Abstract
We consider a free entry, symmetric cost, Cournot equilibrium in a homogeneous product market. There is a parameter A which is the index of the “market size” . If a firm decides to enter, it must incur set-up costs of F. We explore the following question: How do the number of firms, individual output and total output in a free entry equilibrium change with an increase in the market size and entry cost? Conventional wisdom suggests that these should increase with A and decrease with F. However, we show this may not be true. We use the implicit function theorem to provide a complete characterization. We also illustrate our results with specific examples.
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Notes
There is an interesting example in Vives (1999, chapter 4, section 4.3.2, page-109 ), which is somewhat related to our exercise. \(P=A/Q\) and costs \(C(q_{i})=q^{\frac{1}{2}}\) with \(F=0\). At a n-firm symmetric Corunot equilibrum, \(q^{*}=\left[ \frac{2\left( n-1\right) A}{n^{2}}\right] ^{2}\) . Cournot profits per firm is \(\frac{A}{n^{2}}\left[ 2-n\right]\). Clearly, there will be at most two firms in the market for any A (even with zero entry cost).
The structure and assumptions are similar to Dastidar (2010).
It appears that the equilibrium in one-stage and two-stage entry games are related when outputs are strategic substitutes. However, when outputs are strategic complements this relation is not very clear and this aspect needs further investigation. We are very grateful to the anonymous referee for highlighting this point.
In the Mankiw-Whinston model if one takes into account the integer constraint (i.e., the number of firms must be an integer), free entry may be insufficient but never by more than one firm.
See (Seade 1980a) for some related results on stability.
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We dedicate this paper to the memory of Dave Furth. Long back Dave Furth had discussed some of the ideas in the paper with K. G. Dastidar, who fondly remembers him. The authors are deeply indebted to Giacomo Corneo and an anonymous referee for a set of excellent comments.
Appendix
Appendix
Computation of \(\det Z\)
Note that
Hence,
Hence,
From above we get that
Proof of lemma 1
Note that \(P_{1}\left( .\right) <0\) and from (5b) we have \(q^{*}P_{11}\left( n^{e}q^{*},A\right) +2P_{1}\left( n^{e}q^{*},A\right) -C^{\prime \prime }\left( q^{*}\right) <0\). Consequently, by using (9) we get that \(\det Z>0\).\(\blacksquare\)
Proof of Proposition 1
-
(i)
Note that
$$\begin{aligned}&\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial q} &{} \frac{\partial Z_{1}}{\partial A} \\ \frac{\partial Z_{2}}{\partial q} &{} \frac{\partial Z_{2}}{\partial A} \end{array} \right| \nonumber \\&=\det \left| \begin{array}{ll} n^{e}q^{*}P_{11}\left( .\right) +\left( n^{e}+1\right) P_{1}\left( .\right) -C^{\prime \prime }\left( .\right) &{} q^{*}P_{12}\left( .\right) +P_{2}\left( .\right) \\ \left( n^{e}-1\right) q^{*}P_{1}\left( .\right) &{} q^{*}P_{2}\left( .\right) \end{array} \right| \nonumber \\&=q^{*}\left[ \begin{array}{c} n^{e}q^{*}P_{11}\left( .\right) P_{2}\left( .\right) +2P_{1}\left( .\right) P_{2}\left( .\right) \\ -P_{2}\left( .\right) C^{\prime \prime }\left( .\right) -\left( n^{e}-1\right) q^{*}P_{1}\left( .\right) P_{12}\left( .\right) \end{array} \right] \end{aligned}$$(10)From the implicit function theorem we know that
