Comparing Cournot and Bertrand equilibria in an asymmetric duopoly with product R&D

Abstract

We compare the Cournot and Bertrand equilibria in an asymmetric duopoly with product R&D competition. If a firm’s marginal cost is lower than that of its rival, then this firm (its rival) is referred to as the more (less) efficient firm. Under each mode of competition, there are three types of equilibria: blockaded-entry, deterred-entry, and accommodated-entry. Moreover, the presence of R&D investment makes it harder for the less efficient firm to survive. Cournot competition entails a unique equilibrium, whereas Bertrand competition may yield two equilibria. It is harder for the less efficient firm to survive under Bertrand competition than under Cournot competition. Versus Cournot competition, Bertrand competition yields higher industry output, and it shifts production from the less efficient firm to the more efficient firm. This result, together with the known size effect, explains the following three findings. First, the more efficient firm has a normal output ranking, whereas the less efficient firm may demonstrate an output reversal. Second, the more efficient firm may demonstrate a R&D reversal, whereas the less efficient firm has a normal R&D ranking (its Cournot R&D effort exceeds its Bertrand R&D effort). Third, Bertrand competition is more welfare-efficient than Cournot competition.

Appendix

The proof of Fact 4.1

Partially differentiating (3.9) with respect to $\mu _i$ and $\mu _j$, respectively, leads to

$$\begin{aligned} {\partial x_i^C\over \partial \mu _i} ={2\over 4-\gamma ^2}>0,\quad i=1,2;\quad {\partial x_1^C\over \partial \mu _2} ={\partial x_2^C\over \partial \mu _1}={-\gamma \over 4-\gamma ^2}<0. \end{aligned}$$

(8.1)

Differentiating (3.11) with respect to $\mu _i$ yields

$$\begin{aligned} {\partial \phi _i^C\over \partial \mu _i}=2 x_i^C{\partial x_i^C\over \partial \mu _i} >0,\quad i=1,2. \end{aligned}$$

(8.2)

According to (8.1), $\partial x_i^C/ \partial \mu _i$ is a constant. Therefore, partially differentiating (8.2) with respect to $\mu _k$ yields

$$\begin{aligned} {\partial ^2 \phi _i^C\over \partial \mu _k\partial \mu _i}=2{\partial x_i^C\over \partial \mu _i} {\partial x_i^C\over \partial \mu _k},\quad i, k=1,2. \end{aligned}$$

(8.3)

From $R^i(q_1,q_2)=\phi _i^C(\bar{\mu }_1+q_1,\bar{\mu }_2+q_2)$ it follows that $R_i^i=\partial \phi _i^C/ \partial \mu _i$, $R_{ik}^i=\partial ^2 \phi _i^C /[\partial \mu _k\partial \mu _i]$, $i,k=1,2$. Therefore, (8.2) and (8.3), together with (8.1), lead to the desired result. $\square$

The proof of Eq. (4.11)

The two critical constants can be rewritten as $\eta ^{CM}=\gamma /(2-1/k)$ and $\eta ^{CL}=\gamma /\{ 2-4/[(4-\gamma ^2)k]\}$. Accordingly, from $\eta ^C=\gamma / 2$ it follows that $\eta ^{CM}>\eta ^C$. Moreover, from $4/(4-\gamma ^2)>1$ it follows that $\eta ^{CL}>\eta ^{CM}$. In summary, $\eta ^C< \eta ^{CM}<\eta ^{CL}$. It follows that $1/\eta ^{CL} <1/ \eta ^{CM} <1/\eta ^C.\,1/\eta ^{CL}- \eta ^{CL}=(\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2)\times (4-\gamma ^2)^3 / \{2\gamma (4-\gamma ^2)[ (4-\gamma ^2)k- 2]k \}$, which, according to SC and (4.4), is positive. Hence, $\eta ^{CL}<1<1/\eta ^{CL}$. $\square$

The proof of Fact 4.2

Let us begin with presenting two basic facts. As mentioned, ${\varvec{\Pi }}_1^1(0)$ meets the $q_2$-axis at point $(0,q_2^{CL}(0))$ from below. It remains to study how ${\varvec{\Pi }}_1^1(0)$ meets the boundary of $\mathbf{T }$ from above. Since ${\varvec{\Pi }}_1^1(0)$ is a downward-sloping line segment, it meets the $q_1$-axis or $\partial \mathbf{T }^2$:

Fact 8.1

(i) If $\Pi _1^1(q_1^{CL}(0),0)<0$, then there exists a unique $\tilde{q}_1^C\in (0, q_1^{CL}(0))$ such that $\Pi _1^1(\tilde{q}_1^C,0)=0$, and hence, as shown in Fig. 2, ${\varvec{\Pi }}_1^1(0)$ meets the $q_1$-axis at point $(\tilde{q}_1^C,0)$. (ii) If $\Pi _1^1(q_1^{CL}(0),0)\ge 0$, then there exists $\bar{q}_2\ge 0$ such that $\Pi _1^1(q_1^{CL}(\bar{q}_2),\bar{q}_2)=0$, and hence, as shown in Figs. 3 and 4, ${\varvec{\Pi }}_1^1(0)$ meets $\partial \mathbf{T }^2$ at point $(q_1^{CL}(\bar{q}_2),\bar{q}_2)$.

Proof

According to Fact 4.1, $\Pi _1^1(0,0)>0$. Therefore, Part (i) follows from SOC. According to (4.1), $q_1^{CL}(q_2)$ increases with $q_2$. Therefore, SOC and SS together imply that $\Pi _1^1(q_1^{CL}(q_2),q_2)$ decreases with $q_2$. According to Fact 4.1, $\Pi _{11}^1$ and $\Pi _{12}^1$ are constants, and hence $\lim _{q_2 \rightarrow \infty }\Pi _1^1(q_1^{CL}(q_2),q_2)<0$. Part (ii) thus follows. This completes the proof of Fact 8.1. $\square$

We have the following fact regarding ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$:

Fact 8.2

If $q_1^M\ge q_1^{CL}(0)$, then ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ is not empty, and, as shown in Fig. 4, there exists a unique $\hat{q}_2^M\ge 0$ such that $q_1^{CL}(\hat{q}_2^M)=q_1^M$, and ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ meets $\partial \mathbf{T }^2$ at point $(q_1^{CL}(\hat{q}_2^M),\hat{q}_2^M)$. Otherwise, ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ is empty.

Proof

In the $(q_1,q_2)$ space, $\mathbf{N }^2$ is the region lying to the right of $\partial \mathbf{T }^2$ ($\partial \mathbf{T }^2$ included). As mentioned, $\partial \mathbf{T }^2$ meets the $q_1$-axis at $(q_1^{CL}(0),0)$, and $\partial \mathbf{T }^2$ is upward sloping. It follows that $(q_1,q_2)\in \mathbf{N }^2 \Rightarrow q_1\ge q_1^{CL}(0)$. Accordingly, $\{\mathbf{q }\in \mathbf{N }^2:q_1=q_1^M\}\not =\emptyset$ if and only if $q_1^M\ge q_1^{CL}(0)$. This completes the proof of Fact 8.2.

