Innovation investment and subsidy strategy in two-sided market

Abstract

This paper investigates two competitive strategies from two-sides of the e-commerce platform, that is, innovation investment on seller side and product subsidy investment on consumer side. We take competition intensity on seller side into account and analyze how consumer behaviors affect the platform’s strategy under three scenarios: (1) single purchase on single platform(S); (2) single purchase on multi-platforms(M); (3) repeat purchase on single platform (R). The results revel that the innovation investment for sellers is better off in S scenario. However, when the transfer cost is low, taking subsidy strategy is more profitable for the platform in R scenario. If the internal price competition is not sufficiently fierce, subsidy strategy is an efficient approach to reduce the price in M scenario. It is surprising that if the seller’s innovation capability is sufficiently high, the innovation investment strategy dominates no matter what consumer behaviors are. Moreover, how much the platform invests on the seller’s innovation is independent on the consumer’s behavior. These findings have practical managerial insights for the manager of platforms.

Appendices

Appendix 1

Equilibrium Outcomes

2.1 (a) Single purchase from single platform

The seller’s utility functions on platform 1 and platform 2 are given by

$$ U_{s1} = \alpha_{s} \tilde{n}_{b1} + \left( {p - \gamma n_{s1} } \right)n_{b1} - F_{1} + e_{1} - \theta_{j1} $$

and.

$U_{s2} = \alpha_{s} \tilde{n}_{b2} + \left( {p - \gamma n_{s2} } \right)n_{b2} - F_{2} - \theta_{j2}$.

The consumer’s utility function is given by

$$ U_{b1} = w + \alpha_{b} \tilde{n}_{s1} - \left( {p - \gamma n_{s1} } \right) - tx $$

and

$$U_{b2} = w + \alpha_{b} \tilde{n}_{s2} - \left( {p - \gamma n_{s2} - e_{2} } \right) - t\left( {1 - x} \right)$$

An indifferent consumer $\tilde{x}$ is under the equilibrium of $U_{b1} \left( {\tilde{x}} \right) = U_{b2} \left( {\tilde{x}} \right)$ when it accomplishes game equilibrium. According to consumer rational expectation theory, we can get $\tilde{n}_{b1} = n_{b1} = \tilde{x}$. So we get

$$ w + \alpha_{b} \tilde{n}_{s1} - \left( {p - \gamma n_{s1} } \right) - tn_{b1} = w + \alpha_{b} \tilde{n}_{s2} - \left( {p - \gamma n_{s2} - e_{2} } \right) - t\left( {1 - n_{b1} } \right) $$

Then, we get the number of conusmers as

$$ n_{b1} = \frac{1}{2} + \frac{{\left( {\alpha_{b} + \gamma } \right)\left( {n_{s1} - n_{s2} - e_{2} } \right)}}{2t} $$

(27)

and according to $n_{b2} = 1 - n_{b1}$, we can get

$$ n_{b2} = \frac{1}{2} - \frac{{\left( {\alpha_{b} + \gamma } \right)\left( {n_{s1} - n_{s2} - e_{2} } \right)}}{2t} $$

(28)

The demand function on the seller side on platform 1,2 are

$$ n_{s1} = \frac{{\alpha_{s} \tilde{n}_{b1} + \left( {p - \gamma n_{s1} } \right)n_{b1} - F_{1} + e_{1} }}{\theta }, $$

$$ n_{s2} = \frac{{\alpha_{s} \tilde{n}_{b2} + \left( {p - \gamma n_{s2} } \right)n_{b2} - F_{2} }}{\theta } $$

separately.

So

$$ n_{s1} = \frac{{\alpha_{s} \tilde{n}_{b1} + \left( {p - \gamma n_{s1} } \right)\left[ {\frac{1}{2} + \frac{{\left( {\alpha_{b} + \gamma } \right)\left( {n_{s1} - n_{s2} - e_{2} } \right)}}{2t}} \right] - F_{1} + e_{1} }}{\theta } $$

$$ n_{s2} = \frac{{\alpha_{s} \tilde{n}_{b2} + \left( {p - \gamma n_{s2} } \right)\left[ {\frac{1}{2} - \frac{{\left( {\alpha_{b} + \gamma } \right)\left( {n_{s1} - n_{s2} - e_{2} } \right)}}{2t}} \right] - F_{2} }}{\theta } $$

According to $U_{s1} = U_{s2}$, substitute $A1$ and $A2$ into $U_{s1}$ and $U_{s2}$. Then substitute $n_{s1}$ and $n_{s2}$ into $\pi_{1}$ and $\pi_{2}$. First, platform1 determines the innovation level $e_{1}$ and platform2 determines subsidy $e_{2}$. Then, two platforms determine the access fee simultaneously. We use Backward Induction to solve the problem.

