Oligopoly with network effects: firm-specific versus single network

Abstract

We consider symmetric oligopolies with positive network effects where each firm has its own proprietary network, which is incompatible with that of its rivals. We provide minimal conditions for the existence of (non-trivial) symmetric equilibrium in a general setting. We analyze the viability of industries with firm-specific networks and show that the prospects for successful launch decrease with more firms in the market. This is a major reversal from the case of single-network industries. A central part of the paper compares the viability and market performance of industries with compatible and incompatible networks and shows that viability, output, (endogenous) demand, and social welfare are higher for the former. However, the comparison of industry price, profit and consumer surplus requires respective qualifications, of a general nature for the former two but not for the latter. Overall, these results provide theoretical grounding in a general but not universal sense for the conventional view that compatibility leads to superior performance, which was hitherto based on case studies and stylized facts.

Appendix

This appendix contains the proofs of all the results in this paper.

Proof of Theorem 1

• (a) Let the expected size of the network be s for each firm. From Amir and Lambson (2000), we know that, due to (A3), a unique and symmetric Cournot equilibrium exists for all n and s. Let the corresponding per-firm equilibrium output be $q_{n} (s)$. By the upper hemi-continuity of the equilibrium correspondence for strategic games (e.g., Fudenberg and Tirole 1991), $q_{n} (s)$ is upper hemi-continuous as a correspondence. Since it is also single-valued, it must be a continuous function. Finally, by the (smooth) Implicit Function Theorem and the fact that P is $C^{2}$, $q_{n} (s)$ is $C^{1}$ in s.

• (b) Since the set of RECE coincides with the set of fixed points of $q_{n} (s),$ the existence of a RECE follows directly from Brouwer’s fixed- point theorem applied to the function $q_{n} (s)$, which is defined from [0, K] to itself. $\square$

Proof of Lemma 1

By definition, an individual (and hence, industry) output of 0 is a symmetric RECE if $0\in x(0,0)$. This holds if and only if $\pi (0,0,0)\ge \pi (x,0,0)\,\forall x\in [0,K]$, i.e., $0\ge xP(x,0)-cx$ for all $x\in [0,K]$. Moreover, $q_{n}(0)=0$ if and only if $\pi (0,0,0)\ge \pi (x,0,0)$ for all $x\in [0,K]$ if and only if $q_{n+1}(0)=0$. $\square$

The following lemmas will be useful to prove Theorem 2.

Lemma 3

The function $q_{n}(s)$ is differentiable in s, and, if $q_{n}(s)\in (0,K)$ for $s>0$, one has

$$\begin{aligned} \frac{\partial q_{n}(s)}{\partial s}=-\frac{ P_{2}(nq_{n}(s),s)+q_{n}P_{12}(nq_{n}(s),s)}{ (n+1)P_{1}(nq_{n}(s),s)+nq_{n}P_{11}(nq_{n}(s),s)}. \end{aligned}$$

(7)

In particular, if $q_{n}(0)=0,$

$$\begin{aligned} \frac{\partial q_{n}(0)}{\partial s}=-\frac{P_{2}(0,0)}{(n+1)P_{1}(0,0)}. \end{aligned}$$

(8)

Proof of Lemma 3

If $q_{n}(s)$ is interior, it satisfies the first-order condition

$$\begin{aligned} P(nq_{n}(s),s)+q_{n}(s)P_{1}(nq_{n}(s),s)-c=0. \end{aligned}$$

(9)

Since $q_{n}(s)$ is $C^{1}$, one can differentiate both sides of Eq. (9) with respect to s. Reordering terms yields (7). Evaluating at $s=0$ and $q_{n}(0)=0$ yields (8). $\square$

Let $\varPi (z,s)\triangleq \frac{n-1}{n}\left[ \int _{0}^{z}P(t,s)\mathrm{d}t-cz\right] + \frac{1}{n}[zP(z,s)-cz]$, a weighted average of welfare and industry profits when s is the same for all firms. Similar to AL (2011), we have the following result relating, for given s, argmax’s of $\varPi (z,s)$ and symmetric Cournot equilibria.

Lemma 4

If $z^{*}\in \arg \max \{\varPi (z,s),0\le z\le nK\}$, then, $x^{*}\equiv \frac{z^{*}}{n}\in q_{n}(s)$, for all $s\in [0,K]$.

