Net neutrality and asymmetric platform competition

Abstract

In this paper we analyze the interplay between access to the last-mile network and net neutrality in the market for Internet access. We consider two competing Internet service providers (ISPs), which act as platforms between Internet users and content providers (CPs). One of the ISPs is vertically integrated and provides the other (non-integrated) ISP with access to its last-mile network. We study the impact of the access price on the termination fees charged by the ISPs to CPs for carrying their traffic. First, we show that the termination fee set by the integrated ISP decreases with the access price, whereas the termination fee of the non-integrated ISP can either increase or decrease with it. Second, we show that there exists a negative relationship (“waterbed effect”) between the access price and the total termination fee paid by the CPs. As a consequence, it may be socially optimal for the regulator to set the access price above cost when termination fees are left to the market.

Appendices

Appendix A: Proof of Lemma 1

The equilibrium subscription fees are as follows:

$$\begin{aligned} \widetilde{p_{A}}(n_{A},n_{B})= & {} \frac{h}{1+4m}+ \frac{1+2m}{1+4m}c_{d}+\frac{4m(1+2m)}{(1+4m)(3+4m)}c_{n} \nonumber \\&+\, \frac{3+6m}{(1+4m)(3+4m)}a+\frac{2m}{1+4m}v \nonumber \\&+\,\frac{\beta _{C}}{(1+4m)(3+4m)}\left\{ \left[ 1+8m(1+m)\right] n_{A}-\left( 1+2m\right) n_{B}\right\} , \end{aligned}$$

(15)

$$\begin{aligned} \widetilde{p_{B}}(n_{A},n_{B})= & {} \frac{h}{1+4m}+\frac{1+2m}{1+4m}c_{d}+\frac{2m}{(1+4m)(3+4m)}c_{n}\nonumber \\&+\,\frac{3+8m(1+m)}{(1+4m)(3+4m)}a+\frac{2m}{1+4m}v \nonumber \\&+\,\frac{\beta _{C}}{(1+4m)(3+4m)}\left\{ \left[ 1+8m(1+m)\right] n_{B}-\left( 1+2m\right) n_{A}\right\} .\nonumber \\ \end{aligned}$$

(16)

From (15) and (16), we find that $\partial \widetilde{p_{A}}/\partial a=(3+6m)/(3+16m+16m^{2})>0$ and $\partial \widetilde{p_{B}}/\partial a=\left[ 3+8m(1+m)\right] /(3+16m+16m^{2})>0$. Replacing for $\widetilde{p_{A}}$ and $\widetilde{p_{B}}$ into $q_{A}$ and $q_{B}$, we find that $\partial \widetilde{q_{A}}/\partial a=-2m(1+m)/[(3+16m+16m^{2})h]<0$ and $\partial \widetilde{q_{B}}/\partial a=-4m(1+m)(1+2m)/[(3+16m+16m^{2})h]<0$. From (15) and (16), we have $\partial \widetilde{p_{i}}/\partial n_{i}=\beta _{C}\left[ 1+8m(1+m)\right] /\left[ (1+4m)\left( 3+4m\right) \right] >0$ and $\partial \widetilde{q_{i}}/\partial n_{i}=\beta _{C}(1+2m)\left[ 1+8m(1+m)\right] /\left[ 2(1+4m)\left( 3+4m\right) h\right] >0$. For $j\ne i$, we find that $\partial \widetilde{p_{i}}/\partial n_{j}=-\beta _{C}(1+2m)/\left[ (1+4m)\left( 3+4m\right) \right] <0$ and $\partial \widetilde{q_{i}}/\partial n_{j}=-\beta _{C}(1+2m)^{2}/[2(1+4m)(3+4m)h]<0$.

