Content provision strategy selection for a media platform in the presence of traffic revenue

Abstract

Inspired by the diversity of content provision and the importance of traffic revenue in media market, this paper investigates the pricing and advertising decisions for a media platform considering traffic revenue under three content provision strategies: single paid-content-without-ads strategy, single free-content-with-ads strategy, and hybrid content strategy. The optimal solutions among different content provision strategies are compared and analyzed, thereby guiding the media platform and consumers in choosing the best content provision mode, while also providing some valuable insights for the media platform in its decision-making. In presence of traffic revenue, findings show that the hybrid content provision strategy can achieve a win–win situation for the media platform and consumers, in terms of platform profit, consumer surplus, and decision-making speed, whereas the single paid or free content provision strategy cannot achieve such a situation. We also surprisingly find that the media content with higher quality fails in creating more welfare for consumers, when the traffic revenue sensitivity coefficient and the free content’s quality discount factor are high. Numerical experiments verify the theoretical results. We also study two model extensions of endogenous content quality and platforms competition, and find that the major findings of the base model are robust.

Appendix

Proof of Proposition 1

$p_{H}^{*} - p_{P}^{*} = \frac{{m\left( {2\delta m - \left( {1 + \delta } \right)\beta } \right)}}{{2\left( {2\delta m - \beta } \right)}} - \frac{{m\left( {m - \beta } \right)}}{2m - \beta } = - \frac{{m\left( {1 - \delta } \right)\beta^{2} }}{{2\left( {2\delta m - \beta } \right)\left( {2m - \beta } \right)}} < 0$; $r_{H}^{*} - r_{F}^{*} = 0$. Thus, $p_{H}^{*} < p_{P}^{*}$ and $r_{H}^{*} = r_{F}^{*}$. □

Proof of Proposition 2

$q_{H}^{*} - q_{F}^{*} = 0$, $q_{H}^{*} - q_{P}^{*} = \frac{{m\left( {1 - \delta } \right)\beta }}{{\left( {2\delta m - \beta } \right)\left( {2m - \beta } \right)}} > 0$. Thus, $q_{H}^{*} = q_{F}^{*} > q_{P}^{*}$. Similarly, $R_{Ht}^{*} = R_{Ft}^{*} > R_{Pt}^{*}$ can be derived. □

Proof of Proposition 3

$\pi_{H}^{*} - \pi_{F}^{*} = \frac{{m\left( {1 - \delta } \right)}}{4} > 0$, $\pi_{H}^{*} - \pi_{P}^{*} = \frac{{m\left( {1 - \delta } \right)\beta^{2} }}{{4\left( {2\delta m - \beta } \right)\left( {2m - \beta } \right)}} > 0$, $\pi_{P}^{*} - \pi_{F}^{*} = \frac{{m^{2} \left( {1 - \delta } \right)\left( {2\delta m - \left( {1 + \delta } \right)\beta } \right)}}{{2\left( {2\delta m - \beta } \right)\left( {2m - \beta } \right)}} > \frac{{m^{3} \left( {1 - \delta } \right)^{2} \delta }}{{2\left( {2\delta m - \beta } \right)\left( {2m - \beta } \right)}} > 0$ since $\beta < \delta m$. Thus, $\pi_{H}^{*} > \pi_{P}^{*} > \pi_{F}^{*}$. □

Proof of Proposition 4

$CS_{H}^{*} - CS_{F}^{*} = \frac{{m\left( {1 - \delta } \right)}}{8} > 0$.

$CS_{H}^{*} - CS_{P}^{*} = \frac{{3m\beta^{2} \left( {1 - \delta } \right)\left( {12\delta m^{2} - 4m\left( {1 + \delta } \right)\beta + \beta^{2} } \right)}}{{24\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > \frac{{3\delta m^{3} \beta^{2} \left( {1 - \delta } \right)\left( {8 - 3\delta } \right)}}{{24\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > 0$ since $\beta < \delta m$.

