# Private Information Acquisition and Preemption: a Strategic Wald Problem

Guo Bai  PhD student, Department of Economics, University College London.
Address: Drayton House, 30 Gordon Street, London, WC1H 0AX. Email:
<EMAIL_ADDRESS>I am grateful to Martin Cripps, Konrad Mierendorff and
Deniz Kattwinkel for supervision and support.

This paper studies a dynamic information acquisition model with payoff
externalities. Two players can acquire costly information about an unknown
state before taking a safe or risky action. Both information and the action
taken are private. The first player to take the risky action has an advantage
but whether the risky action is profitable depends on the state. The players
face the tradeoff between being first and being right. In equilibrium, for
different priors, there exist three kinds of randomisation: when the players
are pessimistic, they enter the competition randomly; when the players are
less pessimistic, they acquire information and then randomly stop; when the
players are relatively optimistic, they randomly take an action without
acquiring information.

##  1 Introduction

Research and development (R&D) of a new technology is often competitive as
there sometimes exists a first-mover advantage. The first company to verify
the feasibility of the technology and conduct mass production generates more
sales than any late competitor. A company’s R&D hence affects its competitor’s
profit and vice versa. However, companies may not be able to see its
competitor’s breakthrough or their start of mass production. A company may
fall behind without even noticing it. Similarly, the company may not see its
competitor’s breakdown or their silent exit from competition. Then, the
company’s further R&D will lead to either a ‘win’ in a doomed-to-fail project
or a belated exit.

A similar story also applies to other economic or social activities. For
example, researchers compete for novel results. It takes time for a paper to
become public after the researchers find a profound result. Moreover, negative
results are often not reported, hence not observed. It is difficult for the
researchers to know other researchers’ private findings. However, when other
researchers publish their findings first, the novelty disappears. Other
researchers’ early publication will decrease the researchers’ credit received
from the similar findings. By the time the researchers observe the
publication, it is already too late, and the cost associated with the research
has been incurred already. Not being able to see other researchers’ progress
makes it harder for the researchers to evaluate the potential credit they
would get from the project.

Another example is market entry. If a company is the first to enter a new
market, they may capture a higher market share. But before entry, they need to
investigate whether the market is of high or low demand. This investigation
takes time and the preparation of entering a new market is also not
immediately observable. The company may decide to enter a new market with the
belief that they will be the first, but later realise that their competitor
has started the whole process earlier than themselves. Their competitor will
then be the pioneer and the company can only follow. At the time the company
observes the competing company’s decision, it is too late to reverse their own
decision. Because of the delay of observing the competitor’s action, entering
a new market is essentially a private action.

The objective of this paper is to understand the tradeoff between being first
and being right when information acquisition takes time and opponent’s action
taken is not immediately observable. I focus on understanding the players’
equilibrium strategy in the presence of payoff externalities while shutting
down the information externalities. The payoff externalities stem from the
first-mover advantage. One’s early success reduces the opponent’s payoff from
late success. The absence of information externalities is because of the
unobervable actions. There is no information spillover because they do not get
any direct or indirect information from each other.

To model the interactions described above, this paper studies a dynamic model
with private information acquisition and payoff externalities. Both players
can acquire costly information about an unknown payoff-relevant state before
taking an irreversible action. The unknown state is either high or low and is
constant over time. If the state is high, it is optimal for the player to take
the risky action, and if the state is low, it is optimal to take the safe
action. Competition is modelled by the payoff externality. The first player
who stops acquiring information and takes the risky action gets a higher
payoff in both states. For example, two companies competing in the R&D of a
new self-driving car technology. The unknown payoff-relevant state is the
feasibility of this technology. The R&D is abstracted as paying a fixed cost
to acquire information about the technology. The risky action is to start the
mass production of the self-driving cars and the safe action is to exit. The
first to start the mass production will launch the car early and consequently
gain a greater market share. Whether a company starts mass production is not
immediately observable.

Both players have access to conditionally independent signals from an
identical information source. These generate Poisson breakthroughs and
breakdowns. A breakthrough is good news that reveals the high state and a
breakdown is bad news that reveals the low state. I also assume that no news
is good news. In the self-driving car example, this information structure
describes the incremental development with potential breakthroughs and
breakdowns. The company makes incremental improvement, and at the same time,
they can come across conclusive evidence that proves or disproves the
feasibility of the technology.

Both the information acquisition and the action taken are assumed to be
private. This describes an anonymous information acquisition environment: a
player does not see her competitor’s signals or actions, but she is aware of
the competitor’s existence. This shuts down the information externality where
the player learns something from the competitor’s action and focuses on the
competition-generated payoff externality.

My setup is related to the bandit problem, the classical framework studying
R&D races, but is different in the following aspects. In the bandit problem
with the safe and the risky arms, pulling the safe arm is similar to taking
the safe action in this model, while pulling the risky arm is similar to
acquiring information. The payoff generated by the risky arm depends on the
nature of the risky arm, which is similar as the state in this model. The
difference is that in the bandit problem, the player pulling the risky arm
gets information about the arm from the payoff generated by the arm. A high
payoff from the risky arm indicates a good arm. In other words, the payoff is
the information. In contrast, in my model, I separate the information and the
payoff from an action. The breakthroughs, breakdowns, or the lack thereof,
only contain information about the state. If the players want to use the
information, they need to take an action. Acquiring information itself does
not give the player any payoff. Instead, it incurs a positive information
cost. The R&D is modelled as a costly activity that contains information about
the feasibility of the technology, but does not give the company direct
return. The company gets return from the R&D only if they take actions using
the information generated.

This separation of information and the payoffs allows for the discussions not
only on the timing to stop the R&D but also the timing to start the R&D. Take
the self-driving car competition example again. Suppose the feasibility of the
R&D is known, but the profitability is unknown, e.g. the demand for self-
driving cars is uncertain. The companies are interested in when or whether to
start the R&D. Doing R&D is only profitable if the demand is high. Early start
of the R&D means early success and higher profit. Then, companies can pay a
cost to do online surveys to better estimate the demand. Companies doing
online surveys is another example that can be modelled as the costly
information acquisition. The risky action in this setting is to start the R&D
on self-driving cars and the safe action is to abandon this self-driving car
project.

The main result in this paper shows that in equilibrium, players use random
stopping strategy if they acquire information. This is significantly different
from the single DM case. In the single DM case, the deterministic cutoff
strategy is the optimal one: to acquire information if the belief is in the
intermediate range and to take an action if the belief is extreme. In my model
with competition and first-mover advantage, for relatively pessimistic priors,
the players use the random stopping strategy if they acquire information. That
is, they acquire information up to some point and then randomly stop and take
the risky action. This is a result of the interaction of the learning motive
and the preemption motive. At the early stage of information acquisition,
players are uncertain about the state and hence have stronger learning
motives. In addition, since they are relatively pessimistic about the state,
they also believe that the probability of being preempted is low. The value
associated with taking the risky action is large if the state happens to be
high. At this point, acquiring information is a strategic complement. If the
opponent has not stopped and taken the risky action, then, the gain from
acquiring information is high and the player is willing to acquire more
information. The more information the opponent is expected to acquire, the
more information the player is willing to acquire. After the player has been
acquiring information for some time, she becomes more optimistic about the
state (because no news is good news). However, the value associated with
taking the risky action conditional on the high state becomes lower. This is
because the probability of the opponent taking the risky action increases. The
player then has a stronger preemption motive to stop acquiring information and
to preempt the opponent.

When the prior is relatively optimistic, there exists an equilibrium where the
players randomise between the safe and risky action immediately without
acquiring information. The existence of this equilibrium is a result of the
strong preemption motive and the weak learning motive. Players undercut the
time at which the other takes risky action until there is no room for further
preemption. However, the players are not optimistic enough to take the risky
action with probability one. With a positive probability of taking the safe
action, the player secures a zero payoff. At the same time, the opponent’s
learning is deterred. An interesting result is that even when the information
cost is zero, this equilibrium still exists. This is because in my model, the
total information cost consists of the exogenous information cost $c$ and an
endogenous information cost from being preempted. The endogenous information
cost is determined by the opponent’s strategy. When the exogenous information
cost vanishes, the endogenous information cost does not vanish and hence the
total information cost is still positive. This positive total information cost
gives rise to this equilibrium where the players take immediate action at time
zero without acquiring information.

For sufficiently (but not extremely) pessimistic prior, there exists an
asymmetric equilibrium where one player takes the safe action immediately and
the opponent acquires information. Information acquisition here has strategic
substitutes’ property where one player acquires information only if the
opponent does not. The intuition is when the prior is sufficiently
pessimistic, it takes longer for the belief to drift to the random stopping
point. The expected information cost is hence higher. The value associated
with taking the risky action must be high enough to compensate for the higher
information cost so that the player is willing to acquire information. When
the opponent drops out, the value associated with the risky action is the
highest and when the opponent participates in the competition, this value
decreases. In symmetric equilibrium, the player randomises between these two
roles where she mixes between the random stopping strategy and the immediate
safe action at time zero.

#### Related literature

I study strategic dynamic information acquisition in an optimal stopping
framework introduced by Wald (1945, 1947). The single decision maker’s optimal
stopping problem has been studied in both drift-diffusion models and Poisson
models. Fudenberg et al. (2018) study the relationship between decision time
and accuracy of the action in a modified drift-diffusion model. Ke and Villas-
Boas (2019) and Nikandrova and Pancs (2018) investigate how decision maker
optimally acquire information about two different alternatives. The former
considers a drift-diffusion model and the latter considers a Poisson model.
Che and Mierendorff (2019) and Mayskaya (2020) study how a decision maker
optimally choose the bias of the information source in models with Poisson
signals. My model adopts the Poisson signal structure and more importantly,
considers the strategic interaction between players. One way of thinking about
it is that instead of having the cost of information as an exogenous
parameter, it is now endogenous which depends on the opponent’s strategy.

The most closely related papers are Shahanaghi (2022), Ozdenoren et al. (2021)
and Bobtcheff et al. (2021). In the first two papers, the irreversible actions
are observable. In Bobtcheff et al. (2021), the safe action (‘exit’ in their
paper) is observable with a positive probability. I shut down the information
externalities generated by observable actions to focus on the payoff
externalities. Shahanaghi (2022) discusses a dynamic preemption model with
costless information where players have accuracy incentives and credibility
concerns. In equilibrium, players randomise between acquiring more information
and taking an action due to the credibility concern and the observability of
actions. Preemption motives propagate the ex post errors in the actions. This
is different from my paper where the randomisation in equilibrium is mainly
due to the preemption motive. The player in my model has no incentive to
randomise if they were playing alone, which is not the case in Shahanaghi
(2022). Ozdenoren et al. (2021) studies a discrete-time experimentation model
where players have incentives to preempt. It shows that preemption motives
caused by payoff externalities lead to less experimentation. With a discrete-
time setting, they do not have an equilibrium in random stopping. Bobtcheff et
al. (2021) investigates how publicity of actions affects the players’
equilibrium strategy and payoff in a preemption game. In their model, similar
to mine, the players have both the learning motive and the preemption motive.
The difference is that the risky action (‘investment’ in their model) is
always observable while the safe action (‘exit’ in their model) can be
private. They argue that private signals and potential private actions create
a winner’s curse where the first risky action taker believes the opponent may
have received a private breakdown and hence exited. To compensate, the player
acquires more information to be more certain about the project before taking
the risky action. The main difference between this paper and mine is the
unobservability of the irreversible action and the presence of breakthrough.
In my model, both the risky action, safe action and the signals are private.
Not being able to observe the risky action creates a stronger preemption
motive. The players in my model tend to acquire less information because of
this.

Another relevant strand of literature studies static information acquisition
before a game ( see Hellwig and Veldkamp (2009), Yang (2015), Han and
Sangiorgi (2018) and Denti (2019) ). Those papers discuss information
acquisition before a coordination game. When the players play a game with
strategic complements, information choices exhibit strategic complementarity
as well. In my model, information has the features of a strategic complement
but could also be a strategic substitute.

This paper is also related to strategic experimentation literature that model
similar R&D races as a bandit problem (see Bolton and Harris (1999), Keller et
al. (2005), and Keller and Rady (2010)). Besides the differences in the model
setup introduced earlier, there are three other differences. First, strategic
experimentation literature investigates the free-ride problem when information
is a public good. This is mainly due to the observability of the actions as
well as sometimes the signals themselves. In my model, information is private
and information externality does not exist. There is no free ride. Second,
there is no exploration-exploitation tradeoff in my model because the players
only take the action once and it is irreversible. The players must stop
acquiring information and take the risky action in order to exploit the
outcome generated by the risky arm. Third, strategic experimentation
literature studies the intensity of the experimentation. In my model, I assume
the players choose between to acquire information or not, but not the
intensity at each time instant. Since bang-bang solution is normally the
optimal strategy in bandit problems, this simplification appears to be
reasonable.

A distinct but related group of literature is about the equivalence between
static and dynamic information acquisition. Hebert and Woodford (2019) studies
a dynamic rational inattention model and shows that the belief dynamics
generated can resemble either diffusion processes or processes with large
jumps. Morris and Strack (2019) studies what kind of static models with costly
information acquisition has a sequential sampling foundation. In my model,
because of the preemption feature, dynamic information acquisition is
intrinsic as a player’s payoff depends on the order that they act.

##  2 The model

###  2.1 Model setup

There are two players $i\in\\{1,2\\}$. At the beginning of the game, an
unknown, fixed, payoff-relevant state $\omega\in\\{H,L\\}$ is drawn. At any
time $t\in[0,\infty)$, each player can take an irreversible action
$x\in\\{S(afe),R(isky)\\}$, or delay and acquire information about the state.
The irreversible action gives the player an one-off payoff at the moment she
takes the action. Action $S$ yields a payoff which is normalised to be $0$.
Action $R$ payoff depends on the state and whether the player is the first or
second to take $R$. In state $\omega$, the first player to take $R$ gets
$u_{\omega}\in\mathbb{R}$ and the second gets
$u_{\omega}-\bar{\triangle}_{\omega}.$ If the players take $R$ simultaneously,
the payoff is $u_{\omega}-\underline{\triangle}_{\omega}.$ At each time, if
the player delays her action and acquires information, she incurs a positive
information flow cost $c$ per unit of time. I assume no time discounting.
Payoffs satisfy the following two assumptions.

######  Assumption 1.

$\bar{\triangle}_{\omega}>\underline{\triangle}_{\omega}>0$ for
$\omega\in\\{H,L\\}$.

1 says that first $R$ taker gets a higher payoff than the second $R$ taker in
both states. If the players take $R$ simultaneously, the payoff is in between.

######  Assumption 2.

$u_{L}<0<u_{H}-\bar{\triangle}_{H}.$

2 says that $R$ yields a higher payoff than $S$ in state $H$ and a lower
payoff in state $L$. Furthermore, being the second to take $R$ in state $H$ is
still better than taking $S$. Players’ incentive to be the first is increasing
in the value of $\bar{\triangle}_{H}$. The difference between the first and
second $R$ taker payoff describes the intensity of the competition. 2
describes a gentle competition in the sense that being the second to take $R$
in state $H$ is not too bad. This assumption is dropped in Section 5.2, where
the competition is more intense as the second $R$ taker gets a lower payoff
than taking $S$.

Before taking the irreversible action, players have access to costly
information about the state. Information is modelled using Poisson signals. If
a player acquires information for a short time period $dt>0$, then, in state
$H$ ($L$, resp), she receives an $H$-state ($L$-state, resp) revealing signal
with rate $adt$ ($bdt$, resp). Player $i$’s belief $p_{t}^{i}$ is the
probability that the state is $H$. I assume the players have a common prior
$p_{0}$ and that they observe neither the opponent’s action nor their signals.
At each time $t$, players update their beliefs using Bayes’ rule. When a
player acquires information, her belief jumps to $0$ after receiving an
$L$-state revealing signal and jumps to $1$ after receiving an $H$-state
revealing signal. In the absence of the revealing signal, player $i$’s belief
evolves according to

$\displaystyle\frac{dp_{t}^{i}}{dt}=(b-a)p_{t}^{i}(1-p_{t}^{i}).$ (1)

In the following part of the paper, I assume $b>a>0$. In the absence of a
revealing signal, the player’s belief drifts up. This is the ‘no news is good
news’ environment.

###  2.2 Strategies and equilibrium

If a player receives an $H$-state ($L$-state, resp) revealing signal, it is
optimal to stop acquiring the signal and take $R$ ($S$, resp) regardless of
the opponent’s behaviour. Therefore, it is sufficient to describe players’
strategy conditional on no arrival of a revealing signal. A pure strategy
(defined below) specifies the time, $T^{i}$, at which the player stops and
which action, $x^{i}$, they take in the absence of the revealing signals.

######  Definition 1.

Player $i$’s pure strategy $s^{i}$ is defined as
$\left(T^{i},x^{i}\right)\in\mathbb{R}_{+}\times\\{R,S\\}.$ 111 The notation
$\mathbb{R}_{+}$ denotes the set of non-negative real numbers.

