anyway and so his
simultaneous move payoff is the same as his sequential move payoff except
for the delay cost $c$. Thus, there is no pure strategy Nash equilibrium.}

Now we see that there exist parameter sets for which a delay cost can
increase welfare. According to Proposition 3, when duopoly competition is
not severe, more accurate information can lead to inefficient equilibria
because of the informed player's incentive to conceal information. The
corresponding conditions are $\alpha \in \left( \frac{x+y}{2y},1\right) $
and $p\in \left( \frac{y+x-2x\alpha }{2\alpha (y-x)},1\right) $. In Lemma 2,
for these parameter sets of $\alpha $ and $p$, $d_{U}>d_{I}$. Then, if $%
d_{I}<c<d_{U}$, the informed player should act as the leader and the
uninformed player has a chance to observe and imitate the informed player's
choice. Hence, the ex-ante efficiency can be attained. That is, when the
duopoly competition is not severe and the informed player's information is
quite accurate, the moderate but not too high delay cost can resolve the
inefficiency and therefore function as a means to increases welfare.

\subsection{Endogenous Information Quality and Incentive to be Less-informed.%
}

Till now, we have assumed that one player is informed exogenously and the
other is not. In this section, we extend our analysis to the case where the
informed player can decide the information quality endogenously. The main
question we want to ask is whether a player always wants to have the most
precise information. In following, we show that interestingly, sometimes a
player wants to be less-informed although more precise information can be
acquired for free.

Suppose that one player has a chance of being informed without incurring any
information cost. He can not only observe his own private signal, but also
decide its precision. We assume that the other player does not have this
chance of being informed.

From Proposition 2, a player can predict the equilibrium contingent on the
information quality. If he picks the low quality information, i.e., $\frac{1%
}{2}<p<\overline{p}$, there exist multiple equilibria: $%
(t_{U}{}_{1},t_{I})=(t_{2},t_{1})$, $(t_{1},t_{2})$ and mixed equilibrium.
If he picks the high quality information, i.e., $\overline{p}\leq p<1$, in
equilibrium always he should enter in round 2. As $\left. \pi
_{I}^{Mixed}\right\vert _{p<\overline{p}}<\left. \pi
_{I}(t_{U}=t_{1},t_{I}=t_{2})\right\vert _{p<\overline{p}}<\left. \pi
_{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p<\overline{p}}$, we denote $%
(t_{U},t_{I})=(t_{2},t_{1})$ as the first-best outcome, $%
(t_{U},t_{I})=(t_{1},t_{2})$ as the second-best outcome and the mixed
equilibrium as the worst outcome respectively.\footnote{%
For the multiple equilibria of this, both payoff- and risk-dominance
criteria do not yield any dominant equilibrium. Hence we compare each
equilibrium of this case with the equilibrium when $\overline{p}\leq p<1$.}
The analysis yields the following result.

\bigskip

\noindent \textbf{Proposition 4}\newline
\noindent \textit{Let }$\bar{\alpha}=\frac{1}{y}\left( \sqrt{2}\sqrt{xy+y^{2}%
}-y\right) $. \textit{Suppose that a player expects the first-best outcome
when }$\frac{1}{2}<p<\overline{p}$\textit{. If the duopoly competition is
severe, i.e., }$0<\alpha <\bar{\alpha}$\textit{, he will prefer to have
worse information: }$p=\overline{p}-\varepsilon $\textit{\ instead of }$p=%
\overline{p}$\textit{. }

\noindent \textbf{Proof}: \textit{In the appendix.}

\begin{center}
\textit{\TEXTsymbol{<} Figure 2 Here \TEXTsymbol{>}}
\end{center}

Figure 2 describes the informed player's expected payoff as a function of
information quality $p$ when he expects the first-best outcome. Choosing low
quality information is a signal to the uninformed player that "there is not
much for you to learn from me". Moreover if the duopoly competition is
severe, the uninformed player should have less incentive to imitate the
informed player because of a concern of being in the small market together.
In this sense, the interior level of information quality $p=\overline{p}%
-\epsilon $ picked by a player is the best level consistent with not being
mimicked by the uninformed player.\footnote{%
If we assume that the uninformed player deviates from the informed player's
choice when $p=\bar{p}$, the interior information quality picked by the
informed player will be $p=\bar{p}$, which is a minor point.} Figure 2 also
implies that even when the competition is not severe sometimes being
more-informed can make a player worse-off.: when $\alpha >\bar{\alpha}$, if
the current information quality is $p=\overline{p}-\epsilon $, being
more-informed to $\bar{p}\leq p<\dot{p}$ makes a player worse-off.

\begin{center}
\textit{\TEXTsymbol{<} Figure 3 Here \TEXTsymbol{>}}
\end{center}

On the other hand, if a player expects the second-best or worst outcome when 
$\frac{1}{2}<p<\overline{p}$, he picks the most precise information (See
Figure 3). So whether the interior information quality will be picked or not
depends on whether he expects that the equilibrium is more likely to be $%
(t_{U},t_{I})=(t_{2},t_{1})$. If the informed player has a chance to send a
strong signal or a binding threat of entry in round 1, the uninformed player
is more likely to pick round 2 because $\left. \pi _{U}\left(
t_{U}=t_{1},t_{I}=t_{1}\right) \right\vert _{p<\overline{p}}<\left. \pi
_{U}\left( t_{U}=t_{2},t_{I}=t_{1}\right) \right\vert _{p<\overline{p}}$.

In real world usually information is costly to acquire and moreover
acquiring more precise information would be more costly. In the above
analysis, we show that a player can intend to be less-informed although
being more-informed does not incur any cost. Then the incentive to be
less-informed would be more common in real world for the sake of saving a
cost, especially if players face more competitive environment, as predicted
in our model.

\subsection{What If a Monopoly Profit in Small Market is Higher than a
Duopoly Profit in Big Market?}

So far we have assumed that a player would rather be a duopolist in the big
market than a monopolist in the small market. Then what would happen if we
assume instead that $\alpha y<x$?\ \ The equilibrium of this case can be
described as follows.\footnote{%
The detailed analysis is analogous to that used in the previous sections, so
we skip it here.}

\bigskip

\noindent \textbf{Proposition 5}\newline
\textit{Suppose that }$\mathit{\alpha y<x}$. \textit{Then, for all }$p\in
\left( \frac{1}{2},1\right) $\textit{, there exist two pure-strategy
equilibria $\left( t_{U},t_{I}\right) =(t_{2},t_{1}),\left(
t_{1},t_{2}\right) $ and one mixed-strategy equilibrium }$\left( z,w\right) $
\textit{where }$z=\Pr (t_{U}=t_{1})$ \textit{\ and $w=\Pr (t_{I}=t_{1})$.
The values $\left( z,w\right) $ are identical to those found Proposition 2. }

\bigskip

If $\alpha y<x$, the value of being in the big market even as a duopolist
decreases. Especially for the uninformed player, the first mover advantage
of preemption now outweighs the second mover advantage of learning for the
uninformed player. The learning motive, in fact, collapses completely,
because obtaining the benefit of learning requires that the follower become
a duopolist. However it is inferior to being a monopolist even if the
identity of the big market is learned perfectly. For the informed player,
the best case is still to preempt a market his signal says. However, if the
uninformed player already preempted a market his signal says, he has no more
incentive to stick to it because operating as a monopolist in one market,
even though it is a small market, is better. Hence for both players only the
first mover advantage dominates, which induces the same result of the case
where $\alpha y>x$ and information quality is low.

