On the equivalence of quantity competition and supply function competition with sunk costs

Abstract

This paper considers competition in supply functions in a homogeneous goods market in the absence of cost or demand uncertainty. In order to commit to a supply schedule, firms are required to build sufficient capacity to produce any quantity that may be prescribed by that schedule. When the cost of extra capacity (given the level of sales) is strictly positive, any Nash equilibrium outcome of supply function competition is also a Nash equilibrium outcome of the corresponding Cournot game, and vice-versa. Conversely, when the cost-savings from reducing output (given the capacity level) are sufficiently small, any outcome of iterated elimination of weakly dominated strategies in the supply function game is also an outcome of the same process in Cournot, and vice-versa.

Appendix: Proofs

Proof of Proposition 3.1

Suppose $\mathbf {s}^{*}=\left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N}$ is a profile of supply functions, and $p^{*}$ is the associated market-clearing price.We will first prove that for any $i\in N$ such that $x_{i}\left( \mathbf {s} ^{*}\right) <s_{i}^{*}\left( \bar{p}\right)$, it is possible for player i to profitably deviate from her strategy $s_{i}^{*}\left( \cdot \right)$. Suppose that, in addition to $x_{i}\left( \mathbf {s}^{*}\right) <s_{i}^{*}\left( \bar{p}\right)$ for some $i\in N$, we have $S_{\{i\}}^{-}\left( \mathbf {s}^{*},p^{*}\right) <s_{i}^{*}\left( p^{*}\right)$, which implies $S_{N}^{-}\left( \mathbf {s}^{*},p^{*}\right)<D\left( p^{*}\right) <S_{N}\left( \mathbf {s}^{*},p^{*}\right)$. Suppose now player i changes her supply schedule to a $s_{i}^{\prime }\left( \cdot \right)$ such that $s_{i}^{^{\prime }}\left( p\right) =x_{i}\left( \mathbf {s}^{*}\right)$ for $p\le p^{*},$ and $s_{i}^{^{\prime }}\left( p\right) =s_{i}^{*}\left( p\right)$ for $p>p^{*}$. From (3), we know that:

$$\begin{aligned} x_{i}\left( \mathbf {s}^{*}\right) \in \left[ D\left( p^{*}\right) -S_{N\backslash \left\{ i\right\} }\left( \mathbf {s}^{*},p^{*}\right) ,D\left( p^{*}\right) -S_{N\backslash \left\{ i\right\} }^{-}\left( \mathbf {s}^{*},p^{*}\right) \right] \end{aligned}$$

Hence, given the new profile of supply functions $\mathbf {s}^{\prime } =s_{i}^{\prime }\left( \cdot \right) \cup \left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N\backslash \left\{ i\right\} }$, and from the fact that $S_{N\backslash \left\{ i\right\} }^{-}\left( \mathbf {s}^{*},p^{*}\right) =S_{N\backslash \left\{ i\right\} }^{-}\left( \mathbf {s}^{\prime },p^{*}\right)$ and $S_{N\backslash \left\{ i\right\} }\left( \mathbf {s}^{*},p^{*}\right) =S_{N\backslash \left\{ i\right\} }\left( \mathbf {s}^{\prime },p^{*}\right)$ we have:

$$\begin{aligned} S_{N}^{-}\left( \mathbf {s}^{\prime },p^{*}\right)&=S_{N\backslash \left\{ i\right\} }^{-}\left( \mathbf {s}^{*},p^{*}\right) +x_{i}\left( \mathbf {s}^{*}\right) \le D\left( p^{*}\right) \\ S_{N}\left( \mathbf {s}^{\prime },p^{*}\right)&=S_{N\backslash \left\{ i\right\} }\left( \mathbf {s}^{*},p^{*}\right) +x_{i}\left( \mathbf {s}^{*}\right) \geqslant D\left( p^{*}\right) \end{aligned}$$

