Net Neutrality and Universal Service Obligations: It’s All About Bandwidth

Abstract

This paper analyzes whether repealing net neutrality (NN) improves or decreases the capacity of a regulator to make internet service providers (ISPs) extend broadband coverage through universal service obligations (USOs). We model a two-sided market where a monopolistic ISP links content providers (CPs) to end users with a broadband network of a given bandwidth. A regulator determines whether to subject the ISP to NN or to allow it to supply paid priority (P) services to CPs. She can also impose a broadband USO to the ISP: She can mandate the broadband market coverage. We show that the greater is the network bandwidth, the more likely is the repeal of NN to increase ISP profits and social welfare. Regulation can still be necessary, however, as there are bandwidth ranges for which the ISP would benefit from a repeal of NN while such a repeal is detrimental to society.

Appendix

Nomenclature of variables We provide a nomenclature of variables that are used in our model. Depending on the regimes that are studied all of them can be stared, tilded or hatted and adorned with a subscript or superscript letter/number (see Table 2).

Table 2 Nomenclature of variables

Proof of Lemma 1

• (a) Directly for Net Neutrality, $B^{N}(\mu )-R^{N}\left( \mu \right) =aX^{N}\left( \mu \right) >0$. For Prioritization, $B^{P}(\mu )-R^{P}\left( \mu \right) =a\left( X_{1}^{P}\left( \mu \right) +X_{0}^{N}\left( \mu \right) \right) >0$.

• (b) Let us define $v^{i}\left( \mu \right) :=\frac{\partial V^{i}\left( \mu \right) }{\partial \mu },r^{i}\left( \mu \right) :=\frac{ \partial R^{i}\left( \mu \right) }{\partial \mu }$ and $b^{i}\left( \mu \right) :=\frac{\partial B^{i}\left( \mu \right) }{\partial \mu }$. With Net Neutrality we have

  $$\begin{aligned} v^{N}\left( \mu \right) =\left( \alpha +\beta \right) v\left( \mu -1\right) ^{\alpha +\beta -1}=\frac{\alpha +\beta }{\mu -1}V^{N}\left( \mu \right) >0, \end{aligned}$$

  so $v^{N}\left( \mu \right)$ decreases with $\mu$. Moreover $r^{N}\left( \mu \right) =v^{N}\left( \mu \right)$ and $b^{N}\left( \mu \right) =v^{N}\left( \mu \right) +a$, which decreases with $\mu$, and $b^{N}\left( \mu \right) >r^{N}\left( \mu \right)$ for all $\mu \ge 1$. With Prioritization,

  $$\begin{aligned} v^{P}\left( \mu \right) =\frac{H\left( \mu \right) }{\mu \left( \mu -1\right) ((1-\rho )\mu +\rho )}V^{P}(\mu )>0, \end{aligned}$$

  where $H\left( \mu \right) =(2\alpha +\beta )\left( 1-\rho \right) {\mu } ^{2}-\alpha \left( 1-2\rho \right) \mu +\beta \rho >0$, as $H\left( 1\right) =\alpha +\beta >0$; $H^{\prime }\left( 1\right) =2\left( 1-\rho \right) \beta +(3-2\rho )\alpha >0$ and $H^{\prime \prime }\left( \mu \right) =(2\alpha +\beta )\left( 1-\rho \right) >0$. Consequently

  $$\begin{aligned} r^{P}\left( \mu \right) =v^{P}\left( \mu \right) +a\frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \mu }\text { with }\frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \mu }=\frac{2\alpha }{\left( \alpha +\beta \right) }(\mu -1)(1-\rho )>0; \end{aligned}$$

  and

  $$\begin{aligned} b^{P}\left( \mu \right)= & {} v^{P}\left( \mu \right) +a\frac{\partial X^{P}\left( \mu \right) }{\partial \mu } \\ \text { with }\frac{\partial X^{P}\left( \mu \right) }{\partial \mu }= & {} { \frac{\rho \beta }{{\mu }^{2}\left( \alpha +\beta \right) }}+{\frac{\alpha +\beta \left( 1-\rho \right) \ }{\alpha +\beta }+\frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \mu }>\frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \mu }}. \end{aligned}$$

  Indeed $b^{P}\left( \mu \right) >r^{P}\left( \mu \right)$ for all $\mu \ge 1$, as

  $$\begin{aligned} b^{P}\left( \mu \right) -r^{P}\left( \mu \right) =a\left( {\frac{\rho \beta }{{\mu }^{2}\left( \alpha +\beta \right) }}+{\frac{\alpha +\beta \left( 1-\rho \right) \ }{\alpha +\beta }}\right) >0. \end{aligned}$$