\section{The Literature}

A large literature exists on the war of attrition and pre-emption games in
which entry does not result in any learning (see Argenziano \&
Schmidt-Dengler (2008) on pre-emption games, and Brunnermeier \& Morgan
(2010) on clock games). Bouis, Huisman \& Kort (2009) add uncertainty in the
form of changes in market demand, but this is observed even without entry,
and its main interest is in changing the intervals between entry.

The next analytic step is for entry to reveal information. Gal-Or (1987)
shows that a player with superior information on demand may prefer ex ante
to move simultaneously rather than first, to avoid revealing his knowledge
that demand is strong (though Mailath [1993] shows that if the well-
informed player has the \textit{option} of moving first, it will always take
that option, since to delay reveals that it is trying to hide strong
demand). Normann (1997, 2002) also look at what happens when one duopolist
is better informed and he makes his quantity decision. Levin \& Peck (2003)
look at a different kind of asymmetric information: firms differ in their
entry cost, and must decide when to enter a natural monopoly in a variant of
the grab-the-dollar game. In all these models, the players are making
decisions about how hard to compete in one market, not whether to compete at
all, and the quality of the informed player's information is unimportant.

In other models, players are symmetric but the first player's move creates
new information. In Rob (1991), entry reveals informational under payoff
externality, but the players are not asymmetrically informed. The timing of
actions is given exogenously, and the focus is on the second mover's
advantage from learning. Rob does not analyze the possible advantage from
moving second to prevent the other player from learning. Bolton \& Harris
(1999), Hirokawa \& Sasaki (2001) and Hoppe (2000) also look at what happens
when the first mover's move reveals something about the state of the world.
In the present paper, what the second mover gets from observing the first
mover's choice is not direct information about the unknown true state, but
the first mover's information.

Appelbaum \& Lim (1985), Spencer \& Brander (1992), and Somma (1999) also
deal with the topic of market preemption and delay. Their focus is the
trade-off between precommitment and flexibility when uncertainty is resolved
exogenously over time. Delay has option value, because the choice can be
made after uncertainty is resolved. In our model, the uncertainty is not
resolved till both firms make their choices.

Chamley and Gale (1994) and Zhang (1997) discuss strategic delay and the
endogenous timing of action when there are informational externalities but
no payoff externalities. In Chamley and Gale (1994), a player has an
incentive to delay his action for information updating. Zhang (1997) links
this result to informational cascades, finding that the most-informed player
is least willing to wait because he has the least to learn. In these models,
as there are no payoff externalities, each player's main concern is whether
the cost of delay is worth learning other players' information.

Yoon (2009) and Frisell (2003) consider endogenous timing of action game
where there are both informational- and payoff externalities. Yoon (2009)
analyzes a delay race where the worse-informed player delays to learn and a
better-informed player delays to prevent learning. The payoff functions are
different from the present paper's in that the worst outcome is to be
incorrect alone. In our model, of market entry, the worst outcome is for
both players to end up in a small market, having made the same mistake. That
possible payoff is crucial to why the quality of information determines
whether the advantage is to the first or the second mover. Frisell (2003)
asks who will enter as the leader and follower when whether payoff
externality is either positive or negative is given exogenously. The degree
of information superiority does not affect the given payoff externality.
Delay costs are crucial, but delay can be infinitely long. Also the
simultaneous action is not considered. In our paper, on the other hand, one
player cannot simply outwait the other. If he waits, the result can be
simultaneous choices. As a result, a player who moves late must be concerned
about ending up in the same market as his rival, even if he has prevented
purposeful imitation. We find, in contrast to Frisell, that even if industry
profits suffer heavily when both players are in the same market, the
informed player may decide to move first if he is not much better informed.

\section{Concluding Remarks}

Often, a player must make a choice knowing that his choice may be imitated
by another player. This choice might be of a new geographic market, as in
our model, or of a new product, which could be modelled with exactly the
same mathematics. Moving first may or may not deter entry into the market by
the rival player, but it certainly will reveal information. Hence, in a
setting of endogenous timing of entry, the decision on the timing of entry
can be interpreted as the decision on the flow of private information. Of
course, how is revealed information used by the other player affects the
decision on the timing of entry. If the informed player's information is not
strong, the attempt by both players to avoid crowding into one market
results in the pure equilibrium in which they operate as monopolists in
separate markets. On the other hand, if the informed player's information is
relatively valuable, the rival player wants to learn it and imitate his
choice. Hence, an informed player may well choose to delay entry to prevent
imitation, which results in an equilibrium in which no learning is
available. This kind of strategic delay by both players increases the
probability that they end up in the same market. So the good information
that causes the delay can actually end up reducing industry profits. In this
case, however, the gain from preventing the other's learning is less than
the gain from learning. Hence the existence of a delay cost can resolve the
inefficiency caused from the informed player's incentive to conceal his
information. If competition is quite severe, a firm's intention not to be
imitated may increase. Hence, interestingly, a player might choose to refuse
free information--- provided, of course, that his refusal is common
knowledge. He will prefer to have the maximum level of information that
prevents mimicking by his rival.

\pagebreak

\textit{* If requested, some parts of the detailed proofs can be shortened
in the final version. }

\section{Appendix}

In the Appendix, "\textit{I}" denotes the "\textit{informed} player" and " 
\textit{U}" denotes the "\textit{uninformed} player".

\subsection{Proof of Lemma 1}

U has only prior belief about which is a big market if he has no chance to
observe I's choice. In following, $E\pi _{I}\left( a_{I}=\theta \right) $ $%
\left( E\pi _{I}\left( a_{I}\neq \theta \right) \right) $ is I's expected
payoff when he follows (deviates from) his signal. Without loss of
generality, suppose that $\theta =N$. Denote $q$ as I's belief for that U
selects "N". Then $q$ can be interpreted as I's belief for that U selects
what $\theta $ signals. Then 
\begin{eqnarray*}
E\pi _{I}\left( a_{I}=\theta \right) &=&q\left( \sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{I}\left( a_{I}=\theta ,a_{U}=N,w\right)
\right) \\
&&+\left( 1-q\right) \left( \sum_{w\in \{N,S\}}\Pr (\left. w\right\vert
\theta =N)\pi _{I}\left( a_{I}=\theta ,a_{U}=S,w\right) \right) \\
&=&q\left( p\left( \alpha y\right) +(1-p)\left( \alpha x\right) \right)
+(1-q)\left( p\left( y\right) +(1-p)\left( x\right) \right)
\end{eqnarray*}%
\begin{eqnarray*}
E\pi _{I}\left( a_{I}\neq \theta \right) &=&q\left( \sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{I}\left( a_{I}\neq \theta
,a_{U}=N,w\right) \right) \\
&&+\left( 1-q\right) \left( \sum_{w\in \{N,S\}}\Pr (\left. w\right\vert
\theta =N)\pi _{I}\left( a_{I}\neq \theta ,a_{U}=S,w\right) \right) \\
&=&q\left( p\left( x\right) +(1-p)\left( y\right) \right) +(1-q)\left(
p\left( \alpha x\right) +(1-p)\left( \alpha y\right) \right)
\end{eqnarray*}%
and 
\begin{equation*}
E\pi _{I}\left( a_{I}=\theta \right) -E\pi _{I}\left( a_{I}\neq \theta
\right) =\allowbreak q\left( x\alpha -y-x+y\alpha \right) +\left(
x-px+py-y\alpha -px\alpha +py\allowbreak \alpha \right)
\end{equation*}

