Price Competition in the Random Coefficient Attraction Choice Models with Linear Cost

Abstract

We study the pricing game between competing retailers under various random coefficient attraction choice models. We characterize existence conditions and structure properties of the equilibrium. Moreover, we explore how the randomness and cost parameters affect the equilibrium prices and profits under multinomial logit (MNL), multiplicative competitive interaction (MCI) and linear attraction choice models. Specifically, with bounded randomness, for the MCI and linear attraction models, the randomness always reduces the retailer’s profit. However, for the MNL model, the effect of randomness depends on the product’s value gap. For high-end products (i.e., whose value gap is higher than a threshold), the randomness reduces the equilibrium profit, and vice versa. The results suggest high-end retailers in MNL markets exert more effort in disclosing their exact product performance to consumers. We also reveal the effects of randomness on retailers’ pricing decisions. These results help retailers in making product performance disclosure and pricing decisions.

Appendices

Appendix

1 Proof of Theorem 1

Since $\frac{1}{Q_i(p_i, {\varvec{p}}_{-i})}$ is a convex function, we know that

$$\begin{aligned} \frac{\partial ^2 (1/Q_i)}{\partial p_i^2} = -\frac{Q_i\partial ^2 Q_i/\partial p_i^2-2 (\partial Q_i/\partial p_i)^2}{Q_i^3}\geqslant 0. \end{aligned}$$

Therefore, $2Q_i (\partial Q_i/\partial p_i)^2-\partial ^2 Q_i/\partial p_i^2\geqslant 0$. Taking second-order derivative to $R_{i}$ with respect to $Q_i$, we have

$$\begin{aligned} \frac{\partial ^2 R_{i}}{\partial Q_i^2}= & {} 2\frac{\partial p_i}{\partial Q_i}+Q_i\frac{\partial ^2 p_i}{\partial Q_i^2}\\= & {} 2\frac{1}{\partial Q_i/\partial p_i}-Q_i\frac{\partial ^2 Q_i/\partial p_i^2}{(\partial Q_i/\partial p_i)^3}\\= & {} \frac{2 (\partial Q_i/\partial p_i)^2-Q_i\cdot \partial ^2 Q_i/\partial p_i^2}{(\partial Q_i/\partial p_i)^3}\\\leqslant & {} 0. \end{aligned}$$

The last inequality is due to the fact that $Q_i$ is strictly decreasing in $p_i$ hence $\partial Q_i/\partial p_i<0$. From the second-order derivative, we see that $R_{i}$ is concave in $Q_i$. Combining with the monotone mapping between $Q_i$ and $p_i$ that is guaranteed by Assumption 1, we obtain that $R_{i}$ is quasi-concave in $p_i$. Finally, the existence of a Nash equilibrium is derived from the Debreu–Glicksberg–Fan Theorem.

2 Proof of Proposition 1

Condition 1 is

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} p_i}\left( -p_i \frac{a_i'(p_i)}{a_i(p_i)}\right) =-\frac{a_i'+p_i a_i''}{a_i}+\frac{p_i(a_i')^2}{a_i^2}=\frac{1}{a_i^2}\left\{ p_i[(a_i')^2-a_ia_i'']-a_i'a_i\right\} >0. \end{aligned}$$

For Condition 3, we have

$$\begin{aligned} \frac{\partial }{\partial p_i}\frac{1}{a_i(\alpha _i,p_i)} =-\frac{1}{a_i^2}\frac{\partial a_i}{\partial p_i}. \end{aligned}$$

Then,

$$\begin{aligned} \qquad \quad \frac{\partial ^2}{\partial p_i^2}\frac{1}{a_i(\alpha _i,p_i)} =\frac{2}{a_i^3}\left( \frac{\partial a_i}{\partial p_i}\right) ^2-\frac{1}{a_i^2}\frac{\partial ^2 a_i}{\partial p_i^2} =\frac{1}{a_i^3}\left[ 2(\frac{\partial a_i}{\partial p_i})^2-\frac{\partial ^2 a_i}{\partial p_i^2}a_i\right] >0. \end{aligned}$$

From the above results, we conclude that the two conditions each includes a non-empty set of functions that are not covered by the other condition.

3 Proof of Proposition 2

A sufficient condition for Condition 2 is the concavity of $\frac{a_i(\alpha _i,p_i)}{a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)}$ with respect to $p_i$.

$$\begin{aligned}&\quad \frac{\partial }{\partial p_i}\frac{a_i(\alpha _i,p_i)}{a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)}\\&\quad =\frac{\frac{\partial }{\partial p_i}a_i(\alpha _i,p_i)}{a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)} - \frac{a_i(\alpha _i,p_i)\frac{\partial }{\partial p_i}a_i(\alpha _i,p_i)}{\left[ a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)\right] ^2}\\&\quad =\frac{\frac{\partial }{\partial p_i}a_i(\alpha _i,p_i)\left[ a_0+\sum _{l=1,l\ne i}^{n}a_l(\alpha _l,p_l)\right] }{\left[ a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)\right] ^2}.\\&~~~~~\quad \frac{\partial ^2}{\partial p_i^2}\frac{a_i(\alpha _i,p_i)}{a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)}\\&\quad =\frac{\frac{\partial ^2}{\partial p_i^2}a_i(\alpha _i,p_i)\left[ a_0+\sum _{l=1,l\ne i}^{n}a_l(\alpha _l,p_l)\right] }{\left[ a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)\right] ^2}\\&\quad \qquad - 2 \frac{\left[ \frac{\partial }{\partial p_i}a_i(\alpha _i,p_i)\right] ^2\left[ a_0+\sum _{l=1,l\ne i}^{n}a_l(\alpha _l,p_l)\right] }{\left[ a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)\right] ^3}\\&\quad =\frac{a_0+\sum _{l=1,l\ne i}^{n}a_l(\alpha _l,p_l)}{\left[ a_0+\sum _{l=1}^{n}a_l(\alpha _l,p_l)\right] ^3}\\&\quad \qquad \times \left\{ \frac{\partial ^2}{\partial p_i^2}a_i(\alpha _i,p_i)\left[ a_0+\sum _{l=1,l\ne i}^{n}a_l(\alpha _l,p_l)\right] - 2\left[ \frac{\partial }{\partial p_i}a_i(\alpha _i,p_i)\right] ^2\right\} . \end{aligned}$$

4 Proof of Proposition 3

Denote $R_d$ as the profit function under the deterministic model, i.e.,

$$\begin{aligned} R_d(\alpha , p)=\frac{\exp (\alpha -\gamma p)}{a_0+\exp (\alpha -\gamma p)} \cdot (p-c). \end{aligned}$$