We are in a position to study how ${\varvec{\Pi }}_1^1(0)$ meets the boundary of $\mathbf{T }$ from above, and whether ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ is empty. We can show that⁵³

$$\begin{aligned} \Pi _i^i(q_i^{CL}(0),0)\gtreqqless 0, \quad \hbox { if }\bar{\mu }_j \lesseqqgtr \eta ^{CL}\bar{\mu }_i,\quad i=1,2,\end{aligned}$$

(8.4)

$$\begin{aligned} \qquad q_i^M\gtreqqless q_i^{CL}(0), \quad \hbox { if }\bar{\mu }_j \lesseqqgtr \eta ^{CM}\bar{\mu }_i,\quad i=1,2. \end{aligned}$$

(8.5)

In Cases A, B$'$ and C$'$, (4.11) implies that $\bar{\mu }_2$ exceeds not only $\eta ^{CL}\bar{\mu }_1$ but also $\eta ^{CM}\bar{\mu }_1$. Therefore, according to (8.4) and (8.5), $\Pi _1^1(q_1^{CL}(0),0)<0$ and $q_1^M<q_1^{CL}(0)$. From Facts A.1-A.2 it thus follows that ${\varvec{\Pi }}_1^1(0)$ meets the $q_1$-axis and ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ is empty. In Case B, from $\eta ^{CM}\bar{\mu }_1<\bar{\mu }_2\le \eta ^{CL}\bar{\mu }_1$ it follows that $\Pi _1^1(q_1^{CL}(0),0)\ge 0$ and $q_1^M<q_1^{CL}(0)$, implying that ${\varvec{\Pi }}_1^1(0)$ meets $\partial \mathbf{T }^2$ and ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ is empty. In Case C, $\bar{\mu }_2$ falls below not only $\eta ^{CM}\bar{\mu }_1$ but also $\eta ^{CL}\bar{\mu }_1$. Therefore, $\Pi _1^1(q_1^{CL}(0),0)>0$ and $q_1^M\ge q_1^{CL}(0)$, implying that ${\varvec{\Pi }}_1^1(0)$ meets $\partial \mathbf{T }^2$ and ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$ is not empty.

We are in a position to prove Part (iii), the most complex part. Note importantly that, in Case C, $\hat{q}_2^M< \bar{q}_2$,⁵⁴ implying that $\Pi ^1(q_1,q_2)$, regarded as a function of $q_1$, is single-peaked for each given $q_2$. Recall that ${\varvec{\Pi }}_1^1(0)$ meets the $x_2$-axis and $\partial \mathbf{T }^2$ at $(0,q_2^{CL}(0))$ and $(q_1^{CL}(\bar{q}_2),\bar{q}_2)$, respectively. Accordingly, from ${\varvec{\Pi }}_1^1(0)$ is downward sloping it follows that $q_2^{CL}(0)>\bar{q}_2$, which, together with $\hat{q}_2^M<\bar{q}_2$, implies that $\hat{q}_2^M<\bar{q}_2<q_2^{CL}(0)$, and hence there are four cases.

First, assume that $q_2\ge q_2^{CL}(0)$. From $q_2>\hat{q}_2^M$ it follows that firm 1 will choose a point belonging to $\partial \mathbf{T }^2$ if its opportunity set is $\mathbf{N }^2$. Note that $\Pi _1^1(\mathbf{q })\le 0, \forall \mathbf{q }\in \partial \mathbf{T }^1$,⁵⁵ where $\Pi _1^1$ denotes a right-hand derivative. As a result, firm 1 will choose a point belonging to $\partial \mathbf{T }^1$ if its opportunity set is $\tilde{\mathbf{T }}$. This implies that the former best point is dominated by the later best point since $\partial \mathbf{T }^2\subset \tilde{\mathbf{T }}$. Similarly, $\partial \mathbf{T }^1$ is dominated by ${\varvec{\Pi }}_1^1(0;\mathbf{N }^1)$ since $\partial \mathbf{T }^1\subset \mathbf{N }^1$. In summary, if $q_2\ge q_2^{CL}(0)$, then the reaction curve is ${\varvec{\Pi }}_1^1(0;\mathbf{N }^1)$.

Second, assume that $q_2\in (\bar{q}_2,q_2^{CL}(0))$. Again, from $q_2>\hat{q}_2^M$ it follows that firm 1 will choose a point belonging to $\partial \mathbf{T }^2$ if its opportunity set is $\mathbf{N }^2$. It is clear that $\partial \mathbf{T }^2$ is dominated by ${\varvec{\Pi }}_1^1(0)$ since, again, $\partial \mathbf{T }^2\subset \tilde{\mathbf{T }}$. In summary, if $q_2\in (\bar{q}_2,q_2^{CL}(0))$, then the reaction curve is ${\varvec{\Pi }}_1^1(0)$.

Third, assume that $q_2\in (\hat{q}_2^M,\bar{q}_2]$. It is clear that firm 1 will choose a point belonging to $\partial \mathbf{T }^2$ if its opportunity set is $\mathbf{N }^2$. If ${\varvec{\Pi }}_1^1(0)$ meets $\partial \mathbf{T }^2$ at point $(q_1^{CL}(\bar{q}_2),\bar{q}_2)$, then⁵⁶

$$\begin{aligned} \Pi _1^1(\mathbf{q })\ge 0,\quad \forall \mathbf{q }\in \{\mathbf{q }\in \partial \mathbf{T }^2:q_2\le \bar{q}_2\}, \end{aligned}$$

(8.6)

where $\Pi _1^1$ denotes a left-hand derivative. Equation (8.6) implies that firm 1 will choose a point belonging to $\partial \mathbf{T }^2$ if its opportunity set is $\tilde{\mathbf{T }}$. In summary, if $q_2\in (\hat{q}_2^M,\bar{q}_2]$, then the reaction curve is $\partial \mathbf{T }^2$.

Forth, assume that $q_2\in (0,\hat{q}_2^M]$. It is clear that firm 1 will choose a point belonging to $\partial \mathbf{T }^2$ if its opportunity set is $\tilde{\mathbf{T }}$, but this strategy is dominated by the blockaded-entry strategy. Hence, the reaction curve is ${\varvec{\Pi }}_1^1(0;\mathbf{N }^2)$.