As one extreme case, suppose that $q=0$. (That is, I believes that U selects
S when $\theta =N$.) Then,%
\begin{equation*}
\left. E\pi _{I}\left( a_{I}=\theta \right) -E\pi _{I}\left( a_{I}\neq
\theta \right) \right\vert _{q=o}=\allowbreak p\left( y-x-x\alpha +y\alpha
\right) +x-y\alpha >0
\end{equation*}%
for $p\in \left( \frac{1}{2},1\right) $. Therefore, if $q=0$, I follows his
signal $\theta $. As the other extreme case, suppose that $q=1$. (I believes
that U selects N when $\theta =N$.) Then,

\begin{equation*}
\left. E\pi _{I}\left( a_{I}=\theta \right) -E\pi _{I}\left( a_{I}\neq
\theta \right) \right\vert _{q=1}=p\left( y-x-x\alpha +y\alpha \right)
+x\alpha -y
\end{equation*}%
So, if $q=1$, there exists $p^{\ast }=\frac{y-x\alpha }{y-x-x\alpha +y\alpha 
}$ such that if $p\in \left( \frac{1}{2},p^{\ast }\right) $, I deviates from 
$\theta $ and if $p\in \left[ p^{\ast },1\right) $, I follows $\theta $. As $%
E\pi _{I}\left( a_{I}=\theta \right) -E\pi _{I}\left( a_{I}\neq \theta
\right) $ is a linear function of $q$ with negative slope, the above
analysis can be summarized as follows:

\bigskip

\noindent \textbf{Lemma A.1. }\newline
\textit{Suppose that both firms make a choice simultaneously. Let }$q=\Pr
(a_{u}=N)$\textit{\ be I's belief when U has only prior information. In
following, }$q^{\ast }=-\frac{x-px+py-y\alpha -px\alpha +py\allowbreak
\alpha }{x\alpha -y-x+y\alpha }$\textit{. }\newline
\textit{1)\ If }$q<q^{\ast }$\textit{\ (if I believes that U is less likely
to select what }$\theta $\textit{\ signals), I follows his signal.}\newline
\textit{2)\ If }$q>q^{\ast }$\textit{\ (if I believes that U is more likely
to select what }$\theta $\textit{\ signals), }$\exists p^{\ast }$\textit{\
s.t. if }$p\in \left( \frac{1}{2},p^{\ast }\right) $\textit{, I deviates
from his signal and if }$p\in \left( p^{\ast },1\right) $\textit{, I follows
his signal.}\newline
\textit{3)\ If }$q=q^{\ast }$\textit{, I is indifferent between following
and deviating from }$\theta $. \textit{\ }

\bigskip

For the given belief, we derive U's best response and check whether the
given belief is consistent or not. Here, for each given belief, \textit{what
U expects is whether I will follow }$\theta $\textit{\ or not}. \textit{U
cannot expect whether I will go to north or south because }$\theta $\textit{%
\ is private information.} That is, U's belief should be about both true
state and $\theta $. In following, $E\pi _{U}\left( a_{U}=N\right) $ $\left(
E\pi _{U}\left( a_{U}=S\right) \right) $ is U's expected payoff when he
selects N(S).

\bigskip

1)\ If $q<q^{\ast }$ (if U is less likely to select what $\theta $ signals),
I follows $\theta $. Then 
\begin{eqnarray*}
E\pi _{U}\left( a_{U}=N\right) &=&\sum_{w\in \{N,S\}}\sum_{\theta \in
\{N,S\}}\Pr (w,\theta )\pi _{U}\left( a_{I}=\theta ,a_{U}=N,w\right) =\frac{p%
}{2}\alpha y+\frac{1-p}{2}y+\frac{1-p}{2}\alpha x+\frac{p}{2}x \\
E\pi _{U}\left( a_{U}=S\right) &=&\sum_{w\in \{N,S\}}\sum_{\theta \in
\{N,S\}}\Pr (w,\theta )\pi _{U}\left( a_{I}=\theta ,a_{U}=S,w\right) =\frac{p%
}{2}x+\frac{1-p}{2}\alpha x+\frac{1-p}{2}y+\frac{p}{2}\alpha y
\end{eqnarray*}%
Hence, $E\pi _{U}\left( a_{U}=N\right) =E\pi _{U}\left( a_{U}=S\right) $,
which implies that U is indifferent between selecting N and S. Then $%
q<q^{\ast }$\ is a consistent belief.

\bigskip

2)\ If $q>q^{\ast }$ (if U is more likely to select $\theta $ says), the
information quality matters. First, if $p\in \left( \frac{1}{2},p^{\ast
}\right) $, I deviates from his signal $\theta $. Then, $E\pi _{U}\left(
a_{U}=N\right) =E\pi _{U}\left( a_{U}=S\right) $ because 
\begin{eqnarray*}
E\pi _{U}\left( a_{U}=N\right) &=&\sum_{w\in \{N,S\}}\sum_{\theta \in
\{N,S\}}\Pr (w,\theta )\pi _{U}\left( a_{I}\neq \theta ,a_{U}=N,w\right) =%
\frac{p}{2}y+\frac{1-p}{2}\alpha y+\frac{1-p}{2}x+\frac{p}{2}\alpha x \\
E\pi _{U}\left( a_{U}=S\right) &=&\sum_{w\in \{N,S\}}\sum_{\theta \in
\{N,S\}}\Pr (w,\theta )\pi _{U}\left( a_{I}\neq \theta ,a_{U}=S,w\right) =%
\frac{p}{2}\alpha x+\frac{1-p}{2}x+\frac{1-p}{2}\alpha y+\frac{p}{2}y
\end{eqnarray*}%
Second, if $p\in \left( p^{\ast },1\right) $, I follows his signal. We have
already checked in 1) that, in this case, $E\pi _{U}\left( a_{U}=N\right)
=E\pi _{U}\left( a_{U}=S\right) $. So I's belief that $q>q^{\ast }$\ is
consistent.

\bigskip

3)\ If $q=q^{\ast }$, I is indifferent between following or deviating from $%
\theta $. Let's assume that $h$ is the probability that I follows $\theta $
when I is indifferent between following or deviating from $\theta $, i.e., $%
h=\Pr \left( a_{I}=\theta \right) $. Then 
\begin{eqnarray*}
E\pi _{U}\left( a_{U}=N\right) &=&h\left( \sum_{w\in \{N,S\}}\sum_{\theta
\in \{N,S\}}\Pr (w,\theta )\pi _{U}\left( a_{I}=\theta ,a_{U}=N,w\right)
\right) \\
&&+(1-h)\left( \sum_{w\in \{N,S\}}\sum_{\theta \in \{N,S\}}\Pr (w,\theta
)\pi _{U}\left( a_{I}\neq \theta ,a_{U}=N,w\right) \right) \\
&=&h\left( \frac{p}{2}\alpha y+\frac{1-p}{2}y+\frac{1-p}{2}\alpha x+\frac{p}{%
2}x\right) +(1-h)\left( \frac{p}{2}y+\frac{1-p}{2}\alpha y+\frac{1-p}{2}x+%
\frac{p}{2}\alpha x\right) \\
E\pi _{U}\left( a_{U}=S\right) &=&h\left( \sum_{w\in \{N,S\}}\sum_{\theta
\in \{N,S\}}\Pr (w,\theta )\pi _{U}\left( a_{I}=\theta ,a_{U}=S,w\right)
\right) \\
&&+(1-h)\left( \sum_{w\in \{N,S\}}\sum_{\theta \in \{N,S\}}\Pr (w,\theta
)\pi _{U}\left( a_{I}\neq \theta ,a_{U}=S,w\right) \right) \\
&=&h\left( \frac{p}{2}x+\frac{1-p}{2}\alpha x+\frac{1-p}{2}y+\frac{p}{2}%
\alpha y\right) +\left( 1-h\right) \left( \frac{p}{2}\alpha x+\frac{1-p}{2}x+%
\frac{1-p}{2}\alpha y+\frac{p}{2}y\right)
\end{eqnarray*}%
For all $h\in \left[ 0,1\right] $, $E\pi _{U}\left( a_{U}=N\right) =E\pi
_{U}\left( a_{U}=S\right) $, which implies that the belief $q=q^{\ast }$\ is
consistent.\textbf{\ }

