Social responsibility in a bilateral monopoly

Abstract

We work on a linear bilateral monopoly to analyze the effects of firms’ social concern. Both firms in the market, the up-stream manufacturer and the down-stream retailer, can be socially concerned. Firm’s social concern is modeled through a broader firm objective. In addition to their profit both firms also care about a share of consumer surplus. In our two stage game, at first the manufacturer fixes the wholesale price per quantity, which has to be paid by the retailer. Subsequently, the retailer chooses the optimal quantity. First, the game is analyzed for exogenous levels of social concern for both firms. Afterwards, both firms are able to choose endogenously their respective level of social concern. The results show that firm’s social concern increases firm profit for the manufacturer as well as the retailer’s profit. Moreover, the firms’ broader objective function softens the classical double marginalization problem, because in equilibrium all market participants, consumers included, are better off compared to a bilateral monopoly with two pure profit-maximizing firms. Therefore, firms’ social responsibility results in a Pareto improvement.

Appendix

1.1 Proof of Result 1

$$\begin{aligned} w^{*}&= \frac{a(2\!-\!\theta _{m}\!-\!\theta _{r})\!+\!c(2\!-\!\theta _{r})}{4\!-\!\theta _{m}\!-\!2\theta _{r}} \Longleftrightarrow w^{*}\!-\!c\!=\!\frac{a(2\!-\!\theta _{m}\!-\!\theta _{r})\!+\!c(2\!-\!\theta _{r})}{4\!-\!\theta _{m}\!-\!2\theta _{r}}\!-\!c \\&= \Longleftrightarrow w^{*}\!-\!c\!=\!\frac{(a\!-\!c)(2\!-\!\theta _{m}\!-\!\theta _{r})}{4\!-\!\theta _{m}\!-\!2\theta _{r}} \ge 0 \quad \text {for} \, \, \, \theta _{m} \in [0,1] \,\, \, \text {and} \, \, \, \theta _{r} \in [0,1] \end{aligned}$$

$\square$

1.2 Proof of Result 2

1.2.1 Proof of (i)

$$\begin{aligned} \frac{\partial w^{*}}{\partial \theta _{m}}=-\frac{(a-c)(2-\theta _{r})}{(4-\theta _{m}-2 \theta _{r})^{2}}<0 \qquad \text {and} \qquad \frac{\partial w^{*}}{\partial \theta _{r}}=-\frac{(a-c)\, \theta _{m}}{(4-\theta _{m}-2 \theta _{r})^{2}}<0, \end{aligned}$$

whereas the second result can be explained in the following. First, we look at the first-order condition of the first stage given by

$$\begin{aligned} \frac{d v_{m}}{d w}=\frac{\partial v_{m}}{\partial w} + \frac{\partial v_{m}}{\partial q} \frac{\partial q}{\partial w}=0 \,\,\, \forall \,\,\, \theta _r. \end{aligned}$$

Differentiate $\frac{\partial v_{m}}{\partial w}$ w.r.t. $\theta _{r}$ yields

$$\begin{aligned} \frac{\partial ^2 v_{m}}{\partial w \, \partial \theta _r} + \frac{\partial ^2 v_{m}}{\partial w^2} \frac{\partial w}{\partial \theta _r} =0. \end{aligned}$$

By rearranging this we get the following expression

$$\begin{aligned} \frac{\partial w}{\partial \theta _r} =-\frac{\frac{\partial ^2 v_{m}}{\partial w \, \partial \theta _r }}{\frac{\partial ^2 v_{m}}{\partial w^2}} \qquad \text {(implicit function theorem)}. \end{aligned}$$

