The authors are grateful for the detailed and insightful comments and suggestions from referees and the constructive suggestions from the editor. These comments and suggestions have helped us to improve this paper significantly. The first and second author were supported in part by National Natural Science Foundation of China under Grant (No. 71511130129 ). The last author was supported by Netherlands Organisation for Scientific Research under Grant 040.03.021 , 040.21.001 , and National Natural Science Foundation of China under grant 71571125 , 71511130129 , 71711530046 . The corresponding author was partially supported by China Postdoctoral Science Foundation (No. 2017M622487 ). Appendix Proof of Lemma 1 . According to the definition of L i ( x ) , we know that L 1 ( θ q 1 − d 1 − ε ) is jointly convex in ( q 1 , d 1 ) and L 2 ( ( 1 − θ ) q 1 + q 2 − D 2 ) is jointly convex in ( q 1 , q 2 ) . Then, we have the joint concavity of Π ( q 1 , q 2 , d 1 ) . Proof of Proposition 1 . Taking the first-order condition of d 1 for G ( q 1 , d 1 ) , we have (7) ∂ G ∂ d 1 = P ′ ( d 1 ) d 1 + P ( d 1 ) + E θ [ ( e 1 − s 1 ) F 1 ( θ q 1 − d 1 ) ] − e 1 = 0. We obtain d 1 ( q 1 ) by solving (7). Taking the first-order condition of q 2 for H ( q 1 , q 2 ) , we have (8) ∂ H ∂ q 2 = − c 2 − E θ [ ( e 2 − s 2 ) F 2 [ [ ( 1 − θ ) q 1 + q 2 ] ] + e 2 = 0. Solving (8) gives us q 2 ( q 1 ) . Because G ( q 1 , d 1 ) is strictly supermodular in ( q 1 , d 1 ) and H ( q 1 , q 2 ) is strictly submodular in ( q 1 , q 2 ) , d 1 ( q 1 ) is strictly increasing in q 1 while q 2 ( q 1 ) is strictly decreasing in q 1 . Further, by definition, p 1 is strictly decreasing in d 1 . Thus, p 1 ( q 1 ) is strictly decreasing in q 1 . Proof of Proposition 2 . Checking the first derivative of Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) at q 1 = 0 , we have d Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) d q 1 | q 1 = 0 = − c 1 − c T ( 1 − θ ¯ ) + E θ θ [ e 1 − ( e 1 − s 1 ) F 1 ( − d 1 ) ] + E θ ( 1 − θ ) [ e 2 − ( e 2 − s 2 ) F 2 ( q 2 ) ] . For q 1 = 0 , taking the first-order condition of d 1 and q 2 for Π ( q 1 , q 2 , d 1 ) , respectively, we have F 1 ( − d 1 ) = e 1 − ( p m a x + P ′ ( d m i n ) d m i n ) e 1 − s 1 , F 2 ( q 2 ) = e 2 − c 2 e 2 − s 2 . where p m a x = p 1 ( 0 ) and d m i n = d 1 ( 0 ) . By substituting them into d 1 Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) d 1 q 1 | q 1 = 0 , we have d 1 Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) d 1 q 1 | q 1 = 0 ≤ 0 iff p 1 m a x + P ′ ( d m i n ) d m i n ≤ c 1 / θ ¯ − ( c 2 − c T ) ( 1 − θ ¯ ) / θ ¯ . By the concavity of Π , we have q 1 * = 0 and q 2 * = F 2 − 1 ( e 2 − c 2 e 2 − s 2 ) > 0 . Proof of Corollary 1 . Let q 1 = 0 , from (7), we obtain F 1 [ − d 1 ( q 1 ) ] | q 1 = 0 = e 1 − ( P ′ ( d 1 ( q 1 ) ) d 1 ( q 1 ) + P ( d 1 ( q 1 ) ) ) ( e 1 − s 1 ) | q 1 = 0 . So, d 1 ( q 1 ) | q 1 = 0 can be expressed as a function of ( e 1 , s 1 ) . We define d 1 ( q 1 ) | q 1 = 0 = u ( e 1 , s 1 ) . Then, we have d Π d q 1 | q 1 = 0 = − c 1 − c T ( 1 − θ ¯ ) + θ ¯ { P ′ [ d 1 ( q 1 ) ] d 1 ( q 1 ) + P [ d 1 ( q 1 ) ] } + ( 1 − θ ¯ ) c 2 | q 1 = 0 = − c 1 − c T ( 1 − θ ¯ ) + θ ¯ { P ′ [ u ( e 1 , s 1 ) ] u ( e 1 , s 1 ) + P [ u ( e 1 , s 1 ) ] } + ( 1 − θ ¯ ) c 2 . Therefore, d Π d q 1 | q 1 = 0 only depends on e 1 , c 1 , s 1 , c 2 , c T , θ ¯ . Proof of Proposition 3. Differentiating Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) with respect to q 1 , we obtain d Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) d q 1 = − c 1 − c T ( 1 − θ ¯ ) + θ ¯ [ e 1 − ( e 1 − s 1 ) F 1 ( q 1 − d 1 * ( q 1 ) ) ] + ( 1 − θ ¯ ) [ e 2 − ( e 2 − s 2 ) F 2 ( q 1 + q 2 ( q 1 * ) ) ] . By the first-order conditions ∂ Π / ∂ d 1 = 0 and ∂ Π / ∂ q 2 = 0 , we further have (9) P 1 ( d 1 ) + P 1 ′ ( d 1 ) d 1 − [ e 1 − ( e 1 − s 1 ) ( θ ¯ F 1 ( q 1 − d 1 ) + ( 1 − θ ¯ ) F 1 ( − d 1 ) ) ] = 0 , (10) θ ¯ F 2 ( q 2 ) + ( 1 − θ ¯ ) F 2 ( q 1 + q 2 ) = e 2 − c 2 e 2 − s 2 . To find the condition under which q 2 * = 0 , by (10), we have ( 1 − θ ¯ ) F 2 ( q ¯ 1 ) = e 2 − c 2 e 2 − s 2 . Substituting it into d Π q 1 , d 1 * q 1 , q 2 * q 1 d q 1 | q 1 = q ¯ 1 , we have d Π q 1 , d 1 * q 1 , q 2 * q 1 d q 1 | q 1 = q ¯ 1 = − c 1 − c T 1 − θ ¯ + θ ¯ e 1 − e 1 − s 1 F 1 q ¯ 1 − d 1 * q ¯ 1 + 1 − θ ¯ e 2 − e 2 + c 2 ≤ − c 1 − c T 1 − θ ¯ + θ ¯ e 1 − e 2 + c 2 . When e 1 − e 2 < c 1 − c 2 + C T ( 1 − θ ¯ ) θ ¯ , we obtain d 1 Π ( q 1 , d 1 * ( q 1 ) , q 2 * ( q 1 ) ) d 1 q 1 | q 1 = q ¯ 1 < 0 . According to Step 2 of the solution algorithm in Theorem 1 , we conclude that q 2 * > 0 . Proof of Proposition 4. Because of the convexity of L 1 and L 2 , we have Π * ≤ max d 1 , q 1 , q 2 { p 1 E ( D 1 ) − c 1 q 1 − L 1 ( θ ¯ q 1 − d 1 ( p 1 ) − ε ) + p 2 E ( D 2 ) − c 2 q 2 − L 2 ( ( 1 − θ ¯ ) q 1 + q 2 − D 2 ) − c T ( 1 − θ ¯ ) q 1 } = Π c ∗ , where the first inequality is true due to Jensen's inequality. Proof of Proposition 5 . (i) Taking the second derivative of Π p d 2 with respect to d 1 , we have ∂ 2 Π p d 2 ( d 1 | q 1 , q 2 , θ ˆ ) ∂ d 1 2 = P ″ ( d 1 ) d 1 + 2 P ′ ( d 1 ) − ( e 1 − s 1 ) f 1 ( θ ˆ q 1 − d 1 ) . Because d 1 P ( d 1 ) is concave in d 1 under the assumption, we have P ″ ( d 1 ) d 1 + 2 P ′ ( d 1 ) < 0 , leading to ∂ 2 Π p d 2 ( d 1 | q 1 , q 2 , θ ) ∂ d 1 2 < 0 . Therefore, Π p d 2 ( d 1 ) is unimodal in d 1 . The global maximum is given by d ˆ 1 , where d ˆ 1 is the unique solution of (11) ∂ Π p d 2 ( d 1 | q 1 , q 2 , θ ˆ ) ∂ d 1 = P ′ ( d 1 ) d 1 + P ( d 1 ) + ( e 1 − s 1 ) F 1 ( θ ˆ q 1 − d 1 ) − e 1 = 0. (ii) Because Π p d 2 is strictly supermodular in ( d 1 , q 1 ) and ( d 1 , θ ˆ ) , d 1 is strictly increasing in q 1 and θ ˆ . Further, by definition, p 1 is strictly decreasing in d 1 . Thus, p 1 is strictly decreasing in q 1 and θ ˆ . Proof of Proposition 6 . (i) Taking the first and second derivative of Π p d 1 ( q 1 , q 2 ) with respect to q 1 , we have (12) ∂ Π p d 1 q 1 , q 2 ∂ q 1 = ∂ Π p d 1 ∂ E θ d ˆ 1 ∂ E θ d ˆ 1 ∂ q 1 + ∂ Π p d 1 ∂ q 1 , ∂ 2 Π p d 1 q 1 , q 2 ∂ q 1 2 = ∂ 2 Π p d 1 ∂ E θ d ˆ 1 2 ∂ E θ d ˆ 1 ∂ q 1 2 + ∂ Π p d 1 ∂ E θ d ˆ 1 ∂ 2 E θ d ˆ 1 ∂ q 1 2 + ∂ 2 Π p d 1 ∂ q 1 2 . We know ∂ 2 Π p d 2 ∂ d 1 2 < 0 from the proof of Proposition 5 , thus ∂ 2 Π p d 1 ∂ ( E θ d ˆ 1 ) 2 < 0 . Hence, the first part of (12) is negative. According to (11) and ∂ Π p d 1 ( q 1 , q 2 ) ∂ E θ d ˆ 1 = E θ [ P ′ ( d ˆ 1 ) d ˆ 1 + P ( d ˆ 1 ) ] + ( e 1 − s 1 ) E θ F 1 ( θ q 1 − d ˆ 1 ( q 1 ) ) − e 1 , we have ∂ ( Π p d 2 ) ∂ d 1 = 0 from the proof of Proposition 5 , thus the second part of (12) is equal to zero. In addition, the third part of (12) is given by ∂ 2 Π p d 1 ∂ q 1 2 = − ( e 1 − s 1 ) E θ [ θ 2 f 1 ( θ q 1 − d ˆ 1 ) ] − ( e 2 − s 2 ) E θ [ ( 1 − θ ) 2 f 2 ( ( 1 − θ ) q 1 + q 2 ) ] < 0. Therefore, ∂ 2 Π p d 1 ( q 1 , q 2 ) ∂ q 1 2 < 0 . The profit function is concave in q 1 . Further, we have ∂ 2 Π p d 1 ( q 1 , q 2 ) ∂ q 2 2 = − ( e 2 − s 2 ) E θ [ f 2 ( ( 1 − θ ) q 1 + q 2 ) ] < 0 . So the profit function is concave in q 2 . (ii) Since ∂ 2 Π p d 1 ∂ q 1 ∂ q 2 = − ( e 2 − s 2 ) E θ [ ( 1 − θ ) f 2 ( ( 1 − θ ) q 1 + q 2 ) ] < 0 , we can show that Π p d 1 is strictly submodular in ( q 1 , q 2 ) . Thus, q 2 ( q 1 ) is strictly decreasing in q 1 for any given q 1 .
