Comparison between the agency and wholesale model under the e-book duopoly market

Abstract

This paper provides a comprehensive comparison between the agency and wholesale model under the electronic book market duopoly. A model comprised of two retailers and a publisher was established to run the comparison in terms of price, profit, and welfare. We found that although better e-reader offerings may seem favorable in terms of e-book pricing, the actual effect on e-books is limited or even negligible. High wholesale price leads to higher retail prices of e-books and e-readers under the wholesale model than the agency model. Generally speaking, the total profits of e-books returned to the retailers under the wholesale model are lower than those returned under the agency model; conversely, the publisher obtains higher profits under the wholesale model than the agency model despite its power over e-book pricing. We also found that consumers are more likely to derive greater surplus under the wholesale model when the difference in available e-books is relatively narrow. If the wholesale price is low, however, the consumer surplus under the wholesale model is higher than that under the agency model across the board. Further, when the difference in e-books is very small, the total social welfare is minimized under the wholesale model compared to the agency model; low wholesale price but poor substitutability (or high wholesale price with relatively high substitutability) yield favorable social welfare under the wholesale model.

Appendix

Proof of Proposition 1

To maximize the profits of two retailers, we solve $p_{1}^{W}$, $p_{1R}^{W}$, $p_{2}^{W}$, $p_{2R}^{W}$ from $\frac{{\partial \pi_{1}^{W} }}{{\partial p_{1} }} = 0$, $\frac{{\partial \pi_{1}^{W} }}{{\partial p_{1R} }} = 0$, $\frac{{\partial \pi_{2}^{W} }}{{\partial p_{2} }} = 0$, $\frac{{\partial \pi_{2}^{W} }}{{\partial p_{2R} }} = 0$. And we exam both the hesse matrices of $\pi_{1}^{W}$ and $\pi_{2}^{W}$ are $\left( {\begin{array}{*{20}c} { - 2} & { - \frac{1}{2t}} \\ { - \frac{1}{2t}} & { - \frac{1}{t}} \\ \end{array} } \right)$ and this matrix is negative. So, we obtain

$$\begin{aligned} p_{1}^{W} & = \frac{{\left( {1 + 2h + 2w} \right)}}{{2\left( {2 - \alpha } \right)}}{ + }\frac{{\left( {c_{2R} - c_{1R} + v_{d} } \right)}}{{2\left( {3t\alpha + 6t - 1} \right)}}, \\ p_{2}^{W} & = \frac{{\left( {1 + 2h + 2w} \right)}}{{2\left( {2 - \alpha } \right)}} - \frac{{\left( {c_{2R} - c_{1R} + v_{d} } \right)}}{{2\left( {3t\alpha + 6t - 1} \right)}}, \\ p_{1R}^{W} & = \frac{{\left( {2 - \alpha } \right)\left( {c_{1R} + c_{2R} + 2t} \right) + 2w - 2w\alpha - 1 - 2h}}{{2\left( {2 - \alpha } \right)}} - \frac{{t\left( {2 + \alpha } \right)\left( {c_{2R} - c_{1R} - 2v_{d} } \right) + v_{d} }}{{2\left( {3t\alpha + 6t - 1} \right)}}, \\ p_{2R}^{W} & = \frac{{\left( {2 - \alpha } \right)\left( {c_{1R} + c_{2R} + 2t} \right){ + }2w - 2w\alpha - 1 - 2h}}{{2\left( {2 - \alpha } \right)}} + \frac{{t\left( {\alpha + 2} \right)\left( {c_{2R} - c_{1R} - 2v_{d} } \right) + v_{d} }}{{2\left( {3t\alpha + 6t - 1} \right)}}. \\ \end{aligned}$$