A mixed strategy (defined below) specifies the probability that the player
stops before time $t$ conditional on no revealing signal.

######  Definition 2.

Player $i$’s mixed strategy $\gamma^{i}$ is defined as two non-decreasing
measurable functions $\left(\rho^{i},\sigma^{i}\right)$ where
$\rho^{i}:\mathbb{R}_{+}\rightarrow\left[0,1\right]$ and
$\sigma^{i}:\mathbb{R}_{+}\rightarrow\left[0,1\right]$ satisfy
$\rho^{i}\left(t\right)+\sigma^{i}\left(t\right)\leq 1$ for $\forall
t\in\mathbb{R}_{+}$. The first element $\rho^{i}$ is the probability that
player $i$ stops and takes $R$ before or at time $t$ conditional on no
revealing signal. The second element $\sigma^{i}$ is the probability that
player $i$ stops and takes $S$ before or at time $t$ conditional on no
revealing signal.

Before defining the equilibrium, I first write down the player’s expected
payoff. Player $i$’s expected payoff from taking action $x$ at time $t$ in
state $\omega$,
$\mathbb{E}^{\gamma^{j}}\left[u_{\omega}^{x}\mid\omega,t\right],$ depends on
player $j$’s strategy $\gamma^{j}$. The randomness of the payoff from taking
$R$ comes from both player $j$’s strategy and the randomness of the signal.
For example, the opponent using a pure strategy $(0,R)$ induces degenerate
conditional distributions at each time $t$ such that player $i$’s action-$R$
payoff in state $\omega$ is $u_{\omega}-\bar{\triangle}_{\omega}$ with
probability one. Let

$U_{x}^{i,\gamma^{j}}\left(t\right):=\mathbb{E}_{\omega\mid
t}\mathbb{E}^{\gamma^{j}}\left[u_{\omega}^{x}\mid\omega,t\right]$

be player $i$’s expected payoff from taking action $x$ at time $t$, where the
expectation $\mathbb{E}_{\omega\mid t}$ is taken over the distribution of the
state given player $i$’s time $t$ belief. Then, player $i$’s payoff from
taking action $x^{i}$ at time $T^{i}$ is

$\displaystyle\intop_{0}^{T^{i}}\pi_{t}\left(p_{t}^{i}a\mathbb{E}^{\gamma^{j}}\left[u_{H}^{R}\mid
H,t\right]-c\right)dt+\pi_{T^{i}}U_{x^{i}}^{i,\gamma^{j}}\left(t\right)$ (2)

where $\pi_{t}=p_{0}e^{-at}+\left(1-p_{0}\right)e^{-bt}$ is the probability of
no revealing signal up to time $t$. Let
$U^{i,\gamma^{j}}\left(t\right):=\max_{x^{i}}U_{x^{i}}^{i,\gamma^{j}}\left(t\right)$
be player $i$’s payoff associated with taking the optimal irreversible action
$x^{i}$ at time $t$. The following defines the perfect Bayesian equilibrium in
pure strategies. The perfect Bayesian equilibrium in mixed strategies can be
defined in a similar manner.

######  Definition 3.

A perfect Bayesian equilibrium in pure strategies is a strategy profile
$\left(s^{i},s^{-i}\right)$ and beliefs
$\left(\left(p_{t}^{i}\right)_{t\in\left[0,T^{i}\right]},\left(p_{t}^{-i}\right)_{t\in\left[0,T^{-i}\right]}\right)$
such that for $i\in\left\\{1,2\right\\}$,

  1. 1.

$x^{i}\in\arg\max_{\tilde{x}^{i}}U_{\tilde{x}^{i}}^{i,s^{-i}}\left(T^{i}\right);$

  2. 2.

$T^{i}\in\arg\max_{\tilde{T}^{i}}\intop_{0}^{\tilde{T}^{i}}\pi_{t}\left(p_{t}^{i}a\mathbb{E}^{s^{-i}}\left[u_{H}^{R}\mid
H,t\right]-c\right)dt+\pi_{\tilde{T}^{i}}U^{i,s^{-i}}\left(\tilde{T}^{i}\right)$
;

  3. 3.

The belief $p_{t}^{i}$ evolves according to (1).

I seek symmetric equilibria in both pure and mixed strategies. In the
following discussion, the superscript $i$ representing the player is ignored.

The player’s strategy and the random arrival of the signal jointly determine
the distribution over the action taken and timing. Before proceeding, I define
the following two conditional probabilities that characterise the distribution
over action taken and timing. Let $F_{\omega}\left(t\right)$ (
$G_{\omega}\left(t\right)$, resp ) be the probability that the player takes
$R$ ( $S,$ resp ) before or at time $t$ in state $\omega$. Then,
$1-F_{\omega}\left(t\right)-G_{\omega}\left(t\right)\geq 0$ is the probability
that player $i$ continues acquiring information at time $t$ in state $\omega$.

##  3 An illustrative example

In this section, I use a simple two-period model to illustrate the tradeoff
between information acquisition and preemption. I show how players’ learning
motives and preemption motives depend on the opponent’s strategy and the
prior. In equilibrium, mixed strategies create endogenous randomnesses that
either deter learning or prevent the opponent from preemption.

At time $t=0$, players can choose to acquire a signal at cost $c>0$ or to take
one of the actions. At time $t=1$, players have to take one of the actions. If
a player acquires a signal, in state $H$ ($L$, resp), a revealing signal
arrives with probability $a$ ($b$, resp), and the belief $p_{1}$ jumps to $1$
($0$, resp). If she does not receive the revealing signal, then, her belief is
updated to $\frac{p_{1}}{1-p_{1}}=\frac{1-a}{1-b}\frac{p_{0}}{1-p_{0}}$.
Acquiring a signal at time $0$ allows the player to learn the state and hence
take the ‘correct’ action ($R$ in state $H$ and $S$ in state $L$). This gives
the player the ‘learning motive’. Not acquiring a signal, however, secures the
player the first prize. This gives the player the ‘preemption motive’.

The learning motive is stronger when the prior is in the intermediate range
and when the opponent takes $S$ at time $0$. When the prior is in the
intermediate range, the player is uncertain and hence has stronger incentives
to learn. When the opponent takes $S$ at time $0$, the player is the single
decision maker in this game. The payoff associated with taking the correct
action is the highest and hence the value associated with information is
higher at each prior. The preemption motive is stronger when the opponent
acquires a signal at time $0$. This is because when the opponent acquires a
signal, by taking $R$ at time $0$, the player can secure herself the first $R$
taker payoff, while acquiring a signal at time $0$ decreases the probability
of being the first $R$ taker and hence decreases her expected payoff from
taking $R$. The information becomes less valuable. When the opponent takes $R$
at time $0$, the player’s incentive to preempt is less strong. It is then
optimal for the player to acquire a signal for a larger range of priors. This
is because she cannot preempt her opponent only to match their action. This
matching reduces the payoff from acting at time $0$ and correspondingly
increases the payoff of waiting.

Figure 1:  The player’s payoffs given opponent’s different strategies Notes:
This graph is drawn given the following parameter values:
$u_{H}=1$,$u_{L}=-1$, $\bar{\triangle}_{\omega}=0.7$,
$\underline{\triangle}_{\omega}=0.5$, $a=0.6$, $b=0.8$, $c=0.025$. In all
three panels, the red line represents the player’s payoff from taking $R$ at
time $0$, the blue line represents the player’s payoff from acquiring a signal
at time $0$, and the yellow line represents the player’s payoff from taking
$S$ at time $0$.

Figure 1 plots her payoffs associated with taking immediate actions and
acquiring a signal: Panel (a) is the player’s payoffs when the opponent takes
$S$ at time $0$, (b) is when the opponent takes $R$ at time $0$ and (c) is
when the opponent acquires a signal at time $0$. For the opponent’s different
strategies, qualitatively, the player’s best responses have similar
properties: to take an action at time $0$ when the prior belief is extreme and
to acquire a signal when the prior belief is in the intermediate range. The
difference is the range of the priors at which her best response is to acquire
a signal. This range of priors is largest when the opponent takes $S$ at time
$0$ and is smallest when the opponent acquires a signal.

Next, I briefly discuss the possible equilibria: the equilibrium where players
preempt $R$, the equilibrium where players acquire a signal with positive
probability, and the equilibrium where players randomise between two actions
without acquiring a signal.

When the prior is sufficiently high, the preemption motive dominates and hence
there is unravelling. The symmetric equilibrium is such that both players take
immediate $R$ at time $0$ without acquiring any signal. The weak learning
motive and the strong preemption motive work in the same direction which
pushes the players to take $R$ immediately. For intermediate priors, there
exists an asymmetric equilibrium where one player acquires a signal and the
opponent takes $R$ at time $0$. In symmetric equilibrium, the players
randomise over these two roles. This is referred to as a random stopping
strategy: the player randomises between stopping (to take $R$) at time $0$ and
time $1$. By using this strategy, the player creates an endogenous uncertainty
that prevents the opponent from preempting. When a player acquires a signal
with probability one, her opponent has incentives to preempt and it is easy
for them to do so. But when a player randomises, not only the value associated
with preemption is reduced, it is also harder for her opponent to preempt
because of the endogenous randomness. For some intermediate priors, there
exists another symmetric equilibrium where players randomise between two
immediate actions without acquiring any signal. This kind of randomisation
deters learning. For those intermediate priors, players are uncertain about
the state. By simply randomising between $R$ and $S$ at time $0$, the player
‘hedges’ against the uncertainty without paying extra information cost and at
the same time, reduces her opponent’s value associated with learning. 222The
player taking $R$ at time $0$ with a positive probability reduces her
opponent’s expected payoff from taking $R$ at time $1$. When this value is
sufficiently low, the opponent’s learning is deterred.

This two-period example shows the most important tradeoff in the model: the
incentive to learn and the fear of being preempted. However, it can only
discuss whether or not the players acquire information, but not how much
information they get. To understand the optimal quantity of information the
players acquire before taking an action, the dynamic model with multiple
periods is of interest. Next, I analyse the dynamic model introduced in
Section 2.

##  4 Equilibrium analysis

###  4.1 Single decision maker benchmark

Before getting into the detail of the game, as a benchmark, I first consider
the model with one single decision maker (DM). This single DM model is a well-
studied sequential sampling model due to Wald (1945, 1947) where the DM
chooses an optimal stopping time based on the samples she has observed.

The optimal stopping rule depends on the cost of information and the belief.
If the cost of information is too high, then, it is optimal to take an
immediate action based on her prior. In this case, learning about the unknown
state does not give the DM sufficiently high benefit to compensate for the
high cost. When the cost is sufficiently small, it is optimal to acquire
information for intermediate beliefs and to take an immediate action for
extreme beliefs. The DM’s optimal policy at each time $t$ only depends on the
current belief $p_{t}$ but not time $t$ itself. This is because all the past
information is summarised by the belief at time $t$ and the information cost
in the past is sunk. The DM acquires information if the marginal cost $c$ is
smaller than the marginal benefit. The marginal benefit is higher when the DM
is uncertain about the state. Hence, it is optimal to take $S$ if the belief
is sufficiently small, to acquire information if the belief is in an
intermediate range, and to take $R$ if the belief is sufficiently big. This is
summarised in the following proposition.

######  Proposition 1.

There exist cutoffs $\underline{p}<\bar{p}$ and $\bar{c}$ such that when
$c\leq\bar{c}$, given belief $p$, the DM’s optimal policy is to take $S$ if
$p\leq\underline{p}$; to acquire information if $\underline{p}<p<\bar{p}$; and
to take $R$ if $p\geq\bar{p}$.

In the single DM case, the optimal strategy is deterministic: the cutoff where
the DM stops acquiring information is a constant and is uniquely pinned down
by the parameter values. When she stops at the cutoff, the optimal action is
$R$ because she is convinced that the state is more likely to be $H$. When the
DM uses the optimal strategy, if she acquires information, she only makes
mistakes in the low state: she correctly takes $R$ in high state with
probability one and incorrectly takes $R$ in the low state with a positive
probability. This is because when acquiring information, she either receives a
revealing signal and then stop or she acquires information until the belief
drifts up to the upperbound $\bar{p}$. In the high state, the DM takes the
correct action $R$ in both events. In the low state, she takes the incorrect
risky action if she stops at the upperbound $\bar{p}$.

###  4.2 Properties of equilibrium

Now I turn to the model with two players. I establish the properties of the
equilibrium strategy. I explain why there are no jumps in the player’s
equilibrium strategy at all $t>0$. If there is a jump in the player’s
strategy, it only happens at $t=0$. This property is driven by preemption and
the unobservability of actions.

######  Lemma 1.

In equilibrium, $\rho\left(t\right)$, the probability that the player stops
and takes $R$ before or at time $t$ conditional on no revealing signal, is
continuous at all $t>0$.

In the absence of the revealing signal, the player does not take $R$ with
positive probability mass at any time $t>0$. This implies that the single DM’s
deterministic optimal strategy is not part of the equilibrium in the two-
player game. The reason is twofold. First, the jump at any $t>0$ in the
opponent’s strategy gives the player an opportunity to preempt and hence a
profitable deviation. If the opponent takes $R$ at some $\tau>0$ with a
positive probability, then, the player is betteroff stopping at $\tau-dt$
where $dt>0$ is infinitesimal. At time $\tau-dt$, if the player continues
acquiring information for $dt$ longer, she gains nothing but loses the
probability of being the first $R$ taker. Second, the private action and the
private breakdown give rise to winner’s curse. It is more likely for the
player to be the first $R$ taker in state $L$ than in state $H$ because the
opponent might have received the breakdown and dropped out. In other words, it
is more likely to ‘win’ in state $L$. If the opponent takes $R$ at some
$\tau>0$ with a positive probability, then, conditional on no revealing signal
and ‘winning’, the value associated with taking $R$ at time $\tau$ has a
downward jump. It is then not a best response for the player to take $R$ at
time $\tau$.

In equilibrium, $\rho$ can have a jump at $t=0$. This means that in the
absence of the revealing signal, the player takes $R$ with positive
probability mass only at time zero. This is a result of preemption. The
unravelling comes from the player undercutting the time of the opponent taking
$R$ until time zero at which there is no more room for preemption.

###  4.3 Equilibrium strategies

This section introduces five strategies that appear in equilibrium.

#### The Immediate Action Strategy

The Immediate $R$ (Immediate $S$, resp) Strategy is the pure strategy
$\left(0,R\right)$ ($\left(0,S\right)$, resp) where the player does not
acquire information and takes $R$ ($S$, resp) at time $0$.

#### The Immediate Mix Strategy

This is a strategy where the player does not acquire information. She randomly
takes $R$ and $S$ at time zero. The following is the formal definition. The
Immediate Mix Strategy is a mixed strategy
$\left(\rho^{IM},\sigma^{IM}\right)$ where
$\rho^{IM}\left(t\right)=\rho^{IM}\left(0\right)\in(0,1)$ for $\forall t>0,$
$\sigma^{IM}\left(t\right)=\sigma^{IM}\left(0\right)\in(0,1)$ for $\forall
t>0,$ and $\rho^{IM}\left(0\right)+\sigma^{IM}\left(0\right)=1.$

#### The Random Stopping Strategy

This is a strategy where the player acquires information for a period of time
and then stops with a positive rate at each time. The action associated with
stopping is $R$. The following is the definition. The Random Stopping Strategy
is a mixed strategy $\left(\rho^{RS},\sigma^{RS}\right)$ such that
$\rho^{RS}\left(\cdot\right),$ the probability of taking $R$ in the absence of
the revealing signal, weakly increases and $\sigma^{RS},$ the probability of
taking $S$ in the absence of the revealing signal, equals zero for all $t\geq
0$. The first element $\rho^{RS}\left(\cdot\right)$ satisfies
$\rho^{RS}\left(0\right)=0$ and $\rho^{RS}\left(t\right)=1$ for
$t\geq\bar{T}^{RS}$ where $\bar{T}^{RS}>0$. When using this strategy,
conditional on arriving at time $t$ 333That is, the player receives no
revealing signal and the player’s strategy does not prescribe stopping. , the
player stops and takes action $R$ with rate
$\frac{\frac{d\rho^{j}\left(t\right)}{dt}}{1-\rho^{j}\left(t\right)}.$ She
only stops and takes action $S$ after observing a breakdown.

#### The Mixed Learning Strategy

This is a strategy where the player mixes between the Immediate $S$ Strategy
and the Random Stopping Strategy at time $0$. The following is the definition.
The Mixed Learning Strategy is a mixed strategy
$\left(\rho^{ML},\sigma^{ML}\right)$ such that $\rho^{ML}\left(\cdot\right)$
weakly increases and $\sigma^{ML}\left(t\right)=\beta\in\left(0,1\right)$ for
$\forall t$. The probability that the player stops and takes $R$ before time
$t$ (i.e. $\rho^{ML}\left(\cdot\right)$ ) satisfies
$\rho^{ML}\left(0\right)=0$ and $\rho^{ML}\left(t\right)=1-\beta$ for
$t\geq\bar{T}^{ML}$ where $\bar{T}^{ML}>0$.

