Managing consumer privacy risk: The effects of privacy breach insurance

Abstract

Inappropriate management of information security of e-commerce websites exerts a negative effect on consumers’ purchase decisions because it may cause consumer data leakage. To mitigate this negative effect, some e-commerce firms such as Taobao in China voluntarily offer privacy breach insurance to consumers. Even though this fresh management of information security is becoming popular in practice, there are still rare academic studies concerning this hot topic. To fill this gap, this paper constructs a multi-stage game-theoretic model to examine the impacts of privacy breach insurance on a monopoly firm and to identify the market condition under which the firm would like to employ privacy breach insurance. We find that when consumers’ data leakage concern is low and firm’s data leakage possibility is low, privacy breach insurance always encourages more consumers to purchase. Otherwise, privacy breach insurance stimulates purchase only when the compensation of privacy breach insurance remains high. Furthermore, we reveal that the monopoly firm would provide privacy breach insurance when consumers’ data leakage concern remains low. These fresh results suggest that privacy breach insurance acts as an efficient measure e-commerce firms could take to address increasingly serious information security threats.

Appendices

Appendix A

Solving the first-order derivation of function $n^{F}$ with respect to $k$, we can get

$$ \frac{{\partial n^{F} }}{{\partial k}} = \frac{{4\gamma \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } - \left( {2\left( {1 - \alpha + \gamma k} \right)\gamma + 12\gamma \left( {1 + \gamma } \right)\omega } \right)}}{{6\sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } }} $$

(A1)

Letting $X = 2\left( {1 - \alpha + \gamma k} \right)\gamma + 12\gamma \left( {1 + \gamma } \right)\omega$ and $Y = 4\gamma \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega }$, we can get that if $Y - X > 0$,$\frac{{\partial n^{F} }}{{\partial k}} > 0$. Further, we can get

$$ Y^{2} - X^{2} = 12\gamma ^{2} \left[ {\gamma ^{2} k^{2} + 2\gamma \left( {\left( {1 - \alpha } \right) + 6\left( {1 + \gamma } \right)\omega } \right)k + \left( {1 - \alpha - 6\left( {1 + \gamma } \right)\omega } \right)\left( {1 - \alpha + 2\left( {1 + \gamma } \right)\omega } \right)} \right] $$

(A2)

For $X > 0$ and $Y > 0$, we can get that if $Y^{2} - X^{2} > 0$, then $Y - X > 0$.

By analyzing equation (a2), we can find that if.

$1 - \alpha - 6\left( {1 + \gamma } \right)\omega \ge 0$ (a3),

then $Y - X > 0$. When $\omega < \frac{{1 - \alpha }}{6}$ and $\gamma \le \frac{{1 - \alpha }}{{6\omega }} - 1$, equation (a3) holds.

Furthermore, if

$$ 1 - \alpha - 6\left( {1 + \gamma } \right)\omega < 0 $$

(A4)

then when $k > \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\left( {1 + \gamma } \right)\omega } \right) + 4\sqrt {4\left( {1 + \gamma } \right)^{2} \omega ^{2} + \left( {1 - \alpha } \right)\left( {1 + \gamma } \right)\omega } }}{\gamma }$, equation (a2) holds. Analyzing equation (A4), we can get when 1) $\omega \ge \frac{{1 - \alpha }}{6}$; or 2)$\omega < \frac{{1 - \alpha }}{6}$ and $\gamma > \frac{{1 - \alpha }}{{6\omega }} - 1$, equation (A4) holds.

In summary, according to the above results, we can get that.

• (1) if $\omega < \frac{{1 - \alpha }}{6}$ and $\gamma \le \frac{{1 - \alpha }}{{6\omega }} - 1$, then $\frac{{\partial n^{F} }}{{\partial k}} > 0$;

• (2) if $\omega < \frac{{1 - \alpha }}{6}$ and $\gamma > \frac{{1 - \alpha }}{{6\omega }} - 1$, then when $k > k_{0}$, $\frac{{\partial n^{F} }}{{\partial k}} > 0$ and when $k < k_{0}$, $\frac{{\partial n^{F} }}{{\partial k}} < 0$;

• (3) if $\omega \ge \frac{{1 - \alpha }}{6}$, then when $k > k_{0}$, $\frac{{\partial n^{F} }}{{\partial k}} > 0$ and when $k < k_{0}$, $\frac{{\partial n^{F} }}{{\partial k}} < 0$.