We can apply the same procedure to prove Parts (i) and (ii). $\square$

The proof of Proposition 4.2

Parts (ii) and (iii) easily follow from (ii) and (iii) of Lemma 4.1 and Eqs. (3.1), (4.1) and (4.10). It remains to prove Part (i). According to Lemma 4.1-(i), $\mathbf{q }^C$ solves $\Pi _1^1(\mathbf{q })=0$ and $\Pi _2^2(\mathbf{q })=0$, namely,

$$\begin{aligned} q_i^C={ 2(k-{2\over 4-\gamma ^2}) \bar{\mu }_i-\gamma k\bar{\mu }_j \over (\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2)/\Delta },\quad i=1,2. \end{aligned}$$

(8.7)

Note that $\mu _i=\bar{\mu }_i+q_i^C,\,i=1,2$. Therefore, (8.7), together with (3.9), implies that

$$\begin{aligned} x_i^C=(4-\gamma ^2)kq_i^C/4,\quad i=1,2. \end{aligned}$$

(8.8)

According to (4.4), $k-2/(4-\gamma ^2)>0$. The desired results thus straightforwardly follow from (8.7) and (8.8). $\square$

The proof of Corollary 4.1

If the accommodated-entry equilibrium prevails, then the desired result follows from Proposition 4.2-(i). Otherwise, $q_2^C=x_2^C=0$, and the desired result thus straightforwardly follows. $\square$

The proof of Fact 4.5

It suffices to consider Cases A-C. In either of Cases A and B, $\mathbf{q }^C\in \tilde{\mathbf{T }}$, implying that $\mathbf{p }^C-\mathbf{c }$ is given by (3.10). In Case C, where only firm 1 is active, the second-stage equilibrium is characterized by (3.4)–(3.5), implying that $p_1^C-c_1=x_1^C$. $\square$

The proof of Corollary 4.2

It remains to prove Part (iv). If the accommodated-entry equilibrium prevails, then the desired result follows from Part (i). Otherwise, firm 2 is inactive, and hence $p_2^C\le c_2$, together with $p_1^C>c_1$, implies the desired result. $\square$

The proof of Fact 5.1

Partially differentiating (3.12) with respect to $\mu _i$ and $\mu _j$, respectively, leads to

$$\begin{aligned} {\partial x_i^B\over \partial \mu _i}&= {1+1/b\over (1+1/b)^2-\gamma ^2} >0,\quad i=1,2,\end{aligned}$$

(8.9a)

$$\begin{aligned} {\partial x_1^B\over \partial \mu _2}&= {\partial x_2^B\over \partial \mu _1}={-\gamma \over (1+1/b)^2-\gamma ^2} <0. \end{aligned}$$

(8.9b)

Differentiating (3.14) with respect to $\mu _i$ yields

$$\begin{aligned} {\partial \phi _i^B\over \partial \mu _i}= {2\over b} x_i^B{\partial x_i^B\over \partial \mu _i} >0,\quad i=1,2. \end{aligned}$$

(8.10)

According to (8.9a), $\partial x_i^B/ \partial \mu _i$ is a constant. Therefore, partially differentiating (8.10) with respect to $\mu _k$ yields

$$\begin{aligned} {\partial ^2 \phi _i^B\over \partial \mu _k\partial \mu _i}={2\over b}{\partial x_i^B\over \partial \mu _i} {\partial x_i^B\over \partial \mu _k},\quad i,k=1,2. \end{aligned}$$

(8.11)

From $R^i(q_1,q_2)=\phi _i^B(\bar{\mu }_1+q_1,\bar{\mu }_2+q_2)$ it follows that $R_i^i=\partial \phi _i^B/ \partial \mu _i,\,R_{ik}^i=\partial ^2 \phi _i^B/ [ \partial \mu _k\partial \mu _i],\,i,k=1,2$. Therefore, (8.10) and (8.11), together with (8.6), lead to the desired result. $\square$

The proof of Eqs. (5.8)–(5.9)

$\eta ^L-\eta ^{BL}=\gamma ^3 k/\{(2-\gamma ^2)[(4-\gamma ^2)k-2][(2-\gamma ^2)k-1]\}$, which, according to (5.3), is positive. $1/\eta ^B-\eta ^{BL}= 2(k-\hat{r}_{11}^1)/\{\gamma \hat{\Delta }[(4-\gamma ^2)k-2]\}$, which, according to SOC and (5.3), is positive. $\eta ^{BL}$ can be rewritten as $\gamma /\{ (1+1/b)(1-2/[(4-\gamma ^2)k]\}$. According to (5.3), $1>2/[(4-\gamma ^2)k]>0$. Therefore, from $\eta ^B=\gamma /(1+1/b)$ it follows that $\eta ^B<\eta ^{BL}$. Equation (2.4) straightforwardly implies that $\eta ^{BL}>\eta ^{CM}$. This completes the proof of (5.8), which, together with (4.11), implies that $\eta ^C< \eta ^{CM}<\eta ^{BL}$. It remains to prove that $\eta ^{BL}<1/\eta ^{BL}.\,1/\eta ^{BL}-\eta ^{BL}= 2 (\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2) /\{\gamma ^2 \hat{\Delta }[(4-\gamma ^2)k-2]k\}$ which, according to SC and (5.3), is positive. $\square$

The proof of Fact 5.2

In the $(q_1,q_2)$ space, $\mathbf{L }^2$ is the region lying between $\partial \mathbf{V }^2$ and $\partial \mathbf{T }^2$ ($\partial \mathbf{V }^2$ included). As mentioned, $\partial \mathbf{V }^2$ meets the $q_1$-axis at $(q_1^{BL}(0),0)$, and $\partial \mathbf{T }^2$ meets the $q_1$-axis at $(q_1^{CL}(0),0)$. Moreover, both $\partial \mathbf{V }^2$ and $\partial \mathbf{T }^2$ are upward sloping. As a result, $(q_1,q_2)\in \mathbf{L }^2 \Rightarrow q_1\ge q_1^{BL}(0)$. Accordingly, ${\varvec{\Pi }}_1^1(0;\mathbf{L }^2)=\emptyset \Leftrightarrow q_1^L< q_1^{BL}(0)$. Moreover, given that $q_1^L\ge q_1^{BL}(0)$, if $q_1^L\le q_1^{CL}(0)$ (resp. $q_1^L\ge q_1^{CL}(0)$), then ${\varvec{\Pi }}_1^1(0;\mathbf{L }^2)$ meets the $q_1$-axis (resp. $\partial \mathbf{T }^2$) from above. It remains to compare $q_1^L$ with either one of $q^{BL}_1(0)$ and $q^{CL}_1(0)$. We can show that⁵⁷

$$\begin{aligned}&q_i^L \gtreqqless q_i^{BL}(0),\quad \hbox { if }{\bar{\mu }}_j \lesseqqgtr \eta ^L{\bar{\mu }}_i, \end{aligned}$$

(8.12)

$$\begin{aligned}&q_i^L\gtreqqless q_i^{CL}(0),\quad \hbox { if }{\bar{\mu }}_j \lesseqqgtr \eta ^{CM}{\bar{\mu }}_i. \end{aligned}$$

(8.13)