We have Hessian Metric $\left| {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{1}^{2} }}} & {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{1} \partial F_{2} }}} \\ {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{2} \partial F_{1} }}} & {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{2}^{2} }}} \\ \end{array} } \right| > 0$, and $\left| {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{1}^{2} }}} & {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{1} \partial F_{2} }}} \\ {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{2} \partial F_{1} }}} & {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{2}^{2} }}} \\ \end{array} } \right| > 0$, therefore, $\pi_{1}$ and $\pi_{2}$ are concave function on $F_{1}$ and $F_{2}$.

Thus, the optimal access fee under $\frac{{\partial \pi_{i} }}{{\partial F_{i} }} = 0$ is given by

$$ F_{1}^{{s{*}}} = \frac{{k\alpha_{s} + 2p}}{4k} $$

$$ F_{2}^{{s{*}}} = \frac{{k\alpha_{s} + 2p}}{2k - 1}. $$

The optimal investment level for both platforms are

$$ e_{1}^{{s{*}}} = \frac{1}{k}, $$

$$ e_{2}^{{s{*}}} = \frac{{\alpha_{s} + 2p}}{2k - 1}. $$

By maximizing profits for platform 1 and 2, we get the following:

$$ \pi_{1}^{{s{*}}} = \frac{{k^{3} \left( {\alpha_{s} + 2p} \right)^{2} \left[ {2t\left( {\theta + \gamma } \right) - \alpha_{s} \left( {\gamma + \alpha_{d} } \right)} \right]}}{{8\left( {k - 1} \right)^{2} \left( {\theta + \gamma } \right)\left[ {\alpha_{s} \left( {\gamma + \alpha_{d} } \right) - t\left( {\theta + \gamma } \right)} \right]}} $$

$$ \pi_{2}^{{s{*}}} = \frac{{k^{3} \left( {2k - 1} \right)\left( {\alpha_{s} + 2p} \right)\left( {\gamma + \alpha_{d} } \right)\alpha_{s} }}{{16\left( {k - 1} \right)^{2} \left( {\theta + \gamma } \right)[\alpha_{s} \left( {\gamma + \alpha_{d} } \right) - t\left( {\theta + \gamma } \right)]^{2} }}. $$

(b) Single purchase from multi-platform

The seller’s utility functions on platform 1 and platform 2 are given by

$$ U_{s1} = \alpha_{s} \tilde{n}_{b1} + \left( {p - \gamma n_{s1} } \right)n_{b1} - F_{1} + e_{1} - \theta_{j1} $$

and

$$U_{s2} = \alpha_{s} \tilde{n}_{b2} + \left( {p - \gamma n_{s2} } \right)n_{b2} - F_{2} - \theta_{j2}$$

The consumer’s utility function is given by

$$U_{b1} = w + \alpha_{b} \tilde{n}_{s1} - \left( {p - \gamma n_{s1} } \right) - tx,$$

$$ U_{b2} = w + \alpha_{b} \tilde{n}_{s2} - \left( {p - \gamma n_{s2} - e_{2} } \right) - t\left( {1 - x} \right) $$

and

$$ U_{b1,2} = 2w + \alpha_{b} \tilde{n}_{s} - \left( {2p - \gamma n_{s} - e_{2} } \right) - t $$

Given the indifference location between 1 and two platforms is $\tilde{x}_{l}$, the Nash equilibrium is $U_{b1} \left( {\tilde{x}_{l} } \right) = U_{b1,2} \left( {\tilde{x}_{l} } \right)$; given the indifference location in platform 2 or two platforms is $\tilde{x}_{r}$, the Nash equilibrium is $U_{b2} \left( {\tilde{x}_{r} } \right) = U_{b1,2} \left( {\tilde{x}_{r} } \right)$. According to the consumer rational expectation theory, we can get $\tilde{n}_{b1} = n_{b1} = \tilde{x}_{r}$. Using the traditional Hottling model theory and consumer rational expectation hypothesis, we get

$$ n_{b1} = \frac{{w + \alpha_{b} \tilde{n}_{s1} - \left( {p - \gamma n_{s1} } \right)}}{t} $$