Here, $k_{0} = \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\left( {1 + \gamma } \right)\omega } \right) + 4\sqrt {4\left( {1 + \gamma } \right)^{2} \omega ^{2} + \left( {1 - \alpha } \right)\left( {1 + \gamma } \right)\omega } }}{\gamma }$.

Appendix B

Solving the first-order derivation of function $n^{F}$ with respect to $\gamma$, we can get

$$ \frac{{\partial n^{F} }}{{\partial \gamma }} = \frac{{2k\sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } - \left[ {\left( {1 - \alpha + \gamma k} \right)k + 6\left( {1 + 2\gamma } \right)k\omega } \right]}}{{3\sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } }} $$

(A5)

Letting $M = 2k\sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega }$ and $N = \left[ {\left( {1 - \alpha + \gamma k} \right)k + 6\left( {1 + 2\gamma } \right)k\omega } \right]$, we can get that if $M - N > 0$,$\frac{{\partial n^{F} }}{{\partial \gamma }} > 0$. Further, we can get

$$ M^{2} - N^{2} = 3k^{2} \left[ {\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)\gamma ^{2} + 2\left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right)\gamma + \left( {1 - \alpha - 6\omega } \right)\left( {1 - \alpha + 2\omega } \right)} \right] $$

(A6)

For $M > 0$ and $N > 0$, we can get that if $M^{2} - N^{2} > 0$, then $M - N > 0$.

By analyzing equation (A6), we can find that:

• (1) if $\left( {k - 4\omega } \right) \ge 0$ and $\left( {1 - \alpha - 6\omega } \right) > 0$, $M > N$. That is, if $k \ge 4\omega$ and $\omega < \frac{{1 - \alpha }}{6}$,$M > N$;

• (2) if $\left( {k - 4\omega } \right) \ge 0$ and $\left( {1 - \alpha - 6\omega } \right) \le 0$, we have $k \ge 4\omega$ and $\omega \ge \frac{{1 - \alpha }}{6}$. Under this condition, when $\gamma > \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right) + 4\sqrt {\left( {k - 4\omega } \right)\omega \left( { - \left( {1 - \alpha } \right)^{2} + k\left( {1 - \alpha } \right) + 3k\omega } \right)} }}{{\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)}}$, $M > N$; and when $\gamma < \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right) + 4\sqrt {\left( {k - 4\omega } \right)\omega \left( { - \left( {1 - \alpha } \right)^{2} + k\left( {1 - \alpha } \right) + 3k\omega } \right)} }}{{\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)}}$,$M < N$;

• (3) if $\left( {k - 4\omega } \right) < 0$ and $\left( {1 - \alpha - 6\omega } \right) > 0$, we have $k < 4\omega$ and $\omega < \frac{{1 - \alpha }}{6}$. Under this condition, when $\gamma < \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right) - 4\sqrt {\left( {k - 4\omega } \right)\omega \left( { - \left( {1 - \alpha } \right)^{2} + k\left( {1 - \alpha } \right) + 3k\omega } \right)} }}{{\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)}}$, $M > N$; when $\gamma > \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right) - 4\sqrt {\left( {k - 4\omega } \right)\omega \left( { - \left( {1 - \alpha } \right)^{2} + k\left( {1 - \alpha } \right) + 3k\omega } \right)} }}{{\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)}}$,$M < N$;

• (4) if $\left( {k - 4\omega } \right) < 0$ and $\left( {1 - \alpha - 6\omega } \right) \le 0$, we have $k < 4\omega$ and $\omega \ge \frac{{1 - \alpha }}{6}$. Under this condition, $M < N$.

In summary, according to the above results, we can get that.

• (1) if $k \ge 4\omega$ and $\omega < \frac{{1 - \alpha }}{6}$, we have $\frac{{\partial n^{F} }}{{\partial \gamma }} > 0$;

• (2) if $k \ge 4\omega$ and $\omega > \frac{{1 - \alpha }}{6}$, then when $\gamma > \gamma _{1}$, $\frac{{\partial n^{F} }}{{\partial \gamma }} > 0$; and when $\gamma < \gamma _{1}$, $\frac{{\partial n^{F} }}{{\partial \gamma }} < 0$;

• (3) if $k < 4\omega$ and $\omega < \frac{{1 - \alpha }}{6}$, then when $\gamma < \gamma _{2}$, $\frac{{\partial n^{F} }}{{\partial \gamma }} > 0$; and when $\gamma > \gamma _{2}$,$\frac{{\partial n^{F} }}{{\partial \gamma }} < 0$;

• (4) if $k < 4\omega$ and $\omega \ge \frac{{1 - \alpha }}{6}$, we have $\frac{{\partial n^{F} }}{{\partial \gamma }} < 0$.