$CS_{P}^{*} - CS_{F}^{*} = \frac{{2m^{3} \left( {1 - \delta } \right)\left( {4\delta^{2} m^{2} - 4\delta m\left( {1 + \delta } \right)\beta + \left( {1 + \delta^{2} + \delta } \right)\beta^{2} } \right)}}{{4\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }}$,

$f_{1} \left( \beta \right) = \left( {1 + \delta^{2} + \delta } \right)\beta^{2} - 4\delta m\left( {1 + \delta } \right)\beta + 4\delta^{2} m^{2}$, $\frac{{2\delta m\left( {1 + \delta } \right)}}{{1 + \delta^{2} + \delta }} - \delta m = \frac{{\delta m\left( {1 - \delta^{2} + \delta } \right)}}{{1 + \delta^{2} + \delta }} > 0$,

$f_{1} \left( \beta \right)|_{\beta = \delta m} = \delta^{2} m^{2} \left( {1 - 3\delta + \delta^{2} } \right)$: if $0 < \delta < \frac{3 - \sqrt 5 }{2}$, $f_{1} \left( \beta \right)|_{\beta = \delta m} > 0$; if $\frac{3 - \sqrt 5 }{2} < \delta < 1$, $f_{1} \left( \beta \right)|_{\beta = \delta m} < 0$. $\frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }}$ and $\frac{{2\delta m\left( {1 + \delta + \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }}$ are the roots of $f_{1} \left( \beta \right) = 0$.

Thus, when $0 < \delta < \frac{3 - \sqrt 5 }{2}$, $CS_{H}^{*} > CS_{P}^{*} > CS_{F}^{*}$. When $\frac{3 - \sqrt 5 }{2} < \delta < 1$, if $0 < \beta < \frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }}$, then $CS_{H}^{*} > CS_{P}^{*} > CS_{F}^{*}$; if $\frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }} < \beta < \delta m$, then $CS_{H}^{*} > CS_{F}^{*} > CS_{P}^{*}$. □

Proof of Corollary 1

• 1. $\frac{{\partial p_{P}^{*} }}{\partial \beta } = - \frac{{m^{2} }}{{\left( {2m - \beta } \right)^{2} }} < 0$, $\frac{{\partial p_{P}^{*} }}{\partial \beta } - \frac{{\partial p_{H}^{*} }}{\partial \beta } = \frac{{\beta m^{2} \left( {1 - \delta } \right)\left( {4\delta m - \left( {1 + \delta } \right)\beta } \right)}}{{\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > \frac{{\delta \beta m^{3} \left( {1 - \delta } \right)\left( {3 - \delta } \right)}}{{\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > 0$, Thus, $\frac{{\partial p_{H}^{*} }}{\partial \beta } < \frac{{\partial p_{P}^{*} }}{\partial \beta } < 0$.

• 2. $\frac{{\partial r_{F}^{*} }}{\partial \beta } = \frac{{\partial r_{H}^{*} }}{\partial \beta } = - \frac{{\delta^{2} m^{2} }}{{\left( {2\delta m - \beta } \right)^{2} }} < 0$. Thus, $\frac{{\partial r_{F}^{*} }}{\partial \beta } = \frac{{\partial r_{H}^{*} }}{\partial \beta } < 0$.

• 3. $\frac{{\partial q_{F}^{*} }}{\partial \beta } - \frac{{\partial q_{P}^{*} }}{\partial \beta } = \frac{{m\left( {1 - \delta } \right)\left( {4\delta m^{2} - \beta^{2} } \right)}}{{\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > \frac{{\delta m^{3} \left( {1 - \delta } \right)\left( {4 - \delta } \right)}}{{\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > 0$, $\frac{{\partial q_{P}^{*} }}{\partial \beta } = \frac{m}{{\left( {2m - \beta } \right)^{2} }} > 0$, Thus, $\frac{{\partial q_{H}^{*} }}{\partial \beta } = \frac{{\partial q_{F}^{*} }}{\partial \beta } > \frac{{\partial q_{P}^{*} }}{\partial \beta } > 0$.