(29)

$$ n_{b2} = \frac{{w + \alpha_{b} \tilde{n}_{s2} - \left( {p - \gamma n_{s2} } \right) + e_{2} }}{t}, $$

(30)

Based on $n_{b1} + n_{b2} - n_{b1,2} = 1$, we can get

$$ n_{b1,2} = \frac{{2w + \alpha_{b} \tilde{n}_{s} - \left( {2p - \gamma n_{s} } \right) + e_{2} - t}}{t} $$

(31)

According to $U_{s1} = U_{s2}$, substitute $A3,A4$ and $A5$ into $U_{s1}$ and $U_{s2}$, we get $n_{s1}$ and $n_{s2}$.

$$ n_{s1} = \frac{{\alpha_{s} \left( {p - w} \right) - t\left( {e_{1} - F_{1} + p} \right)}}{{\alpha_{s} \left( {\gamma + \alpha_{b} } \right) - t\left( {\theta + \gamma } \right)}}, $$

$$ n_{s2} = \frac{{\alpha_{s} \left( {p - w - e_{2} } \right) - t\left( {p - F_{2} } \right)}}{{\alpha_{s} \left( {\gamma + \alpha_{b} } \right) - t\left( {\theta + \gamma } \right)}}. $$

Then substitute $n_{s1}$ and $n_{s2}$ into $\pi_{1}$ and $\pi_{2}$. First, platform1 determines the innovation level $e_{1}$ and platform2 determines subsidy $e_{2}$. Then, two platforms determine the access fee simultaneously. We use Backward Induction to solve the problem.

We have Hessian Metric $\left| {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{1}^{2} }}} & {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{1} \partial F_{2} }}} \\ {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{2} \partial F_{1} }}} & {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{2}^{2} }}} \\ \end{array} } \right| > 0$, and $\left| {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{1}^{2} }}} & {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{1} \partial F_{2} }}} \\ {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{2} \partial F_{1} }}} & {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{2}^{2} }}} \\ \end{array} } \right| > 0$, therefore, $\pi_{1}$ and $\pi_{2}$ are concave function on $F_{1}$ and $F_{2}$.

Thus, the optimal access fee under $\frac{{\partial \pi_{i} }}{{\partial F_{i} }} = 0$ is given by

$$ F_{1}^{{M{*}}} = \frac{{k\left( {\theta p - \gamma } \right) + w}}{\theta k}, $$

$$ F_{2}^{{M{*}}} = \frac{{k\left( {\theta p - \theta - \gamma } \right) + w}}{\theta k}. $$

The optimal investment level for both platforms are

$$ e_{1}^{{M{*}}} = \frac{1}{k}, $$

$$ e_{2}^{{M{*}}} = \frac{{k\left( {\theta p - \alpha_{b} + \gamma } \right) - w\left( {k\theta + 1} \right)}}{\theta k}. $$

By maximizing profits for platform 1 and 2, we get

$$ \pi_{1}^{{M{*}}} = \frac{{\left[ {2k\left( {\theta p - \gamma } \right) + 2w - \theta } \right]\left[ {\theta k\alpha_{s} \left( {p - w} \right) + t\left( {w - \theta - k\gamma } \right)} \right]}}{{2k^{2} \theta^{2} \left[ {\alpha_{s} \left( {\gamma + \alpha_{b} } \right) - t\left( {\theta + \gamma } \right)} \right]}} $$

$$ \pi_{2}^{{M{*}}} = \frac{{t\left[ {k\left( {\theta p - \theta - \gamma } \right) + w} \right] - \left[ {\theta \left( {w - p} \right) + \alpha_{b} + \gamma } \right]\left[ {k\left( {\theta p - \alpha_{b} - \gamma } \right) - w\left( {k\theta + 1} \right)} \right]}}{{\theta^{2} kt}}. $$

(c) Repeat purchase

The second period utility function of the consumers who purchase from platform 1 in the first period is

$$ U_{{b1\left( 2 \right)}} = \left\{ {\begin{array}{*{20}c} {w + \alpha _{b} \tilde{n}_{{s1\left( 2 \right)}} - \left( {p - \gamma n_{{s1\left( 2 \right)}} } \right) - tx\;platform1} \\ {w + \alpha _{b} \tilde{n}_{{s2\left( 2 \right)}} - \left( {p - \gamma n_{{s2\left( 2 \right)}} - e_{2} } \right) - t\left( {1 - x} \right) - s\;platform2} \\ \end{array} } \right. $$