Appendix B: Equilibrium number of CPs

The equilibrium number of CPs is as follows:

$$\begin{aligned} \frac{n^{*}}{\mu }= & {} -(t_{A}+t_{B})+\frac{(2m+1)\beta _{S}}{4m+1}- \frac{(2m(m+1)\beta _{S})}{h(4m+1)}a+\frac{(2m \beta _{S}(2m+1))}{h(4m+1)}(v-c_{d})\nonumber \\&-\, \frac{2m^{2}\beta _{S}}{h(4m+1)}c_{n}. \end{aligned}$$

(17)

The subgame perfect equilibrium subscription fees $p_{A}^{*}$ and $p_{B}^{*}$ are found by replacing $n_{A}$ and $n_{B}$ for $n^{*}$ into (15) and (16).

Appendix C: Proof of Lemma 3

To begin with, we determine the sign of the indirect effects. For $n_{A}=n_{B}=n$, we have

$$\begin{aligned} \frac{\partial \widetilde{p_{A}}}{\partial n}=\frac{\partial \widetilde{p_{B}}}{ \partial n}=\frac{2m\beta _{C}}{1+4m}>0\;\;\text {and\;}\;\frac{ \partial \widetilde{q_{A}}}{\partial n}=\frac{ \partial \widetilde{q_{B}}}{\partial n}=\frac{m(1+2m)\beta _{C}}{(1+4m)h}>0. \end{aligned}$$

Now, we determine the sign of $\partial n^{*}/\partial a$. Since $n^{*}=\beta _{S}q^{*}-t_{A}-t_{B}$,

$$\begin{aligned} \frac{\partial n^{*}}{\partial a}=\beta _{S}\frac{\partial q^{*}}{\partial a}=\beta _{S}\left( \frac{\partial {\widetilde{q}}}{\partial a}+\frac{\partial {\widetilde{q}}}{\partial n}\frac{\partial n^{*}}{\partial a}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \left( 1-\beta _{S}\frac{\partial {\widetilde{q}}}{\partial n}\right) \frac{\partial n^{*}}{\partial a}=\beta _{S}\frac{\partial {\widetilde{q}}}{\partial a}, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{\partial n^{*}}{\partial a}=\mu \beta _{S}\frac{\partial {\widetilde{q}}}{\partial a}=-\mu \beta _{S}\frac{2m(1+m)}{(1+4m)h}<0. \end{aligned}$$

The access price a has a negative direct effect and a negative indirect effect on the number of users of ISP i. Therefore, $q_{i}^{*}$ decreases with a. By contrast, the access price has a positive direct effect and a negative indirect effect on ISP i’s subscription fee. Therefore, the effect of a on $p_{i}^{*}$ is a priori ambiguous. We find that the equilibrium subscription fees increase with the access price for weak indirect network effects, but increase otherwise. Indeed, from (15), (16), and (17),

$$\begin{aligned} \frac{\partial p_{A}^{*}}{\partial a}=\frac{2m\left( 2m^{2}+6m+3\right) \beta _{C}\beta _{S}-3h(2m+1)}{(4m+3)(2m(2m+1)\beta _{C}\beta _{S}-h(4m+1))} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial p_{B}^{*}}{\partial a}=\frac{2m\left( 6m^{2}+8m+3\right) \beta _{C}\beta _{S}-h\left( 8m^{2}+8m+3\right) }{(4m+3)(2m(2m+1)\beta _{C}\beta _{S}-h(4m+1))}. \end{aligned}$$

If follows that:

• if $\beta _{C}\beta _{S}<h(3+6m)/(2m(3+6m+2m^{2}))$, then $\partial p_{A}^{*}/\partial a>0$ and $\partial p_{B}^{*}/\partial a>0$;

• if $h(3+6m)/(2m(3+6m+2m^{2}))\le \beta _{C}\beta _{S}<h(3+8m+8m^{2})/(2m(3+8m+6m^{2}))$, then $\partial p_{A}^{*}/\partial a\le 0$ and $\partial p_{B}^{*}/\partial a>0$;

• if $h(3+8m+8m^{2})/(2m(3+8m+6m^{2}))\le \beta _{C}\beta _{S}<h(1+4m)/(2m(1+2m))$, then $\partial p_{A}^{*}/\partial a<0$ and $\partial p_{B}^{*}/\partial a<0$.