• 4. $\frac{{\partial \pi_{H}^{*} }}{\partial \beta } - \frac{{\partial \pi_{P}^{*} }}{\partial \beta } = \frac{{\beta m^{2} \left( {1 - \delta } \right)\left( {4\delta m - \left( {1 + \delta } \right)\beta } \right)}}{{2\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > \frac{{\delta \beta m^{3} \left( {1 - \delta } \right)\left( {3 - \delta } \right)}}{{2\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > 0$, $\frac{{\partial \pi_{F}^{*} }}{\partial \beta } = \frac{{\partial \pi_{H}^{*} }}{\partial \beta } = \frac{{\delta^{2} m^{2} }}{{2\left( {2\delta m - \beta } \right)^{2} }} > 0$, $\frac{{\partial \pi_{P}^{*} }}{\partial \beta } = \frac{{m^{2} }}{{2\left( {2m - \beta } \right)^{2} }} > 0$. Thus, $\frac{{\partial \pi_{F}^{*} }}{\partial \beta } = \frac{{\partial \pi_{H}^{*} }}{\partial \beta } > \frac{{\partial \pi_{P}^{*} }}{\partial \beta } > 0$.

• 5. $\frac{{\partial CS_{F}^{*} }}{\partial \beta } = \frac{{\partial CS_{H}^{*} }}{\partial \beta } = \frac{{\delta^{3} m^{3} }}{{\left( {2\delta m - \beta } \right)^{3} }} > 0$, $\frac{{\partial CS_{P}^{*} }}{\partial \beta } = \frac{{m^{3} }}{{\left( {2m - \beta } \right)^{3} }} > 0$.

$\frac{{\partial CS_{H}^{*} }}{\partial \beta } - \frac{{\partial CS_{P}^{*} }}{\partial \beta } = \frac{{m^{3} \left( {\left( {1 - \delta^{3} } \right)\beta^{3} - 6\delta m\left( {1 - \delta^{2} } \right)\beta^{2} + 12\delta^{2} m^{2} \left( {1 - \delta } \right)\beta } \right)}}{{\left( {2\delta m - \beta } \right)^{3} \left( {2m - \beta } \right)^{3} }} = \frac{{m^{3} f_{2} \left( \beta \right)}}{{\left( {2\delta m - \beta } \right)^{3} \left( {2m - \beta } \right)^{3} }}$,

$\frac{{\partial f_{2} \left( \beta \right)}}{\partial \beta } = 3\left( {1 - \delta } \right)f_{1} \left( \beta \right)$, where $f_{1} \left( \beta \right) = \left( {1 + \delta^{2} + \delta } \right)\beta^{2} - 4\delta m\left( {1 + \delta } \right)\beta + 4\delta^{2} m^{2}$ and $\beta \in \left( {0,\delta m} \right)$. When $0 < \delta < \frac{3 - \sqrt 5 }{2}$, we obtain that $\frac{{\partial f_{2} \left( \beta \right)}}{\partial \beta } > 0$, which indicates that $f_{2} \left( \beta \right)$ increases with $\beta \in \left( {0,\delta m} \right)$. When $\frac{3 - \sqrt 5 }{2} < \delta < 1$, if $0 < \beta < \frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }}$, then $\frac{{\partial f_{2} \left( \beta \right)}}{\partial \beta } > 0$, which indicates that $f_{2} \left( \beta \right)$ increases with $\beta$; if $\frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }} < \beta < \delta m$, then $\frac{{\partial f_{2} \left( \beta \right)}}{\partial \beta } < 0$, which indicates that $f_{2} \left( \beta \right)$ decreases with $\beta$. Consequently, when $0 < \delta < \frac{3 - \sqrt 5 }{2}$ and $0 < \beta < \delta m$, since $f_{2} \left( \beta \right) > f_{2} \left( \beta \right)|_{\beta = 0} = 0$, we obtain that $\frac{{\partial CS_{H}^{*} }}{\partial \beta } > \frac{{\partial CS_{P}^{*} }}{\partial \beta }$. When $\frac{3 - \sqrt 5 }{2} < \delta < 1$ and $0 < \beta < \frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }}$, since $f_{2} \left( \beta \right) > f_{2} \left( \beta \right)|_{\beta = 0} = 0$, we obtain that $\frac{{\partial CS_{H}^{*} }}{\partial \beta } > \frac{{\partial CS_{P}^{*} }}{\partial \beta }$. When $\frac{3 - \sqrt 5 }{2} < \delta < 1$ and $\frac{{2\delta m\left( {1 + \delta - \sqrt \delta } \right)}}{{1 + \delta^{2} + \delta }} < \beta < \delta m$, since $f_{2} \left( \beta \right) > f_{2} \left( \beta \right)|_{\beta = \delta m} = \delta^{3} m^{3} \left( {1 - \delta } \right)\left( {\delta^{2} - 5\delta + 7} \right) > 0$, we obtain that $\frac{{\partial CS_{H}^{*} }}{\partial \beta } > \frac{{\partial CS_{P}^{*} }}{\partial \beta }$. Then, $\frac{{\partial CS_{H}^{*} }}{\partial \beta } > \frac{{\partial CS_{P}^{*} }}{\partial \beta }$ for $\beta \in \left( {0,\delta m} \right)$.

