# Microeconomics 2 ```Problem set 2
Microeconomics 2
af Jonas Knudsen
Question 1.
Alice and Bob play a game given in strategic form as follows:
1/2 L
C
R
T (2; 1) (3; 4) (5; 3)
M (4; 3) (1; 2) (2; 1)
B (3; 8) (0; 6) (1; 6)
.
a) Eliminate iteratively all strictly dominated strategies
If we start from the side of player 1, we notice that strategy B is strictly dominated by M
and a rational agent will therefore never play B, it is therefore eliminated.
From player 2's perspective, we notice that strategy R is strictly dominated by C and is
therefore eliminated.
The simpliﬁed game is given by the following matrix.
1/2
L
C
T (2; 1) (3; 4)
M (4; 3) (1; 2)
.
There is no longer any strictly dominated strategies.
b) Find all pure- and mixed strategy Nash equilibria of this game.
Is is noticed that if player 1 plays the strategy T the best response of player 2 will be to
play C.
If player 1 plays the strategy M the best response of player 2 will be to play strategy L.
We quickly notice that both (T ; C) and (M ; L) is pure nash equilibria.
Now for the mixed Nash equilibria.
A mixed strategy is a nash equilibrium if and only if
(p
~ ? )TA(q~? ) &gt;(p
~ )TA(q~? ) 8p
~ 2 S
? T
? T
?
~
~
(p
~ ) B(q ) &gt;(p
~ ) B(q ) 8q
~ 2 S
Where A; B is the payoﬀ matrices for player 1 and player 2 respectively.
1
Therefore, consider the two mixed strategies (1 &iexcl; p; p) and (1 &iexcl; q; q) and the following system
of equations.
@
@
1&iexcl;q
( 1 &iexcl; p p )A
= 0 , ( (2+2 p) (1&iexcl; q)+(3&iexcl;2 p) q) = 0
q
@p
@p
,&iexcl;4q + 2 = 0 , q =
1
2
@
@
1&iexcl; q
( 1 &iexcl; p p )B
= 0 , ( (1+2 p) (1&iexcl; q)+(4&iexcl;2 p) q) = 0
q
@q
@q
,&iexcl;4 p+3 = 0 , p =
3
4
Meaning player 1's mixed strategy is given by 1? = (1 &iexcl; p; p) =
&iexcl;1 1
strategy is given by (1 &iexcl; q; q) = 2 ; 2
&iexcl;1
3
; ,
4 4
and player 2's mixed
Question 2.
Consider the following strategic game:
1/2
L
C
R
T (2; 5) (20; 0) (12; &iexcl;1)
M (4,-5) (8; &iexcl;2) (10; 4)
B (0; 1) (16; 2) (14; &iexcl;5)
.
Is there a mixed strategy Nash equilibrium in which Player 1 uses strategies T and M with nonzero probability and does not use strategy B, and in which Player 2 uses strategies L and C with
Let p1; p2 2 (0; 1)
E(T ; p1) = 2p1 + 20p2 + (1 &iexcl; p1 &iexcl; p2)12 = &iexcl;10p1 + 8p2 + 12
E(M ; p2) = 4p1 + 8p2 + (1 &iexcl; p1 &iexcl; p2)10 = &iexcl;6p1 &iexcl; 2p2 + 10
E(B; p2) = 0p1 + 16p2 + (1 &iexcl; p1 &iexcl; p2)14 = &iexcl;14p1 + 2p2 + 14
He is indiﬀerent between T and M if
&iexcl;10p1 + 8p2 + 12 = &iexcl;6p1 &iexcl; 2p2 + 10 , 2p1 &iexcl; 5p2 = 1
(1)
And he is indiﬀerent between M and B if
&iexcl;6p1 &iexcl; 2p2 + 10 = &iexcl;14p1 + 2p2 + 14 , 2p1 &iexcl; p2 = 1
We now have a system of two equations:
2 &iexcl;5
2 &iexcl;1
p1
p2
0
1
1
1
p1
2 &iexcl;5 &iexcl;1 1
=
,
=
=@ 2 A
1
p2
2 &iexcl;1
1
0
2
(2)
&iexcl;1
1
Meaning player 2 is indiﬀerent between T ; M ; B when (p1; p2; 1 &iexcl; p1 &iexcl; p2)= 2 ; 0; 2 : Meaning that
C is never played. From just this we can already conclude that there does not exist a such mixed
Nash equilibrium as that in the question.
If it is a nash equilibrium an agent must be indiﬀerent between all of his strategies. In this example,
for player 2, he is only indiﬀerent between his strategies when he plays M with probability 0.
Therefore, there does not exist such a Nash equlibrium in this game.
3
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