Here, $\gamma _{1} = \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right) + 4\sqrt {\left( {k - 4\omega } \right)\omega \left( { - \left( {1 - \alpha } \right)^{2} + k\left( {1 - \alpha } \right) + 3k\omega } \right)} }}{{\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)}}$.and $\gamma _{2} = \frac{{ - \left( {\left( {1 - \alpha } \right) + 6\omega } \right)\left( {k - 4\omega } \right) - 4\sqrt {\left( {k - 4\omega } \right)\omega \left( { - \left( {1 - \alpha } \right)^{2} + k\left( {1 - \alpha } \right) + 3k\omega } \right)} }}{{\left( {k + 12\omega } \right)\left( {k - 4\omega } \right)}}$.

Appendix C

Letting $\sqrt {p^{{F*}} } - \sqrt {p^{{N*}} }$, we can get

$$ \sqrt {p^{{F*}} } - \sqrt {p^{{N*}} } = \frac{{\gamma k - \left( {1 - \alpha } \right) + \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } }}{{6\sqrt {\gamma \omega } }} $$

(A7)

Since $\sqrt {p^{{F*}} } > 0$, $p^{{F*}} > 0$,$\sqrt {p^{{N*}} } > 0$ and $p^{{N*}} > 0$, we can get that if $\sqrt {p^{{F*}} } - \sqrt {p^{{N*}} } > 0$,$p^{{F*}} > p^{{N*}}$.

Analyzing equation (A7), we can find that if $\gamma k - \left( {1 - \alpha } \right) \ge 0$, $\sqrt {p^{{F*}} } - \sqrt {p^{{N*}} } > 0$. If $\gamma k - \left( {1 - \alpha } \right) < 0$, we further solving the following equation

$$ \Delta = - \left[ {\gamma k - \left( {1 - \alpha } \right)} \right]^{2} + \left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega $$

(A8)

Solving equation (A8), we can get the following equation

$$ \Delta = 4\gamma k\left( {1 - \alpha } \right) + 12\gamma \left( {1 + \gamma } \right)k\omega $$

(A9)

Clearly, $\Delta > 0$ holds. Therefore, $\sqrt {p^{{F*}} } - \sqrt {p^{{N*}} } > 0$ always holds.

Furthermore, we analyze the effects of privacy breach insurance compensation on price. Solving the first-order derivative of function $p^{{F*}}$ with respect to, we can get

$$ \frac{{\partial p^{{F*}} }}{{\partial k}} = \frac{{\left( {\left( {1 - \alpha + \gamma k} \right) + \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } } \right)}}{{18\omega }}\left( {1 + \frac{{\left( {1 - \alpha + \gamma k} \right) + 6\left( {1 + \gamma } \right)\omega }}{{\sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } }}} \right) $$

(A10)

It is clearly that $\frac{{\partial p^{{F*}} }}{{\partial k}} > 0$. Therefore, $p^{{F*}}$ is increasing in $k$.

Appendix D

The first-order derivation of function $\pi ^{F}$ with respect to $k$ is

$$ \frac{{\partial \pi ^{F} }}{{\partial k}} = \frac{{2\left[ {2\left( {1 + \gamma k} \right) + \alpha - \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } } \right]}}{{108\gamma \omega }}Z $$

(A11)

where

$$ Z = 3\gamma \left[ {\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega + \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } } \right] $$

(A12)

Since $\frac{{2\left[ {2\left( {1 + \gamma k} \right) + \alpha - \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } } \right]}}{{108\gamma \omega }} > 0$, we can get that if $M > 0$,$\frac{{\partial \pi ^{F} }}{{\partial k}} > 0$. Now we analyze equation (A11).