Thus, we can conclude that $\frac{{\partial CS_{F}^{*} }}{\partial \beta } = \frac{{\partial CS_{H}^{*} }}{\partial \beta } > \frac{{\partial CS_{P}^{*} }}{\partial \beta } > 0$. □

Proof of Corollary 2

• 1. $\frac{{\partial p_{P}^{*} }}{\partial m} = \frac{{\beta^{2} - 2\beta m + 2m^{2} }}{{\left( {2m - \beta } \right)^{2} }} > \frac{{m^{2} \left( {1 + \left( {1 - \delta } \right)^{2} } \right)}}{{\left( {2m - \beta } \right)^{2} }} > 0$,

  $\frac{{\partial p_{H}^{*} }}{\partial m} - \frac{{\partial p_{P}^{*} }}{\partial m} = \frac{{\beta^{2} \left( {1 - \delta } \right)\left( {4\delta m^{2} - \beta^{2} } \right)}}{{2\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > \frac{{\delta m^{2} \beta^{2} \left( {1 - \delta } \right)\left( {4 - \delta } \right)}}{{2\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > 0$, Thus, $\frac{{\partial p_{H}^{*} }}{\partial m} > \frac{{\partial p_{P}^{*} }}{\partial m} > 0$.

• 2. $\frac{{\partial r_{F}^{*} }}{\partial m} = \frac{{\partial r_{H}^{*} }}{\partial m} = \frac{{\delta \left( {2\delta^{2} m^{2} - 2\beta \delta m + \beta^{2} } \right)}}{{\left( {2\delta m - \beta } \right)^{2} }} > \frac{{\delta^{3} m^{2} }}{{\left( {2\delta m - \beta } \right)^{2} }} > 0$. Thus, $\frac{{\partial r_{F}^{*} }}{\partial m} = \frac{{\partial r_{H}^{*} }}{\partial m} > 0$.

• 3. $\frac{{\partial q_{H}^{*} }}{\partial m} = \frac{{\partial q_{F}^{*} }}{\partial m} = - \frac{\beta \delta }{{\left( {2\delta m - \beta } \right)^{2} }} < 0$, $\frac{{\partial q_{P}^{*} }}{\partial m} = - \frac{\beta }{{\left( {2m - \beta } \right)^{2} }} < 0$, $\frac{{\partial q_{F}^{*} }}{\partial m} - \frac{{\partial q_{P}^{*} }}{\partial m} = - \frac{{\beta \left( {1 - \delta } \right)\left( {4\delta m^{2} - \beta^{2} } \right)}}{{\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} < 0$, Thus, $\frac{{\partial q_{H}^{*} }}{\partial m} = \frac{{\partial q_{F}^{*} }}{\partial m} < \frac{{\partial q_{P}^{*} }}{\partial m} < 0$.

