A generalized framework for endogenous timing in duopoly games and an application to price-quantity competition

Abstract

This paper extends the analysis of duopoly market by distinguishing two types of competition: (i) the basic form of competition where each firm is unrestricted in its choice of price and quantity and (ii) the non-basic form of competition where firms’ strategic choices over price and quantity are limited a priori. Our analysis focuses on the former rather than the latter. Under a very general setting of concave industrial revenue and asymmetric convex costs, we show that each firm typically makes more profit in the subgame perfect Nash equilibrium (SPNE) of the leader-follower price-quantity competition, one of the basic competition forms, than in the SPNE of the leader-follower price competition and that each firm always makes more profit under simultaneous move price-quantity competition than under simultaneous move price competition. We establish a generalized framework for endogenous timing in duopoly games which is capable of embodying and overcoming the inconsistency across the existing three frameworks in the field. We highlight the advantages of a 3-period general framework.

Appendix

Proof of Lemma 1

Note that the left hand side of equation $\bar{\pi }_k^r (\tau _i )\!=\!\pi _k^*(p_i )$ strictly decreases with $\tau _{i}$ and the right hand side strictly increases with $p_i$ and that $\bar{\pi }_k^r ( 0)\!=\!\pi _k^*( {p_k^m })\ge \pi _k^*( {p_i })\!>\! 0 \!=\!\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} \bar{\pi }_k^r ( {D(p)})$ if $MC_k ( 0)\!<\!p_i \le p_k^m$. Thus, given $p_i$, there is a unique $\tau _i ( {p_i })$ such that $\bar{\pi }_k^r ( {\tau _i })\!=\!\pi _k^*( {p_i })$ and $0\le \tau _i ( {p_i })\!<\!\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} D( p)$. Moreover, $\tau _i ( {p_i })$ strictly decreases with $p_i ;\tau _i ( {p_k^m })\!=\! 0$ and $\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} \tau _i ( p)\!\le \! \mathop {\lim }\nolimits _{p\downarrow MC_k (0)} D( p)$. If $\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} \tau _i ( p)\!<\!\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} D( p)$, then the following conflict emerges: $0\!=\!\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} \pi _k^*( p)\!=\!\bar{\pi }_k^r ( {\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} \tau _i ( p)})>\bar{\pi }_k^r ( {\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} D( p)})\!=\!0$. The inequality is set up owing to that $\bar{\pi }_k^r$ is strictly decreasing.

In addition, $d\bar{\pi }_k^r ( {\tau _i })/d\tau _i =-p_k^r +MC_k ( {D( {p_k^r })-\tau _i })<0$ and

$$\begin{aligned} d\pi _k^*( p)/dp=\left\{ \begin{array}{l} \varphi _k ( p)>0\\ p\dot{D}( p)+D( p)-MC_k ( {D( p)})\dot{D}( p)>0\\ \end{array} \right. \, \text{ if } \, \begin{array}{l} p\le p_k^c\\ p_k^c <p\le p_k^m\\ \end{array} \end{aligned}$$

Thus, Result (1) holds.

Let $\tilde{p}_k$ satisfy $D(\tilde{p}_k ) -\tau _i ( {p_i}) \!=\!\varphi _k (\tilde{p}_k )$ as shown in Fig. 1. If $p_i \!>\!MC_k ( 0)$, then $\tilde{p}_k$ uniquely exists and $\tilde{p}_k \!>\!MC_{k} ( 0)$ owing to $\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} D( p)-\tau _i ({p_i }) > 0~\!=\!\mathop {\lim }\nolimits _{p\downarrow MC_k (0)} \varphi _k (p)$ and $D({p_k^c })\!-\!\tau _i ( {p_i })\!\le \! \varphi _k ( {p_k^c })$. Note that $\left. {\partial \pi _k^r ( {\left. p \right| \tau _i ( {p_i })})/\partial p} \right| _{p=\tilde{p}_k } =D( {\tilde{p}_k })-\tau _i ( {p_i })\!=\!\varphi _k ( {\tilde{p}_k })>0$. Thus, $p_k^r ( {\tau _i ( {p_i })})\!>\!\tilde{p}_k$ and $\pi _k^*( {\tilde{p}_k })<\bar{\pi }_k^r ( {\tau _i ( {p_i })})\!=\!\pi _k^*( {p_i })$. The latter inequality implies that $\tilde{p}_k \!<\! p_{i}$ owing to the strictly increasing $\pi _k^*(p)$ or that Result (2) holds because of the definition of $\tilde{p}_k$.