$\square$

The proof of Fact 5.3

By definition, $\Pi ^1(\mathbf{q }^*)=\pi _1^*$, $\Pi ^1((\bar{\mu }_2+q_2^*)/(\gamma k),q_2^*)=\hat{\pi }_1^*$, $\Pi ^1(q_1^L,0)=\hat{\pi }_1^L$ and $\Pi ^1(\tilde{q}_1^B,0)=\tilde{\pi }_1$ where $\pi _1^*\equiv \phi _1^B(\bar{\mu }_1+q_1^*,\bar{\mu }_2+q_2^*)-k(q_1^*)^2/2$, $\hat{\pi }_1^*\equiv \phi _1^L(\bar{\mu }_1+(\bar{\mu }_2+q_2^*)/(\gamma k),\bar{\mu }_2+q_2^*)-k[(\bar{\mu }_2+q_2^*)/(\gamma k)]^2/2$, $\hat{\pi }_1^L\equiv \phi _1^L(\bar{\mu }_1+q_1^L,\bar{\mu }_2)-k(q_1^L)^2/2$ and $\tilde{\pi }_1\equiv \phi _1^B(\bar{\mu }_1+\tilde{q}_1^B,\bar{\mu }_2)-k(\tilde{q}_1^B)^2/2$. By definition, $\mathbf{q }^*$ belongs to $\mathbf{T }$, and it can be obtained by solving $\Pi _1^1(\mathbf{q })=0$ and $\Pi _2^2(\mathbf{q })=0$:

$$\begin{aligned} q_i^*={(1+1/b)(k-{2\over 4-\gamma ^2}) \bar{\mu }_i-\gamma k \bar{\mu }_j\over (\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2)/\hat{\Delta }} ,\quad i=1,2, \end{aligned}$$

(8.14)

which, together with (3.12) and (3.14), implies

$$\begin{aligned} \pi _1^*={ (k-\hat{r}_{11}^1)k[(1+1/b)(k-{2\over 4-\gamma ^2}) \bar{\mu }_1-\gamma k \bar{\mu }_2]^2\over 2(2-\gamma ^2)(\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2)^2/\hat{\Delta }}. \end{aligned}$$

(8.15)

We next examine whether $\mathbf{q }^*$ is an equilibrium. Equation (8.14), together with Fact 5.2-(i), implies that $q_2^*\gtreqqless \hat{q}_2^L$ if $\bar{\mu }_2\gtreqqless 332.794$. The condition $\bar{\mu }_2\in (\eta ^{BL}\bar{\mu }_1,\eta ^L\bar{\mu }_1]$ means that $\bar{\mu }_2\in (329.67,333.333]$, implying that $q_2^*$ can either exceed or fall below $\hat{q}_2^L$. It follows that $\mathbf{q }^*$ is an equilibrium if $\bar{\mu }_2\ge 332.794$. It remains to analyze those cases with $\bar{\mu }_2\in (329.67, 332.794)$. Equations (5.5) and (5.6) imply that $\hat{\pi }_1^*= 4.81251(535.842-\bar{\mu }_2)(\bar{\mu }_2-48.5268)$, which, together with (8.15), implies that $\pi _1^*-\hat{\pi }_1^*=4.85064(\bar{\mu }_2-331.162)(\bar{\mu }_2-296.3)$. From $\bar{\mu }_2>296.3$ it follows that

$$\begin{aligned} \pi _1^*\lesseqqgtr \hat{\pi }_1^*,\quad \hbox {if }\bar{\mu }_2\lesseqqgtr \eta ^D\bar{\mu }_1 \end{aligned}$$

(8.16)

where $\eta ^D\equiv 0.331162$. In summary, if $\bar{\mu }_2\ge \eta ^D\bar{\mu }_1$, then $\mathbf{q }^*$ is an equilibrium.

It remains to study whether $(q_1^L,0)$ is an equilibrium. Using Fact 5.1 yields that $\tilde{q}_1^B=\hat{r}_0^1/(k-\hat{r}_{11}^1)$, which, together with (3.12) and (3.14), implies that

$$\begin{aligned} \tilde{\pi }_1={[(1+1/b)\bar{\mu }_1 -\gamma \bar{\mu }_2]^2k\over (4-\gamma ^2)^2 (1-\gamma ^2)(k-\hat{r}_{11}^1)}. \end{aligned}$$

(8.17)

Equation (5.5), together with (5.13), implies that

$$\begin{aligned} \hat{\pi }_1^L={[2\gamma k \bar{\mu }_1-(2k-1)\bar{\mu }_2]\bar{\mu }_2\over 2 \gamma ^2k}. \end{aligned}$$

(8.18)

Subtracting (8.18) from (8.17) yields $\tilde{\pi }_1-\hat{\pi }_1^L=3.52773(\bar{\mu }_2-331.42)(\bar{\mu }_2-290.54)$. From $\bar{\mu }_2>290.54$ it follows that

$$\begin{aligned} \tilde{\pi }_1\lesseqqgtr \hat{\pi }_1^L\quad \hbox { if }\bar{\mu }_2\lesseqqgtr \eta ^I\bar{\mu }_1 \end{aligned}$$

(8.19)

where $\eta ^I\equiv 0.33142$. In summary, if $\bar{\mu }_2\le \eta ^I\bar{\mu }_1$, then $(q_1^L,0)$ is an equilibrium. $\square$

The proof of Lemma 5.1

Let us begin with establishing some basic results. As mentioned, (4.6) still holds if $\mathbf{N }^i$ is replaced by $\mathbf{N }^i\cup \mathbf{L }^i$. It follows that

$$\begin{aligned} \mathbf{\Pi }_i^i(0;\mathbf{N }^i\cup \mathbf{L }^i)=\{\mathbf{q }\in \mathbf{N }^i\cup \mathbf{L }^i:q_i=0\}=\{\mathbf{q }\in \mathbf{R }_+^2:q_i=0,q_j\ge q_j^{BL}(0)\}\nonumber \\ \end{aligned}$$

(8.20)

where $\mathbf{\Pi }_i^i(0;\mathbf{N }^i\cup \mathbf{L }^i)\equiv \{\mathbf{q }\in \mathbf{N }^i\cup \mathbf{L }^i:\Pi _i^i(\mathbf{q })=0\}$. Again, as mentioned, (4.8) still holds, implying that Fact 8.2 still holds.