Hence, $p^{*}$ is still the market-clearing price under $\mathbf {s} ^{\prime },$ and $x_{i}\left( \mathbf {s}^{\prime }\right) =x_{i}\left( \mathbf {s}^{*}\right)$. Since $s_{i}^{^{\prime }}\left( \bar{p}\right) =s_{i}^{*}\left( \bar{p}\right)$, the costs incurred by firm i are also the same under $s_{i}^{\prime }\left( \cdot \right)$ as under $s_{i}^{*}\left( \cdot \right)$, and so are the profits. Hence, to show that i can profitably deviate from $s_{i}^{*}\left( \cdot \right)$, we now assume that $s_{i}^{*}\left( \cdot \right)$ is such that $S_{\{i\}}^{-}\left( \mathbf {s}^{*},p^{*}\right) =s_{i}^{*}\left( p^{*}\right)$, because one can always alter $s_{i}^{*}\left( \cdot \right)$ to satisfy this property without changing the profit of firm i. Hence, it remains to consider a situation where:

$$\begin{aligned} S_{\{i\}}^{-}\left( \mathbf {s}^{*},p^{*}\right) =s_{i}^{*}\left( p^{*}\right) =x_{i}\left( \mathbf {s}^{*}\right) <s_{i}^{*}\left( \bar{p}\right) \end{aligned}$$

Suppose then player i switches to an alternative supply schedule $s_{i}^{\prime }\left( \cdot \right)$ such that $s_{i}^{^{\prime }}\left( p\right) =s_{i}^{*}\left( p\right)$ for $p\le p^{*},$ and $s_{i}^{^{\prime }}\left( p\right) =s_{i}^{*}\left( p^{*}\right) =x_{i}\left( \mathbf {s}^{*}\right)$ for $p>p^{*}$. Once again, let $\mathbf {s}^{\prime }=s_{i}^{\prime }\left( \cdot \right) \cup \left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N\backslash \left\{ i\right\} }.$ We then have $S_{N}^{-}\left( \mathbf {s}^{*},p^{*}\right) =S_{N}^{-}\left( \mathbf {s}^{\prime },p^{*}\right)$ and $S_{N}\left( \mathbf {s}^{*},p^{*}\right) =S_{N}\left( \mathbf {s} ^{\prime },p^{*}\right)$, which means $p^{*}$ is still the market-clearing price under $\mathbf {s}^{\prime }$, and $x_{i}\left( \mathbf {s}^{\prime }\right) =x_{i}\left( \mathbf {s}^{*}\right)$. However, we have:

$$\begin{aligned} C\left( x_{i}\left( \mathbf {s}^{\prime }\right) ,s_{i}^{\prime }\left( \bar{p}\right) \right) =C\left( x_{i}\left( \mathbf {s}^{*}\right) ,x_{i}\left( \mathbf {s}^{*}\right) \right) <C\left( x_{i}\left( \mathbf {s}^{*}\right) ,s_{i}^{*}\left( \bar{p}\right) \right) \end{aligned}$$

