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附录:证明 命题3.1的证明: 令m=i,则公司1的利润函数可以写为
πd1=p11-p1am-m2(a1) 由式(a1)可知,πd1是关于p1和m的凹函数,因此对p1和m分别求导可得
πd1p1=ma-2p1a(a2) πd1m=p1a-p1a-2m(a3) 公司1利润函数的海塞矩阵A1为
A1=-2maa-2p1aa-2p1a-2. 由此可知,矩阵A1的一阶主子式为D1=-2ma<0,二阶主子式为D2=4ap1+m-4p21-a2a2;因此,矩阵A1为负定矩阵的充要条件为:D2>0,也即是:a-2ma2命题4.1的证明: 令m=i,则两公司的利润函数分别为
πc1=p11-p1am-m2,p1 和 πc2=0,p1 为了分别最大化两公司的利润,可知纳什均衡存在于p1≥abp2区域里.可知πc1关于p1和m是凹函数,因此,对p1和m求导可得
πc1p1=a-b-2p1+p2ma-b(a6)
和
πc1m=p1a-b-p1+p2a-b-2m(a7) 公司1利润函数的海塞矩阵A2为
A2=-2ma-ba-b-2p1+p2a-ba-b-2p1+p2a-b-2. 由上式可知,矩阵A2的一阶主子式为D1=-2ma-b<0,二阶主子式为D2=a-b4m+4p1-2p2-a-b2-2p1-p22a-b2;因此,矩阵A2为负定矩阵的充要条件为:D2>0,也即,a-b+p2-2m(a-b)2同时,可知πc2是关于p2的凹函数,因此对p2求导可得
πc2p2=bp1-2ap2mb(a-b)(a8) 同时,令式(a6),(a7)和(a8)等于0,联立求解可得均衡决策为:pc1=2aa-b4a-b,pc2=ba-b4a-b和mc=2a2(a-b)4a-b2.由此可知,均衡解服从A2为负定矩阵的充要条件,因此,两公司的最优决策为:pc*1=2aa-b4a-b,pc*2=ba-b4a-b和ic*=4a4a-b24a-b4. 命题5.1的证明 与命题4.1的证明类似,令m=i,n=j.则两公司的利润函数可以写为
πs1=p11-p1ak1m+k2n-m2,p1 和
πs2=0,p1 由式(a9)和(a10)可知,纳什均衡存在于p1≥abp2,同时πs1关于p1和m的凹函数,因此对p1和m求导可得
πs1p1=a-b-2p1+p2(k1m+k2n)a-b(a11)
和
πs1m=k1p1a-b-p1+p2a-b-2m(a12)
公司1利润函数的海塞矩阵为
A3=-2k1m+k2na-bk1a-b-2p1+p2a-bk1a-b-2p1+p2a-b-2 由上式可知,矩阵A3的一阶主子式为D1=-2(k1m+k2n)a-b<0,二阶主子式为D2=2k21a-b2p1-p2-2k1p1-k1p22+4a-bk1m+k2n-a+b2k21a-b2,因此,矩阵A2为负定矩阵的充要条件为D2>0,也即是:k1a-b+p2-(a-b)(k1m+k2n)2k1同时,πs2关于p2和n的凹函数,因此,对p2和n求导可得
πs2p2=bp1-2ap2(k1m+k2n)b(a-b)(a13)
和
πs2n=k2p2bp1-ap2a-b-2n(a14) 公司2利润函数的海塞矩阵为
A4=-2ak1m+k2n(a-b)b-k22ap2-bp1a-bb-k22ap2-bp1a-bb-2. 由上式可知,矩阵A4的一阶主子式为D1=-2a(k1m+k2n)a-b<0,二阶主子式为D2=-2ak2p2-bk2p12+4aba-bk1m+k2na-b2b2,因此,矩阵A2为负定矩阵的充要条件为:D2>0,也即是:bp1k2-2ab(a-b)(k1m+k2n)2k1a 同时令式(a11),(a12),(a13)和(a14)等于0,联立求解得到两公司的均衡决策为:ps1=2aa-b4a-b,ps2=ba-b4a-b,ms=2k1a2(a-b)4a-b2和ns=k2ab(a-b)44a-b2.可知均衡解服从A3和A4为负定矩阵的充要条件;因此,该策略下的两公司最优决策为:ps*1=2aa-b4a-b,ps*2=ba-b4a-b,is*=4k21a4a-b24a-b4和js*=k22a2b2a-b244a-b4.
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