Then, we put them into the profit functions and obtain the profits of publisher, retailer 1 by e-book 1, retailer 1 by e-reader 1, retailer 2 by e-book 2 and retailer 2 by e-reader 2:

$$\begin{aligned} \pi_{p}^{W} & = \frac{{\left( {c - w_{A} } \right)\left( {1 + 2h + 2w\left( {\alpha - 1} \right)} \right)}}{ - 2 + \alpha }, \quad \pi_{1B}^{W} = \frac{{\left[ {\left( {1 - 3t\left( {2 + \alpha } \right)} \right)\left( {2w_{A} - 2w\alpha - 2h - 1} \right) - \left( {\alpha - 2} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right]^{2} }}{{4\left( { - 2 + \alpha } \right)^{2} \left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}, \\ \pi_{1R}^{W} & = \frac{{\left( { - 1 + 3t\left( {2 + \alpha } \right) + \left( {2 + \alpha } \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right)\left[ {\left( {1 - 3t\left( {2 + \alpha } \right)} \right)\left( {\left( {\alpha - 2} \right)\left( {c_{1R} - c_{2R} - v_{d} - 2t} \right) - 1 - 2h + 2w - 2w\alpha } \right) + t\left( {\alpha^{2} - 4} \right)\left( {c_{1R} - c_{2R} - v_{d} } \right)} \right]}}{{4\left( { - 2 + \alpha } \right)\left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}, \\ \pi_{2B}^{W} & \quad = \frac{{\left[ {\left( {1 - 3t\left( {2 + \alpha } \right)} \right)\left( {2w - 2w\alpha - 2h - 1} \right) + \left( {\alpha - 2} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right]^{2} }}{{4\left( { - 2 + \alpha } \right)^{2} \left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}, \\ \pi_{2R}^{W} & \quad = \frac{{\left( { - 1 + 3t\left( {2 + \alpha } \right) + \left( {2 + \alpha } \right)\left( {c_{1R} - c_{2R} - v_{d} } \right)} \right)\left[ {\left( {1 - 3t\left( {2 + \alpha } \right)} \right)\left( {\left( { - 2 + \alpha } \right)\left( {c_{2R} - c_{1R} - 2t + v_{d} } \right) - 2h - 1 + 2w - 2w\alpha } \right) + t\left( {\alpha^{2} - 4} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right]}}{{4\left( { - 2 + \alpha } \right)\left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}. \\ \end{aligned}$$

Substituting $p_{2R}^{W} - p_{1R}^{W} = \frac{{t\left( {\alpha + 2} \right)\left( {c_{2R} - c_{1R} - 2v_{d} } \right) + v_{d} }}{3t\alpha + 6t - 1}$ into the first equation in Eq. (3), according to $0 \le x_{1}^{*} \le 1$, we get $\frac{1}{\alpha + 2} - 3t \le M = c_{2R} - c_{1R} + v_{d} \le 3t - \frac{1}{\alpha + 2}$. Here we assume t is sufficiently large to ensure that $0 < x_{1}^{*} < 1$.

Proof of Proposition 2

To maximize the profits of two retailers, we solve $p_{1}^{A}$, $p_{1R}^{A}$, $p_{2}^{A}$, $p_{2R}^{A}$ from $\frac{{\partial \pi_{P}^{A} }}{{\partial p_{1} }} = 0$, $\frac{{\partial \pi_{P}^{A} }}{{\partial p_{2} }} = 0$, $\frac{{\partial \pi_{1}^{A} }}{{\partial p_{1R} }} = 0$, $\frac{{\partial \pi_{2}^{A} }}{{\partial p_{2R} }} = 0$. And we exam both the Hessian matrices of $\pi_{P}^{A}$, $\pi_{1}^{A}$ and $\pi_{2}^{A}$ are $\left( {\begin{array}{*{20}c} { - 2r} & {2\alpha r} \\ {2\alpha r} & { - 2r} \\ \end{array} } \right)$, $\left( { - \frac{1}{t}} \right)$ and $\left( { - \frac{1}{t}} \right)$ and these matrices are all negative. So, we obtain