If $MC_k ( 0) <p_i <p_k^m$, then $\tau _i ( {p_i }) > 0$ and $(\tilde{p}_k ,p_i ]$ is non-empty. Firm $k$’s profit by naming any price in interval $(\tilde{p}_k ,p_i ]$ with the residual demand $D( p) -\tau _i ( {p_i })$ is always less than $\pi _k^*( {p_i })$, thus $p_k^r ( {\tau _i ( {p_i })})>p_i$ or Result (3) holds.

For Result (4), let $\bar{\pi }_k^{rd} ( {\tau _i ( {p_i })})$ be firm $k$’s maximal profit under the residual demand $D( p) -\tau _i ( {p_i })$ and its cost $C_k^d ( q)$; let $\pi _k ^{*d}( {p_i })$ be firm $i$’s maximum profit by naming price $p_{i}$ with $C_i^d ( q)$ being its cost; and let $q=\varphi _k^d ( p)$ be the inverse function of $q=MC_k^d ( q)$. If $D( {p_i }) -\tau _i ( {p_i })\ge \varphi _k^d ( {p_i })$, then $\tau _i^d ( {p_i })<\tau _i ( p)$ by Result (2). Otherwise let $p_k^r$ be the price at which $\pi _k^r ( {p\left| {\tau _i ( {p_i })} \right. })$ arrives at its maximal. Then $p_k^r >p_i$ and

$$\begin{aligned} \bar{\pi }_k^{rd} ( {\tau _i })&\ge p_k^r \left( {D\left( {p_k^r }\right) -\tau _i }\right) -C_k^d \left( {D\left( {p_k^r }\right) -\tau _i }\right) \\&= \bar{\pi }_k^r ( {\tau _i })-C_k^d \left( {D\left( {p_k^r }\right) -\tau _i }\right) +\,C_k \left( {D\left( {p_k^r }\right) -\tau _i }\right) \\&= \pi _k^*( {p_i })-C_k^d \left( {D\left( {p_k^r }\right) -\tau _i }\right) +C_k \left( {D\left( {p_k^r }\right) -\tau _i }\right) >\pi _k^{*d} ( {p_i }) \end{aligned}$$

The first inequality results from the definition of $\bar{\pi }_k^{rd} ( {\tau _i })$. The last inequality holds because $D( {p_k^r })-\tau _i ( {p_i })<D( {p_i })-\tau _i ( {p_i })<Min\{ {D( {p_i }),\varphi _k^d ( {p_i })} \}$. $\square$

Proof of Theorem 2

Take $\bar{p}$ such that $\bar{p}D( {\bar{p}})<Min\left\{ {\pi _1^*( {p^c}),\pi _2^*( {p^c})} \right\}$. Define $\bar{D}( p)$ equals $D( p)$ if $p<\bar{p}, D( {\bar{p}})( {2-p/\bar{p}})$ if $\bar{p}\le p\le 2\bar{p}$, and 0 if $p>2\bar{p}$. In the same way as proving the existence of the MSNE of the SMPQ competition in Gertner (1986), one can show that the SMPQ competition has an MSNE $( {G_1 ( {p_1 ,q_1 }),G_2 ( {p_2 ,q_2 })})$ in which the industrial demand $D(p)$ is replaced by $\bar{D}(p)$.

Now we show that the MSNE is also the MSNE of the SMPQ competition where the industrial demand is $D(p)$. Assume that firm $k$ takes strategy $G_k ( {p_k ,q_k })$ and firm $i$ takes strategy ($p$, $q)$. Denote firm $i$’s expected profit by $E_i ( {p,q})$ providing that the industrial demand is $D(p)$ and by $\bar{E}_i ( {p,q})$ providing that the industrial demand is $\bar{D}( p)$. (a) If $p\le \bar{p}$, then for any $q, E_i ( {p,q})=\bar{E}_i ( {p,q})$. (b) If $p>\bar{p}$, then for any $q$, $E_i ( {p,q})<\pi _i^*( {p^c})\le E( {\pi _i^s })$. The first inequality results from the definition of $\bar{p}$. The second holds because of Lemma 2. $\square$