Fact 5.1, together with SS, SOC and SC, implies that $\mathbf{\Pi }_i^i(0)$ is an upward sloping line segment, $i=1,2$, and that $\mathbf{\Pi }_1^1(0)$ is steeper than $\mathbf{\Pi }_2^2(0)$. Moreover, $\mathbf{\Pi }_1^1(0)$ meets $\mathbf{\Pi }_1^1(0;\mathbf{N }^1\cup \mathbf{L }^1)$ at point $(0,q_2^{BL}(0))$,⁵⁸ whereas $\mathbf{\Pi }_2^2(0)$ meets $\mathbf{\Pi }_2^2(0;\mathbf{N }^2\cup \mathbf{L }^2)$ at point $(q_1^{BL}(0),0)$. It is easy to show that Fact 8.1 still holds if $q_1^{CL}$, $\tilde{q}_1^C$ and $\partial \mathbf{T }^2$ are replaced by $q_1^{BL}$, $\tilde{q}_1^B$ and $\partial \mathbf{V }^2$, respectively.⁵⁹ According to this fact, it is important to sign $\Pi _i^i(q_i^{BL}(0),0)$:⁶⁰

$$\begin{aligned} \Pi _i^i(q_i^{BL}(0),0)\gtreqqless 0,\quad \hbox { if }\bar{\mu }_j \lesseqqgtr \eta ^{BL}\bar{\mu }_i. \end{aligned}$$

(8.21)

It follows that if $\bar{\mu }_2 > \eta ^{BL}\bar{\mu }_1$, then there exists $0<\tilde{q}_1^B<q_1^{BL}(0)$ such that $\mathbf{\Pi }_1^1(0)$ meets the $q_1$-axis at $(\tilde{q}_1^B,0)$.

The above basic results yield the following fact:

Fact 8.3

If $\eta ^L\bar{\mu }_1< \bar{\mu }_2<\bar{\mu }_1/\eta ^B$, then firm 1’s reaction curve is the union of $\mathbf{\Pi }_1^1(0)$ and $\mathbf{\Pi }_1^1(0;\mathbf{N }^1\cup \mathbf{L }^1)$.

Since Fact 8.3 is a counterpart of Fact 4.2-(i), its proof is omitted. We also have the following fact:

Fact 8.4

(i) If $\eta ^B\bar{\mu }_1<\bar{\mu }_2<\bar{\mu }_1/\eta ^L$, then firm 2’s reaction curve is $\mathbf{\Pi }_2^2(0)\cup \mathbf{\Pi }_2^2(0;\mathbf{N }^2\cup \mathbf{L }^2)$. (ii) If $\eta ^C\bar{\mu }_1<\bar{\mu }_2\le \eta ^B\bar{\mu }_1$, then $\mathbf{\Pi }_2^2(0)=\emptyset$, and firm 2’s reaction curve is simply the whole $q_1$-axis.

Proof

The inequality $\eta ^B\bar{\mu }_1<\bar{\mu }_2<\bar{\mu }_1/\eta ^L$ can be rewritten as $\eta ^L\bar{\mu }_2<\bar{\mu }_1<\bar{\mu }_2/\eta ^B$. Hence, Part (i) follows from Fact 8.3, because the firms’ positions are symmetric. It is proven next that $\eta ^C\bar{\mu }_1<\bar{\mu }_2\le \eta ^B\bar{\mu }_1 \Rightarrow \mathbf{\Pi }_2^2(0)=\emptyset$. From $\bar{\mu }_2\le \eta ^B\bar{\mu }_1$ it follows that there exists $q_2^a\ge 0$ such that $\partial \mathbf{V }^2$ meets the $q_2$-axis at point $(0,q_2^a)$. It is shown next that $\Pi _2^2(0,q_2^a)\le 0$. By definition, $q_2^a$ solves $q_1^{BL}( q_2)=0$. It follows that $q_2^a=\bar{\mu }_1 \gamma /(1+1/b)-\bar{\mu }_2$. According to Fact 5.1, $\Pi _2^2(0,q_2^a)=[ (1+1/b)\bar{\mu }_2-\gamma \bar{\mu }_1]k/(2-\gamma ^2)$. It follows that $\bar{\mu }_2\le \eta ^B\bar{\mu }_1\Rightarrow \Pi _2^2(0,q_2^a)\le 0$. From $\Pi _2^2(0,q_2^a)\le 0$ it follows that $\mathbf{\Pi }_2^2(0)$ is empty because each point belonging to $\mathbf{V }$ lies to the northeast of $(0,q_2^a)$. This completes the proof of Fact 8.4.

Note that, in either one of Cases A1 and A2, $\bar{\mu }_2$ always belongs to $(\eta ^B\bar{\mu }_1,\bar{\mu }_1/\eta ^L)$, and hence Fact 8.4-(i) can be applied. [In Fig. 7, we depict firm 2’s reaction curve for Case A2.] Fact 8.4-(i), together with Fact 8.3, implies that, in Case A1, $\mathbf{q }^*$ is the equilibrium. We next examine Case A2. It is cleat that $\mathbf{q }^*$ is potentially an equilibrium. Fact 8.4-(i), together with Fact 5.2-(ii.a), implies that firm 2’s reaction curve also meet $\mathbf{\Pi }_1^1(0;\mathbf{L }^2)$ at point $(q_1^L,0)$, implying that $(q_1^L,0)$ is potentially an equilibrium. $\square$

The proof of Fact 5.4

Since $(q_1^L,0)$ is an equilibrium, $\Pi ^1(q_1^L,0)\ge \Pi ^1(\tilde{q}_1^B,0)$. We can demonstrate that when $\mathbf{q }$ moves downward and to the right along $\mathbf{\Pi }_1^1(0)$, $\Pi ^1(\mathbf{q })$ increases.⁶¹ It follows that $\Pi ^1(\tilde{q}_1^B,0)>\Pi ^1(\mathbf{q }^*)>\Pi ^1(0,q_2^{BL}(0))$. We can draw two implications from this inequality. First, $\Pi ^1(q_1^L,0)>\Pi ^1(\mathbf{q }^*)$. Second, $\Pi ^1(\mathbf{q }^*)>0$, because $\Pi ^1(0,q_2^{BL}(0))=0$. We can apply the same line of reasoning to prove that $\Pi ^2(\mathbf{q }^*)>0$. It is clear that $\Pi ^2(q_1^L,0)=0$. It thus follows that $\Pi ^2(q_1^L,0)<\Pi ^2(\mathbf{q }^*)$. $\square$

