Alertness, Leadership, and Nascent Market Dynamics

Abstract

In a continuous-time model with uncertain market development, two potential entrants detect a nascent demand only if it reaches a firm-specific threshold. Entry occurs by investing irreversibly before competing in quantities. When leadership in the investment stage implies a first-mover advantage in the market stage, we examine how the firms’ relative “alertness” drives the equilibrium outcomes. If the firms detect the new demand relatively late, the entry strategies and resulting firm values differ qualitatively from those in standard real option games: (1) In case of symmetric detection, the probability of simultaneous entry is nonzero, and can be one, although demand is still nascent. When sequential entry occurs, there is no rent equalization, with the post-entry market advantage, resulting in higher equilibrium expected value to the leader; (2) in case of asymmetric detection, entry is always sequential, and the more alert firm maximizes value by delaying its investment to enter exactly when its short-sighted rival detects demand. The marginal effect of the market advantage on the leader’s equilibrium value increases in the inter-firm alertness differential; and (3) more demand volatility reduces the value differential across firms and makes less likely the impact of imperfect alertness on entry decisions.

Appendices

The first formal treatment of preemption appears in [24] in a deterministic framework, and extended in [32] to a stochastic environment. This appendix is adapted from the recent contribution by [12]. For a more general approach, see [50].

Appendix 1: Strategies

The first formal treatment of preemption appears in [24] in a deterministic framework, and extended in [32] to a stochastic environment. This appendix is adapted from the recent contribution by [12]. For a more general approach, see [50].

The information set for each firm is $h_{t}\doteq \left( Y_{t},(i,j)_{t}\right)$. The state variables $Y_{t}\ge 0$ and $(i,j)_{t}\in \{0,1\}^{2}$ describe the industrywide shock and the firms’ entry status, respectively ($i=0$ if firm f has not entered, and $i=1$ otherwise, while $j=0$ if $-f$ has not entered, and $j=1$ otherwise).

Given a history $h_{t}$ of state variables, denote by $A^{f}\left( h_{t}\right)$, the set of firm f’s pure actions available at any date t that follows detection by both firms (that is, after $\min \{\tau ^{f},\tau _{L}^{*},\bar{\tau }\}$). As a firm faces only two possible actions in the investment stage (it may invest or wait), for firm f, we have $A^{f}\left( h_{t}\right) =\{$invest, wait$\}$ if $i=0$, all j, and $A^{f}\left( h_{t}\right) =\{$wait$\}$ otherwise. The investment process stops as soon as both firms have entered. Then, denote by $\left( \alpha ^{f}\left( h_{t}\right) ,1-\alpha ^{f}\left( h_{t}\right) \right) \in [0,1]^{2}$, a probability distribution over the elements of $A^{f}\left( h_{t}\right)$, with $\alpha ^{f}\left( h_{t}\right)$ defined as the probability of investing at time t (with symmetric notations for firm $-f$).

At any t, firm f chooses a behavioral investment strategy:

Definition 1

A behavioral investment strategy of firm f at time t is a map which assigns to each history $h_{t}$ of state variables a probability distribution $\left( \alpha ^{f}\left( h_{t}\right) ,1-\alpha ^{f}\left( h_{t}\right) \right)$ over the elements of $A^{f}\left( h_{t}\right)$.

For simplicity, in what follows, we slightly abuse terminology by referring to $\alpha _{t}^{f}\doteq \alpha ^{f}\left( h_{t}\right)$ as firm f’s strategy at time t.

To characterize the chosen strategies, suppose that no firm has invested yet at t and consider a $2\times 2$ game $\Gamma$ in which each firm can choose either to invest or to wait.

In strategic form:

In the strategic-form representation of $\Gamma$, the function $V^{f}:[0,1]^{2}\rightarrow R_{+}^{1}$ describes firm f’s present expected value of replaying the same game as a function of the two firms’ respective probabilities of investing, $\alpha _{t}^{f}$ and $\alpha _{t}^{-f}$. With continuous time, the game can be repeated infinitely many times over the interval $[t+\mathrm{d}t]$, with d$t\rightarrow 0$.³⁰ It follows that, whenever $(\alpha _{t}^{f},\alpha _{t}^{-f})\ne (0,0)$, one of the two firms invests with probability 1.