$$\begin{aligned} \frac{\partial n^{e}}{\partial A}=-\frac{\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial q} &{} \frac{\partial Z_{1}}{\partial A} \\ \frac{\partial Z_{2}}{\partial q} &{} \frac{\partial Z_{2}}{\partial A} \end{array} \right| }{\det Z} \end{aligned}$$(11)Since \(\det Z>0\) and \(q^{*}>0\), using (10) and (11) we get that
$$\begin{aligned}&\frac{\partial n^{e}}{\partial A}>0\Leftrightarrow n^{e}q^{*}P_{11}\left( .\right) P_{2}\left( .\right) +2P_{1}\left( .\right) P_{2}\left( .\right) -P_{2}\left( .\right) C^{\prime \prime }\left( .\right) \\&-\left( n^{e}-1\right) q^{*}P_{1}\left( .\right) P_{12}\left( .\right) <0. \end{aligned}$$ -
(ii)
Note that
$$\begin{aligned}&\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial q} &{} \frac{\partial Z_{1}}{\partial F} \\ \frac{\partial Z_{2}}{\partial q} &{} \frac{\partial Z_{2}}{\partial F} \end{array} \right| \nonumber \\&=\det \left| \begin{array}{ll} n^{e}q^{*}P_{11}\left( .\right) +\left( n^{e}+1\right) P_{1}\left( .\right) -C^{\prime \prime }\left( .\right) &{} 0 \\ \left( n^{e}-1\right) q^{*}P_{1}\left( .\right) &{} -1 \end{array} \right| \nonumber \\&=-\left[ n^{e}q^{*}P_{11}\left( .\right) +\left( n^{e}+1\right) P_{1}\left( .\right) -C^{\prime \prime }\left( .\right) \right] \nonumber \\&=-\left[ \left( n^{e}-1\right) \left\{ q^{*}P_{11}\left( .\right) +P_{1}\left( .\right) \right\} +q^{*}P_{11}\left( .\right) +2P_{1}\left( .\right) -C^{\prime \prime }\left( .\right) \right] \nonumber \\&=-\left[ \left( n^{e}-1\right) b+a\right] \end{aligned}$$(12)From the implicit function theorem we know that
$$\begin{aligned} \frac{\partial n^{e}}{\partial F}=-\frac{\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial q} &{} \frac{\partial Z_{1}}{\partial F} \\ \frac{\partial Z_{2}}{\partial q} &{} \frac{\partial Z_{2}}{\partial F} \end{array} \right| }{\det Z} \end{aligned}$$(13)Since \(\det Z>0\), using (12) and (13) we get \(\frac{\partial n^{e}}{\partial F}<0\Leftrightarrow \left[ \left( n^{e}-1\right) b+a\right] <0.\) Using (3) we know that \(\left( n^{e}-1\right) b+a<0\). Hence, \(\frac{\partial n^{e}}{ \partial F}<0\).\(\blacksquare\)
Proof of Proposition 2
-
(i)
Note that
$$\begin{aligned}&\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial A} &{} \frac{\partial Z_{1}}{\partial n} \\ \frac{\partial Z_{2}}{\partial A} &{} \frac{\partial Z_{2}}{\partial n} \end{array} \right| \nonumber \\&=\det \left| \begin{array}{ll} q^{*}P_{12}\left( .\right) +P_{2}\left( .\right) &{} \left( q^{*}\right) ^{2}P_{11}\left( .\right) +q^{*}P_{1}\left( .\right) \\ q^{*}P_{2}\left( .\right) &{} \left( q^{*}\right) ^{2}P_{1}\left( .\right) \end{array} \right| \nonumber \\&=\left( q^{*}\right) ^{3}\left[ P_{1}\left( .\right) P_{12}\left( .\right) -P_{11}\left( .\right) P_{2}\left( .\right) \right] \end{aligned}$$(14)From the implicit function theorem we know that
$$\begin{aligned} \frac{\partial q^{*}}{\partial A}=-\frac{\det \left| \begin{array}{cc} \frac{\partial Z_{1}}{\partial A} &{} \frac{\partial Z_{1}}{\partial n} \\ \frac{\partial Z_{2}}{\partial A} &{} \frac{\partial Z_{2}}{\partial n} \end{array} \right| }{\det Z} \end{aligned}$$(15)Since \(\det Z>0\) and \(q^{*}>0\), using (14) and (15) we get that
$$\begin{aligned} \frac{\partial q^{*}}{\partial A}>0\Leftrightarrow P_{1}\left( .\right) P_{12}\left( .\right) -P_{11}\left( .\right) P_{2}\left( .\right) <0. \end{aligned}$$ -
(ii)
Note that
