Who benefits from offline investment: an analysis of strategic interactions between e-book pricing and bookstores’ investment

Abstract

Due to an increasing public desire for physical experiences, numerous online bookstores have begun to invest in offline bookstores for economic benefits. The investment of e-commerce companies results in additional online sales of both physical and electronic books (e-books) but complicates the publisher’s pricing strategy of e-books. We construct a supply chain comprising a publisher and an online bookstore, and investigate the strategic interactions between two e-books procing models (i.e., wholesale model and the agency model) and the offline investment strategy, which enriches many traditional e-book pricing models. We analytically derive that the offline investment changes the bookstore’s preference for the e-book pricing model. Additionally, in the wholesale model, the publisher prefers the bookstore with low investment efficiency not to invest. Otherwise, the publisher prefers the bookstore to invest. Finally, in the wholesale model, the bookstore always prefers to invest; but in the agency model, the bookstore chooses to invest only if the investment efficiency and revenue sharing ratio satisfy certain conditions.

Appendices

Appendix

Proof of Proposition 1

According to the backward induction, the pricing decisions of p-books and e-books of the online bookstores are first obtained.

The Hessian Matrix $H_{1}$ of $\pi_{bs}^{WN}$ in terms of $\beta$ and $p_{e}$ is obtained through Eq. (4). That is, $H_{1} = \left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta^{2} }}} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta \partial p_{e} }}} \\ {\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e} \partial \beta }}} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e}^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 2p^{2} } & {2p\theta } \\ {2p\theta } & { - 2} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{bs}^{RN} }}{{\partial \beta^{2} }} < 0$. Given that $\left| {H_{1} } \right| = 4p^{2} \left( {1 - \theta^{2} } \right) > 0$, the bookstore’s profit function is a joint concave function with respect to $\beta$ and $p_{e}$. We can get $\left\{ \begin{gathered} \beta \left( {w_{p} ,w_{e} } \right) = \frac{{a + \theta + w_{p} - \theta^{2} w_{p} }}{{2p - 2p\theta^{2} }} \hfill \\ p_{e} \left( {w_{p} ,w_{e} } \right) = \frac{{ - 1 - a\theta - w_{e} + \theta^{2} w_{e} }}{{2\left( { - 1 + \theta^{2} } \right)}} \hfill \\ \end{gathered} \right.$ through $\left\{ \begin{gathered} \frac{{\partial \pi_{bs}^{WN} }}{\partial \beta } = 0 \hfill \\ \frac{{\partial \pi_{bs}^{WN} }}{{\partial p_{e} }} = 0 \hfill \\ \end{gathered} \right.$. Then substituting $\beta \left( {w_{p} ,w_{e} } \right)$ and $p_{e} \left( {w_{p} ,w_{e} } \right)$ into $\pi_{pub}^{WN}$ and the Hessian Matrix $H_{2}$ of $\pi_{pub}^{WN}$ in terms of $w_{p}$ and $w_{e}$ is obtained. That is $H_{2} = \left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{e}^{2} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{e} \partial w_{p} }}} \\ {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p} \partial w_{e} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 1} & \theta \\ \theta & { - 1} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{e}^{2} }} < 0$. Given that $\left| {H_{2} } \right| = 1 - \theta^{2} > 0$, the publisher’s profit function is a joint concave function with respect to $w_{p}$ and $w_{e}$. We get $\left\{ \begin{gathered} w_{e}^{WN} = \frac{ - 1 - 2\theta + a\theta }{{2\left( { - 1 + \theta^{2} } \right)}} \hfill \\ w_{p}^{WN} = \frac{2 - a + \theta }{{2 - 2\theta^{2} }} \hfill \\ \end{gathered} \right.$ through $\left\{ \begin{gathered} \frac{{\partial \pi_{pub}^{WN} }}{{\partial w_{p} }} = 0 \hfill \\ \frac{{\partial \pi_{pub}^{WN} }}{{\partial w_{e} }} = 0 \hfill \\ \end{gathered} \right.$. Finally, substituting $w_{p}$ and $w_{e}$ into $\beta \left( {w_{p} ,w_{e} } \right)$, $p_{e} \left( {w_{p} ,w_{e} } \right)$, we obtained $\left\{ \begin{gathered} p_{e}^{WN} = \frac{{3 + \left( {2 + a} \right)\theta }}{{4 - 4\theta^{2} }} \hfill \\ \beta^{WN} = \frac{2 + a + 3\theta }{{4p\left( {1 - \theta^{2} } \right)}} \hfill \\ \end{gathered} \right.$. Also, we can get $D_{p}^{WN} = \frac{2 - a}{4}$, $D_{e}^{WN} = \frac{1}{4}$, $\pi_{{_{pub} }}^{WN} = \frac{{5 + a^{2} + 4\theta - 2a\left( {2 + \theta } \right)}}{{8 - 8\theta^{2} }}$, $\pi_{bs}^{WN} = \frac{{3 + 3a^{2} - 2a\left( {4 + \theta } \right)}}{{16\left( { - 1 + \theta^{2} } \right)}}$.

