Hotelling’s location model with negative network externalities

Abstract

We study a variation of Hotelling’s location model in which consumers choose between firms based on travel distances as well as the number of consumers visiting each firm. The model in which the network externality is the same for all firms was proposed by Kohlberg (Econ Lett 11:211–216, 1983), who claims that no equilibrium exists for more than two firms. We assume the network effects to be linear and, in contrast to the claim in Kohlberg (Econ Lett 11:211–216, 1983), derive a condition under which a subgame perfect Nash equilibrium exists for four and six firms. Moreover, we show that for more than two firms the equilibrium locations of the firms are different from the equilibrium locations in Hotelling’s location model. Our results suggest that a subgame perfect Nash equilibrium exists if and only if the number of firms is even. We also provide examples of subgame perfect equilibria in which the network externality is different for some of the firms.

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Notes

  1. 1.

    The choice function f is measurable, so the Lebesgue measure of \(f_i^{-1}(x)\) is defined.

  2. 2.

    a satisfies the inequality \(a^2-8a-4\ge 0\).

  3. 3.

    a satisfies the inequality \(a^4-27a^3-84a^2-68a-16\ge 0\).

  4. 4.

    Kohlberg (1982) also proves that there is no equilibrium in this case.

  5. 5.

    Kohlberg (1982) also proves that there is no equilibrium in this case.

  6. 6.

    Kohlberg (1982) also proves that there is no equilibrium in this case.

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Acknowledgements

We would like to thank the associate editor and an anonymous referee for detailed comments and useful suggestions. Financial support from the Graduate School of Business and Economics, Maastricht University, is gratefully acknowledged.

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Correspondence to Marc Schröder.

Appendix

Appendix

Proof

(Lemma 2)

  1. (a)

    Assume that \(x_i=x_j\) for some \(i,j\in N\). Since \(|x_i-z|=|x_j-z|\) for all \(z\in [0,1]\), we can focus on the waiting costs. If \(a_i\cdot s_i(x,f)<a_j\cdot s_j(x,f)\), then a consumer z with \(f(x,z)=j\) could decrease his total costs by visiting firm i. Similar for the reverse inequality. Hence \(C_{z,i}(x,f)=|x_i-z|+a_i\cdot s_i(x,f)=|x_j-z|+a_j\cdot s_j(x,f)=C_{z,j}(x,f)\).

  2. (b)

    Assume that \(x_i<x_j\) for some \(i,j\in N\).

  3. (i)

    Suppose that \(C_{0,i}(x,f)=C_{0,j}(x,f)\). If \(z\in [0,x_i]\), then by assumption, \(C_{z,i}(x,f)=x_i-z+a_i\cdot s_i(x,f)= x_j-z+a_j\cdot s_j(x,f)=C_{z,j}(x,f)\).

If \(z\in (x_i,x_j]\), then

$$\begin{aligned} C_{z,i}(x,f)= & {} z-x_i+a_i\cdot s_i(x,f)>x_i-z+a_i\cdot s_i(x,f)\\= & {} x_j-z+a_j\cdot s_j(x,f)=C_{z,j}(x,f), \end{aligned}$$

where the inequality follows from \(x_i<z\) and the equality by assumption.

If \(z\in (x_j,1]\), then

$$\begin{aligned} C_{z,i}(x,f)&=z-x_i+a_i\cdot s_i(x,f)=z-x_j+x_j-x_i+a_i\cdot s_i(x,f)\\&>z-x_j+x_i-x_j+a_i\cdot s_i(x,f)=z-x_j+a_j\cdot s_j(x,f)=C_{z,j}(x,f), \end{aligned}$$

where the inequality follows from \(x_i<x_j\) and the third equality by assumption.

(ii) Suppose that \(C_{1,i}(x,f)=C_{1,j}(x,f)\). Then, by symmetry of the problem, a similar argument as for (i) holds.

(iii) By (i) and (ii), we can suppose that \(x_i<x_j\), \(C_{0,i}(x,f)\ne C_{0,j}(x,f)\) and \(C_{1,i}(x,f)\ne C_{1,j}(x,f)\). Suppose \(C_{0,i}(x,f)> C_{0,j}(x,f)\). By (i), we have \(C_{z,i}(x,f)>C_{z,j}(x,f)\) for all \(z\in [0,1]\). By condition (ii) of Definition 1, this implies that \(s_i(x)=0\). However, this means that \(C_{0,i}(x,f)< C_{0,j}(x,f)\), which contradicts the assumption. By symmetry, a similar argument holds if \(C_{1,i}(x,f)< C_{1,j}(x,f)\).

So suppose that \(C_{0,i}(x,f)< C_{0,j}(x,f)\) and \(C_{1,i}(x,f)> C_{1,j}(x,f)\). This implies that \(C_{x_i,i}(x,f)<C_{x_i,j}(x,f)\) and \(C_{x_j,i}(x,f)>C_{x_j,j}(x,f)\). But then there is \(y\in (x_i,x_j)\) with \(C_{y,i}(x,f)=C_{y,j}(x,f)\).