The proof of Lemma 5.2

In either one of Cases B and C, $\bar{\mu }_2$ may belong to either $(\eta ^B\bar{\mu }_1,\bar{\mu }_1/\eta ^L)$ or $(\eta ^C\bar{\mu }_1,\eta ^B\bar{\mu }_1]$.⁶² According to Fact 8.4, if $\bar{\mu }_2>\eta ^B\bar{\mu }_1$, then $\mathbf{\Pi }_2^2(0)$ meets the $q_2$-axis and firm 2’s reaction curve is the union of $\mathbf{\Pi }_2^2(0)$ and $\mathbf{\Pi }_2^2(0;\mathbf{N }^2\cup \mathbf{L }^2)$; otherwise, $\mathbf{\Pi }_2^2(0)=\emptyset$ and firm 2’s reaction curve is the whole $q_1$-axis. Because $\mathbf{\Pi }_1^1(0)$ meets $\partial \mathbf{V }^2$, it is always true that $\mathbf{\Pi }_1^1(0)\cap \mathbf{\Pi }_2^2(0)=\emptyset$, regardless of whether $\mathbf{\Pi }_2^2(0)$ is empty or not. It follows that the equilibrium is the point at which firm 1’s reaction curve meets the $q_1$-axis. Hence, firm 1 should be active whereas firm 2 should be inactive. We can show that, in Case B, $\mathbf{\Pi }_1^1(0;\mathbf{L }^2)$ is not empty, and it meets the $q_1$-axis at $(q_1^L,0)$, whereas $\mathbf{\Pi }_1^1(0;\mathbf{N }^2)=\emptyset$, implying that the equilibrium is $(q_1^L,0)$. In Case C, $\mathbf{\Pi }_1^1(0;\mathbf{L }^2)$ is not empty, but it does not meet the $q_1$-axis. Moreover, $\mathbf{\Pi }_1^1(0;\mathbf{N }^2)$ is not empty, and it meets the $q_1$-axis at $(q_1^M,0)$, which is thus the equilibrium. $\square$

The proof of Proposition 5.1

Part (ii) follows form Lemma 5.2-(i) and Eqs. (3.1) and (5.13). It remains to prove Part (i). Since $\mu _i=\bar{\mu }_i+q_i^*$, $i=1,2$, Eq. (8.14), together with (3.12), implies that

$$\begin{aligned} x_i^B=(4-\gamma ^2)kq_i^*/(4-2\gamma ^2),\quad i=1,2. \end{aligned}$$

(8.22)

The desired results thus straightforwardly follow from (8.14) and (8.22). [According to (4.4), $k-2/(4-\gamma ^2)>0$.] $\square$

The proof of Corollary 5.1

If the accommodated-entry equilibrium prevails, then the desired result follows from Proposition 5.1-(i). Otherwise, $q_2^B=x_2^B=0$, and the desired result thereby follows. $\square$

The proof of Fact 5.5

If and only if $\mathbf{q }^B\in \tilde{\mathbf{V }}$, then $\mathbf{p }^B-\mathbf{c}$ is given by (3.13). Part (i) thus follows. In the deterred-entry equilibrium, the second-stage competition is characterized by (3.7), implying that $p_1^B-c_1=\mu _1-x_1^B$. Part (ii) thus follows. $\square$

The proof of Corollary 5.2

It remains to prove Part (ii). If the accommodated-entry equilibrium prevails, then the desired result follows from Part (i). If the deterred-entry equilibrium prevails, then, according to (3.7), $p_2^B=c_2$, implying that $p_1^B-c_1>p_2^B-c_2$. $\square$

The proof of Fact 6.1

According to Fact 4.1 and (4.3), $\tilde{q}_1^C=r_0^1/(k-r_{11}^1)$. Similarly, $\tilde{q}_1^B= \hat{r}_0^1/(k-\hat{r}_{11}^1)$. It is clear that either one of $\tilde{q}_1^C$ and $\tilde{q}_1^B$ decreases with $k$. Moreover,

$$\begin{aligned} \tilde{q}_1^C-\tilde{q}_1^B={2\gamma ^3 (1-\gamma )(k-\check{k}) \bar{\mu }\over (1-\gamma ^2)(4-\gamma ^2)^2(k-r_{11}^1)(k-\hat{r}_{11}^1)} \end{aligned}$$

where $\check{k}\equiv 4(2-\gamma ^2)/[(1-\gamma )(4-\gamma ^2)^2]>0$. Therefore, $\tilde{q}_1^C-\tilde{q}_1^B$ and $k-\check{k}$ share the same sign. Furthermore, it is easily shown that when $k=\check{k}$, $\tilde{q}_1^C=\tilde{q}_1^B=(1/\eta ^P-1)\bar{\mu }$. The desired results thus follow. $\square$

The proof of Proposition 6.2

(i): In Case A, $x_i^C$ and $x_i^B$ ($i=1,2$) are given by (8.8) and (8.22), respectively. Therefore, $x_1^B+x_2^B-x_1^C-x_2^C$ is equal to

$$\begin{aligned} { \gamma ^2(4-\gamma ^2)[(4-\gamma ^2)k-2](\bar{\mu }_1+\bar{\mu }_2)k \over (1+\gamma )(4-\gamma ^2)^3k^2-2(4-\gamma ^2)(8+4\gamma -4\gamma ^2-\gamma ^3)k+8(2-\gamma ^2)}, \end{aligned}$$

whose sign is independent of the degree of asymmetry. The desired result thus follows from the known result stating that $x_i^B>x_i^C$, $i=1,2$ if the two firms are symmetric. In Case B1, $x_i^C$ ($i=1,2$) is given by (8.8), and $x_1^B=x_1^L$ and $x_2^B=0$, implying that $x_1^B+x_2^B-x_1^C-x_2^C$ is equal to

$$\begin{aligned} {2[(4-\gamma ^2)k-2]\over \gamma [(4-\gamma ^2)(2+\gamma )k-4]k}(\bar{\mu }_2-\eta ^{CL}\bar{\mu }_1), \end{aligned}$$

which, according to (5.3), is positive.

(ii): It is clear that $q_1^C=q_1^{CL}(0)$, $q_2^C=0$, $x_1^C=\tilde{x}_1^M$, $x_2^C=0$, $q_1^B=q_1^L$, $q_2^B=0$, $x_1^B=x_1^L$ and $x_2^B=0$. Equation (3.1) thus implies that $x_1^C= (\bar{\mu }_1+q_1^{CL}(0))/ 2$ and $x_1^B=\bar{\mu }_2/ \gamma$. According to (4.1), $q_1^{CL}(0)=\bar{\mu }_2/\eta ^C-\bar{\mu }_1$. It thus follows that $x_1^C=x_1^B$. As a result, $x_1^C+x_2^C=x_1^B+x_2^B$. $\square$

The proof of Proposition 6.3

The following lemma will be used to prove Proposition 6.3: $\square$

Lemma 8.1

Assume that $\bar{\mu }_2\in (\eta ^{BL}\bar{\mu }_1,\bar{\mu }_1/\eta ^{BL})$. There exists $\eta ^E\in (\eta ^{BL},1)$ such that (i) $q_1^C \lesseqqgtr q_1^*$, if $\bar{\mu }_2\lesseqqgtr \eta ^E\bar{\mu }_1$, and (ii) $x_2^C(\bar{\mu }_1+q_1^C,\bar{\mu }_2+q_2^C) \gtreqqless x_2^*$, if $\bar{\mu }_2\lesseqqgtr \eta ^E\bar{\mu }_1$.