Then, taking first-order derivative to $R_d$ with respect to p yields to

$$\begin{aligned} \frac{\partial R_d (\alpha ,p)}{\partial p}=\frac{-\gamma a_0 \exp (\alpha -\gamma p)}{(a_0+\exp (\alpha -\gamma p))^2}(p-c)+\frac{\exp (\alpha -\gamma p)}{a_0+\exp (\alpha -\gamma p)}. \end{aligned}$$

Therefore, since $R_d$ is quasi-concave in p, $p_d^*$ takes the value such that the first-order derivative equals to 0.

$$\begin{aligned} \frac{\partial R_d (\alpha ,p)}{\partial p} \bigg |_{p=p_d^*}=\frac{-\gamma a_0 \exp (\alpha -\gamma p_d^*)}{(a_0+\exp (\alpha -\gamma p_d^*))^2}(p_d^*-c)+\frac{\exp (\alpha -\gamma p_d^*)}{a_0+\exp (\alpha -\gamma p_d^*)} =0. \end{aligned}$$

(A1)

Analogously, $p_r^*$ takes the value such that the first-order derivative of the profit function under the model with random coefficient equals to 0.

$$\begin{aligned}&\quad \, \frac{\partial E_{\delta } (R_d (\alpha +\delta ,p))}{\partial p} \bigg |_{p=p_r^*} =E_{\delta } \frac{\partial R_d (\alpha +\delta ,p)}{\partial p} \bigg |_{p=p_r^*} \\&\quad =\int _{\delta }\frac{-\gamma a_0 \exp (\alpha +\delta -\gamma p_r^*)}{(a_0+\exp (\alpha +\delta -\gamma p_r^*))^2}(p_r^*-c)+\frac{\exp (\alpha +\delta -\gamma p_r^*)}{a_0+\exp (\alpha +\delta -\gamma p_r^*)} \mathrm{d}F(\delta )\\&\quad =0. \end{aligned}$$

Our target is to compare $p_r^*$ with $p_d^*$. From the quasi-concavity of $E_{\delta }R_d (\alpha +\delta ,p)$, we know that if $E_{\delta } \frac{\partial R_d (\alpha +\delta ,p)}{\partial p} \big |_{p=p_d^*}>0$, then $p_r^*>p_d^*$. And if $E_{\delta } \frac{\partial R_d (\alpha +\delta ,p)}{\partial p} \big |_{p=p_d^*}<0$, then $p_r^*<p_d^*$. Therefore, our target reduces to comparing $E_{\delta } \frac{\partial R_d (\alpha +\delta ,p)}{\partial p} \big |_{p=p_d^*}$ with 0. The key for this comparison is to show the concavity/convexity of $\frac{\partial R_d (\alpha ,p)}{\partial p} \big |_{p=p_d^*}$ with respect to $\alpha$ on a small interval containing $\alpha$.

Taking partial derivatives, we have

$$\begin{aligned} \frac{\partial ^2 R_d (\alpha ,p)}{\partial p \partial \alpha } \bigg |_{p=p_d^*} =&\frac{a_0 \gamma \exp (\alpha -\gamma p_d^*)(\exp (\alpha -\gamma p_d^*)-a_0)}{(a_0+\exp (\alpha -\gamma p_d^*))^3}\cdot (p_d^*-c)\\&+\frac{a_0 \exp (\alpha -\gamma p_d^*)}{(a_0+\exp (\alpha -\gamma p_d^*))^2}, \end{aligned}$$

$$\begin{aligned} \frac{\partial ^3 R_d (\alpha ,p)}{\partial p \partial \alpha ^2} \bigg |_{p=p_d^*}= & {} \bigg [\frac{a_0 \gamma (2 \exp (2\alpha -2\gamma p_d^*)-a_0 \exp (\alpha -\gamma p_d^*))(a_0+\exp (\alpha -\gamma p_d^*))}{(a_0+\exp (\alpha -\gamma p_d^*))^4}\nonumber \\&-\frac{3a_0\gamma (\exp (2\alpha -2\gamma p_d^*)-a_0 \exp (\alpha -\gamma p_d^*))\exp (\alpha -\gamma p_d^*)}{(a_0+\exp (\alpha -\gamma p_d^*))^4}\bigg ](p_d^*-c)\nonumber \\&+\frac{a_0 \exp (\alpha -\gamma p_d^*) (a_0-\exp (\alpha -\gamma p_d^*))}{(a_0+\exp (\alpha -\gamma p_d^*))^3}. \end{aligned}$$

(A2)

Bring Eq. (A1) into Eq. (A2), we have

$$\begin{aligned} \frac{\partial ^3 R_d (\alpha ,p)}{\partial p \partial \alpha ^2} \bigg |_{p=p_d^*} =\frac{\exp (2\alpha -2\gamma p_d^*)(3a_0-\exp (\alpha -\gamma p_d^*))}{(a_0+\exp (\alpha -\gamma p_d^*))^3}. \end{aligned}$$

Therefore, if $\exp (\alpha -\gamma p_d^*)<3a_0$, then $\frac{\partial ^3 R_d (\alpha ,p)}{\partial p \partial \alpha ^2} \big |_{p=p_d^*}>0$, and vice versa.

In addition, it is straightforward to check that $\frac{\partial ^3 R_d (\alpha ,p)}{\partial p \partial \alpha ^2}\big |_{p=p_d^*}$ is continuous in $\alpha$. As a consequence, there exists an interval $[\alpha -{\bar{\delta }}_1,\alpha +{\bar{\delta }}_1]$ such that $\frac{\partial R_d (\alpha ,p)}{\partial p}\big |_{p=p_d^*}$ is strictly convex in $\alpha$ for $\alpha \in [\alpha -{\bar{\delta }}_1,\alpha +{\bar{\delta }}_1]$ if $\exp (\alpha -\gamma p_d^*)<3a_0$. By the property of convex function, if the random term $\delta$ supports on $[-{\bar{\delta }}_1,{\bar{\delta }}_1]$, then

$$\begin{aligned} {{E}}_\delta \left( \frac{\partial R_d (\alpha +\delta , p)}{\partial p}\bigg |_{p=p_d^*}\right) >&\frac{\partial R_d ({{E}}_\delta (\alpha +\delta ),p)}{\partial p}\bigg |_{p=p_d^*}\\ =&\frac{\partial R_d (\alpha ,p)}{\partial p}\bigg |_{p=p_d^*}\\ =&0. \end{aligned}$$

Therefore, $p_r^*>p_d^*$ if $\exp (\alpha -\gamma p_d^*)<3a_0$ holds. Similarly, if $\exp (\alpha -\gamma p_d^*)>3a_0$, there exists ${\bar{\delta }}_2$ such that if $\delta$ supports on $[-{\bar{\delta }}_2,{\bar{\delta }}_2]$, then $p_r^*<p_d^*$.

