Course- NEKM31 Microeconomics - Individual Choice & Game Theory::
Exam in Microeconomics - Theory for. Individual Choice and Game Theory. Hans Carlsson and Lars-Gunnar Svensson. February 23, 2008
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1) Consider the game
(1,1 0,0 0,0)
(0,0 2,2 2,2)
(0,0 2,2 2,2)
For what mixed strategies of Bob is the first row of Alice a best
reply? For what mixed strategies of Alice is the second column a best
reply for Bob?
2)Write a two-person game where each player has 4 pure strategies and
where six rounds of iterated elimination of strictly dominated
strategies yields a unique solution.
is there anybody can help me as soon as possible(in 10 hours)? exam is
coming!
I will try to answer rapidly due to your time constraints.
1)
For notation, we suppose Bob can choose strategies s1, s2, and s3;
and, without risk of confusion, we will call Alice's choices the same,
s1,s2, and s3.
Note that the payoffs are the same if Alice and Bob are interchanged,
and note that the second and third rows are the same (as are the
second and third columns).
1a)
In a mixed strategy, Bob and Alice can choose to play moves with a
different probability.
Suppose Bob chooses strategies s1, s2, and s3 with probabilities
p1,p2 and p3.
If Alice chooses strategy 1, her payoff is then
p1*1 + p2*0 + p3 *0 = p1
This is the DEFINITION of payoff in a mixed strategy game, and it's
also common sense. Alice only gets payoff of 1 when Bob chooses s1,
which he only does with probability p1, so on average Alice gets a
payoff of p1 in this case.
If Alice chooses strategy s2, her payoff is then
p1*0+p2*2+p3*2 = 2(p2+p3)=2(1-p1)
By symmetry, her payoff is the same for strategy s3.
Hence, Alice chooses strategy 1 as a best strategy if and only if:
p1>=2(1-p1)
3p1>=2
p1>=2/3
Therefore, Alice choosing strategy 1, the first row, is a best
strategy for Alice if and only if the probability that Bob chooses
strategy 1 in his mixed strategy is greater than or equal to 2/3.
1b)
By symmetry considerations (since the second and third choices are the
same, and Bob and Alice have the same payoff functions), the second
column is a best reply for Bob if and only if the first column is not
a best reply, since
the second column is a best reply if and only if the third column is
also a best reply.
Hence, the second column is a best reply for Bob if and only if
q1<=2/3, where q1 is the probability that Alice chooses strategy 1 in
her mixed strategy. (If it's clearer one can work through the math as
in part a or I can clarify if necessary).
2)
An appropriate normal form matrix satisfying the conditions is:
Bob
(1,1) (2,2) (3,3) (4,4)
Alice (2,2) (3,3) (4,4) (5,5)
(3,3) (4,4) (5,5) (6,6)
(4,4) (5,5) (6,6) (7,7)
At the first round, Alice eliminates strategy 1 (row 1), since it is
strictly dominated.
At the second round, Bob eliminates his first column, since it is
strictly dominated - that is, Bob can always do
better by playing to another columns.
At the third round, Alice eliminates strategy 2, since the second row
is strictly dominated by the third and fourth rows - Alice can always
do better by playing in the 3rd and 4th rows.
At the fourth round, Bob eliminates strategy 2, his second columns.
At the fifth round, Alice eliminates strategy 3, her third row.
At the sixth round, Bob eliminates column 3, his third strategy.
Note that the order of elimination in a particular elimination might
be different.
Remaining is the strategy profiles (row4, column4) with payoff of
7,7.
This is the only Nash equilibrium for the system.
Search Strategy
--------------
For this problem I continued to use the excellent book Game Theory
Evolving, by Herbert Gintis, Princeton, 2000. PLCY 289 MICROECONOMIC THEORY FOR PUBLIC POLICY::
File Format: PDF/Adobe Acrobat - View as HTMLin microeconomic theory -- particularly game theory. These techniques and concepts will be applied to problems such as wage contracts and collective
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Readings in Applied Microeconomic Theory: Market Forces and Solutions Beyond Individual Choice: Teams and Frames in Game Theory
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I was thinking more about problem 2) and I believe I have come up with
more interesting example for you.
The example I gave was a correct solution, but aesthetically it had
certain disadvantages: there were many choices
of strictly dominating strategy to use at each step. A natural
question to ask is: can we find a payoff matrix
such that there is only a SINGLE choice of strictly dominating
strategy to remove at each step in the iterative
removal of dominating strategies?
I found an example; however, it is of course a lot harder to see that
it is correct than the example I gave.
Consider the payoff matrix;
1,4 1,3 1,3 1,3
4,0 2,4 2,3 2,2
4,0 5,2 3,4 3,3
4,0 4,1 4,3 4,4
There are no strictly dominated columns and only the first row is
strictly dominated. So after round 1, Alice play on the first row is
dominated and we can cross it out to reach the matrix:
4,0 2,4 2,3 2,2
4,0 5,2 3,4 3,3
4,0 4,1 4,3 4,4
Here there are no dominated rows and only the first column is strictly
dominated, so Bob's first column play is strictly dominated and can be
removed, and we get the matrix:
2,4 2,3 2,2
5,2 3,4 3,3
4,1 4,3 4,4
Now again, there are no dominated columns and the top row is strictly
dominated by the other rows so we remove it to get:
5,2 3,4 3,3
4,1 4,3 4,4
Now the first column only is strictly dominated so we remove that to
get:
3,4 3,3
4,3 4,4
Now only the first row is strictly dominated, and then only the first
column so we get the Nash equilibrium
(4,4)
That is the ONLY allowed
order of removal is Alice strategy 1, Bob strategy 1, Alice strategy
2, Bob strategy 2, Alice Strategy 3, Bob Strategy 3.
I found this matrix through playing around with the payoff matrix
values.
Again, this clarification is not strictly speaking necessary for
problem 2, I just thought it was a more interesting example
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