Downstream Competition and the Effects of Buyer Power

Abstract

We examine four variations of a model in which oligopolistic retailers compete in a downstream market and one of them is a large retailer that has its own exclusive supplier. We demonstrate that an increase in the buyer power of the large retailer vis-à-vis its supplier leads to a fall in the retail price and an improvement in consumer welfare. More interestingly, we find that the beneficial effects of an increase in buyer power are large when the intensity of downstream competition is low, with the effects being the largest in the case of downstream monopoly.

Appendix: Proofs of Propositions and Corollaries

Proof of Proposition 1

The first statement of this proposition follows from (7). Consumer welfare is measured by consumer surplus, given by

$$CS = \int_{0}^{Q} {P(x)dx} - P(Q)Q.$$

(31)

By Assumption I and Eq. (5), we find that

$$\frac{\partial CS}{\partial \gamma } = - {\kern 1pt} QP'(Q)\frac{\partial Q}{\partial w}\frac{\partial W}{\partial \gamma } > 0.$$

(32)

Social welfare is measured by total surplus, which can be written as

$$TS = \int_{0}^{Q} {P(x)dx} - P(Q)Q + [P(Q) - w]q_{1} + (w - c_{s} )q_{1} + (n - 1)[P(Q) - c]q_{I} .$$

(33)

Differentiating (33) and use the retailers’ optimization conditions, we obtain

$$\frac{\partial TS}{\partial \gamma } = - q_{1} P'(Q)\frac{\partial Q}{\partial w}\frac{\partial W}{\partial \gamma } + (q_{1} - q_{I} )P'(Q)(n - 1)\frac{{\partial q_{I} }}{\partial w}\frac{\partial W}{\partial \gamma } + (w - c_{s} )\frac{{\partial q_{1} }}{\partial w}\frac{\partial W}{\partial \gamma } > 0.$$

(34)

The sign of (34) is determined by (5), $\partial q_{1} /\partial w < 0$, $\partial q_{I} /\partial w > 0$, Assumptions I and III, and $q_{1} \ge q_{I}$ (because $w \le c$). □

Proof of Proposition 2

The first statement of this proposition follows from Assumption II, (5), (6), and (8). Differentiating (31) with respect to $n$, we obtain:

$$\frac{\partial CS}{\partial n} = - {\kern 1pt} QP'(Q)\left[ {\frac{\partial Q}{\partial w}\frac{\partial W}{\partial n} + \frac{\partial Q}{\partial n}} \right] > 0,$$

(35)

the sign of which is determined by (5), (6), and Assumption II. Differentiating (33), we find:

$$\begin{aligned} \frac{\partial TS}{\partial n} = (q_{1} - q_{I} )(n - 1)P'\left[ {\frac{{\partial q_{I} }}{\partial n} + \frac{{\partial q_{I} }}{\partial w}\frac{\partial W}{\partial n}} \right] - {\kern 1pt} P'\left[ {q_{1} \frac{\partial Q}{\partial w}\frac{\partial W}{\partial n} + q_{1} \frac{\partial Q}{\partial n} + q_{1} q_{I} + (n - 1)q_{I}^{2} } \right] \hfill \\ \quad \quad \quad + (w - c_{s} )\left[ {\frac{{\partial q_{1} }}{\partial n} + \frac{{\partial q_{1} }}{\partial w}\frac{\partial W}{\partial n}} \right]. \hfill \\ \end{aligned}$$

(36)

The signs of the first two terms on the RHS of (36) are positive, while the sign of the third term is indeterminate. However, the third term vanishes if $c = c_{s}$ (in which case $w = c_{s}$ by Assumption III). By continuity, $\partial TS/\partial n > 0$ if $c - c_{s}$ is sufficiently small. □

Proof of Proposition 3

From (5), we derive:

$$\frac{\partial p}{\partial w} = P'\frac{\partial Q}{\partial w} = \frac{P'}{{(n - 1)(P' + q_{I} P'') + 2P' + q_{1} P''}} > 0.$$