Finally, from Eq. (A1), we have

$$\begin{aligned} \alpha -\gamma p_d^*+\frac{1}{a_0} \exp (\alpha -\gamma p_d^*)+1=\alpha -\gamma c. \end{aligned}$$

(A3)

Since the left-hand side of the above equation is strictly increasing in $\alpha -\gamma p_d^*$, the inequality $\exp (\alpha -\gamma p_d^*)<3a_0$ is equivalent with $\alpha -\gamma c<4+\log (3a_0)$, and $\exp (\alpha -\gamma p_d^*)>3a_0$ is equivalent with $\alpha -\gamma c>4+\log (3a_0)$. Hence, we have completed the proof.

5 Proof of Proposition 4

Recall that $R_d(\alpha ,p)= \frac{\exp (\alpha -\gamma p)}{a_0+\exp (\alpha -\gamma p)} (p-c)$. Taking second-order derivative to $R_d$ with respect to $\alpha$, we have

$$\begin{aligned} \frac{\partial ^2 R_d(\alpha ,p)}{\partial \alpha ^2} =\frac{a_0 \exp (\alpha -\gamma p)(a_0-\exp (\alpha -\gamma p))}{(a_0+\exp (\alpha -\gamma p))^3}\cdot (p-c). \end{aligned}$$

Therefore, if $a_0-\exp (\alpha -\gamma p_d^*)>0$, then $R_d(\alpha ,p_d^*)$ is strictly convex with respect to $\alpha$ on an interval containing $\alpha$; And if $a_0-\exp (\alpha -\gamma p_d^*)<0$, then $R_d(\alpha ,p_d^*)$ is concave with respect to $\alpha$ on an interval containing $\alpha$. Combining with Eq. (A3), $a_0-\exp (\alpha -\gamma p_d^*)>0$ is equivalent with $\alpha -\gamma c<2+\log a_0$.

Therefore, if $\alpha -\gamma c<2+\log a_0$, by the convexity of $R_d(\alpha ,p_d^*)$, there exists ${\bar{\delta }}_3>0$ such that if $\delta$ supports on $[-{\bar{\delta }}_3,{\bar{\delta }}_3]$,

$$\begin{aligned} R_d(\alpha ,p_d^*) =&R_d(\alpha +E \delta ,p_d^*)\\ <&{{E}}_\delta R_d(\alpha +\delta ,p_d^*)\\ =&R_r({\tilde{\alpha }},p_d^*)\\ \leqslant&R_r({\tilde{\alpha }},p_r^*). \end{aligned}$$

On the other hand, if $\alpha -\gamma c>4+\log 3a_0$, combining with Proposition 3, we have

$$\begin{aligned} a_0-\exp (\alpha -\gamma p_r^*)<a_0-\exp (\alpha -\gamma p_d^*)<0. \end{aligned}$$

Therefore, it is straightforward to show that $R_d(\alpha ,p_r^*)$ is strictly concave with respect to $\alpha$ on an interval containing $\alpha$. Analogously, there exists ${\bar{\delta }}_4>0$ such that if $\delta$ supports on $[-{\bar{\delta }}_4,{\bar{\delta }}_4]$,

$$\begin{aligned} R_d(\alpha ,p_d^*) \geqslant&R_d(\alpha ,p_r^*)\\ >&{{E}}_\delta R_d(\alpha +\delta ,p_r^*)\\ =&R_r({\tilde{\alpha }},p_r^*). \end{aligned}$$

6 Proof of Proposition 5

The main procedure is similar to the proof of Proposition 3. Now, the profit functions have the following form:

$$\begin{aligned} R_d(\alpha ,p)&=\frac{\alpha p^{-\gamma }}{a_0+\alpha p^{-\gamma }} (p-c),\\ R(\alpha ,p)&=\int _{\delta } \frac{(\alpha +\delta ) p^{-\gamma }}{a_0+(\alpha +\delta ) p^{-\gamma }} (p-c)\mathrm{d}F(\delta )={{E}}_\delta R_d(\alpha +\delta ,p). \end{aligned}$$

Taking derivatives, we have

$$\begin{aligned} \frac{\partial R_d(\alpha ,p)}{\partial p}&=\frac{-\gamma \alpha a_0 p^{-\gamma -1}}{(a_0+\alpha p^{-\gamma })^2}\cdot (p-c)+\frac{\alpha p^{-\gamma }}{a_0+\alpha p^{-\gamma }},\nonumber \\ \frac{\partial ^2 R_d(\alpha ,p)}{\partial p \partial a}&=\frac{-\gamma p^{-\gamma -1} a_0 (a_0-\alpha p^{-\gamma })}{(a_0+\alpha p^{-\gamma })^3}\cdot (p-c)+\frac{a_0 p^{-\gamma }}{(a_0+\alpha p^{-\gamma })^2},\nonumber \\ \frac{\partial ^3 R_d(\alpha ,p)}{\partial p \partial a^2}&=\frac{4\gamma a_0^2 p^{-2\gamma -1}-2\gamma a_0 \alpha p^{-3\gamma -1}}{(a_0+\alpha p^{-\gamma })^4}\cdot (p-c)-\frac{2 a_0 p^{-2 \gamma }}{(a_0+\alpha p^{-\gamma })^3}. \end{aligned}$$

(A4)

Since $p_d^*$ satisfies the first-order condition, we can derive that

$$\begin{aligned} p_d^*-c=\frac{(a_0+ \alpha p_d^{*-\gamma }) p_d^*}{\gamma a_0 }. \end{aligned}$$

(A5)

Bring Eq. (A5) into Eq. (A4), we have

$$\begin{aligned} \frac{\partial ^3 R_d(\alpha ,p)}{\partial p \partial a^2}\bigg |_{p=p_d^*} =&\frac{4\gamma a_0^2 p_d^{*-2\gamma -1}-2\gamma a_0 \alpha p_d^{*-3\gamma -1}}{(a_0+\alpha p_d^{*-\gamma })^4}\cdot \frac{(a_0+ \alpha p_d^{*-\gamma }) p_d^*}{\gamma a_0 }\\&-\frac{2 a_0 p_d^{*-2 \gamma }}{(a_0+\alpha p_d^{*-\gamma })^3}\\ =&\frac{2 p_d^{*-2 \gamma }(a_0-\alpha p_d^{*-\gamma })}{(a_0+\alpha p_d^{*-\gamma })^3}. \end{aligned}$$

Therefore, following the same argument as in the proof of Proposition 3, if $\alpha p_d^{*-\gamma }<a_0$, $\frac{\partial R_d(\alpha ,p)}{\partial p }\big |_{p=p_d^*}$ is strictly convex with respect to $\alpha$ on an interval containing $\alpha$, and therefore $p_r^*>p_d^*$; Otherwise, if $\alpha p_d^{*-\gamma }>a_0$, then $p_r^*<p_d^*$.

