Pricing and personal data collection strategies of online platforms in the face of privacy concerns

Abstract

We consider a platform providing free content for users and earning profit from the sale of advertising. The platform can collect and analyze personal data to customize advertisements for individual users. Customization may alleviate users’ aversion to advertising, but it may also raise privacy concerns. Considering the platform as a two-sided market with asymmetric externalities, we investigate the effects of privacy concerns on the platform’s optimal advertising pricing and data collection strategies. We find that it is always beneficial for the platform to collect and analyze personal data. When users attach less concerns on privacy, the price increases with the efficiency of efforts and decreases with the initial nuisance cost of advertisements to users. However, if users attach more concerns on privacy, the price decreases with the efficiency of efforts and increases with the initial nuisance cost of advertisements to users.

Appendix

Proof of Proposition 1

Differentiating Function (8) with respect to $p,e$, we obtain:

$$ \frac{{\partial \pi_{P} }}{\partial p} = \frac{{ - 2p + \alpha \left( {1 - \gamma e^{2} } \right) + \frac{{ke^{2} }}{2}}}{{1 + \alpha \left( {\delta - \beta e} \right)}} $$

(9)

$$ \frac{{\partial \pi_{P} }}{\partial e} = \frac{{p - \frac{{ke^{2} }}{2}}}{{1 + \alpha \left( {\delta - \beta e} \right)}}\left( { - \frac{{ke\left( {\alpha \left( {1 - \gamma e^{2} } \right) - p} \right)}}{{p - \frac{{ke^{2} }}{2}}} + \frac{{ - 2\alpha \gamma e\left( {1 + \alpha \left( {\delta - \beta e} \right)} \right) + \alpha \beta \left( {\alpha \left( {1 - \gamma e^{2} } \right) - p} \right)}}{{1 + \alpha \left( {\delta - \beta e} \right)}}} \right) $$

(10)

By the first order condition $\frac{{\partial \pi_{p} }}{\partial p} = 0,\frac{{\partial \pi_{p} }}{\partial e} = 0$, we have

$$ p^{*} = \frac{{\alpha \left( {1 - \gamma e^{*2} } \right)}}{2} + \frac{{ke^{*2} }}{4} $$

(11)

and $e^{*}$ satisfies

$$ 3\alpha \beta \left( {\frac{k}{4} + \frac{\alpha \gamma }{2}} \right)e^{2} - \left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)e + \frac{{\alpha^{2} \beta }}{2} = 0 $$

(12)

In the quadratic Eq. (12),

$$ \Delta = \left( {k + 2\alpha \gamma } \right)^{2} (1 + \alpha \delta )^{2} - \frac{{3\alpha^{3} \beta^{2} \left( {k + 2\alpha \gamma } \right)}}{2} $$

(13)

And if $k \ge \frac{{2\alpha^{3} \beta^{2} }}{{(1 + \alpha \delta )^{2} }} - 2\alpha \gamma$, it follows that

$$ \left( {k + 2\alpha \gamma } \right)(1 + \alpha \gamma )^{2} - 2\alpha^{3} \beta^{2} \ge 0 $$

(14)

Since $k + 2\alpha \gamma > 0$, we have $\left( {k + 2\alpha \gamma } \right)^{2} (1 + \alpha \gamma )^{2} - 2\alpha^{3} \beta^{2} \left( {k + 2\alpha \gamma } \right) \ge 0$. As a result, $\Delta > 0$, there are two different solutions of Eq. (12).

By solving the quadratic Eq. (12), the two solutions to (12) are as following: $e_{1} = \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - 2\sqrt \Delta }}{{3\alpha \beta \left( {k + 2\alpha \gamma } \right)}}$, $e_{2} = \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) + 2\sqrt \Delta }}{{3\alpha \beta \left( {k + 2\alpha \gamma } \right)}}$.

Next, we will prove that $e_{2}$ is not a feasible solution to the problem.