Proof

(Lemma 3) (i) Suppose that \(s_i(x)=0\) for some \(i\in N\). We show that firm i can increase its market share by a unilateral deviation. Let \(x'_i\) be such that \(\left| \{i\in N\mid x_i=x'_i\}\right| =0\). Since the transportation costs for consumer \(x'_i\) are zero and \(\left| \{i\in N\mid x_i=x'_i\}\right| =0\), we must have \(s_i(x'_i,x_{-i})>0\). This however contradicts condition (i) in Definition 1.

(ii) Suppose that \(x_i\notin [z_{i-1}(x),z_i(x)]\) for some \(i\in N\). We show that firm i can increase its market share by a unilateral deviation. See the figure below. If multiple firms are located at \(x_i\), we consider the firm with the lowest number.

figurea

Since x is an SPE, we have \(C_{z_{i-1}(x),i-1}(x)=C_{z_{i-1}(x),i}(x)\) if \(i>1\) and \(C_{z_i(x),i}(x)=C_{z_i(x),i+1}(x)\). If firm i locates at \(x'_i\), with \(z_i(x)<x'_i<x_i\), then the transportation costs for each consumer \(z\in [z_{i-1}(x),z_i(x)]\) decrease. So in order to offset the decrease in transportation costs, there must be an increase in the market share of firm i. This however contradicts condition (i) in Definition 1.

(iii) We show that \(x_1=s_1(x)\). Symmetry of the problem implies \(x_n=1-s_{n-1}(x)\).

Suppose that \(x_1\ne s_1(x)\). We show that firm 1 can increase its market share by a unilateral deviation. By (ii), we have that \(x_1<s_1(x)\le x_2\). If firm 1 locates at \(x'_1\), with \(x_1<x'_1<s_1(x)\), we observe from \(C_{s_1(x),1}(x)=C_{s_1(x),2}(x)\) that \(s_1(x)-x'_1+a_1\cdot s_1(x)<x_2-s_1(x)+a_2\cdot s_2(x)\). So in order to offset the decrease in transportation costs, there must be an increase in the market share of firm 1. This however contradicts condition (i) in Definition 1.

(iv) We show that \(\left| \{i\in N\mid x_i=x\}\right| \le 2\) for all \(x\in [0,1]\). Suppose \(\left| \{i\in N\mid x_i=x\}\right| >2\) for some \(x\in [0,1]\). We derive a contradiction. Since \(s_i(x)>0\) for all \(i\in N\), there is a firm \(i\in \{i\in N\mid x_i=x\}\) such that \(x_i\notin [z_{i-1}(x),z_i(x)]\). This however contradicts (ii).

(v) Assume that \(x_i=x_j\) for some \(i,j\in N\). Suppose \(a_i<a_j\). We derive a contradiction. Define \(x'=(x_1,\ldots ,x_{i+1},x_i,\ldots ,x_n)\). We obtain the following result.

Claim

x is an SPE if and only if \(x'\) is an SPE.

Proof

Suppose x is an SPE. By Lemma 2 (i), we have \(C_{z,i}(x)= C_{z,i+1}(x)\) for all \(z\in [0,1]\). So in particular for all \(z\in [z_{i-1}(x),z_{i+1}(x)]\). Consider the following choice function

$$\begin{aligned} f(x',z)={\left\{ \begin{array}{ll} i+1&{}\quad \text{ if } z\in [z_{i-1}(x),z_{i-1}(x)+s_{i+1}(x)],\\ i&{}\quad \text{ if } z\in [z_{i-1}(x)+s_{i+1}(x),z_{i+1}(x)],\\ f(x,z)&{}\quad \text{ if } z\in [0,z_{i-1}(x)) \text{ or } z\in (z_{i+1}(x),1]. \end{array}\right. } \end{aligned}$$

So the consumers visiting firm \(i+1\) are on the left of the consumers visiting firm i. Note that \(s_i(x')=s_i(x)\) and \(s_{i+1}(x')=s_{i+1}(x)\) and so \(C_{z,i}(x')= C_{z,i+1}(x')\) for all \(z\in [z_{i-1}(x),z_{i+1}(x)]\). Since x is an SPE, \(x'\) is also an SPE.

Since the converse implication can be proven analogously, this completes the proof of the Claim.

Since \(x_i=x_{i+1}\), we have \(a_i\cdot s_i(x)=a_{i+1}\cdot s_{i+1}(x)\) and thus by assumption \(s_i(x)>s_{i+1}(x)\). We also have \(x_{i+1}=z_i(x)=z_{i-1}(x)+s_i(x)>z_{i-1}(x)+s_{i+1}(x)\), where the first equality follows from (ii). But then (ii) implies that \(x'\) is not an SPE (see the figure below). Hence, by the above Claim, x is no SPE, which is a contradiction.

figureb

Proof

(Theorem 2) The proof below follows the same lines as the proof in Kohlberg (1982). Let x be an SPE with \(x_1<\cdots <x_n\), \(z_1(x)=x_1\), \(z_i(x)\in (x_i,x_{i+1})\) for all \(1<i<n-1\) and \(z_{n-1}(x)=x_n\). Let \(k\in N\) with \(1<k<n\).