The denominator is the SOC and is therefore negative for a maximum. $\frac{\partial w}{\partial \theta _r}<0$ is true if the numerator is negative as well. Differentiate the first-order condition w.r.t. $\theta _r$:

$$\begin{aligned} \frac{d^2 v_{m}}{d w \, d \theta _r}= \underbrace{\frac{\partial v_{m}}{\partial q}}_{b \, q \, \theta _{m}+w-c} \underbrace{\frac{\partial ^2 q}{\partial w \, \partial \theta _r}}_{\frac{-1}{b(2-\theta _{r})^2}}+ \underbrace{\frac{\partial ^2 v_{m}}{\partial q^2}}_{b \, \theta _{m}} \underbrace{\frac{\partial q}{\partial w}}_{\frac{-1}{b(2-\theta _{r})}} \underbrace{\frac{\partial q}{\partial \theta _r}}_{\frac{a-w}{b(2-\theta _{r})^2}} . \end{aligned}$$

From Result 1 we know that $w \ge c$. Therefore, $\frac{\partial v_{m}}{\partial q}>0$ and consequently the first product is negative for $\theta _{m}>0$. The first and third derivative in the second term are positive while the second one is negative. As a result, also the second product is negative. In summary, $\frac{d^2 v_{m}}{d w \, d \theta _r}<0 \implies \frac{\partial w^{*}}{\partial \theta _{r}}<0$. $\square$

1.2.2 Proof of (ii)

$$\begin{aligned} \begin{array}{lll} \frac{\partial q^{*}}{\partial \theta _{m}}=\frac{(a-c)}{b(4-\theta _{m}-2 \theta _{r})^{2}}>0 \qquad \text {and} \qquad \frac{\partial q^{*}}{\partial \theta _{r}}=\frac{2(a-c)}{b(4-\theta _{m}-2 \theta _{r})^{2}}>0 , \\ \frac{\partial p^{*}}{\partial \theta _{m}}=-\frac{(a-c)}{(4-\theta _{m}-2 \theta _{r})^{2}}<0 \qquad \text {and} \qquad \frac{\partial p^{*}}{\partial \theta _{r}}=-\frac{2(a-c)}{(4-\theta _{m}-2 \theta _{r})^{2}}<0 , \\ \frac{\partial CS^{*}}{\partial \theta _{m}}=\frac{(a-c)^{2}}{b(4-\theta _{m}-2 \theta _{r})^{3}}>0 \qquad \text {and} \qquad \frac{\partial CS^{*}}{\partial \theta _{r}}=\frac{2(a-c)^{2}}{b(4-\theta _{m}-2 \theta _{r})^{3}}>0 , \\ \frac{\partial WF^{*}}{\partial \theta _{m}}=\frac{(a-c)^{2}(3-\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2 \theta _{r})^{3}}>0 \qquad \text {and} \\ \frac{\partial WF^{*}}{\partial \theta _{r}}=\frac{2(a-c)^{2}(3-\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2 \theta _{r})^{3}}>0 \end{array} \end{aligned}$$

$\square$

1.2.3 Proof of (iii)

$$\begin{aligned} \frac{\partial q^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial q^{*}}{\partial \theta _{r}} , \qquad \frac{\partial p^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial p^{*}}{\partial \theta _{r}} , \qquad \frac{\partial CS^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial CS^{*}}{\partial \theta _{r}} \,\,\, \text {and} \,\,\, \frac{\partial WF^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial WF^{*}}{\partial \theta _{r}} \end{aligned}$$

$\square$

1.3 Proof of Result 3

1.3.1 Proof of (i)

$$\begin{aligned} \begin{array}{ll} \frac{\partial \pi _{m}^{*}}{\partial \theta _{m}}=-\frac{(a-c)^2 \theta _{m}}{b(4-\theta _{m}-2 \theta _{r})^3}=0 \Longleftrightarrow \theta _{m}(\theta _{r})=0 \\ \frac{\partial \pi _{r}^{*}}{\partial \theta _{r}}=\frac{(a-c)^2 (\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2 \theta _{r})^3}=0 \Longleftrightarrow \theta _{r}(\theta _{m})=\frac{\theta _{m}}{2} \\ \Longleftrightarrow \theta _{m}^{*}=0 \qquad \text {and} \qquad \theta _{r}^{*}=0 \end{array} \end{aligned}$$

