Answer:
So far I am very satisfied.
Zhao Wu
Game Theory HW 04
GAME THEORY HW 04
Answer:
(a) Last Friday, I heard that a group of people (about four) might go to Capen Library to play a game, which required at least 5 people. Then I was considering whether to play with them or just stay at home.
Players: That group of people(A) and Me(B)
Options: Both A and B chose to go or not to
Sequences: A moved first, and B could observe
Payoffs: Both A and B would be happy if participating in the game, and be a little depressed if not go, and be angry if go and there is not enough people.
Extensive-Form
A
GO NOT GO
B B
GO NOT GO GO NOT GO
7 -10 -1 -1
5 -2 -10 -1

Zhao Wu
Game Theory HW 04
Normal-Form
For B, there are 4 strategies named S1,S2,S3,S4.
S1: A Go B Go; A Not Go, B Go.
S2: A Go B Go; A Not Go, B Not Go.
S3: A Go B Not Go; A Not Go, B Go.
S4: A Go B Not Go; A Not Go, B Not Go.
A\B S1 S2
Go 7,5 7,5
S3
-10,-2
S4
-10,-2
Not Go -1,-10 -1, -1 -1 ,-10 -1,-1
(b) As shown in the normal-form, we have three pure NE in this game, that is (A Go,B S1),(A
GO,B S2),and (A Not Go, B S4)
(c) For SPNE:
Extensive-Form
A
GO NOT GO
B B
GO NOT GO GO NOT GO
7 -10 -1 -1
5 -2 -10 -1
We could see from extensive-form that there is only one SPNE in this game, that is (A Go, B GO).
(d) We could see that (A Go,B S1) and (A Go,B S2) describe the same SPNE that is (A GO, B Go).
The third NE(A Not Go, B S4) which is (A Not Go, B Not Go) is actually induced by the non-credible threat (A Go, B Not Go), thus in SPNE we would delete it,

Zhao Wu
Game Theory HW 04
Answer:
Denote the probability that Player 1 will choose Head as x
1
;
Denote the probability that Player 1 will choose Head as x .
2
U
1
 p
 x x
1 2
 x
1
(1
 x
2
)
  x x
1 2
  x
1
)(1
 x
2
  p
 x
1

1(1
 x
1
)] x
1
*
 arg max U
1
 px
2

4 p

2) x
1

1 if x
2

2 p

1
2 p
0 if x
2

2 p

1
2 p
Similarly,
U
2
 x
2
[1 x
1
  x
1
   x
2
 x
1
  x
1

1)] x
2
*
 arg max U
2
 arg max(4 x
1

2) x
2

1 if x
1

0.5
0 if x
1

0.5

Zhao Wu
Game Theory HW 04
We can draw the graph of ( x ,
1
* x ) as below:
2
* i) When 1

1
0 0.5
2 p p
Nash Equilibrium would be ( x ,
1
* x )=(0.5,
2
* 1

1
)
2 p ii) When 1

1
0 0.5
2 p p
Nash Equilibrium would be ( x ,
1
* x )=(0,0)
2
*
As a conclusion, when p>=0.5, player 1 would choose 0.5H+0.5T;When p<0.5, player 1 would always choose T. Thus Mr Zhuang’s Slolution is proved.

Zhao Wu
Game Theory HW 04
Answer ：
1
A B C
2 2 2
R N R N R N
90 100 75 100 40 50
80 60 50 60 80 100
B
C
In this game, there is only one SPNE that is (B,N)
In order to find the NE, I would draw the normal-form.
There would be 3 strategies for Player 1 and 8 strategies for player 2.
S1:if 1 A, then 2 R; if 1 B, then 2 R; if 1 C, then 2 R;
S2:if 1 A, then 2 R; if 1 B, then 2 R; if 1 C, then 2 N;
.
.
.
S1:if 1 A, then 2 N; if 1 B, then 2 N; if 1 C, then 2 N;
1\2
A
S1
90 , 80
S2
90 , 80
S3
90, 80
S4
90, 80
S5 S6 S7 S8
100 ,60 100 ,60 100 ,60 100 ,60
75,50
40,80
75,50 100 , 60 100 , 60 75,50
50, 100 40,80 50, 100 40,80
75,50 100 , 60 100 , 60
50, 100 40,80 50, 100

Zhao Wu
Game Theory HW 04
As shown in the normal-form, there are 6 pure NE in this game. In fact, (B,S3)(B,S4)(B,S7)(B,S8) describe the same SPNE that is (B,N). And (A,S1)(A,S2) is induced by the non-credible threat
(B,R),(C,R).(i.e. If Player 1 consider that Player 2 might choose R, then Player 1 would choose A.
However Player 2 wouldn’t choose R because he is rational.)
Answer:
(a) D choose d—A choose a which makes UA(a,d) gets its optimal value—With the a chosen by A and d chosen by D, UD(a,d) would have a certain value
So the problem to D would be How to choose the value of d so that A will choose such an a that makes UD(a,d) gets its optimal value.
(b) If c= α = β =V=W=1,

U
D
 dW

( d

)
 d

Then ( U
D
) '

W

V
( d
1
 c

1
2 d )
 
if d<=0

U
D
 dW d
 c
 d

=0 if d>0
Then if d<=0, U
D
 d d

1
 d , since it’s required that d>=0, so max U =0;
D
If d>0, U
D
 d d

1
 d <0
As a conclusion, the SPNE for c= α = β =V=W=1 is that d=0 and a=0.

Zhao Wu
Game Theory HW 04
(c) a *

( *

)


( *
 c ) if d

V

 c

U
D
 dW

( d

)
 d

( U
D
) '

W

( *
 c )
(1
 d d
*
* 1
 c 2
)
SPNE : (
( *

)


( *
 c ) , arg max d * 0
(

( *

)
 d *

))
=0 if d

V

 c

U
D
 dW d
 c
 d

( U
D
) '

Wc
( *
 c )
2
 d *

0
SPNE : (0, arg max( d d *
 c
 d d) The NE for the simultaneous is (
( *
 c )


( *
 c ) ,
( *
 c )


( *
 c ) ).
Conclustion: SPNE

NE
0
Answer:
First, we denote any possible equilibrium of the simultaneous game as ( a
C *
, d
C *
), denote the equilibrium of the sequential game as (
S *
,
S * a d ).
Now the second mover’s best response sets are always a singleton, which means the second mover’s best response to
C * d is unique.
Since the first mover always has the option to choose
C * d even in the sequential game which would “force” second mover to choose
C * a in order to get the NE. As a result, the first mover in

Zhao Wu
Game Theory HW 04 the sequential equilibrium can always have at least the same utility as in the simultaneous game if d
S *  d
C *
. Therefore, if d
S *  d
C *
, then it be that U
D a
S * d
S *
( , )
 C * C *
U
D
( a , d ) otherwise the first mover could just simply choose
C * d .
Prooved.