To simplify the condition $\alpha p_d^{*-\gamma }<a_0$ one step further, revisit Eq. (A5), we have

$$\begin{aligned} \alpha c^{-\gamma }=\bigg (\frac{\gamma a_0}{(\gamma -1) a_0-\alpha p_d^{*-\gamma }}\bigg )^{\gamma }\alpha p_d^{*-\gamma }. \end{aligned}$$

(A6)

If $\gamma \leqslant 2$, it is easy to note that $\alpha p_d^{*-\gamma }<a_0$ for sure, otherwise the right-hand side of Eq. (A6) would be negative and cannot equal to the left-hand side of Eq. (A6). Therefore, $p_r^*>p_d^*$ for sure if $\gamma \leqslant 2$. On the other side, if $\gamma >2$, since the right-hand side of Eq. (A6) is increasing in $\alpha p_d^{*-\gamma }$, the equivalent condition can be written as:

$$\begin{aligned} \alpha p_d^{*-\gamma }<a_0 \iff \alpha c^{-\gamma }<\left( \frac{\gamma }{\gamma -2}\right) ^{\gamma } a_0, \end{aligned}$$

which complete the proof.

7 Proof of Proposition 6

Taking second-order derivative to $R_d(\alpha , p)$ with respect to $\alpha$, we have

$$\begin{aligned} \frac{\partial ^2 R_d(\alpha , p)}{\partial \alpha ^2}=\frac{-2 a_0 p^{-2 \gamma }}{(a_0+\alpha p^{-\gamma })^3}(p-c), \end{aligned}$$

that is negative for either $p=p_d^*$ or $p=p_r^*$. This implies that $R_d(\alpha ,p_r^*)$ is strictly concave with respect to $\alpha$. Therefore,

$$\begin{aligned} R_r({\tilde{\alpha }},p_r^*)={{E}}_\delta R_d(\alpha +\delta , p_r^*) <R_d(\alpha +E \delta , p_r^*)=R_d(\alpha , p_r^*) \leqslant R_d(\alpha , p_d^*) \end{aligned}$$

that complete the proof.

8 Proof of Proposition 7

The proof approach is similar to the proof of Proposition 3. Now taking derivaties, we have

$$\begin{aligned} R_d(\alpha ,p)&=\frac{\alpha -\gamma p}{a_0+\alpha -\gamma p} (p-c),\nonumber \\ \frac{\partial R_d(\alpha ,p)}{\partial p}&=\frac{-\gamma a_0}{(a_0+\alpha -\gamma p)^2}\cdot (p-c)+\frac{\alpha -\gamma p}{a_0+\alpha -\gamma p}, \end{aligned}$$

(A7)

$$\begin{aligned} \frac{\partial ^2 R_d(\alpha ,p)}{\partial p \partial \alpha }&=\frac{2\gamma a_0}{(a_0+\alpha -\gamma p)^3}\cdot (p-c)+\frac{ a_0}{(a_0+\alpha -\gamma p)^2},\nonumber \\ \frac{\partial ^3 R_d(\alpha ,p)}{\partial p \partial \alpha ^2}&=\frac{-6\gamma a_0}{(a_0+\alpha -\gamma p)^4}\cdot (p-c)+\frac{-2 a_0}{(a_0+\alpha -\gamma p)^3}. \end{aligned}$$

(A8)

Therefore, since $p_d^*$ satisfies the first-order condition, i.e., makes Eq. (A7) equals to zero, we can take $p_d^*$ into Eq. (A8) and obtain that

$$\begin{aligned} \frac{\partial ^3 R_d(\alpha ,p)}{\partial p \partial \alpha ^2}\bigg |_{p=p_d^*}&=\frac{-6\gamma a_0}{(a_0+\alpha -\gamma p_d^*)^4}\frac{(a_0+\alpha -\gamma p_d^*)(\alpha -\gamma p_d^*)}{\gamma a_0}\\&-\frac{2 a_0}{(a_0+\alpha -\gamma p_d^*)^3}\\&=\frac{-2 a_0-6(\alpha -\gamma p_d^*)}{(a_0+\alpha -\gamma p_d^*)^3}\\&<0. \end{aligned}$$

By the similar argument as that in the proof of Proposition 3, we know that $\frac{\partial R_d(\alpha ,p)}{\partial p}\big |_{p=p_d^*}$ is strictly concave in $\alpha$ and therefore $\frac{\partial R_r({\tilde{\alpha }},p)}{\partial p}\big |_{p=p_d^*}<\frac{\partial R_d(\alpha ,p)}{\partial p}\big |_{p=p_d^*}=0$. Combining with the quasi-concavity of $R(\alpha , p)$ with respect to p, we know that $p_r^* < p_d^*$. The only requirement is that the attraction factor has to be positive, i.e., $\alpha +\delta -\gamma p_d^*>0$.

9 Proof of Proposition 8

Taking second-order derivative to $R_d$ with respect to $\alpha$, we have

$$\begin{aligned} \frac{\partial ^2 R_d(\alpha ,p)}{\partial \alpha ^2}&=\frac{-2 a_0}{(a_0+\alpha -\gamma p)^3}(p-c) <0 \end{aligned}$$

for either $p=p_r^*$ or $p=p_d^*$, which implies the concavity of $R_d(\alpha , p)$ with respect to $\alpha$.