###  4.4 Main results

The theorem below lists the symmetric equilibria given different priors. Let
$\frac{p^{M}}{1-p^{M}}:=\frac{-\left(u_{L}-\underline{\triangle}_{L}\right)}{u_{H}-\underline{\triangle}_{H}}$,
$\frac{p^{L}}{1-p^{L}}:=\frac{-u_{L}}{u_{H}}$ and
$\frac{\tilde{p}}{1-\tilde{p}}:=\frac{-bu_{L}-c}{a\bar{\triangle}_{H}+c}$ be
three prior cutoffs. Both $p^{M}$ and $p^{L}$ are positive as $u_{L}$ is
negative. A sufficiently small information cost $c$ guarantees that
$p^{M}<\tilde{p}$

######  Theorem 1.

When $b<2a$ and $c$ is sufficiently small, there exist cutoffs
$\underline{p}<p^{*}<p^{L}<p^{M}<\tilde{p}$ such that:

If $p_{0}\leq\underline{p}$, there exists an equilibrium where both players
use the Immediate $S$ Strategy;

If $\underline{p}<p_{0}<p^{*}$, there exists an equilibrium where both players
use the Mixed Learning Strategy;

If $p^{*}\leq p_{0}<\tilde{p}$, there exists an equilibrium where both players
use the Random Stopping Strategy;

If $p^{L}<p_{0}<p^{M}$, there exists an equilibrium where both players use the
Immediate Mix Strategy;

If $p_{0}\geq p^{M}$, there exists an equilibrium where both players use the
Immediate $R$ Strategy.

The lowest cutoff $\underline{p}$ is the lowerbound in Proposition 1, below
which the single DM’s optimal strategy is to take $S$ immediately. The cutoff
$p^{*}$ is a fixed point and the detail can be found in Appendix D. The cutoff
$p^{L}$ ($p^{M}$, resp) is the belief at which the player is indifferent
between taking $S$ and taking immediate $R$ if she is the first $R$ taker (if
she and the opponent take $R$ simultaneously, resp).

When the prior is extreme, both players use the Immediate $R$/$S$ Strategy in
equilibrium. It seems to be intuitive as the players’ learning motive is weak
due to the relatively small value of information. But this is not the whole
story. The subtlety is the role of the preemption motive. When the prior is
sufficiently small, the preemption motive is absent because $S$ is the optimal
action that gives the player the same payoff. The weak learning motive is
indeed the only reason why the players take immediate $S$. However, when the
prior is high, the preemption motive is the main reason why the players take
immediate $R$ with a positive probability. To understand the existence of the
equilibria where the players use the immediate $R$ or the immediate mix
strategy, consider a static game where the players do not have access to
information 444This is a game where both players have two actions: action $R$
and action $S$. In state $\omega$, if both players take action $R$ (action
$S$, resp), the payoff is $u_{\omega}-\underline{\triangle}_{\omega}$ ($0$,
resp) for each of them. If one player takes action $R$ and the other player
takes action $S$, the player who takes action $R$ gets a payoff of
$u_{\omega}$ and the player who takes action $S$ gets a payoff of $0$..
Consider the sufficiently high prior such that there exists an equilibrium
where the players take $R$ with a positive probability in this static game
555That, when the belief is higher than $p^{L}$. . Then, even if the players
are now given access to costly information, it is still optimal to take an
immediate action. This is because when the prior is sufficiently high, the
preemption motive is strong and the learning motive is weak. In the event that
the opponent takes immediate $R$, if the player also takes immediate $R$, she
then gets the simultaneous-move payoff. If the player acquires information for
an infinitesimal time period $dt$, then, the loss from becoming a second $R$
taker is strictly positive but the gain from acquiring information is
negligible. The positive probability of the opponent taking immediate $R$
induces a downward jump in the player’s expected payoff from taking $R$. This
deters learning.

######  Lemma 2.

When $c$ is sufficiently small, $p^{M}<\bar{p}$.

The cutoff $\bar{p}$ is the upperbound in Proposition 1: the belief above
which the player takes $R$ immediately. Lemma 2 says that the range of priors
where the players take immediate $R$ is larger when the players have the
incentive to preempt. The preemption motive leads to less information
acquisition and more immediate risky action for relatively optimistic prior
beliefs.

The intuition of the equilibrium where the players use the Random Stopping
Strategy can be explained by the tradeoff between learning and the preemption
motive. When the prior is in the lower intermediate range, the player has the
learning motive because the value of information is high. If a player were
playing this game alone, the optimal strategy is to acquire information until
the belief drifts up to an upperbound and then takes $R$ (see Proposition 1).
Before she stops acquiring information, the distribution of the time at which
she takes $R$ is the distribution of the Poisson breakthrough, which is
continuous. At the time the belief drifts up to the upperbound, according to
the strategy, the player stops and takes $R$. Therefore, the induced
distribution of the time at which she takes $R$ has an atom at the time when
the belief drifts up to the upperbound. If player $i$ were playing against
player $j$ who uses this strategy, player $i$ would preempt player $j$. To
avoid being preempted by the opponent, what the player could do is to
randomise at which time she stops and takes action $R$. This randomisation
will make the opponent’s marginal cost and marginal benefit (explain later)
from acquiring information the same and hence eliminate the incentive to
preempt. The indifference condition gives rise to an ODE that the equilibrium
strategy satisfies which is concluded in Lemma 6 in the appendix. Here in the
main text, I give expressions of the marginal cost and marginal benefit from
acquiring information and show how they are determined by the opponent’s
strategy. Suppose player $j$ uses the mixed strategy
$\left(\rho,\sigma\right)$ defined in Definition 2. Suppose $\rho$ and
$\sigma$ are differentiable. Then,

$\displaystyle F_{H}\left(t\right)$
$\displaystyle=\intop_{0}^{t}\left[e^{-as}\left(1-\rho\left(s\right)\right)\left(a+\frac{\frac{d\rho\left(s\right)}{ds}}{1-\rho\left(s\right)}\right)\right]ds$
$\displaystyle=1-e^{-at}\left(1-\rho\left(t\right)\right)$

is the probability that player $j$ takes $R$ before time $t$ in state $H$ and

$\displaystyle F_{L}\left(t\right)$
$\displaystyle=\intop_{0}^{t}\left[e^{-bs}\left(1-\rho\left(s\right)\right)\frac{\frac{d\rho\left(s\right)}{ds}}{1-\rho\left(s\right)}\right]ds$
$\displaystyle=\intop_{0}^{t}\left[e^{-bs}\frac{d\rho\left(s\right)}{ds}\right]ds.$

is the probability that player $j$ takes $R$ before time $t$ in state $L$. At
time $t$, player $i$ is indifferent between taking $R$ now and $dt$ later if

$\displaystyle
c+p_{t}\frac{dF_{H}\left(t\right)}{dt}\bar{\triangle}_{H}+\left(1-p_{t}\right)\frac{dF_{L}\left(t\right)}{dt}\bar{\triangle}_{L}=\left(1-p_{t}\right)b\left[-u_{L}+F_{L}\left(t\right)\bar{\triangle}_{L}\right].$
(3)

The left-hand side of (3) is the marginal cost of acquiring information for
$dt$ longer at time $t$ and the right-hand side is the marginal benefit. The
cost consists of the direct information cost $c$ and the indirect cost from
being preempted. Intuitively, in state $\omega$, the opponent preempts the
player in this short time period $dt$ with probability
$\frac{dF_{\omega}\left(t\right)}{dt}.$ The benefits of acquiring information
comes from the breakdown because it corrects the player’s action from $R$ to
$S$ in state $L$. The player gets the payoff of zero instead of the negative
payoff associated with $R$ in state $L$. This indifference condition gives an
ODE that the equilibrium strategy satisfies. Lemma 6 and Lemma 14 in Appendix
D characterises the conditions that the equilibrium strategy satisfies.

######  Corollary 1.

When $c$ is sufficiently small, there exist multiple equilibria if
$\underline{p}<p_{0}<p^{M}$.

When $\underline{p}<p_{0}<\tilde{p}$, in the equilibrium where the players use
the Random Stopping Strategy, the players both acquire information up to some
time $\hat{T}$ and then randomly stop at each time greater than $\hat{T}$. The
time $\hat{T}$ at which the player starts the randomisation is not unique and
this leads to multiple equilibria. To get the intuition, suppose player $j$
uses the Random Stopping Strategy such that she starts the random stopping at
some time $\hat{T}>0$. To have both players using this strategy as an
equilibrium, before time $\hat{T}$, player $i$ must strictly prefer to acquire
information given that the opponent acquires information. That is, given the
opponent is still acquiring information, the marginal cost from acquiring
information for $dt$ longer must be smaller than the marginal benefit. As a
result, the latest time instant at which the player starts randomising is the
first time instant at which the marginal cost of acquiring information equals
the marginal benefit given that the opponent acquires information. Before this
time point, if the opponent acquires the information, the marginal benefit of
acquiring information exceeds the cost. The player acquires information. After
this time point, even though the player knows the opponent is still acquiring
information, the value associated with information is too small to keep the
player engaged. After time $\hat{T}$, player $i$ is indifferent between taking
$R$ at each time $t>\hat{T}$, and she must prefer to randomise instead of
taking $S$. Therefore, the earliest time instant at which the player is
willing to start the random stopping is the earliest time instant such that
taking $R$ gives her a payoff of at least zero. Before this time instant, the
player is not optimistic enough about the state to take $R$. After this time
instant, the player matches the opponent’s action. If the opponent’s strategy
is to randomise, the player also randomises.

The intuition behind the multiplicity is the strategic complementarity of
information. Information is a strategic complement at time $t<\hat{T}$ because
when the opponent is still acquiring information, the value of information is
relatively high. The player hence is also willing to acquire more information.
If the opponent starts randomising, then, the value associated with acquiring
information becomes lower. The player then matches the time at which the
opponent starts randomising. This strategic complementarity also explains the
existence of multiple equilibria when $p_{0}\in\left[p^{L},p^{M}\right)$. For
this range of priors, according to Theorem 1, there exist both kinds of
equilibria where the players use the Random Stopping Strategy and where the
players use the Immediate Mix Strategy. In other words, there simultaneously
exist the equilibria where the players acquire information and where the
players do not acquire information.

The existence of the equilibrium where both players use the Random Stopping
Strategy requires the player to start acquiring information at time $0$. If
the prior is relatively low, then, the expected duration of acquiring
information before stopping is long and the cost of delay is high. In
addition, the players are pessimistic about the state being high. Therefore,
the players do not have strong incentives to start acquiring information and
hence this symmetric equilibrium where both players acquire information do not
exist. However, there exists asymmetric equilibrium where one player acquires
information and the other takes the immediate safe action. This is because
when the player faces a longer potential delay of action, the expected benefit
associated with taking $R$ must be higher to compensate for the expected
information cost. The benefit associated with taking $R$ depends on the
opponent’s strategy. If the opponent’s takes $S$ at time $0$, then, the
player’s gain from information is higher. This higher gain can incentivise the
player to start learning at time $0$. However, if the opponent acquires
information, then, the player gains less from taking $R$ because there is a
positive probability that she can only get the lower payoff. This gain may not
be sufficient to incentivise the player to start acquiring information at time
$0$. In this case, information is a strategic substitute. In the symmetric
equilibrium, the players mix between two strategies at time $0$: the Immediate
$R$ Strategy and the Random Stopping Strategy. Compared to the case that the
opponent uses Random Stopping Strategy, the opponent takes $R$ with a lower
probability. This in turn increases the player’s gain from taking $R$ and
hence increases the gain from acquiring information. Then, the player is
incentivised to start learning at a lower prior.

##  5 Information cost and the competition

In this section, I discuss the effects of a vanishing information cost $c$ and
the effects of the intensity of competition.

###  5.1 Vanishing information cost $c$

When the information cost vanishes, the players acquire information for a
larger range of priors but not for all the priors. In the equilibrium where
the players use the random stopping strategy, they acquire information for a
longer duration before randomisation. When they randomise, they stop and take
$R$ at a higher rate.

The cost of acquiring information consists of two parts: the exogenous
information cost $c$ and the endogenous cost from being preempted. This is the
main difference between the two-player case and the single DM case. In the
single DM case, since the exogenous information cost $c$ is the total cost of
acquiring information, when $c$ vanishes, the DM always acquires information
until she is certain. In contrast, in the two-player game, the total
information cost does not vanish with the exogenous cost $c$. When the prior
is high, the player believes that the state is very likely to be $H$ and
infers that the probability of being preempted is high. Thus, the endogenous
cost is high for optimistic priors even when $c$ vanishes. As a result, when
$c$ vanishes, the equilibrium such that both players acquire information
exists for a larger range of priors, but does not exist for extremely high
priors. However, when the prior is low, the player is pessimistic about the
state and hence she believes that it is not likely that the opponent takes
$R$. In addition, for low priors, if the player does not acquire information,
the optimal action is to take $S$, the payoff of which does not depend on the
opponent’s action. Thus, the endogenous information cost is low when the prior
is close to zero. As a result, for low priors, the total information cost
vanishes with $c$ and the player is willing to acquire information for low
priors. This property is summarised in the following proposition.

######  Proposition 2.

The cutoff $\tilde{p}$ decreases in $c$. When $c\rightarrow 0$,
$\underline{p}\rightarrow 0$ and
$\tilde{p}\rightarrow\frac{\frac{b}{a}\frac{-u_{L}}{\bar{\triangle}_{H}}}{1+\frac{b}{a}\frac{-u_{L}}{\bar{\triangle}_{H}}}\in\left(0,1\right)$.

For the equilibrium where the players use the random stopping strategy, when
$c$ vanishes, the effects are twofold. First, the players acquire information
for a longer duration before they start the randomisation stage. Second, when
the players just enter the randomisation stage, they stop and take $R$ at a
higher rate. The first is intuitive as when $c$ decreases, information becomes
cheaper and hence players are willing to acquire information for a longer
period. The intuition for the second effect is related to the indifference
condition when the players randomise. In equilibrium, the players randomise
between acquiring information and taking $R$ because the marginal cost and
marginal benefit from acquiring information are the same. The marginal cost
from acquiring information consists of the exogenous and the endogenous
information cost. When the player just enters the randomisation stage, as
shown in (3), the marginal benefit is not affected by the exogenous
information cost $c$ and the players’ stopping rate. The marginal benefit
depends only on the cumulative density of the opponent taking $R$. According
the equilibrium strategy, since the opponent has not started the random
stopping yet, the marginal benefit only depends on the arrival of the
revealing signals but not player’s stopping rate. The endogenous information
cost however, increases in the opponent’s stopping rate. If the players stop
and take $R$ at a higher rate, the endogenous information cost is higher. As a
result, when the players just enters the randomisation stage, if the exogenous
information cost $c$ vanishes, the marginal benefit is unchanged while the
marginal cost decreases because of the vanishing information cost $c$. To have
the indifference condition hold, the players stop and take $R$ at a higher
rate to increase the endogenous information cost.

###  5.2 Competition

The competition in this game comes from the payoff difference between the
first and second $R$ taker. The intensity of the competition increases in the
value of the payoff difference. In all the previous discussion, the payoff
difference satisfies 2. That is, the second $R$ taker gets a higher payoff
than taking $S$ in state $H$. With this assumption, after the player learns
the state is high, taking $R$ is a dominant strategy. To understand the effect
of an intense competition, I drop 2 and impose 3.

######  Assumption 3.

$u_{L}<u_{H}-\bar{\triangle}_{H}<0$.

3 says that in state $H$, the payoff of the second $R$ taker is smaller than
the safe action payoff. This assumption explicitly imposes a ‘loser gets
punished’ condition 666Note that given 2, the player may eventually be
punished due to the total information cost. The difference here is that the
player explicitly knows that the second prize from taking $R$ is worse than
taking $S$.. Given 3, whether the player takes $R$ after learning the state is
$H$ depends on her belief on the opponent’s strategy. Take an extreme case as
an example. If the player believes that the opponent takes $R$ immediately,
then, taking $S$ is the optimal action regardless of the state. Information
becomes worthless because learning the state does not change the player’s
action and $S$ always gives the player a constant payoff. Then it is never
optimal for the player to acquire information. This deters information
acquisition completely.