Appendix D: Proof of Proposition 1

1.1 Expressions for welfare

We denote by $w_{S}$ the welfare on the service side, and by $w_{C}$ the welfare on the content side. Total welfare is then $w\equiv w_{S}+w_{C}$. We have

$$\begin{aligned} w_{S}= & {} \intop _{0}^{{\hat{x}}^{*}}(v+\beta _{C}n^{*}-hx)dx+\intop _{{\hat{x}}^{*}}^{1}(v+\beta _{C}n^{*}-h(1-x))dx+ \\&\quad m\intop _{0}^{{\hat{x}}_{A}^{*}}(v+\beta _{C}n^{*}-hx)dx+m\intop _{0}^{{\hat{x}}_{B}^{*}}(v+\beta _{C}n^{*}-hx)dx+n^{*}(t_{A}+t_{B})-cq^{*}, \end{aligned}$$

and

$$\begin{aligned} w_{C}=\intop _{0}^{n^{*}}(\beta _{S}q^{*}-t_{A}-t_{B}-y)dy. \end{aligned}$$

We find that

$$\begin{aligned} w_{S}= & {} (v+\beta _{C}n^{*}-c)q^{*}-(({\hat{x}}_{A}^{*})^{2}+({ \hat{x}}_{B}^{*})^{2})mh/2\nonumber \\&-(({\hat{x}}^{*})^{2}+(1-{\hat{x}}^{*})^{2})h/2+ n^{*}(t_{A}+t_{B}) \end{aligned}$$

(18)

and that

$$\begin{aligned} w_{C}=\beta _{S}q^{*}n^{*}-(t_{A}+t_{B})n^{*}-(n^{*})^{2} /2=(n^{*})^{2}/2. \end{aligned}$$

(19)

1.2 Variations of welfare with termination fees

From (18) and (19), using the fact that $v+\beta _{C}n^{*}-h{\hat{x}}_{i}^{*}=p_{i}^{*}$ and that $\partial p_{B}^{*}/\partial t_{i}=\partial p_{A}^{*}/\partial t_{i}$, after some computations we find that

$$\begin{aligned} \frac{\partial w}{\partial t_{A}}{=}\beta _{C}q^{*}\frac{\partial n^{*}}{\partial t_{A}}+(t_{A}+t_{B})\frac{\partial n^{*}}{\partial t_{A}}+\beta _{S}n^{*}\frac{\partial q^{*}}{\partial t_{A}}+\frac{m}{h}\underset{i=A,B}{\sum }(p_{i}^{*}-c) \left( \beta _{C}-\frac{\partial \widetilde{p_{i}}}{\partial n^{e}}\right) \frac{\partial n^{*}}{\partial t_{A}}. \end{aligned}$$

First, we have $\partial n^{*}/\partial t_{A}<0$ and $\partial q^{*}/\partial t_{A}<0$ from Lemma 2. Second, we find that $\beta _{C}-(\partial \widetilde{p_{i}}/\partial n^{e})=(1+2m)\beta _{C}/(1+4m)>0$. Therefore, $\partial w/\partial t_{A}<0$. Similarly, we have $\partial w/\partial t_{B}<0$.

1.3 Variations of welfare with access price

First, we have $\partial w_{C}/\partial a=n^{*}\partial n^{*}/\partial a<0.$ Second, from (18) and (19), computations show that

$$\begin{aligned} \frac{\partial w_{S}}{\partial a}= & {} \beta _{C}q^{*}\frac{\partial n^{*}}{\partial a}+m(p_{A}^{*}-c)\frac{\partial {\hat{x}}_{A}^{*}}{\partial a}+m(p_{B}^{*}-c)\frac{\partial {\hat{x}}_{B}^{*}}{\partial a}+ \nonumber \\&\quad (t_{A}+t_{B})\frac{\partial n^{*}}{\partial a}-\left( {\hat{x}}^{*}-\frac{1}{2}\right) \left( \frac{\partial p_{B}^{*}}{\partial a}-\frac{\partial p_{A}^{*}}{\partial a}\right) . \end{aligned}$$

(20)