Proof of Corollary 1

(i) We can obtain $\frac{{\partial \pi_{pub}^{WN} }}{\partial \theta } = \frac{{2 + 5\theta + a^{2} \theta + 2\theta^{2} - a\left( {1 + 4\theta + \theta^{2} } \right)}}{{4\left( { - 1 + \theta^{2} } \right)^{2} }}$. Note $f\left( \theta \right) = 2 + 5\theta + a^{2} \theta + 2\theta^{2} - a\left( {1 + 4\theta + \theta^{2} } \right)$, $\frac{\partial f\left( \theta \right)}{{\partial \theta }} = 5 + a^{2} + 4\theta - a\left( {4{ + 2}\theta } \right)$. From $\frac{{\partial^{{2}} f\left( \theta \right)}}{{\partial \theta^{{2}} }} = {4} - {2}a \ge 0$, we can get $\frac{\partial f\left( \theta \right)}{{\partial \theta }}$ increases with $\theta$. $\frac{\partial f\left( \theta \right)}{{\partial \theta }}\left| {_{\theta = 0} } \right. = \left( {a - 2} \right)^{2} + 1 > 0$, so $\frac{\partial f\left( \theta \right)}{{\partial \theta }} > 0$. $f\left( \theta \right)\left| {_{\theta = 0} } \right. = 2 - a > 0$, so $f\left( \theta \right) \ge 0$. Finally, we get $\frac{{\partial \pi_{pub}^{WN} }}{\partial \theta } > 0$.

(ii) We can obtain $\frac{{\partial \pi_{bs}^{WN} }}{\partial \theta } = \frac{{ - 3\theta - 3a^{2} \theta + a\left( {1 + 8\theta + \theta^{2} } \right)}}{{8\left( { - 1 + \theta^{2} } \right)^{2} }}$. Note $g\left( \theta \right) = - 3\theta - 3a^{2} \theta + a\left( {1 + 8\theta + \theta^{2} } \right)$, $\frac{\partial g\left( \theta \right)}{{\partial \theta }} = - 3 - 3a^{2} + a\left( {8 + 2\theta } \right)$. From $\frac{{\partial^{2} g\left( \theta \right)}}{{\partial \theta^{2} }} = 2a \ge 0$, we can get $\frac{\partial g\left( \theta \right)}{{\partial \theta }}$ increases with $\theta$. When $\frac{\partial g\left( \theta \right)}{{\partial \theta }} = 0$, $\theta^{*} = \frac{{3 - 8a + a^{2} }}{2a}$. So when $\theta < \theta^{*}$, $\frac{\partial g\left( \theta \right)}{{\partial \theta }}$ decreases with $\theta$; when $\theta > \theta^{*}$, $\frac{\partial g\left( \theta \right)}{{\partial \theta }}$ increases with $\theta$. $\frac{\partial g\left( \theta \right)}{{\partial \theta }}\left| {_{{\theta = \theta^{*} }} } \right. = \frac{{3\left( { - 1 + a} \right)^{2} \left( {a - 3} \right)\left( { - 3a + 1} \right)}}{4a}$. ① when $0 < a < \frac{1}{3}$, from $g\left( \theta \right) = 0$,we get $\theta^{^{\prime}} = \frac{{3 - 8a + 3a^{2} - \sqrt 3 \sqrt {3 - 16a + 26a^{2} - 16a^{3} + 3a^{4} } }}{2a} < 1$, $\theta^{^{\prime\prime}} = \frac{{3 - 8a + 3a^{2} + \sqrt 3 \sqrt {3 - 16a + 26a^{2} - 16a^{3} + 3a^{4} } }}{2a} > 1$(Not with in the parameter $\theta$ constraint). Thus when $\theta < \theta^{^{\prime}}$, $\pi_{bs}^{WN}$ increases; when $\theta^{^{\prime}} < \theta < 1$, $\pi_{bs}^{WN}$ decreases. ② when $\frac{1}{3} < a < 2$, $\frac{\partial g\left( \theta \right)}{{\partial \theta }}\left| {_{{\theta = \theta^{*} }} } \right. > 0$, so $g\left( \theta \right) > 0$, $\pi_{bs}^{WN}$ increases with $\theta$.

Proof of Proposition 2

Similar to Proof of Proposition 1, the bookstore’s profit function $\pi_{bs}^{AN} \left( \beta \right)$ is concave in $\beta$, that is, $\frac{{\partial^{2} \pi_{bs}^{AN} \left( \beta \right)}}{{\partial \beta^{2} }} = - 2p^{2} < 0$. We obtained the optimal $\beta \left( {w_{p} ,p_{e} } \right) = \frac{{\alpha + 2\theta p_{e} - \gamma \theta p_{e} + w_{p} }}{2p}$ with $\frac{{\partial \pi_{bs}^{AN} \left( \beta \right)}}{\partial \beta } = 0$. By substituting $\beta = \beta \left( {w_{p} ,p_{e} } \right)$ into $\pi_{pub}^{AN} \left( {w_{p} ,p_{e} } \right)$, we obtained the Hessian Matrix $H_{3}$ of $\pi_{bs}^{RN}$ in terms of $w_{p}$ and $p_{e}$. That is $H_{3} = \left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p} \partial p_{e} }}} \\ {\frac{{\partial^{2} \pi_{pub} }}{{\partial p_{e} \partial w_{p} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial p_{e}^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 1} & {\gamma \theta } \\ {\gamma \theta } & { - \gamma \left( {2 + \left( { - 2 + \gamma } \right)\theta^{2} } \right)} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }} < 0$. Given that $\left| {H_{3} } \right| = 2\gamma \left( {1 - \theta^{2} } \right) > 0$, the publisher’s profit function is a joint concave function with respect to $w_{p}$ and $p_{e}$. We obtained the optimal $\left\{ \begin{gathered} w_{p}^{AN} = \frac{1}{2}\left( {\alpha + \frac{{\gamma \theta \left( {1 + \alpha \theta } \right)}}{{1 - \theta^{2} }}} \right) \hfill \\ p_{e}^{AN} = \frac{1 + \alpha \theta }{{2\left( {1 - \theta^{2} } \right)}} \hfill \\ \end{gathered} \right.$ with $\left\{ \begin{gathered} \frac{{\partial \pi_{pub}^{AN} }}{{\partial w_{p} }} = 0 \hfill \\ \frac{{\partial \pi_{pub}^{AN} }}{{\partial p_{e} }} = 0 \hfill \\ \end{gathered} \right.$. Finally, substituting $w_{p}$ and $p_{e}$ into $\beta \left( {w_{p} ,p_{e} } \right)$, we obtained $\beta^{AN} = \frac{{3a + 2\theta - a\theta^{2} }}{{4p\left( {1 - \theta^{2} } \right)}}$, also, we can get $D_{p}^{AN} = \frac{a}{4}$, $D_{e} = \frac{1}{4}\left( {2 + a\theta } \right)$, $\pi_{{_{pub} }}^{AN} = \frac{{2\gamma + 4a\gamma \theta + a^{2} \left( {1 - \left( {1 - 2\gamma } \right)\theta^{2} } \right)}}{{8\left( {1 - \theta^{2} } \right)}}$, $\pi_{bs}^{AN} = \frac{{4\left( { - 1 + \gamma } \right) + 8a\left( { - 1 + \gamma } \right)\theta + a^{2} \left( { - 1 + \left( { - 3 + 4\gamma } \right)\theta^{2} } \right)}}{{16\left( { - 1 + \theta^{2} } \right)}}$.