Therefore, with a positive attraction factor, i.e., $\alpha +\delta >0$,

$$\begin{aligned} R_r({\tilde{\alpha }},p_r^*)={{E}}_\delta R_d(\alpha +\delta ,p_r^*)<R_d(\alpha ,p_r^*)\leqslant R_d(\alpha ,p_d^*). \end{aligned}$$

10 Proof of Theorem 2

Unimodality for $R_1$:

$$\begin{aligned} R_1= & {} \ln \left( \frac{\mathrm{e}^{\alpha _1+b-r_1p_1}+K}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K}\right) (p_1-c_1) \\ \frac{\partial R_1}{\partial p_1}= & {} \ln \left( \frac{\mathrm{e}^{\alpha _1+b-r_1p_1}+K}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K}\right) +\left[ \frac{-r_1\mathrm{e}^{\alpha _1+b-r_1p_1}}{\mathrm{e}^{\alpha _1+b-r_1p_1}+K}-\frac{-r_1\mathrm{e}^{\alpha _1-b-r_1p_1}}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K}\right] (p_1-c_1)\\= & {} \ln \left( 1+\frac{\mathrm{e}^{\alpha _1+b-r_1p_1}-\mathrm{e}^{\alpha _1-b-r_1p_1}}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K}\right) \\&+\frac{Kr_1(\mathrm{e}^{\alpha _1-b-r_1p_1}-\mathrm{e}^{\alpha _1+b-r_1p_1})}{(\mathrm{e}^{\alpha _1+b-r_1p_1}+K)(\mathrm{e}^{\alpha _1-b-r_1p_1}+K)}(p_1-c_1)\\= & {} \frac{(\mathrm{e}^{\alpha _1-b-r_1p_1}-\mathrm{e}^{\alpha _1+b-r_1p_1})}{(\mathrm{e}^{\alpha _1-b-r_1p_1}+K)}\left[ -\frac{\ln (1+f(p_1))}{f(p_1)}+Kr_1\frac{(p_1-c_1)}{\mathrm{e}^{\alpha _1+b-r_1p_1}+K}\right] \\= & {} \frac{(\mathrm{e}^{\alpha _1-b-r_1p_1}-\mathrm{e}^{\alpha _1+b-r_1p_1})}{(\mathrm{e}^{\alpha _1+b-r_1p_1}+K)}\left[ -\frac{\ln (1+g(p_1))}{g(p_1)}+Kr_1\frac{(p_1-c_1)}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K}\right] . \end{aligned}$$

We know that $-\frac{\ln (1+x)}{x}$ is increasing in x. We want $g(p_1)$ to be increasing too.

Let $f(p_1)=\frac{\mathrm{e}^{\alpha _1+b-r_1p_1}-\mathrm{e}^{\alpha _1-b-r_1p_1}}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K}$. $g(p_1)=\frac{\mathrm{e}^{\alpha _1-b-r_1p_1}-\mathrm{e}^{\alpha _1+b-r_1p_1}}{\mathrm{e}^{\alpha _1+b-r_1p_1}+K}$,

$$\begin{aligned} g'(p_1)= & {} \frac{-r_1\mathrm{e}^{\alpha _1-b-r_1p_1}+r_1\mathrm{e}^{\alpha _1+b-r_1p_1}}{\mathrm{e}^{\alpha _1+b-r_1p_1}+K} +\frac{r_1\mathrm{e}^{\alpha _1+b-r_1p_1}(\mathrm{e}^{\alpha _1-b-r_1p_1}-\mathrm{e}^{\alpha _1+b-r_1p_1})}{(\mathrm{e}^{\alpha _1+b-r_1p_1}+K)^2}\\= & {} \frac{Kr_1(\mathrm{e}^{\alpha _1+b-r_1p_1}-\mathrm{e}^{\alpha _1-b-r_1p_1})}{(\mathrm{e}^{\alpha _1+b-r_1p_1}+K)^2}>0. \end{aligned}$$

We want $f(p_1)$ to be increasing.

$$\begin{aligned} f'(p_1)= & {} \frac{-r_1\mathrm{e}^{\alpha _1+b-r_1p_1}+r_1\mathrm{e}^{\alpha _1-b-r_1p_1}}{\mathrm{e}^{\alpha _1-b-r_1p_1}+K} +\frac{r_1\mathrm{e}^{\alpha _1-b-r_1p_1}(\mathrm{e}^{\alpha _1+b-r_1p_1}-\mathrm{e}^{\alpha _1-b-r_1p_1})}{(\mathrm{e}^{\alpha _1-b-r_1p_1}+K)^2}\\= & {} \frac{Kr_1(\mathrm{e}^{\alpha _1+b-r_1p_1}-\mathrm{e}^{\alpha _1-b-r_1p_1})}{(\mathrm{e}^{\alpha _1-b-r_1p_1}+K)^2}>0. \end{aligned}$$

Unimodality for $R_2$:

Denote $a_1^L=\mathrm{e}^{\alpha _1-b-r_1p_1}$ and $a_1^R=\mathrm{e}^{\alpha _1+b-r_1p_1}$,

$a_2=\mathrm{e}^{\alpha _2-r_2p_2}$ and $a_2'=-r_2a_2$.

$$\begin{aligned} R_2= & {} \frac{a_2}{a_0+a_2}\ln \left( \frac{(a_0+a_1^L+a_2)a_1^R}{(a_0+a_1^R+a_2)a_1^L}\right) (p_2-c_2),\\ \frac{\partial R_2}{\partial p_2}= & {} \frac{a_0a_2'}{(a_0+a_2)^2}\ln \left( \frac{(a_0+a_1^L+a_2)a_1^R}{(a_0+a_1^R+a_2)a_1^L}\right) (p_2-c_2)\\&+\frac{a_2}{a_0+a_2}(\frac{a_2'}{a_0+a_1^L+a_2}-\frac{a_2'}{a_0+a_1^R+a_2})(p_2-c_2)\\&+\frac{a_2}{a_0+a_2}\ln \left( \frac{(a_0+a_1^L+a_2)a_1^R}{(a_0+a_1^R+a_2)a_1^L}\right) \\= & {} -\frac{a_0a_2'}{(a_0+a_2)^2}\ln \left( 1+\frac{(a_0+a_2)(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)a_1^R}\right) (p_2-c_2)\\&-\frac{a_2a_2'}{a_0+a_2}\frac{(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)(a_0+a_1^R+a_2)}(p_2-c_2)\\&-\frac{a_2}{a_0+a_2}\ln \left( 1+\frac{(a_0+a_2)(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)a_1^R}\right) \\= & {} \frac{a_2}{a_0+a_2}h(p_2)\left\{ r_2\frac{a_0}{a_0+a_2}\frac{\ln (1+h(p_2))}{h(p_2)}(p_2-c_2)\right. \\&\left. +\frac{r_2a_2a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)}(p_2-c_2) -\frac{\ln (1+h(p_2))}{h(p_2)}\right\} , \end{aligned}$$

where $h(p_2)=\frac{(a_0+a_2)(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)a_1^R}<0$ and is increasing in $p_2$.

$$\begin{aligned} h'(p_2)= & {} \frac{a_2'(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)a_1^R}-\frac{a_2'(a_0+a_2)(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)^2a_1^R}\\= & {} \frac{a_2'a_1^L(a_1^L-a_1^R)}{(a_0+a_1^L+a_2)^2a_1^R}>0. \end{aligned}$$

Let $f(p_2)=\frac{a_0}{a_0+a_2}\frac{\ln (1+h(p_2))}{h(p_2)} +\frac{a_2a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)}$.