$\square$

1.3.2 Proof of (ii)

Using backward induction, the manufacturer chooses its level of social concern to maximize its profit by

$$\begin{aligned} \frac{\partial \pi _{m}^{*}}{\partial \theta _{m}}=-\frac{(a-c)^2 \theta _{m}}{b(4-\theta _{m}-2 \theta _{r})^3}=0 \Longleftrightarrow \theta _{m}(\theta _{r})=0 \Longleftrightarrow \theta _{m}^{*}=0. \end{aligned}$$

By inserting $\theta _{m}(\theta _{r})=0$ into the value of the retailer we get the following expression, optimization problem and equilibrium levels of social concern:

$$\begin{aligned} \pi _{r}=\frac{(a-c)^2(1-\theta _{r})}{4b(2-\theta _{r})^2} \Longleftrightarrow \frac{\partial \pi _{r}}{\partial \theta _{r}}=\frac{-(a-c)^2\theta _{r}}{4b(2-\theta _{r})^3} \Longleftrightarrow \theta _{r}^{*}=0. \end{aligned}$$

$\square$

1.3.3 Proof of (iii)

Again, by using backward induction, at first, the retailer chooses the profit-maximizing level of social concern.

$$\begin{aligned} \frac{\partial \pi _{r}^{*}}{\partial \theta _{r}}=\frac{(a-c)^2 (\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2 \theta _{r})^3}=0 \Longleftrightarrow \theta _{r}(\theta _{m})=\frac{\theta _{m}}{2} \end{aligned}$$

Inserting $\theta _{r}(\theta _{m})=\frac{\theta _{m}}{2}$ into the manufacturer’s profit function results in the following expression:

$$\begin{aligned} \pi _{m}^{*}(\theta _{m})=\frac{(a-c)^2(4-3 \theta _{m})}{8b(2-\theta _{m})^2} . \end{aligned}$$

After the retailer has chosen the firm’s optimal level of social concern, the manufacturer chooses its profit-maximizing level of social concern:

$$\begin{aligned} \frac{\partial \pi _{m}^{E}(\theta _{m})}{\partial \theta _{m}}=\frac{(a-c)^2(3 \theta _{m}-2)}{8b(2-\theta _{m})^3}=0 \Longleftrightarrow \theta _{m}^{*}=\frac{2}{3} \end{aligned}$$

The sequential choice of firms’ profit-maximizing level of social concern, whereas the manufacturer determines before the retailer does, yields the following equilibrium values of $\theta _{i}$:

$$\begin{aligned} \theta _{m}^{*}=\frac{2}{3} \qquad \text {and} \qquad \theta _{r}^{*}=\frac{1}{3} . \end{aligned}$$

The firms’ commitment effect through its objective function in the case of sequential choice is explainable by a strategic effects analysis. First, we take a look at the retailer. Its strategic effects are given by

$$\begin{aligned} \frac{d \pi _{r}}{d \theta _{r}}= \underbrace{\frac{\partial \pi _{r}}{\partial \theta _{r}}}_{=0}+ \underbrace{\frac{\partial \pi _{r}}{\partial w}}_{\frac{-2(a-w)(1-\theta _{r})}{b(2-\theta _{r})^2}} \underbrace{\frac{\partial w}{\partial \theta _{r}}}_{\frac{-(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}}+ \underbrace{\frac{\partial \pi _{r}}{\partial q}}_{a-2 \, b \, q - w} \underbrace{\frac{\partial q}{\partial w}}_{\frac{-1}{b(2-\theta _{r})}} \underbrace{\frac{\partial w}{\partial \theta _{r}}}_{\frac{-(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}}. \end{aligned}$$

