Bargaining power and firm profits in asymmetric duopoly: an inverted-U relationship

Abstract

This paper investigates the effects of bargaining power on downstream firms’ profits. Consider a vertically related industry consisting of one upstream and two downstream firms, the latter having different marginal costs. Each pair bargains over a linear wholesale price, and then the downstream firms engage in Cournot competition. We show that the inefficient downstream firm may benefit from an increase in the bargaining power of the upstream firm. Furthermore, we obtain similar results when each downstream firm trades with its exclusive upstream agent, under non-linear demand function, or when downstream firms compete in price.

Appendix: Proofs

Proof of Lemma 1

Differentiating $w_1^{*}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d w_1^{*}}{d \theta }=\frac{(4+\theta ^2)(a-z_1)+4\theta (a-z_2)+(2-\theta )^2(z_2-z_1)}{(4-\theta ^2)^2}>0. \end{aligned}$$

(29)

Differentiating $w_2^{*}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d w_2^{*}}{d \theta }&=\frac{(4+\theta ^2)[(2+\theta )a-(4-\theta )z_2+2(1-\theta )z_1]+\theta (4-\theta ^2)(a+z_2-2z_1)}{2(4-\theta ^2)^2}\nonumber \\&>0. \end{aligned}$$

(30)

Similarly, we have

$$\begin{aligned} \frac{d w_1^{*}}{d \theta }-\frac{d w_2^{*}}{d \theta }=\frac{3(z_2-z_1)}{(2+\theta )^2}>0, \end{aligned}$$

(31)

which proves the lemma. $\square$

Proof of Proposition 1

Differentiating $q_1^{*}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d q_1^{*}}{d \theta }=-\frac{(4+\theta ^2)(a-z_1)+4\theta (a-z_2)+4(2-\theta )^2(z_2-z_1)]}{3 (4-\theta ^2)^2}<0. \end{aligned}$$

(32)

Similarly, differentiating $q_2^{*}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d q_2^{*}}{d \theta }=\frac{-(2+\theta )^2a-4(4-5\theta + \theta ^2) z_1+ (20-16\theta +5\theta ^2) z_2}{3 (4-\theta ^2)^2}. \end{aligned}$$

(33)

This expression is positive if and only if

$$\begin{aligned} z_2>\frac{(2+\theta )^2a+4(4-5\theta + \theta ^2) z_1}{20-16\theta +5\theta ^2}. \end{aligned}$$

(34)

We now show that the interval between this lower bound and the upper bound of the interior solution condition is nonempty. Comparing these two values, we have

$$\begin{aligned}&\frac{(2+\theta )(4-3\theta )a+(8-8\theta +3\theta ^2)z_1}{2(8-5\theta )}-\frac{(2+\theta )^2a+4(4-5\theta + \theta ^2) z_1}{20-16\theta +5\theta ^2} \nonumber \\&\quad =\frac{3(2-\theta )(2+\theta )(8-16\theta +5\theta ^2)(a-z_1)}{2(8-5\theta )(20-16\theta +5\theta ^2)}>0, \nonumber \\&\quad \Leftrightarrow \theta <\frac{2}{5}(4-\sqrt{6}) \approx 0.62, \end{aligned}$$

(35)

which gives the condition for Proposition 1. This completes the proof of the proposition. $\square$

Proof of Proposition 2

Differentiating $Q^{*}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d Q^{*}}{d \theta }=-\frac{2a-z_1-z_2}{3(2-\theta )^2}<0. \end{aligned}$$

(36)

Similarly, we have

$$\begin{aligned} \frac{d \left( \pi _{D1}^{*}+\pi _{D2}^{*}\right) }{d \theta }= & {} \left[ \frac{d p^{*}}{d \theta }-\frac{d w_1^{*}}{d \theta }\right] q_1^{*}+(p^{*}-c_1)\frac{d q_1^{*}}{d \theta }+\left[ \frac{d p^{*}}{d \theta }-\frac{d w_2^{*}}{d \theta }\right] q_2^{*} \nonumber \\&+\,(p^{*}-c_2)\frac{d q_2^{*}}{d \theta } \end{aligned}$$

(37)

$$\begin{aligned}= & {} \frac{d q_1^{*}}{d \theta }q_1^{*}+(p^{*}-c_1)\frac{d q_1^{*}}{d \theta }+\frac{d q_2^{*}}{d \theta }q_2^{*}+(p^{*}-c_2)\frac{d q_2^{*}}{d \theta } \end{aligned}$$

(38)

$$\begin{aligned}= & {} \frac{d q_1^{*}}{d \theta }\left( p^{*}-c_1+q_1^{*}\right) +\frac{d q_2^{*}}{d \theta }\left( p^{*}-c_2+q_2^{*}\right) . \end{aligned}$$

(39)

Note that $p^{*}-c_1+q_1^{*}>p^{*}-c_2+q_2^{*}>0$. Since $\left| \frac{d q_1^{*}}{d \theta }\right| >\frac{d q_2^{*}}{d \theta }$, the sign of Eq. (39) becomes negative. $\square$

Proof of Lemma 2

Differentiating $w_i^{**}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d w_i^{**}}{d \theta }=\frac{[4(a-2z_i+z_j)+\theta (a-2z_j+z_i)](16+\theta ^2)}{(16-\theta ^2)^2}>0, \quad i\ne j. \end{aligned}$$

(40)

Similarly, we have

$$\begin{aligned} \frac{d w_1^{**}}{d \theta }-\frac{d w_2^{**}}{d \theta }=\frac{12(z_2-z_1)}{(4+\theta )^2}>0, \end{aligned}$$

(41)

which proves the lemma. $\square$

Proof of Proposition 3

Differentiating $q_1^{**}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d q_1^{**}}{d \theta }=-\frac{4 \left[ (4+\theta )^2a-(80-32\theta +5 \theta ^2) z_1+4 (16-10 \theta +\theta ^2) z_2\right] }{3 (16-\theta ^2)^2}. \end{aligned}$$

(42)

The numerator of the expression can be written as follows:

$$\begin{aligned} 4\left[ (16+\theta ^2)(a-z_1)+8\theta (a-z_2)+(64-32\theta +4\theta ^2)(z_2-z_1)\right] >0, \end{aligned}$$

(43)

which proves the first part of the proposition.