Let \(s_i^L\), \(s_i^R\), \(z_i^L\) and \(z_i^R\) for all \(i\in N\) denote the left and right partial derivative of \(s_i(x)\) and \(z_i(x)\), respectively, with respect to \(x_k\).

We will prove by induction that \(\frac{s_i^L}{z_i^L}<\frac{s_i^ R}{z_i^R}\) for all \(1<i<k\). If firm k locates marginally to the left, then by taking the derivative of the equation for firm 1 and 2, see Eq. (1), we get

$$\begin{aligned} -z_1^L+a\cdot s_1^L=-z_1^L+a\cdot s_2^L. \end{aligned}$$

If firm k locates marginally to the right, then we get

$$\begin{aligned} z_1^R+a\cdot s_1^R=-z_1^R+a\cdot s_2^R. \end{aligned}$$

Since \(s_1^L=z_1^L\) and \(s_1^R=z_1^R\), we have

$$\begin{aligned} \frac{s_2^L}{z_1^L}=1<1+2/a=\frac{s_2^R}{z_1^R}, \end{aligned}$$

and since \(z_2^L=z_1^L+s_2^L\) and \(z_2^R=z_1^R+s_2^R\), we conclude that

$$\begin{aligned} \frac{s_2^L}{z_2^L}=\frac{1}{\left( \frac{s_2^L}{z_1^L}\right) ^{-1}+1}<\frac{1}{\left( \frac{s_2^R}{z_1^R}\right) ^{-1}+1}=\frac{s_2^R}{z_2^R}. \end{aligned}$$

Now suppose that \(\frac{s_i^L}{z_i^L}<\frac{s_i^ R}{z_i^R}\) for all i with \(1<i<k-1\), we show that the inequality also holds for firm \(i+1\). If firm k locates marginally to the left, we get

$$\begin{aligned} z_i^L+a\cdot s_i^L=-z_i^L+a\cdot s_{i+1}^L. \end{aligned}$$

If firm k locates marginally to the right, we get

$$\begin{aligned} z_i^R+a\cdot s_i^R=-z_i^R+a\cdot s_{i+1}^R. \end{aligned}$$

Since

$$\begin{aligned} \frac{s_{i+1}^L}{z_i^L}=\frac{s_i^L}{z_i^L}+2/a, \end{aligned}$$

and

$$\begin{aligned} \frac{s_{i+1}^R}{z_i^R}=\frac{s_i^R}{z_i^R}+2/a, \end{aligned}$$

we conclude that

$$\begin{aligned} \frac{s_{i+1}^L}{z_{i+1}^L}&=\frac{1}{\left( \frac{s_{i+1}^L}{z_i^L}\right) ^{-1}+1}=\frac{1}{\left( \frac{s_i^L}{z_i^L}+2/a\right) ^{-1}+1}\\&<\frac{1}{\left( \frac{s_i^R}{z_i^R}+2/a\right) ^{-1}+1}=\frac{1}{\left( \frac{s_{i+1}^R}{z_i^R}\right) ^{-1}+1}=\frac{s_{i+1}^R}{z_{i+1}^R}, \end{aligned}$$

where the first equality follows from \(z_{i+1}^L=z_i^L+s_{i+1}^L\), the inequality from the induction hypothesis, and the last equality from \(z_{i+1}^R=z_i^R+s_{i+1}^R\).

Moreover \(\frac{s_i^L}{z_i^L}>0\) for all \(1<i<k\) as both \(s_i\) and \(z_i\) decrease once firm k moves marginally to the left.

By symmetry we obtain that \(0<\frac{s_i^R}{-z_{i-1}^R}<\frac{s_i^L}{-z_{i-1}^L}\) for all \(k<i<n\).

Also, if firm k moves marginally to the left, then by taking the derivative of the equation for firm \(k-1\) and k, see Eq. (1), we get

$$\begin{aligned} z_{k-1}^L+a\cdot s_{k-1}^L=1-z_{k-1}^L+a\cdot s_k^L, \end{aligned}$$

and for firm k and \(k+1\), see Eq. (1), we get

$$\begin{aligned} z_k^L-1+a\cdot s_k^L=-z_k^L+a\cdot s_{k+1}^L. \end{aligned}$$

Adding these two equations, and using that \(z_k^L-z_{k-1}^L=s_k^L\) and that in equilibrium \(s_k^L\ge 0\) yields

$$\begin{aligned} a\cdot s_{k-1}^L+a\cdot s_{k+1}^L=(2+2a)\cdot s_k^L\ge 0. \end{aligned}$$