$$\begin{aligned} \frac{\partial R_2}{\partial p_2}= & {} \frac{a_2}{a_0+a_2}h(p_2)\left\{ r_2(p_2-c_2)f(p_2)-\frac{\ln (1+h(p_2))}{h(p_2)}\right\} . \end{aligned}$$

In the RHS of the above equation, $-\frac{\ln (1+h(p_2))}{h(p_2)}$ is negative and increasing. $f(p_2)$ is always positive. We want $f(p_2)$ to be increasing as well. If so, then $R_2$ is unimodal in $p_2$.

$$\begin{aligned} f'(p_2)= & {} -\frac{a_0a_2'}{(a_0+a_2)^2}\frac{\ln (1+h(p_2))}{h(p_2)} +\frac{a_0}{a_0+a_2}\frac{h'(p_2)}{(1+h(p_2))h(p_2)}\\&-\frac{a_0}{a_0+a_2}\frac{h'(p_2)\ln (1+h(p_2))}{h^2(p_2)} +\frac{a_2'a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)}\\&-\frac{a_2'a_2a_1^R}{(a_0+a_2)^2(a_0+a_1^R+a_2)}-\frac{a_2'a_2a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)^2}\\= & {} -\frac{a_0a_2'}{(a_0+a_2)^2}\frac{\ln (1+h(p_2))}{h(p_2)}-\frac{a_0}{a_0+a_2}\frac{h'(p_2)\ln (1+h(p_2))}{h^2(p_2)}\\&+\frac{a_0}{a_0+a_2}\frac{a_2'a_1^L}{\frac{(a_0+a_1^R+a_2)a_1^L}{(a_0+a_1^L+a_2)a_1^R}(a_0+a_2)(a_0+a_1^L+a_2)}\\&-\frac{a_2'a_2a_1^R}{(a_0+a_2)^2(a_0+a_1^R+a_2)}\\&+\frac{a_2'a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)}-\frac{a_2'a_2a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)^2}\\= & {} -\frac{a_0a_2'}{(a_0+a_2)^2}\frac{\ln (1+h(p_2))}{h(p_2)}-\frac{a_0}{a_0+a_2}\frac{h'(p_2)\ln (1+h(p_2))}{h^2(p_2)}\\&+\frac{a_2'a_1^R}{(a_0+a_2)^2(a_0+a_1^R+a_2)}[a_0-a_2+a_0+a_2]\\&-\frac{a_2'a_2a_1^R}{(a_0+a_2)(a_0+a_1^R+a_2)^2}\\= & {} -\frac{a_0a_2'}{(a_0+a_2)^2}\frac{\ln (1+h(p_2))}{h(p_2)}-\frac{a_0}{a_0+a_2}\frac{h'(p_2)\ln (1+h(p_2))}{h^2(p_2)}\\&+\frac{a_2'a_1^R}{(a_0+a_2)^2(a_0+a_1^R+a_2)^2}[2a_0(a_0+a_1^R+a_2)-a_2(a_0+a_2)]. \end{aligned}$$

We know that $\ln (1+x)\leqslant x$ for $x\in (-1,+\infty )$. Therefore, $\frac{\ln (1+x)}{x}\geqslant 1$, for $x\in (-1,0)$.

$$\begin{aligned} f'(p_2)\geqslant & {} -\frac{a_0a_2'}{(a_0+a_2)^2}-\frac{a_0}{a_0+a_2}\frac{h'(p_2)}{h(p_2)}\\&+\frac{a_2'a_1^R}{(a_0+a_2)^2(a_0+a_1^R+a_2)^2}[2a_0(a_0+a_1^R+a_2)-a_2(a_0+a_2)]\\= & {} -\frac{a_2'}{(a_0+a_2)^2}\bigg \{a_0-\frac{a_0a_1^L}{(a_0+a_1^L+a_2)}+\frac{a_0a_1^R}{(a_0+a_1^R+a_2)}\\&-\frac{a_1^R}{(a_0+a_1^R+a_2)^2}[a_0(a_0+a_1^R)-a_2^2]\bigg \}\\= & {} -\frac{a_2'}{(a_0+a_2)^2}\bigg \{\frac{a_0(a_1^R-a_1^L)(a_0+a_2)}{(a_0+a_1^R+a_2)(a_0+a_1^L+a_2)}\\&+\frac{1}{(a_0+a_1^R+a_2)^2}[a_0(a_0+a_1^R+a_2)^2-a_0^2a_1^R-a_0(a_1^R)^2+a_2^2a_1^R]\bigg \}\\\geqslant & {} 0. \end{aligned}$$

11 Proof of Table 1

Following the same approach as in the proof of Proposition 3, we first determine the sign of $\frac{\partial R_{ir}(p_1,p_2)}{\partial p_i}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}$ based on the concavity/convexity of $\frac{\partial R_{ir}(p_1,p_2)}{\partial p_i}$ with respect to $\alpha _1$. In the following derivation, we assume that the partial differentiation and the integration on $\delta$ are interchangeable. Taking second-order derivative of $\frac{\partial R_{id}(p_1,p_2)}{\partial p_i}$ with respect to $\alpha _1$, we have

$$\begin{aligned} \frac{\partial ^3 R_{1d}(p_1,p_2)}{\partial \alpha _1^2 \partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}&= \frac{a_{1d}^{*2}}{(a_0+a_{1d}^*+a_{2d}^*)^3}(3(a_0+a_{2d}^{*})-a_{1d}^{*}),\\ \frac{\partial ^3 R_{2d}(p_1,p_2)}{\partial \alpha _1^2 \partial p_2}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}&= \frac{a_{1d}^{*}a_{2d}^{*2}}{(a_0+a_{1d}^*)(a_0+a_{1d}^*+a_{2d}^*)^3}(3a_{1d}^{*}-(a_0+a_{2d}^{*})). \end{aligned}$$

Therefore, if $a_{1d}^{*}<3(a_0+a_{2d}^{*})$, then $\frac{\partial R_{1d}(p_1,p_2)}{\partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}$ is convex in $\alpha _1$. As a consequence, there exists a support interval of $\delta$ that ensures the convexity of $\frac{\partial R_{1r}(p_1,p_2)}{\partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}$. With this support interval of $\delta$,

$$\begin{aligned} \frac{\partial R_{1r}(p_1,p_2)}{\partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}&=\frac{\partial {{E}}_\delta R_{1d}(p_1,p_2, \alpha _1+\delta , \alpha _2)}{\partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*} \\&>\frac{\partial R_{1d}(p_1,p_2,\alpha _1+E\delta , \alpha _2)}{\partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*} \\&=0. \end{aligned}$$

In this way, we are able to determine the signs of $\frac{\partial R_{1r}(p_1,p_2)}{\partial p_1}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}$ and $\frac{\partial R_{2r}(p_1,p_2)}{\partial p_2}\big |_{p_1=p_{1d}^*,p_2=p_{2d}^*}$ when $a_{1d}^*$ belongs to different intervals, as given in Table 1.