• (c) We have

  $$\begin{aligned} \frac{\partial ^{2}R^{N}\left( \mu \right) }{\partial a\partial \mu }=0< \frac{\partial ^{2}B^{N}\left( \mu \right) }{\partial a\partial \mu }=1, \end{aligned}$$

  and

  $$\begin{aligned} 0<\frac{\partial ^{2}R^{P}\left( \mu \right) }{\partial a\partial \mu }= \frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \mu }<\frac{ \partial ^{2}B^{N}\left( \mu \right) }{\partial a\partial \mu }=\frac{ \partial X^{P}\left( \mu \right) }{\partial \mu }. \end{aligned}$$

Proof of Lemma 2

• (a) From (5) and (11), we form

  $$\begin{aligned} \Delta X\equiv X^{P}\left( \mu \right) -X^{N}\left( \mu \right) =\Delta X\equiv X^{P}\left( \mu \right) -X^{N}\left( \mu \right) , \end{aligned}$$

  which writes

  $$\begin{aligned} \Delta X=\left( \alpha \left( 1-\rho \right) \mu -\beta \rho \right) \frac{ \left( \mu -1\right) ^{2}}{\left( \alpha +\beta \right) \mu }\lesseqqgtr 0 \text { as }\mu \lesseqqgtr \mu _{X}\equiv \frac{\beta }{\alpha }\frac{\rho }{ 1-\rho }, \end{aligned}$$

  (23)

  with $\mu _{X}>\frac{\rho }{1-\rho }$ as $\beta >\alpha$. Note that $\mu _{X}>1$ iff $\frac{\beta }{\alpha +\beta }>\rho$. So if $\rho \ge \frac{ \beta }{\alpha +\beta }$ then $\Delta X\ge 0$ for all $\mu >1.$

• (b) From (12) and (13), we form

  $$\begin{aligned} \frac{V^{P}}{V^{N}}=\frac{((1-\rho )\mu +\rho )^{\alpha +\beta }}{\mu ^{\beta }}, \end{aligned}$$

  so that

  $$\begin{aligned} V^{P}\lesseqqgtr V^{N}\text { as }G\left( \mu \right) \equiv \frac{\ln ((1-\rho )\mu +\rho )}{\ln \mu -\ln ((1-\rho )\mu +\rho )}\lesseqqgtr \frac{ \beta }{\alpha }. \end{aligned}$$

  (24)

  Indeed, $G\left( \mu \right)$ is well defined as for all $\mu >1$ and $\rho \in ]0,1[$ we have $\mu>(1-\rho )\mu +\rho >1$. Note that one can also write $G\left( \mu \right) =\left( g\left( \mu \right) -1\right) ^{-1}$, where $g\left( \mu \right) =\frac{\ln \mu }{\ln ((1-\rho )\mu +\rho )}>1$ is decreasing and convex with respect to $\mu >1$. As $G^{\prime }\left( \mu \right) =-G^{2}\left( \mu \right) g^{\prime }\left( \mu \right) >0$ then $G\left( \mu \right)$ is a strictly increasing and concave function of $\mu$ such that $\lim _{\mu \rightarrow 1}G\left( \mu \right) =\frac{1-\rho }{\rho }$ and $\lim _{\mu \rightarrow \infty }G\left( \mu \right) =\infty$. So, if $\frac{\beta }{\alpha +\beta }>\rho$, there exists $\mu _{V}:G\left( \mu _{V}\right) =\frac{\beta }{\alpha }\Leftrightarrow \frac{\rho }{1-\rho } G\left( \mu _{V}\right) =\mu _{X}>1$. Moreover as $\lim _{\mu \rightarrow 1}G^{\prime }\left( \mu \right) =-\left( \frac{1-\rho }{\rho }\right) ^{2}\left( -\frac{1}{2}\frac{\rho }{1-\rho }\right) <\frac{1-\rho }{\rho }$ and by concavity of $G\left( \mu \right)$, we have

  $$\begin{aligned} 1\le \frac{\rho }{1-\rho }G\left( \mu \right) \le \mu \text { for }\mu \ge 1. \end{aligned}$$

  (25)

  This implies that $\mu _{V}>\frac{\rho }{1-\rho }G\left( \mu _{V}\right) =\mu _{X}$.

• (c) From (23) we see straightforwardly that $\frac{\partial \mu _{X}}{\partial \rho }>0$. Differentiating $G\left( \mu _{V}\right) = \frac{\beta }{\alpha }$ with respect to $\rho$ yields $G^{\prime }\left( \mu _{V}\right) \frac{\partial \mu _{V}}{\partial \rho }=-\frac{\partial G\left( \mu \right) }{\partial \rho }$. As $\frac{\partial G\left( \mu \right) }{ \partial \rho }=$ $\frac{\left( \mu -1\right) \ln \mu }{((1-\rho )\mu +\rho )\left( \ln ((1-\rho )\mu +\rho )-\ln \mu \right) ^{2}}<0$; this implies $\frac{\partial \mu _{V}}{\partial \rho }>0$.