• 4. $\frac{{\partial \pi_{P}^{*} }}{\partial m} - \frac{{\partial \pi_{H}^{*} }}{\partial m} = \frac{{\beta^{2} \left( {1 - \delta } \right)\left( {4\delta m^{2} - \beta^{2} } \right)}}{{4\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > \frac{{\delta m^{2} \beta^{2} \left( {1 - \delta } \right)\left( {4 - \delta } \right)}}{{4\left( {2\delta m - \beta } \right)^{2} \left( {2m - \beta } \right)^{2} }} > 0$, $\frac{{\partial \pi_{H}^{*} }}{\partial m} - \frac{{\partial \pi_{F}^{*} }}{\partial m} = \frac{1 - \delta }{4} > 0$, $\frac{{\partial \pi_{F}^{*} }}{\partial m} = \frac{{\delta^{2} m\left( {\delta m - \beta } \right)}}{{\left( {2\delta m - \beta } \right)^{2} }} > 0$. Thus, $\frac{{\partial \pi_{P}^{*} }}{\partial m} > \frac{{\partial \pi_{H}^{*} }}{\partial m} > \frac{{\partial \pi_{F}^{*} }}{\partial m} > 0$.

• 5. $\frac{{\partial CS_{F}^{*} }}{\partial m} = \frac{{\delta^{3} m^{2} \left( {2\delta m - 3\beta } \right)}}{{2\left( {2\delta m - \beta } \right)^{3} }}$, Thus, if $\beta < \frac{2\delta m}{3}$, then $\frac{{\partial CS_{F}^{*} }}{\partial m} > 0$; if $\frac{2\delta m}{3} < \beta < \delta m$, then $\frac{{\partial CS_{F}^{*} }}{\partial m} < 0$.

$\frac{{\partial CS_{P}^{*} }}{\partial m} = \frac{{m^{2} \left( {2m - 3\beta } \right)}}{{2\left( {2m - \beta } \right)^{3} }}$. Thus, when $0 < \delta < \frac{2}{3}$, $\frac{{\partial CS_{P}^{*} }}{\partial m} > 0$. When $\frac{2}{3} < \delta < 1$, if $0 < \beta < \frac{2m}{3}$, then $\frac{{\partial CS_{P}^{*} }}{\partial m} > 0$; if $\frac{2m}{3} < \beta < \delta m$, then $\frac{{\partial CS_{P}^{*} }}{\partial m} < 0$.

$\frac{{\partial CS_{H}^{*} }}{\partial m} = \frac{{\left( {\delta - 1} \right)\beta^{3} - 6\delta m\left( {\delta - 1} \right)\beta^{2} - 12\delta^{2} m^{2} \beta + 8\delta^{3} m^{3} }}{{8\left( {2\delta m - \beta } \right)^{3} }}$,

$f_{3} \left( \beta \right) = \left( {\delta - 1} \right)\beta^{3} - 6\delta m\left( {\delta - 1} \right)\beta^{2} - 12\delta^{2} m^{2} \beta + 8\delta^{3} m^{3}$.

Since $\frac{{\partial f_{3} \left( \beta \right)}}{\partial \beta } = 3\left( {\delta - 1} \right)\beta^{2} - 12\delta m\left( {\delta - 1} \right)\beta - 12\delta^{2} m^{2} < 0$ if $\beta < \delta m$, $f_{3} \left( \beta \right)$ decreases with $\beta \in \left( {0,\delta m} \right)$. $f_{3} \left( \beta \right)|_{\beta = \delta m} = \delta^{3} m^{3} \left( {1 - 5\delta } \right)$: if $0 < \delta < \frac{1}{5}$, $f_{3} \left( \beta \right)|_{\beta = \delta m} > 0$; if $\frac{1}{5} < \delta < 1$, $f_{3} \left( \beta \right)|_{\beta = \delta m} < 0$. $\frac{{2\delta m\left( {\left( {1 - \delta } \right)\left( {\delta_{1} + \delta } \right) - \delta_{1}^{2} } \right)}}{{\delta_{1} \left( {1 - \delta } \right)}} \in \left( {0,\delta m} \right)$ is the root of $f_{3} \left( \beta \right) = 0$, where $\delta_{1} = \left( {\delta \left( {1 - \delta } \right)^{2} \left( {1 + \delta_{11} } \right)} \right)^{\frac{1}{3}}$ and $\delta_{11} = \sqrt {\frac{1}{1 - \delta }}$.

