1
Lecture 1
1.1
Mixed Strategies, Dominance and Never a Best
Response
1. Defn: A normal-from (or strategic-form) game consists of three components:
a set of players N ,
for each player i 2 N , a set of strategies Ai ,
for each player i 2 N , a utility function ui : A ! < that maps
each outcome a (a1 ; :::; an ) to a real number.
2. In a simultaneous-move game like the battle of the sexes the set of
strategies is the same as the set of feasible actions. But the concept
of strategy is applicable to games that involve multiple stages as well.
Example: Sequential Battle of the Sexes. Here, the second mover has
two feasible actions but four strategies. The point is: a strategy is what
one can do, and what one can do depends on what one knows as well
as the set of actions one can choose from.
3. Defn: A mixed strategy for player i, i , is a probability dist. over Ai ,
and i (x) is the probability that player i chooses strategy x 2 Ai . We
use i or Ai to denote the set of mixed strategies.
4. Eg. Suppose Ai = fa1i ; a2i ; a3i ; a4i g. Then a mixed strategy can be denoted by a vector, like, (0:1; 0:2; 0:3; 0:4). The pure strategy of choosing
a1i is denoted by (1; 0; 0; 0).
5. Defn: Each mixed strategy pro…le
( 1 ; :::; n ) induces a prob dist.
over A. The expected utility of player i is
X
( j2N j (aj )) ui (a) .
ui ( ) =
a2A
6. Defn: A strategy ai is dominant for player i if for all a
a0i 2 Ai =ai
ui (ai ; a i ) > ui (a0i ; a i ) :
i
2A
i
and
If a strategy dominates all a0i 2 Ai =ai , it dominates all mixed strategies
i 2 i =ai .
1
7. Defn: A strategy is strictly (weakly) dominated if there exists
such that for all i 2 i
ui ( i0 ;
i)
> ( ) ui ( i ;
0
i
6=
i
i) :
Note that 1) a mixed strategy is strictly (weakly) dominated if any
pure strategy in its support is strictly (weakly) dominated. 2) A pure
strategy may be strictly dominated by a mixed strategy and not by any
pure strategy.
8. For example:
a 1
0
b 0
1
c 0:3 0:3
c is dominated by mixing 50-50 a and b.
9. Defn: a mixed strategy
i 2
i such that
i
is never a best response if there exists no
ui ( i ;
for all
0
i
2
i)
ui ( i0 ;
i)
i.
10. It is obvious that a strictly dominated strategy is never a best response.
The question is whether there are strategies that are never a best response but yet not dominated.
11. Proposition 1. In two-person games, a strategy is strictly dominated if
and only if it is never a best response.
12. The proposition is true only when the number of players is two. When
there are more than two, the in the above proof need not be a mixedstrategy pro…le. In general a undominated strategy need not be a best
response.
13. The proof is provided at the end of this section. Here, I provide an
outline of the argument. Consider the following game:
l
r
a 1
0
b 0
1
c 0:3 0:3
d 0:6 0:5
2
(I only include the payo¤ to the row player.) Since the column player
has only two strategies, we can plot the payo¤ in a diagram. In Figure
1 each dot represents the payo¤ for one of the pure strategies of the row
player. A mixed strategy is a convex combination of pure strategies.
Hence, the entire payo¤ set, denoted by A, is the convex hull of the
four points. Each point in the set correspond to the payo¤ of a mixed
strategy.
14. A mixed strategy is not dominated if it is on the north-east frontier
of the payo¤ set. For example d is not dominated— the set of payo¤s
that strictly dominates d, denoted by B, and the feasible payo¤ set is
disjoint. Since both sets are convex, there exists a hyperplane h passing
through d that separates the two sets. From college algebra, we know
that each hyperplane is characterized by a normal vector = ( 1 ; 2 )
and a constant c such that for all x 2 <2
8
< > c if x 2 A
:x = c if x 2 h :
:
< c if x 2 B
Finally, notes that has positive slope, so that each component of
is non-negative, and we can normalize so that 1 + 2 = 1. It then
follows that d is a best response against .