The second period utility function of the consumers who purchase from platform 2 in the first period is

$$ U_{{b_{{2\left( 2 \right)}} }} = \left\{ {\begin{array}{*{20}c} {w + \alpha _{b} \tilde{n}_{{s_{{1\left( 2 \right)}} }} - \left( {p - \gamma n_{{s1\left( 2 \right)}} } \right) - tx - s\;platform1} \\ {w + \alpha _{b} \tilde{n}_{{s_{{2\left( 2 \right)}} }} - \left( {p - \gamma n_{{s_{{2\left( 2 \right)}} }} - e_{2} } \right) - t\left( {1 - x} \right)\;platform2} \\ \end{array} } \right. $$

The seller’s utility function in the second period is

$$ U_{{s_{{\left( 2 \right)}} }} = \left\{ {\begin{array}{*{20}c} {\alpha \tilde{n}_{{b_{{1\left( 2 \right)}} }} + \left( {p - \gamma n_{{s_{{1\left( 2 \right)}} }} } \right)n_{{b_{{1\left( 2 \right)}} }} - F_{{1\left( 2 \right)}} + e_{1} \;platform1} \\ {\alpha \tilde{n}_{{b_{{2\left( 2 \right)}} }} + \left( {p - \gamma n_{{s_{{2\left( 2 \right)}} }} } \right)n_{{b_{{2\left( 2 \right)}} }} - F_{{2\left( 2 \right)}} \;platform2} \\ \end{array} } \right. $$

According to consumer rational expectation theory, the number of $\eta$ percent consumers who choose platform 1 and 2 in the second period are

$$ n_{b1\left( 2 \right)}^{\eta } = \eta \frac{{\left( {\alpha_{b} + \gamma } \right)\left( {n_{s1\left( 2 \right)} - n_{s2\left( 2 \right)} } \right) - e_{2} + t}}{2t} + \frac{\eta s}{{2t}}\left( {n_{b1\left( 2 \right)} - n_{b2\left( 2 \right)} } \right) $$

(32)

$$ n_{b2\left( 2 \right)}^{\eta } = \eta \frac{{\left( {\alpha_{b} + \gamma } \right)\left( {n_{s2\left( 2 \right)} - n_{s1\left( 2 \right)} } \right) + e_{2} + t}}{2t} + \frac{\eta s}{{2t}}\left( {n_{b2\left( 2 \right)} - n_{b1\left( 2 \right)} } \right) $$

(33)

We have $\left( {1 - \eta } \right)$ percent consumers keeping the same choice for two periods. Therefore, the number of consumers who choose platform1 and platform 2 in the second period are as follows.

$$ n_{b1\left( 2 \right)}^{1 - \eta } = \left( {1 - \eta } \right)n_{b1\left( 1 \right)} $$

$$ n_{b2\left( 2 \right)}^{1 - \eta } = \left( {1 - \eta } \right)n_{b2\left( 1 \right)} $$

The number of consumers who choose platform 1 and platform 2 in the second period are

$$ n_{b1\left( 2 \right)} = n_{b1\left( 2 \right)}^{\eta } + n_{b1\left( 2 \right)}^{1 - \eta } $$

$$ n_{b2\left( 2 \right)} = n_{b2\left( 2 \right)}^{\eta } + n_{b2\left( 2 \right)}^{1 - \eta } $$

We assume consumers are shortsighted and they are not concerned about the impact of their decision in the first period on the second period. Let $\sigma$ denotes the discount factor of future profit. Then the platform i’s profit function in the first period is

$$ \pi_{i\left( 1 \right)} = F_{i\left( 1 \right)} n_{si\left( 1 \right)} + \sigma \pi_{i\left( 2 \right)} $$

The number of consumers who choose platform 1 and platform 2 in the second period are shown in equations $A6$ and $A7$, respectively. The platform determines the innovation level and the access fee at the same time. In the first period, platform 1,2 determines the innovation levels and the access fee at the same time. The profit functions of platform 1 and platform 2, respectively, are as follows:

$$ \pi_{{1_{\left( 1 \right)} }} = \left( {F_{{1_{\left( 1 \right)} }} - \frac{1}{2}ke_{1\left( 1 \right)}^{2} } \right)n_{{s_{1\left( 1 \right)} }} + \sigma \pi_{{1_{\left( 2 \right)} }} $$

$$ \pi_{{2_{\left( 1 \right)} }} = F_{{2_{\left( 1 \right)} }} n_{{s_{2\left( 1 \right)} }} - e_{2\left( 1 \right)}^{{}} n_{{b_{2\left( 1 \right)} }} + \sigma \pi_{{2_{\left( 2 \right)} }} $$