\bigskip

Finally, it can be checked that, I's belief $q\in \left[ 0,1\right] $\ is a
consistent belief. $\square $

\subsection{Proof of Proposition 1}

First, consider the case where the entry is simultaneous. In this case, U
chooses randomly between N and S because he has only prior belief. Now, as
for the best response of I, without loss of generality, assume that $\theta
=N$. Then,%
\begin{align}
E\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) & =\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\left[ .5\pi _{I}\left( \cdot ,a_{U}=N\right)
+.5\pi _{I}\left( \cdot ,a_{U}=S\right) \right]  \tag{A1} \\
& =\allowbreak \frac{1}{2}\left( x-px+py\right) \left( \alpha +1\right) 
\notag
\end{align}%
\begin{eqnarray}
E\pi _{I}\left( a_{I}\neq \theta ,a_{U},w\right) &=&\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\left[ .5\pi _{I}\left( \cdot ,a_{U}=N\right)
+.5\pi _{I}\left( \cdot ,a_{U}=S\right) \right]  \TCItag{A2} \\
&=&\frac{1}{2}\left( y+px-py\right) \left( \alpha +1\right)  \notag
\end{eqnarray}%
where $(A1)$ is the expected payoff when informed player uses his signal and 
$(A2)$ is the one when he deviates from it. Then:%
\begin{equation}
E\pi _{I}\left[ a_{I}=N,a_{U},w\right] -E\pi _{I}\left[ a_{I}=S,a_{U},w%
\right] =\allowbreak \frac{1}{2}\left( y-x\right) \left( 2p-1\right) \left(
\alpha +1\right) >0  \tag{A3}
\end{equation}%
So, I's best response is $a_{I}=N$. This implies that if $\theta =S$, it
should be that $a_{I}=S$. Hence, I's best response is to use his signal.

Second, consider the case where the entry is sequential.

1)\ What if U moves first? There are two cases depending on whether I's
signal equals U's action or not: $\theta =a_{U}$ and $\theta \neq a_{U}$. 
\newline
\noindent Case i: $\theta =a_{U}$. Without loss of generality, assume that $%
\theta =a_{U}=N$. Then, under the posterior beliefs $\Pr (\left.
w=N\right\vert \theta =N)=p$ and $\Pr (\left. w=S\right\vert \theta =N)=1-p$,%
\begin{align}
E\pi _{I}(a_{I}& =\theta ,a_{U},w)=\sum_{w\in \{N,S\}}\Pr (\left.
w\right\vert \theta =N)\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) =p\left(
\alpha y\right) +(1-p)\left( \alpha x\right)  \tag{A4} \\
E\pi _{I}(a_{I}& \neq \theta ,a_{U},w)=\sum_{w\in \{N,S\}}\Pr (\left.
w\right\vert \theta =N)\pi _{I}\left( a_{I}\neq \theta ,a_{U},w\right)
=p\left( x\right) +(1-p)\left( y\right)  \tag{A5}
\end{align}%
Then, $p\gtreqless \frac{\left( x\alpha -y\right) }{\allowbreak \left(
\alpha +1\right) \left( x-y\right) }=\overline{p}\Longrightarrow
(A4)\gtreqless (A5)$ where $\overline{p}\in \left( \frac{1}{2},1\right) $. 
\newline
Case ii: $\theta \neq a_{U}$. Without loss of generality, assume that $%
\theta =N$ and $a_{U}=S$. Then, under the posterior beliefs $\Pr (\left.
w=N\right\vert \theta =N)=p$ and $\Pr (\left. w=S\right\vert \theta =N)=1-p$,%
\begin{align}
E\pi _{I}(a_{I}& =\theta ,a_{U},w)=\sum_{w\in \{N,S\}}\Pr (\left.
w\right\vert \theta =N)\pi _{U}\left( a_{I}=\theta ,a_{U},w\right)
=py+\left( 1-p\right) x  \tag{A6} \\
E\pi _{I}(a_{I}& \neq \theta ,a_{U},w)=\sum_{w\in \{N,S\}}\Pr (\left.
w\right\vert \theta =N)\pi _{U}\left( a_{I}\neq \theta ,a_{U},w\right)
=p\left( \alpha x\right) +(1-p)\left( \alpha y\right)  \tag{A7}
\end{align}%
Then for all $p\in \left( \frac{1}{2},1\right) $, $(A6)>(A7)$. \newline
Therefore, I's best response is as follows: If $p>\frac{\left( x\alpha
-y\right) }{\allowbreak \left( \alpha +1\right) \left( x-y\right) }=%
\overline{p}$, regardless of U's choice in round 1, I uses his signal. On
the other hand, when $p<\overline{p}$, if $\theta =a_{U}$, he deviates from
his signal and if $\theta \neq a_{U}$, he uses his signal.

2) What if I moves first? This is the harder case, because what U observes
is $a_{I}$, not $\theta $. As $\theta $ is private information, U does not
know whether I follows his signal or not in deciding a location. Here, note
that whether $\theta =N$ or $\theta =S$, both cases are ex-ante symmetric.
Hence, intuitively a pooling strategy or semi-separating strategy cannot
constitute equilibrium. The following analysis shows that the separating
strategy which constitutes equilibrium is the one which implies that I
follows his signal in selecting location.

U must assign belief $\lambda $ for that $a_{I}=\theta $. In a pure strategy
equilibrium, this belief is $\lambda \in \{0,1\}$. As a first step in
looking at beliefs, suppose that U's belief $\lambda $ does equal zero or
one. Suppose U believes $a_{I}=\theta $, i.e., $\lambda =1$. Without loss of
generality, let $a_{I}=N$. U's posterior beliefs are $\Pr (\left.
w=N\right\vert \theta =N)=p$ and $\Pr (\left. w=S\right\vert \theta =N)=1-p$
, so%
\begin{eqnarray}
E\pi _{U}\left( a_{I}=\theta ,a_{U},w\right) &=&\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}=a_{U},w\right) =p\left(
\alpha y\right) +(1-p)\left( \alpha x\right)  \TCItag{A8} \\
E\pi _{U}\left( a_{I}\neq \theta ,a_{U},w\right) &=&\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{U}\left( a_{U}\neq a_{I},w\right)
=p\left( x\right) +(1-p)\left( y\right)  \TCItag{A9}
\end{eqnarray}%
Next, suppose U believes that $a_{I}\neq \theta $, i.e., $\lambda =0$. Then,
U's posterior beliefs are $\Pr (\left. w=N\right\vert \theta =S)=1-p$ and $%
\Pr (\left. w=S\right\vert \theta =S)=p$, so%
\begin{eqnarray}
E\pi _{U}\left( a_{I}=\theta ,a_{U},w\right) &=&\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =S)\pi _{U}\left( a_{I}=a_{U},w\right) =\left(
1-p\right) \left( \alpha y\right) +p\left( \alpha x\right)  \TCItag{A10} \\
E\pi _{U}\left( a_{I}\neq \theta ,a_{U},w\right) &=&\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =S)\pi _{U}\left( a_{I}\neq a_{U},w\right)
=\left( 1-p\right) \left( x\right) +p\left( y\right)  \TCItag{A11}
\end{eqnarray}%
More generally, I might mix, so U's belief that $a_{I}=\theta $ would be $%
\lambda \in \lbrack 0,1]$. Then%
\begin{eqnarray}
&&E\pi _{U}\left( a_{I}=\theta ,a_{U},w\right)  \TCItag{A12} \\
&=&\lambda \left( \sum_{w\in \{N,S\}}\Pr (\left. w\right\vert \theta =N)\pi
_{U}\left( a_{I}=a_{U},w\right) \right) +(1-\lambda )\left( \sum_{w\in
\{N,S\}}\Pr (\left. w\right\vert \theta =S)\pi _{U}\left(
a_{I}=a_{U},w\right) \right)  \notag
\end{eqnarray}%
\begin{eqnarray}
&&E\pi _{U}\left( a_{I}\neq \theta ,a_{U},w\right)  \TCItag{A13} \\
&=&\lambda \left( \sum_{w\in \{N,S\}}\Pr (\left. w\right\vert \theta =N)\pi
_{U}\left( a_{U}\neq a_{I},w\right) \right) +(1-\lambda )\left( \sum_{w\in
\{N,S\}}\Pr (\left. w\right\vert \theta =S)\pi _{U}\left( a_{I}\neq
a_{U},w\right) \right)  \notag
\end{eqnarray}%
Consequently,%
\begin{equation}
\lambda \gtreqqless \lambda ^{\ast }\allowbreak \Longrightarrow E\pi
_{U}\left( a_{U}=a_{I},w\right) \gtreqqless E\pi _{U}\left( a_{U}\neq
a_{I},w\right)  \tag{A14}
\end{equation}%
where $\lambda ^{\ast }=\frac{\left( x-px+py-y\alpha -px\alpha +py\alpha
\right) }{\allowbreak \left( y-x\right) \left( 2p-1\right) \left( \alpha
+1\right) }$. We will state and prove Lemma A.1, included only here in the
Appendix to help prove Proposition 1.