Proof of Proposition 1

Note that for all $\mu >0$, $\Delta B\left( \mu \right) =\Delta V\left( \mu \right) +a\Delta X_{0}\left( \mu \right) +a\Delta X_{1}\left( \mu \right) <\Delta V\left( \mu \right) +a\Delta X_{0}\left( \mu \right) =\Delta R\left( \mu \right)$ since $\Delta X_{1}\left( \mu \right) <0$. When $\alpha <\beta$, since $\Delta V\left( \mu \right) \lesseqqgtr 0$ as $\mu \lesseqqgtr \mu _{V}$, $a\Delta X_{0}\left( 1\right) =0,a\lim _{\mu \rightarrow \infty }\Delta X_{0}\left( \mu \right) \rightarrow \infty$ and, $\forall \mu ,$ there exists a $\mu _{R}<\mu _{V}$ such that $\Delta R\left( \mu \right) =\Delta V\left( \mu \right) +a\Delta X_{0}\left( \mu \right) \lesseqqgtr 0$ as $\mu \lesseqqgtr \mu _{R}$. As a result $\Delta R\left( \mu \right)$ is locally increasing around $\mu =\mu _{R}:\frac{\partial \Delta R\left( \mu _{R}\right) }{ \partial \mu }>0$, then $\frac{\partial \mu _{R}}{\partial \rho }=- \frac{\partial \Delta R\left( \mu \right) }{\partial \rho }\Big / \frac{ \partial \Delta R\left( \mu \right) }{\partial \mu }>0$ since $\frac{ \partial \Delta R\left( \mu \right) }{\partial \rho }=\frac{\partial \Delta V\left( \mu \right) }{\partial \rho }+a\frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \rho }<0$ because

$$\begin{aligned} \frac{\partial \Delta X_{0}\left( \mu \right) }{\partial \rho }= & {} -\frac{ \alpha }{\alpha +\beta }\left( \mu -1\right) ^{2}<0;\quad \text {and } \\ \frac{\partial \Delta V\left( \mu \right) }{\partial \rho }= & {} -\left( \alpha +\beta \right) ~\left( \mu -1\right) ~((1-\rho )\mu +\rho )^{\alpha +\beta -1}V^{N}(\mu )<0. \end{aligned}$$

Similarly, since $\Delta R\lesseqqgtr 0$ as $\mu \lesseqqgtr \mu _{R}$, $a\left( \Delta X\left( 1\right) \right) =0$, $\lim _{\mu \rightarrow \infty }a\left( \Delta X\left( \mu \right) \right) \rightarrow \infty$, and $\Delta B\left( \mu \right) <0=\Delta R\left( \mu \right)$ at $\mu =\mu _{R}$, there exists a $\mu _{B}>\mu _{R}$ such that $\Delta B\left( \mu \right) \lesseqqgtr 0$ as $\mu \lesseqqgtr \mu _{B}$. As $\mu _{X}<\mu _{V}$, $\Delta B\left( \mu _{X}\right) =\Delta V\left( \mu _{X}\right) <0$, so that $\mu _{B}>\mu _{X}$, and $\Delta B\left( \mu _{V}\right) =\Delta X\left( \mu _{V}\right) >0$, so that $\mu _{B}<\mu _{V}$. With the use of same arguments as above, $\frac{\partial \mu _{B}}{\partial \rho }=- \frac{\partial \Delta B\left( \mu \right) }{\partial \rho }Big/ \frac{\partial \Delta B\left( \mu \right) }{\partial \mu }>0$ as $\frac{\partial \Delta X\left( \mu \right) }{\partial \rho }=-\left( \alpha \mu +\beta \right) <0$. $\square$

Proof of Proposition 2

Assume that $\mu <\mu _{R}.$ From Lemma 1, $\Delta R<0$ so that $n_{I}^{P}<n_{I}^{N}$. $\Delta R<0$ also implies that $\Pi ^{P}\left( n_{I}^{P},\mu \right) <\Pi ^{N}\left( n_{I}^{P},\mu \right)$; since profits $\Pi ^{i}\left( n,\mu \right)$ are strictly concave in n, $n_{I}^{N}$ is a unique maximum to $\Pi ^{N}$ and $\Pi ^{N}\left( n_{I}^{P},\mu \right) <\Pi ^{N}\left( n_{I}^{N},\mu \right)$. We thus have $\Pi ^{P}\left( n_{I}^{P},\mu \right) <\Pi ^{N}\left( n_{I}^{N},\mu \right) .$ The proof is similar for $\mu \ge \mu _{R}$. $\square$