15. Separating Hyperplane Theorem (Mas-Colell, Whinston, and Green
pp.948): Suppose that the convex sets A, B <N are disjoint. Then
there is p 2 <N with p 6= 0, and a value c 2 <, such that p:x c for
every x 2 A and p:y c for every y 2 B: That is, there is a hyperplane
that separates A and B, leaving A and B on di¤erent sides of it.
16. Proof of Proposition 1 (never best response=strictly dominated): We
will show that if ai is not strictly dominated, it must be a best response
against some j . Suppose #Sj = m. For each ai 2 Ai , de…ne
v(ai )
(ui (ai ; a1j ); ui (ai ; a2j ); :::; ui (ai ; am
j )):
The k component of vector v(ai ) is the payo¤ for player i when he
cofv(x) : x 2
chooses ai and his opponent chooses akj . De…ne V ai
m
Ai =ai g and W
fx 2 R jx > v(ai )g: If ai is not strictly dominated,
then V ai \ W = ;: By the supporting hyperplane theorem, there
3
exists 2 Rm with k k = 1 such that 8x 2 V ai and 8y 2 W; :y
:v (ai )
:x. All components of must be non-negative. If i < 0
then :y < c for any c if i-th component of y is su¢ ciently large and
all other components of y is bounded from above. Hence, ai is a best
response to the mixed strategy j
.
1.2
Nash Equilibrium
1. Defn: A mixed strategy pro…le
i 2 N and for all i0 2 i
ui ( i ;
is a Nash Equilibrium (NE) if for all
i)
ui ( i0 ;
i) .
2. Some …nite games like Matching Pennies have no pure strategy equilibrium, but every …nite game has a mixed strategy NE.
3. Issue 1: The Nash equilibrium implicitly assumes that players knows
that each player is to play the equilibrium. Given this knowledge,
no player wants to deviate. So, there is a sort of circularity in this
concept— the players behave in this way because they are supposed to
behave in this way.
4. The mutual understanding required by Nash equilibrium can be justi…ed in several ways: the players may reach a self-enforcing agreement
to play this way through pregame communication. For examples: you
may agree with a friend to meet at a particular restaurant for dinner. 2. A steady-state convention may be a result of some dynamic
learningnevolutionary process. Examples: we usually takes nodding
your head to mean yes and shaking your head means no. 3. In coordination games, certain equilibrium just “stands out”as a focal point.
5. Issue 2: Note that a player is willing to mix between two strategies if
she is indi¤erent between them. (We use this condition to …nd equilibrium strategies.) One may wonder why should a player mix speci…cally
according to the equilibrium probabilities?
6. There are several responses to this questions. See OR for details.
7. Issue 3: Nash equilibrium requires each strategy only be a weak best
response, not strict. It only requires that no player has any incentive
4
to deviate, it doesn’t require that the players are strictly better o¤
following the equilibrium. In the following game:
a
b
c 0; 1 1; 1
d 100; 0 1; 0
there are two equilibria: (c; b) and (d; a). One could argue that in the
…rst equilibrium, the row player should play d rather than c, because
d is as good as c against b and strictly better against a. But this
argument requires an argument that goes beyond Nash equilibrium—
in the equilibrium (c; b), the row player is certain that the column player
is going to play b, it would not matter that d would do better against
a.
8. We may argue that there is always some very small probability that
the row player may make a “mistake” and choose a even when he
intends to choose b. One way to formalize this idea is to require that
an equilibrium be “robust,” meaning that it remains an equilibrium
even when the game is “slightly”changed.
9. If we replace the weak inequality sign in the de…nition with the strict
inequality sign, it is called a strong Nash equilibrium. The problem
is that strong Nash equilibrium may not exist; e.g.; there is none in
Matching Pennies.
1.3
Common Knowledge of Rationality
1. N.E. requires that each player has the correct expectation of his opponents’ strategies. This is unrealistic in one-shot games without prior
communication.
2. But even when we are not sure about players’expectations, we know
that a rational player will always choose a best response against some
belief. Hence, if a player is con…dent that his opponent is rational, he
should choose a strategy that is a best response against strategies of
his opponent that are themselves best responses (to his strategies).
3. If it is common knowledge that both players are rational, we can continue this process inde…nitely. The set of strategies that survive this
process is called rationalizable.