If $\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega \ge 0$, $\frac{{\partial \pi ^{F} }}{{\partial k}} > 0$ holds. Solving $\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega \ge 0$, we can get the following results:

• (1) if $\omega \le \frac{{1 + 2\alpha }}{6}$ and $\gamma \le \frac{{1 + 2\alpha }}{{6\omega }} - 1$, $\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega \ge 0$ holds;

• (2) if 1)$\omega \le \frac{{1 + 2\alpha }}{6}$ and $\gamma > \frac{{1 + 2\alpha }}{{6\omega }} - 1$; or 2) $\omega > \frac{{1 + 2\alpha }}{6}$, when $k \ge - \frac{{1 + 2\alpha - 6\left( {1 + \gamma } \right)\omega }}{\gamma }$, $\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega \ge 0$ holds.

If $\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega < 0$, we can further let

$$ Z1 = - \left[ {\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega } \right]^{2} + \left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega $$

(A13)

Since $\left( {1 + \gamma k} \right) + 2\alpha - 6\left( {1 + \gamma } \right)\omega < 0$ and $\sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } > 0$, we can get that if $Z1 > 0$, $Z > 0$ holds. Next, we analyze equation (A13). By solving the equation, we can get the following results:

• (1)If $\omega$ and $\gamma$ satisfy the following conditions: 1)$\frac{{1 + 2\alpha }}{6} < \omega \le \frac{{2 + \alpha }}{6}$ and $\gamma \le \frac{{2 + \alpha }}{{6\omega }} - 1$; or 2) $\omega \le \frac{{1 + 2\alpha }}{6}$ and $\frac{{1 + 2\alpha }}{{6\omega }} - 1 < \gamma \le \frac{{2 + \alpha }}{{6\omega }} - 1$, then when $k < - \frac{{1 + 2\alpha - 6\left( {1 + \gamma } \right)\omega }}{\gamma }$, $Z1 > 0$;

• (2) if $\omega$ and $\gamma$ satisfy the following conditions: 1) $\omega > \frac{{2 + \alpha }}{6}$; or 2) $\omega \le \frac{{2 + \alpha }}{6}$ and $\gamma > \frac{{2 + \alpha }}{{6\omega }} - 1$, then when $- \frac{{\left[ { - \alpha + 2\left( {1 + \gamma } \right)\omega } \right]\left[ {\left( {2 + \alpha } \right) - 6\left( {1 + \gamma } \right)\omega } \right]}}{{2\gamma \left[ { - \alpha + 4\left( {1 + \gamma } \right)\omega } \right]}} < k < - \frac{{1 + 2\alpha - 6\left( {1 + \gamma } \right)\omega }}{\gamma }$, $Z1 > 0$; and when $k < - \frac{{\left[ { - \alpha + 2\left( {1 + \gamma } \right)\omega } \right]\left[ {\left( {2 + \alpha } \right) - 6\left( {1 + \gamma } \right)\omega } \right]}}{{2\gamma \left[ { - \alpha + 4\left( {1 + \gamma } \right)\omega } \right]}}$, $Z1 < 0$.

In summary, according to the above results, we can get that.

• (1) if $\omega \le \frac{{1 + 2\alpha }}{6}$ and $\gamma \le \frac{{1 + 2\alpha }}{{6\omega }} - 1$, we can get $\frac{{\partial \pi ^{F} }}{{\partial k}} > 0$;

• (2) if $\frac{{1 + 2\alpha }}{6} < \omega \le \frac{{2 + \alpha }}{6}$ and $\gamma \le \frac{{2 + \alpha }}{{6\omega }} - 1$, we can get $\frac{{\partial \pi ^{F} }}{{\partial k}} > 0$;

• (3) if $\omega \le \frac{{1 + 2\alpha }}{6}$ and $\frac{{1 + 2\alpha }}{{6\omega }} - 1 < \gamma \le \frac{{2 + \alpha }}{{6\omega }} - 1$, we can get $\frac{{\partial \pi ^{F} }}{{\partial k}} > 0$;

• (4) if 1)$\omega > \frac{{2 + \alpha }}{6}$; or 2) $\omega \le \frac{{2 + \alpha }}{6}$ and $\gamma > \frac{{2 + \alpha }}{{6\omega }} - 1$, we can get that.

  • 1)when $k > - \frac{{\left[ { - \alpha + 2\left( {1 + \gamma } \right)\omega } \right]\left[ {\left( {2 + \alpha } \right) - 6\left( {1 + \gamma } \right)\omega } \right]}}{{2\gamma \left[ { - \alpha + 4\left( {1 + \gamma } \right)\omega } \right]}}$,$\frac{{\partial \pi ^{F} }}{{\partial k}} > 0$;

  • 2)when $k < - \frac{{\left[ { - \alpha + 2\left( {1 + \gamma } \right)\omega } \right]\left[ {\left( {2 + \alpha } \right) - 6\left( {1 + \gamma } \right)\omega } \right]}}{{2\gamma \left[ { - \alpha + 4\left( {1 + \gamma } \right)\omega } \right]}}$,$\frac{{\partial \pi ^{F} }}{{\partial k}} < 0$.