Proof of Proposition 3

As $B^{i}(\mu )>R^{i}(\mu )$ for all $\mu$, by definitions (15) and (16) of coverages, we have the first result. Now, consider the case where $\mu <\mu _{B}$. From Lemma 1, $\Delta B<0,$ so that $n_{*}^{P}<n_{*}^{N}$. Moreover, $\Delta B<0$ also implies that $W^{P}\left( n_{*}^{P},\mu \right) <W^{N}\left( n_{*}^{P},\mu \right)$; since welfare $W^{i}\left( n,\mu \right)$ is strictly concave in n, $n_{*}^{N}$ is a unique maximum to $W^{N}$ and $W^{N}\left( n_{*}^{P},\mu \right) <W^{N}\left( n_{*}^{N},\mu \right)$. We thus have $W^{P}\left( n_{*}^{P},\mu \right) <W^{N}\left( n_{*}^{N},\mu \right)$. The proof is similar for $\mu \ge \mu _{B}$. $\square$

Proof of Proposition 4

Note that for all $\mu$, coverage solutions are given by $N\left( x,\mu \right) =\frac{x}{c\mu }$, which is an increasing function in x. So we have:

$$\begin{aligned}{} & {} W^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right) -W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right) \\{} & {} \quad =N\left( R^{P}\left( \mu \right) ,\mu \right) B^{P}\left( \mu \right) \\{} & {} \qquad -C\left( n_{I}^{P}\left( \mu \right) ,\mu \right) -N\left( R^{N}\left( \mu \right) ,\mu \right) B^{N}\left( \mu \right) +C\left( n_{I}^{N}\left( \mu \right) ,\mu \right) . \end{aligned}$$

At $\mu = \mu _{R},$ we have:

$$\begin{aligned}{} & {} W^{P}\left( n_{I}^{P}\left( \mu _{R}\right) ,\mu _{R}\right) -W^{N}\left( n_{I}^{N}\left( \mu _{R}\right) ,\mu _{R}\right) \\{} & {} \quad <\left[ N\left( R^{P}\left( \mu _{R}\right) ,\mu _{R}\right) -N\left( R^{N}\left( \mu _{R}\right) ,\mu _{R}\right) \right] B^{N}\left( \mu _{R}\right) \\{} & {} \qquad -C\left( n_{I}^{P}\left( \mu _{R}\right) ,\mu _{R}\right) +C\left( n_{I}^{N}\left( \mu _{R}\right) ,\mu _{R}\right) =0, \end{aligned}$$

where the inequality comes from the fact that $B^{P}\left( \mu _{R}\right) <B^{N}\left( \mu _{R}\right)$ and the equality, from the fact that $R^{P}\left( \mu _{R}\right) =R^{N}\left( \mu _{R}\right)$ and $n_{I}^{P}\left( \mu _{R}\right) =n_{I}^{N}\left( \mu _{R}\right)$. Similarly, at $\mu =\mu _{B}$, we have:

$$\begin{aligned}{} & {} W^{P}\left( n_{I}^{P}\left( \mu _{B}\right) ,\mu _{B}\right) -W^{N}\left( n_{I}^{N}\left( \mu _{B}\right) ,\mu _{B}\right) \\{} & {} \quad>N\left( R^{N}\left( \mu _{B}\right) ,\mu _{B}\right) \left( B^{P}\left( \mu _{B}\right) -B^{N}\left( \mu _{B}\right) \right) +C\left( n_{I}^{N}\left( \mu _{B}\right) ,\mu _{B}\right) >0, \end{aligned}$$

where the first inequality comes from the fact that $R^{P}\left( \mu _{B}\right) >R^{N}\left( \mu _{B}\right)$ and the second, from the fact that $B^{P}\left( \mu _{B}\right) =B^{N}\left( \mu _{B}\right)$. By continuity, there exists a $\tilde{\mu }_{0}\in (\mu _{R},\mu _{B})$ such that $W^{P}\left( n_{I}^{N}\left( \tilde{\mu }_{0}\right) ,\tilde{\mu } _{0}\right) =W^{N}\left( n_{I}^{P}\left( \tilde{\mu }_{0}\right) ,\tilde{\mu } _{0}\right)$. Moreover, from Propositions 1 and 2, for $\mu<$ $\mu _{R}<\mu _{B}$, $n_{*}^{N}\left( \mu \right)>n_{I}^{N}\left( \mu \right) >n_{I}^{P}\left( \mu \right)$ and $W^{P}\left( n,\mu \right) -W^{N}\left( n,\mu \right) <0$, so that we have

$$\begin{aligned} W^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right)<W^{N}\left( n_{I}^{P}\left( \mu \right) ,\mu \right)<W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right) <W^{N}\left( n_{*}^{N}\left( \mu \right) ,\mu \right) . \end{aligned}$$