In this game, firm f’s expected payoff is given by the recursive expression

$$\begin{aligned} V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})= & {} \alpha _{t}^{f}\alpha _{t}^{-f}S(y)+\alpha _{t}^{f}(1-\alpha _{t}^{-f})L(y)+(1-\alpha _{t}^{f})\alpha _{t}^{-f}F^{*}\left( y\right) \\&+\,(1-\alpha _{t}^{f})(1-\alpha _{t}^{-f})V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f}), \end{aligned}$$

where $F^{*}(y)$, L(y), and S(y) are defined in (4), (5), and (6). Here the displayed expression of $V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})$ is firm f’s value for a history $h_{t}$, with a symmetric expression for $-f$. With a Markov refinement, only the current state levels are used, for any information set, in the definition of the value functions $F^{*}(y)$, L(y), and S(y). This leads to the following solution concept:

Definition 2

At any $t\ge 0$ , given $h_{t}$, a pair of behavioral strategies $(\hat{\alpha } _{t}^{f},\hat{\alpha }_{t}^{-f})$ is a Markov Nash equilibrium of $\Gamma$ if $V^{f}(\hat{\alpha }_{t}^{f},\hat{\alpha }_{t}^{-f})\ge V^{f}(\alpha _{t}^{f},\hat{\alpha }_{t}^{-f})$ and $V^{-f}(\hat{ \alpha }_{t}^{f},\hat{\alpha }_{t}^{-f})\ge V^{-f}(\hat{\alpha } _{t}^{f},\alpha _{t}^{-f})$, all $\alpha _{t}^{f},\alpha _{t}^{-f}$ .

For subgame perfection, the condition must hold at each point in time.

Definition 3

A collection of behavioral strategies $\{(\hat{\alpha }_{t}^{f},\hat{\alpha } _{t}^{-f})\}_{t\ge 0}$ is a subgame perfect Markov Nash equilibrium if for every $t\in [0,\infty )$ the pair of behavioral strategies $(\hat{\alpha }_{t}^{f},\hat{\alpha }_{t}^{-f})$ is a Markov Nash equilibrium of $\Gamma$.

The next step is to compute equilibrium strategies. After a reorganization of terms in the expression of firm f’s expected payoff, define

$$\begin{aligned} V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})=\left\{ \begin{array}{l@{\quad }l} \frac{\alpha _{t}^{f}\alpha _{t}^{-f}S(y)+\alpha _{t}^{f}(1-\alpha _{t}^{-f})L(y)+(1-\alpha _{t}^{f})\alpha _{t}^{-f}F^{*}\left( y\right) }{ \alpha _{t}^{f}+\alpha _{t}^{-f}-\alpha _{t}^{f}\alpha _{t}^{-f}} &{} \text {if }(\alpha _{t}^{f},\alpha _{t}^{-f})\ne (0,0) \\ \frac{L(y)+F^{*}\left( y\right) }{2} &{} \text {if }\alpha _{t}^{f}=\alpha _{t}^{-f}=0 \end{array} \right. , \end{aligned}$$

which describes firm f’s objective for a given $h_{t}$ as a function of its own strategy $\alpha _{t}^{f}$ and of the rival’s strategy $\alpha _{t}^{-f}$.³¹

The first-order derivative of $V^{f}$ in $\alpha _{t}^{f}$ is

$$\begin{aligned} \frac{\partial V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}=\alpha _{t}^{-f}\frac{\alpha _{t}^{-f}\left( S(y)-L(y)\right) +L(y)-F^{*}\left( y\right) }{\left( \alpha _{t}^{f}+\alpha _{t}^{-f}-\alpha _{t}^{f}\alpha _{t}^{-f}\right) ^{2}}, \end{aligned}$$

(15)

with a symmetric expression for $-f$. The sign of the derivative in (15) is the same as the sign of the numerator, which depends on the comparison of S(y) with L(y), and of L(y) with $F^{*}(y)$. If the numerator is zero, the value $V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})$ does not depend on $\alpha _{t}^{f}$, and there is no cost to specify that firm f plays the same strategy as what is obtained when the numerator is either negative or positive.

We already know from Lemma 2 that $S(y)<(=)F^{*}(y)$ if and only $y<(=) \tilde{y}\in \left( y_{p},y_{F}^{*}\right)$, and from the definitions in (5, 6) that $S(y)\le L(y)$ for all y (the market profit to the leader is $\pi _{10}\ge \hat{\pi }$ when $y<y_{F}^{*}$ and is followed by $\pi _{L}\ge \hat{\pi }$ when $y\ge y_{F}^{*}$, while the profit is $\hat{\pi }$ all along in case of simultaneous entry). It remains to consider the partition of the support of y in three intervals, as follows.

