39. (1) T is player A’s dominate strategy if and only if a?e and c?g; L is player B’s dominate strategy iff. b?d and f?h. Therefore we should have:
a?e, c?g, b?d and f?h
(2) As (T, L) is Nash equilibrium, we know that given A choosing T, B will choose L, and vice verse. For this to hold; we need b?d and a?e
(3) Similarly, for (B, R) to be Nash equilibrium, we need h?f and g?c. Therefore if (T, L) and (B, R) are both Nash equilibrium, a-h should satisfy:
a?e, g?c, b?d and h?f 40. The payoff of the game is:
The Penality Taker
Left Right
Left (1,0) (0,1)
The Goalkeeper
Right (0,1) (1,0)
Easy to check, there is no pure strategy Nash equilibrium in this game. Suppose in the equilibrium, the mixed strategy of the goalkeeper is (p, 1-p), which should equalize two expected payoffs of the penality taker: takertakerUleft?0?p?1?(1?p)?Uright?1?p?0?(1?p)? p?0.5 Suppose in the equilibrium, the mixed strategy of the penality taker is (q, 1-q), which should equalize two expected payoffs of the goalkeeper: keeperkeeperUleft?1?q?0?(1?q)?Uright?0?q?1?(1?q)? q?0.5 Therefore the mixed equilibrium is: {(0.5, 0.5); (0.5, 0.5)} The payoff of each player is;
Utaker?0.25?0?0.25?1?0.25?1?0.25?0?0.5
Utaker?0.25?1?0.25?0?0.25?0?0.25?1?0.5
第五部分 一般均衡理论
41. （1）
max xAyAs..tpxxA?pyyA?3px?2py
解出：xA?3px?2py2px2px，yA?3px?2py2py类似的，xB?px?6py，yB?px?6py2py又有均衡条件：xA?xB?3?1，yA?yB?2?6
所以，
4px?8py2px?4，
4px?8py2py?8，所以，px?2py
代入可知：xA?2，yA?4，xB?2，yB?4。 （2）
max xAyAs..t?4?xA??8?yA??uB
L?xAyA???uB??4?xA??8?yA??
f.o.c. yA???yA?8??0，xA???xA?4??0 得到：yA?2xA
42．（1）
max xA?yAs..tpxxA?pyyA?3px?2py
f.o.c.1??px?0,1??py?0
存在内点解时：px?py
max xByBs..tpxxB?pyyB?px?6py
解出：xB?3.5，yB?3.5。 通过均衡条件：xA?0.5，yB?4.5。 （2）
max xByBs..t?4?xB???8?yB??uA
L?xByB???uA??4?xB???8?yB??
f.o.c. yB???0，xB???0
所以，在存在内点解时，xB?yB。
43．（1）由uA可知，A的最优解为xA?yA。所以，xA?yA?再由uB可知,xB?3px?2pypx?py。
px?6py2px，yB?px?6py2py。
3?33py 233?133?1517?33所以，xB?, xB?，xA?yA?。
444（2）因为A的偏好是完全互补的，所以，只要约束是凸的，其最优解均为xA?yA，仅定义在xA??0,4?，yA??0,4?。
由均衡条件可知：px?
44.
2）
45.（1）
（2）