Dividing by \(z_{k-1}^L>0\) and using that \(z_{k-1}^L\ge z_k^L\) and \(s_{k+1}^L\le 0\) yields

$$\begin{aligned} a\cdot \frac{s_{k-1}^L}{z_{k-1}^L}+a\cdot \frac{s_{k+1}^L}{z_k^L}\ge 0, \end{aligned}$$

and thus

$$\begin{aligned} \frac{s_{k-1}^L}{z_{k-1}^L}\ge \frac{s_{k+1}^L}{-z_{k}^L}. \end{aligned}$$

By symmetry we obtain that

$$\begin{aligned} \frac{s_{k-1}^R}{z_{k-1}^R}\le \frac{s_{k+1}^R}{-z_k^R}. \end{aligned}$$

Combining all these inequalities yields

$$\begin{aligned} \frac{s_{k-1}^L}{z_{k-1}^L}<\frac{s_{k-1}^R}{z_{k-1}^R}\le \frac{s_{k+1}^R}{-z_k^R}<\frac{s_{k+1}^L}{-z_{k}^L}\le \frac{s_{k-1}^L}{z_{k-1}^L}. \end{aligned}$$

Hence we obtained a contradiction and can conclude that there is no SPE in which \(x_1<\cdots <x_n\), \(z_1(x)=x_1\), \(z_i(x)\in (x_i,x_{i+1})\) for all \(1<i<n-1\) and \(z_{n-1}(x)=x_n\).

Proof

(Theorem 4) By Lemma 3 (iv) and symmetry, we distinguish the following four cases.

(1) Assume that \(x_1=x_2<x_3=x_4\). By Lemma 3 (iii), we obtain the figure below.

figurec

Solving the equalities corresponding to the above figure

$$\begin{aligned} a\cdot s_1(x)&=a\cdot s_2(x)\\ s_2(x)+a\cdot s_2(x)&=s_3(x)+a\cdot s_3(x)\\ a\cdot s_3(x)&=a\cdot (1-z_3(x)), \end{aligned}$$

yields \(s(x)=\left( \frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right) \) and thus \(x=\left( \frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4}\right) \). Suppose firm 2 locates at \(x'_2=\frac{1}{2}\). Solving

$$\begin{aligned} s_1(x'_2,x_{-2})-\frac{1}{4}+a\cdot s_1(x'_2,x_{-2})&=\frac{1}{2}-s_1(x'_2,x_{-2})+a\cdot s_2(x'_2,x_{-2})\\ z_2(x'_2,x_{-2})-\frac{1}{2}+a\cdot s_2(x'_2,x_{-2})&=\frac{3}{4}-z_2(x'_2,x_{-2})+a\cdot s_3(x'_2,x_{-2})\\ a\cdot s_3(x'_2,x_{-2})&=a\cdot (1-z_3(x'_2,x_{-2})), \end{aligned}$$

yields

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{8+15a+4a^2}{8\cdot (4+7a+2a^2)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(2) Assume that \(x_1<x_2=x_3<x_4\). By Lemma 3 (ii) and (iii), we obtain the figure below.

figured

Solving the equalities corresponding to the above figure yields \(s(x)= \left( \frac{1+a}{2+4a},\frac{a}{2+4a},\frac{a}{2+4a},\frac{1+a}{2+4a}\right) \) and thus \(x=\left( \frac{1+a}{2+4a},\frac{1}{2},\frac{1}{2},\frac{1+3a}{2+4a}\right) \). Suppose firm 2 locates as leftmost firm at \(x'_2=s_2(x'_2,x_{-2})\). Then

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{(2+a)\cdot (2+3a+2a^2)}{2\cdot (4+16a+17a^2+4a^3)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a\ge 2\cdot (2+\sqrt{5})\).

To prove that the strategy profile is an SPE, it is sufficient to check that firm 1 does not want to deviate to the right and firm 2 does not want to deviate to the left.

By Lemma 3 (iii), firm 1 has no incentive to deviate in between \(x_1\) and \(x_2\). By symmetry and since there is an additional firm at \(x_4\), firm 1 has also no incentive to deviate to the right of \(\frac{1}{2}\).

Since market shares are piecewise linear in the location of firm 2, we show that firm 2 has no incentive to locate at \(x'_2=x_1\):

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{(2+a)\cdot (2+5a+4a^2)}{4\cdot (1+2a)\cdot (4+7a+2a^2)}. \end{aligned}$$

Since \(s_2(x'_2,x_{-2})<s_2(x)\) if \(a\ge 2\cdot (2+\sqrt{5})\), firm 2 has no incentive to deviate. Since there is also no incentive to locate as leftmost firm, firm 2 has no profitable deviation.

(3) Assume that \(x_1=x_2<x_3<x_4\). By Lemma 3 (ii) and (iv), we obtain the figure below. Note that you can show that there is no SPE with \(z_2(x)<x_3\).

figuree

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_2(x)=\frac{a^2}{1+3a+4a^2}. \end{aligned}$$

Suppose firm 2 locates at \(x'_2=x_3\). Then

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{a\cdot (4+7a+4a^2)}{2\cdot (3+2a)\cdot (1+3a+4a^2)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(4) Assume that \(x_1<x_2<x_3<x_4\).Footnote 4 We show the proof if \(x_2<z_2(x)< x_3\). Other cases, \(z_2(x)=x_2\) and \(z_2(x)=x_3\), can be solved analogously.