Case 1: :

$y<y_{p}$. We know from above that $S(y)\le L(y)$, and from (5) and (6) that $L\left( y\right) <F^{*}\left( y\right)$. It follows from (15) that $\frac{\partial V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}<0$ and $\frac{\partial V^{-f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}<0$, for all $(\alpha _{t}^{f},\alpha _{t}^{-f})$, implying that f and $-f$ maximize $V^{f}$ and $V^{-f}$, respectively, by choosing the lowest possible probability of investing. Therefore, the unique equilibrium is $\hat{\alpha } _{t}^{f}=\hat{\alpha }_{t}^{-f}=0$.

Case 2: :

$y_{p}\le y<y_{F}^{*}$. Again $S(y)\le L(y)$ for all $y\le y_{F}^{*}$, and we know from (5-6) that $L\left( y\right) >\left( =\right) F^{*}\left( y\right)$ if $y_{p}<\left( =\right) y<y_{F}^{*}$. Therefore, the sign of the derivative in (15) is a priori indeterminate. There are now two situations that depend on the level of y relative to $\tilde{y}$.

First, if $y<\left( =\right) \tilde{y}$, we have $S\left( y\right) <\left( =\right) F^{*}\left( y\right)$ from Lemma 2. Moreover, either $y_{p}\le y<\tilde{y}<y_{F}^{*}$ and $S\left( y\right) <L\left( y\right)$, or $y_{p}<y=\tilde{y}<y_{F}^{*}$ and $S\left( y\right) =F^{*}\left( y\right) <L\left( y\right)$, so in both cases $S\left( y\right) <L\left( y\right)$. This leads to three possibilities that depend on the level of $\alpha _{t}^{-f}$:

• (i) Suppose that $0\le \alpha _{t}^{-f}<\frac{L(y)-F^{*}\left( y\right) }{L(y)-S(y)}$, implying that $\frac{\partial V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}\ge 0$, for all $\alpha _{t}^{f}$. Then firm f maximizes $V^{f}$ by choosing $\alpha _{t}^{f}=1$. Hence $\frac{\partial V^{-f}(1,\alpha _{t}^{-f})}{\partial \alpha _{t}^{-f}}=S(y)-F^{*}\left( y\right) <\left( =\right) 0$ for all $y<\left( =\right) \tilde{y}$, so that $-f$ maximizes $V^{-f}$ with $\alpha _{t}^{-f}=0$, and the solution is asymmetric.

• (ii) Suppose that $\frac{L(y)-F^{*}\left( y\right) }{L(y)-S(y)} <\alpha _{t}^{-f}\le 1$, implying that $\frac{\partial V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}<0$, for all $\alpha _{t}^{f}$. Then firm f maximizes $V^{f}$ by choosing $\alpha _{t}^{f}=0$. This implies in turn that $\frac{\partial V^{-f}(0,\alpha _{t}^{-f})}{ \partial \alpha _{t}^{-f}}=0$, for all $\alpha _{t}^{-f}>0$ sufficiently high to satisfy the initial supposition, leading to an asymmetric solution.

• (iii) The last possibility is $\alpha _{t}^{-f}=\frac{L(y)-F^{*}\left( y\right) }{L(y)-S(y)}$, implying that $\frac{\partial V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}=0$, all $\alpha _{t}^{f}\in [0,1]$. The latter continuum renders admissible the unique symmetric strategy

  $$\begin{aligned} \hat{\alpha }_{t}^{f}=\hat{\alpha }_{t}^{-f}=\frac{L(y)-F^{*}(y)}{L(y)-S(y) }, \end{aligned}$$

  (16)

  so that firm values are $V^{f}(\hat{\alpha }_{t}^{f},\hat{\alpha } _{t}^{-f})=V^{-f}(\hat{\alpha }_{t}^{f},\hat{\alpha }_{t}^{-f})=L(y)=F^{*}(y)>S(y)$.

