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


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.

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  1. Schön, C.: On the product line selection problem under attraction choice models of consumer behavior. Eur. J. Oper. Res. 206(1), 260–264 (2010)

    Article  MathSciNet  Google Scholar 

  2. Kamakura, W.A., Russell, G.J.: A probabilistic choice model for market segmentation and elasticity structure. J. Mark. Res. 26(4), 379–390 (1989)

    Article  Google Scholar 

  3. Chintagunta, P.K., Jain, D.C., Vilcassim, N.J.: Investigating heterogeneity in brand preferences in logit models for panel data. J. Mark. Res. 28, 417–428 (1991)

    Article  Google Scholar 

  4. Aksoy-Pierson, M., Allon, G., Federgruen, A.: Price competition under mixed multinomial logit demand functions. Manag. Sci. 59(8), 1817–1835 (2013)

    Article  Google Scholar 

  5. Luce, R.D.: Individual Choice Behavior: A Theoretical Analysis. Wiley, New York (1959)

    MATH  Google Scholar 

  6. Bell, D.E., Keeney, R.L., Little, J.D.C.: A market share theorem. J. Mark. Res. 12(2), 136–141 (1975)

    Article  Google Scholar 

  7. McFadden, D., et al.: Conditional logit analysis of qualitative choice behavior. In: Zarembka, P. (ed.) Frontiers in Econometrics, pp. 105–142. Academic, London (1973)

    Google Scholar 

  8. Urban, G.L.: A mathematical modeling approach to product line decisions. J. Mark. Res. 6, 40–47 (1969)

    Article  Google Scholar 

  9. Hanssens, D.M., Parsons, L.J., Schultz, R.L.: Market Response Models: Econometric and Time Series Analysis, vol. 12. Springer Science & Business Media, Berlin (2003)

    Google Scholar 

  10. Xie, Y., Xie, L., Lu, M., Yan, H.: Performance-price-ratio utility: market equilibrium analysis and empirical calibration studies. Prod. Oper. Manag. 30, 1442–1456 (2020)

    Article  Google Scholar 

  11. Rossi, P.E., Allenby, G.M.: A Bayesian approach to estimating household parameters. J. Mark. Res. 30, 171–182 (1993)

    Article  Google Scholar 

  12. Hanson, W., Martin, K.: Optimizing multinomial logit profit functions. Manag. Sci. 42(7), 992–1003 (1996)

    Article  Google Scholar 

  13. Song, J.-S., Xue, Z.: Demand management and inventory control for substitutable products. SSRN Electr. J.

  14. Li, H., Huh, W.T.: Pricing multiple products with the multinomial logit and nested logit models: concavity and implications. Manuf. Serv. Oper. Manag. 13(4), 549–563 (2011)

    Article  Google Scholar 

  15. Gallego, G., Wang, R.: Multiproduct price optimization and competition under the nested logit model with product-differentiated price sensitivities. Oper. Res. 62(2), 450–461 (2014)

    Article  MathSciNet  Google Scholar 

  16. Li, G., Rusmevichientong, P., Topaloglu, H.: The \(d\)-level nested logit model: assortment and price optimization problems. Oper. Res. 63(2), 325–342 (2015)

    Article  MathSciNet  Google Scholar 

  17. Alptekinoğlu, A., Semple, J.H.: The exponomial choice model: a new alternative for assortment and price optimization. Oper. Res. 64(1), 79–93 (2016)

    Article  MathSciNet  Google Scholar 

  18. Huh, W.T., Li, H.: Pricing under the nested attraction model with a multistage choice structure. Oper. Res. 63(4), 840–850 (2015)

    Article  MathSciNet  Google Scholar 

  19. Gallego, G., Ratliff, R., Shebalov, S.: A general attraction model and sales-based linear program for network revenue management under customer choice. Oper. Res. 63(1), 212–232 (2015)

    Article  MathSciNet  Google Scholar 

  20. Topkis, D.M.: Equilibrium points in nonzero-sum \(n\)-person submodular games. SIAM J. Control. Optim. 17(6), 773–787 (1979)

    Article  MathSciNet  Google Scholar 

  21. Vives, X.: Oligopoly Pricing: Old Ideas and New Tools. MIT Press, Cambridge (2001)

    Google Scholar 

  22. Mizuno, T.: On the existence of a unique price equilibrium for models of product differentiation. Int. J. Ind. Organ. 21(6), 761–793 (2003)

    Article  Google Scholar 

  23. Caplin, A.: Aggregation and imperfect competition: on the existence of equilibrium. Econom. J. Econom. Soc. 59, 25–59 (1991)

    MathSciNet  MATH  Google Scholar 

  24. Gallego, G., Huh, W.T., Kang, W., Phillips, R.: Price competition with the attraction demand model: existence of unique equilibrium and its stability. Manuf. Serv. Oper. Manag. 8(4), 359–375 (2006)

    Article  Google Scholar 

  25. Yang, L., Guo, P., Wang, Y.: Service pricing with loss-averse customers. Oper. Res. 66(3), 761–777 (2018)

    Article  MathSciNet  Google Scholar 

  26. Federgruen, A., Ming, H.: Sequential multiproduct price competition in supply chain networks. Oper. Res. 64(1), 135–149 (2016)

    Article  MathSciNet  Google Scholar 

  27. Wang, R.: What is the impact of nonrandomness on random choice models? Manuf. Serv. Oper. Manag. (2021)

  28. Debreu, G.: A social equilibrium existence theorem. Proc. Natl. Acad. Sci. 38(10), 886–893 (1952)

    Article  MathSciNet  Google Scholar 

  29. Fan, K.: Fixed-point and minimax theorems in locally convex topological linear spaces. Proc. Natl. Acad. Sci. 38(2), 121–126 (1952)

    Article  MathSciNet  Google Scholar 

  30. Glicksberg, I.L.: A further generalization of the Kakutani fixed point theorem, with application to Nash equilibrium points. Proc. Am. Math. Soc. 3(1), 170–174 (1952)

    MathSciNet  MATH  Google Scholar 

  31. Bensoussan, A., Xie, Y., Yan, H.: Joint inventory-pricing optimization with general demands: an alternative approach for concavity preservation. Prod. Oper. Manag. 28(9), 2390–2404 (2019)

    Article  Google Scholar 

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Correspondence to Yang-Yang Xie.

Additional information

This work was partially supported by the National Natural Science Foundation of China (No. 72001198 and Nos. 71991464/71991460), the Fundamental Research Funds for the Central Universities (No. WK2040000027), the National Key R&D Program of China (Nos. 2020AAA0103804/ 2020AAA0103800), USTC (University of Science and Technology of China) Research Funds of the Double First-Class Initiative (No. YD2040002004), Collaborative Research Fund (No. C1143-20G), and General Research Fund (No. 115080/17).



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

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


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

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.

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

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

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

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

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

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

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

Since \(p_d^*\) satisfies the first-order condition, we can derive that

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

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

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

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

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

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:

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

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.

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Feng, XY., Xie, YY. & Yan, HM. Price Competition in the Random Coefficient Attraction Choice Models with Linear Cost. J. Oper. Res. Soc. China 10, 623–658 (2022).

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