Proof of Lemma 3

There is a unique $p_i^n$ such that $E( {\pi _i^s })=\pi _i^*( {p_i^n })$, where $E(\pi _i^s)$ is the expected profit of firm $i$ in the MSNE of the SMPQ competition. Then firm $i$ does not choose any strategy ($p,\,q)$ with $p<p_i^n$. Denote by $\left[ {p_i^{\min } ,\;p_i^{\max } } \right]$ the support set of distribution $\mu _i^*( {p_i })$, which implies that $p_i^n \le p_i^{\min } \le p_k^m$. It is safe to assume that $p_1^{\min } \le p_2^{\min }$. Then $p_1^n \le p_1^{\min } \le p_2^{\min }$ as Fig. 2 shows. We affirm that $p_1^n =p_1^{\min } =p_2^{\min }$. Otherwise $p_1^n <p_2^{\min }$. Then for any $\tilde{p}\in ( {p_1^n ,p_2^{\min } })$, $E_1 (\tilde{p},\;Min\{D(\tilde{p}),\;\varphi _1 (\tilde{p})\})=\pi _1^*( {\tilde{p}})>\pi _1^*(p_1^n )=E(\pi _1^{s} )$, this leads to a contradiction. For the symmetric reason, that $p_2^n <p_2^{\min } =p_1^{\min }$ can yields a contradiction. So $p_2^n =p_2^{\min } =p_1^{\min } =p_1^n \equiv p^{\min }$. In addition, $p^{\min }\ge p^c$. We affirm that $p^{\min }>p^c$. Otherwise, the SMPQ competition has a PSNE in which firm $i$’s strategy is ($p^{c}$, $\varphi _{i}(p^{c}))$ for $i$ = 1 and 2. This is in contradiction to that the SMPQ competition does not have a PSNE. Therefore Lemma 3 holds. $\square$

Proof of Lemma 4

Let $(G_1 (p_1 ,\;q_1 ),\;G_2 (p_2 ,\;q_2 ))$ be the accumulative distribution functions associated with the MSNE of the SMPQ competition. Assume that firm $i$ takes strategy ($p$, $q)$. Let $g_k (q_k )=G_k (p,q_k )-\mathop {\lim }\nolimits _{p^{\prime }\uparrow p} G_k ( p^{\prime },q_{k})$. If $\mu _i^*(p_i )$ is continuous at $p_i =p$, then $g_k (q_k )\equiv 0$. Thus,

$$\begin{aligned} \frac{\partial E_i (p,q)}{\partial q}&= \left( 1-\mu _k^*(p)\right) p+p\int _0^{+\infty } {h(q_k ,q)dg_k (q_k )}\\&\quad +p\mathop {\lim }\nolimits _{p^{\prime }\uparrow p} G_k ( p^{\prime },D(p)-q)-MC_i (q) \end{aligned}$$

with

$$\begin{aligned} h(q_k ,q)=\left\{ \begin{array}{l} 1 \\ D(p)q_{k}/(q+q_k )^{2}\\ \end{array} \right. \quad \text{ if } \quad \begin{array}{l} q+q_k \le D(p)\\ q+q_k >D(p)\\ \end{array} \end{aligned}$$

The first item on the right-hand side of the equation is independent of $q$, the second and third are decreasing with $q$, and the last one is strictly increasing with $q$. Thus, there exists a unique $q_i^*(p)$ at which $E_i (p,q)$ reaches its maximum. $\square$

Proof of Lemma 5

Assume that $p_k^{tp} \ge p_i^{tp}$. For any $p$, equation $pq_k^E-C_k ( {q_k^E })=\pi _k^*( {p^{\min }})$ can uniquely determine $q_k^E$, with $q_k^E =q_k^E (p)$. Then $q_k^*(p)\ge q_k^E (p)$ holds almost everywhere (a.e.) under firm $k$’s strategy in the MSNE.

• (a) If $\mu _i^*(p)$ is discontinuous at $p=p^{\min }$, then for any $q>D(p^{\min })-\varphi _i (p^{\min }),$

  $$\begin{aligned} \mathop {\lim }\limits _{p\downarrow p^{\min }} E_k (p,q)=\left( 1-\mu _i^*(p^{\min })\right) p^{\min }q-C_k (q)<\pi _k^*(p^{\min }) \end{aligned}$$

  and for any $q\le D(p^{\min })-\varphi _i (p^{\min })$,

  $$\begin{aligned} \mathop {\lim }\limits _{p\downarrow p^{\min }} E_k (p,q)=p^{\min }q-C_k (q)<\pi _k^*( {p^{\min }}) \end{aligned}$$

  As a result, the infimum of the support set of $\mu _k^*(p_k )$ is greater than $p^{\min }$, which is a contradiction.

• (b) If $p^{\min }<p_k^{tp}$, then $\bar{\pi }_k^r ( {\varphi _i ({p^{\min }})})>\pi _k^*( {p^{\min }})$. Thus, for any sufficiently small $\varepsilon >$ 0, the curve $q_k^E (p)$ intersects the curve $D( p)-\varphi _i ( {p^{\min }})-\varepsilon$. Denote the price at the lowest intersection point of the two curves by $p_k^s (\varepsilon )$. For any $p\in \left[ {p^{\min },p_k^s (\varepsilon )} \right] ,\varphi _i ( {p^{\min }})+\varepsilon <D(p).$

We affirm that for any $p\in (p^{\min },p_k^s (\varepsilon )],\,q_i^*(p)\le \varphi _i (p^{\min })+\varepsilon$ holds almost everywhere (a.e.) under firm $i$’s mixed strategy in the MSNE, which implies that $q_i^*(p)\le \varphi _i ( {p^{\min }})$ if $p<\mathop {\lim }\nolimits _{\,\varepsilon \downarrow 0} p_k^s (\varepsilon )$. Otherwise, there is a $p_i \le p_k^s (\varepsilon )$ such that $E_i ( {p_i ,q_i^*(p_i )})=E_i ( {\pi _i^s })$ and $q_i^*(p)>\varphi _i ( {p^{\min }})+\varepsilon$. We consider 2 cases.