\bigskip

\noindent \textbf{Lemma A.1.} \textit{Suppose that I chose a location as the
leader.}\newline
\textit{a)\ Suppose that } $p\in \left( \frac{1}{2},\overline{p} \right) $ 
\textit{. Then, for all }$\lambda \in \left[ 0,1\right] $\textit{, U
diverges from I's choice.}\newline
\textit{b)\ Suppose that } $p\in \left( \overline{p},1\right) $\textit{.
Then, there exists } $\lambda ^{\ast }\in \left( 0,1\right) $\textit{\ such
that if } $\lambda \in (\lambda ^{\ast },1]$\textit{, U imitates I's choice,
if }$\lambda \in \lbrack 0,\lambda ^{\ast })$\textit{\ , U diverges from I's
choice, and if } $\lambda =\lambda ^{\ast }$\textit{, U is indifferent
between imitating and diverging.}

\noindent \textbf{Proof of Lemma A.1.} We start by checking whether $\lambda
^{\ast }\in \left( 0,1\right) $ or not. First, for $\lambda ^{\ast }$, if $%
p\gtreqless \frac{y\alpha -x}{\allowbreak \left( y-x\right) \left( \alpha
+1\right) }$, $\lambda ^{\ast }\gtreqless 0$. However, $\frac{y\alpha -x}{%
\allowbreak \left( y-x\right) \left( \alpha +1\right) }-\frac{1}{2}=\frac{%
\left( \alpha -1\right) \left( x+y\right) }{2\left( \alpha +1\right) \left(
y-x\right) }<0$. So, for $p\in \left( \frac{1}{2},1\right) $ , $\lambda
^{\ast }>0$. Also, from $\lambda ^{\ast }-1=\frac{\left( y+px-py-x\alpha
+px\alpha -py\alpha \right) }{\left( \alpha +1\right) \left( 2p-1\right)
\left( y-x\right) }$, if $p\gtreqless \frac{y-\alpha x}{\left( y-x\right)
\left( \alpha +1\right) }$, $\lambda ^{\ast }\gtreqless 1$ where $\frac{%
y-\alpha x}{\left( y-x\right) \left( \alpha +1\right) }\in \left( \frac{1}{2}%
,1\right) $. Therefore, we can summarize as follows: a) If $p\in \left( 
\frac{1}{2},\frac{y-\alpha x}{\left( y-x\right) \left( \alpha +1\right) }%
\right) $ then for all $\lambda \in \left[ 0,1\right] $, $E\pi
_{U}(a_{U}=a_{I})<E\pi _{U}(a_{U}\neq a_{I})$. b)\ If $p\in \left( \frac{%
y-\alpha x}{\left( y-x\right) \left( \alpha +1\right) },1\right) $ then if $%
\lambda \gtreqless \lambda ^{\ast }$, $E\pi _{U}(a_{U}=a_{I})\gtreqless E\pi
_{U}(a_{U}\neq a_{I})$. $\square $

Let us now return to I's best response, which we can derive using Lemma A.1.
In following, we denote $\frac{y-\alpha x}{\left( y-x\right) \left( \alpha
+1\right) }\equiv \bar{p}$. Without loss of generality, assume $\theta =N$.
Then, I's posterior beliefs are $\Pr (\left. w=N\right\vert \theta =N)=p$
and $\Pr (\left. w=S\right\vert \theta =N)=1-p$.

First, assume that $p\in \left( \frac{1}{2},\overline{p}\right) $. In this
case, U chooses a location different from I's choice for all $\lambda \in
\lbrack 0,1]$. Then%
\begin{align}
E\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) & =\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}=\theta \neq
a_{U},w\right) =p\left( y\right) +(1-p)\left( x\right)  \tag{A15} \\
E\pi _{I}\left( a_{I}\neq \theta ,a_{U},w\right) & =\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}\neq \theta
=a_{U},w\right) =p\left( x\right) +(1-p)\left( y\right)  \tag{A16}
\end{align}%
and%
\begin{equation}
E\pi _{I}(a_{I}=\theta ,a_{U},w)-E\pi _{I}(a_{I}\neq \theta
,a_{U},w)=\allowbreak \left( y-x\right) \left( 2p-1\right) >0  \tag{A17}
\end{equation}%
Thus, I's best response is to choose the location following his signal,
which is consistent to U's belief that $\lambda \in \lbrack 0,1]$.