In a less extreme case, information acquisition is not necessarily deterred at
the beginning. Instead, there exists an upperbound on the duration of
information acquisition. Suppose the opponent uses some well-behaved strategy
such that she acquires information at time $0$ with probability one. Then, the
player’s expected payoff from taking $R$ conditional on state $H$ is the
highest at time 0 and then decreases. This is because the player gets the
first $R$ taker payoff for certain at time $0$ and as time passes, the
probability of the opponent taking $R$ increases. Since the second $R$ prize
is negative, there exists a time instant at which the player’s expected payoff
from taking $R$ conditional on state $H$ decreases to zero. Then, despite
learning the state or not, at and after this time instant, the player’s
optimal action is $S$. The player hence gains nothing from acquiring
information after this time instant. However, this fact that information
becomes worthless after certain time instant does not deter information
acquisition at the beginning. When the player just starts acquiring
information, the probability of being preempted is low. She puts a higher
weight on the first prize conditional on state $H$. The negative second prize
does not matter too much. Information acquisition is not deterred as long as
the expected payoff from taking $R$ conditional on state $H$ is positive. Once
at the time instant such that $R$ gives the player an expected payoff of zero
in state $H$, information acquisition stops and $S$ will be taken. This puts
an upperbound on the duration of information acquisition. Since the player
takes the safe action that creates no preemption motive when stopping, there
is no random stopping in equilibrium. As a result, there exists a pure
strategy equilibrium where in the absence of the revealing signal, the players
acquire information up to some time and then take the safe action. This result
is summarised in the following proposition.

######  Proposition 3.

Suppose the payoff satisfies 1 and 3. There exists a prior cutoff $p^{NR}$ and
a non-negative upperbound
$T^{PS}=\frac{1}{a}\log\frac{\bar{\triangle}_{H}}{-\left(u_{H}-\bar{\triangle}_{H}\right)}$
such that for the prior $p_{0}\in\left(p^{NR},\tilde{p}\right),$ there exists
a symmetric equilibrium where the players use the pure strategy
$\left(T^{PS},S\right).$

This proposition says that when the second prize for $R$ is negative, for
intermediate priors, there exists an equilibrium where the players acquires
information for a maximum of $T^{PS}$ time. Before time $T^{PS}$, the players
stop and take an action only after receiving the revealing signal. The maximum
time $T^{PS}$ is independent of the prior.

The existence of such pure strategy equilibrium is the main difference between
the cases with the positive and negative second prize in state $H$. The
intuition is that when the second $R$ prize in state $H$ is negative, the
longer the player acquires information, the less attractive $R$ becomes. When
the player becomes sufficiently certain that the state is $H$, she is also
convinced that $R$ has been taken. At the end of the information acquisition
stage, being sufficiently optimistic is associated with taking $S$. However,
in the positive second prize case, being sufficiently optimistic is associated
with taking $R$, which generates the preemption motive and hence the
equilibrium in the random stopping strategy. In equilibrium, the maximum
duration $T^{PS}$ is independent of the prior. Given that the opponent takes
$R$ after receiving the $H$-state revealing signal, $T^{PS}$ is the time at
which the expected payoff from taking $R$ conditional on state $H$ decreases
to zero. Since $T^{PS}$ is pinned down by the payoff in state $H$, it is
independent of the prior.

Next I discuss the effect of a more intense competition while 2 holds. In this
case, when the difference between the first and second prize from taking $R$
increases, the equilibrium where both players acquire information exists for a
smaller range of priors. This is intuitive because the endogenous cost
associated with acquiring information increases as the game becomes more
competitive. Then, the equilibrium where the players acquire information is
harder to be sustained for the lower priors.

The discussion above is about the case when competition becomes more intense.
The other limiting case is when there is no competition. That is, when the
payoff difference between the first and second $R$ taker is zero. The payoff
from taking $R$ does not depend on the opponent’s action. Then, there is no
strategic interaction between the two players. The equilibrium in this
limiting case is such that both players use the single DM optimal strategy.

##  6 Extensions

###  6.1 More than two players

In this section, I generalise the original model to $N>2$ players. I assume
that the first player to take $R$ gets the first prize and all other $R$
takers get the second prize. I use this generalisation to show that as the
number of players increases, the range of priors where the learning
equilibrium can exist shrinks.

Increasing the number of players intensifies the competition in the game. In
Section 5.2, I discussed the competition in terms of the payoff difference
between the first and second prize associated with $R$. This section further
discusses the effect of competition in terms of number of players in the game.
I show that the existence of the learning equilibrium requires the competition
to be not too intense. The payoff difference between the first and the second
prize must decrease as fast as $\frac{1}{N-1}$ to guarantee the existence of
the learning equilibrium.

Since I assume that the first $R$ taker gets the first prize and all other $R$
takers get the second prize, the player only cares about whether the first $R$
prize has been taken or not. Because of this, from the player’s point of view,
the remaining $N-1$ players essentially act as one big opponent. Let
$\frac{\tilde{p}_{N}}{1-\tilde{p}_{N}}:=\frac{-bu_{L}-c}{\left(N-1\right)\bar{\triangle}_{H}a+c}$
be a prior cutoff. Since $u_{L}<0$, when $c$ is sufficiently small,
$\tilde{p}_{N}$ is positive. The following proposition characterises a
necessary condition for the existence of the learning equilibrium where the
players use the random stopping strategy.

######  Proposition 4.

For $N>2$, there exists an equilibrium where the players use the random
stopping strategy only if $p_{0}<\tilde{p}_{N}.$ The cutoff $\tilde{p}_{N}$
decreases in $N$. When $N\rightarrow\infty$, $\tilde{p}_{N}\rightarrow 0$.

The learning equilibrium can exist only if the prior is sufficiently low and
this prior cutoff decreases in the number of players. When the number of
players goes to infinity, the cutoff $\tilde{p}_{N}$ approaches zero and the
potential learning region vanishes.

######  Corollary 2.

When $N\rightarrow\infty$, if $\bar{\triangle}_{N}$ decreases as fast as
$\frac{1}{N-1}$, then, $\tilde{p}_{N}$ is finite.

When the number of players increases to infinity, if the prize difference
$\bar{\triangle}_{H}$ decreases as fast as $\frac{1}{N-1}$, then, the
potential learning region still exists. The intuition is that the learning
equilibrium can exist only if the competition is not too intense. If there are
a lot of players competing, then, the payoff difference cannot be too big.

###  6.2 Observable actions

The irreversible risky and safe actions are assumed to be private in this
paper. This allows me to focus on the role of payoff externalities. In this
extension, I assume that after a player takes an irreversible action, it is
immediately observed by the opponent. The public actions generate information
externalities as observing no action taken is itself informative. The purpose
of this extension is to show that the existence of the learning equilibrium
found in Theorem 1 is robust to some exposure to information externalities. To
be more specific, I define the mimicking and random stopping strategy (MRSS)
and show that there exists a symmetric equilibrium where the players use this
strategy.

######  Definition 4.

The mimicking and random stopping strategy (MRSS) is a strategy such that

  1. 1.

After receiving the $H$-state revealing signal, the player stops and takes $R$
immediately;

  2. 2.

After receiving the $L$-state revealing signal, the player stops and takes $S$
immediately;

  3. 3.

After receiving no revealing signal and observing no action taken, the player
stops and takes $R$ at each time $t$ with a rate $h\left(t\right)>0;$

  4. 4.

After receiving no revealing signal and observing the opponent taking $S$, the
player stops and takes $S$;

  5. 5.

After receiving no revealing signal and observing the opponent taking $R$, the
player uses the single DM optimal strategy.

The MRSS and the random stopping strategy are similar in the sense that after
observing no private revealing signal, the player randomly stops and takes
$R$. The conditions for the existence of the symmetric equilibrium where the
players use MRSS is presented in the following proposition.

######  Proposition 5.

Suppose $\bar{\triangle}_{H}=\bar{\triangle}_{L}$ and $c$ sufficiently small
such that $p^{L}<\tilde{p}.$ If $p^{L}<p_{0}<\tilde{p}$, then, there exists an
equilibrium where the players use MRSS.

Proposition 5 suggests that when there are information externalities, there
still exists the equilibrium where the players randomly stop and take $R$
after no revealing signal. The reason is twofold. First, the observability of
the actions does not eliminate the preemption motive. After the history of no
action taken, if the player’s strategy prescribes taking $R$ with positive
probability mass, then, the opponent has an incentive to preempt. Second,
random stopping reduces the information involved in the player’s action. The
action taken (or no action taken) contains the player’s private information.
It is essentially an additional signal and the informativeness of it depends
on the player’s strategy. For example, if the player only stops acquiring
information after receiving a revealing signal, then, her action perfectly
reveals her private information. In this case, this additional signal is very
informative for the opponent. However, if the player stops acquiring
information and takes an action randomly, then, her action contains less
private information. Information externalities enhance the player’s incentive
to stop randomly.

##  7 Conclusion

In this paper, I study a model in which the players can acquire costly private
information before taking an irreversible private action. Acquiring
information takes time which allows the player to take the ‘correct’ action
but increases the probability of being preempted. With the assumption that
there is a first-mover advantage associated with the risky action, I find that
in the equilibrium where the players acquire information, they become more
optimistic that the state is high but at the same time, the conditional
expected payoff from taking the risky action decreases. Because of the
interaction between the learning motive and the preemption motive, the players
randomly stop and take the risky action. This result is significantly
different from the single decision maker case where the optimal strategy is
deterministic.

In my model, depending on the prior, the players’ decisions on acquiring
information can be both strategic substitutes or strategic complements. The
strategic substitutability prevails when players have pessimistic priors and
it induces players’ initial mix between acquiring information and immediate
exit. The strategic complementarity prevails when players are relatively
optimistic and it gives rise to the multiple equilibria where players use the
random stopping strategy. In addition, I find that the equilibrium where the
players do not acquire information can exist for not only extreme priors but
also a large range of intermediate priors. This stems from the presence of the
endogenous information cost. Such equilibrium exists for arbitrarily small
exogenous information cost.

## References

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##  Appendix A Formulation of the player’s problem

This section formulates one player’s problem given opponent’s strategies.
Given player $j$’s strategy $\gamma^{j}$, player $i$’s best-reply problem is
to choose a time $T^{i}$ with value function
$\hat{V}^{i,\gamma^{j}}:\mathbb{R}_{+}\rightarrow\mathbb{R}$ such that

$\displaystyle\hat{V}^{i,\gamma^{j}}\left(0\right)$
$\displaystyle:=\max_{T^{i}}\intop_{0}^{T^{i}}\pi_{t}\left[p_{t}^{i}a\mathbb{E}^{\gamma^{j}}\left[u_{H}^{R}\mid
H,t\right]-c\right]dt+\pi_{T^{i}}U^{i,\gamma^{j}}\left(T^{i}\right)$ (4)
$\displaystyle\text{s.t. }\frac{dp_{t}^{i}}{dt}=(b-a)p_{t}^{i}(1-p_{t}^{i}),$

where $\mathbb{E}^{\gamma^{j}}\left[u_{H}^{R}\mid H,t\right]$ and
$U^{i,\gamma^{j}}\left(T^{i}\right)$ are as defined in Section 2.2. The
Hamilton-Jacobi-Bellman (HJB) equation for player $i$’s problem is the
following differential equation in
$V^{i,\gamma^{j}}:\mathbb{R}_{+}\rightarrow\mathbb{R},$ where

$\displaystyle\max\left\\{\overbrace{p_{t}^{i}a\left[\mathbb{E}^{\gamma^{j}}\left[u_{H}^{R}\mid
H,t\right]-V^{i,\gamma^{j}}\left(t\right)\right]+\left(1-p_{t}^{i}\right)b\left[-V^{i,\gamma^{j}}\left(t\right)\right]}^{\text{A}}+\overbrace{\frac{dV^{i,\gamma^{j}}\left(t\right)}{dt}}^{\text{B}}-c,\right.$
$\displaystyle\left.U^{i,\gamma^{j}}\left(t\right)-V^{i,\gamma^{j}}\left(t\right)\right\\}$
$\displaystyle=0.$ (5)

The interpretation of (A) is that at time $t$, player $i$ chooses between to
continue acquiring information or stopping. She acquires information if the
marginal gain is greater than the marginal cost. Otherwise, she stops and gets
the payoff $U^{i,\gamma^{j}}\left(t\right)$. The marginal gain consists of the
expected gain from receiving the revealing signal (labelled as A in (A)) plus
the rate of change of the value (labelled as B in (A)).

Given opponent’s strategy $\gamma^{j}$, if the player’s problem is well-
behaved, then, the value function $V^{i,\gamma^{j}}\left(t\right)$ is a
classical solution to the HJB equation (A). The player’s best response can
then be characterised correspondingly. However, in our problem, (A) is not
well-behaved because given the opponent’s strategy $\gamma^{j}$,
$U^{i,\gamma^{j}}\left(t\right)$ has a kink and may not be continuous.

##  Appendix B Proof of Proposition 1

When player $i$ is a single DM, her problem is

$\displaystyle\hat{V}\left(p_{0}\right):=\max_{T}\intop_{0}^{T}\pi_{t}\left[p_{t}au_{H}+\left(1-p_{t}\right)bu^{S}-c\right]dt+\pi_{T}U\left(T\right).$

The HJB equation is

$\displaystyle\max\left\\{p_{t}a\left[u_{H}-V\left(t\right)\right]+\left(1-p_{t}\right)b\left[u^{S}-V\left(t\right)\right]+\frac{dV\left(t\right)}{dt}-c,\right.$
$\displaystyle\left.U\left(t\right)-V\left(t\right)\right\\}=0$

Since the argument $t$ only enters the equation via $p_{t}$, I use $p$ instead
of $t$ as the state variable. Then HJB equation becomes

$\displaystyle\max\left\\{pa\left[u_{H}-V\left(p\right)\right]+\left(1-p\right)b\left[u^{S}-V\left(p\right)\right]+\frac{dV\left(p\right)}{dp}\frac{dp}{dt}-c,\right.$
$\displaystyle\left.U\left(p\right)-V\left(p\right)\right\\}=0$ (6)

To find player $i$’s value function, I construct a candidate value function
and show it is a viscosity solution of (B).

If the learning region (the range of beliefs at which the DM acquires the
signal) exists, the value function is a solution of the ordinary differential
equation

$\displaystyle
pa\left[u_{H}-V\left(p\right)\right]+\left(1-p\right)b\left[u^{S}-V\left(p\right)\right]+\frac{dV\left(p\right)}{dp}\left(b-a\right)p\left(1-p\right)=c.$
(7)

The free boundary solution to (7) is

$\displaystyle
V^{L}\left(p\right)=p\left[u_{H}-\frac{c}{a}\right]+\left(1-p\right)\left[u_{L}-\frac{c}{b}\right]+K\left(\frac{p}{1-p}\right)^{\frac{b}{b-a}}\left(1-p\right)$

where $K$ is a constant. Suppose the learning region is
$\left(\underline{p},\bar{p}\right)$. Value matching and smooth pasting pin
down the value of $\bar{p}$ and $K$. Then, value matching pins down
$\underline{p}$. We have

$\displaystyle\frac{\bar{p}}{1-\bar{p}}=\frac{u^{S}-u_{L}-\frac{c}{b}}{\frac{c}{b}}:=\bar{L},$
$\displaystyle
K=\frac{c}{b}\left(\frac{b}{a}-1\right)\bar{L}^{1-\frac{b}{b-a}}$

and $\underline{L}:=\frac{\underline{p}}{1-\underline{p}}$ satisfies

$\displaystyle\left[u_{H}-u^{S}-\frac{c}{a}\right]\underline{L}+K\underline{L}^{\frac{b}{b-a}}=\frac{c}{b}.$
(8)

Let $\hat{p}$ be a belief cutoff at which the player is indifferent between
$R$ and $S$. The existence of the learning requires $\bar{p}>\hat{p}$. That
is,

$c<b\frac{\left(-u_{L}\right)u_{H}}{u_{H}-u_{L}}:=\bar{c}.$

Outside the learning region, the value function satisfies
$V\left(p\right)=U\left(p\right)$. When $c<\bar{c}$, the candidate value
function is

$\displaystyle
V\left(p\right)=\begin{cases}p\left[u_{H}-\frac{c}{a}\right]+\left(1-p\right)\left[u_{L}-\frac{c}{b}\right]+K\left(\frac{p}{1-p}\right)^{\frac{b}{b-a}}\left(1-p\right)&p\in\left(\underline{p},\bar{p}\right)\\\
U\left(p\right)&\text{o.w.}\end{cases}.$

This candidate has a kink at $\underline{p}$ and is differentiable everywhere
else. I next show that it is a viscosity solution of (B). Let

$\displaystyle
H\left(p,V\left(p\right),V^{\prime}\left(p\right)\right):=pa\left[u_{H}-V\left(p\right)\right]+\left(1-p\right)b\left[u^{S}-V\left(p\right)\right]+V^{\prime}\left(p\right)\left(b-a\right)p\left(1-p\right)-c.$

For the points where $V\left(p\right)$ is differentiable, I show that (1) if
$p>\bar{p}$, then,
$H\left(p,V\left(p\right),V^{\prime}\left(p\right)\right)\leq 0$; (2) if
$p\in\left(\underline{p},\bar{p}\right]$, then, $V\left(p\right)\geq
U\left(p\right)$; (3) if $p<\underline{p}$, then
$H\left(p,V\left(p\right),V^{\prime}\left(p\right)\right)\leq 0$. At the point
where $V\left(p\right)$ is not differentiable, that is, at $p=\underline{p}$,
I show that $H\left(p,V\left(p\right),z\right)\geq 0$ for $z\in D^{+}$ where
$D^{+}=\emptyset$ (ignore) and (4) $H\left(p,V\left(p\right),z\right)\leq 0$
for $z\in D^{-}$ where
$D^{-}=\left[\frac{dU_{S}\left(p\right)}{dp}\mid_{p=\underline{p}},\frac{dV^{L}\left(p\right)}{dp}\mid_{p=\underline{p}}\right]$.