$$\begin{aligned} p_{1}^{A} & = \frac{{2c\left( {1 - \alpha } \right) + r\left( {2h + 1} \right)}}{{4r\left( {1 - \alpha } \right)}} + \frac{{\left( {v_{d} + c_{2R} - c_{1R} } \right)}}{{2\left[ {6t\left( {1 + \alpha } \right) - 1 + r} \right]}}, \\ p_{2}^{A} & = \frac{{2c\left( {1 - \alpha } \right) + r\left( {1 + 2h} \right)}}{{4r\left( {1 - \alpha } \right)}} - \frac{{v_{d} + c_{2R} - c_{1R} }}{{2\left( {6t\left( {1 + \alpha } \right) - 1 + r} \right)}}, \\ p_{1R}^{A} & = \frac{{\left( {r - 1} \right)\left( {r\left( {2h + 1} \right) - 2c\left( {\alpha - 1} \right)} \right) - 2r\left( {\alpha - 1} \right)\left( {c_{2R} + c_{1R} + 2t} \right)}}{{4r\left( {1 - \alpha } \right)}} \\&\quad + \frac{{v_{d} }}{2} - \frac{{t\left( {\alpha + 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)}}{{\left( {6t\left( {1 + \alpha } \right) - 1 + r} \right)}}, \\ p_{2R}^{A} & = \frac{{\left( {r - 1} \right)\left( {r\left( {2h + 1} \right) - 2c\left( {\alpha - 1} \right)} \right) - 2r\left( {\alpha - 1} \right)\left( {c_{1R} + c_{2R} + 2t} \right)}}{{4r\left( {1 - \alpha } \right)}} \\&\quad - \frac{{v_{d} }}{2} + \frac{{t\left( {\alpha + 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)}}{{\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)}}. \\ \end{aligned}$$

Then, we put them into the profit functions and obtain the profits of publisher, retailer 1 by e-book 1, retailer 1 by e-reader 1, retailer 2 by e-book 2 and retailer 2 by e-reader 2:

$$\begin{aligned} \pi_{1B}^{A} & = \frac{{\left( {r - 1} \right)\left[ {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)\left[ {r\left( {2h + 1} \right) - 2c\left( {\alpha - 1} \right)} \right] - 2r\left( {\alpha - 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right]M_{1} }}{{16r^{2} \left( { - 1 + \alpha } \right)\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}, \\ \pi_{1R}^{A} & = - \frac{{\left( {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right) + 2\left( {1 + \alpha } \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right)\left\{ {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)N + 4rt\left( {\alpha^{2} - 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right\}}}{{8r\left( { - 1 + \alpha } \right)\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}, \\ \pi_{2B}^{A} & = \frac{{\left( {r - 1} \right)\left[ {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)\left( {r\left( {2h + 1} \right) - 2c\left( {\alpha - 1} \right)} \right) + 2r\left( {\alpha - 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right]M_{2} }}{{16r^{2} \left( { - 1 + \alpha } \right)\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}, \\ \pi_{2R}^{A} & = - \frac{{\left( {r - 1 + 6t\left( {1 + \alpha } \right) + 2\left( {1 + \alpha } \right)\left( {c_{1R} - c_{2R} - v_{d} } \right)} \right)\left\{ {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)N + 4rt\left( {\alpha^{2} - 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right\}}}{{8r\left( { - 1 + \alpha } \right)\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}, \\ \end{aligned}$$

where

$$\begin{aligned} M_{1} & = \left[ {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)\left( {r\left( {2h + 1} \right) + 2c\left( {\alpha - 1} \right)} \right) + 2r\left( {1 + \alpha } \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right], \\ M_{2} & = \left[ {\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)\left( {r\left( {2h + 1} \right) + 2c\left( {\alpha - 1} \right)} \right) - 2r\left( {1 + \alpha } \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)} \right], \\ N & = \left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)\left[ {\left( {r - 1} \right)\left( {r\left( {2h + 1} \right) - 2c\left( {\alpha - 1} \right)} \right) + 2r\left( {\alpha - 1} \right)\left( {c_{1R} - c_{2R} - v_{d} - 2t} \right)} \right], \\ \pi_{p}^{A} & = \frac{{\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)^{2} \left[ {r\left( {2h + 1} \right) + 2c\left( {\alpha - 1} \right)} \right]^{2} - 4r^{2} \left( {\alpha^{2} - 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{8r\left( {1 - \alpha } \right)\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}. \\ \end{aligned}$$

Proof of Proposition 3

Let $\Delta_{B}^{P} = p_{1}^{W} { + }p_{2}^{W} - p_{1}^{A} - p_{2}^{A} = - \frac{{r\left( {4w\left( { - 1 + \alpha } \right) + \alpha + 2h\alpha } \right) + 2c\left( {2 - 3\alpha + \alpha^{2} } \right)}}{{2r\left( { - 2 + \alpha } \right)\left( { - 1 + \alpha } \right)}} > 0$, we obtain $w > \frac{{\alpha r\left( {1 + 2h} \right) + 2c\left( {1 - \alpha } \right)\left( {2 - \alpha } \right)}}{{4r\left( {1 - \alpha } \right)}}$.