Proof of Proposition 3

The bookstore’s profit function $\pi_{bs}^{WN} \left( {\beta ,p_{e} ,s} \right)$ is concave in $\beta$, $p_{e}$, $s$, that is $H_{4} = \left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta^{2} }}} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta \partial p_{e} }}} & {\frac{{\partial^{2} \pi_{bs} }}{\partial \beta \partial s}} \\ {\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e} \partial \beta }}} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e}^{2} }}} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e} \partial s}}} \\ {\frac{{\partial^{2} \pi_{bs} }}{\partial s\partial \beta }} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial s\partial p_{e} }}} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial s^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 2p^{2} } & {2p\theta } & {p\lambda } \\ {2p\theta } & { - 2} & \lambda \\ {p\lambda } & \lambda & { - k} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta^{2} }} = - 2p^{2} < 0$, $\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta^{2} }}\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e}^{2} }} - \frac{{\partial^{2} \pi_{bs} }}{{\partial \beta \partial p_{e} }}\frac{{\partial^{2} \pi_{bs} }}{{\partial p_{e} \partial \beta }} = 4p^{2} \left( {1 - \theta^{2} } \right) > 0$, and $\left| {H_{4} } \right| = - 4kp^{2} + 4kp\theta^{2} + 2p\lambda^{2} + 4p^{2} \theta \lambda^{2} + 2p^{2} \lambda^{2} < 0$ must be met. We get the optimal $\left\{ \begin{gathered} p_{e} \left( {w_{p} ,w_{e} } \right) = \frac{{ - 2k - 2ak\theta + \lambda^{2} - a\lambda^{2} + \left( {1 + \theta } \right)\left( {2k\left( { - 1 + \theta } \right) + 3\lambda^{2} } \right)w_{e} + \left( {1 + \theta } \right)\lambda^{2} w_{p} }}{{4\left( {1 + \theta } \right)\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}} \hfill \\ \beta \left( {w_{p} ,w_{e} } \right) = \frac{{ - 2ak - 2k\theta - \lambda^{2} + a\lambda^{2} + \left( {1 + \theta } \right)\lambda^{2} w_{e} + \left( {1 + \theta } \right)\left( {2k\left( { - 1 + \theta } \right) + 3\lambda^{2} } \right)w_{p} }}{{4p\left( {1 + \theta } \right)\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}} \hfill \\ s\left( {w_{p} ,w_{e} } \right) = - \frac{{\lambda \left( {1 + a + \left( { - 1 + \theta } \right)w_{e} + \left( { - 1 + \theta } \right)w_{p} } \right)}}{{2\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}} \hfill \\ \end{gathered} \right.$ through $\left\{ \begin{gathered} \frac{{\partial \pi_{bs}^{WN} }}{{\partial p_{e} }} = 0 \hfill \\ \frac{{\partial \pi_{bs}^{WN} }}{\partial \beta } = 0 \hfill \\ \frac{{\partial \pi_{bs}^{WN} }}{\partial s} = 0 \hfill \\ \end{gathered} \right.$. By substituting $p_{e} \left( {w_{p} ,w_{e} } \right)$, $\beta \left( {w_{p} ,w_{e} } \right)$, $s\left( {w_{p} ,w_{e} } \right)$ into $\pi_{pub}^{WN}$, then we obtained the Hessian Matrix $H_{5}$ of $\pi_{pub}^{WN}$ in terms of $w_{p}$ and $w_{e}$. That is $H_{5} { = }\left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p} \partial w_{e} }}} \\ {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{e} \partial w_{p} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{e}^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - \frac{{2k\left( { - 1 + \theta } \right) + \left( {1 + \theta } \right)\lambda^{2} }}{{2\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}}} & {\frac{{2k\left( { - 1 + \theta } \right)\theta + \left( {1 + \theta } \right)\lambda^{2} }}{{2\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}}} \\ {\frac{{2k\left( { - 1 + \theta } \right)\theta + \left( {1 + \theta } \right)\lambda^{2} }}{{2\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}}} & { - \frac{{2k\left( { - 1 + \theta } \right) + \left( {1 + \theta } \right)\lambda^{2} }}{{2\left( {k\left( { - 1 + \theta } \right) + \lambda^{2} } \right)}}} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }} = \frac{{2k\left( { - 1 + \theta } \right) + \left( {1 + \theta } \right)\lambda^{2} }}{{2\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}} < 0$ and $\left| {H_{5} } \right| = \frac{{k\left( {1 - \theta } \right)^{2} \left( {1 + \theta } \right)}}{{k\left( {1 - \theta } \right) - \lambda^{2} }} > 0$ must be contained, that is,$k\left( {1 - \theta } \right) - \lambda^{2} > 0$, $2k\left( { - 1 + \theta } \right) + \left( {1 + \theta } \right)\lambda^{2} < 0$. The publisher’s profit function is a joint concave function with respect to $w_{p}$ and $w_{e}$, we get the optimal equilibrium $\left\{ \begin{gathered} w_{p}^{WI} = \frac{a + \theta }{{2 - 2\theta^{2} }} \hfill \\ w_{e}^{WI} = \frac{1 + a\theta }{{2 - 2\theta^{2} }} \hfill \\ \end{gathered} \right.$. Finally, substituting $w_{p}^{WI}$ and $w_{e}^{WI}$ into $p_{e} \left( {w_{p} ,w_{e} } \right)$, $\beta \left( {w_{p} ,w_{e} } \right)$, $s\left( {w_{p} ,w_{e} } \right)$,we obtained $p_{e}^{WI} = \frac{{6k\left( {1 - \theta } \right)\left( {1 + a\theta } \right) + \left( { - 5 + a + \theta - 5a\theta } \right)\lambda^{2} }}{{8\left( {1 - \theta^{2} } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$, $\beta^{WI} = \frac{{6k\left( {1 - \theta } \right)\theta + \left( {1 - 5\theta } \right)\lambda^{2} + a\left( {6k\left( {1 - \theta } \right) - \left( {5 - \theta } \right)\lambda^{2} } \right)}}{{8p\left( {1 - \theta^{2} } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$, $s^{WI} = \frac{{\left( {1 + a} \right)\lambda }}{{4\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$ also, we can get $D_{p}^{WI} = \frac{{\lambda^{2} + a\left( {2k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}{{8\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$, $D_{e}^{WI} = \frac{{2k\left( {1 - \theta } \right) - \left( {1 - a} \right)\lambda^{2} }}{{8\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$, $\pi_{pub}^{WI} = \frac{{2k\left( {1 + a^{2} + 2a\theta } \right) + \left( {1 - a} \right)^{2} \lambda^{2} }}{{16\left( {1 + \theta } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$, $\pi_{bs}^{WI} = \frac{{2k\left( {1 + a^{2} + 2a\theta } \right) - \left( {1 - a} \right)^{2} \lambda^{2} }}{{32\left( {1 + \theta } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$.