Given assumption (A1) holds and using $x_{i}\left( \mathbf {s}^{*}\right) <s_{i}^{*}\left( \bar{p}\right)$, this means costs are lower under $s_{i}^{\prime }\left( \cdot \right)$ than under $s_{i}^{*}\left( \cdot \right)$, making the profit of firm i larger in the former case. Hence, any player $i\in N$ who sets a supply schedule $s_{i}^{*}\left( \cdot \right)$ such that $x_{i}\left( \mathbf {s}^{*}\right) <s_{i}^{*}\left( \bar{p}\right)$, can benefit by unilaterally deviating to an alternative supply schedule $s_{i}^{\prime }\left( \cdot \right)$ such that $x_{i}\left( \mathbf {s}^{\prime }\right) =s_{i}^{^{\prime }}\left( \bar{p}\right)$. Consequently, any Nash equilibrium strategy profile $\mathbf {s}^{*}$ must satisfy $x_{i}\left( \mathbf {s}^{*}\right) =s_{i}^{*}\left( \bar{p}\right) \quad$ for all $i\in N$.Suppose then $q_{NE}^{*}$ is a Cournot Nash equilibrium quantity, i.e. we have $q_{NE}^{*}=q_{BR}\left( \left( n-1\right) q_{NE}^{*}\right)$, where $q_{BR}\left( Q\right)$ is the Cournot best-response to an aggregate quantity Q produced by all other players. Consider a profile of supply schedules $\mathbf {s}^{*}=\left\{ s_{j} ^{*}\left( \cdot \right) \right\} _{j\in N}$ such that for every $j\in N$ we have $s_{j}^{*}\left( p\right) =q_{NE}^{*}$ for all $p\geqslant 0$. This means $\pi _{j}\left( \mathbf {s}^{*}\right) =P\left( n~q_{NE}^{*}\right) q_{NE}^{*}-C\left( q_{NE}^{*}\right)$ for all $j\in N$, i.e. profits are equal to the Cournot Nash equilibrium ones. Thus, an optimal deviation by player i from $s_{i}^{*}\left( \cdot \right)$ must entail a supply schedule $s_{i}^{\prime }\left( \cdot \right)$ such that $x_{i}\left( \mathbf {s}^{\prime }\right) =s_{i}^{^{\prime }}\left( \bar{p}\right)$, where again $\mathbf {s}^{\prime }=s_{i}^{\prime }\left( \cdot \right) \cup \left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N\backslash \left\{ i\right\} }$. The resulting profit would be $\pi _{i}\left( \mathbf {s}^{\prime }\right) =P\left( \left( n-1\right) q_{NE}^{*}\right) s_{i}^{^{\prime }}\left( \bar{p}\right) -C\left( s_{i}^{^{\prime }}\left( \bar{p}\right) \right)$, i.e. it would equal the Cournot profit given quantity $s_{i}^{^{\prime }}\left( \bar{p}\right)$ when others produce $\left( n-1\right) q_{NE}^{*}$ in total. Thus, it cannot exceed the profit resulting from $s_{i}^{*}\left( \cdot \right)$ due to $s_{i}^{*}\left( \bar{p}\right) =q_{BR}\left( \left( n-1\right) q_{NE}^{*}\right)$ being the Cournot best-response quantity. As a result, the Nash equilibrium outcome of the Cournot game is also a NE outcome of the supply-function competition game.Conversely, suppose a profile of supply schedules $\mathbf {s}^{*}=\left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N}$ is a NE of the supply-function competition game, which means it must satisfy

$$\begin{aligned} x_{i}\left( \mathbf {s}^{*}\right) =s_{i}^{*}\left( \bar{p}\right) \text { and }\pi _{i}\left( \mathbf {s}^{\prime }\right) =P\left( {\textstyle \sum \nolimits _{j\in N}} s_{j}^{*}\left( \bar{p}\right) \right) s_{i}^{*}\left( \bar{p}\right) -C\left( s_{i}^{*}\left( \bar{p}\right) \right) \quad \text { for all }i\in N \end{aligned}$$

Suppose further that we do not have $s_{i}^{*}\left( \bar{p}\right) =q_{NE}^{*}$ for all $i\in N,$ where $q_{NE}^{*}$ is a Cournot Nash equilibrium quantity. It must then be the case that for some $i\in N$ and some $q^{\prime }\ne s_{i}^{*}\left( \bar{p}\right)$ we have:

$$\begin{aligned} P\left( q^{\prime }+ {\textstyle \sum \nolimits _{j\in N/\{i\}}} s_{j}^{*}\left( \bar{p}\right) \right) q^{\prime }-C\left( q^{\prime }\right) >P\left( {\textstyle \sum \nolimits _{j\in N}} s_{j}^{*}\left( \bar{p}\right) \right) s_{i}^{*}\left( \bar{p}\right) -C\left( s_{i}^{*}\left( \bar{p}\right) \right) \end{aligned}$$

Hence, by deviating from $s_{i}^{*}\left( \cdot \right)$ to a $s_{i}^{\prime }\left( \cdot \right)$ such that $s_{i}^{\prime }\left( p\right) =q^{\prime }$ for all $p\geqslant 0,$ player i can increase its payoff in the supply-function game, i.e. $\mathbf {s}^{*}$ is not a Nash equilibrium if it does not implement a Cournot Nash equilibrium outcome.