Only in (iii) we have symmetry. From here on, we focus on this case. Then (16) implies that $\hat{\alpha }_{t}^{f}=\hat{\alpha }_{t}^{-f}$ describes a nonzero instantaneous probability of simultaneous entry for all $y\in (y_{p},y_{F}^{*})$, with $\lim _{y\rightarrow y_{p}}\hat{\alpha } _{t}^{f}=\lim _{y\rightarrow y_{p}}\hat{\alpha }_{t}^{-f}=0$, and $\left. \hat{ \alpha }_{t}^{f}\right| _{y=\tilde{y}}=\left. \hat{\alpha }_{t}^{-f}\right| _{y=\tilde{y}}=1$.

Second, if $y>\tilde{y}$, $S(y)>F^{*}\left( y\right)$ implies $\alpha _{t}^{-f}\left( S(y)-L(y)\right) +L(y)-F^{*}\left( y\right) >0$, so that $\frac{\partial V^{f}(\alpha _{t}^{f},\alpha _{t}^{-f})}{\partial \alpha _{t}^{f}}\ge 0$ from (15), all $\alpha _{t}^{-f}\ge 0$. Then firm f maximizes $V^{f}$ with $\alpha _{t}^{f}=1$. Hence, $\frac{ \partial V^{-f}(1,\alpha _{t}^{-f})}{\partial \alpha _{t}^{-f}}=S(y)-F^{*}\left( y\right) >0$, so that $-f$ maximizes $V^{-f}$ with $\alpha _{t}^{-f}=1$, leading to a symmetric solution:

$$\begin{aligned} \hat{\alpha }_{t}^{f}=\hat{\alpha }_{t}^{-f}=1. \end{aligned}$$

(17)

Case 3: :

$y_{F}^{*}\le y$. From (4), the follower enters at y “immediately after” the leader (hence the flow profit term remains $\pi _{F}\le \pi _{L}$), and $F^{*}(y)\le S(y)\le L(y)$, with equality only if instantaneous profits are symmetric ($\phi =\gamma =\lambda$). Therefore, as in the previous case (with $y>\tilde{y}$), the symmetric solution is (17).

Apeendix 2: Proof of Lemma 1

The value function $F^{*}(y)$, which is strictly increasing with y, is also strictly convex as

$$\begin{aligned} \frac{\mathrm{d}^{2}F^{*}(y)}{\mathrm{d}y^{2}}=\frac{I\beta }{y^{2}}\left( \frac{y}{ y_{F}^{*}}\right) ^{\beta }>0, \end{aligned}$$

(18)

all $y>0$, while L(y) is strictly (weakly) concave if and only if $\pi _{L}<(=)\pi _{10}$ because

$$\begin{aligned} \frac{\mathrm{d}^{2}L(y)}{\mathrm{d}y^{2}}=\frac{\beta \left( \beta -1\right) }{y^{2}}\left( \frac{y}{y_{F}^{*}}\right) ^{\beta }\frac{\pi _{L}-\pi _{10}}{r-\alpha } y_{F}^{*}<(=)0, \end{aligned}$$

(19)

all $y>0$. It follows that $F^{*}(y)-L(y)=0$ may admit up to two roots. Consider first the benchmark case $\phi =\lambda$ (Cournot), for which $\hat{y}_{F}^{*}\doteq \left. y_{F}^{*}\right| _{\phi =\lambda }$ is a root. Then $\hat{F}^{*}(0)=0>\hat{L}(0)=-I$, and

$$\begin{aligned} \left. \frac{\mathrm{d}\hat{F}^{*}(y)}{\mathrm{d}y}\right| _{y=\hat{y}_{F}^{*}}= \frac{\hat{\pi }}{r-\alpha }>\left. \frac{\mathrm{d}\hat{L}(y)}{\mathrm{d}y}\right| _{y= \hat{y}_{F}^{*}}=\frac{\hat{\pi }}{r-\alpha }-\beta \frac{\pi _{10}-\hat{ \pi }}{r-\alpha }, \end{aligned}$$

(20)

together are sufficient to conclude that there exists another positive root $\hat{y}_{p}<(=)\hat{y}_{F}^{*}$ for any $\gamma <(=)1$. Then, consider the more general case $\phi \le \lambda$. Define ${\mathcal {F}}\doteq F^{*}-\hat{F}^{*}$ and ${\mathcal {L}}\doteq L-\hat{L}$, to compare the slopes of $F^{*}(y)$ and L(y) with the slopes of $\hat{F}^{*}(y)$ and $\hat{L}(y)$, respectively. As $\beta >1$, we find

$$\begin{aligned} \frac{\mathrm{d}{\mathcal {F}}(y)}{\mathrm{d}y}=\frac{I}{y}\frac{\beta }{\beta -1}\left[ \left( \frac{y}{\hat{y}_{F}^{*}}\right) ^{\beta }-\left( \frac{y}{y_{F}^{*}} \right) ^{\beta }\right] <0 \end{aligned}$$