Step (1): When $p>\bar{p}$, we have $V\left(p\right)=U_{R}\left(p\right)$ and
$H\left(p,U_{R}\left(p\right),U_{R}^{\prime}\left(p\right)\right)<0$ if and
only if $p>\bar{p}$.

Step (2): It can be shown that $V^{L}\left(p\right)$ is convex. At
$p=\bar{p}$, we have $V^{L}\left(p\right)=U_{R}\left(p\right)$. If we decrease
$p$ by a little bit, $U_{R}\left(\cdot\right)$ decreases faster than
$V^{L}\left(\cdot\right)$. Therefore, we have $V^{L}\left(p\right)\geq
U_{R}\left(p\right)$ for $p\leq\bar{p}$. At $p=\underline{p}$, we have
$V^{L}\left(\underline{p}\right)=U_{S}\left(\underline{p}\right)$ and
$\frac{dV^{L}\left(p\right)}{dp}>\frac{dU_{S}\left(p\right)}{dp}$. As a
result, we have $V^{L}\left(p\right)\geq U_{S}\left(p\right)$ for
$p>\underline{p}$.

Step (3): If $p<\underline{p}$, then, $V\left(p\right)=U_{S}\left(p\right)$.
To have
$H\left(p,U_{S}\left(p\right),\frac{dU_{S}\left(p\right)}{dp}\right)\leq 0$,
we need $p\leq\frac{\frac{c}{a}}{u_{H}-u^{S}}$. It can be shown that
$\underline{p}<\frac{\frac{c}{a}}{u_{H}-u^{S}}$. As a result, if
$p<\underline{p}$, then
$H\left(p,V\left(p\right),V^{\prime}\left(p\right)\right)\leq 0$.

Step (4): Since $H\left(p,V\left(p\right),z\right)$ is increasing in $z$ and
we have
$H\left(\underline{p},V\left(\underline{p}\right),\frac{dV^{L}\left(p\right)}{dp}\mid_{p=\underline{p}}\right)=0$,
it is true that $H\left(p,V\left(p\right),z\right)\leq 0$ for $\forall
z\in\left[\frac{dU_{S}\left(p\right)}{dp}\mid_{p=\underline{p}},\frac{dV^{L}\left(p\right)}{dp}\mid_{p=\underline{p}}\right]$.

To conclude, if $c<\bar{c}$, the DM’s optimal strategy is to acquire the
signal if the belief if in the range $\left(\underline{p},\bar{p}\right)$, to
take action $R$ if $p\geq\bar{p}$ and to take action $S$ if
$p\leq\underline{p}$. If $c\geq\bar{c}$, the DM’s optimal strategy is to take
an action without acquiring the signal.

##  Appendix C Proof of Lemma 1

Consider a strategy $\gamma=\left(\rho,\sigma\right)$ as defined in Definition
2. Suppose $\rho$ is discontinuous at some $\tau>0$ and continuous everywhere
else. Since by definition $\rho$ is right-continuous and weakly increasing,
this implies that
$\rho\left(\tau\right)>\lim_{t\rightarrow\tau_{-}}\rho\left(t\right).$ Then,
the induced probability that the player takes $R$ before or at time $t$ in
state $\omega$, $F_{\omega}^{\gamma}\left(t\right),$ 777The superscript
indicates the strategy that this function is induced from. is right continuous
such that
$F_{\omega}^{\gamma}\left(\tau\right)>\lim_{t\rightarrow\tau_{-}}F_{\omega}^{\gamma}\left(t\right)$
for $\forall\omega\in\\{H,L\\}$. Let $M_{\omega}$ denote the mass that
$F_{\omega}^{\gamma}\left(\tau\right)$ places on $\tau$. Consider a deviation
$\gamma^{D}=\left(\rho^{D},\sigma\right)$ such that it is identical to
$\gamma$ except that the mass that $\rho$ places on $\tau$ is shifted to
$\tau-\triangle$. There exists a $\triangle>0$ such that given the opponent
uses the $\gamma$ strategy, using $\gamma^{D}$ gives the player a higher
payoff than $\gamma$.

Suppose the opponent uses the $\gamma$ strategy described above. Consider the
history that the player receives no revealing signal until time
$\tau-\triangle$ where $\triangle>0$. The total gain from acquiring
information for $\triangle$ time longer is

$\displaystyle
p_{\tau-\triangle}a\triangle\left[u_{H}-F_{H}^{\gamma}\left(\tau\right)\bar{\triangle}_{H}\right]+\left[1-p_{\tau-\triangle}a\triangle-\left(1-p_{\tau-\triangle}\right)b\triangle\right]U_{R}\left(\tau\right)-U_{R}\left(\tau-\triangle\right)$

where

$\displaystyle
U_{R}\left(t\right)=\begin{cases}p_{t}\left[u_{H}-F_{H}^{\gamma}\left(t\right)\bar{\triangle}_{H}\right]+\left(1-p_{t}\right)\left[u_{L}-F_{L}^{\gamma}\left(t\right)\bar{\triangle}_{L}\right]&t<\tau\\\
p_{t}\left[u_{H}-F_{H}^{\gamma}\left(t\right)\bar{\triangle}_{H}-\left(1-F_{H}^{\gamma}\left(t\right)\right)\underline{\triangle}_{H}\right]+\left(1-p_{t}\right)\left[u_{L}-F_{L}^{\gamma}\left(t\right)\bar{\triangle}_{L}-\left(1-F_{L}^{\gamma}\left(t\right)\right)\underline{\triangle}_{L}\right]&t=\tau\\\
\end{cases}.$

When $\triangle>0$ approaches zero, this gain approaches

$p_{\tau}\left[-M_{H}\bar{\triangle}_{H}-\left(1-F_{H}^{\gamma}\left(\tau\right)\right)\underline{\triangle}_{H}\right]+\left(1-p_{\tau}\right)\left[-M_{L}\bar{\triangle}_{L}-\left(1-F_{L}^{\gamma}\left(\tau\right)\right)\underline{\triangle}_{L}\right],$

which is negative due to the mass point $M_{\omega}$ and the tie-breaking at
time $\tau$. Therefore, there is always a deviation to put the mass places on
$\tau$ to $\tau-\triangle$.

##  Appendix D Proof of Theorem 1

Let
$\frac{p^{M}}{1-p^{M}}:=\frac{-u_{L}+\underline{\triangle}_{L}}{u_{H}-\underline{\triangle}_{H}}$,
$\frac{p^{L}}{1-p^{L}}:=\frac{-u_{L}}{u_{H}}$ and
$\frac{\tilde{p}}{1-\tilde{p}}:=\frac{-bu_{L}-c}{a\bar{\triangle}_{H}+c}$ be
three prior cutoffs. As $u_{L}<0$, when $c<-bu_{L}$, all of the three cutoffs
are positive. The cutoff $\underline{p}$ is defined in Proposition 1 and
$p^{*}$ is a fixed point that will be defined later in step 5.

The method to find the symmetric equilibrium is ‘guess and verify’. The
following outlines the steps of the proof.

  1. 1.

I show that there exists an equilibrium where both players use the pure
strategy $\left(0,S\right)$ when $p_{0}\leq\underline{p}$.

  2. 2.

I show that there exists an equilibrium where both players use the pure
strategy $\left(0,R\right)$ when $p_{0}\geq p^{M}$.

  3. 3.

I show that there exists an equilibrium where both players use the Immediate
Mix with No Learning Strategy when $p^{L}<p_{0}<p^{M}$.

  4. 4.

I show that there exists an equilibrium where both players use a Randomised
Stopping Time Strategy when $p^{L}<p_{0}<\tilde{p}.$

  5. 5.

I show that there exists an equilibrium where both players use a Randomised
Stopping Time Strategy when $p^{*}<p_{0}<p^{L}.$

  6. 6.

I show that there exists an equilibrium where both players use the Mixed
Learning Strategy when $\underline{p}<p_{0}<p^{*}$.

#### Step 1

When $p_{0}\leq\underline{p}$, Proposition 1 implies that a single DM takes
$S$ immediately. If a player takes $S$ immediately, the other player is the
single DM in the game. Given the player takes $S$ immediately, the opponent’s
best response is to take $S$. Therefore, when $p_{0}\leq\underline{p}$, both
players taking $S$ immediately is an equilibrium. This is summarised in the
following lemma.

######  Lemma 3.

If $p_{0}\leq\underline{p}$, then, there exists an equilibrium where both
players use the strategy $\left(0,S\right)$.

#### Step 2

I prove the following lemma.

######  Lemma 4.

If $p_{0}\geq p^{M}$, there exists an equilibrium where both players use the
pure strategy $\left(0,R\right)$

###### Proof.

Suppose player $j$ uses the strategy $\left(0,R\right)$. I check whether
player $i$ wants to stop at time $0$ and take action $R$ or to acquire the
signal for $dt$ longer. If player $i$ takes the immediate $R$ action at time
$0$, the payoff is

$U_{R}^{IR}\left(0\right)=p_{0}\left(u_{H}-\underline{\triangle}_{H}\right)+\left(1-p_{0}\right)\left(u_{L}-\underline{\triangle}_{L}\right).$

If the player takes action $R$ at time $t>0$, the payoff is

$U_{R}^{IR}\left(t\right)=p_{t}\left(u_{H}-\bar{\triangle}_{H}\right)+\left(1-p_{t}\right)\left(u_{L}-\bar{\triangle}_{L}\right)$

At time $0$, the gain from acquiring the signal for $dt$ longer is

$\displaystyle
p_{0}adt\left(u_{H}-\bar{\triangle}_{H}\right)+\left[1-p_{0}adt-\left(1-p_{0}\right)bdt\right]U_{R}^{IR}\left(0+dt\right)-U_{R}^{IR}\left(0\right).$
(9)

When $dt\rightarrow 0$, (9) tends to something negative, which is smaller than
the cost of information. Therefore, if the opponent stops at time $0$, player
$i$ prefers taking action $R$ at time $0$ to acquiring the signal for $dt$
longer. At time $0$, player $i$ prefers action $R$ to action $S$ at time $0$
if $U_{R}^{IR}\left(0\right)\geq 0.$ Since $p_{0}\geq p^{M}$, the inequality
$U_{R}^{IR}\left(0\right)\geq 0$ holds. ∎

#### Step 3

I show the following lemma.

######  Lemma 5.

If $p^{L}<p_{0}<p^{M}$, then, there exists an equilibrium where both players
use the Immediate Mix Strategy.

###### Proof.

Suppose player $j$ uses the strategy $\left(\rho^{IM},\sigma^{IM}\right)$ such
that $\rho^{IM}\left(t\right)=m$ for $\forall t\in\left[0,\infty\right)$ and
$\sigma^{IM}\left(t\right)=1-m$ for $\forall t\in\left[0,\infty\right)$. Then,
player $i$’s payoff from taking action $R$ at time $0$ is

$\displaystyle U_{R}^{IM}\left(0\right)$
$\displaystyle=m\left[p_{0}u_{H}+\left(1-p_{0}\right)u_{L}\right]+\left(1-m\right)\left[p_{0}\left(u_{H}-\underline{\triangle}_{H}\right)+\left(1-p_{0}\right)\left(u_{L}-\underline{\triangle}_{L}\right)\right]$
$\displaystyle=p_{0}\left[mu_{H}+\left(1-m\right)\left(u_{H}-\underline{\triangle}_{H}\right)\right]+\left(1-p_{0}\right)\left[mu_{L}+\left(1-m\right)\left(u_{L}-\underline{\triangle}_{L}\right)\right].$

Player $i$’s payoff from taking action $R$ at time $t>0$ is

$\displaystyle U_{R}^{IM}\left(t\right)$
$\displaystyle=m\left[p_{t}u_{H}+\left(1-p_{t}\right)u_{L}\right]+\left(1-m\right)\left[p_{t}\left(u_{H}-\bar{\triangle}_{H}\right)+\left(1-p_{t}\right)\left(u_{L}-\bar{\triangle}_{L}\right)\right]$
$\displaystyle=p_{t}\left[u_{H}-\left(1-m\right)\bar{\triangle}_{H}\right]+\left(1-p_{t}\right)\left[u_{L}-\left(1-m\right)\bar{\triangle}_{L}\right].$

At time $t=0$, player $i$ does not want to acquire the information due to the
same reasoning as in step 2.

Next I show that for the prior $p_{0}\in\left(p^{L},p^{M}\right)$, there
exists $m\in\left(0,1\right)$ such that player $i$ is indifferent between
taking action $R$ and action $S$ at time $t=0$. The indifference requires
$0=U_{R}\left(0\right)$. That is,

$\displaystyle\frac{p_{0}}{1-p_{0}}$
$\displaystyle=-\frac{\left[mu_{L}+\left(1-m\right)\left(u_{L}-\underline{\triangle}_{L}\right)\right]}{\left[mu_{H}+\left(1-m\right)\left(u_{H}-\underline{\triangle}_{H}\right)\right]}.$
(10)

For any $p_{0}\in\left(p^{L},p^{M}\right)$, there exists
$m\in\left(0,1\right)$ such that (10) holds. ∎

#### Step 4

This step shows the existence of the equilibrium where players use the
randomised stopping time strategy. I show that there exists an equilibrium
where both players take action $R$ at each time $t\geq 0$ with positive rate.
Suppose cost $c$ is sufficiently small such that $p^{L}<\tilde{p}$. Let
$L_{t}=\frac{p_{t}}{1-p_{t}}$ be the likelihood ratio. I prove the following
lemma.

######  Lemma 6.

Suppose $c$ is sufficiently small and $b<2a.$ If $p^{L}<p_{0}<\tilde{p},$
then, there exists an equilibrium in mixed strategies
$\left(\rho,\sigma\right)$ such that $\sigma\left(t\right)=0$ for $\forall
t\in\mathbb{R}_{+}$ and $\rho:\mathbb{R}_{+}\rightarrow\left[0,1\right]$
satisfies the following conditions:

  1. 1.

(Initial condition) $\rho\left(0\right)=0$.

  2. 2.

(Increasing condition) There exists a $T>0$ such that for
$t\in\left[0,T\right]$, $\rho$ is a solution to the following differential
equation

$\displaystyle\frac{d\rho\left(t\right)}{dt}$
$\displaystyle=\frac{b\left(-u_{L}\right)-c-L_{t}c+b\bar{\triangle}_{L}\intop_{0}^{t}e^{-bs}\frac{d\rho\left(s\right)}{ds}ds-
L_{t}\bar{\triangle}_{H}ae^{-at}\left(1-\rho\left(t\right)\right)}{L_{t}\bar{\triangle}_{H}e^{-at}+\bar{\triangle}_{L}e^{-bt}}$
(11)

with the initial condition $\rho\left(0\right)=0$ where $\frac{d\rho}{dt}>0$
for $\forall t\in\left[0,T\right]$ and $\rho\left(T\right)=1$.

  3. 3.

(Terminal condition) $\rho\left(t\right)=1$ for $\forall t>T.$

In words, the equilibrium strategy is that the players stop and take action
$R$ at each time with positive rate.

###### Proof.

Fix player $j$’s strategy $\left(\rho,\sigma\right)$, I first show that at
each time $t$, player $i$ is indifferent between taking action $R$ now or $dt$
later if $\rho$ satisfies (11). Then Lemma 7 shows that when $c$ is
sufficiently small and $b<2a$, if $p_{0}<\tilde{p}$, we can find a $\rho$
function such that $\frac{d\rho}{dt}>0$. Last, I show that if $p_{0}>p^{L}$,
player $i$ prefers acquiring information to taking action $S$.