To show that $\partial w_{S}/\partial a<0$, first note that $\partial n^{*}/\partial a<0$ from Lemma 3. Since the number of CPs and the subscription fees decrease with a, then $\partial {\hat{x}}_{i}^{*}/\partial a<0.$ Finally, we have $\partial \widetilde{p_{B}}/\partial a>\partial \widetilde{p_{A}}/\partial a$ from “Appendix A”, and $\partial \widetilde{p_{B}}/\partial n^{e}=\partial \widetilde{p_{A}}/\partial n^{e}$. Therefore, $\partial p_{B}^{*}/\partial a>\partial p_{A}^{*}/\partial a$. It follows that the term $-({\widetilde{x}}^{*}-1/2)\left( \partial p_{B}^{*}/\partial a-\partial p_{A}^{*}/\partial a\right)$ is negative, since ${\widetilde{x}}^{*}>1/2$. This shows that $\partial w_{S}/\partial a<0$.

Appendix E: Proof of Lemma 4

We first determine the second-order conditions (i), and then study under which condition the termination fees are strategic substitutes or complements (ii).

(i) The first-order conditions of profit maximization with respect to termination fees are

$$\begin{aligned} \frac{\partial \pi _{A}^{*}}{\partial t_{A}}=q_{A}^{*}\frac{\partial p_{A}^{*}}{\partial t_{A}}+(p_{A}^{*}-c)\frac{\partial q_{A}^{*}}{\partial t_{A}}+(a-c_{n})\frac{\partial q_{B}^{*}}{\partial t_{A}}+\frac{\partial n^{*}}{\partial t_{A}}t_{A}+n^{*}=0, \end{aligned}$$

(21)

and

$$\begin{aligned} \frac{\partial \pi _{B}^{*}}{\partial t_{B}}=q_{B}^{*}\frac{\partial p_{B}^{*}}{\partial t_{B}}+(p_{B}^{*}-a-c_{d})\frac{\partial q_{B}^{*}}{\partial t_{B}}+\frac{\partial n^{*}}{\partial t_{B}}t_{B}+n^{*}=0. \end{aligned}$$

(22)

Using (21) and (22) and the fact that $n^{*}$, $p_{i}^{*}$, and $q_{i}^{*}$ are linear in the termination fees, we find that the second-order conditions are satisfied if and only if

$$\begin{aligned} \frac{\partial ^{2}\pi _{A}^{*}}{\partial t_{A}^{2}}=\frac{\partial ^{2}\pi _{B}^{*}}{\partial t_{B}^{2}}=2\left( \frac{\partial q_{A}^{*}}{\partial t_{A}}\frac{\partial p_{A}^{*}}{\partial t_{A}}+\frac{\partial n^{*}}{\partial t_{A}}\right) \equiv Z<0. \end{aligned}$$

(23)

Let $X\equiv (\partial p_{A}^{*}/\partial t_{A})(\partial q_{A}^{*}/\partial t_{A})$ and $Y\equiv \partial n^{*}/\partial t_{A}$. From Lemma 2, we have $X>0$ and $Y<0$. Therefore, we have $Z=2(X+Y)<0$ iff $X<-Y$.

(ii) Now, assume that the SOC holds, i.e., $Z<0$. Using (21) and (22) and the fact that $\partial p_{i}^{*}/\partial t_{i}=\partial p_{j}^{*}/\partial t_{i}$, $\partial q_{i}^{*}/\partial t_{i}=\partial q_{j}^{*}/\partial t_{i}$, and $\partial n^{*}/\partial t_{A}=\partial n^{*}/\partial t_{B}$, we have, from the implicit function theorem,

$$\begin{aligned} \frac{\partial t_{A}^{BR}}{\partial t_{B}}=\frac{\partial t_{B}^{BR}}{\partial t_{A}}=-\frac{1}{Z}\left( 2\frac{\partial q_{A}^{*}}{\partial t_{A}}\frac{\partial p_{A}^{*}}{\partial t_{A}}+\frac{\partial n^{*}}{\partial t_{A}}\right) =\frac{2X+Y}{-Z}. \end{aligned}$$

(24)

The termination fees are strategic substitutes if $\partial t_{A}^{BR}/\partial t_{B}<0$, i.e., if $X<-Y/2$, and strategic complements otherwise.