Proof of Proposition 4

The bookstore’s profit function $\pi _{{bs}}^{{AI}} \left( {\beta ,s} \right)$ is concave in $\beta$ and $s$, that is, $H_{6} = \left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta^{2} }}} & {\frac{{\partial^{2} \pi_{bs} }}{\partial \beta \partial s}} \\ {\frac{{\partial^{2} \pi_{bs} }}{\partial s\partial \beta }} & {\frac{{\partial^{2} \pi_{bs} }}{{\partial s^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 2p^{2} } & {p\lambda } \\ {p\lambda } & { - k} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{bs} }}{{\partial \beta^{2} }} = - 2p^{2} < 0$, $\left| {H_{6} } \right| = 2k - \lambda^{2} > 0$ must be contained. We get the optimal.

$\left\{ \begin{gathered} \beta \left( {w_{p} ,p_{e} } \right) = \frac{{ak + 2k\theta p_{e} - k\gamma \theta p_{e} + \lambda^{2} p_{e} - \gamma \lambda^{2} p_{e} + kw_{p} - \lambda^{2} w_{p} }}{{p\left( {2k - \lambda^{2} } \right)}} \hfill \\ s\left( {w_{p} ,p_{e} } \right) = \frac{{\lambda \left( {a + \left( {2\left( {1 + \theta } \right) - \gamma \left( {2 + \theta } \right)} \right)p_{e} - w_{p} } \right)}}{{2k - \lambda^{2} }} \hfill \\ \end{gathered} \right.$ with $\left\{ \begin{gathered} \frac{{\partial \pi_{bs}^{AI} }}{\partial \beta } = 0 \hfill \\ \frac{{\partial \pi_{bs}^{AI} }}{\partial s} = 0 \hfill \\ \end{gathered} \right.$. By substituting $\beta \left( {w_{p} ,p_{e} } \right)$, $s\left( {w_{p} ,p_{e} } \right)$, into $\pi_{pub}^{AI}$, then we obtained the Hessian Matrix $H_{7}$ of $\pi_{pub}^{AN}$ in terms of $w_{p}$ and $P_{e}$. That is $H_{7} = \left( {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p} \partial p_{e} }}} \\ {\frac{{\partial^{2} \pi_{pub} }}{{\partial p\partial w_{p} }}} & {\frac{{\partial^{2} \pi_{pub} }}{{\partial p_{e}^{2} }}} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - \frac{2k}{{2k - \lambda^{2} }}} & {\frac{{2k\gamma \theta - \left( { - 1 + 2\gamma } \right)\left( {1 + \theta } \right)\lambda^{2} }}{{2k - \lambda^{2} }}} \\ {\frac{{2k\gamma \theta - \left( { - 1 + 2\gamma } \right)\left( {1 + \theta } \right)\lambda^{2} }}{{2k - \lambda^{2} }}} & { - \frac{{2\gamma \left( {k\left( {2 + \left( { - 2 + \gamma } \right)\theta^{2} } \right) + \left( { - 3 + 2\gamma } \right)\left( {1 + \theta } \right)\lambda^{2} } \right)}}{{2k - \lambda^{2} }}} \\ \end{array} } \right)$, where $\frac{{\partial^{2} \pi_{pub} }}{{\partial w_{p}^{2} }} = - \frac{2k}{{2k - \lambda^{2} }} < 0$, $\left| {H_{7} } \right| = - \frac{{\left( {1 + \theta } \right)\left( {8k^{2} \gamma \left( { - 1 + \theta } \right) + 4k\gamma \left( {3 + \theta - 2\gamma \left( {1 + \theta } \right)} \right)\lambda^{2} + \left( {1 - 2\gamma } \right)^{2} \left( {1 + \theta } \right)\lambda^{4} } \right)}}{{\left( { - 2k + \lambda^{2} } \right)^{2} }} > 0$ must be contained, that is,$8k^{2} \gamma \left( { - 1 + \theta } \right) + 4k\gamma \left( {3 + \theta - 2\gamma \left( {1 + \theta } \right)} \right)\lambda^{2} + \left( {1 - 2\gamma } \right)^{2} \left( {1 + \theta } \right)\lambda^{4} < 0$. We obtained $\left\{ \begin{gathered} w_{p}^{AI} = \frac{{ - \gamma \left( {a\lambda^{4} (1 - 2\gamma )(1 + \theta ) + 4ak^{2} \left( {\gamma \theta^{2} - \theta^{2} + 