Assume \(x_2<z_2(x)< x_3\). Solving the equalities yields

$$\begin{aligned} s_2(x)=\frac{a^2+a^3-(1+2a-a^2)\cdot x_2+(1+4a+3a^2)\cdot x_3}{2\cdot (1+2a)\cdot (1+3a+a^2)}. \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

Proof

(Theorem 5) By Lemma 3 (iv) and symmetry, we distinguish the following nine cases.

(1) Assume that \(x_1=x_2<x_3=x_4<x_5=x_6\). By Lemma 3 (iii), we obtain the figure below.

figuref

Solving the equalities corresponding to the above figure

$$\begin{aligned} a\cdot s_1(x)&=a\cdot s_2(x)\\ s_2(x)+a\cdot s_2(x)&=s_3(x)+a\cdot s_3(x)\\ a\cdot s_3(x)&=a\cdot s_4(x)\\ s_4(x)+a\cdot s_4(x)&=s_5(x)+a\cdot s_5(x)\\ a\cdot s_5(x)&=a\cdot (1-z_5(x)), \end{aligned}$$

yields \(s(x)=\left( \frac{1}{6},\frac{1}{6},\frac{1}{6},\frac{1}{6},\frac{1}{6},\frac{1}{6}\right) \) and thus \(x=\left( \frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{1}{2},\frac{5}{6},\frac{5}{6}\right) \). Suppose firm 2 locates at \(x'_2=\frac{1}{5}\). Solving

$$\begin{aligned} s_1(x'_2,x_{-2})-\frac{1}{6}+a\cdot s_1(x'_2,x_{-2})&=\frac{1}{5}-s_1(x'_2,x_{-2})+a\cdot s_2(x'_2,x_{-2})\\ z_2(x'_2,x_{-2})-\frac{1}{5}+a\cdot s_2(x'_2,x_{-2})&=\frac{1}{2}-z_2(x'_2,x_{-2})+a\cdot s_3(x'_2,x_{-2})\\ a\cdot s_3(x'_2,x_{-2})&=a\cdot s_4(x'_2,x_{-2})\\ z_4(x'_2,x_{-2})-\frac{1}{2}+a\cdot s_4(x'_2,x_{-2})&=\frac{5}{6}-z_4(x'_2,x_{-2})+a\cdot s_5(x'_2,x_{-2})\\ a\cdot s_5(x'_2,x_{-2})&=a\cdot (1-z_5(x'_2,x_{-2})) \end{aligned}$$

yields

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{(8+5a)\cdot (20+33a+6a^2)}{60\cdot (16+36a+21a^2+3a^3)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(2) Assume that \(x_1=x_2<x_3=x_4<x_5<x_6\). By Lemma 3 (iii), we obtain the figure below. Note that you can show that there is no SPE with \(z_4(x)<x_5\).

figureg

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_2(x)=\frac{a^2}{1+3a+6a^2}. \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(3) Assume that \(x_1=x_2<x_3<x_4=x_5<x_6\). By Lemma 3 (iii), we obtain the figure below. Note that you can show that there is no SPE with \(z_4(x)<x_5\).

figureh

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s(x)=\frac{1+a}{4+6a}. \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(4) Assume that \(x_1=x_2<x_3<x_4<x_5=x_6\). By Lemma 3 (iii), we obtain the figure below. Note that you can show that there is no SPE with \(z_2(x)<x_3\), or with \(z_4(x)>x_4\).

figurei

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_2(x)=\frac{a}{2+6a}. \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. This deviation is always profitable.

(5) Assume that \(x_1<x_2=x_3<x_4=x_5<x_6\). By Lemma 3 (iii), we obtain the figure below.

figurej

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s(x)=\frac{a}{2+6a}. \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(6) Assume that \(x_1<x_2<x_3=x_4<x_5<x_6\). By Lemma 3 (iii), we obtain the figure below.

figurek

Solving the equalities corresponding to the above figure yields \(s(x)=\big (\frac{1+2a+a^2}{2+6a+6a^2},\frac{a+a^2}{2+6a+6a^2}, \frac{a^2}{2+6a+6a^2},\frac{a^2}{2+6a+6a^2},\frac{a+a^2}{2+6a+6a^2}, \frac{1+2a+a^2}{2+6a+6a^2}\big )\) and \(x=\Big (\frac{1+2a+a^2}{2+6a+6a^2},\frac{1+3a+2a^2}{2+6a+6a^2}, \frac{1}{2},\frac{1}{2},\frac{1+3a+4a^2}{2+6a+6a^2},\frac{1+4a+5a^2}{2+6a+6a^2}\Big )\).

To prove that the strategy profile is an SPE, it is sufficient to check that firm 1 does not want to deviate to the right, firm 2 does not want to deviate to the left nor the right, and firm 3 does not want to deviate to the left.