Next, let $p\in \left( \overline{p},1\right) $. First, suppose $\lambda
>\lambda ^{\ast }$, so U imitates I. Then:%
\begin{align}
E\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) & =\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}=\theta =a_{U},w\right)
=p\left( \alpha y\right) +(1-p)\left( \alpha x\right)  \tag{A18} \\
E\pi _{I}\left( a_{I}\neq \theta ,a_{U},w\right) & =\sum_{w\in \{N,S\}}\Pr
(\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}=a_{U}\neq \theta
,w\right) =p\left( \alpha x\right) +(1-p)\left( \alpha y\right)  \tag{A19}
\end{align}%
and:%
\begin{equation}
E\pi _{I}(a_{I}=\theta ,a_{U},w)-E\pi _{I}(a_{I}\neq \theta
,a_{U},w)=\allowbreak \alpha \left( y-x\right) \left( 2p-1\right) >0 
\tag{A20}
\end{equation}%
Hence, I's best response is to choose a location following the signal, which
is consistent to U's belief that $\lambda >\lambda ^{\ast }$. \newline
Second, suppose $\lambda <\lambda ^{\ast }$, so U diverges from I's choice.
Then, from (A15) - (A16), I chooses a location following the signal.
However, this is inconsistent to U's belief that $\lambda <\lambda ^{\ast }$
. Hence, this case is excluded.\newline
Third, suppose $\lambda =\lambda ^{\ast }$, so , U is indifferent between
imitating and diverging from $a_{I}$. Suppose $\sigma _{U}$ is the
probability that U imitates I's choice. Then:%
\begin{eqnarray}
&&E\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) -E\pi _{I}\left( a_{I}\neq
\theta ,a_{U},w\right)  \TCItag{A21} \\
&=&\sigma \left( \sum_{w\in \{N,S\}}\Pr (\left. w\right\vert \theta =N)\pi
_{U}\left( a_{I}=a_{U},w\right) \right) +(1-\sigma )\left( \sum_{w\in
\{N,S\}}\Pr (\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}\neq
a_{U},w\right) \right)  \notag \\
&&-\left( \sigma \left( \sum_{w\in \{N,S\}}\Pr (\left. w\right\vert \theta
=N)\pi _{U}\left( a_{I}=a_{U},w\right) \right) +(1-\sigma )\left( \sum_{w\in
\{N,S\}}\Pr (\left. w\right\vert \theta =N)\pi _{U}\left( a_{I}\neq
a_{U},w\right) \right) \right)  \notag \\
&=&\allowbreak \left( y-x\right) \left( 2p-1\right) \left( \alpha \sigma
-\sigma +1\right)  \notag
\end{eqnarray}%
It can be checked that $E\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) =E\pi
_{I}\left( a_{I}\neq \theta ,a_{U},w\right) $ only at $\sigma =\frac{1}{%
1-\alpha }$. But $\frac{1}{1-\alpha }\notin \left[ 0,1\right] $ for $%
0<\alpha <1$. Hence, there exists no $\sigma \in \left[ 0,1\right] $ which
yields $E\pi _{I}\left( a_{I}=\theta ,a_{U},w\right) =E\pi _{I}\left(
a_{I}\neq \theta ,a_{U},w\right) $.

Finally, I's best response in round 1 is to act so as to reveal his signal
perfectly. Since U's belief must be correct in equilibrium, it must be $
\lambda =1$ and his strategy must be to diverge from $a_{I}$ if $p<\overline{
p}$ and to imitate $a_{I}$ if $p>\overline{p}$, as stated in Proposition 1. $
\square $

\subsection{Proof of Proposition 2}

\noindent Denote $z=\Pr (t_{U}=t_{1})$ \textit{\ }and\textit{\ }$w=\Pr
(t_{I}=t_{1})$. Also, let $E_{i}(t_{i}=t_{k})$ denote $i$'s expected payoff
when he acts at round $k$, where $i\in \{I,U\}$ and $k\in \{1,2\}$.

(1) Consider the case in which $\frac{1}{2}<p<\overline{p}$. From Table 2:%
\begin{align}
E_{U}\left[ t_{U}=t_{1}\right] & =w\allowbreak \left( -\frac{\left(
\allowbreak p\left( y-x\right) \left( 1-\alpha \right) -y-x\alpha \right) }{2%
}\right) +(1-w)\left( \frac{1}{2}\left( x+y\right) \right)  \tag{A22} \\
E_{U}\left[ t_{U}=t_{2}\right] & =w\allowbreak \left( y+px-py\right)
+(1-w)\left( \allowbreak -\frac{\left( \allowbreak p\left( y-x\right) \left(
1-\alpha \right) -y-x\alpha \right) }{2}\right)  \tag{A23}
\end{align}%
Thus:%
\begin{equation}
E_{U}\left[ t_{U}=t_{1}\right] -E_{U}\left[ t_{U}=t_{2}\right] =w\left(
x\alpha -\frac{1}{2}y-\frac{1}{2}x-px\alpha +py\alpha \right) +\left( \frac{1%
}{2}\right) \left( x-px+py\right) \left( 1-\alpha \right)  \tag{A24}
\end{equation}%
For $p\in \left( \frac{1}{2},\overline{p}\right) $, $x\alpha -\frac{1}{2}y-%
\frac{1}{2}x-px\alpha +py\alpha <0$. Hence U's best response for given $w$
is:%
\begin{equation}
w<w^{\ast }\Longrightarrow z=1\text{, }w=w^{\ast }\Longrightarrow z\in
\lbrack 0,1]\text{, }w>w^{\ast }\Longrightarrow z=0  \tag{A25}
\end{equation}%
where $w^{\ast }=\frac{\left( x-px+py\right) \allowbreak \left( \alpha
-1\right) }{\left( 2x\alpha -y-x-2px\alpha +2py\alpha \right) }\in \left(
0,1\right) $. Returning to Table 2 for $I$'s payoffs:%
\begin{align}
E_{I}\left[ t_{I}=t_{1}\right] & =z\allowbreak \left( \frac{1}{2}\left(
x-px+py\right) \left( \alpha +1\right) \right) +(1-z)\left( x-px+py\right) 
\tag{A26} \\
E_{I}\left[ t_{I}=t_{2}\right] & =z\allowbreak \left( \frac{1}{2}\left(
x+y\right) \right) +(1-z)\left( \frac{1}{2}\left( x-px+py\right) \left(
\alpha +1\right) \right)  \tag{A27}
\end{align}%
Thus:%
\begin{equation}
E_{I}\left[ t_{I}=t_{1}\right] -E_{I}\left[ t_{I}=t_{2}\right] =\allowbreak
z\left( \allowbreak x\alpha -\frac{1}{2}y-\frac{1}{2}x-px\alpha +py\alpha
\right) +\frac{1}{2}\left( x-px+py\right) \left( 1-\alpha \right)  \tag{
A28}
\end{equation}%
Equation (A28) is identical to (A24). Hence, I's best response for $z$ is
the same as U's one for $w$:%
\begin{equation}
z<z^{\ast }\Longrightarrow \text{ }w=1\text{, }z=z^{\ast }\Longrightarrow
w\in \lbrack 0,1]\text{, }z>z^{\ast }\Longrightarrow w=0  \tag{A29}
\end{equation}%
Finally, the intersection of the players' best response functions (A25) and
(A29) yields that $\left( z,w\right) =(0,1),(1,0)$ and $(z^{\ast },w^{\ast
}) $ -- there exist two pure strategy equilibria $\left( t_{U},t_{I}\right)
=(t_{2},t_{1}),\left( t_{1},t_{2}\right) $\ and one mixed strategy
equilibrium $\left( z,w\right) =\left( \frac{\left( x-px+py\right)
\allowbreak \left( \alpha -1\right) }{\left( 2x\alpha -y-x-2px\alpha
+2py\alpha \right) },\frac{\left( x-px+py\right) \allowbreak \left( \alpha
-1\right) }{\left( 2x\alpha -y-x-2px\alpha +2py\alpha \right) }\right) $.