• (i) If $\mu _k^*(p)$ is continuous at $p=p_i$, then

  $$\begin{aligned} \partial E_i (p_i ,q)/\partial q|_{q\;=\;\varphi _i (p^{\min })\!+\!\varepsilon }\! =\!\left( 1-\mu _k^*({p_i })\right) p_i\! -\!MC_k \left( \varphi _i ( {p^{\min }})\!+\!\varepsilon \right) \ge 0\qquad \end{aligned}$$

  (1)

  The equation holds because $q_k^*(p)\ge q_k^E (p)$ holds almost everywhere under firm $k$’s mixed strategy in the MSNE. The inequality results from that given $p$, $\partial E_i (p,q)/\partial q$ strictly decreases with $q$ as shown in Lemma 4 and that $q_i^*(p)\ge \varphi _i (p^{\min })+\varepsilon$. Thus,

  $$\begin{aligned} E_i (p,\bar{q})>( {1-\mu _k^*(p)})p\bar{q}-C_k (\bar{q})\ge MC_k (\bar{q})\bar{q}-C_k (\bar{q})>\pi _i^*( {p^{\min }}) \end{aligned}$$

  (2)

  where $\bar{q}=\varphi _i (p^{\min })+\varepsilon$. The first inequality in (2) results from the definition of $E_i (p,\bar{q})$. The second holds because of the inequality in (1). The last inequality holds because $qMC_k (q)-C_k (q)$ strictly increases with $q$. However, (2) is in conflict with $E_i (p,\bar{q})\le \pi _i^*( {p^{\min }})$.

• (ii) If $\mu _k^*( {p_k })$ is discontinuous at $p_k =p_i$, say, firm $k$ takes strategy $( {p_i ,q_i^*(p_i )})$ with a positive probability, then $q_k^*( {p_i })\ge q_k^E ( {p_i })$. Thus, for any $q>\varphi _i (p^{\min })+\varepsilon ,\,E_i (p_i ,q)<\mathop {\lim }\nolimits _{p^{\prime }\uparrow p_i } E_i (p^\prime ,q)\le \pi _i^*( {p^{\min }})$.

Owing to that $\mu _i^*(p)$ is continuous from the right, take an $\varepsilon >0$ such that for a sufficiently small $\delta > 0,$ $\mu ^*_i(p^s_k(\varepsilon ))-\mu ^*_i (\lim _{h\downarrow 0}p^s_k(h))<\delta$. Then for an $x\in (0,\varepsilon )$,

$$\begin{aligned}&E_k(p^s_k(\varepsilon ), q^E_k(p^s_k(\varepsilon ))+x)\\&\quad \ge p^s_k(\varepsilon )q^E_k(p^s_k(\varepsilon ))+(1-\delta )p^s_k(\varepsilon )x-C_k(q^E_k(p^s_k(\varepsilon ))+x)\\&\quad > p^s_k(\varepsilon )q^E_k(p^s_k(\varepsilon ))-C_k(q^E_k(p^s_k(\varepsilon )))=\pi ^*_k(p^{\min }). \end{aligned}$$

The first inequality results from the definition of $E_k(p,q)$. The second inequality holds owing to $p^s_k(\varepsilon )>\lim _{h\downarrow 0} p^s_k(h)\gg MC_k(q^E_k(p^s_k(\varepsilon )))$. Thus, firm $k$ can make a profit greater than $\pi ^*_k(p^{\min })$ by naming price $p^s_k(\varepsilon )$ and choosing an output slightly greater than $q^E_k(p^s_k(\varepsilon ))$.

Thus, firm $k$ can make a profit greater than $\pi _k^*( {p^{\min }})$ by naming price $p_k^{s*}$ and choosing an output slightly greater than $q_k^E ( {p_k^{s*} })$. $\square$

Proof of Lemma 6

It is safe to assume that $Max( {p_1^c ,p_2^c })=p_1^c$. Assume that $p^{\min }>p_1^c$. For any $p\ge p^{\min }$, define $\mu _i ( p)=(\pi _k^*(p)-\pi _k^*( {p^{\min }}))/( {pD( p)})$. We proceed through the following 4 steps to prove the lemma.