(37)

Differentiating (37) with respect to $w$ and $n$ (respectively), we find

$$\frac{{\partial^{2} p}}{{\partial w^{2} }} = \frac{{\left[ {P''(P''Q - P') - P'P'''Q} \right] \cdot \left[ {\partial Q/\partial w} \right]}}{{\left[ {(n - 1)(P' + q_{I} P'') + 2P' + q{}_{1}P''} \right]^{2} }} \ge 0;$$

(38)

$$\frac{{\partial^{2} p}}{\partial n\partial w} = \frac{{\left[ {P''(P''Q - P') - P'P'''Q} \right] \cdot \left[ {\partial Q/\partial n} \right] - P'^{2} }}{{\left[ {(n - 1)(P' + q_{I} P'') + 2P' + q{}_{1}P''} \right]^{2} }} < 0.$$

(39)

Now differentiate (7) to obtain:

$$\frac{{\partial^{2} p}}{\partial n\partial \gamma } = \left[ {\frac{{\partial^{2} p}}{{\partial w^{2} }} \cdot \frac{\partial W}{\partial n} + \frac{{\partial^{2} p}}{\partial n\partial w}} \right]\frac{\partial W}{\partial \gamma } + \frac{\partial p}{\partial w} \cdot \frac{{\partial^{2} W}}{\partial n\partial \gamma }.$$

(40)

The sign of the first term on the RHS of (40) is positive by (38), (39) and Assumptions I and II. The sign of the second term is positive as well if $\partial^{2} W/\partial n\partial \gamma > 0$. Hence the sign of (40) is positive under the same condition. □

Proof of Corollary 1

Follows from (16) and Proposition 3. □

Proof of Corollary 2

Using the linear demand function, we write the total surplus as:

$$TS = \int_{0}^{Q} {(a - bx)dx - } (n - 1)cq_{I} - c_{s} q_{1} .$$

(41)

Differentiate (41) to obtain:

$$\frac{\partial TS}{\partial n} = \frac{{(a - 2c + w^{N} )(p - 2c + c_{s} )}}{{b(n + 1)^{2} }} - \frac{{(p - c + nc - nc_{s} )}}{b(n + 1)}\frac{\partial W}{\partial n},$$

(42)

which is positive if $p - 2c + c_{s} > 0$. Using (11), we can show that the latter holds for all $\gamma \in (0,\,1)$ if $c - c_{s} < (a - c)/(n + 2)$. □

Proof of Proposition 4

To simplify presentation, define $Z \equiv (2 - \theta )[2 + (n - 1)\theta ]$. Using (19), (20) and (23), we find:

$$\frac{{\partial p_{1}^{C} }}{\partial \gamma } = \frac{2 - \theta + \theta (n - 1)(1 - \theta )}{Z}\frac{{\partial W^{C} }}{\partial \gamma } < 0;\quad \frac{{\partial p_{I}^{C} }}{\partial \gamma } = \frac{\theta }{Z}\frac{{\partial W^{C} }}{\partial \gamma } < 0.$$

(43)

From (23), we obtain:

$$\frac{{\partial^{2} W^{C} }}{\partial n\partial \gamma } = \frac{{\theta (2 - \theta )\left( {a - c} \right)}}{{2(2 + \theta n - 2\theta )^{2} }} > 0.$$

(44)

Then (43)–(44) imply

$$\frac{{\partial^{2} p_{1}^{C} }}{\partial n\partial \gamma } = \frac{1}{{Z^{2} }}\left( {\frac{{\partial^{2} W^{C} }}{\partial n\partial \gamma }Z\left[ {2 - \theta + \theta \left( {n - 1} \right)(1 - \theta )} \right] - (2 - \theta )\theta^{2} \frac{{\partial W^{C} }}{\partial \gamma }} \right) > 0;$$