Rewriting the equation above results in the following expression

$$\begin{aligned} \frac{d \pi _{r}}{d \theta _{r}}= \underbrace{\frac{\partial \pi _{r}}{\partial \theta _{r}}}_{=0}+ \underbrace{\frac{\partial w}{\partial \theta _{r}}}_{\frac{-(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}} \left[ \underbrace{\frac{\partial \pi _{r}}{\partial w}+\frac{\partial \pi _{r}}{\partial q} \, \frac{\partial q}{\partial w}}_ {\frac{-(a-w)(4-3\theta _{r})+2b \, q (2-\theta _{r})}{b(2-\theta _{r})^2}} \right] , \end{aligned}$$

whereas $\frac{\partial w}{\partial \theta _{r}}<0$ for $\theta _{m}>0$. By inserting $q^{*}$ and $w^{*}$ into $\frac{-(a-w)(4-3\theta _{r})+2b \, q (2-\theta _{r})}{b(2-\theta _{r})^2}$ we get $\frac{-(a-c)(2-3\theta _{r})}{b(2-\theta _{r})(4- \theta _{m} -2 \theta _{r})}$. We know from the previous result that $\theta _{r}^{*}=\frac{1}{3}$. Therefore, $(\frac{\partial \pi _{r}}{\partial w}+\frac{\partial \pi _{r}}{\partial q} \, \frac{\partial q}{\partial w})$ becomes negative. The overview of the retailer’s strategic effects which explains the retailer’s incentive to choose $\theta _{r}>0$ is summarized below

$$\begin{aligned} \frac{d \pi _{r}}{d \theta _{r}}= \underbrace{\frac{\partial \pi _{r}}{\partial \theta _{r}}}_{=0}+ \underbrace{\underbrace{\frac{\partial w}{\partial \theta _{r}}}_{<0} \left[ \underbrace{\frac{\partial \pi _{r}}{\partial w}+\frac{\partial \pi _{r}}{\partial q} \, \frac{\partial q}{\partial w}}_ {<0} \right] }_{>0} >0 . \end{aligned}$$

In the next step, we care about the manufacturer’s strategic effects which are given by

$$\begin{aligned} \frac{d \pi _{m}}{d \theta _{m}}&= \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{m}}}_{=0}+ \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{r}}}_{=\frac{(a-c)(4-3\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2\theta _{r})^3}} \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{=\frac{1}{2}}+ \underbrace{\frac{\partial \pi _{m}}{\partial w}}_{=\frac{a+c-2w}{b(2-\theta _{r})}} \underbrace{\frac{\partial w}{\partial \theta _{r}}}_{=\frac{-(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}} \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{=\frac{1}{2}}+ \\&+\underbrace{\frac{\partial \pi _{m}}{\partial q}}_{=w-c} \underbrace{\frac{\partial q}{\partial w}}_{=\frac{-1}{b(2-\theta _{r})}} \underbrace{\frac{\partial w}{\partial \theta _{r}}}_{=\frac{-(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}} \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{=\frac{1}{2}} . \end{aligned}$$

By rewriting we get

$$\begin{aligned} \frac{d \pi _{m}}{d \theta _{m}}= \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{m}}}_{=0}+ \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{r}}}_{=\frac{1}{2}} \left[ \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{r}}}_{=\frac{(a-c)(4-3\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2\theta _{r})^3}}+ \underbrace{\frac{\partial w}{\partial \theta _{r}}}_{=\frac{-(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}} \left[ \underbrace{\frac{\partial \pi _{m}}{\partial w}+\frac{\partial \pi _{m}}{\partial q} \frac{\partial q}{\partial w}}_ {=\frac{(a+2c-3w)}{b(2-\theta _{r})}} \right] \right] \end{aligned}$$

We know that $\theta _{r}(\theta _{m})=\frac{\theta _{m}}{2}$. By replacing $\theta _{r}$ with $\frac{\theta _{m}}{2}$ in $\frac{(a-c)(4-3\theta _{m}-2\theta _{r})}{b(4-\theta _{m}-2\theta _{r})^3}$ we get $\frac{(a-c)^2(1-\theta _{m})}{2b(2-\theta _{m})}>0$ for $\theta _{m}<1$. Moreover, $-\frac{(a-c)\theta _{m}}{(4-\theta _{m}-2\theta _{r})^2}<0$ for $\theta _{m}<1$. Additionally, $\frac{(a+2c-3w)}{b(2-\theta _{r})}<0$ in equilibrium for $w=w^{**}$, $\theta _{m}^{*}=\frac{2}{3}$ and $\theta _{r}^{*}=\frac{1}{3}$. The summary of the manufacturer’s strategic effect analysis is given by the following expression