Step 1: (a) $\mu _i (p)$ is not constant in any interval; (b) $( {1-\mu _i (p)})p>MC_k ( {D(p)})$; (c) $\mu _i (p)< \text{ a } \text{ constant }<1$.

Proof:

• (a) If $\mu _i (p)\equiv \alpha$ (a constant), then $(1-\alpha )pD(p)-C(D(p))-\pi _k^*(p^{\min })\equiv 0$ or $(1-\alpha )\dot{D}(p)(MR(p)-MC_k (D(p)))\equiv 0$. However, the left-hand side of the latter equation is not identical to 0 in any interval.

• (b) The result (b) is equivalent to $\pi _k^*(p^{\min })>MC_k (D(p))D(p)-C_k (D(p))$. Note that the left-hand side of this inequality is strictly decreasing with $p$ for any $p>p_k^c$. So the inequality holds.

• (c) $\mu _i (p)<1-\pi _i^*( {p^{\min }})/( {pD(p)})$. Because $pD(p)$ has an upper bound, the result (c) holds.

Step 2: There is an $\varepsilon >$ 0 such that for any $p\in \left[ {p^{\min }, p^{\min }+ \varepsilon } \right)$ and $i\in \left\{ {1, 2} \right\}$, (i) $q_i^*(p)=D(p)$; (ii)$\mu _i^*(p)$ is strictly increasing and continuous in [$p^{\min }$, $p^{\min }+\varepsilon )$; (iii) $E_i (p,D(p))=E( {\pi _i^s })$; and (iv) $\mu _i^*(p)=\mu _i (p)$.

Proof:

• (i) There is an $\varepsilon >$ 0 such that for any $p\in \left[ {p^{\min }, p^{\min }+ \varepsilon } \right)$ and $i\in \left\{ {1, 2} \right\}$, $(1-\mu _i^*(p))p>p_1^c$ owing to that both $\mu _1^*$ and $\mu _2^*$ are continuous at $p^{\min }$. Thus, for any $q\le D(p)$, $\partial E_i (p,q)/\partial q\ge (1-\mu _k^*(p))p-MC_k (q)>p_1^c -MC_k (q)>0$. Therefore, $q_i^*(p)=D( p)_{.}$

• (ii) If $\mu _i^*(p)$ is constant in [$p^{\prime }$, $p^{\prime }$+), where $p^{\prime }$+ is greater than $p^{\prime }$, it is safe to assume that $p^\prime =Min\{\left. p \right| \mu _i^*(p)=\mu _i^*(p^\prime )\}$. Then $E_k ( {p^\prime ,D( {p^\prime })})=E( {\pi _k^s })$. For any $p\in \left[ {p^\prime ,p^\prime +} \right)$

  $$\begin{aligned} E_k (p,D(p))=\left( 1-\mu _i^*(p^\prime )\right) pD(p)-C_k (D(p)), \end{aligned}$$

  (3)

  Note that $E_k (p,D(p))\le E(\pi _k^s )=E_k (p^\prime ,D(p^\prime ))$, i.e. $E_k (p,D(p))$ reaches its maximum at $p^{\prime }$. Thus,

  $$\begin{aligned} dE_k (p,D(p))/dp=\dot{D}(p)\;\left( (1-\mu _i^*(p^{\prime }))MR\;(p)-MC_k (D(p))\right) <0. \end{aligned}$$

  The inequality holds because marginal revenue MR($p)$ strictly increasing with $p$ and $\dot{D}(p)<0$. Namely, for any $p > p^{\prime }, E_k (p,D(p))<E(\pi _k^s )=E_k (p^\prime ,D(p^\prime ))$. Moreover, for any ($p, q)$ with $p > p^{\prime }$,

  $$\begin{aligned} E_k (p,q)\!\le \! \left( 1\!-\!\mu _i^*(p^\prime )\right) pq\!-\!C_k (q)\!\le \! \left( 1\!-\!\mu _i^*(p^\prime )\right) pD(p)\!-\!C_k (D(p))\!<\!E\left( {\pi _k^s }\right) \!\!\!\!\!\nonumber \\ \end{aligned}$$

  (4)