$\square$

Proof of Proposition 3.2

Consider a strategy profile $\mathbf {s} ^{*}=\left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N}$ such that for some $i\in N$ we have $x_{i}\left( \mathbf {s}^{*}\right) <s_{i}^{*}\left( \bar{p}\right)$, and a strategy $s_{i}^{\prime }\left( \cdot \right)$ such that $s_{i}^{\prime }\left( p\right) =s_{i}^{*}\left( \bar{p}\right)$ for all $p\geqslant 0$. In the first part of the proof, we will show that under the condition stated in the proposition we then have $\pi _{i}\left( \mathbf {s}^{\prime }\right) >\pi _{i}\left( \mathbf {s}^{*}\right)$, where $\mathbf {s}^{\prime }=s_{i}^{\prime }\left( \cdot \right) \cup \left\{ s_{j}^{*}\left( \cdot \right) \right\} _{j\in N\backslash \left\{ i\right\} }$.Observe first that $x_{i}\left( \mathbf {s}^{\prime }\right) =s_{i}^{*}\left( \bar{p}\right)$ based on (3), since $S_{\{i\}}^{-}\left( \mathbf {s} ^{\prime },p\right) =S_{\{i\}}\left( \mathbf {s}^{\prime },p\right) =s_{i}^{*}\left( \bar{p}\right)$ for any $p>0$ (note that we must have $p^{*}\left( \mathbf {s}^{\prime }\right) >0$, since condition (5) together with assumption (A2) imply $P^{\prime }\left( nq^{c}\right) q^{c}+P\left( nq^{c}\right) >0$, and so $P\left( nq^{c}\right) >0$). Let $q_{i}=s_{i}^{*}\left( \bar{p}\right)$ and $Q_{-i}=D\left( p^{*}\right) -s_{i}^{*}\left( \bar{p}\right) ,$ where $p^{*}$ is the market-clearing price associated with $\mathbf {s}^{\prime }$, i.e. one that satisfies $D\left( p^{*}\right) \in \left[ S_{N}^{-}\left( \mathbf {s}^{\prime },p^{*}\right) ,S_{N}\left( \mathbf {s}^{\prime },p^{*}\right) \right]$. We then have:

$$\begin{aligned} \pi _{i}\left( \mathbf {s}^{\prime }\right) =\pi _{i}^{c}\left( q_{i} ,Q_{-i}\right) =P\left( q_{i}+Q_{-i}\right) q_{i}-C\left( q_{i}\right) \end{aligned}$$

The market-clearing price under $\mathbf {s}^{*}$ cannot be smaller than $p^{*},$ since $S_{N}^{-}\left( \mathbf {s}^{*},p\right) \le S_{N} ^{-}\left( \mathbf {s}^{\prime },p\right)$ and $S_{N}\left( \mathbf {s} ^{*},p\right) \le S_{N}\left( \mathbf {s}^{\prime },p\right)$ for all $p\geqslant 0$. Hence, as supply schedules are non-decreasing, the demand allocated to other players cannot be smaller than $Q_{-i}$. This means:

$$\begin{aligned} \pi _{i}\left( \mathbf {s}^{*}\right) \le \hat{\pi }_{i}\left( \hat{q} _{i},Q_{-i}\right) =P\left( \hat{q}_{i}+Q_{-i}\right) \hat{q}_{i}-C\left( \hat{q}_{i},q_{i}\right) \quad \text {for}~\hat{q}_{i}=x_{i}\left( \mathbf {s} ^{*}\right) \end{aligned}$$

We proceed to show that the RHS of the above inequality is strictly smaller than $\pi _{i}^{c}\left( q_{i},Q_{-i}\right)$. A sufficient condition for this is that $\partial \hat{\pi }_{i}/\partial \hat{q}_{i}$ is non-negative for all and strictly positive for some $\hat{q}_{i}\in \left[ x_{i}\left( \mathbf {s}^{*}\right) ,q_{i}\right]$. We have:

$$\begin{aligned} \partial \hat{\pi }_{i}/\partial \hat{q}_{i}=P^{\prime }\left( \hat{q}_{i} +Q_{-i}\right) \hat{q}_{i}+P\left( \hat{q}_{i}+Q_{-i}\right) -C_{x}\left( \hat{q}_{i},q_{i}\right) \end{aligned}$$

and, by virtue of Assumptions (1) and (A2):

$$\begin{aligned} P^{\prime }\left( \hat{q}_{i}+Q_{-i}\right) \hat{q}_{i}+P\left( \hat{q} _{i}+Q_{-i}\right) -C_{x}\left( \hat{q}_{i},q_{i}\right) \geqslant P^{\prime }\left( nq^{c}\right) q^{c}+P\left( nq^{c}\right) -C_{x}\left( q^{c},q^{c}\right) \end{aligned}$$

where the inequality is strict for $\hat{q}_{i}\in \left[ x_{i}\left( \mathbf {s}^{*}\right) ,q_{i}\right)$, and the RHS of the inequality is non-negative by virtue of the condition imposed in the proposition. Thus, we have shown that $\pi _{i}\left( \mathbf {s}^{\prime }\right) >\pi _{i}\left( \mathbf {s}^{*}\right)$.