12 Proof of Proposition 9

Taking the uniform distribution function into Eq. (4), we have

$$\begin{aligned} {\tilde{R}}_1(p_1,p_2)&= \frac{1}{2b}\int _{-b}^{b} \frac{{\tilde{a}}_1}{a_0+{\tilde{a}}_1+a_2} (p_1-c_1) \mathrm{d}\delta \\&= \frac{1}{2b}\int _{\mathrm{e}^{\alpha _1-b-\gamma _1p_1}}^{\mathrm{e}^{\alpha _1+b-\gamma _1p_1}} \frac{1}{a_0+{\tilde{a}}_1+a_2} (p_1-c_1) \mathrm{d} {\tilde{a}}_1\\&=\frac{1}{2b} \ln \bigg (\frac{a_0+\mathrm{e}^{\alpha _1+b-\gamma _1p_1}+\mathrm{e}^{\alpha _2-\gamma _2p_2}}{a_0+\mathrm{e}^{\alpha _1-b-\gamma _1p_1}+\mathrm{e}^{\alpha _2-\gamma _2p_2}}\bigg )(p_1-c_1),\\ {\tilde{R}}_2(p_1,p_2)&= \frac{1}{2b}\int _{-b}^{b} \frac{a_2}{a_0+{\tilde{a}}_1+a_2} (p_1-c_1) \mathrm{d}\delta \\&= \frac{1}{2b}\int _{\mathrm{e}^{\alpha _1-b-\gamma _1p_1}}^{\mathrm{e}^{\alpha _1+b-\gamma _1p_1}} \frac{a_2}{{\tilde{a}}_1(a_0+{\tilde{a}}_1+a_2)} (p_2-c_2) \mathrm{d} {\tilde{a}}_1\\&=\frac{1}{2b}\frac{a_2}{a_0+a_2} \ln \bigg (\frac{(a_0+\mathrm{e}^{\alpha _1-b-\gamma _1p_1}+\mathrm{e}^{\alpha _2-\gamma _2p_2})\mathrm{e}^{\alpha _1+b-\gamma _1p_1}}{(a_0+\mathrm{e}^{\alpha _1+b-\gamma _1p_1}+\mathrm{e}^{\alpha _2-\gamma _2p_2})\mathrm{e}^{\alpha _1-b-\gamma _1p_1}}\bigg )(p_2-c_2). \end{aligned}$$

Therefore, $({\tilde{p}}_1^*,{\tilde{p}}_2^*)$ takes the value that satisfies the first-order condition:

$$\begin{aligned} \frac{\partial {\tilde{R}}_1(p_1,p_2)}{\partial p_1}\bigg |_{p_1={\tilde{p}}_1^*,p_2={\tilde{p}}_2^*}&\!=\frac{1}{2b}\int _{-b}^{b}\left[ \frac{-\gamma _1{\tilde{a}}_1^*(a_0+a_2^*)}{(a_0+{\tilde{a}}_1^*+a_2^*)^2}(p_1-c_1) +\frac{{\tilde{a}}_1^*}{a_0+{\tilde{a}}_1^*+a_2^*}\right] \mathrm{d}\delta \\&=0,\\ \frac{\partial {\tilde{R}}_2(p_1,p_2)}{\partial p_2}\bigg |_{p_1={\tilde{p}}_1^*,p_2={\tilde{p}}_2^*}&\!=\frac{1}{2b}\int _{-b}^{b}\left[ \frac{-\gamma _2a_2^*(a_0+{\tilde{a}}_1^*)}{(a_0+{\tilde{a}}_1^*+a_2^*)^2}(p_2-c_2) +\frac{a_2^*}{a_0+{\tilde{a}}_1^*+a_2^*}\right] \mathrm{d}\delta \\&=0. \end{aligned}$$

Here, ${\tilde{a}}_1^*=\exp (\alpha _1+\delta -\gamma _1 {\tilde{p}}_1^*)$ and $a_2^*=\exp (\alpha _2-\gamma _2 {\tilde{p}}_2^*)$. For simplicity of the expressions, denote $a_{1L}^*=\exp (\alpha _1-b-\gamma _1 {\tilde{p}}_1^*)$, $a_{1R}^*=\exp (\alpha _1+b-\gamma _1 {\tilde{p}}_1^*)$, $\varSigma _L^*=a_0+a_{1L}^*+a_2^*$, $\varSigma _R^*=a_0+a_{1R}^*+a_2^*$. Then, the first-order conditions yield to the following equation set:

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( \frac{1}{\varSigma _L^*}-\frac{1}{\varSigma _R^*}\right) ({\tilde{p}}^*_1-c_1)=\frac{\ln (\varSigma _R^*)-\ln (\varSigma _L^*)}{\gamma _1 (a_0+a_2^*)},\\ \frac{a_0}{a_0+a_2^*}\cdot \left( \ln \frac{a_{1R}^*\varSigma _L^*}{a_{1L}^*\varSigma _R^*}+a_2^*(\frac{1}{\varSigma _L^*}-\frac{1}{\varSigma _R^*})\right) \cdot ({\tilde{p}}^*_2-c_2)=\frac{1}{\gamma _2} \ln \frac{a_{1R}^*\varSigma _L^*}{a_{1L}^*\varSigma _R^*}. \end{array}\right. } \end{aligned}$$

(A9)

In the following analysis, we assume that b is sufficiently small and use Taylor expansion to further simplify the equation set. Denote that $a_{1}^*=\exp (\alpha _1-\gamma _1 {\tilde{p}}_1^*)$ and $\varSigma ^*=a_0+a_1^*+a_2^*$. Some useful results are listed as follows:

$$\begin{aligned} a_{1L}^*&=\left( 1-b+\frac{b^2}{2}\right) a_{1}^*+o(b^2),\\ a_{1R}^*&=\left( 1+b+\frac{b^2}{2}\right) a_{1}^*+o(b^2),\\ \varSigma _L^*&=\varSigma ^* -\left( b-\frac{b^2}{2}\right) a_{1}^*+o(b^2),\\ \varSigma _R^*&=\varSigma ^* +\left( b+\frac{b^2}{2}\right) a_{1}^*+o(b^2). \end{aligned}$$

Now, the first equation in (A9) can be simplified in the following way.