(21)

if $\phi <\gamma$, and

$$\begin{aligned} \frac{\mathrm{d}{\mathcal {L}}(y)}{\mathrm{d}y}=\frac{\beta }{r-\alpha }\left[ \left( \frac{y}{ \hat{y}_{F}^{*}}\right) ^{\beta -1}\left( \pi _{10}-\hat{\pi }\right) -\left( \frac{y}{y_{F}^{*}}\right) ^{\beta -1}\left( \pi _{10}-\pi _{L}\right) \right] >0 \end{aligned}$$

(22)

if $\gamma <\lambda$, all $y>0$. In the Stackelberg scenario, the slope of the second entrant’s value function is strictly lower than in Cournot, whereas the slope of the first entrant’s value function is higher. Together with $F^{*}\left( 0\right) =\hat{F}^{*}\left( 0\right) =0$ and $L\left( 0\right) =\hat{L}\left( 0\right) =-I$, this leads to the conclusion that $F^{*}$ and L intersect at a lower y than $\hat{F}^{*}$ and $\hat{L}$. This establishes that $y_{p}<\hat{y}_{p}$. In addition, it is sufficient to check that $F^{*}(y_{F}^{*})<\lim _{y\uparrow y_{F}^{*}}L(y)$ when $\phi <\lambda$ to conclude that $F^{*}$ and L cannot intersect for any $y>y_{p}$ on which the value functions are defined. $\square$

Appendix 3: Proof of Lemma 2

Set $\phi =\gamma$. From (4) and (6), the slope of $F^{*}(y)$ is (weakly) lower than the slope of S(y) for all $y<\left( =\right) y_{F}^{*}$ (with $\phi /\gamma =1$ implying $y_{F}^{*}=\hat{y} _{F}^{*}$ here). Moreover, $F^{*}(0)=0>S(0)=-I$ and $F^{*}(y_{F}^{*})=S(y_{F}^{*})$ imply that $F^{*}(y)>\left( =\right) S(y)$ for all $y<\left( =\right) y_{F}^{*}$. It is immediate to check that for all $y\le y_{F}^{*}$, (i) the slope of $F^{*}(y)$ is monotone decreasing when $\phi$ departs from $\gamma$ toward 0, while S(y) is unchanged, and (ii) the slope of S(y) is monotone increasing when $\gamma$ departs from $\phi$ toward 1, while $F^{*}(y)$ is unchanged. It follows that there exists a unique $\tilde{y}$ in $(0,y_{F}^{*}]$ such that $F^{*}(y)>\left( =\right) S(y)$ if and only if $y<\left( =\right) \tilde{y}$, where $\tilde{y}$ is strictly increasing in $\phi$ and decreasing in $\gamma$, and with $\lim _{\phi /\gamma \uparrow 1}\tilde{y}=\hat{y}_{F}^{*}$.

Next, observe from (5) that $L^{*}(y)$ is strictly concave, and from (6) that S(y) is linear, for all $y<\left( =\right) y_{F}^{*}$, so that $L^{*}(y)-S(y)=0$ has at most two roots. We know that $L^{*}(0)=S(0)$, and from (4–6) that $L^{*}(y_{F}^{*})>(=)S(y_{F}^{*})$ if and only if $\lambda /\gamma >(=)1$ . Therefore, $L^{*}(y)>S(y)$, all $y\in (0,y_{F}^{*})$, including $\tilde{y}$, hence $L^{*}(\tilde{y})>S(\tilde{y})=F^{*}(\tilde{y})$. Then on the same interval $(0,y_{F}^{*})$, it is sufficient to recall that $L^{*}(y)>F^{*}(y)$ if and only if $y>y_{p}$ to obtain $\tilde{y }>y_{p}$. Finally, $F^{*}(y)$ is monotone in $\phi$ with $\lim _{\phi \downarrow 0}F^{*}(y)=0$, and $S\left( y\right)$ remains constant, so when $\phi$ approaches 0 the root $\tilde{y}$ converges to the solution to $S\left( y\right) =0$, that is $\frac{r-\alpha }{\hat{\pi }}I$. $\square$