As a result, observe that we can conduct IEWDS in $\mathcal {G}$ until we end up with a restricted game $\mathcal {G} ^{^{\prime }}$ such that:

• 1. for any strategy profile $\mathbf {s}^{*}$ that is part of $\mathcal {G}^{^{\prime }}$ we have $x_{i}\left( \mathbf {s}^{*}\right) =s_{i}\left( \bar{p}\right)$ for all $i\in N$

• 2. for any $q\in \left[ 0,q^{c}\right] ,i\in N$ there exists a $s_{i}\left( \cdot \right)$ in $\mathcal {G}^{^{\prime }}$ such that $s_{i}\left( \bar{p}\right) =q$

In other words, we then have $\mathcal {G}^{^{\prime }}$ $\equiv \,\mathcal {G}_{C}$. Suppose then a quantity $q_{i}^{\prime }$ weakly dominates $q_{i}$ in $\mathcal {G}_{C}$ for some player i, and as such $q_{i}$ can be eliminated, resulting in a restricted game $\mathcal {G}_{C}^{^{\prime }}$. This means we can equally eliminate all such $s_{i}\left( \cdot \right)$ in $\mathcal {G}^{^{\prime }}$ that satisfy $s_{i}\left( \bar{p}\right) =q_{i},$ as each of them is weakly dominated in $\mathcal {G}^{^{\prime }}$ by any $s_{i}^{^{\prime }}\left( \cdot \right)$ in $\mathcal {G}^{^{\prime }}$ that satisfies $s_{i}^{\prime }\left( \bar{p}\right) =q_{i}^{\prime }$. Eliminating all such $s_{i}\left( \cdot \right)$ results in a further restricted game $\mathcal {G}^{^{\prime \prime }}$, where $\mathcal {G}^{^{\prime \prime }}\equiv \,\mathcal {G}_{C}^{^{\prime }}$. Thus, we can continue to apply the same reasoning to eliminate any further strategies from $\mathcal {G}_{C}^{^{\prime }}$, and, correspondingly, from $\mathcal {G}^{^{\prime \prime }}$. In the end of the IEWDS process, we end up with two restricted games, $\mathcal {G} _{C}^{^{\prime \prime }}$ and $\mathcal {G}^{^{\prime \prime \prime }},$ such that $\mathcal {G}^{^{\prime \prime \prime }}\equiv$ $\mathcal {G}_{C}^{^{\prime \prime } }$. Consequently, for any restricted game obtained from $\mathcal {G}_{C}$ by IEWDS, an equivalent game can be obtained from $\mathcal {G}$ by the same process.

We proceed to show the converse, i.e. that for any restricted game obtained from $\mathcal {G}$ by IEWDS, there exists an equivalent restricted game obtained from $\mathcal {G}_{C}$ by the same process. By virtue of what was shown above, any restricted game $\mathcal {G}^{*}$ that remains after completing the process of IEWDS in $\mathcal {G}$, must satisfy $x_{i}\left( \mathbf {s}^{*}\right) =s_{i}^{*}\left( \bar{p}\right)$ for all $i\in N$ and any strategy profile $\mathbf {s}^{*}$ in $\mathcal {G}^{*}$.