By Lemma 3 (iii), firm 1 has no incentive to deviate in between \(x_1\) and \(x_2\). Since market shares are piecewise linear in the location of firm 1, firm 1 has no incentive to deviate in between \(x_2\) and \(x_3\). By symmetry and since there are two additional firms at \(x_5\) and \(x_6\), firm 1 has also no incentive to deviate to the right of \(\frac{1}{2}\).

Since market shares are piecewise linear in the location of firm 2, we show that firm 2 has no incentive to locate at \(x'_2=x_1\):

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{(1+a)\cdot (16+56a+72a^2+39a^3+6a^4)}{4\cdot (1+3a+3a^2) \cdot (16+36a+21a^2+3a^3)}. \end{aligned}$$

Then we show that firm 2 has no incentive to locate as leftmost firm at \(x''_2=s_2(x''_2,x_{-2})\):

$$\begin{aligned} s_2(x''_2,x_{-2})=\frac{(1+a)\cdot (16+72a+124a^2+105a^3+44a^4+6a^5)}{4\cdot (1+3a+3a^2)\cdot (16+72a+100a^2+47a^3+6a^4)}. \end{aligned}$$

Finally we show that firm 2 has no incentive to locate as rightmost firm at \(x'''_2=1-s_2(x'''_2,x_{-2})\):

$$\begin{aligned} s_2(x'''_2,x_{-2})=\frac{(1+a)\cdot (16+72a+124a^2+101a^3+40a^4+6a^5)}{2\cdot (1+3a+3a^2)\cdot (16+72a+100a^2+49a^3+6a^4)}. \end{aligned}$$

Since \(s_2(x'_2,x_{-2})<s_2(x)\), \(s_2(x''_2,x_{-2})<s_2(x)\) and \(s_2(x'''_2,x_{-2})<s_2(x)\) if \(a\ge 29.8873\), firm 2 has no incentive to deviate.

Since market shares are piecewise linear in the location of firm 3, we show that firm 3 has no incentive to locate at \(x'_3=x_2\):

$$\begin{aligned} s_3(x'_3,x_{-3})=\frac{a\cdot (2+3a)\cdot (8+18a+13a^2+2a^3)}{4\cdot (1+3a+3a^2) \cdot (12+32a+22a^2+3a^3)}. \end{aligned}$$

Then we show that firm 3 has no incentive to locate at \(x''_3=s_1(x''_3,x_{-3})+s_3(x''_3,x_{-3})\):

$$\begin{aligned} s_3(x''_3,x_{-3})=\frac{a\cdot (16+76a+132a^2+111a^3+47a^4+6a^5)}{4\cdot (1+3a+3a^2)\cdot (4+24a+42a^2+24a^3+3a^4)}. \end{aligned}$$

Finally we show that firm 3 has no incentive to locate as leftmost firm at \(x'''_3=s_3(x'''_3,x_{-3})\):

$$\begin{aligned} s_3(x'''_3,x_{-3})=\frac{(1+2a)\cdot (4+6a+a^2)\cdot (4+14a+19a^2+11a^3+3a^4)}{2\cdot (1+3a+3a^2)\cdot (16+96a+204a^2+184a^3+65a^4+6a^5)}. \end{aligned}$$

Since \(s_3(x'_3,x_{-3})<s_3(x)\), \(s_3(x_{-3},x''_3)\le s_3(x)\) and \(s_3(x'''_3,x_{-3})<s_3(x)\) if \(a\ge 29.8873\), firm 3 has no incentive to deviate.

(7) Assume that \(x_1=x_2<x_3<x_4<x_5<x_6\). By Lemma 3 (iii), we obtain the figure below. Note that you can show that there is no SPE with \(z_2(x)<x_3\), or with \(z_3(x)<x_4\), or with \(z_4(x)<x_5\).

figurel

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_2(x)=\frac{a^4}{1+5a+10a^2+10a^3+6a^4}. \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. This deviation is always profitable.

(8) Assume that \(x_1<x_2=x_3<x_4<x_5<x_6\). By Lemma 3 (iii), we obtain the figure below. Note that you can show that there is no SPE with \(z_3(x)<x_4\), or with \(z_4(x)<x_5\).

figurem

Solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_3(x)=\frac{(1+a)\cdot a^2\cdot a^3}{1+4a+7a^2+6a^3}. \end{aligned}$$

Suppose firm 3 locates an arbitrary small \(\epsilon >0\) to the right. This deviation is always profitable.

(9) Assume that \(x_1<x_2<x_3<x_4<x_5<x_6\).Footnote 5 We show the proof if \(x_2<z_2(x)<x_3\), \(x_3<z_3(x)<x_4\) and \(x_4<z_4(x)<x_5\). Other cases can be solved analogously.