Similarly, differentiating $q_2^{**}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d q_2^{**}}{d \theta }=-\frac{4 \left[ (4+\theta )^2a-(80-32\theta +5 \theta ^2) z_2+4 (16-10 \theta +\theta ^2) z_1\right] }{3 (16-\theta ^2)^2}. \end{aligned}$$

(44)

This expression is positive if and only if

$$\begin{aligned} z_2>\frac{(4+\theta )^2a+4 (16-10 \theta +\theta ^2) z_1}{80-32\theta +5 \theta ^2}. \end{aligned}$$

(45)

We now show that the interval between this lower bound and the upper bound of the interior solution condition is nonempty.

$$\begin{aligned}&\frac{(4+\theta )a+2(2-\theta )z_1}{8-\theta }-\frac{(4+\theta )^2a+4 (16-10 \theta +\theta ^2) z_1}{80-32\theta +5 \theta ^2} \nonumber \\&\quad =\frac{6(32-16\theta -2\theta ^2+\theta ^3)(a-z_1)}{(8-\theta )(80-32\theta +5 \theta ^2)}>0. \end{aligned}$$

(46)

This completes the proof of the proposition. $\square$

Proof of Lemma 3

Differentiating $w_1^{B}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d w_i^{B}}{d \theta }&=\frac{2[(2-\gamma ^2)A_i+\theta \gamma A_j][4(2-\gamma ^2)^2-\theta ^2 \gamma ^2]+2\theta ^2 \gamma [2(2-\gamma ^2)A_i+\theta \gamma A_j]}{[4(2-\gamma ^2)^2-\theta ^2 \gamma ^2]^2}\nonumber \\&>0. \end{aligned}$$

(47)

Similarly, we have

$$\begin{aligned} \frac{d w_1^{B}}{d \theta }-\frac{d w_2^{B}}{d \theta }=\frac{2(1+\gamma )(2-\gamma )(2-\theta \gamma -\gamma ^2)(z_2-z_1)}{[4(2-\gamma ^2)^2-\theta ^2 \gamma ^2]^2}>0, \end{aligned}$$

(48)

which proves the lemma. $\square$

Proof of Proposition 4

Differentiating $q_1^{B}$ with respect to $\theta$, we have

$$\begin{aligned} \frac{d q_1^{B}}{d \theta }= & {} \frac{1}{(1-\gamma ^2)(4-\gamma ^2)} \left[ -(2-\gamma ^2)\frac{d w_1^{B}}{d \theta } +\gamma \frac{d w_2^{B}}{d \theta } \right] \end{aligned}$$

(49)

$$\begin{aligned}&<\frac{1}{(1-\gamma ^2)(4-\gamma ^2)} \left[ -(2-\gamma ^2)\frac{d w_1^{B}}{d \theta } +\gamma \frac{d w_1^{B}}{d \theta } \right] \end{aligned}$$

(50)

$$\begin{aligned}= & {} \frac{1}{(1-\gamma ^2)(4-\gamma ^2)} \left[ -(1-\gamma )(2+\gamma ) \frac{d w_1^{B}}{d \theta }\right] <0. \end{aligned}$$

(51)

Note that we used Lemma 3 to get the inequality on the second line.

We can relatively easy derive the sufficient condition for the result. Applying the mean-value theorem, for some $\hat{\theta }\in (0,1)$, we have

$$\begin{aligned} \frac{q_2^{B}(1)-q_2^{B}(0)}{1-0}=\frac{d q_2^{B}(\hat{\theta })}{d \theta }. \end{aligned}$$

(52)

The condition that the left-hand side of the above expression is positive is

$$\begin{aligned} q_2^{B}(1)-q_2^{B}(0)>0 \Leftrightarrow \alpha <\frac{(3-\gamma ^2)(4-3\gamma ^2)\gamma z_1-2(2-\gamma ^2)^3 z_2}{(2-\gamma -\gamma ^2)^2(4+\gamma -2\gamma ^2)}. \end{aligned}$$

(53)

We can confirm that the interval between the above threshold value of $\alpha$ and that of the interior solution condition is nonempty. That is,

$$\begin{aligned}&\frac{(3-\gamma ^2)(4-3\gamma ^2)\gamma z_1-2(2-\gamma ^2)^3 z_2}{(2-\gamma -\gamma ^2)^2(4+\gamma -2\gamma ^2)} \nonumber \\&\qquad -\,\frac{[8+2\gamma ^4-(8+\theta )\gamma ^2]z_2-(2-\theta )(2-\gamma ^2)\gamma z_1}{(2-\gamma -\gamma ^2)(4+\theta \gamma -2\gamma ^2)} \nonumber \\&\quad = \frac{(z_2-z_1)\gamma (2+\gamma -\gamma ^2)[2(1+\theta )(4-\gamma ^2)(1-\gamma ^2)+(2+\theta )\gamma ^2]}{(2-\gamma -\gamma ^2)^2(4+\gamma -2\gamma ^2)(4+\theta \gamma -2\gamma ^2)}>0.\nonumber \\ \end{aligned}$$

(54)

$\square$