Pricing and market conduct in a vertical relationship

Abstract

We consider a vertical relationship where an upstream monopolist supplies input to downstream duopolistic firms. Under the assumption that downstream firms produce under a soft capacity restriction, we show that the balance between price and quantity in downstream firms’ strategy is endogenous. In this way, the monopolist’s charge for input co-determines downstream market conduct. We spell out some consequences of this, for example, that an increase of downstream capacity costs can result in increased output. We discuss other implications in relation to pass-through and incidence of cost changes.

Appendix

Proof of Proposition 1

Downstream firm i maximizes aggregate profits, given by $\pi ^{i}=p^{i}q^{i}-TC_q^i -TC_k^i$, by setting price and capacity. With respect to $TC_q^i$, if the firm produces at or below the capacity, the marginal cost is $\omega ;$ if the firm produces beyond capacity, marginal cost increases to $\omega +\theta$. That is, in the price stage, $TC_q^i$ is:

$$\begin{aligned} TC_q^i =\left\{ {{ \begin{array}{ll} \omega q^{i}, &{} \quad q^{i}\le k^{i} \\ \left( {\omega +\theta } \right) q^{i}, &{} \quad q^{i}>k^{i} \\ \end{array}}} \right. \end{aligned}$$

(6)

and $TC_k^i =\psi q^{i}$. There is a disincentive to produce beyond capacity because of the unfavorable change of marginal cost. In this way, it is possible for firms to sustain a price above the marginal cost in spite of competing in prices. Hence, capacity choice can be used strategically to soften competition in the market stage because there is a difference between “short-run” and “long-run” marginal costs. If firms produce at their capacity, as happens in equilibrium, the profit function becomes $\pi ^{i}=\left( {p^{i}-\psi -\omega } \right) q^{i}$.

In the second stage, firms engage in Bertrand competition given the capacities that have been decided upon. The Bertrand response functions are the solutions to $argmax_{p^{i}} p^{i}q^{i}-TC_q^i, i=1, 2:$

$$\begin{aligned} r^{i}\left( {p^{j}, x^{i}} \right) =1/2\left( {\left( {\left( {1-d} \right) a+dp^{j}} \right) +x^{i}} \right) , \quad i, j=1, 2, i\ne j, \end{aligned}$$

(7)

where $x^{i}=\omega$ for $q^{i}\le k^{i}$ and $x^{i}=\omega +\theta$ for $q^{i}>k^{i}$. Equation (7) defines firm i’s best response to the price charged by firm j. Notice that the best response function “jumps” when output reaches capacity.

Equation (7) defines firms’ best responses dependent on capacity and the price set by the other firms. To find the reaction functions, define the prices for firm i, which ensures that the firm produces exactly at its capacity, $k^{i}$, given the price charged by firm $j, p^{j};$ that is ${\Phi }^{i}\left( {p^{j},k^{i}} \right) =\left( {a+dp^{j}-D^{-1}k^{i}} \right)$. Suppose that ${\Phi }^{i}\left( {p^{j},k^{i}} \right) <r^{i}\left( {p^{j},x^{i}} \right)$, that is, the price firm i ideally likes to set is in excess of the price which just makes sure that the firm utilizes capacity. In other words, the marginal-revenue curve of firm i cuts the marginal-cost curve given by $\omega$ for $q^{i}\le k^{i}$. This is clearly feasible and the firm’s reaction function is:

$$\begin{aligned} R^{i}\left( {p^{j},k^{i}} \right) =1/2\left( {\left( {\left( {1-d} \right) a+dp^{j}} \right) +\omega } \right) . \end{aligned}$$

(8)

In the intermediate case, firm $i^{\prime }\hbox {s}$ marginal-revenue curve cuts the marginal-cost curve where marginal costs jump from $\omega$ to $\omega +\theta$. That is, the firm ideally sets a price that results in production at the capacity limit. Clearly, it can charge a higher price and produce less, or it can charge a lower price and sell more. That is, $r^{i}\left( {p^{j},\omega +\theta } \right)>{\Phi }^{i}\left( {p^{j},k^{i}} \right) >r^{i}\left( {p^{j},\omega } \right)$. In this situation, the firm sets a price given by ${\Phi }^{i}\left( {p^{j},k^{i}} \right)$, hence the reaction function is:

$$\begin{aligned} R^{i}\left( {p^{j},k^{i}} \right) =\left( {a+dp^{j}-D^{-1}k^{i}} \right) . \end{aligned}$$

(9)

Finally, consider the parameter configuration that satisfies ${\Phi }^{i}\left( {p^{j},k^{i}} \right) >r^{i}\left( {p^{j},\omega +\theta } \right)$, that is, firm $i^{\prime }\hbox {s}$ marginal-revenue curve cuts the marginal-cost curve given by $\omega +\theta$ for some output in excess of capacity, $q^{i}\ge k^{i}$. Thus, the firm’s reaction function is:

$$\begin{aligned} R^{i}\left( {p^{j},k^{i}} \right) =1/2\left( {\left( {\left( {1-d} \right) a+dp^{j}} \right) +\left( {\omega +\theta } \right) } \right) . \end{aligned}$$

(10)

All in all, we have:

$$\begin{aligned} R^{i}\left( {p^{j},k^{i}} \right) = \begin{array}{l} 1/2\left( {\left( {\left( {1-d} \right) a+dp^{j}} \right) +\omega } \right) ,{\Phi }^{i}\left( {p^{j},k^{i}} \right) <r^{i}\left( {p^{j},x^{i}} \right) , \\ \left( {a+dp^{j}-D^{-1}k^{i}} \right) ,r^{i}\left( {p^{j},\omega +\theta } \right)>{\Phi }^{i}\left( {p^{j},k^{i}} \right)>r^{i}\left( {p^{j},\omega } \right) , \\ 1/2\left( {\left( {\left( {1-d} \right) a+dp^{j}} \right) +\left( {\omega +\theta } \right) } \right) ,{\Phi }^{i}\left( {p^{j},k^{i}} \right) >r^{i}\left( {p^{j},\omega +\theta } \right) . \end{array} \end{aligned}$$