  The first inequality in (4) results from the definition and $q_i^*(p)=D(p)$ for any $p\le p^\prime$. That $( {1-\mu _i^*( {p^\prime })})p>p_k^c$ implies that the second term strictly increases with $q$. So the second inequality holds. The last inequality holds because of the Eq. (3) and the inequality $E_k (p,D(p))<E( {\pi _k^s })$. Thus, firm $k$ does not choose the strategy ($p$, $q)$ with $p > p^{\prime }$ or $\mu _k^*(p^\prime )=1$. This conflicts with that $( {1-\mu _i^*( {p^\prime })})p^\prime >p_1^c$. If $\mu _i^*(p)$ is discontinuous at $p^{\prime \prime }$, then for any $q$, $\mathop {\lim }\nolimits _{p\downarrow p^{\prime \prime }} E_k (p,q)<\mathop {\lim }\nolimits _{p\uparrow p^{\prime \prime }} E_k (p,q)\le E( {\pi _k^s })$. Thus, $\mu _k^*(p)$ is constant on [$p^{\prime \prime }$, $p^{\prime \prime }$+) and this contradicts the result we just confirmed.

• (iii) If $E_i (p,D(p))<E( {\pi _i^s })$, owing to the result (ii) of Step 2, $E_i (p^{\prime \prime \prime \prime },D(p^{\prime \prime \prime \prime })<E( {\pi _i^s })$ for any p$^{\prime \prime \prime \prime }$ in a neighborhood of p, which implies that $\mu _i^*(p)$ is constant in the neighborhood. Therefore, for any $p\in \left[ {p^{\min },p^{\min }+ \varepsilon } \right)$ and $i\in \left\{ {1, 2} \right\}$, $E_i (p,D(p))=E( {\pi _i^s })$, which can immediately yield $\mu _i^*(p)=\mu _i (p)$ by applying the results (i) and (ii) of Step 2.

Step 3: Let $p^{x}$ be the maximum of all $p^{\prime \prime \prime }$ such that for any $p\in \left[ {p^{\min },p^{\prime \prime \prime }} \right)$ and $i\in \left\{ {1, 2} \right\}$, $\mu _i^*(p)=\mu _i (p)$. Then $p^{x }<$ +$\infty$ because of the result (c) of Step 1 and $\mu _i^*(p)$ is strictly increasing and continuous on [$p^{\min }$, $p^{x})$ owing to the result (a) of Step 1. Moreover, for $p\in \left[ {p^{\min },p^x} \right)$, $q_i^*(p)=D(p)$ and $E_i (p,D(p))=\pi _i^*(p^{\min })$. In fact,

$$\begin{aligned} \partial E_i (p,q)/\partial q\!\ge \! (1\!-\!\mu _i (p))p\!-\!MC_i (q)\!\ge \! (1\!-\!\mu _i (p))p\!-\!MC_i (D(p))\!>\!0 \end{aligned}$$

(5)

The first inequality in (5) results from the definition of $E_{i }(p, q)$. The last holds owing to the result (b) of Step 1.

• (a) If $\mu _k^*(p^x)\ne \mu _k ( {p^x}),\mu _k^*(p)$ is discontinuous at $p^{x}$, and $E_k (p^x,q_k^*(p^x))=\pi _k^*(p^{\min })$. By the definition of the left-hand side of this equation, we have that $q_k^*(p^x)=D( {p^x})$ and $\mu _i^*(p)$ is continuous at $p^{x}$ or $\mu _i^*( {p^x})=\mu _i ( {p^x})< 1$. Moreover, for any $q$,

  $$\begin{aligned} \mathop {\lim }\limits _{p\downarrow p^x} E_i (p,q)<\mathop {\lim }\limits _{p\uparrow p^x} E_i (p,q)\le \pi _i^*(p) \end{aligned}$$

  Hence $\mu _i^*(p)$ is constant on [$p^{x}, p^{x}$+). Using the same argument in the proof of the result (ii) of Step 2, we arrive at that $\mu _k^*(p^x)=1$. This implies that firm $k$’s profit is identical to 0 by taking any strategy ($p, q)$ with $p > p^{x }$or $\mu _i^*(p^x)=1$, which contradicts $\mu _i^*(p^x)<1$.

• (b) If $\mu _i^*(p^x)=\mu _i (p^x)$ for $i=1$ or 2, then $\mu _i^*(p)$ is continuous at $p^{x}$ for $i\in \left\{ {1, 2} \right\}$. There is an $\varepsilon >$ 0 such that for any $p\in \left[ {p^x,p^x+ \varepsilon } \right)$ and $i\in \left\{ {1, 2} \right\} _{,} (1-\mu _i^*(p))p>MC_k (D(p))$ owing to the result (b) of Step 1. As a result, the inequality (5) still holds and $q_k^*(p)=D(p)$. Using the same argument in the proof of the results (ii) and (iii) of Step 2 with $(1-\mu _i^*(p^{\prime }))p^{\prime }>p_1^c$ being replaced by $(1-\mu _i^*(p^{\prime }))p^{\prime }>MC_k (D(p^{\prime })))$, we obtain that $\mu _i^*(p)=\mu _i (p)$ for $i=1$ or 2. This leads to a contradiction.$\square$