To sum up, termination fees are strategic substitutes if $X\in \left[ 0,-Y/2\right]$, and strategic complements if $X\in \left( -Y/2,-Y\right)$. We find that $X<-Y/2$ iff $h>{\widetilde{h}}$ with ${\widetilde{h}}\equiv 2m(1+2m)(\beta _{S}+2m(\beta _{C}+2\beta _{S}))\beta _{C}/(1+4m)^{2}$ and $X<-Y$ iff $h>{\widehat{h}}$ with ${\widehat{h}}\equiv (2m(1+2m)(\beta _{S}+m(\beta _{C}+4\beta _{S}))\beta _{C}/(1+4m)^{2}$. It follows that the termination fees are strategic complements for $h\in \left( {\widehat{h}},{\widetilde{h}}\right)$ and strategic substitutes for $h\ge {\widetilde{h}}$. Finally, note that ${\widetilde{h}}$ increases in $\beta _{S}$ and $\beta _{C}$. Therefore, for a given value of h, the termination fees are strategic substitutes if $\beta _{S}$ and/or $\beta _{C}$ are small enough.

Appendix F: Proof of Lemma 5

We find that $\partial t_{A}^{BR}/\partial a=AT$ and $\partial t_{B}^{BR}/\partial a=BT$, where

$$\begin{aligned} A\equiv & {} -h,(3+4m(3+4m))\beta _{C}-h(1+4m)(3+4m)\beta _{S}\\&+\,6m( \beta _{C}+2\beta _{C}m)^{2}\beta _{S}+2m(1+2m)(3+4m)\beta _{C}\beta _{S}^{2}, \\ \quad B\equiv & {} -4m^{2}(1+2m)\beta _{S}\beta _{C}^{2}-h(1+4m)(3+4m) \beta _{S}\\&+\,2m(1+2m)(4h+(3+4m)\beta _{C}\beta _{S}^{2}), \end{aligned}$$

and

$$\begin{aligned} T\equiv \frac{m(1+m)}{h(3+4m)\left( h(1+4m)^{2}-2m(1+2m)(cm+ \beta _{S}+4m\beta _{S})\beta _{C}\right) }. \end{aligned}$$

From the SOC, T is positive. Furthermore, A is decreasing in h and negative at $h={\widetilde{h}}$. Therefore, $\partial t_{A}^{BR}/\partial a<0$. We also have $B>0$ at $h={\widetilde{h}}$ and $\partial B/\partial h=8m(\beta _{C}+2m\beta _{C}-2\beta _{S}(m+1))-3\beta _{S}$, which is positive iff $\beta _{C}>\beta _{S}(4m+1)(4m+3)/\left[ 8m(2m+1)\right] \equiv {\widetilde{\beta }}_{C}$. Therefore, if $\beta _{C}>{\widetilde{\beta }}_{C}$, then $B>0$ always holds and if $\beta _{C}\le {\widetilde{\beta }}_{C}$, we have $B>0$ for low values of h and $B\le 0$ otherwise. If follows that $\partial t_{B}^{BR}/\partial a>0$ if $\beta _{C}>{\widetilde{\beta }}_{C}$, or if $\beta _{C}\le {\widetilde{\beta }}_{C}$ and h is low enough. Otherwise, $\partial t_{B}^{BR}/\partial a<0$.

Finally, we find that

$$\begin{aligned} \frac{\partial t_{B}^{BR}}{\partial a}-\frac{\partial t_{A}^{BR}}{\partial a}=(8m+3)\beta _{C}\left[ h(1+4m)-2\beta _{C}\beta _{S}m(2m+1)\right] T, \end{aligned}$$

which is positive from the SOC. Therefore, an increase in the access price decreases more the best response of ISP A than the best response of ISP B.