Assume \(x_2<z_2(x)<x_3\), \(x_3<z_3(x)<x_4\) and \(x_4<z_4(x)<x_5\). Solving the equalities yields

$$\begin{aligned}&s_2(x)=\\&\frac{a^4{+}a^5{-}(4{+}16a{+}15a^2{-}2a^3{-}3a^4)\cdot x_2{+}(4{+}22a{+}41a^2{+}30a^3{+}7a^4)\cdot x_3{+}(2a{+}9a^2{+}12a^3{+}5a^4)\cdot x_4{+}(a^2{+}4a^3{+}3a^4)\cdot x_5}{2\cdot (2{+}6a{+}3a^2)\cdot (2{+}8a{+}8a^2{+}a^3)} \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

Proof

(Theorem 7) Assume that \(a>0\). By Lemma 3 (iv) and symmetry, we distinguish the following five cases.

(1) Assume that \(x_1=x_2<x_3<x_4=x_5\). By Lemma 3 (ii) and (iv), we obtain the figure below.

figuren

We consider two subcases.

If \(z_2(x)=x_3\), then solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_2(x)=\frac{a}{3+5a}. \end{aligned}$$

Suppose firm 2 locates at \(x'_2=x_3\). Then

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{a\cdot (4+5a)\cdot (4+6a+a^2)}{(3+5a)\cdot (16+44a+32a^2+5a^3)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

If \(z_2(x)<x_3<z_3(x)\), then solving the equalities corresponding to the above figure yields

$$\begin{aligned} s_2(x)=\frac{(3+5a)x_3+a+a^2}{9+18a+5a^2}. \end{aligned}$$

Suppose firm 2 locates at \(x'_2=x_2+\epsilon \), where \(\epsilon >0\) but arbitrary small. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(2) Assume that \(x_1=x_2<x_3=x_4<x_5\). By Lemma 3 (ii) and (iv), we obtain the figure below.

figureo

Solving the equalities corresponding to the above figure

$$\begin{aligned} s_2(x)=\frac{a}{1+5a}. \end{aligned}$$

Suppose firm 2 locates at \(x'_2=z_2(x'_2,x_{-2})\). Then

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{a\cdot (2+a)\cdot (3+5a)}{(1+a)\cdot (1+5a)\cdot (6+5a)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(3) Assume that \(x_1<x_2=x_3<x_4<x_5\). By Lemma 3 (ii) and (iv), we obtain the figure below.

figurep

Solving the equalities corresponding to the above figure

$$\begin{aligned} s_3(x)=\frac{a^2}{1+4a+5a^2}. \end{aligned}$$

Suppose firm 3 locates at \(x'_3=x_4\). Then

$$\begin{aligned} s_3(x'_3,x_{-3})=\frac{a\cdot (8+22a+20a^2+5a^3)}{(1+4a+5a^2)\cdot (12+20a+5a^2)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(4) Assume that \(x_1=x_2<x_3<x_4<x_5\). By Lemma 3 (ii) and (iv), we obtain the figure below.

figureq

Solving the equalities yields

$$\begin{aligned} s_2(x)=\frac{a^3}{1+4a+6a^2+5a^3}. \end{aligned}$$

Suppose firm 2 locates at \(x'_2=x_3\). Then

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{a^2\cdot (6+10a+5a^2)}{(8+5a)\cdot (1+4a+6a^2+5 a^3)}. \end{aligned}$$

For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

(5) Assume that \(x_1<x_2<x_3<x_4<x_5\).Footnote 6 We show the proof if \(x_2<z_2(x)<x_3\) and \(x_3<z_3(x)<x_4\). Other cases can be solved analogously.

Assume \(x_2<z_2(x)<x_3\) and \(x_3<z_3(x)<x_4\). Solving the equalities yields

$$\begin{aligned} s_2(x)=\frac{a^3-(2+4a-2a^2)\cdot x_2+(2+7a+5a^2)\cdot x_3+(a+3a^2)\cdot x_4}{(2+5a)\cdot (2+5a+a^2)} \end{aligned}$$

Suppose firm 2 locates an arbitrary small \(\epsilon >0\) to the right. For this deviation to be unprofitable, we must have \(a=0\). This contradicts the assumption that \(a>0\).

Proof

(Lemma 4) Assume \(x_1<x_2<x_3\). By Lemma 3 (iii), we have \(x_1=s_1(x)\) and \(x_3=1-s_3(x)\). See Fig. 6.