Proof of Theorem 4

Let $c(t)$ be strictly increasing. (1) if $t_{1 }< t_{2}$ and ($t_{1}, t_{2})$ is on the SPNE path of the ($T, l, c(t))$ framework, then ($t_{1}, t_{2})$ = (0, 1). In fact, given that firm 2 chooses $t_{2}$ at stage 1, firm 1 prefers time 0 to $t_{1}$ if $t_{1}>$ 0 because firm 1’s profit is fixed but its time cost decreases. Thus, $t_{1 }$ = 0. Given that firm 1 chooses time 0 at stage 1, firm 2 prefers time 1 to $t_{2}$ if $t_{2 }>$ 1 for the same reason. (2) For the symmetric reason, if $t_{1 }> t_{2}$ and ($t_{1}$, $t_{2})$ is on the SPNE path of the ($T, l, c(t))$ framework, then ($t_{1}, t_{2})$ = (1, 0). Thus, only (0, 0), (1, 1), ..., ($T-$1, $T-$1), (1, 0) and (0, 1) may be on the SPNE path of the $T$-stage framework.

Writing out the condition under which ($j, j)$ ($j$ = 1, 2, ..., $T-$1) is on the SPNE path of the ($T, l, c(t))$ framework and then letting $l$ approach to 0 lead to the condition that $L_{i }<$ 0 and $F_{i }\le$ 0 for $i$ = 1 and 2. On the other hand, if $L_{i }<$ 0 and $F_{i }\le$ 0, there is a sufficiently small $l$ such that only ($j$, $j)$ ($j$ = 0, 1, 2, ..., $T-$1) is on the SPNE path of the ($T, l, c(t))$ framework. Thus, the result (1) holds. In the same way we can confirm that the results (2) and (3) hold. $\square$

Proof of Theorem 6

Define $\bar{p}=Max\{\bar{p}^1,\bar{p}^2\}$, where $D(\bar{p}^i)=\varphi _i (p^c)$. Define $\bar{D}(p)$ equals $D(p)$ if $p<\bar{p}$, equals $D(\bar{p})(2-p/\bar{p})$ if $\bar{p}\le p\le 2\bar{p}$, and equals 0 if $p>2\bar{p}$. Then the simultaneous move price competition has an MSNE $(g_1 (p_1 ),g_2 (p_2 ))$ in which the industrial demand $D(p)$ is replaced by $\bar{D}(p)$. Let $[\bar{p}_i^n ,\bar{p}_i^x ]$ be the support set of $g_{i}(p_{i})$ and $E(\pi _i^{sp} )$ be firm $i$’s expected profit in the MSNE. Denote firm $i$’s expected profit by $\bar{E}_i (p)$ in corresponding with firm $k$ taking strategy $g_k (p_k )$ and firm $i$ naming price $p$.

• (1) It is easy to show that for firm $i$, any price less than $p^{c}$ is dominated by price $p^{c}$. Thus, $\bar{p}_i^n \ge p^c$ and $E(\pi _i^{sp} )\ge \pi _i^*(p^c)$.

• (2) If $\bar{p}_1^x =\bar{p}_2^x$, then either firm 1 or firm 2 names price $\bar{p}_1^x$ with a probability of 0. Otherwise both firm name price $\bar{p}_1^x$ with a positive probability, which implies that $\bar{E}_i (\bar{p}_1^x )=E(\pi _i^{sp} )$. However, $\mathop {\lim }\nolimits _{p\uparrow \bar{p}_1^x } \bar{E}_i (p)-\bar{E}_i (\bar{p}_1^x )>$ 0, which implies that there is a $p_i$ such that $\bar{E}_i (p_i )>\bar{E}_i (\bar{p}_1^x )=E(\pi _i^{sp} )$. A contradiction emerges.