If $e = e_{2}$, then

$$ p - \frac{{ke^{2} }}{2} = \frac{\alpha }{2} - \frac{{\left( {k + 2\alpha \gamma } \right)e^{2} }}{4} = \frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} }}{{9\alpha^{2} \beta^{2} }} - \frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }} $$

(15)

Since

$$ \left( {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} }}{{9\alpha^{2} \beta^{2} }}} \right)^{2} - \left( {\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }}} \right)^{2} = \frac{2}{{9\alpha \beta^{2} }}\left( {2\alpha^{3} \beta^{2} - \left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} } \right) $$

(16)

And from Eq. (14), it follows that $\frac{2}{{9\alpha \beta^{2} }}\left( {2\alpha^{3} \beta^{2} - \left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} } \right) \le 0$. Hence, Eq. (16) is non-positive, $\left( {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}} \right)^{2} - \left( {\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }}} \right) \le 0$. Thus, $\left| {\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }}} \right| \ge \left| {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}} \right|$. Because $\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }} > 0$, $\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }} \ge \left| {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}} \right| \ge \frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}$. Hence, from Eq. (15), we have $p - \frac{{ke^{2} }}{2} \le 0$. This implies that the profit of the platform from each advertisement is nonpositive. Obviously, it is unreasonable.

If $e = e_{1}$,

$$ p - \frac{{ke^{2} }}{2} = \frac{\alpha }{2} - \frac{{\left( {k + 2\alpha \gamma } \right)e^{2} }}{4} = \frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} }}{{9\alpha^{2} \beta^{2} }} + \frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }} $$

(17)

Because

$$ \left( {\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }}} \right)^{2} - \left( {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} }}{{9\alpha^{2} \beta^{2} }}} \right)^{2} = \frac{2}{{9\alpha \beta^{2} }}\left( {\left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} - 2\alpha^{3} \beta^{2} } \right) $$

(18)

And from Eq. (14), it follows that $\frac{2}{{9\alpha \beta^{2} }}\left( {\left( {k + 2\alpha \gamma } \right)(1 + \alpha \delta )^{2} - 2\alpha^{3} \beta^{2} } \right) \ge 0$. Hence, Eq. (18) is nonnegative. Namely, $\left| {\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }}} \right| \ge \left| {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}} \right|$. Because of $\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }} > 0$, $\frac{{2\left( {1 + \alpha \delta } \right)\sqrt \Delta }}{{9\alpha^{2} \beta^{2} }} \ge \left| {\frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}} \right| \ge \frac{2\alpha }{3} - \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)^{2} }}{{9\alpha^{2} \beta^{2} }}$. As a result, from Eq. (17), $p - \frac{{ke^{2} }}{2} >$ $0$ must hold, and $p > 0$. And from Eq. (13), we have $e > 0$.

Moreover,

$$ \alpha \left( {1 - \gamma e^{2} } \right) - p = \frac{{\alpha \left( {1 - \gamma e^{2} } \right)}}{2} - \frac{{ke^{2} }}{4} = p - \frac{{ke^{2} }}{2} > 0 $$

(19)

Besides,

$$ 1 + \alpha \left( {\delta - \beta e} \right) = \frac{1 + \alpha \delta }{3} + \frac{2\sqrt \Delta }{{3\left( {k + 2\alpha \gamma } \right)}} > 0 $$

(20)

Therefore, $\pi_{P} = \left( {p - \frac{{ke^{2} }}{2}} \right)\frac{{\alpha \left( {1 - \gamma e^{2} } \right) - p}}{{1 + \alpha \left( {\delta - \beta e} \right)}} > 0$. Thus, we have $e^{*} = e_{1} = \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - 2\sqrt \Delta }}{{3\alpha \beta \left( {k + 2\alpha \gamma } \right)}}$, and $p^{*} = \frac{{\alpha \left( {1 - \gamma e^{*2} } \right)}}{2} + \frac{{ke^{*2} }}{4}$. And we can prove that $0 < N_{A}^{*} < 1$, and $0 < N_{C}^{*} < 1$.

From Eq. (19) and Eq. (20), we have $1 + \alpha \left( {\delta - \beta e^{*} } \right) > 0$, $\alpha \left( {1 - \gamma e^{*2} } \right) - p > 0$. Combining with Eq. (6), we obtain $N_{A}^{*} > 0$.

From Eq. (5), we know $N_{A}^{*} = \alpha N_{C}^{*} - p^{*}$. Because we have proofed $N_{A}^{*} > 0$, it follows that $\alpha N_{C}^{*} - p^{*} > 0$. Hence, $N_{C}^{*} > \frac{{p^{*} }}{\alpha } > 0$.