Fig. 6
figure6

Three firms at different locations

Solving the equalities corresponding to the above figure

$$\begin{aligned} a_1\cdot s_1(x)&=x_2-x_1+a_2\cdot s_2(x)\\ x_3-x_2+a_2\cdot s_2(x)&=a_3\cdot (1-z_2(x)), \end{aligned}$$

yields

$$\begin{aligned} s_2(x)=\frac{a_3+a_1a_3+(a_1-a_3)\cdot x_2}{1+a_1+2a_2+a_3+a_1a_2+a_1a_3+a_2a_3}. \end{aligned}$$

Suppose firm 2 locates at \(x'_2=x_1\). Then

$$\begin{aligned}&s_2(x'_2,x_{-2})\\&\quad =\frac{a_1\cdot (2a_3+2a_1a_3+4a_2a_3+a_1a_2a_3+a_3^2+a_1a_3^2+a_2a_3^2+(2+a_1+4a_2+a_3)\cdot x_2)}{(1+a_1+2a_2+a_3+a_1a_2+a_1a_3+a_2a_3)\cdot (2a_1+2a_2+a_1a_2+a_1a_3+a_2a_3)}. \end{aligned}$$

Under the assumption that \(x_1<x_2<x_3\),

$$\begin{aligned} s_2(x)\ge s_2(x'_2,x_{-2}) \Rightarrow a_3\le a_1-2. \end{aligned}$$

Suppose firm 2 locates at \(x''_2=x_3\). Then

$$\begin{aligned}&s_2(x''_2,x_{-2})\\&\quad =\frac{a_3\cdot (2+3a_1{+}4a_2{+}a_3+a_1^2+4a_1a_2+a_1^2a_2{+}2a_1a_3+a_1^2a_3+a_1a_2a_3-(2+a_1+4a_2+a_3)\cdot x_2)}{(1+a_1+2a_2+a_3+a_1a_2+a_1a_3+a_2a_3)\cdot (2a_2+2a_3+a_1a_2+a_1a_3+a_2a_3)}. \end{aligned}$$

Under the assumption that \(x_1<x_2<x_3\),

$$\begin{aligned} s_2(x)\ge s_2(x''_2,x_{-2}) \Rightarrow a_3\ge a_1+2. \end{aligned}$$

Since \(a_3\le a_1-2\) and \(a_3\ge a_1+2\) are incompatible, firm 2 always has a profitable deviation.

Proof

(Theorem 9) Let x be an SPE. By Lemma 3 (iii), we have \(x_1=s_1(x)\) and \(x_3=1-s_3(x)\). Lemma 4 implies \(x_1=x_2\) or \(x_2=x_3\). We show the analysis for \(x_1=x_2\). A symmetric argument can be given for \(x_2=x_3\).

Let \(x_1=x_2\). By Lemma 3 (iv), we have \(a_1=a_2\) and thus \(s_1(x)=s_2(x)\). See Fig. 7.

Fig. 7
figure7

Two out of three firms locate at the same location

Solving the equalities corresponding to the above figure

$$\begin{aligned} a_2\cdot s_1(x)&=a_2\cdot s_2(x)\\ s_2(x)+a_2\cdot s_2(x)&=a_3\cdot (1-z_2(x)), \end{aligned}$$

yields \(s(x)=\left( \frac{a_3}{1+a_2+2a_3},\frac{a_3}{1+a_2+2a_3}, \frac{1+a_2}{1+a_2+2a_3}\right) \) and thus \(x=\Big (\frac{a_3}{1+a_2+2a_3},\frac{a_3}{1+a_2+2a_3},\frac{2a_3}{1+a_2+2a_3}\Big )\). Note that \(a_3>0\), since otherwise \(s_1(x)=s_2(x)=0\) and then firm 1 has an incentive to locate at 1.

Suppose firm 2 locates at \(x'_2=x_3\). Then

$$\begin{aligned} s_2(x'_2,x_{-2})=\frac{a_3\cdot (2+3a_2+a_3+a_2^2+2a_2a_3)}{(1+a_2+2a_3) \cdot (2a_2+2a_3+a_2^2+2a_2a_3)}. \end{aligned}$$

So

$$\begin{aligned} s_2(x)\ge s_2(x'_2,x_{-2}) \Rightarrow a_3\ge a_2+2. \end{aligned}$$

To prove that the strategy profile is an SPE, it is sufficient to check that firm 2 does not want to deviate to the right, and firm 3 does not want to deviate to the left.

Since market shares are piecewise linear in the location of firm 2, firm 2 has no incentive to locate at \(x'_2\), with \(x_2<x'_2\le x_3\). Since \(1-s_2(x'_2,x_{-2})<x_3\) if \(x'_2=x_3\), there is also no incentive for firm 2 to locate as the rightmost firm.

Observe that firm 3 could only improve by being the leftmost firm. Suppose firm 3 locates as leftmost firm at \(x'_3=s_3(x'_3,x_{-3})\). Then

$$\begin{aligned} s_3(x'_3,x_{-3})=\frac{(1+a_2)\cdot (a_2+2a_3)}{(1+a_2+2a_3)\cdot (2+a_2+2a_3)}. \end{aligned}$$

Since \(s_3(x'_3,x_{-3})<s_3(x)\) if \(a_3\ge a_2+2\), firm 3 has no incentive to deviate.

Hence the strategy profile is an SPE.

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Peters, H., Schröder, M. & Vermeulen, D. Hotelling’s location model with negative network externalities. Int J Game Theory 47, 811–837 (2018). https://doi.org/10.1007/s00182-018-0615-0

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Keywords

  • Hotelling’s location model
  • Network externalities
  • Subgame perfect Nash equilibrium