Clearly, the reaction-functions’ slopes lie in the range ]0, 1] and is why the price sub-game allows for equilibrium in pure strategies. Thus, Proposition 1 in Maggi (1996) applies. $\square$

Proof of Proposition 2

Consider the case where the marginal cost is strictly higher than $\hat{\omega } -\left( {\theta -\psi } \right)$, but lower than $\hat{\omega }$, meaning that the marginal cost equals $MR^{b}$ as well as $MR^{c}:$ one that induces Cournot competition among downstream duopolistic firms and one that results in Bertrand competition among the downstream duopolistic firms.

When there is Bertrand competition in the downstream duopoly, we can write the first-order conditions for maximizing profit as $\psi =a-\theta -\left( {1+d} \right) Q^{b}$, or:

$$\begin{aligned} \upsilon =\omega ^{b}-1/2\left( {1+d} \right) \left( {2-d} \right) Q^{b}. \end{aligned}$$

(11)

Similarly, for Cournot competition in the downstream duopoly, $\upsilon =a-\psi -\left( {2+d} \right) Q^{c}$, or:

$$\begin{aligned} \upsilon =\omega ^{c}-1/2\left( {2+d} \right) Q^{c}. \end{aligned}$$

(12)

This allows us to write the monopolist’s profits as $\pi ^{b}=1/2\left( {1+d} \right) \left( {2-d} \right) Q^{{b}^{2}}$ and $\pi ^{c}=1/2\left( {2+d} \right) Q^{{c}^{2}}$, respectively. It is straightforward that $\pi ^{c} \gtrless \pi ^{b}$ when:

$$\begin{aligned} Q^{c} \gtrless \sqrt{\left( {1+d} \right) \cdot {2-d}/{2+d}\cdot }Q^{b}. \end{aligned}$$

(13)

Using that $Q^{c}=2\left( {2+d} \right) ^{-1}\left( {a-\psi -\omega } \right)$ and $Q^{b}=2\left( {\left( {1+d} \right) \left( {2-d} \right) } \right) ^{-1}\left( {a-\theta -\omega } \right)$, this condition becomes:

$$\begin{aligned} a-\psi -\omega ^{c}\gtrless \sqrt{\left( {2+d} \right) \left( {\left( {1+d} \right) \left( {2-d} \right) } \right) ^{-1}}\cdot \left( {a-\theta -\omega ^{b}} \right) . \end{aligned}$$

(14)

Now, the profit-maximizing input price is $\omega ^{c}=1/2\left( {a-\psi +\upsilon } \right)$ when there is Cournot competition among the downstream duopolists, and $\omega ^{b}=1/2\left( {a-\psi +\theta } \right)$ when there is Bertrand competition among the downstream duopolists. Using this we have:

$$\begin{aligned} a-\psi \gtrless \sqrt{.}\left( {a-\theta } \right) +\left( {1-\sqrt{.}} \right) \upsilon , \end{aligned}$$

(15)

where $\sqrt{.}=\sqrt{\left( {2+d} \right) \left( {\left( {1+d} \right) \left( {2-d} \right) } \right) ^{-1}}$. The right-hand side is decreasing in $\psi$.

Equation (15) is, of course, $\pi ^{c}-\pi ^{b}\gtrless 0$ and the left-hand side is increasing in $\upsilon$.

First, take $\upsilon =\hat{\omega }$ and consider the following variant of Eq. (15):

$$\begin{aligned} a-\psi -\hat{\omega } >\sqrt{.}\left( {a-\theta -\hat{\omega }} \right) . \end{aligned}$$

(16)

Using Eq. (4) and rewriting this simplifies to:

$$\begin{aligned} 2+d>\sqrt{.}\left( {2+d-d^{2}} \right) , \end{aligned}$$

(17)

or:

$$\begin{aligned} 1>\sqrt{\left( {2+d-d^{2}} \right) \left( {2+d} \right) ^{-1}}, \end{aligned}$$

(18)

which is satisfied. Hence, $\tilde{\upsilon } \rightarrow \hat{\omega }$ implies $\pi ^{c}>\pi ^{b}$.

Second, take the lowest feasible value of $\psi$, that is $\psi =\omega \left( {\hat{\theta }} \right) -\left( {\hat{\theta } -c_0} \right)$. Using this, consider the following variant of Eq. (15):

$$\begin{aligned} 2+d<\sqrt{.}\left( {2+d-d^{2}} \right) +d^{2}\left( {1-\sqrt{.}} \right) \end{aligned}$$

(19)

or:

$$\begin{aligned} 1<\sqrt{\left( {2+d-d^{2}} \right) \left( {2+d} \right) ^{-1}}+d^{2}\left( {2+d} \right) ^{-1}\left( {1-\sqrt{\left( {2+d-d^{2}} \right) ^{-1}\left( {2+d} \right) }} \right) .\nonumber \\ \end{aligned}$$

(20)

Some straightforward manipulation gives:

$$\begin{aligned} \left( {2+d-d^{2}} \right) \left( {2+d} \right) ^{-1}<\sqrt{\left( {2+d-d^{2}} \right) \left( {2+d} \right) ^{-1}}, \end{aligned}$$

(21)

which is satisfied. Hence, $\tilde{\psi } \rightarrow \omega \left( {\hat{\theta }} \right) -\left( {\hat{\theta } -c_0} \right)$ implies $\pi ^{c}<\pi ^{b}$. $\square$