Consider then the first round of IEWDS in $\mathcal {G}$ after which for some $q\in \left[ 0,q^{c}\right] ,i\in N$ there exists no $s_{i}\left( \cdot \right)$ in the resulting restricted game $\mathcal {G}^{^{\prime }}$ such that $s_{i}\left( \bar{p}\right) =q$. In particular, this means strategy $s_{i}\left( \cdot \right)$ such that $s_{i}\left( p\right) =q$ for all $p\geqslant 0$ must have already been eliminated, being weakly dominated by some other strategy $s_{i}^{\prime }\left( \cdot \right)$. Thus, it must also have been weakly dominated by a strategy $s_{i}^{\prime \prime }\left( \cdot \right)$ such that $s_{i} ^{\prime \prime }\left( p\right) =s_{i}^{\prime }\left( \bar{p}\right)$ for all $p\geqslant 0$. This implies $\pi _{i}\left( s_{i}^{\prime \prime }\left( \cdot \right) ,\mathbf {s}_{-i}\right) \geqslant \pi _{i}\left( s_{i}\left( \cdot \right) ,\mathbf {s}_{-i}\right)$ for all $\mathbf {s}_{-i}=\left\{ s_{j}\left( \cdot \right) \right\} _{j\in N/\{i\}}$ such that for every $j\in N/\{i\}$ we have $s_{j}\left( p\right) =q^{\prime }$ for all $p\geqslant 0$ and some $q^{\prime }\in \left[ 0,q^{c}\right]$. This in turn means that $\pi _{i}^{c}\left( s_{i}^{\prime }\left( \bar{p}\right) ,Q_{-i}\right) \geqslant \pi _{i}^{c}\left( q,Q_{-i}\right)$ for all $Q_{-i}\in \left[ 0,\left( n-1\right) q^{c}\right]$.

Under condition (1), the last inequality must be strict for at least some $Q_{-i}\in \left[ 0,\left( n-1\right) q^{c}\right]$, i.e. quantity q must be weakly dominated by quantity $s_{i}^{\prime }\left( \bar{p}\right)$ in $\mathcal {G}_{C}$. Thus, one could eliminate q from $\mathcal {G}_{C}$ to obtain a restricted game $\mathcal {G}_{C}^{\prime }$ such that $\mathcal {G}_{C}^{\prime }\equiv \mathcal {G}^{\prime }$.

One could then apply the same reasoning again, and consider the first round of IEWDS in $\mathcal {G}$ after which for some $i\in N,q^{\prime }\in \left[ 0,q^{c} \right] ,q^{\prime }\ne q$ there exists no $s_{i}\left( \cdot \right)$ in the resulting restricted game $\mathcal {G}^{^{\prime \prime }}$ such that $s_{i}\left( \bar{p}\right) =q^{\prime }$. Repeating the same steps would show that it is then possible to remove $q^{\prime }$ from the set of strategies available to player i in $\mathcal {G}_{C}^{\prime },$ to obtain a further restricted game $\mathcal {G}_{C}^{\prime \prime }$ such that $\mathcal {G}_{C}^{\prime \prime }\equiv \mathcal {G}^{\prime \prime }$.

The process could be repeated until such time that it is impossible to eliminate any further strategies from $\mathcal {G}$, and $\mathcal {G}^{*}$ is the restricted game that remains. Correspondingly, there will then exist a $\mathcal {G}_{C}^{*}$, obtained from $\mathcal {G}_{C}$ by IEWDS, such that $\mathcal {G}^{*}\equiv \mathcal {G}_{C}^{*}$. Clearly, there may not exist two quantities $q,q^{\prime }$ in $\mathcal {G}_{C}^{*}$ such that q weakly dominates $q^{\prime }$ in $\mathcal {G}_{C}^{*}$. If this was the case, then a strategy $s_{i}\left( \cdot \right)$ such that $s_{i}\left( p\right) =q$ for all $p\geqslant 0$ would weakly dominate a strategy $s_{i}^{\prime }\left( \cdot \right)$ such that $s_{i}^{\prime }\left( p\right) =q^{\prime }$ for all $p\geqslant 0$. As both $s_{i}\left( \cdot \right)$ and $s_{i}^{\prime }\left( \cdot \right)$ would be part of $\mathcal {G}^{*}$ by virtue of $\mathcal {G}^{*}\equiv \mathcal {G}_{C}^{*}$, this would then contradict the fact that $\mathcal {G}^{*}$ is what remains after the IEWDS process is complete. Thus, we have shown that for any restricted game obtained from $\mathcal {G}$ by IEWDS, there exists an equivalent restricted game obtained from $\mathcal {G}_{C}$ by the same process. $\square$