We have Hessian Metric $\left| {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{1}^{2} }}} & {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{1} \partial F_{2} }}} \\ {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{2} \partial F_{1} }}} & {\frac{{\partial^{2} \pi_{1} }}{{\partial F_{2}^{2} }}} \\ \end{array} } \right| > 0$, and $\left| {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{1}^{2} }}} & {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{1} \partial F_{2} }}} \\ {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{2} \partial F_{1} }}} & {\frac{{\partial^{2} \pi_{2} }}{{\partial F_{2}^{2} }}} \\ \end{array} } \right| > 0$, therefore, $\pi_{1}$. and $\pi_{2}$ are concave function on $F_{1}$ and $F_{2}$.

For the second period, we get the optimal innovation level of platform 1,2 as

$$ e_{1\left( 2 \right)}^{{R{*}}} = \frac{1}{k} $$

$$ e_{2\left( 2 \right)}^{{R{*}}} = \frac{s\eta }{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}} $$

Given the same utility for sellers when they join two platforms under equilibrium, and $\frac{{\partial \pi_{i\left( 2 \right)} }}{{\partial F_{i\left( 2 \right)} }} = 0$, we can get the access fee as following:

$$ F_{1\left( 2 \right)}^{{R{*}}} - F_{2\left( 2 \right)}^{{R{*}}} = \frac{{\eta \left[ {s\left( {\alpha_{s} + 2p} \right) + s^{2} \eta - \left( {k + 3} \right)} \right]}}{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}} $$

And the optimal profit is

$$ \frac{\eta \left[ \left( \alpha_{s} + 2p \right){\mathcal{R}} -k(p+\alpha_{s} ) \left( \gamma + \alpha_b \right) \right]}{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}$$

where ${\mathcal{R}} = 3s^{2} - \left( {\gamma + \alpha_{b} } \right)s.$

For the first period, the optimal innovation level of platform 1,2 are

$$ e_{1\left( 1 \right)}^{{R{*}}} = \frac{1}{k}, $$

$$ e_{2\left( 1 \right)}^{{R{*}}} = \frac{{\alpha_{s} + 2p}}{{\left( {2k - 1} \right)\eta }}. $$

The difference of optimal access fees of platforms ($i = 1,2$) is

$$ F_{1\left( 1 \right)}^{{R{*}}} - F_{2\left( 1 \right)}^{{R{*}}} = \frac{{\eta \left[ {s\left( {\alpha_{s} + 2p} \right) + k\left( {p + \alpha_{s} } \right)\left( {\gamma + \alpha_{b} } \right)} \right]}}{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}} $$

Appendix 2

5.1 Proof of Propositions

5.1.1 Proof of Proposition 1

From expressions of ${F}_{1}^{s*}$ and ${F}_{2}^{s*}$, we have ${F}_{1}^{s*}-{F}_{2}^{s*}=-\frac{\left(k{\alpha }_{s}+2p\right)\left(2k+1\right)}{4k\left(2k-1\right)}<0$, which implies that ${F}_{1}^{s*}<{F}_{2}^{s*}.$

According to ${\pi }_{1}^{s*}$ and${\pi }_{2}^{s*}$, we set ${\mathcal{H}} = \frac{{\pi_{1}^{{s{*}}} }}{{\pi_{2}^{{s{*}}} }} = \frac{{2\left( {\alpha_{s} + 2p} \right)\left[ {2t\left( {\theta + \gamma } \right) - \alpha_{s} \left( {\gamma + \alpha_{d} } \right)} \right]\left[ {\alpha_{s} \left( {\gamma + \alpha_{d} } \right) - t\left( {\theta + \gamma } \right)} \right]}}{{\left( {2k - 1} \right)\left( {\gamma + \alpha_{d} } \right)\alpha_{s} }}$. If $p > \tilde{p}$, where $\tilde{p} = \frac{{\left( {2k - 1} \right)\left( {\gamma + \alpha_{d} } \right)\alpha_{s} }}{{4\left[ {2t\left( {\theta + \gamma } \right) - \alpha_{s} \left( {\gamma + \alpha_{d} } \right)} \right]\left[ {\alpha_{s} \left( {\gamma + \alpha_{d} } \right) - t\left( {\theta + \gamma } \right)} \right]}} - \frac{{\alpha_{s} }}{2}$, we obtain ${\mathcal{H}} > 1,$ which implies that $\pi_{1}^{{s{*}}}$ > $\pi_{2}^{{s{*}}}$.