Suppose player $j$ uses the mixed strategy $\left(\rho,\sigma\right)$. This
strategy induces

$\displaystyle F_{H}\left(t\right)$
$\displaystyle=\intop_{0}^{t}\left[e^{-as}\left(1-\rho\left(s\right)\right)\left(a+\frac{\frac{d\rho\left(s\right)}{ds}}{1-\rho\left(s\right)}\right)\right]ds$
$\displaystyle=1-e^{-at}\left(1-\rho\left(t\right)\right)$

and

$\displaystyle F_{L}\left(t\right)$
$\displaystyle=\intop_{0}^{t}\left[e^{-bs}\left(1-\rho\left(s\right)\right)\frac{\frac{d\rho\left(s\right)}{ds}}{1-\rho\left(s\right)}\right]ds$
$\displaystyle=\intop_{0}^{t}\left[e^{-bs}\frac{d\rho\left(s\right)}{ds}\right]ds.$

Player $i$’s payoff from taking action $R$ at time $t$ is

$U_{R}\left(t\right):=p_{t}\left[u_{H}-F_{H}\left(t\right)\bar{\triangle}_{H}\right]+\left(1-p_{t}\right)\left[u_{L}-F_{L}\left(t\right)\bar{\triangle}_{L}\right].$

The equilibrium condition is that given player $j$’s strategy, player $i$ is
indifferent between taking action $R$ and acquiring the signal at each time
instant. That is,

$\displaystyle
p_{t}adt\left[u_{H}-F_{H}\left(t+dt\right)\bar{\triangle}_{H}\right]+\left[1-p_{t}adt-\left(1-p_{t}\right)bdt\right]U_{R}\left(t+dt\right)-U_{R}\left(t\right)=cdt.$

The interpretation is that the marginal cost and marginal benefit associated
with acquiring information for $dt$ longer are the same. When $dt\rightarrow
0$, we have the following differential equation

$\displaystyle c$
$\displaystyle=\left(1-p_{t}\right)b\left[-u_{L}+F_{L}\left(t\right)\bar{\triangle}_{L}\right]-p_{t}\frac{dF_{H}\left(t\right)}{dt}\bar{\triangle}_{H}-\left(1-p_{t}\right)\frac{dF_{L}\left(t\right)}{dt}\bar{\triangle}_{L}$

After plugging in the expressions of $F_{H}\left(t\right)$ and
$F_{L}\left(t\right),$ we have

$\displaystyle\frac{d\rho\left(t\right)}{dt}$
$\displaystyle=\frac{b\left(-u_{L}\right)-c-L_{t}c+b\bar{\triangle}_{L}\intop_{0}^{t}e^{-bs}\frac{d\rho\left(s\right)}{ds}ds-
L_{t}\bar{\triangle}_{H}ae^{-at}\left(1-\rho\left(t\right)\right)}{L_{t}\bar{\triangle}_{H}e^{-at}+\bar{\triangle}_{L}e^{-bt}}.$

######  Lemma 7.

The differential equation (11) with initial condition $\rho\left(0\right)=0$
has a unique solution defined for all $t\in\left[0,\infty\right)$.

###### Proof.

Let $A\left(t\right):=e^{-bt}\rho\left(t\right)$. Then, we have
$\frac{dA}{dt}=-bA+e^{-bt}\frac{d\rho}{dt}$. Use the formula that
$L_{t}=L_{0}e^{\left(b-a\right)t}$, (11) can be rewritten as

$\displaystyle\left(\frac{dA}{dt}+bA\right)e^{bt}$
$\displaystyle=\frac{b\left(-u_{L}\right)-c-cL_{0}e^{\left(b-a\right)t}+b\bar{\triangle}_{L}\left(A+b\intop_{0}^{t}A\left(s\right)ds\right)-L_{0}\bar{\triangle}_{H}ae^{\left(b-2a\right)t}\left(1-e^{bt}A\right)}{L_{0}\bar{\triangle}_{H}e^{\left(b-2a\right)t}+\bar{\triangle}_{L}e^{-bt}}.$
(12)

Let $z\left(t\right):=\intop_{0}^{t}A\left(s\right)ds$. Then, the differential
equation can be written as

$\displaystyle
z^{\prime\prime}+g_{1}\left(t\right)z^{\prime}+g_{2}\left(t\right)z=g_{3}\left(t\right)$
(13)

where

$\displaystyle g_{0}\left(t\right)$
$\displaystyle=\frac{1}{L_{0}\bar{\triangle}_{H}e^{2\left(b-a\right)t}+\bar{\triangle}_{H}},$
$\displaystyle g_{1}\left(t\right)$
$\displaystyle=b-\frac{b\bar{\triangle}_{L}+L_{0}\bar{\triangle}_{H}ae^{2\left(b-a\right)t}}{g_{0}\left(t\right)},$
$\displaystyle g_{2}\left(t\right)$
$\displaystyle=-\frac{b^{2}\bar{\triangle}_{L}}{g_{0}\left(t\right)},$
$\displaystyle g_{3}\left(t\right)$
$\displaystyle=\frac{b\left(-u_{L}\right)-c-cL_{0}e^{\left(b-a\right)t}-L_{0}\bar{\triangle}_{H}ae^{\left(b-2a\right)t}}{g_{0}\left(t\right)}.$

Since $g_{0}\left(t\right)$, $g_{1}\left(t\right)$, $g_{2}\left(t\right)$ and
$g_{3}\left(t\right)$ are continuous on $\left[0,\infty\right)$, (13) with
initial conditions $z\left(0\right)=0$ and $z^{\prime}\left(0\right)=0$ has a
unique solution defined for all $t$ on $\left[0,\infty\right)$. The existence
of $z\left(\cdot\right)$ implies the existence of $\rho\left(\cdot\right)$. ∎

######  Lemma 8.

$\frac{d\rho}{dt}\mid_{t=0}>0$ iff $p_{0}<\tilde{p}.$

###### Proof.

$\displaystyle\frac{d\rho}{dt}\mid_{t=0}$
$\displaystyle=\frac{b\left(-u_{L}\right)-c-L_{0}c-L_{0}\bar{\triangle}_{H}a}{L_{0}\bar{\triangle}_{H}+\bar{\triangle}_{L}}$
$\displaystyle>0$

if and only if

$L_{0}=\frac{p_{0}}{1-p_{0}}<\frac{b\left(-u_{L}\right)-c}{a\bar{\triangle}_{H}+c}=\frac{\tilde{p}}{1-\tilde{p}}.$

∎

This lemma implies that if $p_{0}<\tilde{p},$ then, $\frac{d\rho}{dt}>0$ at
the neighbourhood of $0$. The following lemma and its proof shows that under
some conditions, if $\rho\left(\cdot\right)$ increases at the neighbourhood of
$0$ and the value of $\rho$ is smaller than 1, then, it continues increasing.

######  Lemma 9.

Suppose $b<2a$ and $c$ sufficiently small. If
$\rho\left(t\right)\in\left[0,1\right)$ for $\forall t\in\left[0,\tau\right]$
and $\frac{d\rho}{dt}>0$ $\forall t\in\left[0,\tau\right),$ then,
$\frac{d\rho}{dt}>0$ at $t=\tau$.

###### Proof.

If $b<2a$, $c$ sufficiently small, and $\frac{d\rho}{dt}>0$ $\forall
t\in\left[0,\tau\right),$ then,

$\displaystyle\Phi\left(t\right):=\frac{b\left(-u_{L}+\bar{\triangle}_{L}\intop_{0}^{t}e^{-bs}\frac{d\rho}{ds}ds\right)-c-L_{0}e^{\left(b-a\right)t}c}{\bar{\triangle}_{H}ae^{\left(b-2a\right)t}\left(1-\rho\left(t\right)\right)}$

increases in $t$ for $\forall t\in\left[0,\tau\right).$ Since
$\rho\left(t\right)\in\left[0,1\right)$ for $\forall t\in\left[0,\tau\right),$
then, $\Phi\left(t\right)>L_{0}$ for $\forall t\in\left[0,\tau\right)$ iff
$\frac{d\rho}{dt}>0$ for $\forall t\in\left[0,\tau\right).$ Next, I show by
contradiction that if $\frac{d\rho}{dt}>0$ for $\forall
t\in\left[0,\tau\right)$ and $\rho\left(t\right)\in\left[0,1\right)$ for
$\forall t\in\left[0,\tau\right],$ then, it must be that $\frac{d\rho}{dt}>0$
at $t=\tau$. Suppose $\frac{d\rho}{dt}=0$ at $t=\tau$. Then, it must be that
$\Phi\left(\tau\right)=L_{0}.$ However, we know that $\Phi\left(t\right)$
increases in $t$ for $\forall t\in\left[0,\tau\right)$ and
$\Phi\left(t\right)>L_{0}$ for $\forall t\in\left[0,\tau\right)$. Then, it
cannot be that $\Phi\left(\tau\right)=L_{0}.$ There is a contradiction and
hence $\frac{d\rho}{dt}\neq 0$ at $t=\tau$. Since $\rho$ is continuous, it
cannot be that $\frac{d\rho}{dt}>0$ at $t=\tau$. ∎

Next I show that given Lemma 7 and Lemma 9, $\rho\left(t\right)=1$ will be
reached in a finite time.

######  Lemma 10.

Given Lemma 7 and Lemma 9, there exists a $T<\infty$ such that
$\rho\left(T\right)=1.$

###### Proof.

If $\frac{d\rho}{dt}>0$ and $\rho\left(t\right)\in\left[0,1\right)$, then

$\displaystyle\Phi\left(t\right)$
$\displaystyle<\frac{b\left(-u_{L}+\bar{\triangle}_{L}\rho\left(t\right)\right)-c-L_{0}e^{\left(b-a\right)t}c}{\bar{\triangle}_{H}ae^{\left(b-2a\right)t}\left(1-\rho\left(t\right)\right)}$
$\displaystyle<\frac{b\left(-u_{L}+\bar{\triangle}_{L}\right)-c-L_{0}e^{\left(b-a\right)t}c}{\bar{\triangle}_{H}ae^{\left(b-2a\right)t}\left(1-\rho\left(t\right)\right)}.$

As $t\rightarrow\infty$, we have $\Phi\left(t\right)<0$. If
$\Phi\left(t\right)<0$ and $\rho\left(t\right)<1$ when $t\rightarrow\infty$,
then, $\frac{d\rho}{dt}<0$ as $t\rightarrow\infty$. There is a contradiction.
Therefore, it must be that $\rho\left(t\right)>1$ for some $t$. Because of
continuity, there exists a $T$ such that $\rho\left(T\right)=1$. ∎

######  Lemma 11.

If $p^{L}<p_{0}$ and $\rho\left(t\right)$ satisfies the conditions in Lemma 6,
then, $\sigma\left(t\right)=0$ for $\forall t\in\left[0,T\right]$.

###### Proof.

Let $V\left(t\right)$ be the value associated with the mixed strategy
characterised in Lemma 6. Then,

$V\left(t\right)=U_{R}\left(t\right)-ct=U_{R}\left(0\right)$

because the player’s payoffs are the same at each time when she is randomising
between continuing and stopping (to take action $R$). Since $p^{L}<p_{0},$ we
have $U_{R}\left(0\right)>0$. Hence, at each $t<T$, we have
$U_{R}\left(t\right)>0$. As a result, $\sigma\left(t\right)=0$ for $\forall
t\in\left[0,T\right]$. ∎

This completes the proof of Lemma 6.

∎

#### Step 5

In the previous step, I consider the situation that the player starts
randomisation at time $t=0$. However, it is possible that the player strictly
prefers to acquire information for a certain time period and then starts
randomisation. This step shows the existence of the equilibrium where players
use a mixed strategy $\left(\hat{\rho},\hat{\sigma}\right)$ such that (1)
$\hat{\rho}\left(t\right)=0$ for $t\leq\hat{T}$, (2)
$\frac{d\hat{\rho}\left(t\right)}{dt}\mid_{t\geq\hat{T}}>0,$ (3)
$\hat{T}\in\left[\hat{T}_{l},\hat{T}_{r}\right]$ and (4)
$\hat{\sigma}\left(t\right)=0$ for $\forall t\geq 0$. That is, a randomised
stopping time strategy such that the players acquire information with
probability one until some time $\hat{T}>0$ and then start randomising between
stopping and acquiring information. I first define three parameters $T_{l},$
$T_{r},$ $p^{*}$ and show their existence. Then, I show the conditions that
$\hat{\rho}$ satisfies in equilibrium and under what conditions such
equilibrium exists.

Let

$\displaystyle
T_{r}:=\min\left\\{t:\frac{p_{0}}{1-p_{0}}e^{\left(b-a\right)t}=\frac{b\left(-u_{L}\right)-c}{c+ae^{-at}\bar{\triangle}_{H}}\right\\}$
(14)

and

$\displaystyle
T_{l}:=\min\left\\{t:\frac{p_{0}}{1-p_{0}}e^{\left(b-a\right)t}=\frac{-u_{L}}{u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H}}\right\\}.$
(15)

######  Lemma 12.

If $p_{0}<\tilde{p}$, then, $T_{r}>0$ exists. If $p_{0}<p^{L}$, then,
$T_{l}>0$ exists.

Let

$\displaystyle\frac{p^{*}\left(t\right)}{1-p^{*}\left(t\right)}:=\frac{\frac{c}{b}}{u_{H}-\frac{1}{2}\bar{\triangle}_{H}-\frac{c}{a}+J_{2}\left(t\right)}$
(16)

where

$J_{2}\left(t\right)=\left[\frac{1}{2}e^{-at}\bar{\triangle}_{H}+\frac{c}{a}-\left(-u_{L}-\frac{c}{b}\right)\frac{1}{L_{t}}\right]e^{-at}.$

Let

$\underline{p}^{*}\left(p_{0}\right):=p^{*}\left(T_{r}\right)$

be $p^{*}\left(t\right)$ evaluated at $t=T_{r}$. I denote it as
$\underline{p}^{*}\left(p_{0}\right)$ because $T_{r}$ depends on $p_{0}$. Let
$p^{*}$ be the fixed point such that
$p^{*}=\underline{p}^{*}\left(p^{*}\right).$

######  Lemma 13.

Suppose $c$ is sufficiently small. There exists a $p^{*}<p^{L}$ such that
$p^{*}=\underline{p}^{*}\left(p^{*}\right).$ We have $p^{*}<p_{0}$ if and only
if $\underline{p}^{*}\left(p_{0}\right)<p_{0}.$

###### Proof.

When $0<p_{0}<p^{L},$ $\underline{p}^{*}\left(\cdot\right)$ decreases in
$p_{0}$. When $p_{0}\rightarrow 0,$
$p_{0}<\underline{p}^{*}\left(p_{0}\right)<p^{L}.$ When $p_{0}\rightarrow
p^{L}$ and $c$ sufficiently small, we have
$\underline{p}^{*}\left(p_{L}\right)<\underline{p}^{*}\left(0\right)<p^{L}.$
Therefore, there exists $p^{*}<p^{L}$ such that
$p^{*}=\underline{p}^{*}\left(p^{*}\right)$. Since
$\underline{p}^{*}\left(\cdot\right)$ decreases in $p_{0}$, $p^{*}<p_{0}$ if
and only if $\underline{p}^{*}\left(p_{0}\right)<p_{0}.$ ∎

Next, I show the existence of the equilibrium and the conditions $\hat{\rho}$
satisfies in equilibrium. Suppose player $j$ uses the mixed strategy
$\left(\hat{\rho},\hat{\sigma}\right)$ such that (1)
$\hat{\rho}\left(t\right)=0$ for $t\leq\hat{T}$, (2)
$\frac{d\hat{\rho}\left(t\right)}{dt}\mid_{t\geq\hat{T}}>0,$ (3)
$\hat{T}\in\left[\hat{T}_{l},\hat{T}_{r}\right]$ and (4)
$\hat{\sigma}\left(t\right)=0$ for $\forall t\geq 0$. Given the assumption
that $\hat{\rho}\left(t\right)=0$ for $t\leq\hat{T}$ and
$\hat{\rho}\left(t\right)>0$ for $t>\hat{T}$, consider time $\hat{T}$ as a new
time $0$ and denote the time line starting from $\hat{T}$ as $\tau$. Let
$\tau:=t-\hat{T}$ and
$\lambda\left(\tau\right):=\hat{\rho}\left(\tau+\hat{T}\right)$. Let
$q_{\tau}=p_{\tau+\hat{T}}$ be the belief and let
$\hat{U}_{R}\left(\tau\right)$ be player $i$’s payoff from taking action $R$
at time $\tau$, then,

$\displaystyle\hat{U}_{R}\left(\tau\right)=$ $\displaystyle
q_{\tau}\left[\left(1-\hat{F}_{H}\left(\tau\right)\right)\eta+\hat{F}_{H}\left(\tau\right)\left(u_{H}-\bar{\triangle}_{H}\right)\right]+\left(1-q_{\tau}\right)\left[u_{L}-\hat{F}_{L}\left(\tau\right)\bar{\triangle}_{L}\right]$

where

$\hat{F}_{H}\left(\tau\right)=1-e^{-a\tau-a\hat{T}}+e^{-a\tau-a\hat{T}}\lambda\left(\tau\right),$
$\hat{F}_{L}\left(\tau\right)=\int_{0}^{\tau+\hat{T}}e^{-bs}\lambda\left(s\right)ds$

and $\eta:=u_{H}-\left(1-e^{-a\hat{T}}\right)\bar{\triangle}_{H}.$ The
intuition is that when considering time $\hat{T}$ as a new time $0$, player
$i$’s problem is essentially the same as in step 4 with $\eta$ being the
payoff for the first action $R$ taker instead of $u_{H}$.

