Non-neutrality Pushed by Big Content Providers

Abstract

Major content/service providers are publishing grades they give to Internet Service Providers (ISPs) about the quality of delivery of their content. The goal is to inform customers about the “best” ISPs. But this could be an incentive for, or even a pressure on, ISPs to differentiate service and provide a better quality to those big content providers in order to be more attractive. This fits the network neutrality debate, but instead of the traditional vision of ISPs pressing content providers, we face here the opposite situation, still possibly at the expense of small content providers though. This paper designs a model describing the various actors and their strategies, analyzes it using non-cooperative game theory tools, and quantifies the impact of those advertised grades with respect to the situation where no grade is published. We illustrate that a non-neutral behavior, differentiating traffic, is not leading to a desirable situation.

A Proof of Proposition 1

Proof

A user $\theta$ strictly prefers ISP A over B if and only if

$$\begin{aligned} \frac{\theta C'_{A,1}+ (1-\theta )C'_{A,2}}{m_A}>\frac{\theta C'_{B,1}+(1-\theta )C'_{B,2}}{m_B}, \end{aligned}$$

or equivalently $\theta \varDelta >\delta$ with $\left\{ \begin{array}{l} \varDelta :=\frac{C'_{A,1}-C'_{A,2}}{m_A} - \frac{C'_{B,1}-C'_{B,2}}{m_B}\\ \delta :=\frac{C'_{B,2}}{m_B} - \frac{C'_{A,1}}{m_A}. \end{array}\right.$

Therefore, we can distinguish three possible types of equilibria.

• (a) In an equilibrium with $\varDelta >0$, all users with $\theta >\delta / \varDelta$ select ISP A, all users with $\theta <\delta / \varDelta$ prefer ISP B, and the set of users indifferent between A and B is of measure zero. Hence, for a type-$a$ equilibrium the masses of users with each ISP are of the form $m_A = 1-F(\theta ^*_{a})$ and $m_B=F(\theta ^*_{a})$ with F the cdf of user-specific values $\theta$, and $\theta ^*_{a}$ the corresponding value of $\delta / \varDelta$ at this equilibrium. A user with sensitivity $\theta ^*_{a}$ should be indifferent between both ISPs, i.e.,

  $$ \frac{\theta ^*_{a} C'_{A,1}+ (1-\theta ^*_{a})C'_{A,2}}{1-F(\theta ^*_{a})}-\frac{\theta ^*_{a} C'_{B,1}+(1-\theta ^*_{a})C'_{B,2}}{F(\theta ^*_{a})}=0. $$

  When $\theta$ follows a uniform distribution over [0, 1], then $F(x)=x$ and the previous equality becomes

  $$\begin{aligned} \frac{\theta ^*_{a} }{1-\theta ^*_{a}}(C'_{A,1}-C'_{A,2})+C'_{B,2}-C'_{B,1}-\frac{C'_{B,2}}{\theta ^*_{a}}+\frac{C'_{A,2}}{1-\theta ^*_{a}}=0. \end{aligned}$$

  (4)

  With non-blocking ISP strategies, $C'_{A,1}$ and $C'_{B,2}$ are strictly positive, and (4) has a unique solution in (0, 1). Summarizing, we have a type-$a$ equilibrium if the corresponding $\varDelta$ is strictly positive, i.e., if in addition to (4) we have

  $$\begin{aligned} \frac{C'_{A,1}-C'_{A,2}}{1-\theta ^*_{a}} > \frac{C'_{B,1}-C'_{B,2}}{\theta ^*_{a}}. \end{aligned}$$

  (5)

  In particular, (5) implies that $(C'_{A,2}-C'_{A,1})\frac{\theta ^*_{a}}{1-\theta ^*_{a}}<C'_{B,2}-C'_{B,1}$, which plugged into (4) gives

  $$\begin{aligned} \frac{\theta ^*_{a}}{1-\theta ^*_{a}}>\frac{C'_{B,2}}{C'_{A,2}}, \end{aligned}$$

  (6)

  or equivalently

  $$\begin{aligned} \frac{1}{\theta ^*_{a}} < 1+\frac{C'_{A,2}}{C'_{B,2}}. \end{aligned}$$

  (7)

  Finally, re-writing (4) as

  $$\begin{aligned} C'_{A,1}\frac{\theta ^*_{a} }{1-\theta ^*_{a}}-C'_{B,2}\frac{1}{\theta ^*_{a}}+C'_{A,2}+C'_{B,2}-C'_{B,1}=0 \end{aligned}$$

  (8)

  and plugging (6) and (7), we get

  $$\begin{aligned} 0< & {} C'_{A,1} \frac{C'_{B,2}}{C'_{A,2}} -C'_{B,2} \left( 1+\frac{C'_{A,2}}{C'_{B,2}}\right) +C'_{A,2}+C'_{B,2}-C'_{B,1}\\&=\quad C'_{A,1} \frac{C'_{B,2}}{C'_{A,2}} -C'_{B,1} \end{aligned}$$

  or $\frac{C'_{A,1}}{C'_{A,2}} > \frac{C'_{B,1}}{C'_{B,2}}$. As a result, a type-$a$ equilibrium, which is unique from (4) can only exist when $\frac{C'_{A,1}}{C'_{A,2}} > \frac{C'_{B,1}}{C'_{B,2}}$.