Further, from Eq. (19), we have $p^{*} < \alpha \left( {1 - \gamma e^{2} } \right)$. Then, we obtain $\frac{{1 - \gamma e^{2} + p\left( {\delta - \beta e} \right)}}{{1 + \alpha \left( {\delta - \beta e} \right)}} < \frac{{1 - \gamma e^{2} + \alpha \left( {1 - \gamma e^{2} } \right)\left( {\delta - \beta e} \right)}}{{1 + \alpha \left( {\delta - \beta e} \right)}} = 1 - \gamma < 1$. Hence, from Eq. (7), we prove $N_{C}^{*} < 1$.

$$ N_{A}^{*} = \frac{{2\alpha \left( {1 - \gamma e^{*2} } \right) - ke^{*2} }}{{4\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right)}} $$

(21)

Because

$$ 4\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right) - \left( {2\alpha \left( {1 - \gamma e^{*2} } \right) - ke^{*2} } \right) = 4\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right) + ke^{*2} + 2\alpha \gamma e^{*2} - 2\alpha $$

(22)

where $ke^{*2} + 2\alpha \gamma e^{*2} - 2\alpha = \frac{{8\left( {1 + \alpha \delta } \right)\left( {\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta } \right) + 12\alpha^{3} \beta^{2} }}{{9\alpha^{2} \beta^{2} }}$. From Eq. (13), we can find $\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) > \sqrt \Delta$. Hence, $ke^{*2} + 2\alpha \gamma e^{*2} - 2\alpha > 0$. And from Eq. (20), we have $1 + \alpha \left( {\delta - \beta e^{*} } \right) > 0$. Therefore, $4\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right) + ke^{*2} + 2\alpha \gamma e^{*2} - 2\alpha > 0$. Based on Eq. (22), we can know $4\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right) > 2\alpha \left( {1 - \gamma e^{*2} } \right) - ke^{*2}$. Namely, $\frac{{2\alpha \left( {1 - \gamma e^{*2} } \right) - ke^{*2} }}{{4\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right)}} < 1$. From Eq. (21), we prove $N_{A}^{*} < 1$.

Next, we will verify if the second order conditions hold for $p^{*}$ and $e^{*}$. The second-order derivatives of the platform’s profit function are as follows:

$$ \frac{{\partial^{2} \pi_{P} }}{{\partial p^{2} }} = - \frac{2}{{1 + \alpha \left( {\delta - \beta e} \right)}} $$

$$ \frac{{\partial^{2} \pi_{P} }}{{\partial e^{2} }} = - \frac{1}{{1 + \alpha \left( {\delta - \beta e} \right)}}\left( {\left( {\frac{{\alpha \left( {1 - \gamma e^{2} } \right)}}{2} - \frac{{ke^{2} }}{4}} \right)\left( {k + 2\alpha \gamma } \right) - 4k\alpha \gamma e^{2} } \right) $$

$$ \frac{{\partial^{2} \pi_{P} }}{\partial e\partial p} = \frac{{\left( {k - 2\alpha \gamma } \right)e}}{{1 + \alpha \left( {\delta - \beta e} \right)}} $$

From Eq. (20), we have $1 + \alpha \left( {\delta - \beta e^{*} } \right) > 0$. Thus, $\frac{{\partial^{2} \pi_{P} }}{{\partial p^{2} }} = - \frac{2}{{1 + \alpha \left( {\delta - \beta e} \right)}} < 0$. And

$$ \frac{{\partial^{2} \pi_{P} }}{{\partial p^{2} }}\frac{{\partial^{2} \pi_{P} }}{{\partial e^{2} }} - \left( {\frac{{\partial^{2} \pi_{P} }}{\partial e\partial p}} \right)^{2} = \frac{k + 2\alpha \gamma }{{\left( {1 + \alpha \left( {\delta - \beta e} \right)} \right)^{2} }}\left( {\alpha - \frac{{3\left( {k + 2\alpha \gamma } \right)e^{2} }}{2}} \right) $$

(23)

Since $\alpha - \frac{{3\left( {k + 2\alpha \gamma } \right)e^{2} }}{2} = \frac{8\sqrt \Delta }{{3\alpha \beta^{2} \left( {k + 2\alpha \gamma } \right)}} > 0$, we have $\frac{{\partial^{2} \pi_{P} }}{{\partial p^{2} }}\frac{{\partial^{2} \pi_{P} }}{{\partial e^{2} }} - \left( {\frac{{\partial^{2} \pi_{P} }}{\partial e\partial p}} \right)^{2} > 0$. As a result, the Hessian matrix is a negative definite matrix. And the profit function of the platform has the maxima on $p^{*} = \frac{{\alpha \left( {1 - \gamma e^{*2} } \right)}}{2} + \frac{{ke^{*2} }}{4}$, $e^{*} = \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - 2\sqrt \Delta }}{{3\alpha \beta \left( {k + 2\alpha \gamma } \right)}}$.□