5
4. Example 1: Consider the following game (Gibbons pp. 6) :
Player 2
L
M R
Player 1 U 1,0 1,2 0,1
D 0,3 0,1 2,0
5. Note that R is strictly dominated by M for Player 2. Now, if Player 1
knows that Player 2 is rational, then Player 1 knows that Player 2 will
never choose M. If M is eliminated, then D becomes dominated by U.
Now, if Player 2 knows Player 1 knows that Player 2 is rational, then
Player 2 knows that Player 1 will not choose D. In that case, Player 2
should choose M.
6. Formally, if player j thinks that player i is rational, player j expects
player i to choose a strategy from the set
n
o
Y
0
0
;
)
8
2
A=
2
j9
2
s:t:
u
(
;
)
u
(
i
i :
i
i
i
j6=i j
i i
i
i i
i
7. But player j does not know exactly which strategy player i may choose.
It is possible, for example, that player j thinks that there is 10% chance
that player i’s strategy is a and 90% chance that player i’s strategy is
b. In a …nite game, this is equivalent (in terms of payo¤) to player j
expecting player i to play a mixed strategy of choosing a with probability 0.1 and b with probability 0.9. Hence, if player j is rational and
he knows that player i is rational, then he may choose a strategy that
is a best response against a player i’s strategy in the convex hull of A.
8. Defn. Let e 0i
en
i
f
i
i
2 e ni 1 j9
and, for each i, de…ne recursively
i
2
j6=i co(
e n 1 ) s:t: Ui ( i ;
j
i)
Ui ( i0 ;
i)
8
e n is the set of strategies that we think player i may choose if every
i
player knows that every player knows that.....every player is rational
for n times. The set of rationalizable strategies for player i is de…ned
en
as: Ri \1
n=0 i :
9. R = (R1 ; :::; Rn ) captures the restriction imposed by common knowledge of rationality. See OR ch.5 for a formal defn. of common knowledge.
6
0
i
2 e ni 1 g:
10. Note that the de…nition we use here is di¤erent from that of OR in that
we require that i be a best response to some mixed strategies. This
rules out any correlation between strategies of the other players. This
is the way Bernheim and Pearce originally de…ne the concept, and it is
consistent with the (implicit) assumption that it is common knowledge
that players choose strategies independently. If players have access to
random devices, then correlated rationalizaibility (as de…ned in OR) is
the proper solution conception.
11. In two person games a strategy is never a best response i¤ it is strictly
dominated. Hence, in two person games, iterated deletion of neverbest-responses is the same as the convex hull of pure strategies that
survive iterated deletion of strictly dominated strategies. In n person
games, n > 2, a never-best-response may be undominated. In that
case, iterated deletion of strictly dominated strategies is equivalent to
correlated rationalizability and is a weaker than rationalizability (in the
sense that the set of rationalizable strategies is a subset of the strategies
that survives iterated deletion of strictly dominated strategies).
12. In our example (U,M) is also the unique Nash equilibrium. In general,
any strategy is part of a Nash equilibrium is rationalizable. But the
converse is not true. Iterated deletion of strictly dominated strategies
is strictly weaker than Nash.
13. Thm 1. Non-emptiness. Ri is non-empty and contains at least one pure
strategy.
14. Outline of the proof: e ni are closed for all n, and countable intersections
of closed sets in non-empty. (Another way to see that is to recognize
that N.E are always contained in R. Thus, R is non-empty whenever
equilibrium exists.) If i 2 Ri , then all pure strategies that i put
positive probability on must also belong to Ri .
15. Find the set of pure strategies that survive iterated deletion of strictly
dominated strategies.
b1 b2 b3 b4
a1 1,4 4,1 3,3 4,2
a2 2,3 6,0 2,2 2,2
a3 3,1 2,7 2,2 2,4
a4 6,6 3,5 1,3 1,4
7
1.3.1
Example: Cournot Oligopoly
1. There are two identical …rms in a market. Let q1 and q2 denote the
outputs of …rms 1 and 2, respectively. The market demand is: 1 q1 q2 .
The marginal cost of production is c.