This proves that $W^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right) -W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right) <0$ for $\mu <\mu _{R}$. Identically, for $\mu>$ $\mu _{B}>\mu _{R}$, $n_{I}^{P}\left( \mu \right) >n_{I}^{N}\left( \mu \right)$ and $W^{P}\left( n,\mu \right) -W^{N}\left( n,\mu \right) >0$ so

$$\begin{aligned} W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right)<W^{P}\left( n_{I}^{N}\left( \mu \right) ,\mu \right)<W^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right) <W^{P}\left( n_{*}^{P}\left( \mu \right) ,\mu \right) . \end{aligned}$$

This proves that $\tilde{\mu }_{0}$ is unique and $W^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right) \lesseqqgtr W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right)$ as $\mu \lesseqqgtr \tilde{\mu } _{0}$.

Proof of Proposition 5

At the second stage, the ISP chooses the regime independently of coverage, so that the regime is N if $\mu \le \mu _{R}$ and P if $\mu >\mu _{R}.$ If $\mu \le \mu _{R}$ or $\mu >\mu _{B}$, the regime chosen by the ISP is also the regime that is preferred by the regulator, so that the regulator can impose the welfare-maximizing coverage. If $\mu \in (\mu _{R},\mu _{B}]$, the ISP chooses P while N is the welfare-maximizing regime, so that the regulator chooses $n_{*}^{P}<n_{*}$. $\square$

Proof of Proposition 6

First, we prove that there exists a $\tilde{\mu }_{1}<\mu _{R}$ such that $B^{P}\left( \tilde{\mu } _{1} \right) -R^{N}\left( \tilde{\mu }_{1} \right) =\Delta V\left( \tilde{\mu } _{1} \right) +aX^{P}\left( \tilde{\mu }_{1} \right) =0$, and $n_{*}^{P} = n_{I}^{N}$. Indeed, consider a $\mu <\mu _{R}$ and $\Delta V(\mu )+aX^{P}(\mu )>\Delta V(\mu )+a\Delta X_{0}(\mu )=\Delta R(\mu )$. Now, since $\Delta V(\mu )<0$ for $1< \mu <\mu _{R},$ $\Delta V\left( 1\right) +aX^{P}(1)=0$, $\Delta V(\mu _{R})+aX^{P}\left( \mu _{R}\right) =B^{P}\left( \mu _{R} \right) -R^{N}\left( \mu _{R} \right) >\Delta R\left( \mu _{R}\right) =0$ and the fact that $X^{P}\left( \mu \right) =\frac{ \mu ^{2}-1 }{2\left( \alpha +\beta \right) }\left( \alpha +\frac{\beta }{\mu }\right)$ is strictly increasing, as $X^{P \prime }\left( \mu \right) =\frac{\beta +2\alpha \mu ^{3}+\beta \mu ^{2}}{\mu ^{2}}>0$. This completes the proof. $\square$

Second, we turn to cases (a)-(e).

• (a) if $\mu <\tilde{\mu }_{1},$ then $R^{N}\left( \mu \right) >B^{P}\left( \mu \right)$ so that $n_{I}^{N}>n_{*}^{P}$, which implies that $n_{*}^{N}>n_{I}^{N}>n_{*}^{P}>n_{I}^{P}$. As $n_{I}^{N}>n_{I}^{P}$, this also implies that N is chosen under UM and USO. Moreover, since $\mu <\mu _{R}$, we have $R^{N}\left( \mu \right) >R^{P}\left( \mu \right) ,$ so that

  $$\begin{aligned} W^{N}\left( n_{I}^{N},\mu \right) >W^{P}\left( n_{I}^{P},\mu \right) , \end{aligned}$$

  and N is chosen under TMR. Since $\mu <\mu _{B}$, we have $B^{N}\left( \mu \right) >B^{P}\left( \mu \right) ,$ so that

  $$\begin{aligned} W^{N}\left( n_{*}^{N},\mu \right) >W^{P}\left( n_{*}^{P},\mu \right) , \end{aligned}$$

  so that N is chosen under FR.

• (b) if $\tilde{\mu }_{1}\le \mu <\mu _{R}$ then $B^{P}\left( \mu \right) \ge R^{N}\left( \mu \right)$ so that $n_{*}^{P}\ge n_{I}^{N}$. Since $\mu <\mu _{B}$, $n_{*}^{N}>n_{*}^{P}$ and since $\mu <\mu _{R},$ $n_{I}^{N}>n_{I}^{P}$, We thus have that

  $$\begin{aligned} n_{*}^{N}>n_{*}^{P}\ge n_{I}^{N}>n_{I}^{P}. \end{aligned}$$

  Since $\mu<\mu _{B}<\mu _{B}$, arguments in (a) apply to show that N is chosen under all regulatory frameworks.