1} \right) + ak\lambda^{2} \left( { - 2\gamma \theta^{2} + 4\gamma \theta + 4\gamma + \theta^{2} - 5\theta - 6} \right) + \left( {2k - \lambda^{2} } \right)\left( {2\gamma \theta k - (2\gamma - 1)(\theta + 1)\lambda^{2} } \right)} \right)}}{{\left( {1 + \theta } \right)A}} \hfill \\ p_{e}^{AI} = \frac{{ - k\left( {4k\gamma \left( {1 + a\theta } \right) + \lambda^{2} \left( { - 2\gamma + a\left( {1 + \theta - 2\gamma \theta } \right)} \right)} \right)}}{{\left( {1 + \theta } \right)A}} \hfill \\ \end{gathered} \right.$ through $\left\{ \begin{gathered} \frac{{\partial \pi_{pub}^{AI} }}{{\partial w_{p} }} = 0 \hfill \\ \frac{{\partial \pi_{pub}^{AI} }}{{\partial p_{e} }} = 0 \hfill \\ \end{gathered} \right.$. Finally, substituting $w_{p}^{AI}$ and $p_{e}^{AI}$ into $\beta \left( {w_{p} ,p_{e} } \right)$, $s\left( {w_{p} ,p_{e} } \right)$,we obtained $\beta^{AI} = \frac{\begin{gathered} a\left( {\gamma \lambda^{4} (2\gamma - 1)(1 + \theta ) + 2\gamma k^{2} \left( {\theta^{2} - 3} \right) - k\lambda^{2} \left( {2\gamma^{2} \left( {\theta^{2} + 4\theta + 3} \right) - \gamma \left( {3\theta^{2} + 9\theta + 8} \right) + \theta (1 + \theta )} \right)} \right) \hfill \\ + \gamma \left( {\lambda^{4} (1 - 2\gamma )(1 + \theta ) - 4\theta k^{2} + k\lambda^{2} (4\gamma \theta + 4\gamma - \theta - 3)} \right) \hfill \\ \end{gathered} }{(1 + \theta )pA}$, $s^{AI} = \frac{{2k\lambda \gamma \left( {2\gamma - 2 - a\left( {1 + \theta - 2\gamma \theta } \right)} \right) + \lambda^{3} \left( {1 - 2\gamma } \right)\left( {\gamma + a\left( { - 1 + \gamma - \theta + 2\gamma \theta } \right)} \right)}}{A}$,also, we can get $D_{p}^{AI} = \frac{{ - \lambda^{2} k\gamma + ak\gamma \left( {2k\left( { - 1 + \theta } \right) + \lambda^{2} \left( {2 + \theta - 2\gamma - 2\gamma \theta } \right)} \right)}}{A}$, $D_{e}^{AI} = \frac{\begin{gathered} \lambda^{4} (1 - a)\left( {2\gamma^{2} - 3\gamma + 1} \right)(\theta + 1) + 2\gamma k^{2} (\theta - 1)(a\theta + 2) + \hfill \\ k\lambda^{2} \left( {\gamma (3\theta + 6 - 4\gamma \theta - 4\gamma ) - a\left( {2\gamma^{2} \theta^{2} + 2\gamma^{2} \theta - 3\gamma \theta^{2} - 4\gamma \theta + 2\gamma + \theta^{2} - 1} \right)} \right) \hfill \\ \end{gathered} }{A}$, $D_{e}^{AI} = \frac{\begin{gathered} \lambda^{4} (1 - a)\left( {2\gamma^{2} - 3\gamma + 1} \right)(\theta + 1) + 2\gamma k^{2} (\theta - 1)(a\theta + 2) + \hfill \\ k\lambda^{2} \left( {\gamma (3\theta + 6 - 4\gamma \theta - 4\gamma ) - a\left( {2\gamma^{2} \theta^{2} + 2\gamma^{2} \theta - 3\gamma \theta^{2} - 4\gamma \theta + 2\gamma + \theta^{2} - 1} \right)} \right) \hfill \\ \end{gathered} }{A}$, $\pi_{pub}^{AI} = - \frac{{k\gamma \left( {\gamma \left( {2k - \lambda^{2} } \right) + a\left( {4k\gamma \theta + \lambda^{2} \left( {1 + \theta - 2\gamma \theta } \right)\lambda^{2} } \right) + a^{2} \left( {k\left( {1 + \theta^{2} \left( { - 1 + 2\gamma } \right)} \right) + \lambda^{2} \left( { - 1 + \gamma - \theta + 2\gamma \theta } \right)} \right)} \right)}}{{\left( {1 + \theta } \right)A}}$, where $A = 8k^{2} \gamma \left( { - 1 + \theta } \right) + 4k\gamma \lambda^{2} \left( {3 + \theta - 2\gamma - 2\gamma \theta } \right) + \lambda^{4} \left( {1 - 2\gamma } \right)^{2} \left( {1 + \theta } \right)$.