2. Firm 1’s pro…t function is given by
1
Di¤erentiate
for …rm 1:
1
= (1
q1
q2
c)q1 :
with respect to q1 , we obtain the best response function
c q2
:
2
Similarly, the best response function for …rm 2 is
q1 (q2 ) =
q2 (q1 ) =
1
1
c
2
q1
:
3. Solving the equations, we obtain the Nash equilibrium:
q1 = q2 =
1
c
3
:
4. Let’s consider iterated deletion of strictly dominated strategies. Since
q1 is strictly decreasing in q2 , a rational …rm will never produce more
than q1 (0) = 1 2 c . Thus, the set of strategies that remain after one
round of deletion, Q1 , is given by 0; 1 2 c : Now if a …rm expects its
competitor is rational and will not produce more than 1 2 c , it will not
produce less than q ( 1 2 c ) = 1 4 c . You can see it from the diagram that
the process will converge to the unique Nash equilibrium.
i
h
5. More formally: If Qn is equal to q n ; q n , then Qn+1 will be equal to
h
i
h
i
qn 1 c
1 c qn 1 c qn
qn
1 c
n+2
;
and
Q
will
be
equal
to
+
;
+
. Thus,
2
2
4
4
4
4
we have the following recursive relations:
q n+2 =
and
q n+2 =
8
1
c
4
1
c
4
+
+
qn
4
qn
:
4
Working recursively backward, we obtain :
qn =
1
and
qn =
Set q 0 = 0 and q 0 =
q1 = 1 3 c :
1 c
2
1
n
1
n
n
1
n
2
cX
1i
12
+ q0
4 i=0 4
4
2
cX
1i
12
+ q0 :
4 i=0 4
4
and let n goes to in…nity. We obtain q 1 =
6. The argument will not work when there are three …rms because …rm
1 may expect …rms 2 and 3 each produces an output of 1 2 c and, in
that case, …rm 1 best response is q (1 c) = 0. Thus, no strategy will
be deleted after the …rst round and any output between 0 and 1 2 c is
rationalizable.
1.3.2
Order Independence
1. An important property of rationalizability is that it is order independence—
it does not matter which order we take to do the deletion. If some strategy should be eliminated in round k, then were it put back in the set
would also be eliminated in each subsequent round j, for all j k:The
solution concept would not make sense if this does not hold. Why?
2. Thm 4. Let An denote the set of strategies that survive n round of
iterated deletion of dominated strategies. If ai is strictly dominated
with respect to An , then it is strictly dominated with respect to Am ,
for all m > n:
3. Proof. A pure strategy ai is strictly dominated with respect to An
means that there exists some in 2 co(Ani ) such that
ui (
n
i ; a i)
> ui (ai ; a i ) 8a
i
2 An i :
n
Now, for all m n, if in 2 co (Am
i ), then i strictly dominates ai in
round m as co(Am
co(Ani ). If in 2
= co (Am
i )
i ) ; then it must be that
in some round k between n and m; some pure strategies in the support
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of in are strictly dominated by some mixed strategies in co(Aki 1 ).
Construct a new strategy ik by replacing each of the pure strategies
eliminated by the mixed strategies that dominates it. Since each of
these mixed strategies is in co(Aki 1 ), ik is also in co(Aki 1 ). Moreover,
since strict dominance is transitive, ik strictly dominates ai . If ik is
in co (Am
i ), the proof is done. If not, repeat the same procedure until
round m is reached.
1.4
Rationality and Coherent Theories
1. (I have made up the terminologies in this part. Gul uses di¤erent
terminologies but his are less convenient.)
2. Nash equilibrium is a theory or prediction about the outcome of a
strategic situation. Here we consider some general properties of any
re…nements of common knowledge of rationality.
3. Defn. A collection of mixed strategies X
(X1 ; :::; Xn )
are called self-justi…able if 8 i and 8 i 2 Xi ; 9 i 2 X i s.t.
:::: n
2 br( i ):
1
i
4. The de…nition says that if X is self-justi…able then every strategy in it
can be justi…ed as a best response to some other strategy in X:
5. Any Nash equilibrium is self-justi…able. More generally, any solution concepts that is consistent with rational behavior should be selfjusti…able.
6. Defn. A collection of mixed strategies X
(X1 ; :::; Xn )
are called self-contained if 8 i and 8 i 2 X i s.t. i 2 br(
1
i)
::::
Xi .
n
7. A set of strategies are self-contained if it includes all best responses
against itself. A solution concept that yields self-contained predictions exhausts all implications of rational behavior— in the sense that
if everyone behave according to the way the solution concept predicts,
then no rational person would want to behave di¤erently.