Proof of Lemma 4

Since the cost function is linear, it is convex. By AL (2011), Lemma 14, for any $n\in N$ and $s\in [0,K]$, if $z^{*}\in \arg \max \{\varPi (z,s),0\le z\le nK\}$, then, $z^{*}\in Q_{n}(s)$, where $Q_{n}(s)$ is the total output equilibrium correspondence for a given s. Then, by symmetry, $z^{*}\in Q_{n}(s)$ implies that $x^{*}\equiv \frac{z^{*}}{n}\in q_{n}(s)$. $\square$

Proof of Theorem 2

• (a) If the trivial outcome is not part of the equilibrium set, Theorem 1 guarantees there is a symmetric RECE with strictly positive individual output.

• (b) Parts (b) and (c) use the following argument. By the proof of Theorem 1, $q_{n}(s)$ is $C^{1}$ and maps [0,K] into itself. In addition, suppose that there exists $s^{\prime }\in (0,K)$ such that $q_{n}(s^{\prime })>s^{\prime }$, then, by Brouwer’s fixed-point theorem, it exists at least one fixed point, say $s^{\prime \prime }$, such that $s^{\prime \prime }>s^{\prime }$ and hence $s^{\prime \prime }>0$, i.e., there exists a non-trivial symmetric RECE, $s^{\prime \prime }$. Therefore, we only need to show that such $s^{\prime }$ exists; to this end, it suffices to have $q_{n}^{\prime }(0)>1$, since it implies that there is a small $\epsilon >0$ for which $q_{n}(\epsilon )>\epsilon$. By hypothesis, $q_{n}(0)=0$ and by Lemma 3, $q_{n}^{\prime }(0)>1$ if $(n+1)P_{1}(0,0)+P_{2}(0,0)>0$, which proves our result.

• (c) The third condition in this part implies that $\varPi _{1}(z,s)\ge 0$ for some $s\in (0,K]$ and for all $z\le ns$, i.e., there exist $s\in (0,K]$ and $z^{\prime }\ge ns$ such that $\varPi (z^{\prime },s)\ge \varPi (z,s)$ for all $z\le ns$. Hence, the largest argmax of $\varPi (z,s)$, say $z^{*}$, must be greater than or equal to ns, i.e., $z^{*}\ge ns$ and $z^{*}/n\ge s$. By Lemma 4, $z^{*}/n\in q_{n}(s)$ so there is an $s\in (0,K]$ such that an element of $q_{n}(s)$ is greater or equal than s. By the argument in part (b), it follows that a non-trivial symmetric equilibrium exists for n firms.

$\square$

Proof of Proposition 1

From Amir and Lambson (2000, Theorem 2.3), $q_{n}(s)$ is decreasing in n for each fixed s when P is log-concave. (This is just the condition for the regular Cournot game to be of strategic substitutes, for each fixed s.) Consequently, the viability of the industry decreases in n (since the critical mass increases). $\square$

Proof of Lemma 2

By Lemma 1 in AL (2011), and (A1)–(A4), every selection of the best-response correspondence z(y, s) strictly increases in both y and s. (This follows from a strengthening of Topkis’s Theorem, see Amir 1996b or Topkis 1998, p. 79.) Then, the correspondence

$$\begin{aligned}&B_{s}{:}\,[0,(n-1)K]\rightarrow 2^{[0,(n-1)K]},\\&y\rightarrow \frac{n-1}{n}z(y,s) \end{aligned}$$

has a unique fixed point, which corresponds to the symmetric Cournot equilibrium (Amir and Lambson 2000). By Milgrom and Roberts (1990), such fixed point, say $y_{n}(s)$, increases in s. Hence, by symmetry and z(y, s) increasing in s, the function (by Proof of Theorem 1) $q_{n}{:}\,[0,K]\longrightarrow [0,K]$, $q_{n}(s)=y_{n}(s)/(n-1)$, is increasing in s. $\square$

Proof of Proposition 3

By Proposition 1, if an industry with n incompatible networks is viable, the industry with one incompatible network is too. In other words, ${q}_{1}(s)$ has a fixed point different than zero which is also a non-trivial RECE for the monopolist with complete compatibility. Hence, the viability of the n-oligopoly with incompatible networks implies the viability of a monopolist in the single-network model. By AL (2011, Theorem 7), the viability of the industry with complete compatibility increases in n, which completes our proof. $\square$

Proof of Proposition 4

• (a) The (largest) RECE of the oligopoly with incompatible networks is the (largest) fixed point of ${q}_{n}(s)$, say ${q}_{n}(s^{\prime })=s^{\prime }$ . Hence, $\overline{x}_{n}^{I}=s^{\prime }.$

On the other hand, the (largest) RECE of the oligopoly with a single-network may be seen as the (largest) intersection point of ${q}_{n} (s)$ with the line (through the origin) s / n, say ${q}_{n} (s^{ \prime \prime }) =s^{ \prime \prime }/n$. Hence, $\overline{x}_{n}^{C} ={q}_{n} (s^{ \prime \prime }).$

Since ${q}_{n} (s)$ is increasing in s, it is easy to see that $\overline{x }_{n}^{C} ={q}_{n} (s^{ \prime \prime }) =s^{ \prime \prime }/n \ge \overline{x}_{n}^{ I} ={q}_{n} (s^{ \prime }) =s^{ \prime }$.