• (c) if $\mu _{R}\le \mu <\tilde{\mu }_{0}$ then $W^{N}\left( n_{I}^{N},\mu \right) >W^{P}\left( n_{I}^{P},\mu \right) ,$ so that N is chosen under TMR. Since $\mu \ge \mu _{R}$, $n_{I}^{P}\ge n_{I}^{N}$ and $\Pi ^{P}\left( n_{I}^{P},\mu \right) \ge \Pi ^{N}\left( n_{I}^{N},\mu \right)$. Since $\mu <\mu _{B}$, $n_{*}^{N}>n_{*}^{P}$ and $W^{N}\left( n_{*}^{N},\mu \right) >W^{P}\left( n_{*}^{P},\mu \right) .$ We thus have that

  $$\begin{aligned} n_{*}^{N}>n_{*}^{P}>n_{I}^{P}\ge n_{I}^{N}, \end{aligned}$$

  and that P is chosen under UM and USO, while N is chosen under FR.

• (d) $\tilde{\mu }_{0}\le \mu <\mu _{B}$, then $W^{P}\left( n_{I}^{P},\mu \right) >W^{N}\left( n_{I}^{N},\mu \right)$, so that P is chosen under TMR. Since $\mu _{R}>\mu >\mu _{B}$, coverages are set as in (c). Arguments in (c) apply to show that P is chosen under UM and USO, while N is chosen under FR.

• (e) if $\mu \ge \mu _{B}$ then $n_{*}^{P}\ge n_{*}^{N}$ and $W^{P}\left( n_{*}^{P},\mu \right) \ge W^{N}\left( n_{*}^{N}\text {,} \mu \right)$ and P is chosen under FR. Since $\mu >\tilde{\mu }_{0}$, $n_{*}^{P}>n_{I}^{N}$ and since $\mu >\mu _{R}$, $n_{I}^{P}>n_{I}^{N}$ and

  $$\begin{aligned} n_{*}^{P}\ge n_{*}^{N}>n_{I}^{P}>n_{I}^{N}, \end{aligned}$$

  and P is also chosen under UM, USO and TMR.

Proof of Proposition 7

For $\mu \in \left( \tilde{\mu }_{0},\mu _{B}\right)$, we have $n_{U}=n_{*}^{P}>n_{T}=n_{I}^{P}$ so this yields $W^{P}\left( n_{*}^{P}\left( \mu \right) ,\mu \right) >W^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right)$, which proves the second part. If $\mu \in (\mu _{R},\tilde{\mu }_{0}]$, from Proposition 6, $n_{*}^{N}>n_{*}^{P}>n_{I}^{P}\ge n_{I}^{N}$ and if $\mu =\tilde{\mu }_{0}:W^{P}\left( n_{*}^{P}\left( \tilde{\mu }_{0}\right) ,\tilde{\mu }_{0}\right)>W^{P}\left( n_{I}^{P}\left( \tilde{\mu }_{0}\right) ,\tilde{\mu }_{0}\right) =W^{N}\left( n_{I}^{P}\left( \tilde{\mu }_{0}\right) ,\tilde{\mu }_{0}\right) >W^{N}\left( n_{I}^{N}\left( \tilde{\mu }_{0}\right) ,\tilde{\mu }_{0}\right)$. Now we have

$$\begin{aligned}{} & {} W^{P}\left( n_{*}^{P}\left( \mu \right) ,\mu \right) -W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right) \\{} & {} \quad =\Pi ^{P}\left( n_{*}^{P}\left( \mu \right) ,\mu \right) -\Pi ^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right) +a\left( X_{1}^{P}\left( \mu \right) -X_{1}^{N}\left( \mu \right) \right) \\{} & {} \quad <\Pi ^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right) -\Pi ^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right) +a\left( X_{1}^{P}\left( \mu \right) -X_{1}^{N}\left( \mu \right) \right) , \end{aligned}$$

as $\Pi ^{P}\left( n_{I}^{P}\left( \mu \right) ,\mu \right) >\Pi ^{P}\left( n,\mu \right)$ for all n. So if $\mu =\mu _{R}$

$$\begin{aligned} W^{P}\left( n_{*}^{P}\left( \mu _{R}\right) ,\mu _{R}\right) -W^{N}\left( n_{I}^{N}\left( \mu _{R}\right) ,\mu _{R}\right)<a\left( X_{1}^{P}\left( \mu _{R}\right) -X_{1}^{N}\left( \mu _{R}\right) \right) <0, \end{aligned}$$