After time $\hat{T}$, the equilibrium condition requires player $i$ being
indifferent between acquiring the information and taking action $R$ at each
time instant $\tau>0$. That is,

$\displaystyle c=$
$\displaystyle\left(1-q_{\tau}\right)b\left[-\left(u_{L}-\hat{F}_{L}\left(\tau\right)\bar{\triangle}_{L}\right)\right]$
$\displaystyle-
q_{t}\frac{d\hat{F}_{L}\left(\tau\right)}{d\tau}\left(\eta-\left(u_{H}-\bar{\triangle}_{H}\right)\right)-\left(1-q_{\tau}\right)\frac{d\hat{F}_{L}\left(\tau\right)}{d\tau}\bar{\triangle}_{L}.$

After plugging in the expressions of $\hat{F}_{H}\left(t\right)$ and
$\hat{F}_{H}\left(t\right),$ we have

$\displaystyle\frac{d\lambda\left(\tau\right)}{d\tau}=$
$\displaystyle\frac{b\left(-u_{L}\right)-c-\hat{L}_{\tau}c+b\bar{\triangle}_{L}\intop_{0}^{\tau}e^{-bs}\frac{d\lambda\left(s\right)}{ds}ds-\hat{L}_{\tau}\bar{\triangle}_{H}ae^{-a\tau}\left(1-\lambda\left(t\right)\right)}{\hat{L}_{\tau}\left(\eta-\left(u_{H}-\bar{\triangle}_{H}\right)\right)e^{-a\tau}+\bar{\triangle}_{L}e^{-b\tau}}$
(17)

where $\hat{L}_{\tau}=\frac{q_{\tau}}{1-q_{\tau}}$. To have
$\frac{d\lambda\left(s\right)}{ds}>0$, we need

$\displaystyle\hat{L}_{\tau}$
$\displaystyle<\frac{b\left(-u_{L}\right)-c+b\bar{\triangle}_{L}\intop_{0}^{\tau}e^{-bs}\frac{d\lambda\left(s\right)}{ds}ds}{c+\left(\eta-\left(u_{H}-\bar{\triangle}_{H}\right)\right)ae^{-a\tau}\left(1-\lambda\left(\tau\right)\right)}.$

That is,

$\displaystyle L_{0}e^{\left(b-a\right)t}$
$\displaystyle<\frac{b\left(-u_{L}\right)-c+b\bar{\triangle}_{L}\intop_{0}^{t}e^{-bs}\frac{d\hat{\rho}\left(s\right)}{ds}ds}{c+\bar{\triangle}_{H}ae^{-at}\left(1-\hat{\rho}\left(t\right)\right)}$
(18)

for all $t>0$. Equation (18) is derived by substituting in $\tau=t-\hat{T}$,
$\lambda\left(\tau\right)=\hat{\rho}\left(t\right)$ and
$\eta=u_{H}-\left(1-e^{-a\hat{T}}\right)\bar{\triangle}_{H}$. The existence of
an increasing function $\hat{\rho}\left(\cdot\right)$ has been shown in Lemma
7. that satisfies (18).

Next, I characterise the condition that $\hat{T}$ satisfies in equilibrium. I
am going to show that there exists an interval
$\left[\hat{T}_{l},\hat{T}_{r}\right]$ such that an equilibrium exists when
$\hat{T}\in\left[\hat{T}_{l},\hat{T}_{r}\right]$. The idea is that before time
$\hat{T}$, player $i$ must strictly prefer to acquire the signal and after
time $\hat{T}$, player $i$ is indifferent between acquiring the signal and
taking action $R$ at each time instant. To have the player strictly prefer to
acquire the signal before time $\hat{T}$, we need the marginal cost smaller
than the marginal benefit associated with acquiring the signal. That is,

$\displaystyle
L_{0}e^{\left(b-a\right)t}<\frac{b\left(-u_{L}\right)-c}{c+ae^{-at}\bar{\triangle}_{H}}$
(19)

for all $t\leq\hat{T}$. The upperbound $\hat{T}_{r}$ is the first time the
marginal cost of acquiring the signal exceeds the marginal benefit. Given
Lemma 12, if $p_{0}<\tilde{p}$, then, $\hat{T}_{r}=T_{r}>0$ exists.

The lowerbound of $\hat{T}$ is the earliest time point at which the player is
willing to start randomising. That is, if player $j$ starts randomising at
time $\hat{T}$, player $i$ must prefer to start randomising at time $\hat{T}$
instead of taking action $S$. At time $\hat{T}$, the value associated with
randomisation is the same as the value associated with taking action $R$
because of the opponent’s randomisation. Therefore, in order to have player
$i$ prefer randomisation to taking action $S$ at time $\hat{T}$, we need

$\displaystyle
p_{\hat{T}}\left[u_{H}-\left(1-e^{-a\hat{T}}\right)\bar{\triangle}_{H}\right]+\left(1-p_{\hat{T}}\right)u_{L}$
$\displaystyle\geq 0.$

That is,

$\displaystyle L_{0}e^{\left(b-a\right)\hat{T}}$
$\displaystyle\geq\frac{-u_{L}}{u_{H}-\left(1-e^{-a\hat{T}}\right)\bar{\triangle}_{H}}.$
(20)

The lowerbound of $\hat{T}$ is the smallest $\hat{T}$ such that inequality
(20) holds. Given Lemma 12, if $p_{0}<p^{L}$, then, $\hat{T}_{l}=T_{l}>0.$ If
$p_{0}\geq p^{L}$, then, $\hat{T}_{l}=0.$

The following lemma characterises the equilibrium when
$p^{*}<p_{0}<\tilde{p}$.

######  Lemma 14.

Suppose $b<2a$ and $c$ is sufficiently small. If $p^{*}<p_{0}<\tilde{p}$,
there exists an equilibrium in mixed strategies
$\left(\hat{\rho},\hat{\sigma}\right)$ such that
$\hat{\sigma}\left(t\right)=0$ for $\forall t\in\mathbb{R}_{+}$ and
$\hat{\rho}\left(\cdot\right)$ satisfies the following conditions:

  1. 1.

$\hat{\rho}\left(t\right)=0$ for $t\leq\hat{T}$

  2. 2.

$\hat{\rho}\left(t\right)\in\left(0,1\right]$ for
$t\in\left(\hat{T},\bar{T}\right]$ and
$\hat{\lambda}\left(\tau\right)=\hat{\rho}\left(\tau+\hat{T}\right)$ is a
solution to the differential equation (17) with initial condition
$\hat{\lambda}\left(0\right)=0$ where $\frac{d\hat{\rho}}{dt}>0$ for $\forall
t\in\left[\hat{T},\bar{T}\right]$

  3. 3.

$\hat{\rho}\left(t\right)=1$ for $\forall t>\bar{T}$

  4. 4.

$\hat{T}\in\left[\hat{T}_{l},\hat{T}_{r}\right]$ where $\hat{T}_{r}=T_{r}$,
$\hat{T}_{l}=T_{l}$ if $p_{0}<p^{L}$ and $\hat{T}_{l}=0$ if $p_{0}\geq p^{L}$.

###### Proof.

I have shown the existence of an increasing $\hat{\rho}$ function, and (17)
guarantees that the players stop and take action $R$ with a positive rate at
each time $t\in\left[\hat{T},\bar{T}\right]$. I have also derived the
conditions for $\hat{T}$. What left to show is the equilibrium exists when
$p_{0}>p^{*}$. Given Lemma 13, $p^{*}<p_{0}$ if and only if
$\underline{p}^{*}\left(p_{0}\right)<p_{0}.$ What left to show is the
equilibrium exists when $\underline{p}^{*}\left(p_{0}\right)<p_{0}.$. Suppose
player $j$ uses $\left(\hat{\rho},\hat{\sigma}\right)$ strategy described in
the lemma. Since I have discussed what happens after time $\hat{T}$, I will
characterise player $i$’s value $W\left(\cdot\right)$ at time $t<\hat{T}$. The
HJB equation is

$\displaystyle\max\left\\{p_{t}a\left[u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H}-W\left(t\right)\right]\right.$
$\displaystyle+\left(1-p_{t}\right)b\left[-W\left(t\right)\right]-c+W^{\prime}\left(t\right),$
$\displaystyle\left.U\left(t\right)-W\left(t\right)\right\\}$
$\displaystyle=0$

If the learning region exists, in the learning region, we have

$\displaystyle W\left(t\right)=$ $\displaystyle
p_{t}\left(\left(u_{H}-\bar{\triangle}_{H}\right)-\frac{c}{a}+\frac{1}{2}\bar{\triangle}_{H}e^{-at}\right)+\left(1-p_{t}\right)\left(-\frac{c}{b}\right)+p_{t}e^{at}J_{2}$

where $J_{2}$ is a constant. At time $\hat{T}$, we have
$W\left(\hat{T}\right)=U\left(\hat{T}\right)=U_{R}\left(\hat{T}\right)$. This
pins down $J_{2}$ as a function of $\hat{T}$, where

$J_{2}\left(\hat{T}\right)=\left[\frac{1}{2}e^{-a\hat{T}}\bar{\triangle}_{H}+\frac{c}{a}-\left(-u_{L}-\frac{c}{b}\right)\frac{1}{L_{\hat{T}}}\right]e^{-a\hat{T}}.$

Player 1 acquires the signal at time $t=0$ if $W\left(0\right)\geq 0$, which
requires

$\displaystyle
L_{0}\geq\frac{\frac{c}{b}}{u_{H}-\frac{1}{2}\bar{\triangle}_{H}-\frac{c}{a}+J_{2}\left(\hat{T}\right)}:=\underline{L}\left(\hat{T}\right):=\frac{p^{*}\left(\hat{T}\right)}{1-p^{*}\left(\hat{T}\right)}.$
(21)

The discussion shows that given the opponent uses the strategy described in
the lemma, player $i$’s best response is to use the Randomised Stopping Time
Strategy if $p_{0}\geq p^{*}\left(\hat{T}\right).$

Since $\hat{T}$ can be any value in the interval
$\left[\hat{T}_{l},\hat{T}_{r}\right]$, I use this to characterise the
lowerbound of the prior $p_{0}$ such that the equilibrium described in Lemma
14 exists.

######  Lemma 15.

$\underline{p}^{*}\left(p_{0}\right)\leq p^{*}\left(\hat{T}\right)$ for
$\hat{T}\in\left[\hat{T}_{l},\hat{T}_{r}\right]$.

Lemma 15 is true because when $\hat{T}\leq\hat{T}_{r}$,
$J_{2}\left(\cdot\right)$ increases in its argument. Since
$\frac{p^{*}\left(\hat{T}\right)}{1-p^{*}\left(\hat{T}\right)}$ decreases in
$J_{2}$, given $\hat{T}\leq\hat{T}_{r}$, we have

$\underline{p}^{*}\left(p_{0}\right)\leq p^{*}\left(\hat{T}\right).$

∎

Lemma 15 implies that for any $p_{0}>\underline{p}^{*}\left(p_{0}\right)$,
there exists an equilibrium where the players use the Randomised Stopping Time
Strategy as described in Lemma 14.

#### Step 6

Suppose player $j$ uses the Mixed Learning Strategy
$\left(\rho^{ML},\sigma^{ML}\right)$ such that (1)
$\sigma^{ML}\left(t\right)=\beta>0$ for $\forall t\geq 0$, (2)
$\rho^{ML}\left(t\right)\in\left(0,1\right]$ for
$t\in\left(\hat{T}^{\beta},\bar{T}^{\beta}\right],$ (3)
$\rho^{ML}\left(t\right)=1-\beta$ for $\forall t>\bar{T}^{\beta},$ and (4)
$\hat{T}^{\beta}\in\left[\hat{T}_{l}^{\beta},\hat{T}_{r}^{\beta}\right]$ where

$\hat{T}^{\beta}_{r}:=\min\left\\{t:L_{0}e^{\left(b-a\right)t}=\frac{b\left(-u_{L}\right)-c}{c+\left(1-\beta\right)ae^{-at}\bar{\triangle}_{H}}\right\\}.$

and

$\hat{T}_{l}^{\beta}:=\min\left\\{t:L_{0}e^{\left(b-a\right)t}=\frac{-u_{L}}{e^{-aT}u_{H}-\left(1-\beta\right)\left(e^{-at}\right)\bar{\triangle}_{H}}\right\\}.$

That is, she uses the Immediate action $S$ Strategy and the Randomised
Stopping Time Strategy with probability $\beta\in\left(0,1\right)$ and
$1-\beta$ such that $\hat{T}^{\beta}$ is the time at which player $j$ starts
randomising conditional on she uses the Randomised Stopping Time Strategy.

Given the assumption that $\rho^{ML}\left(t\right)=0$ for
$t\leq\hat{T}^{\beta}$ and $\rho^{ML}\left(t\right)>0$ for
$t>\hat{T}^{\beta}$, consider time $\hat{T}^{\beta}$ as a new time $0$ and
denote the time line starting from $\hat{T}^{\beta}$ as $\tau$. Let
$\tau:=t-\hat{T}^{\beta}$ and
$\lambda^{ML}\left(\tau\right):=\rho^{ML}\left(\tau+\hat{T}\right)$. Let
$q_{\tau}^{ML}:=p_{\tau+\hat{T}}$ be the belief and let
$L_{\tau}^{ML}:=\frac{q_{\tau}^{ML}}{1-q_{\tau}^{ML}}$ be the likelihood
ratio. Following the same discussion as in step 5, at time
$t\geq\hat{T}^{\beta},$ that is, $\tau\geq 0$, player $i$ is indifferent
between acquiring information and taking action $R$ if

$\displaystyle\frac{d\lambda^{ML}\left(\tau\right)}{d\tau}=$
$\displaystyle\frac{b\left(-u_{L}\right)-c-L_{\tau}^{ML}c+b\bar{\triangle}_{L}\intop_{0}^{\tau}e^{-bs}\frac{d\lambda^{ML}\left(s\right)}{ds}ds-
L_{\tau}^{ML}\bar{\triangle}_{H}ae^{-a\tau}\left(1-\lambda^{ML}\left(t\right)\right)}{L_{\tau}^{ML}\left(\eta^{ML}-\left(u_{H}-\bar{\triangle}_{H}\right)\right)e^{-a\tau}+\bar{\triangle}_{L}e^{-b\tau}}$
(22)

where
$\eta^{ML}:=u_{H}-\left(1-\beta\right)\left(1-e^{-a\hat{T}}\right)\bar{\triangle}_{H}.$
The intuition is that when considering time $\hat{T}^{\beta}$ as a new time
$0$, player $i$’s problem is essentially the same as in step 5 with
$\eta^{ML}$ being the payoff for the first action $R$ taker instead of $\eta$.
The existence of an increasing function $\rho^{ML}\left(\cdot\right)$ can be
shown following the same logic as in the proof of Lemma 7.

Next, I characterise the upperbound and lowerbound of $\hat{T}^{\beta}$.
Following similar argument as in step 5, the upperbound $\hat{T}^{\beta}_{r}$
is the first time the marginal cost of acquiring information exceeds the
marginal benefit. At time $0\leq t<\hat{T}^{\beta}$, given player $j$’s
strategy, player $i$’s payoff associated with taking action $R$ at time $t$ is

$\displaystyle U_{R}^{\beta}\left(t\right)=$
$\displaystyle\beta\left[p_{t}u_{H}+\left(1-p_{t}\right)u_{L}\right]$
$\displaystyle+\left(1-\beta\right)\left[p_{t}\left(u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H}\right)+\left(1-p_{t}\right)u_{L}\right].$

That is,

$\displaystyle U_{R}^{\beta}\left(t\right)=$ $\displaystyle p_{t}\left[\beta
u_{H}+\left(1-\beta\right)\left(u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H}\right)\right]+\left(1-p_{t}\right)u_{L}.$

The marginal cost of acquiring information is smaller than the marginal
benefit if

$\displaystyle\frac{b\left(-u_{L}\right)-c}{c+\left(1-\beta\right)ae^{-at}\bar{\triangle}_{H}}$
$\displaystyle>L_{0}e^{\left(b-a\right)t}$

Then, if
$L_{0}<\frac{b\left(-u_{L}\right)-c}{c+\left(1-\beta\right)a\bar{\triangle}_{H}}$,
there exists a $\bar{T}^{\beta}_{r}>0$ such that

$\hat{T}^{\beta}_{r}:=\min\left\\{t:L_{0}e^{\left(b-a\right)t}=\frac{b\left(-u_{L}\right)-c}{c+\left(1-\beta\right)ae^{-at}\bar{\triangle}_{H}}\right\\}.$

The lowerbound of $\hat{T}^{\beta}_{l}$ is the earliest time point at which
the player is willing to start randomising instead of taking action $S$ given
player $j$’s strategy. Following similar argument as in step 5, if
$p_{0}<p^{L}$, then, there exists a $\bar{T}^{\beta}_{l}>0$ such that

$\hat{T}_{l}^{\beta}:=\min\left\\{t:L_{0}e^{\left(b-a\right)t}=\frac{-u_{L}}{e^{-aT}u_{H}-\left(1-\beta\right)\left(e^{-at}\right)\bar{\triangle}_{H}}\right\\}.$

The following lemma characterises the equilibrium when
$\underline{p}<p_{0}<p^{*}$.

tbc

######  Lemma 16.

Suppose $b<2a$ and $c$ is sufficiently small. If $\underline{p}<p_{0}<p^{*}$,
then, there exists an equilibrium where both players use the Mixed Learning
Strategy $\left(\rho^{ML},\sigma^{ML}\right)$ such that

  1. 1.