• (3) We affirm that if $\bar{p}_1^x >\bar{p}_2^x$ or “$\bar{p}_1^x =\bar{p}_2^x$ and firm 2 names price $\bar{p}_1^x$ with a probability of 0”, then (a) $\bar{E}_1 (\bar{p}_1^x )=E(\pi _1^{sp} )$ and (b) $\bar{p}_1^x <\bar{p}$. For the result (a), (i) If firm 1 names price $\bar{p}_1^x$ with a positive probability, the affirmation holds. (ii) If firm 1 names price $\bar{p}_1^x$ with a probability of 0, then for any $\tilde{p}<\bar{p}_1^x$, there is $\hat{p}\in (\tilde{p},\bar{p}_1^x )$ such that $\bar{E}_1 (\hat{p})=E(\pi _1^{sp} )$ because otherwise the supremum of the support set of firm 1’s mixed strategy in the MSNE must be less than $\bar{p}_1^x$. Thus, there is a strictly increasing sequence $\{\hat{p}_n \}_{n=1}^{+\infty }$ such that $\bar{E}_1 (\hat{p}_n )=E(\pi _1^{sp} )$ for any $n$ and $\mathop {\lim }\nolimits _{n\rightarrow +\infty } \hat{p}_n =\bar{p}_1^x$, which yields that $\bar{E}_1 (\bar{p}_1^x )=\mathop {\lim }\nolimits _{n\rightarrow +\infty } \bar{E}_1 (\hat{p}_n )=E(\pi _1^{sp} )$. For the result (b), $\pi _1^*(p^c)\le E(\pi _1^{sp} )=\bar{E}_1 (\bar{p}_1^x )\le \pi _1^r (\left. {\bar{p}_1^x } \right| \varphi _2 (p^c))$. The last inequality holds because firm 2’s output is greater than $\varphi _2 (p^c)$ if it names a price in $[\bar{p}_2^n ,\bar{p}_2^x ]$ and firm 1 names price $\bar{p}_1^x$. Note that for any $p\ge \bar{p}$, $\pi _1^r (\left. p \right| \varphi _2 (p^c))\equiv 0$. Thus $\bar{p}_2^x \le \bar{p}_1^x <\bar{p}$.

• (4) Symmetrically, if $\bar{p}_1^x <\bar{p}_2^x$ or “$\bar{p}_1^x =\bar{p}_2^x$ and firm 1 names price $\bar{p}_1^x$ with a probability of 0”, then $\bar{p}_2^x <\bar{p}$.

In summary, $\bar{p}>Max\{\bar{p}_1^x ,\bar{p}_2^x \}$ and $Min\{\bar{p}_1^n ,\bar{p}_2^n \}\ge p^c$.

Now we affirm that $(g_1 (p_1), g_2 (p_2 ))$ is an MSNE of simultaneous move price competition providing that the industrial demand is $D(p)$. Denote by $E_{i}(p)$ firm $i$’s expected profit where firm $i$ names price $p$ and firm $k$ takes strategy $g_{k }(p_{k})$. Then for any $p\le \bar{p}, E_i (p)=\bar{E}_i (p)$; and for any $p>\bar{p}, E_i (p)=0=\bar{E}_i (p)$. Thus, the affirmation holds. $\square$

Proof of Proposition 4

It is equivalent to the following result. Assume that both firms take price-quantity strategy and that the SPNE of the $-$3-period (or 3-period) framework under endogenous timing exists, then the profit of firm $i$ ($i$ = 1, 2) is greater than $\pi _i^*(Max\{p_1^{tp} ,p_2^{tp} \})$.

By the result 2 of Theorem 5, if ($-$1, 0) is on the SPNE path, then $L_{1 }>0,$ or equivalently $p_1^*\tau _1 (p_1^*)-C_1 (\tau _1 (p_1^*))>E(\pi _1^s )\ge \pi _1^*(Max\{p_1^{tp} ,p_2^{tp} \})$, where $p_1^*\in P_1^*$ and the last inequality results from Theorem 3. Thus, $p_1^*>Max\{p_1^{tp} ,p_2^{tp} \}$, which implies that firm 2’s profit is greater than $\pi _2^*(Max\{p_1^{tp}, p_2^{tp} \})$.

Symmetrically, if (0, $-$1) is on the SPNE path, Proposition 4 is true.

If (2, 2) is on the SPNE path, then $\pi _i^*(p^{\min })=E(\pi _i^s )>\pi _i^*(p_k^*)$ for $i$ = 1, 2. The equality results from Theorem 3 and the inequality holds because of the result 2 of Theorem 5. Thus, $p^{\min }>Max\{p_1^*,p_2^*\}>Max\{p_1^{tp} ,p_2^{tp} \}$, where the first inequality results from the fact that $\pi _i^*(p)$ is strictly increasing on (0, $p_i^m )$ and the last results from Proposition 1. Hence, Proposition 4 is true.

If (0, 0) is on the SPNE path, then

$$\begin{aligned} \pi _i^*(p^{\min })=E\left( \pi _i^s \right) \ge p_i^*\tau _i \left( p_i^*\right) -C_i \left( \tau _i \left( p_i^*\right) \right) >\pi _i^*\left( p_i^{tp} \right) . \end{aligned}$$

The first inequality results from the result 2 of Theorem 5 and the last results from Proposition 1. Thus, $p^{\min }>Max\{ {p_1^{tp} ,p_2^{tp} }\}$ and Proposition 4 holds. $\square$