Proof of Proposition 2

Because $e^{*} = \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - 2\sqrt \Delta }}{{3\alpha \beta \left( {k + 2\alpha \gamma } \right)}}$, deriving the above expression with respect to $\gamma$, we have $\frac{{\partial e^{*} }}{\partial \gamma } = - \frac{{\alpha^{3} \beta }}{{\left( {k + 2\alpha \gamma } \right)\sqrt \Delta }}$. Since $\alpha ,\beta ,k,\gamma$ are all positive, thus, $\frac{{\partial e^{*} }}{\partial \gamma } < 0$.

Besides, $\frac{{\partial p^{*} }}{\partial \gamma } = - \frac{{\alpha e^{*2} }}{2} - \frac{{\left( {2\alpha \gamma - k} \right)e^{*} }}{2}\frac{{\partial e^{*} }}{\partial \gamma }$. If $\gamma < \frac{k}{2\alpha }$, it follows that $2\alpha \gamma - k < 0$. In addition, $\frac{{\partial e^{*} }}{\partial \gamma } < 0$, we obtain $\frac{{\left( {2\alpha \gamma - k} \right)e^{*} }}{2}\frac{{\partial e^{*} }}{\partial \gamma } > 0$. Thus, $- \frac{{\alpha e^{*2} }}{2} - \frac{{\left( {2\alpha \gamma - k} \right)e^{*} }}{2}\frac{{\partial e^{*} }}{\partial \gamma } < 0$. Namely, $\frac{{\partial p^{*} }}{\partial \gamma } < 0$.□

Proof of Proposition 3

$\frac{{\partial e^{*} }}{\partial \beta } = \frac{{2\left( {1 + \alpha \delta } \right)\left( {\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta } \right)}}{{3\alpha \beta^{2} \sqrt \Delta }}$. From Eq. (13), we have $\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta > 0$. Therefore, $\frac{{\partial e^{*} }}{\partial \beta } > 0$.

Besides, $\frac{{\partial p^{*} }}{\partial \beta } = \frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2}\frac{{\partial e^{*} }}{\partial \beta }$. If $\gamma < \frac{k}{2\alpha }$, we have $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2} > 0$. Because $\frac{{\partial e^{*} }}{\partial \beta } > 0$, we obtain $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2}\frac{{\partial e^{*} }}{\partial \beta } > 0$. Finally, $\frac{{\partial p^{*} }}{\partial \beta } > 0$. However, if $\gamma \ge \frac{k}{2\alpha }$, we have $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2} \le 0$. Because $\frac{{\partial e^{*} }}{\partial \beta } > 0$, and $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2}\frac{{\partial e^{*} }}{\partial \beta } \le 0$., $\frac{{\partial p^{*} }}{\partial \beta } \le 0$.□

Proof of Proposition 4

. By $\frac{{\partial e^{*} }}{\partial \delta } = \frac{{2\left( {\sqrt \Delta - \left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right)} \right)}}{3\beta \sqrt \Delta }$ and Eq. (13), we have $\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta > 0$. Therefore,$\frac{{\partial e^{*} }}{\partial \delta } < 0$.

Besides, $\frac{{\partial p^{*} }}{\partial \delta } = \frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2}\frac{{\partial e^{*} }}{\partial \delta }$. If $\gamma < \frac{k}{2\alpha }$, we have $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2} > 0$. Because $\frac{{\partial e^{*} }}{\partial \delta } < 0$, it follows that $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2}\frac{{\partial e^{*} }}{\partial \delta } < 0$, so $\frac{{\partial p^{*} }}{\partial \delta } < 0$. However, if $\gamma \ge \frac{k}{2\alpha }$, we have $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2} \le 0$. Since $\frac{{\partial e^{*} }}{\partial \delta } < 0$, we have $\frac{{e^{*} \left( {k - 2\alpha \gamma } \right)}}{2}\frac{{\partial e^{*} }}{\partial \delta } \ge 0$, $\frac{{\partial p^{*} }}{\partial \delta } \ge 0$.□