$\sigma^{ML}\left(t\right)=\beta>0$ for $\forall t\geq 0$.

  2. 2.

$\rho^{ML}\left(t\right)=0$ for $t<\hat{T}^{\beta}$.

  3. 3.

$\rho^{ML}\left(t\right)\in\left(0,1-\beta\right]$ for
$t\in\left(\hat{T}^{\beta},\bar{T}^{\beta}\right]$ and
$\lambda^{ML}\left(\tau\right)=\rho^{ML}\left(\tau+\hat{T}\right)$ is a
solution to the differential equation (22) with initial condition
$\lambda^{ML}\left(0\right)=0$ where $\frac{d\rho^{ML}}{dt}>0$ for $\forall
t\in\left[\hat{T}^{\beta},\bar{T}^{\beta}\right]$.

  4. 4.

$\rho^{ML}\left(t\right)=1-\beta$ for $\forall t>\bar{T}^{\beta}.$

  5. 5.

$\hat{T}^{\beta}\in\left[\hat{T}_{l}^{\beta},\hat{T}_{r}^{\beta}\right].$

###### Proof.

Suppose player $j$ uses the Mixed Learning Strategy
$\left(\rho^{ML},\sigma^{ML}\right)$.

I have shown that at time $t\geq\hat{T}^{\beta}$, player $i$ is indifferent
between taking action $R$ and acquiring information for $dt$ longer and that
at time $0<t<\hat{T}^{\beta}$, player $i$ prefers acquiring information to
taking action $R$. What left is to show that if
$\underline{p}<p_{0}<\underline{p}^{*}$, there exists a
$\beta\in\left(0,1\right)$ such that player $i$ is indifferent between the
Immediate action $S$ Strategy and the Randomised Stopping Time Strategy at
time $0$. Given Lemma 13, I show that such $\beta\in\left(0,1\right)$ exists
if $\underline{p}<p_{0}<\underline{p}^{*}\left(p_{0}\right)$.

At time $t<\hat{T}^{\beta}$, player $i$’s value associated with acquiring
information is

$\displaystyle W^{\beta}\left(t\right)=$ $\displaystyle
p_{t}\left[u_{H}-\left(1-\beta\right)\bar{\triangle}_{H}+\frac{1}{2}\left(1-\beta\right)\bar{\triangle}_{H}e^{-at}-\frac{c}{a}\right]$
$\displaystyle-\left(1-p_{t}\right)\frac{c}{b}+p_{t}e^{at}H\left(\hat{T}^{\beta}\right)$

where

$H\left(\hat{T}^{\beta}\right)=\left[\frac{1}{2}\left(1-\beta\right)\bar{\triangle}_{H}e^{-a\hat{T}^{\beta}}+\frac{c}{a}-\frac{1}{L_{\hat{T}^{\beta}}}\left(-u_{L}-\frac{c}{b}\right)\right]e^{-a\hat{T}^{\beta}}$

is a constant. The value of acquiring the information at time $t=0$ equals the
value associated with taking action $S$ without acquiring the signal if
$W^{\beta}\left(0\right)=0.$ That is,

$\displaystyle L_{0}$
$\displaystyle=\frac{\frac{c}{b}}{u_{H}-\frac{1}{2}\bar{\triangle}_{H}+\frac{1}{2}\beta\bar{\triangle}_{H}-\frac{c}{a}+H\left(\hat{T}^{\beta}\right)}:=\underline{L}^{\beta}\left(\hat{T}^{\beta}\right).$

Then, $\beta$ can be pinned down by
$L_{0}=\underline{L}^{\beta}\left(\hat{T}^{\beta}\right)$.

######  Lemma 17.

When $\underline{p}<p_{0}<\underline{p}^{*}\left(p_{0}\right)$, there exists a
$\beta\in\left(0,1\right)$ such that
$L_{0}=\underline{L}^{\beta}\left(\hat{T}^{\beta}_{r}\right)$.

###### Proof.

The proof uses the intermediate value theorem. When $\beta=0$, we have
$\underline{L}^{\beta=0}\left(\hat{T}^{\beta=0}_{r}\right)=\underline{L}\left(\hat{T}_{r}\right).$
Since $p_{0}<\underline{p}^{*}\left(p_{0}\right),$ we have
$\underline{L}^{\beta=0}\left(\hat{T}^{\beta=0}_{r}\right)=\underline{L}\left(\hat{T}_{r}\right)>L_{0}.$

Next, I show that when $\beta=1$, we have
$\underline{L}^{\beta=1}\left(\hat{T}^{\beta=1}_{r}\right)<\underline{L}.$
When $\beta=1$, $\hat{T}^{\beta=1}_{r}$ satisfies

$L_{0}e^{\left(b-a\right)\hat{T}^{\beta=1}_{r}}=\frac{b\left(-u_{L}\right)-c}{c}=\bar{L}.$

Then,

$\underline{L}^{\beta=1}\left(\hat{T}_{r}^{\beta=1}\right)=\frac{\frac{c}{b}}{u_{H}-\frac{c}{a}+H\left(\hat{T}_{r}^{\beta=1}\right)}$

where
$H\left(\hat{T}_{r}^{\beta=1}\right)=\left(\frac{c}{a}-\frac{c}{b}\right)e^{-a\hat{T}_{r}^{\beta}}.$
Then, we have

$\underline{L}^{\beta=1}\left(\hat{T}_{r}^{\beta=1}\right)\left[u_{H}-\frac{c}{a}\right]+\left(\frac{c}{a}-\frac{c}{b}\right)e^{-a\hat{T}_{r}^{\beta}}\underline{L}^{\beta=1}\left(\hat{T}_{r}^{\beta=1}\right)=\frac{c}{b}.$

Since $L_{0}e^{\left(b-a\right)\hat{T}^{\beta=1}_{r}}=\bar{L},$ from the proof
of proposition 1 we know that

$\underline{L}\left[u_{H}-\frac{c}{a}\right]+\left(\frac{c}{a}-\frac{c}{b}\right)e^{-a\hat{T}_{r}^{\beta}}\left(\frac{\underline{L}}{L_{0}}\right)^{\frac{a}{b-a}}\underline{L}=\frac{c}{b}.$

Since $\underline{p}<p_{0}$, we have
$\underline{L}^{\beta=1}\left(\hat{T}^{\beta=1}_{r}\right)<\underline{L}$ and
thus $\underline{L}^{\beta=1}\left(\hat{T}^{\beta=1}_{r}\right)<L_{0}.$

Since
$\underline{L}^{\beta=1}\left(\hat{T}^{\beta=1}_{r}\right)<L_{0}<\underline{L}^{\beta=0}\left(\hat{T}^{\beta=0}_{r}\right)$
there exists a $\beta\in\left(0,1\right)$ such that
$L_{0}=\underline{L}^{\beta}\left(\hat{T}^{\beta}_{r}\right)$. ∎

Given Lemma 17, we know that when
$\underline{p}<p_{0}<\underline{p}^{*}\left(p_{0}\right)$, there exists at
least a pair $(\beta\in(0,1),\hat{T}_{r}^{\beta})$ such that the players play
the Immediate action $S$ strategy with probability $\beta$ and player the
Randomised Stopping Time Strategy with probability $1-\beta$. ∎

##  Appendix E Proof of Lemma 2 and Proposition 2

By definition,
$\frac{p^{M}}{1-p^{M}}:=\frac{-\left(u_{L}-\underline{\triangle}_{L}\right)}{u_{H}-\underline{\triangle}_{H}}$
and $\frac{\bar{p}}{1-\bar{p}}=\frac{-u_{L}-\frac{c}{b}}{\frac{c}{b}}.$ When
$\frac{c}{b}<\frac{-u_{L}\left(u_{H}-\underline{\triangle}_{H}\right)}{u_{H}-\underline{\triangle}_{H}-\left(u_{L}-\underline{\triangle}_{L}\right)},$
we have $p^{M}<\bar{p}.$ Lemma 2 is shown.

The cutoff $\underline{p}$ satisfies

$\displaystyle\left[u_{H}-\frac{c}{a}\right]\frac{\underline{p}}{1-\underline{p}}+K\left(\frac{\underline{p}}{1-\underline{p}}\right)^{\frac{b}{b-a}}=\frac{c}{b}$
(23)

where
$K=\frac{c}{b}\left(\frac{b}{a}-1\right)\left(\frac{\bar{p}}{1-\bar{p}}\right)^{1-\frac{b}{b-a}}.$
When $c\rightarrow 0$, $\bar{p}\rightarrow 1$ and $\underline{p}\rightarrow
0$.

When $c\rightarrow 0$, the cutoff
$\frac{\tilde{p}}{1-\tilde{p}}=\frac{-bu_{L}-c}{a\bar{\triangle}_{H}+c}\rightarrow\frac{-bu_{L}}{a\bar{\triangle}_{H}}$
and hence
$\tilde{p}\rightarrow\frac{\frac{b}{a}\frac{-u_{L}}{\bar{\triangle}_{H}}}{1+\frac{b}{a}\frac{-u_{L}}{\bar{\triangle}_{H}}}\in\left(0,1\right).$

##  Appendix F Proof of Proposition 3

Suppose player $j$ uses the pure strategy $\left(T^{PS},S\right)$ where
$T^{PS}=\frac{1}{a}\log\frac{\bar{\triangle}_{H}}{-\left(u_{H}-\bar{\triangle}_{H}\right)}.$
I show that when the conditions in Proposition 3 are satisfied, player $i$’s
best respond is to use the same pure strategy. The method is ‘guess and
verify’. Given that player $j$ uses the pure strategy $\left(T^{PS},S\right)$,
player $i$’s time $t$ payoff from taking $R$ is

$\displaystyle
U_{R}\left(t\right):=\begin{cases}p_{t}\left[u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H}\right]+\left(1-p_{t}\right)u_{L}&t<T^{PS}\\\
p_{t}\left[u_{H}-\left(1-e^{-aT^{PS}}\right)\bar{\triangle}_{H}\right]+\left(1-p_{t}\right)u_{L}&t\geq
T^{PS}\end{cases}.$

Let

$U\left(t\right):=\max\left\\{U_{R}\left(t\right),0\right\\}$

be player $i$’s time $t$ payoff if she stops and takes an irreversible action.
Let $W\left(\cdot\right)$ be the value function associated with player $i$’s
best response. Then, the value function satisfies the following HJB equation

$\displaystyle\max\left\\{p_{t}a\left[\max\left\\{u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H},0\right\\}-W\left(t\right)\right]-\left(1-p_{t}\right)bW\left(t\right)+W^{\prime}\left(t\right)-c,\right.$
$\displaystyle\left.U\left(t\right)-W\left(t\right)\right\\}$
$\displaystyle=0.$ (24)

Let

$\displaystyle\frac{p^{NR}}{1-p^{NR}}:=\frac{\frac{c}{b}}{\frac{1}{2}\left(2u_{H}+\bar{\triangle}_{H}\right)-\frac{c}{a}-u^{S}+\psi\left(T^{PS}\right)}$

where

$\displaystyle\psi\left(t\right):=\left[\frac{1}{2}e^{-at}\bar{\triangle}_{H}+\frac{c}{a}-\left(-u_{L}-\frac{c}{b}\right)\frac{1-p_{t}}{p_{t}}\right]e^{-at}$

I prove the following lemma.

######  Lemma 18.

When $c$ sufficiently small, if $p_{0}\in\left(p^{NR},\tilde{p}\right)$, the
value function $W\left(\cdot\right)$ is

$\displaystyle W\left(t\right)=\begin{cases}W_{L}\left(t\right)&t<T^{PS}\\\
0&t\geq T^{PS}\end{cases}$

where

$\displaystyle
W_{L}\left(t\right)=p_{t}\left(\left(u_{H}^{R}-\bar{\triangle}_{H}\right)-\frac{c}{a}+\frac{1}{2}\bar{\triangle}_{H}e^{-at}\right)-\left(1-p_{t}\right)\frac{c}{b}+p_{t}e^{at}\psi\left(T^{PS}\right).$

###### Proof.

Let

$\displaystyle H\left(t,W\left(t\right),W^{\prime}\left(t\right)\right):=$
$\displaystyle
p_{t}a\left[\max\left\\{u_{H}-\left(1-e^{-at}\right)\bar{\triangle}_{H},0\right\\}-W\left(t\right)\right]$
$\displaystyle-\left(1-p_{t}\right)bW\left(t\right)+W^{\prime}\left(t\right)-c.$

I show that $W\left(t\right)$ is a viscosity solution of the HJB equation (F).
For all the points where $W\left(t\right)$ is differentiable: if $t<T^{PS}$,
then $W\left(t\right)\geq U\left(t\right)$; if $t>T^{PS}$, then
$H\left(t,W\left(t\right),W^{\prime}\left(t\right)\right)\leq 0$. At the point
where $W\left(t\right)$ is not differentiable, that is, $t=T^{PS}$, I show
that $H\left(t,W\left(t\right),z\right)\geq 0$ for $z\in D^{+}$ where
$D^{+}=\left[W_{L}^{\prime}\left(T^{PS}\right),0\right]$ and
$H\left(t,W\left(t\right),z\right)\leq 0$ for $z\in D^{-}$ where
$D^{-}=\emptyset$.

#### Step 1

For $t<T^{PS}$,

$\displaystyle W\left(t\right)-U\left(t\right)$
$\displaystyle=W_{L}\left(t\right)-U_{R}\left(t\right)$
$\displaystyle=p_{t}\left[-\frac{1}{2}\left(\bar{\triangle}_{H}\right)e^{-at}+e^{at}\psi\left(T^{PS}\right)-\frac{c}{a}\right]+\left(1-p_{t}\right)\left(-u_{L}-\frac{c}{b}\right).$

When $p_{0}<\tilde{p}$, $\psi\left(t\right)$ increases in $t$. Then,

$\displaystyle W\left(t\right)-U\left(t\right)$
$\displaystyle>p_{t}\left[-\frac{1}{2}\left(\bar{\triangle}_{H}\right)e^{-at}+e^{at}\psi\left(t\right)-\frac{c}{a}\right]+\left(1-p_{t}\right)\left(-u_{L}-\frac{c}{b}\right)=0$

#### Step 2

When $t>T^{PS}$, $W\left(t\right)=0$ and $W^{\prime}\left(t\right)=0.$
Therefore, $H\left(t,W\left(t\right),W^{\prime}\left(t\right)\right)=-c<0.$

#### Step 3

I first show that for sufficiently small $c$,
$W_{L}^{\prime}\left(T^{PS}\right)<0.$ We have

$\displaystyle W_{L}^{\prime}\left(t\right)=$
$\displaystyle\frac{dp_{t}}{dt}\left[u_{H}-\bar{\triangle}_{H}-\frac{c}{a}+\frac{1}{2}\bar{\triangle}_{H}e^{-at}\right]-p_{t}\frac{1}{2}a\bar{\triangle}_{H}e^{-at}$
$\displaystyle+\frac{dp_{t}}{dt}\frac{c}{b}+\frac{dp_{t}}{dt}e^{at}\psi\left(T^{PS}\right)+p_{t}ae^{at}\psi\left(T^{PS}\right)$

and if
$\frac{c}{a}<\left(-u_{L}-\frac{c}{b}\right)\frac{1-p_{T^{PS}}}{p_{T^{PS}}}$,

$\displaystyle W_{L}^{\prime}\left(T^{PS}\right)=$
$\displaystyle\frac{dp_{t}}{dt}\mid_{t=T^{PS}}\left[\frac{c}{b}-\left(-u_{L}-\frac{c}{b}\right)\frac{1-p_{T^{PS}}}{p_{T^{PS}}}\right]+p_{T^{PS}}a\left[\frac{c}{a}-\left(u^{S}-u_{L}-\frac{c}{b}\right)\frac{1-p_{T^{PS}}}{p_{T^{PS}}}\right]$
$\displaystyle<0.$

When $t=T^{PS}$, $H\left(t,W\left(t\right),W^{\prime}\left(t\right)\right)=0$.
Since $H\left(t,W\left(t\right),z\right)$ increases in $z$, we have
$H\left(t,W\left(t\right),z\right)\geq 0$ for $z\in D^{+}$.

∎

##  Appendix G Multi-player extension: proof of Proposition 4

In this extension, I generalise the two-player model to a multi-player model.
There are $N>2$ players in this model. The first $R$ taker gets the first
prize $u_{\omega}$ in state $\omega\in\left\\{H,L\right\\}$ and all other $R$
takers get the second prize $u_{\omega}-\bar{\triangle}_{\omega}$. In case of
the simultaneous move, the payoff is assumed to be the convex combination of
the first and second prizes. I focus on discussing the conditions for the
existence of the learning equilibrium where the players use the random
stopping strategy.