Appendix E

(1) Equilibriums when the firm not providing privacy breach insurance.

Privacy information provision. In the fourth stage, we consider how much privacy consumers will disclose to the firm. Given firm’s product price $p^{N}$, the consumer maximizes his/her utility which implying the optimal privacy provision

$$ y^{N} = \frac{v}{{2\gamma \omega }} $$

(A14)

Participate Decision. In the third stage, consumers decide whether to participate in the market. The consumer participates in the market if

$$ vy^{N} - \omega \gamma \left( {y^{N} } \right)^{2} - p^{N} \ge 0 $$

(A15)

Based on equation (A14) and equation (A15), and given firm’s product price $p^{N}$, we can derive consumers participate when

$$ v \ge \sqrt {4\gamma \omega p^{N} } $$

(A16)

and as a result, the number of participating is

$$ n^{N} = 1 - \sqrt {4\gamma \omega p^{N} } $$

(A17)

Product price. In the second stage, the firm maximizes its profit by setting the product price. The first-order of derivative of function $\pi ^{N}$ with respect to $p^{N}$ is.

$$ \frac{{\partial \pi ^{N} }}{{\partial p^{N} }} = 1 - \alpha - 3\sqrt {\omega \gamma } \left( {p^{N} } \right)^{{\frac{1}{2}}} + \gamma c\sqrt {\gamma \omega } \left( {p^{N} } \right)^{{ - \frac{1}{2}}} $$

(A18)

Letting $\frac{{\partial \pi ^{N} }}{{\partial p^{N} }} = 0$, we derive the equilibrium product price as

$$ p^{{N*}} = \frac{{\left( {1 - \alpha + \sqrt {\left( {1 - \alpha } \right)^{2} + 12\gamma ^{2} \omega c} } \right)^{2} }}{{36\gamma \omega }} $$

(A19)

According the equilibrium product price, we can further get the equilibrium number of consumers in the market as follows

$$ n^{N} = 1 - \frac{{1 - \alpha + \sqrt {\left( {1 - \alpha } \right)^{2} + 12\gamma ^{2} \omega c} }}{{\text{3}}} $$

(A20)

(2) Equilibriums when the firm providing privacy breach insurance.

Privacy information provision. In the fourth stage, we consider the decision problem that a consumer decides how much privacy he/she will disclose to the firm. The consumer maximizes his/her utility, given the firm’s product price $p^{F}$. We can derive that the privacy provision is

$$ y^{F} = \frac{{v + \gamma k}}{{2\gamma \omega }} $$

(A21)

Participate Decision. In the third stage, consumers decide whether to participate in the market. The consumer can get a utility of zero when existing the market, implying it participates in the market if

$$ vy^{F} - \omega \gamma \left( {y^{F} } \right)^{2} - p^{F} + \gamma y^{F} k \ge 0 $$

(A22)

Based on equation (a21) and equation (a22), and given firm’s product price $p^{F}$, we can find consumers participates when

$$ v \ge \sqrt {4\gamma \omega p^{F} } - \gamma k $$

(A23)

implying the number of participating consumers

$$ n^{F} = 1 - \sqrt {4\gamma \omega p^{F} } + \gamma k $$

(A24)

Product price. In the second stage, the firm maximizes its profit by setting the product price. The first-order of derivative of function $\pi ^{F}$ with respect to $p^{F}$ is

$$ \frac{{\partial \pi ^{F} }}{{\partial p^{F} }} = 1 - \alpha - 3\sqrt {\gamma \omega p^{F} } + \gamma k + \left( {\left( {1 + \gamma } \right)k + \gamma c} \right)\sqrt {\gamma \omega } \left( {p^{F} } \right)^{{ - 1/2}} $$

(A25)

Letting $\frac{{\partial \pi ^{F} }}{{\partial p^{F} }} = 0$, we derive the equilibrium product price as

$$ p^{{F*}} = \frac{{\left( {1 - \alpha + \gamma k + \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {\left( {1 + \gamma } \right)k + \gamma c} \right)\omega } } \right)^{2} }}{{36\gamma \omega }} $$