In addition, it’s worth noting that by $\frac{{\partial \Delta_{B}^{P} }}{\partial \alpha } = \frac{2w}{{\left( {\alpha - 2} \right)^{2} }} - \frac{1 + 2h}{2}\frac{{2 - \alpha^{2} }}{{\left( {\alpha - 1} \right)^{2} \left( {\alpha - 2} \right)^{2} }} = \frac{{2w\left( {\alpha - 1} \right)^{2} - \left( {1/2 + h} \right)\left( {2 - \alpha^{2} } \right)}}{{\left( {\alpha - 1} \right)^{2} \left( {\alpha - 2} \right)^{2} }}$, we can find $\Delta_{B}^{P}$ is increasing in α in $\left( {0,\frac{{4w - \sqrt {4w\left( {1{ + }2h} \right) + 2\left( {1 + 2h} \right)^{2} } }}{4w + 1 + 2h}} \right)$ while decreasing in $\left[ {\frac{{4w - \sqrt {4w\left( {1{ + }2h} \right) + 2\left( {1 + 2h} \right)^{2} } }}{4w + 1 + 2h},1} \right)$. And when α = 0, we have $\Delta_{B}^{P} = \frac{rw - c}{r}$. Since the royalty rate generally is less than r, we suggest that $\frac{rw - c}{r} > 0$. When $\alpha \to 1$, we can suggest $\Delta_{B}^{P} \to - \infty$. According to Zero-point Theorem, we consider there must be a unique solution $\alpha^{*} > \frac{{4w - \sqrt {4w\left( {1{ + }2h} \right) + 2\left( {1 + 2h} \right)^{2} } }}{4w + 1 + 2h}$ to make $\Delta_{B}^{P} |_{{\alpha = \alpha^{*} }} { = 0}$ holds. By solving $\Delta_{B}^{P} \ge 0$ in α, we also get $\frac{{6c - 4rw - r\left( {1 + 2h} \right) - \sqrt \Delta }}{4c} < \alpha < \frac{{6c - 4rw - r\left( {1 + 2h} \right) + \sqrt \Delta }}{4c}$, where $\Delta = \left( {4rw - 6c + r\left( {1 + 2h} \right)} \right)^{2} { + }32c\left( {rw - c} \right)$. In conclusion, we find $0 < \alpha < \alpha^{*} = \frac{{6c - 4rw - r\left( {1 + 2h} \right) + \sqrt \Delta }}{4c} < 1$.

Proof of Proposition 4

Let $\Delta_{R}^{P} = p_{1R}^{W} { + }p_{2R}^{W} - p_{1R}^{A} - p_{2R}^{A} = \frac{{r\left( {\left( {1 + 2h} \right)r\left( {\alpha - 2} \right) + 4w\left( {\alpha - 1} \right)^{2} + \alpha + 2h\alpha } \right) - 2c\left( {r - 1} \right)\left( {2 - 3\alpha + \alpha^{2} } \right)}}{{2r\left( {\alpha - 2} \right)\left( {\alpha - 1} \right)}} \ge 0$, we obtain $w > \frac{{r^{2} \left( {1 + 2h} \right)\left( {2 - \alpha } \right) - 2c\left( {1 - r} \right)\left( {\alpha - 1} \right)\left( {\alpha - 2} \right) - \alpha r\left( {1 + 2h} \right)}}{{4r\left( {\alpha - 1} \right)^{2} }}$.

Proof of Corollary 1

For α, we have

$$\begin{aligned} \frac{{\partial \Delta_{R}^{P} }}{\partial \alpha } & = - \frac{{r\left( {1 + 2h} \right)}}{{2\left( {\alpha - 1} \right)^{2} }} - \frac{2w}{{\left( {\alpha - 2} \right)^{2} }} + \frac{1 + 2h}{2}\frac{{2 - \alpha^{2} }}{{\left( {\alpha - 1} \right)^{2} \left( {\alpha - 2} \right)^{2} }} \\ & = \frac{{ - \left( {\frac{{\left( {1 + 2h} \right)\left( {1 + r} \right)}}{2} + 2w} \right)\alpha^{2} + \left[ {2r\left( {1 + 2h} \right) + 4w} \right]\alpha - \left( {1 + 2h} \right)\left( {2r - 1} \right) - 2w}}{{\left( {\alpha - 1} \right)^{2} \left( {\alpha - 2} \right)^{2} }} \\ \end{aligned}$$