Proof of Proposition 5

$\frac{{\partial e^{*} }}{\partial k} = - \frac{{\alpha^{2} \beta }}{{2\left( {k + 2\alpha \gamma } \right)\sqrt \Delta }}$. Because $\alpha ,\beta ,k,\gamma$ are all positive, $\frac{{\partial e^{*} }}{\partial k} < 0$.□

Proof of Proposition 6

. By $\frac{{\partial e^{*} }}{\partial \alpha } = \frac{{2\left( {k + 2\alpha \gamma } \right)\left( {\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta } \right) + \frac{{3k\alpha^{3} \beta^{2} }}{2}}}{{3\alpha^{2} \beta \left( {k + 2\alpha \gamma } \right)\sqrt \Delta }}$, and Eq. (13), we have $\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta > 0$. Thus, $2\left( {k + 2\alpha \gamma } \right)\left( {\left( {k + 2\alpha \gamma } \right)\left( {1 + \alpha \delta } \right) - \sqrt \Delta } \right) > 0$. And because the denominator is positive, $\frac{{\partial e^{*} }}{\partial \alpha } > 0$.□

Proof of Proposition 7

According to the Envelope Theorem, we have

$$ \frac{{\partial \pi_{P}^{*} }}{\partial \gamma } = \frac{{\partial \pi_{P} }}{\partial \gamma }|_{{p = p^{*} ,e = e^{*} }} = - \left( {p^{*} - \frac{{ke^{*2} }}{2}} \right)\frac{{\alpha e^{*2} }}{{1 + \alpha \left( {\delta - \beta e^{*} } \right)}} $$

$$ \frac{{\partial \pi_{P}^{*} }}{\partial \alpha } = \frac{{\partial \pi_{P} }}{\partial \alpha }|_{{p = p^{*} ,e = e^{*} }} = \left( {p^{*} - \frac{{ke^{*2} }}{2}} \right)\frac{{\alpha \left( {1 - \gamma e^{*2} } \right) - p^{*} + p^{*} \left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right)}}{{\alpha \left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right)^{2} }} $$

$$ \frac{{\partial \pi_{P}^{*} }}{\partial \beta } = \frac{{\partial \pi_{P} }}{\partial \beta }|_{{p = p^{*} ,e = e^{*} }} = \left( {p^{*} - \frac{{ke^{*2} }}{2}} \right)\frac{{\left( {\alpha \left( {1 - \gamma e^{*2} } \right) - p^{*} } \right)\alpha e^{*2} }}{{\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right)^{2} }} $$

$$ \frac{{\partial \pi_{P}^{*} }}{\partial k} = \frac{{\partial \pi_{P} }}{\partial k}|_{{p = p^{*} ,e = e^{*} }} = - \frac{{e^{*2} }}{2}\frac{{\alpha \left( {1 - \gamma e^{*2} } \right) - p^{*} }}{{1 + \alpha \left( {\delta - \beta e^{*} } \right)}} $$

$$ \frac{{\partial \pi_{P}^{*} }}{\partial \delta } = \frac{{\partial \pi_{P} }}{\partial \delta }|_{{p = p^{*} ,e = e^{*} }} = - \left( {p^{*} - \frac{{ke^{*2} }}{2}} \right)\frac{{\left( {\alpha \left( {1 - \gamma e^{*2} } \right) - p^{*} } \right)\alpha }}{{\left( {1 + \alpha \left( {\delta - \beta e^{*} } \right)} \right)^{2} }} $$

From Eq. (19), it follows that $p^{*} - \frac{{ke^{*2} }}{2} > 0$ and $\alpha \left( {1 - \gamma e^{*2} } \right) - p^{*} > 0$. From Eq. (20), we have $1 + \alpha \left( {\delta - \beta e^{*} } \right) > 0$. As a result, $\frac{{\partial \pi_{P}^{*} }}{\partial \gamma } < 0$, $\frac{{\partial \pi_{P}^{*} }}{\partial \alpha } > 0$, $\frac{{\partial \pi_{P}^{*} }}{\partial \beta } > 0$, $\frac{{\partial \pi_{P}^{*} }}{\partial k} < 0$, $\frac{{\partial \pi_{P}^{*} }}{\partial \delta } < 0$.□