Appendix G: Proof of Proposition 2

From “Appendices E” and “F”, we infer the derivatives of the equilibrium termination fees with respect to the access price. Indeed, for all $i,j\in \{A,B\}$ and $i\ne j$,

$$\begin{aligned} \frac{\partial t_{i}^{*}}{\partial a}=\left( \frac{\partial t_{i}^{BR}}{\partial a}+\frac{\partial t_{i}^{BR}}{\partial t_{j}}\frac{\partial t_{j}^{BR}}{\partial a}\right) /\left( 1-\frac{\partial t_{i}^{BR}}{\partial t_{j}}\frac{\partial t_{j}^{BR}}{\partial t_{i}}\right) . \end{aligned}$$

We calculate $\partial t_{A}^{*}/\partial a$ and find that it is equal to a fraction. Its denominator is positive from the second-order condition for the termination fees. The derivative of the numerator with respect to h is equal to $-2m(1+m)(1+4m)\left[ 2(3+8m(2+3m))\beta _{C}+3\beta _{S}+16m(1+m)\beta _{S}\right]$, and therefore it is negative, which shows that the numerator is decreasing in h. Since the numerator is strictly negative at $h={\widetilde{h}}$, it is negative for all h. Therefore, we have $\partial t_{A}^{*}/\partial a<0$.

We proceed in the same way for $\partial t_{B}^{*}/\partial a$. We find that it is equal to a fraction whose denominator is positive from the second-order condition. The numerator is strictly positive at $h={\widetilde{h}}$. It is increasing in h if $\beta _{C}>(1+4m)(3+4m)\beta _{S}/[3+4m(7+12m)]$. If this condition holds, the numerator is always positive and $\partial t_{B}^{*}/\partial a>0$. If the condition on $\beta _{C}$ does not hold, the numerator is positive for low values of h, and negative otherwise. In this case, we have $\partial t_{B}^{*}/\partial a>0$ for low values of h, and $\partial t_{B}^{*}/\partial a<0$ for high values of h.

Finally, we find that

$$\begin{aligned} t_{A}^{*}-t_{B}^{*}=-\frac{2(a-c_{n})m(1+m)(3+8m)\beta _{C}}{h(1+4m)(3+4m)}. \end{aligned}$$

Therefore, we have $t_{A}^{*}<t_{B}^{*}$ if $a>c_{n}$, and $t_{A}^{*}=t_{B}^{*}$ if $a=c_{n}$.

Appendix H: Proof of Proposition 3

We find that

$$\begin{aligned} \frac{d(t_{A}^{*}+t_{B}^{*})}{da}=\frac{2m(1+m)((\beta _{C}+2 \beta _{S}+8m\beta _{S})h-2m(1+2m)(\beta _{C}+2\beta _{S}) \beta _{C}\beta _{S})}{(2m(1+2m)(3\beta _{S}+4m( \beta _{C}+3\beta _{S}))\beta _{C}-3(1+4m)^{2}h)h}. \end{aligned}$$

The second-order condition for the termination fee-setting game (i.e., $h>{\widehat{h}}$; see “Appendix E”) implies that the denominator is negative. The numerator is increasing in h, and it is strictly positive at $h={\widetilde{h}}$; hence it is positive for all h. This proves that $d(t_{A}^{*}+t_{B}^{*})/da<0$.

Appendix I: Proof of Proposition 4

From the definitions of welfare on the service and the content side (see “Appendix D”, (18) and (19)), we obtain

$$\begin{aligned} \frac{dW}{da}=(n^{**}+\beta _{C}q^{**}+t^{*})\frac{\partial n^{**}}{\partial a}+m(p_{A}^{**}-c)\frac{\partial {\hat{x}}_{A}^{**}}{\partial a}+m(p_{B}^{**}-c)\frac{\partial {\hat{x}}_{B}^{**}}{\partial a}+ \end{aligned}$$

$$\begin{aligned} \frac{\partial t*}{\partial a}n^{**}-\left( {\hat{x}}^{**}-\frac{1}{2}\right) \left( \frac{\partial p_{B}^{**}}{\partial a}-\frac{\partial p_{A}^{**}}{\partial a}\right) , \end{aligned}$$

(25)

where the superscript “**” indicates that the prices, quantities, and locations of the indifferent consumers are evaluated at the equilibrium termination fees.