5.2 Proof of Proposition 2

• (1) According to expression of $n_{b1}$ and $n_{b2}$, there exist $\hat{\gamma }$, when $\gamma < \hat{\gamma }$, then $n_{b1} < n_{b2}$; while when $\gamma > \hat{\gamma }$, then $n_{b1} > n_{b2}$.

• (2) From $F_{1}^{M*} and F_{2}^{M*}$, it’s obvious that $F_{1}^{M*} > F_{2}^{M*}$.

• (3) According to $\pi_{1}^{M*}$ and $\pi_{2}^{M*}$, we have

  $$ {\mathcal{B}} = \pi_{1}^{M*} - \pi_{2}^{M*} = \frac{{t\left[ {2k\left( {\theta p - \gamma } \right) + 2w - \theta } \right]\left[ {\theta k\alpha_{s} \left( {p - w} \right) + t\left( {w - \theta - k\gamma } \right)} \right] - A}}{{2k^{2} \theta^{2} \left[ {\alpha_{s} \left( {\gamma + \alpha_{b} } \right) - t\left( {\theta + \gamma } \right)} \right]t}}, $$

where $A = 2k\left[ {\alpha_{s} \left( {\gamma + \alpha_{b} } \right) - t\left( {\theta + \gamma } \right)} \right]\left\{ {t\left[ {k\left( {\theta p - \theta - \gamma } \right) + w} \right] - \left[ {\theta \left( {w - p} \right) + \alpha_{b} + \gamma } \right]\left[ {k\left( {\theta p - \alpha_{b} - \gamma } \right) - w\left( {k\theta + 1} \right)} \right]} \right\}$.

Thus, there exist $\hat{k}$, when $k > \hat{k}$, we have ${\mathcal{B}} < 0$, which implies that when innovation capability is not so high, subsidy strategy dominates.

5.3 Proof of Proposition 3

(1) According to $\pi_{1\left( 2 \right)}^{R*} - \pi_{2\left( 2 \right)}^{R*} = \frac{{\eta \left[ {\left( {\alpha_{s} + 2p} \right)\left( {3s^{2} - \left( {\gamma + \alpha_{b} } \right)s} \right) - k\left( {p + \alpha_{s} } \right)\left( {\gamma + \alpha_{b} } \right)} \right]}}{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}},$ which is concave function on s. Thus, there exist $\hat{s}$, when $s < \hat{s}$, we get $\pi_{1\left( 2 \right)}^{R*} < \pi_{2\left( 2 \right)}^{R*}$; when $s > \hat{s}$, we get $\pi_{1\left( 2 \right)}^{R*} < \pi_{2\left( 2 \right)}^{R*}$.

(2) It’s obvious that $e_{1\left( 1 \right)}^{R*} = e_{1\left( 2 \right)}^{R*} = \frac{1}{k}$.

(3) Since $F_{1\left( 1 \right)}^{R*} - F_{2\left( 1 \right)}^{R*} = \frac{{\eta \left[ {s\left( {\alpha_{s} + 2p} \right) + k\left( {p + \alpha_{s} } \right)\left( {\gamma + \alpha_{b} } \right)} \right]}}{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}} > 0$, thus platform 1 charges higher access fee than platform 2. From $F_{1\left( 2 \right)}^{R*} - F_{2\left( 2 \right)}^{R*} = \frac{{\eta \left[ {s\left( {\alpha_{s} + 2p} \right) + s^{2} \eta - \left( {k + 3} \right)} \right]}}{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}}$, there exist $\tilde{s}$, when $s < \tilde{s}$, we get $\frac{{\eta \left[ {s\left( {\alpha_{s} + 2p} \right) + s^{2} \eta - \left( {k + 3} \right)} \right]}}{{\left[ {2\left( {1 - \eta } \right) - 2t + k\left( {\gamma + \alpha_{b} } \right)} \right]}} < 0.$

5.4 Proof of Proposition 4

According to $e_{i}^{S*} = e_{i}^{M*} = e_{i}^{R*} = \frac{1}{k}$, it’s easy to get this result.