and we find $\Delta_{R}^{P}$ is decreasing in α in $\left( {0,\frac{{4w + 2r\left( {1{ + }2h} \right) - \sqrt {4w\left( {1{ + }2h} \right)\left( {1 - r} \right) + 2\left( {1 + 2h} \right)^{2} \left( {1 - r} \right)} }}{{4w + \left( {1{ + }2h} \right)\left( {1 + r} \right)}}} \right)$ while increasing in $\left[ {\frac{{4w + 2r\left( {1{ + }2h} \right) - \sqrt {4w\left( {1{ + }2h} \right)\left( {1 - r} \right) + 2\left( {1 + 2h} \right)^{2} \left( {1 - r} \right)} }}{{4w + \left( {1{ + }2h} \right)\left( {1 + r} \right)}},1} \right)$ (The other root $\frac{{4w + 2r\left( {1{ + }2h} \right) + \sqrt {4w\left( {1{ + }2h} \right)\left( {1 - r} \right) + 2\left( {1 + 2h} \right)^{2} \left( {1 - r} \right)} }}{{4w + \left( {1{ + }2h} \right)\left( {1 + r} \right)}} > 1$ because $4w + 2r\left( {1{ + }2h} \right) < 4w + \left( {1{ + }2h} \right)\left( {1 + r} \right)$, therefore, we abandon it). Further, when $\alpha = 0$,$\Delta_{R}^{P} = w + \frac{{c\left( {1 - r} \right)}}{r} - \frac{{r\left( {1 + 2h} \right)}}{2} > 0$. When $\alpha \to 1$, $\Delta_{R}^{P} \to + \infty$. By solving $\Delta_{R}^{P} \le 0$ in α, we also get $\frac{{6c\left( {r - 1} \right) + r\left( {\left( {1 + r} \right)\left( {2h{ + }1} \right) - 8w} \right) + \sqrt \Delta }}{{4c\left( {r - 1} \right) - 8rw}} \le \alpha \le \frac{{6c\left( {r - 1} \right) + r\left( {\left( {1 + r} \right)\left( {2h{ + }1} \right) - 8w} \right) - \sqrt \Delta }}{{4c\left( {r - 1} \right) - 8rw}}$, where $\Delta = \left[ {\left( {1 + 2h} \right)\left( {3 - r} \right)r - 2c\left( {1 - r} \right)} \right]^{2} - 8\left( {1 + 2h} \right)r^{2} \left( {1 - r} \right)\left( {1 + 2h + 2w} \right)$. However, we find $\frac{\partial \Delta }{\partial r} = 2\left( {1 + 2h} \right)r\left( {\left( {1 + 2h} \right)\left( {1 + r} \right)\left( {1 + 2r} \right) + 8\left( {3r - 2} \right)w} \right) - 8c^{2} \left( {1 - r} \right) - 4c\left( {1 + 2h} \right)\left( {3 + r\left( {3r - 8} \right)} \right) > 0$ and $\frac{\partial \Delta }{\partial w} = 16\left( {1 + 2h} \right)\left( {r - 1} \right)r^{2} < 0$ in general hold. So, as long as r is not too high and w is not too low, such as $r \le 0.7$ and $w \ge 10$, whether the e-book difference between the two retailers is big or not, e-reader price under the wholesale model is higher than under the agency model because of $\Delta < 0$. Otherwise, when $\hbox{max} \left\{ {0,\frac{{6c\left( {r - 1} \right) + r\left( {\left( {1 + r} \right)\left( {2h{ + }1} \right) - 8w} \right) + \sqrt \Delta }}{{4c\left( {r - 1} \right) - 8rw}}} \right\} \le \alpha \le \frac{{6c\left( {r - 1} \right) + r\left( {\left( {1 + r} \right)\left( {2h{ + }1} \right) - 8w} \right) - \sqrt \Delta }}{{4c\left( {r - 1} \right) - 8rw}}$, where $\Delta = \left[ {\left( {1 + 2h} \right)\left( {3 - r} \right)r - 2c\left( {1 - r} \right)} \right]^{2} - 8\left( {1 + 2h} \right)r^{2} \left( {1 - r} \right)\left( {1 + 2h + 2w} \right)$, then $\Delta_{R}^{P} \le 0$ holds.