$$\begin{aligned} {\tilde{p}}_1^*-c_1&=\frac{\varSigma _L^*\varSigma _R^*\ln \frac{\varSigma _R^*}{\varSigma _L^*}}{\gamma _1(a_0+a_2^*)(a_{1R}^*-a_{1L}^*)}\\&=\frac{\varSigma _L^*\varSigma _R^* \left( \frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*}-\frac{1}{2}\left( \frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*}\right) ^2 +\frac{1}{3}\left( \frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*}\right) ^3+o\left( \left( \frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*}\right) ^3\right) \right) }{\gamma _1(a_0+a_2^*)(a_{1R}^*-a_{1L}^*)}\\&=\frac{\varSigma _R^*-\frac{1}{2}\varSigma _R^*\frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*} +\frac{1}{3}\varSigma _R^*\left( \frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*}\right) ^2+o\left( \left( \frac{a_{1R}^*-a_{1L}^*}{\varSigma _L^*}\right) ^2\right) }{\gamma _1(a_0+a_2^*)}\\&=\frac{\varSigma ^*+\frac{a_{1}^*(3\varSigma ^*-4a_{1}^*)}{6\varSigma ^*}b^2+o(b^2)}{\gamma _1(a_0+a_2^*)}\\&=\frac{\varSigma ^*}{\gamma _1(a_0+a_2^*)}+\frac{a_{1}^*(3(a_0+a_{2}^*)-a_{1}^*)}{6\gamma _1(a_0+a_2^*)\varSigma ^*}b^2+o(b^2). \end{aligned}$$

In the same way but with more effort taking, the second equation in (A9) is simplified as follows:

$$\begin{aligned} {\tilde{p}}_2^*-c_2= & {} \frac{1}{\gamma _2}\frac{1}{\frac{a_0}{a_0+a_2^{*}}+\frac{a_2^*(a_{1R}^*-a_{1L}^*)}{\varSigma _L^*\varSigma _R^* \ln \frac{a_{1R}^*\varSigma _L^{*}}{a_{1L}^*\varSigma _{R}^{*}}}}\\= & {} \frac{1}{\gamma _2}\frac{1}{\frac{a_0}{a_0+a_2^*}+\frac{a_2^*(a_{1R}^*-a_{1L}^*)^{}}{\varSigma _L^*\varSigma _R^* \left( \frac{(a_{1R}^*-a_{1L}^*)(a_0+a_2^*)}{a_{1L}^*\varSigma _R^*} -\frac{1}{2}\left( \frac{(a_{1R}^*-a_{1L}^*)(a_0+a_2^*)}{a_{1L}^*\varSigma _R^*}\right) ^2 +\frac{1}{3}\left( \frac{(a_{1R}^*-a_{1L}^*)(a_0+a_2^*)}{a_{1L}^*\varSigma _R^*}\right) ^3+o\left( \left( a_{1R}^*-a_{1L}^*\right) ^3\right) \right) ^{}}}\\= & {} \frac{1}{\gamma _2} \frac{1}{\frac{a_0}{a_0+a_2^*}+\frac{a_2}{ \frac{\varSigma ^*(a_0+a_2^*)}{a_1^*}+\left( b+\frac{b^2}{2}\right) \frac{(a_0+a_2^*)^2}{a_1^*} -\frac{(b+2b^2)(a_0+a_2^*)^{2^{}}}{a_1^*}+\frac{2b^2(a_0+a_2^*)^2}{\varSigma ^*} +\frac{4(a_0+a_2^*)^3}{3a_1^*\varSigma ^*}b^2+o(b^2)}}\\= & {} \frac{1}{\gamma _2} \frac{1}{\frac{a_0}{a_0+a_2^*}+\frac{a_2}{\frac{\varSigma ^*(a_0+a_2^*)}{a_1^*} +\frac{(a_0+a_2^*)^{2^{}}(3a_1^*-a_0-a_2^*)}{6a_1^*\varSigma ^*} b^2+o(b^2)}}\\= & {} \frac{1}{\gamma _2} \frac{1}{ \frac{a_0}{a_0+a_2^*}+\frac{a_1^*a_2^*}{\varSigma ^*(a_0+a_2^*)^{}}\left( 1-\frac{(a_0+a_2^*)(3a_1^*-a_0-a_2^*)}{6\varSigma ^{*2}}b^2+o(b^2)\right) ^{} }\\= & {} \frac{1}{\gamma _2}\frac{\varSigma ^*}{a_0+a_1^*}+\frac{a_1^*a_2^*(3a_1^*-a_0-a_2^*)}{6(a_0+a_1^*)^2\varSigma ^*}b^2+o(b^2). \end{aligned}$$

Substitute $\varSigma ^*$ with $a_0+a_1^*+a_2^*$, the proof is done.

13 Proof of Proposition 10

The two thresholds for the deterministic model satisfy $a_{1d}^*=\frac{a_0+a_{2d}^*}{3}$ and $a_{1d}^*=3(a_0+a_{2d}^*)$, respectively. Take case 1 as an example, i.e., the threshold for $p_{1r}^*>p_{1d}^*$ and $p_{2r}^*<p_{2d}^*$. With $p_{1r}^*>p_{1d}^*$ and $p_{2r}^*<p_{2d}^*$, and that $a_{i}=\exp (\alpha -\gamma _i p_i)$ is decreasing in $p_i$, we obtain $a_{1r}^*<a_{1d}^*$ and $a_{2r}^*>a_{2d}^*$, and hence

$$\begin{aligned} \frac{a_{2r}^*}{a_0+a_{1r}^*}>\frac{a_{2d}^*}{a_0+a_{1d}^*}. \end{aligned}$$

If $a_{1r}^*>\frac{a_0+a_{2r}^*}{3}$, for b that is small enough,

$$\begin{aligned} p_{2r}^*-c_2= & {} \frac{1}{\gamma _2}\frac{a_0+a_{1r}^*+a_{2r}^*}{a_0+a_{1r}^*} +\frac{1}{\gamma _2}\frac{a_{1r}^* a_{2r}^* (3a_{1r}^*-(a_0+a_{2r}^*))}{6(a_0+a_{1r}^*)^2(a_0+a_{1r}^*+a_{2r}^*)} b^2+o(b^2)\\> & {} \frac{1}{\gamma _1}\frac{a_0+a_{1d}^*+a_{2d}^*}{a_0+a_{1d}^*}\\= & {} p_{2d}^*-c_2, \end{aligned}$$

which contradicts $p_{2r}^*<p_{2d}^*$. Hence, we obtain there exists $l_1<\frac{1}{3}$, such that $a_{1r}^*=\frac{a_0+a_{2r}^*}{l_1}$. Likewise, we are able to obtain the results for cases 2 and 3.