Proof

According to (6.1), from $\bar{\mu }_2\in (\eta ^{BL}\bar{\mu }_1,\bar{\mu }_1/\eta ^{BL})$ it follows that $\bar{\mu }_2\in (\eta ^{CL}\bar{\mu }_1,\bar{\mu }_1/\eta ^{CL})$. Lemma 4.1-(i) thus implies that $\mathbf{q }^C$ belongs to $\mathbf{T }$, and hence it is given by (8.7). In this section, we let $\det [\Pi _{ij}^i]_C$ denote $\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2$ in (8.7). Similarly, $\det [\Pi _{ij}^i]_B$ stands for $\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2$ in (8.14). Note that both $\det [\Pi _{ij}^i]_C$ and $\det [\Pi _{ij}^i]_B$ are positive. We can use (8.14) and (8.7) to show that $\partial (q_1^C-q_1^*)/ \partial \bar{\mu }_2= 2 \gamma ^3 \{\cdot \} / \{ (1-\gamma ^2)(4 -\gamma ^2)^5k\times \det [\Pi _{ij}^i]_C\times \det [\Pi _{ij}^i]_B \}$, which is positive, because (5.3) guarantees that $\{\cdot \}=(4 -\gamma ^2)^3k^2-8 (4 -\gamma ^2)(2 -\gamma ^2)k+8 (2 -\gamma ^2)>0$. We can use (8.8) and (8.22) to establish that $\partial (x_2^C-x_2^B)/ \partial \bar{\mu }_2 =(4-\gamma ^2)k/4\times \partial [q_2^C-2q_2^*/(2-\gamma ^2)]/\partial \bar{\mu }_2$, which is negative, because $\partial (q_2^C-q_2^*)/\partial \bar{\mu }_2$ is negative⁶³ and $\partial q_2^*/\partial \bar{\mu }_2$ is positive. Furthermore, $q_1^C=q_1^*= 4(2-\gamma ^2)[ (4-\gamma ^2)k-2]\bar{\mu }_1/\{ \cdot \}$ and $x_2^C=x_2^B= \gamma (4 -\gamma ^2)^2k^2\bar{\mu }_1/\{\cdot \} ,$ if $\bar{\mu }_2=\eta ^E\bar{\mu }_1$ where $\eta ^E\equiv \gamma (4-\gamma ^2)^2[(4-\gamma ^2)k-2]k/\{\cdot \}$. We can show that $\eta ^{BL}<\eta ^E<1$.⁶⁴ This completes the proof of Lemma 8.1.

Since Bertrand competition yields an accommodated-entry equilibrium, according to Lemmas 5.1 and 5.2, $\bar{\mu }_2$ should belong to $(\eta ^{BL}\bar{\mu }_1,\bar{\mu }_1]$ and $\mathbf{q }^B=\mathbf{q }^*$. Parts (iii) and (iv) thus follow straightforwardly from Lemma 8.1. The counterpart for firm 2 of Lemma 8.1-(i) states that $q_2^C \lesseqqgtr q_2^*$, if $\bar{\mu }_1\lesseqqgtr \eta ^E\bar{\mu }_2$. From $\bar{\mu }_2\in (\eta ^{BL}\bar{\mu }_1,\bar{\mu }_1)$ it follows that $\bar{\mu }_2<\bar{\mu }_1$, which, together with $\eta ^E<1$, implies that $\bar{\mu }_1> \eta ^E\bar{\mu }_2$. Part (i) thus follows. Part (ii) similarly follows from the counterpart for firm 2 of Lemma 8.1-(ii). $\square$

The proof of Corollary 6.1

According to Proposition 6.3, $\eta ^E<1$, which, together with $\bar{\mu }_2\le \eta ^E\bar{\mu }_1$, implies that $\bar{\mu }_2<\bar{\mu }_1$. Therefore, Corollaries 4.1 and 5.1 imply that $q_1^C>q_2^C$, $q_1^B>q_2^B$, $x_1^C>x_2^C$ and $x_1^B>x_2^B$. Parts (i) and (iii) thus follow straightforwardly from Proposition 6.3. It is clear that (ii) follows from (i). Moreover, either one of (iv) and (v) follows from (iii). $\square$

The proof of Proposition 6.4

From (8.7) it follows that

$$\begin{aligned} q_1^C-q_2^C={ 2(k-{2\over 4-\gamma ^2}) +\gamma k \over (\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2)/\Delta }(\bar{\mu }_1-\bar{\mu }_2). \end{aligned}$$

From (8.14) it follows that

$$\begin{aligned} q_1^*-q_2^*= {(1+1/b)(k-{2\over 4-\gamma ^2}) +\gamma k \over (\Pi _{11}^1\Pi _{22}^2-\Pi _{12}^1\Pi _{21}^2)/\hat{\Delta }}(\bar{\mu }_1-\bar{\mu }_2). \end{aligned}$$

$(q_1^*-q_2^*)-(q_1^C-q_2^C)=2 \gamma ^3 (4 -\gamma ^2)k(\bar{\mu }_1-\bar{\mu }_2)/[\cdot ]$ and $(x_1^B-x_2^B)-(x_1^C-x_2^C)=\gamma ^2 (4 -\gamma ^2) \{(4 -\gamma ^2)k-2 \}k(\bar{\mu }_1-\bar{\mu }_2)/[\cdot ]$ where $[\cdot ]=(4 -\gamma ^2)^3 (1-\gamma )k^2-2 (4 -\gamma ^2)(8 -4 \gamma -4 \gamma ^2+\gamma ^3)k + 8 (2\ -\gamma ^2)$, which is guaranteed by (5.4) to be positive. Moreover, according to (5.3), $\{\cdot \}$ is positive. Parts (i) and (ii) thus follow. In order to prove (iii) it suffices to establish that $x_1^B/x_2^B>x_1^C/x_2^C$. We can show that $x_1^B/x_2^B-x_1^C/x_2^C= \gamma ^3 (4-\gamma ^2)[ (4-\gamma ^2)k-2]k(\bar{\mu }_1^2-\bar{\mu }_2^2)/\{2[ (4-\gamma ^2)k-2]\bar{\mu }_2-\gamma (4-\gamma ^2)k\bar{\mu }_1\}/ \{(2 -\gamma ^2)[ (4-\gamma ^2)k-2]\bar{\mu }_2-\gamma (4-\gamma ^2)k\bar{\mu }_1 \}$, which, according to (5.3), is positive. $\square$