Proof of Lemma 1

Firstly, we can get $\Delta \pi_{pub}^{WN - AN} = \pi_{pub}^{WN} - \pi_{pub}^{AN} = \frac{{ - 2\gamma \left( {1 + a\theta } \right)^{2} + 5 - 4a + 4\theta - 2a\theta + a^{2} \theta^{2} }}{{8\left( {1 - \theta^{2} } \right)}}$, then we denote $f\left( \gamma \right) = - 2\gamma \left( {1 + a\theta } \right)^{2} + 5 - 4a + 4\theta - 2a\theta + a^{2} \theta^{2}$, thus when $f\left( \gamma \right) > 0$, $\pi_{pub}^{WN} > \pi_{pub}^{AN}$; $f\left( \gamma \right) < 0$, $\pi_{pub}^{WN} < \pi_{pub}^{AN}$. In a word, we need to figure out the relationship between $f\left( \gamma \right)$ and 0. We can see $\frac{\partial f\left( \gamma \right)}{{\partial \gamma }} = - 2\left( {1 + a\theta } \right)^{2} < 0$, so we have two scenarios to discuss: (1) $f\left( \gamma \right) < 0$; (2) $f\left( \gamma \right) > 0$. From $f\left( 0 \right) = 5 - 4a + 4\theta - 2a\theta + a^{2} \theta^{2}$, we get $\frac{\partial f\left( 0 \right)}{{\partial a}} = - 4 - 2\theta + 2a\theta^{2} < 0$, when $f\left( 0 \right) = 0$, $a_{2}^{*} = \frac{{2 + \theta - 2\sqrt {1 + \theta - \theta^{2} - \theta^{3} } }}{{\theta^{2} }}$ ($a_{2}^{*}$ increases with $\theta$, when $a_{2}^{*} = 2$,$\theta = \frac{\sqrt 3 }{2}$). When $f\left( 0 \right) > 0$, $f\left( \gamma \right) = 0$$\gamma_{1}^{*} = \frac{{5 - 4a + 4\theta - 2a\theta + a^{2} \theta^{2} }}{{2\left( {1 + a\theta } \right)^{2} }}$($\gamma_{1}^{*}$ decreases with $a$, when $\gamma_{1}^{*} = 1$, $a_{1}^{*} = \frac{{ - 2 - 3\theta + 2\sqrt {\left( {1 + \theta } \right)^{3} } }}{{\theta^{2} }}$). Next we consider the following three cases.

Case A $f\left( 0 \right) < 0$, that is, $0 < \theta < \frac{\sqrt 3 }{2}$, $a_{2}^{*} < a < 2$.

We can prove that in this case $f\left( \gamma \right)$ is always less than 0, that is, the publisher prefers the agency model.

Case B $f\left( 0 \right) > 0$ and $\gamma_{1}^{*} > 1$, that is $0 < \theta < 1$, $0 < a < a_{1}^{*}$.

We can prove that in this case $f\left( \gamma \right)$ is always greater than 0 in the range $0 < \gamma < 1$, the publisher prefers the wholesale model always.

Case C $f\left( 0 \right) > 0$ and $0 < \gamma_{1}^{*} < 1$, that is $0 < \theta < \frac{\sqrt 3 }{2}$, $a_{1}^{*} < a < a_{2}^{*}$ and $\frac{\sqrt 3 }{2} < \theta < 1$, $a_{1}^{*} < a < 2$.

We can prove in this case when $0 < \gamma < \gamma_{1}^{*}$,$f\left( \gamma \right) > 0$,the publisher chooses the wholesale model, if $\gamma_{1}^{*} < \gamma < 1$, $f\left( \gamma \right) < 0$, the publisher prefers the agency model.