since $X_{1}^{P}\left( \mu \right) -X_{1}^{N}\left( \mu \right) =-\frac{ \beta }{2}\frac{\left( \mu -1\right) ^{2}}{\mu \left( \alpha +\beta \right) } <0$. By continuity, there exists a $\tilde{\mu }_{u}\in (\mu _{R},\tilde{\mu } _{2})$ such that $W^{P}\left( n_{*}^{P}\left( \tilde{\mu }_{u}\right) , \tilde{\mu }_{u}\right) =W^{N}\left( n_{I}^{N}\left( \tilde{\mu }_{u}\right) , \tilde{\mu }_{u}\right)$ and $W^{P}\left( n_{*}^{P}\left( \mu \right) ,\mu \right) \lesseqqgtr W^{N}\left( n_{I}^{N}\left( \mu \right) ,\mu \right)$ as $\mu \lesseqqgtr \tilde{\mu }_{u}$. $\square$

Proof of Proposition 8

First, let us give a characterization of $M_{*}^{i}$. Inserting first-best coverages $n_{*}^{i}\left( \mu \right) =\frac{B^{i}\left( \mu \right) }{c\mu }$ into the ISP participation constraint (17) implies that $\Pi ^{i}\left( n_{*}^{i}\left( \mu \right) ,\mu \right) =R^{i}\left( \mu \right) -\frac{1}{2}B^{i}\left( \mu \right) \ge 0$. This can also be written as $\frac{1}{2}\left( V^{i}\left( \mu \right) -aZ^{i}\left( \mu \right) \right) \ge 0$, where $Z^{N}\left( \mu \right) =X^{N}\left( \mu \right)$ and $Z^{P}\left( \mu \right) =X^{P}\left( \mu \right) -2X_{0}^{N}\left( \mu \right) =\frac{1}{2}\left( \mu -1\right) \frac{\alpha \mu ^{2}+\left( \beta -3\alpha \right) \mu +\beta }{\mu \left( \alpha +\beta \right) }>0$.²¹ Here $Z^{N}\left( \mu \right)$ is increasing and linear, and $Z^{P}\left( \mu \right)$ is increasing and concave for all $\mu >1$. Then for $a=0$, we have $\frac{1}{2}V^{i}\left( \mu \right) -aZ^{i}\left( \mu \right) >0$ and

$$\begin{aligned} \lim _{a\rightarrow +\infty }\left( V^{i}\left( \mu \right) -aZ^{i}\left( \mu \right) \right) =-Z^{i}\left( \mu \right) \lim _{a\rightarrow +\infty }\left( a\right) =-\infty . \end{aligned}$$

So there exists a unique $a_{\pi }^{i}:\frac{1}{2}\left( V^{i}\left( \mu \right) -aZ^{i}\left( \mu \right) \right) =0$. More precisely

$$\begin{aligned} a_{\pi }^{N}= & {} \frac{V^{N}\left( \mu \right) }{X^{N}\left( \mu \right) } =v\left( \mu -1\right) ^{\alpha +\beta -1}>0;\quad \text {and } \\ a_{\pi }^{P}= & {} \frac{V^{P}\left( \mu \right) }{Z^{P}\left( \mu \right) } =v\left( \alpha +\beta \right) \left( \frac{1}{2}\right) ^{\alpha +\beta -1} \frac{\left( \mu -1\right) ^{\alpha +\beta -1}\left( \mu +1\right) ^{\alpha +\beta }\mu ^{-(1+\beta )}}{\alpha \mu ^{2}+\left( \beta -3\alpha \right) \mu +\beta }>0. \end{aligned}$$

When $\alpha +\beta <1$, these price thresholds are decreasing functions of $\mu$ with $\lim _{\mu \rightarrow +\infty }a_{\pi }^{i}=0$, so that for all a, there exists a unique $\bar{\mu }_{*}^{i}:a_{\pi }^{i}=a$. Then $\bar{\mu }_{*}^{i}$ becomes the inverse function of $a_{\pi }^{i}$, defined from $\mathbb {R}_{+}$ to $[1,+\infty )$ and it is decreasing in a. As a result the ISP participation constraint (17) is satisfied for $\mu \le \bar{\mu }_{*}^{i}$ and $M_{*}^{i}=[1,\bar{\mu }_{*}^{i}]$. Note that for a given $\mu$

$$\begin{aligned} a_{\pi }^{P}<a_{\pi }^{N}, \end{aligned}$$

as for all $\mu >1$ we have

$$\begin{aligned} \left( \alpha +\beta \right) \left( \frac{1}{2}\right) ^{\alpha +\beta -1} \frac{\left( \mu +1\right) ^{\alpha +\beta }\mu ^{-(1+\beta )}}{\alpha \mu ^{2}+\left( \beta -3\alpha \right) \mu +\beta }<\frac{1}{\mu ^{2}}<1. \end{aligned}$$

This implies that $\bar{\mu }_{*}^{N}$ and $\bar{\mu }_{*}^{P}$ cannot intersect for any a.