It follows that industry output goes the same way, i.e., $\overline{z} _{n}^{C}=n\overline{x}_{n}^{C}\ge n\overline{x}_{n}^{I}=\overline{z}_{n}^{I} .$

• (b) The endogenous inverse demand functions in Cases C and I are, respectively, $P(\cdot ,n\bar{x}_{n}^{C})$ and $P(\cdot ,\bar{x}_{n}^{I})$ . The conclusion follows from part (a).

• (c) For this proof and the proof of Proposition 5(a), we introduce an auxiliary parametrized Cournot oligopoly with an exogenous demand shifter $\alpha$. Thus, we consider the profit function

  $$\begin{aligned} \varPi (x,\alpha )=x[P(x+y,\alpha )-c] \end{aligned}$$

  (10)

  and re-frame the questions as ones of comparative statics with respect to $\alpha$ in this Cournot game.

Now, to show that $P_{n}^{C}=P(n\bar{x}_{n}^{C},n\bar{x}_{n}^{C})\ge P_{n}^{I}=P(n\bar{x}_{n}^{I},\bar{x}_{n}^{I})$ we show that the Cournot equilibrium price in (10), call it $p(\alpha ),$ is increasing in $\alpha$, under condition (4). This clearly implies our desired conclusion here, i.e., $P_{n}^{C}=p(n\bar{x}_{n}^{C})\ge p(\bar{x} _{n}^{I})\ge P_{n}^{I},$ in light of part (a).

The first-order condition for a Cournot equilibrium in (10) may be rewritten as (with $z(\alpha )$ as equilibrium total output)

$$\begin{aligned} nP(z(\alpha ),\alpha )+z(\alpha )P_{1}(z(\alpha ),\alpha )-nc=0. \end{aligned}$$

(11)

Since there is a unique Cournot equilibrium under our assumptions, we can differentiate with respect to $\alpha$ throughout (11) and collect terms to obtain

$$\begin{aligned} z^{\prime }(\alpha )=-\frac{nP_{2}(z(\alpha ),\alpha )+zP_{12}(z(\alpha ),\alpha )}{(n+1)P_{1}(z(\alpha ),\alpha )+zP_{11}(z(\alpha ),\alpha )}. \end{aligned}$$

(12)

Since $p(\alpha )=P(z(\alpha ),\alpha )$, we have $p^{\prime }(\alpha )=P_{1}(z(\alpha ),\alpha )z^{\prime }(\alpha )+P_{2}(z(\alpha ),\alpha )$. Substituting (12) yields upon collection of terms that (with all P terms evaluated at $(z(\alpha ),\alpha )$)

$$\begin{aligned} p^{\prime }(\alpha )=-\frac{-P_{1}P_{2}+z(P_{1}P_{12}-P_{2}P_{11})}{ (n+1)P_{1}+zP_{11}}=-\frac{\varDelta _{4}}{(n+1)P_{1}+zP_{11}}. \end{aligned}$$

The denominator $(n+1)P_{1}+zP_{11}<0$ by the stability (in the sense of best-reply Cournot dynamics) of the unique Cournot equilibrium here; therefore, $p^{\prime }(\alpha )$ has the same sign as $\varDelta _{4}$, and is thus $\ge 0.$ $\square$

Proof of Proposition 5

• (a) The proof here follows the same idea as the previous proof [part (c)] of using the parametrized Cournot oligopoly with profit function given by (10). We show that the Cournot equilibrium profit $\pi (\alpha )$ is increasing in $\alpha$, under condition (5). This clearly implies our desired conclusion for reasons similar to those of the previous proof.

To this end, differentiate $\pi (\alpha )=x(\alpha )[P(z(\alpha ),\alpha )-c]$ with respect to $\alpha$ and then use the first- order condition for a Cournot equilibrium to simplify to

$$\begin{aligned} \pi ^{\prime }(\alpha )=-x\frac{ -P_{1}(2P_{2}+xP_{12})+z(P_{1}P_{12}-P_{2}P_{11})}{(n+1)P_{1}+zP_{11}}=-x \frac{\varDelta _{5}(z)}{(n+1)P_{1}+zP_{11}} \end{aligned}$$

Since the denominator $(n+1)P_{1}+zP_{11}<0$ by Cournot stability, the desired conclusion follows.