Thus, when $0 < \delta < \frac{1}{5}$, $\frac{{\partial CS_{H}^{*} }}{\partial m} > 0$. When $\frac{1}{5} < \delta < 1$, if $0 < \beta < \frac{{2\delta m\left( {\left( {1 - \delta } \right)\left( {\delta_{1} + \delta } \right) - \delta_{1}^{2} } \right)}}{{\delta_{1} \left( {1 - \delta } \right)}}$, then $\frac{{\partial CS_{H}^{*} }}{\partial m} > 0$; if $\frac{{2\delta m\left( {\left( {1 - \delta } \right)\left( {\delta_{1} + \delta } \right) - \delta_{1}^{2} } \right)}}{{\delta_{1} \left( {1 - \delta } \right)}} < \beta < \delta m$, then $\frac{{\partial CS_{H}^{*} }}{\partial m} < 0$, where $\delta_{1} = \left( {\delta \left( {1 - \delta } \right)^{2} \left( {1 + \delta_{11} } \right)} \right)^{\frac{1}{3}}$ and $\delta_{11} = \sqrt {\frac{1}{1 - \delta }}$. □

Proof of Corollary 3

• 1. $\frac{{\partial p_{H}^{*} }}{\partial \delta } = \frac{{m\beta^{2} }}{{2\left( {2\delta m - \beta } \right)^{2} }} > \frac{{\partial p_{P}^{*} }}{\partial \delta } = 0$. Thus, $\frac{{\partial p_{H}^{*} }}{\partial \delta } > \frac{{\partial p_{P}^{*} }}{\partial \delta } = 0$.

• 2. $\frac{{\partial r_{F}^{*} }}{\partial \delta } = \frac{{\partial r_{H}^{*} }}{\partial \delta } = \frac{{m\left( {2\delta^{2} m^{2} - 2\beta \delta m + \beta^{2} } \right)}}{{\left( {2\delta m - \beta } \right)^{2} }} > \frac{{\delta^{2} m^{3} }}{{\left( {2\delta m - \beta } \right)^{2} }} > 0$. Thus, $\frac{{\partial r_{F}^{*} }}{\partial \delta } = \frac{{\partial r_{H}^{*} }}{\partial \delta } > 0$.

• 3. $\frac{{\partial q_{H}^{*} }}{\partial \delta } = \frac{{\partial q_{F}^{*} }}{\partial \delta } = - \frac{\beta m}{{\left( {2\delta m - \beta } \right)^{2} }} < 0$, $\frac{{\partial q_{P}^{*} }}{\partial \delta } = 0$. Thus, $\frac{{\partial q_{H}^{*} }}{\partial \delta } = \frac{{\partial q_{F}^{*} }}{\partial \delta } < \frac{{\partial q_{P}^{*} }}{\partial \delta } = 0$.

• 4. $\frac{{\partial \pi_{F}^{*} }}{\partial \delta } = \frac{{\delta m^{2} \left( {\delta m - \beta } \right)}}{{\left( {2\delta m - \beta } \right)^{2} }} > 0$, $\frac{{\partial \pi_{P}^{*} }}{\partial \delta } = 0$, $\frac{{\partial \pi_{H}^{*} }}{\partial \delta } = - \frac{{m\beta^{2} }}{{4\left( {2\delta m - \beta } \right)^{2} }} < 0$. Thus, $\frac{{\partial \pi_{F}^{*} }}{\partial \delta } > \frac{{\partial \pi_{P}^{*} }}{\partial \delta } = 0$, $\frac{{\partial \pi_{H}^{*} }}{\partial \delta } < 0$.