(45)

$$\frac{{\partial^{2} p_{I}^{C} }}{\partial n\partial \gamma } = \frac{1}{{Z^{2} }}\left( {\frac{{\partial^{2} W^{C} }}{\partial n\partial \gamma }\theta Z - (2 - \theta )\theta^{2} \frac{{\partial W^{C} }}{\partial \gamma }} \right) > 0.$$

(46)

□

Proof of Proposition 5

Define

$$T \equiv [1 + \theta (n - 2)][2 + \theta (n - 2)] - \theta^{2} (n - 1);$$

(47)

$$Y \equiv (1 - \theta )[2 + \theta (2n - 3)]a + \theta (n - 1)[1 + \theta (n - 2)]c.$$

(48)

It is easy to verify that $T > 0$ and that $Y/T > c$. Then the wholesale price in (29) can be written as $W^{B} (\gamma ,n) = (1 - \gamma )Y/T + \gamma \,c_{s}$. Since $Y/T > c$ and $c > c_{s}$, we conclude that $W^{B} (\gamma ,n) > c_{s}$ for $\gamma \in (0,\;1)$. The threshold $\gamma_{L}^{B}$ is obtained by solving $(1 - \gamma )Y/T + \gamma c_{s} = c$, which yields:

$$\gamma_{L}^{B} = \frac{Y - Tc}{{Y - Tc_{s} }}.$$

(49)

Differentiating (29), we find:

$$\frac{{\partial W^{B} (\gamma ,n)}}{\partial \gamma } = - \frac{Y}{T} + c_{s} < - c + c_{s} < 0.$$

(50)

From (50), we obtain

$$\frac{{\partial^{2} W(\gamma ,n)}}{\partial n\partial \gamma } = \frac{(1 - \theta )\theta }{{T^{2} }}G > 0,$$

(51)

where $G \equiv [\theta^{2} (2n^{2} - 6n + 5) + \theta (4n - 7) + 2]a - [\theta^{2} (2n^{2} - 6n + 5) + \theta (2n - 3)]c,$ which is positive because $a > c$.

Define $S \equiv 2[1 + \theta (n - 2)][2 + \theta (n - 2)] - \theta^{2} (n - 1)$. Differentiating (26), we obtain

$$\frac{{\partial p_{1}^{B} }}{\partial \gamma } = \frac{{\theta^{2} (n - 2)^{2} + 3\theta (n - 2) + 2}}{S} \cdot \frac{\partial W(\gamma ,n)}{\partial \gamma } < 0.$$

(52)

Then from (50)–(52), we find:

$$\frac{{\partial^{2} p_{1}^{B} }}{\partial n\partial \gamma } = \frac{\theta }{{S^{2} T^{2} }}\left\{ {GS(1 - \theta )[\theta^{2} (n - 2)^{2} + 3\theta (n - 2) + 2] + (Y - Tc_{s} )T\theta [\theta^{2} (n - 2)n + 3\theta - 2]} \right\},$$

(53)

which is clearly positive if $\theta^{2} (n - 2)n + 3\theta - 2 \ge 0$. In the case where $\theta^{2} (n - 2)n + 3\theta - 2 < 0$, we can show that $(1 - \theta )[\theta^{2} (n - 2)^{2} + 3\theta (n - 2) + 2] > - \theta [\theta^{2} (n - 2)n + 3\theta - 2]$. Moreover, it can be verified that $G > (Y - Tc_{s} )$ and $S > T$. These imply that (53) is positive in this case as well.

Regarding the price of the retailers other than R ₁, we obtain from their best response functions at stage 2

$$p_{I}^{B} = \frac{{(1 - \theta )a + [1 + \theta (n - 2)]c + \theta p_{1}^{B} }}{2 + \theta (n - 2)}.$$

(54)

Differentiating (54) and using (52)–(53), we can verify that $\partial p_{I}^{B} /\partial \gamma < 0$ and $\partial^{2} p_{I}^{B} /\partial n\partial \gamma > 0$. □