(2) Consider the case in which $\overline{p}\leq p<1$. Table 3 gives $U$'s
payoffs as:%
\begin{align}
E_{U}\left[ t_{U}=t_{1}\right] & =-\frac{\left( \allowbreak p\left(
y-x\right) \left( 1-\alpha \right) -y-x\alpha \right) }{2}  \tag{A30} \\
E_{U}\left[ t_{U}=t_{2}\right] & =w\left( \allowbreak \alpha \left(
x-px+py\right) \right) +(1-w)\left( -\frac{\left( \allowbreak p\left(
y-x\right) \left( 1-\alpha \right) -y-x\alpha \right) }{2}\right)  \tag{
A31}
\end{align}%
Thus:%
\begin{equation}
E_{U}\left[ t_{U}=t_{1}\right] -E_{U}\left[ t_{U}=t_{2}\right] =\frac{1}{2}%
w\left( \allowbreak p\left( -\left( y-x\right) \left( \alpha +1\right)
\right) +y-x\alpha \right)  \tag{A32}
\end{equation}%
Note that for $\overline{p}\leq p<1$, $p\left( -\left( y-x\right) \left(
\alpha +1\right) \right) +y-x\alpha <0$. So U's best response for $w$ is:%
\begin{equation}
w<0\Longrightarrow \text{ }z=1,w=0\Longrightarrow z\in \lbrack
0,1],w>0\Longrightarrow z=0  \tag{A33}
\end{equation}%
Next, Table 3 gives $I$'s payoffs as:%
\begin{align}
E_{I}\left[ t_{I}=t_{1}\right] =& \allowbreak z\left( \frac{1}{2}\left(
x-px+py\right) \left( \alpha +1\right) \right) +(1-z)\left( \alpha \left(
x-px+py\right) \right) \text{ }  \tag{A34} \\
E_{I}\left[ t_{I}=t_{2}\right] =& \allowbreak \frac{1}{2}\left(
x-px+py\right) \left( \alpha +1\right)  \tag{A35}
\end{align}%
Thus:%
\begin{equation}
E_{I}\left[ t_{I}=t_{1}\right] -E_{I}\left[ t_{I}=t_{2}\right] =z\left(
\allowbreak \frac{1}{2}\left( x-px+py\right) \left( 1-\alpha \right) \right)
+\frac{1}{2}\left( x-px+py\right) \left( \alpha -1\right)  \tag{A36}
\end{equation}%
Hence, I's best response to the uninformed player is:%
\begin{equation}
z>1\Longrightarrow w=1,\text{ }z=1\Longrightarrow w\in \lbrack 0,1],\text{ }%
z<1\Longrightarrow w=0  \tag{A37}
\end{equation}%
The intersections of both players' best response functions (A33) and (A37)
yield that $z\in \lbrack 0,1]$ and $w=0$. $\square $

\subsection{Proof of Proposition 3}

\noindent \noindent Case 1: When $\bar{p}\leq p<1$: In equilibrium, $%
t_{I}=t_{2}$ and $z\in \lbrack 0,1]$ where $z=\Pr (t_{U}=t_{1})$. From Table
3:%
\begin{align}
\sum_{i\in \{U,I\}}\pi _{i}\left( t_{1},t_{1}\right) & =\sum_{i\in
\{U,I\}}\pi _{i}\left( t_{1},t_{2}\right) =\sum_{i\in \{U,I\}}\pi _{i}\left(
t_{2},t_{2}\right) =\allowbreak \frac{1}{2}\left( x+y+2x\alpha -2px\alpha
+2py\alpha \right)  \tag{A38} \\
\sum_{i\in \{U,I\}}\pi _{i}\left( t_{2},t_{1}\right) & =2\alpha \left(
x-px+py\right)  \tag{A39}
\end{align}%
The computation of (A38) and (A39) yields the following result: A) Suppose
that $\frac{x+y}{2y}<\alpha <1$. Then, if $\bar{p}<p<\frac{y+x-2x\alpha }{%
2\alpha (y-x)}$, $(A38)>(A39)$ but if $\frac{y+x-2x\alpha }{2\alpha (y-x)}%
<p<1$ , $(A38)<(A39)$. B) Suppose that $\frac{x}{y}<\alpha <\frac{x+y}{2y}$.
Then, for all $p\in \left( \bar{p},1\right) $ $(A38)>(A39)$. First, assume
that $\frac{x+y}{2y}<\alpha <1$. If $\overline{p}<p<\frac{y+x-2x\alpha }{%
2\alpha (y-x)}$, the ex-ante efficient case is $\left( t_{U},t_{I}\right)
=\left( t_{1},t_{1}\right) $, $\left( t_{1},t_{2}\right) $ or $\left(
t_{2},t_{2}\right) $. If U uses a pure strategy, i.e., $z\in \{0,1\}$, the
outcome is $\left( t_{1},t_{2}\right) $ or $\left( t_{2},t_{2}\right) $,
which is ex-ante efficient. If U uses a mixed strategy, i.e., $z\in \left(
0,1\right) $, then $E\pi _{U}=-\frac{\left( \allowbreak p\left( y-x\right)
\left( 1-\alpha \right) -y-x\alpha \right) }{2}$ and $E\pi _{I}=\frac{\left(
x-px+py\right) \left( \alpha +1\right) }{2}$. So $E\pi _{U}+E\pi _{I}=\frac{1%
}{2}\left( x+y+2x\alpha -2px\alpha +2py\alpha \right) $, which is also
ex-ante efficient. Therefore, all equilibria are ex-ante efficient. On the
other hand, if $\frac{y+x-2x\alpha }{2\alpha (y-x)}<p<1$, the ex-ante
efficient case is the one where $\left( t_{U},t_{I}\right) =\left(
t_{2},t_{1}\right) $. Since $t_{I}=t_{2}$ in equilibrium, the equilibrium is
not ex-ante efficient. Second, assume that $\frac{x}{y}<\alpha <\frac{x+y}{2y%
}$. Then, for all $p\in \left( \frac{1}{2},1\right) $, the ex-ante efficient
case is $\left( t_{U},t_{I}\right) =\left( t_{1},t_{1}\right) $, $\left(
t_{1},t_{2}\right) $ or $\left( t_{2},t_{2}\right) $. From the analysis for
the case where $\frac{x+y}{2y}<\alpha <1$ and $\bar{p}<p<\frac{y+x-2x\alpha 
}{2\alpha (y-x)}$, all equilibria are ex-ante efficient.

Case 2: When $\frac{1}{2}<p<\bar{p}$: Recall that if $\frac{1}{2}<p<%
\overline{p}$, the equilibrium is $\left( t_{U},t_{I}\right) =\left(
t_{1},t_{2}\right) $, $\left( t_{2},t_{1}\right) $ and the mixed strategy
equilibrium $\left( z^{\ast },w^{\ast }\right) $ where $z=\Pr (t_{U}=t_{1})$
and $w=\Pr (t_{I}=t_{1})$. From Table 2, 
\begin{align}
\sum_{i\in \{U,I\}}\pi _{i}\left( t_{1},t_{1}\right) & =\sum_{i\in
\{U,I\}}\pi _{i}\left( t_{2},t_{2}\right) =\allowbreak \frac{1}{2}\left(
x+y+2x\alpha -2px\alpha +2py\alpha \right) \text{ }  \tag{A40} \\
\sum_{i\in \{U,I\}}\pi _{i}\left( t_{1},t_{2}\right) & =\sum_{i\in
\{U,I\}}\pi _{i}\left( t_{2},t_{1}\right) =x+y  \tag{A41}
\end{align}%
Comparison of (A40) and (A41) yields that for all $p\in \left( \frac{1}{2},%
\overline{p}\right) $, $(A40)<(A41)$. Therefore, for the ex-ante efficiency,
the players must enter sequentially. Hence the pure-strategy equilibria $%
\left( t_{U},t_{I}\right) =(t_{2},t_{1}),\left( t_{1},t_{2}\right) $ are
ex-ante efficient. As for the mixed strategy equilibrium, the computation
yields that:%
\begin{align}
& \left( E\pi _{U}+E\pi _{I}\right) -\left( x+y\right)   \tag{A42} \\
& =\frac{p^{2}\allowbreak \left( 2\left( y-x\right) ^{2}\left( \alpha
^{2}+1\right) \right) +\allowbreak p\left( \allowbreak 2\left( y-x\right)
\left( x-y-x\alpha -y\alpha +2x\alpha ^{2}\right) \right) +\left(
\allowbreak x^{2}-2xy\alpha +y^{2}-2x^{2}\alpha +2x^{2}\alpha ^{2}\right) }{%
2\left( 2x\alpha -y-x-2px\alpha +2py\alpha \right) }  \notag
\end{align}%
For $p\in \left( \frac{1}{2},\overline{p}\right) $, the denominator is
negative. The numerator is a convex function of $p$ and it attains the
minimum of $\frac{\left( x+y\right) ^{2}\left( \alpha -1\right) ^{2}}{%
2\left( \alpha ^{2}+1\right) }>0$. Therefore, for all $p\in \left( \frac{1}{2%
},\overline{p}\right) $, $E\pi _{U}+E\pi _{I}<x+y$, which means that the
mixed strategy equilibrium is ex-ante inefficient.