The first term on the second line of (25) is negative. It corresponds to the decrease in the ISPs’ total termination revenue, which follows from the decrease in the total termination fee (see Proposition 3). The second term on the second line represents a consumption misallocation effect on the service side and is also negative. Indeed, from the definition of the equilibrium subscription and termination fees, we find that $p_{B}^{**}-p_{A}^{**}=2(a-c_{n})m/(3+4m)$ and, therefore, ${\hat{x}}^{**}\ge 1/2$ and $\partial p_{B}^{**}/\partial a-\partial p_{A}^{**}/\partial a>0$. The first term on the first line of (25) is positive iff $\partial n^{**}/\partial a>0$. Finally, at $a=c_{n}$, we have $p_{A}^{**}=p_{B}^{**}$ and, therefore, $m(p_{A}^{**}-c)(\partial {\hat{x}}_{A}^{**}/\partial a)+m(p_{B}^{**}-c)(\partial {\hat{x}}_{B}^{**}/\partial a)>0$ iff $\partial q^{**}/\partial a>0$. This proves that, to have $dW/da>0$ in the neighborhood of $a=c_{n}$, it is necessary that either $\partial n^{**}/\partial a>0$ or $\partial q^{**}/\partial a>0$.

We find that

$$\begin{aligned} \frac{\partial q^{**}}{\partial a}=\frac{2m(m+1)(2m(2m+1)(\beta _{C}+2\beta _{S}) \beta _{C}-3h(4m+1))}{h\left( 3h(4m+1)^{2}-2m(2m+1)(4m( \beta _{C}+3\beta _{S})+3\beta _{S})\beta _{C}\right) } \end{aligned}$$

and

$$\begin{aligned} \frac{\partial n^{**}}{\partial a}=\frac{2m(m+1)((1+4m)\beta _{S}-\beta _{C})}{2m(2m+1)(4m(\beta _{C}+3\beta _{S})+3 \beta _{S})\beta _{C}-3h(4m+1)^{2}}. \end{aligned}$$

It follows that $\partial q^{**}/\partial a>0$ iff $4\beta _{C}\beta _{S}+2\beta _{C}^{2}>3h\left( \frac{2}{2m+1}+\frac{1}{m}\right)$. We also find that $\partial n^{**}/\partial a>0$ iff $\beta _{C}>(1+4m)\beta _{S}$.

Fig. 4

Range of parameters in $(h,\beta _{C})$ space that satisfy assumptions

Fig. 5

Socially-optimal access price as a function of $\beta _{C}$

Appendix J: Proof of Proposition 5

For our numerical simulations, we make the following assumptions. To ensure that all users located on the Hotelling line subscribe to an ISP for a wide range of parameter values, we assume that the utility derived from Internet access is high enough, namely $v=5$. We normalize the network effects originating from the service side to $\beta _{S}=1$ and let the network effects originating from the content side $\beta _{C}$ vary between 0 and 3 / 2. We set the mass of Internet users in the hinterlands to $m=1$ and let the transportation cost h vary between 1 and 5. Finally, we assume that the marginal cost of the last-mile network and the retail marginal cost are equal to zero: $c_{n}=c_{d}=0$.

In Fig. 4, the colored area represents the range of parameters such that Assumption 2 is satisfied, welfare is concave in the access price, and the subscription fees, the termination fees, the number of CPs, the number of Internet users, and the utility of the users located on the Hotelling line are non-negative. We observe that $h=3$ and $\beta _{C}\in [0,1.15]$ belongs to this range of parameters and use these specific parameter values to draw Fig. 5.

Figure 5 shows the socially-optimal access price as a function of the degree of externality $\beta _{C}$ for the case studied in the main model and for the extension presented in Sect. 6 (Internet fragmentation). The two cases are represented by the plain curve and the dashed curve, respectively. As for the main case, we observe that the socially-optimal access price is higher than the marginal cost $c_{n}=0$ for all $\beta _{C}$ between 0 and about 1.12, which proves Proposition 5.