Proof of Proposition 5

We get

$$\begin{aligned} \Delta_{B}^{\pi } & = \pi_{1B}^{A} { + }\pi_{2B}^{A} - \pi_{1B}^{W} - \pi_{2B}^{W} = \frac{{\left( {1 + 2h} \right)^{2} \left( {r - 1} \right)}}{{8\left( {\alpha - 1} \right)}} - \frac{{\left( {1 + 2h + 2w\left( {\alpha - 1} \right)} \right)^{2} }}{{2\left( {\alpha - 2} \right)^{2} }} \\ & \quad - \frac{{c^{2} \left( {r - 1} \right)\left( {\alpha - 1} \right)}}{{2r^{2} }} + \frac{{\left( {1 - r} \right)\left( {\alpha + 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{2\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)^{2} }} - \frac{{\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{2\left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}. \\ \end{aligned}$$

Generally speaking, since t is quite larger than other variables, we can omit $\frac{{\left( {1 - r} \right)\left( {\alpha + 1} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{2\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)^{2} }} - \frac{{\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{2\left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}$ with a negligible effect on $\Delta_{B}^{\pi }$. As a result, we suggest that $\Delta_{B}^{\pi } { = }\frac{{\left( {1 + 2h} \right)^{2} \left( {r - 1} \right)}}{{8\left( {\alpha - 1} \right)}} - \frac{{\left( {1 + 2h + 2w\left( {\alpha - 1} \right)} \right)^{2} }}{{2\left( {\alpha - 2} \right)^{2} }} - \frac{{c^{2} \left( {r - 1} \right)\left( {\alpha - 1} \right)}}{{2r^{2} }}$. By solving $\Delta_{B}^{\pi } > 0$, we obtain $r^{2} \left[ {\frac{{\left( {1 + 2h} \right)^{2} }}{{4\left( {\alpha - 1} \right)^{2} }} - \frac{{\left( {1 + 2h{ + }2\left( {\alpha - 1} \right)w_{A} } \right)^{2} }}{{\left( {r - 1} \right)\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)^{2} }}} \right] > c^{2}$.

Proof of Corollary 2

Let $\Delta_{B}^{\pi } = 0$, we get $w^{1} = \frac{1 + 2h + K}{{2\left( {1 - \alpha } \right)}}$, $w^{2} = \hbox{max} \left\{ {0,\frac{{1{ + }2h - K}}{{2\left( {1 - \alpha } \right)}}} \right\}$, where $K = \left( {2 - \alpha } \right)\sqrt {\frac{1 - r}{1 - \alpha }} \sqrt {\frac{{\left( {1 + 2h} \right)^{2} }}{4} - \frac{{c^{2} \left( {1 - \alpha } \right)^{2} }}{{r^{2} }}}$. Obviously, we need $\frac{{\left( {1 + 2h} \right)^{2} }}{4} - \frac{{c^{2} \left( {1 - \alpha } \right)^{2} }}{{r^{2} }} > 0$ (i.e., $\frac{1 + 2h}{2} > \frac{{c\left( {1 - \alpha } \right)}}{r}$). Thus, there must be w make agency model generates more profit in $\left( {\hbox{max} \left\{ {0,\frac{{1{ + }2h - K}}{{2\left( {1 - \alpha } \right)}}} \right\},} \right.\left. {\frac{{1{ + }2h + K}}{{2\left( {1 - \alpha } \right)}}} \right)$ with $\frac{1 + 2h}{2} > \frac{{c\left( {1 - \alpha } \right)}}{r}$. However, if $\frac{1 + 2h}{2} < \frac{{c\left( {1 - \alpha } \right)}}{r}$, there is no possibility that e-book profit under the agency model is higher than under the wholesale model.