• (b) We can treat similarly the case when $\varDelta <0$, by exchanging the roles of ISPs A and B with respect to the previous case. Hence, such an equilibrium is of the form $m_A=\theta ^*_{b}, m_B=1-\theta ^*_{b}$ with $\theta ^*_{b}$ the unique solution in (0, 1) of

  $$\begin{aligned} \frac{\theta ^*_{b} C'_{B,1}}{1-\theta ^*_{b}}-\frac{(1-\theta ^*_{b})C'_{A,2}}{\theta ^*_{b}}+C'_{B,2}-C'_{A,1}=0, \end{aligned}$$

  (9)

  and such a (unique) equilibrium can only exist when $\frac{C'_{A,1}}{C'_{A,2}} < \frac{C'_{B,1}}{C'_{B,2}}$.

• (c) Finally, at an equilibrium with $\varDelta =0$, all users must be indifferent between the ISPs (otherwise they all prefer the same, while its competitor has no demand (no congestion) and thus infinite quality, a contradiction). Therefore, for all $\theta$ we have $\frac{\theta C'_{A,1}+ (1-\theta )C'_{A,2}}{m_A} = \frac{\theta C'_{B,1}+(1-\theta )C'_{B,2}}{m_B}$, which implies, using $m_A+m_B=1$, that we have

  $$\begin{aligned} \left\{ \begin{array}{l}\frac{C'_{A,1}}{m_A}=\frac{C'_{B,1}}{1-m_A}\\ \frac{C'_{A,2}}{m_A}=\frac{C'_{B,2}}{1-m_A}\end{array}\right. \end{aligned}$$

  In particular, such an equilibrium can only exist when $\frac{C'_{A,1}}{C'_{A,2}} = \frac{C'_{B,1}}{C'_{B,2}}$ (by looking at ratios of left-hand sides and right-hand sides in the above system of equations), and while the specific user choices are not unique, solving the equation(s) leads to unique values of $m_A$ and $m_B$, given as in (3).

Regrouping the three cases establishes the uniqueness of a user equilibrium as stated in the proposition. To establish existence, we reason on the relative values of $\frac{C'_{A,1}}{C'_{A,2}}$ and $\frac{C'_{B,1}}{C'_{B,2}}$.

• First, if $\frac{C'_{A,1}}{C'_{A,2}} = \frac{C'_{B,1}}{C'_{B,2}}$ then one can check that $m_A$ and $m_B$ given as in (3) is indeed an equilibrium.

• Second, if $\frac{C'_{A,1}}{C'_{A,2}} > \frac{C'_{B,1}}{C'_{B,2}}$, assume that we do not have a type-$a$ equilibrium, i.e., that the solution $\theta ^*_{a}$ of (4) does not give a strictly positive $\varDelta$ (or in other words, (5) is not satisfied). We prove below that this leads to a contradiction, following steps close to those of the uniqueness proof. Indeed, (5) not being satisfied means $\frac{\theta ^*_{a}}{1-\theta ^*_{a}}(C'_{A,1}-C'_{A,2}) \le C'_{B,1}-C'_{B,2}$, which when plugged into (4) gives

  $$\begin{aligned} \frac{\theta ^*_{a}}{1-\theta ^*_{a}}\le \frac{C'_{B,2}}{C'_{A,2}} \qquad \text { and } \qquad \frac{1}{\theta ^*_{a}} \ge 1+\frac{C'_{A,2}}{C'_{B,2}}. \end{aligned}$$

  Like for the uniqueness proof, plugging those inequalities into (8) leads to $\frac{C'_{A,1}}{C'_{A,2}} \le \frac{C'_{B,1}}{C'_{B,2}}$, a contradiction with our starting assumption. Hence, when $\frac{C'_{A,1}}{C'_{A,2}} > \frac{C'_{B,1}}{C'_{B,2}}$ there exists a type-$a$ equilibrium.

• Third, if $\frac{C'_{A,1}}{C'_{A,2}} < \frac{C'_{B,1}}{C'_{B,2}}$, by exchanging the roles of A and B we also have existence, of a type-$b$ equilibrium.