$$\begin{aligned} \frac{d \pi _{m}}{d \theta _{m}}= \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{m}}}_{=0}+ \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{>0} \left[ \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{r}}}_{>0}+ \underbrace{\underbrace{\frac{\partial w}{\partial \theta _{r}}}_{<0} \left[ \underbrace{\frac{\partial \pi _{m}}{\partial \theta _{r}}+\frac{\partial \pi _{m}}{\partial q} \frac{\partial q}{\partial w}}_{<0} \right] }_{>0} \right] >0. \end{aligned}$$

Furthermore, the strategic effect analysis explains one finding made in Result 2. There, we have shown that the impact of the manufacturer’s social concern is only half of the retailer’s one. In detail, $\frac{\partial q^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial q^{*}}{\partial \theta _{r}} , \frac{\partial p^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial p^{*}}{\partial \theta _{r}} , \frac{\partial CS^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial CS^{*}}{\partial \theta _{r}} \,\,\, \text {and} \,\,\, \frac{\partial WF^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial WF^{*}}{\partial \theta _{r}}$. The consideration of the strategic effects yields the following further insights:

$$\begin{aligned} \begin{array}{ll} \frac{\partial q^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial q^{*}}{\partial \theta _{r}}=\frac{\partial q^{*}}{\partial \theta _{r}} \, \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{\frac{1}{2}} , \qquad \frac{\partial p^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial p^{*}}{\partial \theta _{r}}=\frac{\partial p^{*}}{\partial \theta _{r}} \, \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{\frac{1}{2}} , \qquad \\ \frac{\partial CS^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial CS^{*}}{\partial \theta _{r}}=\frac{\partial CS^{*}}{\partial \theta _{r}} \, \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{\frac{1}{2}} \qquad \text {and} \qquad \frac{\partial WF^{*}}{\partial \theta _{m}}=\frac{1}{2} \, \frac{\partial WF^{*}}{\partial \theta _{r}}=\frac{\partial WF^{*}}{\partial \theta _{r}} \, \underbrace{\frac{\partial \theta _{r}}{\partial \theta _{m}}}_{\frac{1}{2}} . \end{array} \end{aligned}$$

$\square$

1.4 Proof of Result 4

$$\begin{aligned} \begin{array}{ll} \pi _{m}^{**}-\pi _{m}^{*}\Big |_{\theta _{m}=\theta _{r}=0}=\frac{9(a-c)^2}{64b}+F-\left( \frac{(a-c)^2}{8b}+F \right) =\frac{(a-c)^2}{64b}&{}>0 \\ \pi _{r}^{**}-\pi _{r}^{*}\Big |_{\theta _{m}=\theta _{r}=0}=\frac{3(a-c)^2}{32b}-F-\left( \frac{(a-c)^2}{16b}-F \right) =\frac{(a-c)^2}{32b}&{}>0 \\ PS^{**}-PS^{*}\Big |_{\theta _{m}=\theta _{r}=0}=\frac{15(a-c)^2}{64b}-\frac{3(a-c)^2}{16b}=\frac{3(a-c)^2}{64b}&{}>0 \\ CS^{**}-CS^{*}\Big |_{\theta _{m}=\theta _{r}=0}=\frac{9(a-c)^2}{128b}-\frac{(a-c)^2}{32b}=\frac{5(a-c)^2}{128b}&{}>0 \\ WF^{**}-WF^{*}\Big |_{\theta _{m}=\theta _{r}=0}=\frac{39(a-c)^2}{128b}-\frac{7(a-c)^2}{32b}=\frac{11(a-c)^2}{128b}&{}>0 \end{array} \end{aligned}$$

$\square$