• (b) Consider the following inequalities

  $$\begin{aligned} CS_{n}^{C}-CS_{n}^{I}= & {} \int _{0}^{n{x}_{n}^{C}}[P(t,n{x}_{n}^{C})-P(n{x} _{n}^{C},n{x}_{n}^{C})]\;\;\mathrm{d}t\\&-\,\int _{0}^{n{x}_{n}^{I}}[P(t,{x}_{n}^{I})-P(n{x} _{n}^{I},{x}_{n}^{I})]\mathrm{d}t \\\ge & {} \int _{0}^{n{x}_{n}^{I}}[P(t,n{x}_{n}^{C})-P(n{x}_{n}^{C},n{x} _{n}^{C})]\;\;\mathrm{d}t\\&-\,\int _{0}^{n{x}_{n}^{I}}[P(t,{x}_{n}^{I})-P(n{x}_{n}^{I},{x} _{n}^{I})]\;\;\mathrm{d}t \\\ge & {} \int _{0}^{n{x}_{n}^{I}}[P(t,n{x}_{n}^{C})-P(n{x}_{n}^{C},n{x} _{n}^{C})]\;\;\mathrm{d}t\\&-\,\int _{0}^{n{x}_{n}^{I}}[P(t,{x}_{n}^{I})-P(n{x}_{n}^{C},{x} _{n}^{I})]\;\;\mathrm{d}t\ge 0. \end{aligned}$$

The first and second inequalities follow from the facts that ${x} _{n}^{C}\ge {x}_{n}^{I}$ and $P_{1}(z,s)<0$. The last one is implied by the submodularity of P, i.e., the assumption that $P_{12}(z,s)\le 0$. To see this, notice that $t\in [0,n{x}_{n}^{I}]$ and $n{x}_{n}^{C}\ge n{x} _{n}^{I}$ imply that $n{x}_{n}^{C}\ge t$, so that $P_{12}(z,s)\le 0$ and $n {x}_{n}^{C}\ge {x}_{n}^{I}$ imply that $P(t,n{x}_{n}^{C})-P(t,{x} _{n}^{I})\ge P(n{x}_{n}^{C},n{x}_{n}^{C})-P(n{x}_{n}^{C},{x}_{n}^{I})$ for all $t\in [0,n{x}_{n}^{I}]$.

• (c) The social welfare function for any per-firm output x and expected size of the network s with n symmetric firms is given by

  $$\begin{aligned} V_{n}(x,s)=\int _{0}^{nx}P(t,s)\;\mathrm{d}t-ncx, \end{aligned}$$

  which is a concave function with respect to x since $\frac{\partial ^{2}V_{n}(x,s)}{\partial x^{2}}=n^{2}P_{1}(nx,s)<0$, by (A1). Then, we have that at any regular equilibrium

  $$\begin{aligned} W_{n}^{C}-W_{n}^{I}= & {} \left\{ \int _{0}^{n{x}_{n}^{C}}P(t,n{x} _{n}^{C})\;\mathrm{d}t-nc{x}_{n}^{C}\right\} -\left\{ \int _{0}^{n{x}_{n}^{I}}P(t,{x} _{n}^{I})\;\mathrm{d}t-nc{x}_{n}^{I}\right\} \\\ge & {} \left\{ \int _{0}^{n{x}_{n}^{C}}P(t,n{x}_{n}^{C})\;\mathrm{d}t-nc{x} _{n}^{C}\right\} -\left\{ \int _{0}^{n{x}_{n}^{I}}P(t,n{x}_{n}^{C})\;\mathrm{d}t-nc{x} _{n}^{I}\right\} \\= & {} V_{n}({x}_{n}^{C},n{x}_{n}^{C})-V_{n}({x}_{n}^{I},n{x}_{n}^{C}) \\\ge & {} \frac{\partial V_{n}({x}_{n}^{C},n{x}_{n}^{C})}{\partial x}({x} _{n}^{C}-{x}_{n}^{I}) \\= & {} n[P(n{x}_{n}^{C},n{x}_{n}^{C})-c]({x}_{n}^{C}-{x}_{n}^{I})\ge 0. \end{aligned}$$

The first inequality follows by $n{x}_{n}^{C}\ge {x}_{n}^{I}$ and $P_{2}(z,s)>0$: the second one, by concavity of $V_{n}(\cdot ,s)$, and the last one, because $P(n{x}_{n}^{C},n{x}_{n}^{C})\ge c$ and ${x}_{n}^{C}\ge { x}_{n}^{I}$. This completes the proof of the Proposition. $\square$