Then the above analysis can be summarized as follows:

\bigskip

\noindent \textbf{Result}: \newline
\textit{1) Suppose that duopoly profit is low relative to monopoly profit,
so }$\frac{x}{y}<\alpha <\frac{x+y}{2y}$\textit{. }\newline
\textit{\noindent 1-1) If }$p<\overline{p}$\textit{, both pure strategy
equilibria are efficient and the mixed strategy equilibrium is not. (A1 in
Figure 1) \newline
1-2) If }$\overline{p}<p$\textit{, all equilibria are efficient. (A2 in
Figure 1) \newline
\noindent 2) Suppose that duopoly profit is high relative to monopoly
profit, so }$\frac{x+y}{2y}<\alpha $\textit{. }\newline
\textit{2-1) If }$p<\overline{p}$\textit{, both pure strategy equilibria are
efficient but the mixed strategy equilibrium is not. (A5 in Figure 1) 
\newline
2-2) If }$\overline{p}<p<\frac{y+x-2x\alpha }{2\alpha (y-x)}$\textit{, all
equilibria are efficient. (A4 in Figure 1) \newline
2-3) If }$\frac{y+x-2x\alpha }{2\alpha (y-x)}<p$\textit{\ , all equilibria
are inefficient. (A3 in Figure 1) }

\bigskip

So we proved Proposition 3. $\square $

\subsection{Proof of Lemma 2}

Player U gains from a delay by being able to observe I's choice. Hence, $%
d_{U}=\pi _{u}^{F}-\pi _{u}^{L}=\pi _{u}^{F}-\pi _{u}^{S}$. Player I's gain
from a delay is from preventing U from observing his choice. Hence, $%
d_{I}=\pi _{I}^{F}-\pi _{I}^{L}=\pi _{I}^{S}-\pi _{I}^{L}$. Then,%
\begin{equation}
d_{U}-d_{I}=p\left( y\alpha -x\alpha \right) +x\alpha -\frac{1}{2}y-\frac{1}{%
2}x  \tag{A46}
\end{equation}%
So, $p\gtreqless \frac{\left( x+y-2x\alpha \right) }{2\left( y-x\right)
\alpha }\Longrightarrow $ $d_{U}\gtreqless d_{I}$. It is easily verified
that: $\frac{\left( x+y-2x\alpha \right) }{2\left( y-x\right) \alpha }>\frac{%
1}{2}$ and $\frac{\left( x+y-2x\alpha \right) }{2\left( y-x\right) \alpha }>%
\overline{p}\equiv \frac{y-x\alpha }{\left( y-x\right) \left( \alpha
+1\right) }$. However, $\frac{\left( x+y-2x\alpha \right) }{2\left(
y-x\right) \alpha }-1=-\frac{\left( 2y\alpha -y-x\right) }{2\left(
y-x\right) \alpha }$. Here, $2y\alpha -y-x$ is an increasing function of $%
\alpha $. Also, from the condition that $\alpha y>x$, we know $\frac{x}{y}%
<\alpha $. Then, $\left. 2y\alpha -y-x\right\vert _{\alpha =\frac{x}{y}}<0$
and $\left. 2y\alpha -y-x\right\vert _{\alpha =1}>0$. Hence, there exists $%
\alpha ^{\ast }=\frac{x+y}{2y}\in \left( \frac{x}{y},1\right) $ such that if 
$\alpha \in \left( \frac{x}{y},\alpha ^{\ast }\right) $, $\frac{\left(
x+y-2x\alpha \right) }{2\left( y-x\right) \alpha }>1$ and if $\alpha \in
\left( \alpha ^{\ast },1\right) $, $\frac{\left( x+y-2x\alpha \right) }{%
2\left( y-x\right) \alpha }<1$. Finally, if $\alpha \in \left( \frac{x}{y}%
,\alpha ^{\ast }\right) $, $d_{U}<d_{I}$ for all $p\in \left( \bar{p}%
,1\right) $. On the other hand, if $\alpha \in \left( \alpha ^{\ast
},1\right) $, for $p\in \left( \bar{p},\frac{\left( x+y-2x\alpha \right) }{%
2\left( y-x\right) \alpha }\right) $, $d_{U}<d_{I}$ and if $p\in \left( 
\frac{\left( x+y-2x\alpha \right) }{2\left( y-x\right) \alpha },1\right) $, $%
d_{U}>d_{I}$. $\square $

\subsection{Proof of Proposition 4}

Note that $\left. \pi _{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p<\bar{p}%
}=\left( x-px+py\right) $ and $\left. \pi _{I}\left( p\right) \right\vert
_{p>\bar{p}}=\frac{\left( x-px+py\right) \left( \alpha +1\right) }{2}$.
Then, a) Both $\left. \pi _{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p<\bar{p%
}}$ and $\left. \pi _{I}\left( p\right) \right\vert _{p>\bar{p}}$ are
monotone increasing functions, b) $\left. \pi
_{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p=\frac{1}{2}}=\frac{1}{2}\left(
x+y\right) $, c)\ $\left. \pi _{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p=%
\bar{p}}=\frac{\left( x+y\right) }{\left( \alpha +1\right) }$, d)\ $\left.
\pi _{I}\left( p\right) \right\vert _{p=\frac{y-x\alpha }{\left( y-x\right)
\left( \alpha +1\right) }}=\allowbreak \frac{1}{2}\left( x+y\right) $, e)\ $%
\left. \pi _{I}\left( p\right) \right\vert _{p=1}=\allowbreak \frac{1}{2}%
\left( \alpha +1\right) y$, f)\ $\left. \pi
_{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p=\bar{p}}>\left. \pi _{I}\left(
p\right) \right\vert _{p=\bar{p}}$, and g)\ $\left. \pi
_{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p=\bar{p}}-\left. \pi _{I}\left(
p\right) \right\vert _{p=1}=\allowbreak -\frac{\left( 2y\alpha -y-2x+y\alpha
^{2}\right) }{2\left( \alpha +1\right) }$. In g), if we let $A\left( \alpha
\right) \equiv $ $2y\alpha -y-2x+y\alpha ^{2}$, it is a convex function and
attains the Min value at $\alpha =-1$ where $A\left( \alpha =-1\right) <0$.
Also $A\left( \alpha =0\right) <0$ and $A\left( \alpha =1\right) >0$. Hence
there exists $\bar{\alpha}$ such that if $0<\alpha <\bar{\alpha}$, $\left.
\pi _{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p=\bar{p}}>\left. \pi
_{I}\left( p\right) \right\vert _{p=1}$ and if $\bar{\alpha}<\alpha <1$, $%
\left. \pi _{I}(t_{U}=t_{2},t_{I}=t_{1})\right\vert _{p=\bar{p}}<\left. \pi
_{I}\left( p\right) \right\vert _{p=1}$. Therefore, if $0<\alpha <\bar{\alpha%
}$, he pick $p=\overline{p}-\epsilon $ and if $\bar{\alpha}<\alpha <1$, he
picks $p=1-\epsilon $. Here, $\bar{\alpha}=\frac{1}{y}\left( \sqrt{2}\sqrt{%
xy+y^{2}}-y\right) $. Figure 2 represents this result. $\square $

\pagebreak

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