The proof of Proposition 6.5

(i): Since Bertrand competition yields a deterred-entry equilibrium and Cournot competition yields an accommodated-entry equilibrium, according to Lemmas 4.1, 5.1, 5.2, $\bar{\mu }_2\in (\eta ^{CL}\bar{\mu }_1,\eta ^L\bar{\mu }_1]$. Since Bertrand competition yields a deterred-entry equilibrium, $q_1^B=q_1^L ,~q_2^B=0,~ x_1^B=x_1^L$ and $x_2^B=0$. Since Cournot competition yields an accommodated-entry equilibrium, (8.7) is valid. We can use (5.13) and (8.7) to show that $\partial (q_1^C-q_1^B)/ \partial \bar{\mu }_2 =-[(4 -\gamma ^2)(16 -4 \gamma ^2+\gamma ^4)k^2-16 (4 -\gamma ^2)k+16 ]/\{ \cdot \}$, which, according to (5.3), is negative,where $\{\cdot \}=\gamma (4 -\gamma ^2)^3k\times \det [\Pi _{ij}^i]_C.$ Moreover, $x_1^B-x_1^C=-\{ 4[ (4-\gamma ^2)k-2 ]^2k( \eta ^{CL}\bar{\mu }_1-\bar{\mu }_2)\}/ \{\cdot \}$, which is positive since $\bar{\mu }_2>\eta ^{CL}\bar{\mu }_1$. Furthermore, $q_1^C=q_1^B$ when $\bar{\mu }_2=\eta ^e\bar{\mu }_1$ where

$$\begin{aligned} \eta ^e\equiv {8 \gamma [ (4 -\gamma ^2)k-2]k \over (4 -\gamma ^2)(16 - \gamma ^2+\gamma ^4)k^2-16 (4 -\gamma ^2)k+16 }. \end{aligned}$$

We can show that $\eta ^{CL}<\eta ^e<\eta ^{BL}$.⁶⁵

(ii): Equations (8.12) and (8.13) imply that $q_1^{BL}(0)<q_1^L<q_1^{CL}(0)$. Therefore, the desired result follows from the proof of Proposition 6.2-(ii). $\square$

The proof of Corollary 6.2

Part (i) follows from Proposition 6.2; Part (ii) from Proposition 6.5-(ii). $\square$

The proof of Proposition 6.6

Since the Cournot equilibrium is an interior equilibrium, $p_2^C>c_2$. In Case B1, $p_2^B=c_2$, implying that $p_2^C>p_2^B$. It remains to study Case A. Let $p_i^*$ ($i=1,2$) represent the price of good $i$ when Bertrand competition entails an accommodated-entry equilibrium. If $\bar{\mu }_2=\eta ^{BL}\bar{\mu }_1$, then $\mathbf{\Pi }_1^1(0)$ and $\mathbf{\Pi }_2^2(0)$ meet at point $(q_1^{BL}(0),0)$, i.e., $\mathbf{q }^*=(q_1^{BL}(0),0)$, implying that $p_2^*=c_2$ and hence $p_2^C>p_2^*$. We can show that $\partial (p_i^C-p_i^*)/\partial \bar{\mu }_2=\gamma ^3[(4-\gamma ^2 )k-2]^2k^2/[(4-\gamma ^2)^3(1-\gamma ^2)\times \det [\Pi _{ij}^i]_C\times \det [\Pi _{ij}^i]_B]>0$, $\forall \bar{\mu }_2\in [\eta ^{BL}\bar{\mu }_1,\bar{\mu }_1]$. It follows that if $\bar{\mu }_2\in [\eta ^{BL},\bar{\mu }_1]$, then $p_2^C>p_2^*$. $\square$

The proof of Fact 6.2

Since $\bar{\mu }_2=\eta ^{BL}\bar{\mu }_1$, Cournot competition entails the accommodated-entry equilibrium. Accordingly, (3.10), together with (8.7)–(8.8), implies that $p_1^C\doteq 0.4952(k-0.7279)(k-0.4209)k\bar{\mu }_1/[(k-0.4267)(k-0.5333)(k-0.7111)]$. Similarly, (3.13), together with (8.14) and (8.22), implies that $p_1^*\doteq 1.60714k\bar{\mu }_1/(3.75k-2)$. From (3.7) it follows that $p_1^L=\bar{\mu }_1+q_1^L-x_1^L$ which , together with (3.1) and (5.13), implies that $p_1^L\doteq (0.14297+1.6071 k)\bar{\mu }_1/(3.75k-2)$. With some subtractions, we have

$$\begin{aligned} \left\{ \begin{array}{l} p_1^L-p_1^*\doteq 0.0381\bar{\mu }_1/(k-0.5333)\\ p_1^*-p_1^C\doteq 0.6667(0.8250-k)(k-0.3940)k\bar{\mu }_1/[(k-0.7111)\\ \qquad \qquad \qquad \times (k-0.5333)(k-0.4267)] \\ p_1^L-p_1^C\doteq 0.6667(0.9680-k)(0.179-0.8224k+k^2) \bar{\mu }_1/[(k-0.7111)\\ \qquad \qquad \qquad \times (k-0.5333)(k-0.4267)] \end{array} \right. \end{aligned}$$

According to footnote 34, from $\gamma =0.5$ it follows that $\tilde{r}+\hat{r} =0.8$. Equation (5.11) thus means that $k>0.8$. It follows that $p_1^L>p_1^*.$ Moreover, the denominators of $p_1^*-p_1^C$ and $p_1^L-p_1^C$ are positive. We can show that $0.179-0.8224k+k^2>0$. Hence, the desired result follows from

$$\begin{aligned} \left\{ \begin{array}{l} p_1^*\gtreqqless p_1^C\Leftrightarrow k\lesseqqgtr \hat{k}\equiv 0.8250,\\ p_1^L\gtreqqless p_1^C\Leftrightarrow k\lesseqqgtr \tilde{k}\equiv 0.9680. \end{array} \right. \end{aligned}$$

$\square$

The proof of footnote 50

We can show that if $\bar{\mu }_2=\eta ^{BL}\bar{\mu }_1$, then

$$\begin{aligned} p_1^C-p_1^*={ (4-\gamma ^2) \gamma ^2\delta (k)k\bar{\mu }_1\over (2-\gamma ^2)[(4-\gamma ^2)k-2] \{ \cdot \} } , \end{aligned}$$

where $\delta (k)=(4-\gamma ^2)^2(2- \gamma ^2 )k^2-8(4-\gamma ^2)k+8$ and $\{\cdot \} =(4-\gamma ^2)^3k^2-16(4-\gamma ^2)k+16.$ Equation (5.3) implies that $\{\cdot \}$ is positive. It follows that $p_1^C-p_1^*$ and $\delta (k)$ share the same sign. It is clear that $\delta ''(k)>0$. Note that, according to (5.11), $k$ should exceed $\tilde{r}+\hat{r}$. Let $k^\delta$ denote the $k$ which solves $\delta '(k)=0$. From $\tilde{r}+\hat{r}-k^\delta =\gamma (2-2\gamma +\gamma ^2)/[(4-\gamma ^2)(1-\gamma )(2-\gamma ^2)]>0$ it follows that $\tilde{r}+\hat{r}>k^\delta$. Accordingly, $\delta (k)>\delta (\tilde{r}+\hat{r})$ if $k> \tilde{r}+\hat{r}$. We can show that $\delta (\tilde{r}+\hat{r})$ equals to $-\gamma ^2(2-4\gamma +\gamma ^2)/(1-\gamma )^2$, which is positive if $\gamma \in (0.5857 ,1).$ $\square$