(A26)

According to the equilibrium product price, we can further get the equilibrium number of consumers in the market

$$ n^{F} = \frac{{2\left( {1 + \gamma k} \right) + \alpha - \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {\left( {1 + \gamma } \right)k + \gamma c} \right)\omega } }}{3} $$

(A27)

Appendix F

Comparing the profits in the two scenarios, we can get

$$ \begin{gathered} \pi ^{F} - \pi ^{N} = \frac{{\left[ {2\left( {1 + \gamma k + \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } } \right) + \alpha } \right]\left[ {2\left( {1 + \gamma k} \right) + \alpha - \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } } \right]^{2} }}{{108\gamma \omega }} \hfill \\ \begin{array}{*{20}c} {\begin{array}{*{20}c} {} & {} \\ \end{array} } & {} \\ \end{array} - \frac{{\left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} }}{{108\gamma \omega }} \hfill \\ \end{gathered} $$

(A28)

For $0 \le n^{F} = \frac{{2\left( {1 + \gamma k} \right) + \alpha - \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } }}{3} \le 1$, we can get that $2\left( {1 + \gamma k} \right) + \alpha \ge \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega }$ and $2\left( {1 + \gamma k} \right) + \alpha - 3 \le \sqrt {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega }$.

Hence, we can get

$$ \pi ^{F} - \pi ^{N} \le \frac{1}{{108\gamma \omega }}\left\{ \begin{gathered} \left[ {2\left( {1 + \gamma k} \right) + \alpha } \right]^{3} - \left( {2\left( {1 + \gamma k} \right) + \alpha } \right)\left[ {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } \right] \hfill \\ - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} \hfill \\ \end{gathered} \right\} $$

(A29)

and

$$ \pi ^{F} - \pi ^{N} \ge \frac{1}{{108\gamma \omega }}\left\{ \begin{gathered} \left[ {2\left( {1 + \gamma k} \right) + \alpha } \right]^{3} + \left[ {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } \right]\left[ { - \left( {2\left( {1 + \gamma k} \right) + \alpha } \right) - 6} \right] \hfill \\ - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} \hfill \\ \end{gathered} \right\} $$

(A30)

Analyzing equation (A29), we can get that if

$$ \left[ {2\left( {1 + \gamma k} \right) + \alpha } \right]^{3} - \left( {2\left( {1 + \gamma k} \right) + \alpha } \right)\left[ {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } \right] - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} < 0 $$

(A31)

then $\pi ^{F} < \pi ^{N}$ holds. Here, we can easily get that

$$ \left[ {2\left( {1 + \gamma k} \right) + \alpha } \right]^{3} - \left( {2\left( {1 + \gamma k} \right) + \alpha } \right)\left( {1 - \alpha + \gamma k} \right)^{2} - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} > 0 $$

(A32)

Based on equation (A32) and solving equation (A31), we can get the following results:

$$ \omega > \frac{{3\left( {1 + \gamma k} \right)\left( {2\left( {1 + \gamma k} \right) + \alpha } \right)\left( {1 + \gamma k + 2\alpha } \right) - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} }}{{12\gamma \left( {1 + \gamma } \right)k\left( {2\left( {1 + \gamma k} \right) + \alpha } \right)}}: = \omega _{H} $$

Analyzing equation (A30), we can get that if

$$ \left[ {2\left( {1 + \gamma k} \right) + \alpha } \right]^{3} + \left[ {\left( {1 - \alpha + \gamma k} \right)^{2} + 12\gamma \left( {1 + \gamma } \right)k\omega } \right]\left[ { - \left( {2\left( {1 + \gamma k} \right) + \alpha } \right) - 6} \right] - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} > 0 $$

(A33)

then $\pi ^{F} > \pi ^{N}$ holds.

Solving equation (A33), we can get when.

$$ \omega < \frac{{3\left( {1 + \gamma k} \right)\left( {2\left( {1 + \gamma k} \right) + \alpha } \right)\left( {1 + \gamma k + 2\alpha } \right) - 6\left( {1 - \alpha + \gamma k} \right)^{2} - \left( { - \alpha + 4} \right)\left( {2\alpha + 1} \right)^{2} }}{{12\gamma \left( {1 + \gamma } \right)k\left[ {\left( {2\left( {1 + \gamma k} \right) + \alpha } \right) + 6} \right]}}: = \omega _{L} $$

equation (A33) holds.