Proof of Proposition 6

We get

$$\begin{aligned} \Delta_{R}^{\pi } & = \pi_{1}^{A} 2 { + }\pi_{2}^{A} 2 - \pi_{1}^{W} 2 - \pi_{2}^{W} 2 = \frac{{c\left( {r - 1} \right)}}{2r} - w - \frac{{\left( {1 + 2h + 2w} \right)}}{{2\left( {\alpha - 2} \right)}} - \frac{{\left( {1 + 2h} \right)\left( {r - 1} \right)}}{{4\left( { - 1 + \alpha } \right)}} \\ & \quad + \frac{{\left( {1 - r} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} \left( {r - 1 + 3t\left( {1 + \alpha } \right)} \right)}}{{9t\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)^{2} }} + \frac{{\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} \left( {2 - 3t\left( {2 + \alpha } \right)} \right)}}{{18t\left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}. \\ \end{aligned}$$

Generally speaking, since t is quite larger than other variables, we can omit $\frac{{\left( {1 - r} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} \left( {r - 1 + 3t\left( {1 + \alpha } \right)} \right)}}{{9t\left( {r - 1 + 6t\left( {1 + \alpha } \right)} \right)^{2} }} + \frac{{\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} \left( {2 - 3t\left( {2 + \alpha } \right)} \right)}}{{18t\left( {1 - 3t\left( {2 + \alpha } \right)} \right)^{2} }}$ with a negligible effect on $\Delta_{R}^{\pi }$. As a result, we suggest that $\Delta_{R}^{\pi } { = }\frac{{c\left( {r - 1} \right)}}{2r} + \frac{{\left( {1 + 2h + 2w\left( {\alpha - 1} \right)} \right)}}{{2\left( {2 - \alpha } \right)}} - \frac{{\left( {1 + 2h} \right)\left( {r - 1} \right)}}{{4\left( {\alpha - 1} \right)}}$. By solving $\Delta_{R}^{\pi } < 0$, we obtain $\frac{r}{r - 1}\left[ {\frac{{1 + 2h + 2\left( {\alpha - 1} \right)w}}{{\left( {\alpha - 2} \right)}} + \frac{{\left( {1 + 2h} \right)\left( {r - 1} \right)}}{{2\left( {\alpha - 1} \right)}}} \right] > c$.

Proof of Proposition 7

We get

$$\begin{aligned} \Delta_{P}^{\pi } & = \pi_{P}^{A} - \pi_{P}^{W} = \frac{{4w\left( {1 - \alpha } \right)\left( {w - c} \right) - \left( {1 + 2h} \right)\left( {2w - c\alpha } \right)}}{{2\left( {2 - \alpha } \right)}} \\ & \quad + \frac{{r\left( {1 + 2h} \right)^{2} }}{{8\left( {1 - \alpha } \right)}} + \frac{{c^{2} \left( {1 - \alpha } \right)}}{2r} + \frac{{r\left( {2 - r} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{12t\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}. \\ \end{aligned}$$

Generally speaking, since t is quite larger than other variables, we can omit $\frac{{r\left( {2 - r} \right)\left( {c_{2R} - c_{1R} + v_{d} } \right)^{2} }}{{12t\left( { - 1 + r + 6t\left( {1 + \alpha } \right)} \right)^{2} }}$ with a negligible effect on $\Delta_{P}^{\pi }$. As a result, we suggest that $\Delta_{B}^{\pi } { = }\frac{{4w\left( {1 - \alpha } \right)\left( {w - c} \right) - \left( {1 + 2h} \right)\left( {2w - c\alpha } \right)}}{{2\left( {2 - \alpha } \right)}} + \frac{{r\left( {1 + 2h} \right)^{2} }}{{8\left( {1 - \alpha } \right)}} + \frac{{c^{2} \left( {1 - \alpha } \right)}}{2r}$. By solving $\Delta_{B}^{\pi } < 0$, we obtain $w^{1} = \frac{{r\left( {1 + 2h + 2c\left( {1 - \alpha } \right)} \right) - \sqrt \Delta }}{{4r\left( {1 - \alpha } \right)}} < w < w^{2} = \frac{{r\left( {1 + 2h + 2c\left( {1 - \alpha } \right)} \right) + \sqrt \Delta }}{{4r\left( {1 - \alpha } \right)}}$, where $\Delta = \left( {1 + 2h} \right)^{2} r^{2} \left( {1 + r\left( {\alpha - 2} \right)} \right) + 4c\left( {1 + 2h} \right)r^{2} \left( {\alpha - 1} \right)^{2} + 4c^{2} r\left( {\alpha - 1} \right)^{2} \left( {r + \alpha - 2} \right)$.