Summarizing the above results, we can conclude the publisher’s preferences as follows:

$$\left\{ \begin{gathered} 0 < \theta < \frac{\sqrt 3 }{2}\left\{ \begin{gathered} 0 < a < a_{1}^{*} {\text{ RN}} \hfill \\ a_{1}^{*} < a < a_{2}^{*} \, \left\{ \begin{gathered} 0 < \gamma < \gamma_{1}^{*} {\text{ RN}} \hfill \\ \gamma_{1}^{*} < \gamma < 1{\text{ AN}} \hfill \\ \end{gathered} \right. \hfill \\ a_{2}^{*} \, < a < 2{\text{ AN}} \hfill \\ \end{gathered} \right. \hfill \\ \frac{\sqrt 3 }{2} < \theta < 1\left\{ \begin{gathered} 0 < a < a_{1}^{*} {\text{ RN}} \hfill \\ a_{1}^{*} \, < a < 2 \, \left\{ \begin{gathered} 0 < \gamma < \gamma_{1}^{*} {\text{ RN}} \hfill \\ \gamma_{1}^{*} < \gamma < 1{\text{ AN}} \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.$$

where $a_{1}^{*} = \frac{{ - 2 - 3\theta + 2\sqrt {\left( {1 + \theta } \right)^{3} } }}{{\theta^{2} }}$, $a_{2}^{*} = \frac{{2 + \theta - 2\sqrt {\left( {1 - \theta } \right)\left( {1 + \theta } \right)^{2} } }}{{\theta^{2} }}$, $\gamma_{1}^{*} = \frac{{5 - 4a + 4\theta - 2a\theta + a^{2} \theta^{2} }}{{2\left( {1 + a\theta } \right)^{2} }}$.

Proof of Lemma 2

Similar to Proof of Lemma 1, denote $\Delta \pi_{bs}^{WN - AN} = \pi_{bs}^{WN} - \pi_{bs}^{AN} = \frac{{4\gamma \left( {1 + a\theta } \right)^{2} - \left( {7 + a\left( { - 8 + 6\theta } \right) + a^{2} \left( {4 + 3\theta^{2} } \right)} \right)}}{{16\left( {1 - \theta^{2} } \right)}}$, then we need to figure out the relationship between $g\left( \gamma \right) = 4\gamma \left( {1 + a\theta } \right)^{2} - \left( {7 + a\left( { - 8 + 6\theta } \right) + a^{2} \left( {4 + 3\theta^{2} } \right)} \right)$ and 0. We can obtain $\frac{\partial g\left( \gamma \right)}{{\partial \gamma }} = 4\left( {1 + a\theta } \right)^{2} > 0$, then we get $g\left( 0 \right) = - \left( {7 + a\left( { - 8 + 6\theta } \right) + a^{2} \left( {4 + 3\theta^{2} } \right)} \right) < 0$. With $g\left( \gamma \right) = 0$, we can obtain $\gamma_{2}^{*} = \frac{{7 + a\left( { - 8 + 6\theta } \right) + a^{2} \left( {4 + 3\theta^{2} } \right)}}{{4\left( {1 + a\theta } \right)^{2} }}$. For the range of $0 < \gamma < 1$, we need to analyze function $\gamma_{2}^{*}$. Moreover, one can show $\frac{{\partial \gamma_{2}^{*} }}{\partial a} = \frac{{2\left( { - 1 + a} \right)\left( {1 + \theta } \right)}}{{\left( {1 + a\theta } \right)^{3} }}$, when $0 < a < 1$, $\frac{{\partial \gamma_{2}^{*} }}{\partial a} < 0$, and when $1 < a < 2$, $\frac{{\partial \gamma_{2}^{*} }}{\partial a} > 0$ and $\gamma_{2}^{*} \left| {_{a = 1} } \right. = 0.75$. With $\gamma_{2}^{*} = 1$, we get $a = \frac{1}{2 + \theta } < 1$ or $\frac{3}{2 - \theta } > 1$($\frac{3}{2 - \theta }$ increases with $\theta$, when $\frac{3}{2 - \theta } = 2$,$\theta = 0.5$). Next we consider the following three cases.

Case A $0 < \theta < 0.5$, $a_{3}^{*} = \frac{1}{2 + \theta }$, $a_{4}^{*} = \frac{3}{2 - \theta }$.

When $0 < \theta < 0.5$, $0 < a < a_{3}^{*} \,$ and $a_{4}^{*} < a < 2$, the bookstore prefers the agency model and when $a_{3}^{*} < a < a_{4}^{*}$, she need to observe the threshold $\gamma_{2}^{*}$, when $0 < \gamma < \gamma_{2}^{*}$, she profits more from the agency model, when $\gamma_{2}^{*} < \gamma < 1$, she prefers the wholesale model.

Case B $0.5 < \theta < 1$,$a_{3}^{*} = \frac{1}{2 + \theta }$.

When $0.5 < \theta < 1$,$0 < a < a_{3}^{*} \,$, the bookstores choose the agency model and when $a_{3}^{*} < a < 2$, the results are similar to the above Case A situation $a_{3}^{*} < a < a_{4}^{*}$.

Summarizing the above results, we can conclude the bookstore’s preferences as follows:

$$\left\{ \begin{gathered} 0 < \theta < 0.5\left\{ \begin{gathered} 0 < a < a_{3}^{*} {\text{ AN}} \hfill \\ a_{3}^{*} < a < a_{4}^{*} \, \left\{ \begin{gathered} 0 < \gamma < \gamma_{2}^{*} {\text{ AN}} \hfill \\ \gamma_{2}^{*} < \gamma < 1{\text{ RN}} \hfill \\ \end{gathered} \right. \hfill \\ a_{4}^{*} < a < 2{\text{ AN}} \hfill \\ \end{gathered} \right. \hfill \\ 0.5 < \theta < 1\left\{ \begin{gathered} 0 < a < a_{3}^{*} {\text{ AN }} \hfill \\ a_{3}^{*} < a < 2 \, \left\{ \begin{gathered} 0 < \gamma < \gamma_{2}^{*} {\text{ AN}} \hfill \\ \gamma_{2}^{*} < \gamma < 1{\text{ RN}} \hfill \\ \end{gathered} \right. \, \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.$$

where $a_{3}^{*} = \frac{1}{2 + \theta }$, $a_{4}^{*} = \frac{3}{2 - \theta }$, $\gamma_{2}^{*} = \frac{{7 + a\left( { - 8 + 6\theta } \right) + a^{2} \left( {4 + 3\theta^{2} } \right)}}{{4\left( {1 + a\theta } \right)^{2} }}$.