• 5. $\frac{{\partial CS_{P}^{*} }}{\partial \delta } = 0$. $\frac{{\partial CS_{F}^{*} }}{\partial \delta } = \frac{{\delta^{2} m^{3} \left( {2\delta m - 3\beta } \right)}}{{2\left( {2\delta m - \beta } \right)^{3} }}$, Thus, if $\beta < \frac{2\delta m}{3}$, then $\frac{{\partial CS_{F}^{*} }}{\partial \delta } > 0$; if $\frac{2\delta m}{3} < \beta < \delta m$, then $\frac{{\partial CS_{F}^{*} }}{\partial \delta } < 0$. $\frac{{\partial CS_{H}^{*} }}{\partial \delta } = - \frac{{m\beta^{2} \left( {6\delta m - \beta } \right)}}{{8\left( {2\delta m - \beta } \right)^{3} }} < 0$ since $\beta < \delta m$. Thus, $\frac{{\partial CS_{H}^{*} }}{\partial \delta } < 0$. □

1.1 Proof of endogenous content quality

All decisions are decided by the media platform, and thus whether the game sequence is simultaneous or sequential makes no mathematical difference. For convenience, we use the sequential game. Thus, based on the equilibrium profit with exogenous content quality, we can further derive the optimal content quality via the first-order derivative condition. The optimal content quality among different strategies are as follows:${m}_{P}^{E*}=\beta$,${m}_{F}^{E*}=\frac{\beta }{\delta }$,${m}_{H}^{E*}=\frac{\beta \left(1+\sqrt{\delta }\right)}{2\delta }$. substituting the optimal content quality into the equilibrium solutions with exogenous content quality, the equilibrium profit and consumer surplus with endogenous content quality are as follows. Strategy P: $\pi_{P}^{E*} = \frac{\beta }{2}$, $CS_{P}^{E*} = \frac{\beta }{2}$. Startegy F: $\pi_{F}^{E*} = \frac{\beta }{2}$, $CS_{F}^{E*} = \frac{\beta }{2}$. Strategy H: $\pi_{H}^{E*} = \frac{{\beta \left( {1 + \sqrt \delta } \right)^{2} }}{8\delta }$, $CS_{H}^{E*} = \frac{{\beta \left( {1 + \sqrt \delta } \right)^{2} }}{8\delta }$.

$m_{F}^{E*} - m_{H}^{E*} = \frac{{\beta \left( {1 - \sqrt \delta } \right)}}{2\delta } > 0$, $m_{H}^{E*} - m_{P}^{E*} = \frac{{\beta \left( {1 + \sqrt \delta - 2\delta } \right)}}{2\delta } > 0$ since $0 < \delta < 1$. Thus, $m_{F}^{E*} > m_{H}^{E*} > m_{P}^{E*}$, which proves Proposition 5.

$\pi_{H}^{E*} - \pi_{P}^{E*} = \frac{{\beta \left( {1 + 2\sqrt \delta - 3\delta } \right)}}{8\delta } > 0$ since $0 < \delta < 1$. Thus, we can conclude that $\pi_{H}^{E*} > \pi_{P}^{E*} = \pi_{F}^{E*}$, $CS_{H}^{E*} > CS_{P}^{E*} > CS_{F}^{E*}$, which proves Proposition 6. □

1.2 Proof of platforms competition

With the goal of maximizing profit, Media platform A and media platform B compete with a Nash game and simultaneously make their decisions. Hence, by solving equations $\frac{\partial {\pi }^{A}}{\partial {r}^{A}}=\frac{\partial {\pi }^{A}}{\partial {p}^{A}}=\frac{\partial {\pi }^{B}}{\partial {p}^{B}}=0$, we can obtain the equilibrium decisions for two competing media platforms: ${r}^{A*}=\frac{m\left(m-\beta \right)}{2m-\beta }$, ${p}^{A*}=\frac{m\left(m-\beta \right)}{2m-\beta }$, ${p}^{B*}=\delta m$. Then, the equilibrium profits and consumer surplus are as follows: ${\pi }^{A*}=\frac{{m}^{2}}{2\left(2m-\beta \right)}$, ${\pi }^{B*}=0$, $C{S}^{*}=\frac{{m}^{3}}{2{\left(2m-\beta \right)}^{2}}$. As a result, when the platform B using strategy P competes with the platform A using strategy H, the platform A uses the joint decisions of price and ads to attract all customers, leaving the platform B with zero order (${q}_{P}^{B*}=0$) and zero profit (${\pi }^{B*}=0$). Proposition 7 (a) can be easily obtained. The proof of Proposition 7 (b) is similar to that of Corollary 3 (5). □