$$\begin{aligned}&\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial F} &{} \frac{\partial Z_{1}}{\partial n} \\ \frac{\partial Z_{2}}{\partial F} &{} \frac{\partial Z_{2}}{\partial n} \end{array} \right| \nonumber \\&=\det \left| \begin{array}{ll} 0 &{} \left( q^{*}\right) ^{2}P_{11}\left( .\right) +q^{*}P_{1}\left( .\right) \\ -1 &{} \left( q^{*}\right) ^{2}P_{1}\left( .\right) \end{array} \right| \nonumber \\&=q^{*}\left[ q^{*}P_{11}\left( .\right) P_{1}\left( .\right) \right] \end{aligned}$$(16)From the implicit function theorem we know that
$$\begin{aligned} \frac{\partial q^{*}}{\partial F}=-\frac{\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial F} &{} \frac{\partial Z_{1}}{\partial n} \\ \frac{\partial Z_{2}}{\partial F} &{} \frac{\partial Z_{2}}{\partial n} \end{array} \right| }{\det Z} \end{aligned}$$(17)Since \(\det Z>0\) and \(q^{*}>0\), using (16) and (17) we get that \(\frac{ \partial q^{*}}{\partial F}>0\Leftrightarrow q^{*}P_{11}\left( .\right) +P_{1}\left( .\right) <0\).\(\blacksquare\)
Proof of Proposition 3
-
(i)
Note that using (11) and (15) and some routine computations we get
$$\begin{aligned} \frac{\partial \left( n^{e}q^{*}\right) }{\partial A}&=n^{e}\frac{ \partial q^{*}}{\partial A}+q^{*}\frac{\partial n^{e}}{\partial A} \nonumber \\&=-\frac{1}{\det Z}\left[ n^{e}\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial A} &{} \frac{\partial Z_{1}}{\partial n} \\ \frac{\partial Z_{2}}{\partial A} &{} \frac{\partial Z_{2}}{\partial n} \end{array} \right| +q^{*}\det \left| \begin{array}{ll} \frac{\partial Z_{1}}{\partial q} &{} \frac{\partial Z_{1}}{\partial A} \\ \frac{\partial Z_{2}}{\partial q} &{} \frac{\partial Z_{2}}{\partial A} \end{array} \right| \right] \nonumber \\&=-\frac{\left( q^{*}\right) ^{2}}{\det Z}\left[ q^{*}P_{1}\left( .\right) P_{12}\left( .\right) +2P_{1}\left( .\right) P_{2}\left( .\right) -P_{2}\left( .\right) C^{\prime \prime }\left( .\right) \right] \end{aligned}$$(18)Since \(\det Z>0\) and \(q^{*}>0\), from (18) we have
$$\begin{aligned} \frac{\partial \left( n^{e}q^{*}\right) }{\partial A}>0\Leftrightarrow \left[ q^{*}P_{1}\left( .\right) P_{12}\left( .\right) +2P_{1}\left( .\right) P_{2}\left( .\right) -P_{2}\left( .\right) C^{\prime \prime }\left( .\right) \right] <0. \end{aligned}$$ -
(ii)
Note that using (13) and (17) we get
$$\begin{aligned} \frac{\partial \left( n^{e}q^{*}\right) }{\partial F}&=n^{e}\frac{ \partial q^{*}}{\partial F}+q^{*}\frac{\partial n^{e}}{\partial F} \nonumber \\&=-\frac{1}{\det Z}\left[ n^{e}\det \left| \begin{array}{cc} \frac{\partial Z_{1}}{\partial F} &{} \frac{\partial Z_{1}}{\partial n} \\ \frac{\partial Z_{2}}{\partial F} &{} \frac{\partial Z_{2}}{\partial n} \end{array} \right| +q^{*}\det \left| \begin{array}{cc} \frac{\partial Z_{1}}{\partial q} &{} \frac{\partial Z_{1}}{\partial F} \\ \frac{\partial Z_{2}}{\partial q} &{} \frac{\partial Z_{2}}{\partial F} \end{array} \right| \right] \nonumber \\&=-\frac{q^{*}}{\det Z}\left[ -P_{1}\left( .\right) +C^{\prime \prime }\left( .\right) \right] \end{aligned}$$(19)Since \(\det Z>0\) and \(q^{*}>0\), from (19) we have \(\frac{\partial \left( n^{e}q^{*}\right) }{\partial F}<0\Leftrightarrow \left[ -P_{1}\left( .\right) +C^{\prime \prime }\left( .\right) \right] >0\). Note that from Assumption (4) we get that \(\left[ -P_{1}\left( .\right) +C^{\prime \prime }\left( .\right) \right] >0\). Hence, \(\frac{\partial \left( n^{e}q^{*}\right) }{\partial F}<0\).\(\blacksquare\)
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Dastidar, K.G., Marjit, S. Market size, entry costs and free entry Cournot equilibrium. J Econ 136, 97–114 (2022). https://doi.org/10.1007/s00712-021-00769-1
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DOI: https://doi.org/10.1007/s00712-021-00769-1