Proof of Lemma 3

The difference profit of the publisher without and with the bookstore’s investment under the wholesale model is $\Delta \pi_{pub}^{WN - WI} = \frac{{8k\left( {1 - a} \right)\left( {1 - \theta } \right) - \left( {3 - a} \right)^{2} \lambda^{2} }}{{16\left( {1 - \theta } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$. In Sect. 5.1, we have got the condition $k > \frac{{\lambda^{2} }}{1 - \theta }$ that ensures $16\left( {1 - \theta } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right) > 0$. When $1 < a < 2$, $8k\left( {1 - a} \right)\left( {1 - \theta } \right) - \left( {3 - a} \right)^{2} \lambda^{2} < 0$ is obvious. So we have $\pi_{pub}^{WN} > \pi_{pub}^{WI}$. When $0 < a < 1$, with $8k\left( {1 - a} \right)\left( {1 - \theta } \right) - \left( {3 - a} \right)^{2} \lambda^{2} = 0$, we get $k_{2}^{*} = \frac{{\left( {3 - a} \right)^{2} \lambda^{2} }}{{8\left( {1 - a} \right)\left( {1 - \theta } \right)}} > k_{1}^{*} = \frac{{\lambda^{2} }}{{\left( {1 - \theta } \right)}}$. Thus if $k > k_{2}^{*}$,$\Delta \pi_{pub}^{WN - WI} > 0$, he prefers the publisher not to invest in the offline bookstores; if $k_{1}^{*} < k < k_{2}^{*}$, $\Delta \pi_{pub}^{WN - WI} < 0$, he would like the bookstore to invest.

We also can get $\Delta \pi_{bs}^{WN - WI} = \frac{{8k\left( {1 - a} \right)^{2} \left( { - 1 + \theta } \right) - \lambda^{2} \left( {a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta } \right)}}{{32\left( {1 - \theta^{2} } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right)}}$. Obviously, $32\left( {1 - \theta^{2} } \right)\left( {k\left( {1 - \theta } \right) - \lambda^{2} } \right) > 0$, $8k\left( {1 - a} \right)^{2} \left( { - 1 + \theta } \right) < 0$. If $\lambda^{2} \left( {a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta } \right) > 0$, then $\Delta \pi_{bs}^{WN - WI}$ will less than zero; if $\lambda^{2} \left( {a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta } \right) < 0$, we should compare $8k\left( {1 - a} \right)^{2} \left( { - 1 + \theta } \right)$ with $\lambda^{2} \left( {a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta } \right)$. Then we can see $f\left( a \right) = a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta$ is a quadratic function of $a$, the axis of symmetry is $a = \frac{9 + \theta }{{7 - \theta }} > 1$,$f\left( 0 \right) = - 7 + \theta < 0$, $f\left( 2 \right) = 9\theta + 1 > 0$. With $f\left( a \right) = 0$, we get $a_{5}^{*} = \frac{{ - 9 - \theta + 4\sqrt 2 \sqrt {1 + \theta } }}{ - 7 + \theta } < 1$. When $a > a_{5}^{*}$, $f\left( a \right) > 0$, $\lambda^{2} f\left( a \right) > 0$$\Delta \pi_{bs}^{WN - WI} < 0$, the bookstore prefers to invest. When $0 < a < a_{5}^{*}$, $8k\left( {1 - a} \right)^{2} \left( { - 1 + \theta } \right) - \lambda^{2} \left( {a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta } \right)$ decreases with respect to $k$ monotonically, so with $8k\left( {1 - a} \right)^{2} \left( { - 1 + \theta } \right) - \lambda^{2} \left( {a^{2} \left( { - 7 + \theta } \right) + 2a\left( {9 + \theta } \right) - 7 + \theta } \right) = 0$, we obtained $k_{3}^{*} = \frac{{\lambda^{2} }}{1 - \theta }*\frac{{a^{2} \left( {7 - \theta } \right) - 2a\left( {9 + \theta } \right) + 7 - \theta }}{{8\left( {1 - a} \right)^{2} }}$. ${{\partial \frac{{a^{2} \left( {7 - \theta } \right) - 2a\left( {9 + \theta } \right) + 7 - \theta }}{{8\left( {1 - a} \right)^{2} }}} \mathord{\left/ {\vphantom {{\partial \frac{{a^{2} \left( {7 - \theta } \right) - 2a\left( {9 + \theta } \right) + 7 - \theta }}{{8\left( {1 - a} \right)^{2} }}} {\partial a}}} \right. \kern-\nulldelimiterspace} {\partial a}} = \frac{{\left( {1 + a} \right)\left( {1 + \theta } \right)}}{{2\left( { - 1 + a} \right)^{3} }} < 0$, $k_{3}^{*} \left| {_{a = 0} } \right. = \frac{{\lambda^{2} }}{1 - \theta }*\frac{7 - \theta }{8} < \frac{{\lambda^{2} }}{1 - \theta }$. Thus summarizing the above results, we can conclude the bookstore’s preferences when $k > \frac{{\lambda^{2} }}{1 - \theta }$, she always prefers to invest.