Appendix 4: Proof of Proposition 3

The probability that only one firm, say f, enters as a leader, while $-f$ waits and enters as a follower, is given by $p_{L}^{f}=\alpha _{t}^{f}\left( 1-\alpha _{t}^{-f}\right) +\left( 1-\alpha _{t}^{f}\right) \left( 1-\alpha _{t}^{-f}\right) p_{L}^{f}$. For all $(\alpha _{t}^{f},\alpha _{t}^{-f})\ne (0,0)$, a reorganization of terms gives

$$\begin{aligned} p_{L}^{f}=\alpha _{t}^{f}\frac{1-\alpha _{t}^{-f}}{\alpha _{t}^{f}+\alpha _{t}^{-f}-\alpha _{t}^{f}\alpha _{t}^{-f}}. \end{aligned}$$

(23)

Next, the probability of simultaneous entry is $p_{S}=\alpha _{t}^{f}\alpha _{t}^{-f}+\left( 1-\alpha _{t}^{f}\right) \left( 1-\alpha _{t}^{-f}\right) p_{S}$. For all $(\alpha _{t}^{f},\alpha _{t}^{-f})\ne (0,0)$, a reorganization of terms gives

$$\begin{aligned} p_{S}=\frac{\alpha _{t}^{f}\alpha _{t}^{-f}}{\alpha _{t}^{f}+\alpha _{t}^{-f}-\alpha _{t}^{f}\alpha _{t}^{-f}}. \end{aligned}$$

(24)

$\square$

In a symmetric equilibrium $\alpha _{t}=\alpha _{t}^{f}=\alpha _{t}^{-f}$, (23) and (24) simplify to $p_{L}^{f}=\frac{ 1-\alpha _{t}}{2-\alpha _{t}}$ and $p_{S}=\frac{\alpha _{t}}{2-\alpha _{t}}$ . It remains to plug in the latter expressions the equilibrium strategies $\left( \hat{\alpha }_{t}^{f},\hat{\alpha }_{t}^{-f}\right)$ derived in Appendix 1. Again there are two cases that depend on the level of y relative to $\tilde{y}:$

• If $y_{p}\le y<\tilde{y}$, the symmetric equilibrium strategies $\hat{\alpha }_{t}^{f}=\hat{\alpha }_{t}^{-f}=\frac{ L(y)-F^{*}(y)}{L(y)-S(y)}$ in (16) give

  $$\begin{aligned} p_{L}^{f}(y)=\frac{F^{*}(y)-S(y)}{L(y)+F^{*}(y)-2S(y)}\qquad \text {and}\qquad p_{S}(y)=\frac{L(y)-F^{*}(y)}{L(y)+F^{*}(y)-2S(y)}. \end{aligned}$$

  (25)

  Then $L(y_{p})=F^{*}(y_{p})\Rightarrow p_{L}^{f}(y_{p})=\frac{1}{2} >p_{S}(y_{p})=0$, and $L(y)>F^{*}(y)\Rightarrow p_{S}(y)>0$ for all $y\in (y_{p},\tilde{y})$.

• If $\tilde{y}\le y<y_{F}^{*}$, the symmetric equilibrium strategies $\hat{\alpha }_{t}^{f}=\hat{\alpha }_{t}^{-f}=1$ in (17) lead to

  $$\begin{aligned} p_{L}^{f}(y)=p_{F}^{-f}(y)=0\qquad \text {and}\qquad p_{S}(y)=1. \end{aligned}$$

  (26)

Next, suppose that $Y_{t}$ reaches $y_{S}$, with $y_{p}<y_{S}<\tilde{y}$, at a given date $t>0$. Immediately after detection, by continuity $y_{p}<\lim _{\Delta t\rightarrow 0}Y_{t+\Delta t}=y_{S}<\tilde{y}$. Then from (25), we have $\lim _{\Delta t\rightarrow 0}p_{L}^{f}(Y_{t+\Delta t})=p_{L}^{f}(y_{S})=\frac{F^{*}(y_{S})-S(y_{S}) }{L(y_{S})+F^{*}(y_{S})-2S(y_{S})}$, and $\lim _{\Delta t\rightarrow 0}p_{S}(Y_{t+\Delta t})=p_{S}(y_{S})=\frac{L(y_{S})-F^{*}(y_{S})}{ L(y_{S})+F^{*}(y_{S})-2S(y_{S})}$, which together give (8a) and (8b) in Proposition 3. By the same token (26) leads to (9).$\square$