Second, define just for the purpose $\bar{\mu }_{*}^{i}$ as functions of a, and let us define the following ad-price threshold

$$\begin{aligned} a_{B}^{P}:\bar{\mu }_{*}^{P}(a_{B}^{P})=\mu _{B}. \end{aligned}$$

It exists and is unique as $\bar{\mu }_{*}^{P}\left( a\right)$ is a decreasing one-to-one mapping from $\mathbb {R}_{+}$ to $[1,+\infty )$, but $\mu _{B}$ is from $\mathbb {R}_{+}$ to $[\mu _{X},\mu _{V}]$. Then by definition, we have, when $a=a_{B}^{P}:n_{*}^{P}\left( \bar{\mu }_{*}^{P}(a_{B}^{P})\right) =n_{*}^{N}\left( \mu _{B}\right) =n_{*}^{P}\left( \mu _{B}\right)$ and $\Pi ^{N}\left( n_{*}^{N}\left( \mu _{B})\right) ,\mu _{B}\right) <\Pi ^{P}\left( n_{*}^{N}\left( \mu _{B})\right) ,\mu _{B}\right) =\Pi ^{P}\left( n_{*}^{P}\left( \mu _{B})\right) ,\mu _{B}\right) =0$ as, using Proposition 1, we have that for all $\left( n,\mu \right) :\Pi ^{P}\left( n,\mu \right) >\Pi ^{N}\left( n,\mu \right)$ if $\mu >\mu _{R}$, as $\Delta \Pi =n\Delta R$. Here, $\mu =\mu _{B}>\mu _{R}$, so the ISP participation constraint for $i=N$ is violated: this implies that $\bar{\mu }_{*}^{P}(a_{B}^{P})>\bar{\mu } _{*}^{N}(a_{B}^{P})$. As $\bar{\mu }_{*}^{N}(a)$ and $\bar{\mu }_{*}^{P}(a)$ cannot intersect, this proves that $\bar{\mu }_{*}^{N}<\bar{\mu } _{*}^{P}$. $\square$

Proof of Proposition 9

From (5), (11) and ( 20), we obtain:

$$\begin{aligned} m_{n}^{\prime }= & {} \frac{1}{2}\left( \frac{aX_{f}^{i}}{F}\right) ^{\frac{1}{2 }}n^{-\frac{1}{2}}>0; \\ m_{\mu }^{\prime }= & {} \frac{1}{2}\left( \frac{an}{F}\right) ^{\frac{1}{2} }\left( X_{f}^{i}\right) ^{-\frac{1}{2}}\left( X_{f}^{i}\right) ^{\prime }>0;\quad \text {and } \\ m_{\mu n}^{\prime \prime }= & {} \frac{1}{4}\left( \frac{an}{F}\right) ^{-\frac{ 1}{2}}\left( X_{f}^{i}\right) ^{-\frac{1}{2}}\left( X_{f}^{i}\right) ^{\prime }>0. \end{aligned}$$

Moreover, since

$$\begin{aligned} \Delta m\left( n,\mu \right) =\left( \frac{an}{F}\right) ^{\frac{1}{2} }\left( \sqrt{X_{f}^{P}\left( \mu \right) }-\sqrt{X_{f}^{N}\left( \mu \right) }\right) <0, \end{aligned}$$

we obtain:

$$\begin{aligned} \Delta m\left( n,\mu \right)= & {} \left( \frac{na}{F}\right) ^{\frac{1}{2} }\left( \sqrt{X_{f}^{P}\left( \mu \right) }-\sqrt{X_{f}^{N}\left( \mu \right) }\right) \\= & {} \left( \frac{na}{F}\right) ^{\frac{1}{2}}\left( \sqrt{ \frac{1}{2}\frac{\beta }{\alpha +\beta }\frac{\left( \mu ^{2}-1\right) }{\mu }}-\sqrt{\frac{\beta }{\alpha +\beta }\left( \mu -1\right) }\right)<0 \\ \frac{\partial \left( \Delta m\left( n,\mu \right) \right) }{\partial n}= & {} \frac{1}{2} n^{-1}\Delta m<0 \\ \frac{\partial ^{2}\left( \Delta m\left( n,\mu \right) \right) }{\partial n^{2}}= & {} -\frac{1}{4} n^{-2}\Delta m>0. \end{aligned}$$

$\square$