8. If a set of strategies is not self-contained, then it imposes additional
restrictions beyond common knowledge of rationality.
9. A Nash equilibrium may not be self-contained (e.g the one in Matching
Pennies); a strong Nash equilibrium is.
10
10. Thm 2. R is the largest set of strategy pro…les that are self-justi…able.
11. The procedure introduced in de…nition 1 eliminates all strategies that
are not best response iteratively. The theorem says that a strategy that
is not eliminated by this procedure must be justi…able by some other
strategies that are also not eliminated by this procedure.
12. Outline of the proof: Any X that is justi…able cannot be eliminated by
the procedure, and thus must belong to R: To see that R is justi…able,
note that the procedure must stop after a …nite number of iterations.
13. Thm 3. R is self-contained.
14. Proof: If X is self-justi…able and i is a best response to some strategy
in X i , then f i g [ X must be self-justi…able: If R is not self-contained,
then it could not the largest set of self-justi…able strategies.
15. This simple theorem means that R exhausts all implications of common
knowledge of rationality.
1.5
Dominance and Iterated Deletion of Weakly Dominated Strategies
1. We have already seen that Nash equilibrium does not rule out weakly
dominated strategies. For the same reason, they are also not ruled out
by rationalizability. For example:
l
r
a 0; 1 1; 0 :
b 2; 0 1; 0
2. In this game, l and r are both best response against b, and a and b are
both best response against r. But r weakly dominates l, and b weakly
dominates a.
3. There seems to be a strong intuition ruling out weakly dominated
strategies. We would think that the likely outcome is r; b.
4. The intuition is formalized by assuming that each player believes that
there is some " > 0 probability (for some arbitrarily small ") that his
11
opponent may be irrational and would play any strategy with strictly
positive probability. We then carry out the iteration process for the
strategies of the rational players. The outcome is the set obtained by
deleting all weakly dominated strategies and then deleting all strictly
dominated strategies iteratively. (Gul, 1996 and slightly di¤erently
Dekel and Fudenberg, 1990)
5. It is important to note that only strictly dominated strategies are eliminated after the …rst round, and so the procedure does not justify iterated deletion of weakly dominated strategies.
6. For example
U
D
L
C
R
10000; 2 1; 2 1; 3 :
0; 5
1; 5 0; 1
7. In this example: there are three pure strategy equilibria: (D; L), (D; C),
(U; R). All strategies are rationalizable.
8. L is weakly dominated by C. After deleting L, D is weakly dominated
by U . So the outcome surviving iterated deletion of weakly dominated
strategies is (U; R). But I will argue that (D; C) is not unreasonable.
The standard reason for deleting D is that even though it is a weak
response against C is that it is worse for the row player against R. But
if the row player thinks there is always a chance that the column will
play R, then he might also think there is a small chance that he may
play L in which case U will be a terrible choice.
9. Furthermore, deletion of weakly dominated strategies is based on the
idea that there is always some small probability that any strategy may
be chosen (with " probability). On the other hand, the idea of iterated
deletion is based on the idea that strategies deleted in earlier rounds
will never be chosen for sure. In this sense, iterated deletion of weakly
dominated strategies combine two con‡icting ideas.
10. Another problem is that the iteration of weakly dominated strategies
does not satisfy order-independence. For example, in the …rst game
if we …rst eliminates l, then a and b will both survive. On the other
hand, if we …rst eliminates a, then both l and r survive.
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1.6
Level-k Models
1. Nash and IDSDS are based on di¤erent logic. IDSDS does not require
that the players know that the equilibrium is going to be played, so it
requires less coordination. However, common knowledge of rationality
is itself a very strong assumption. Nash equilibrium, on the other hand,
only requires that the players have mutual knowledge of rationality. See
OR Ch. 5. We will return to the relation between rationalizability and
Nash equilibrium next week.
2. Consider the p-beauty contest game: There are n players; each is to
choose a number between 0 and 100. The person whose number is
closest to p times the group mean wins a prize. If more than one
person chooses the closest number, then every one who does share the
prize with equal probability.
3. If p < 1, then the unique rationalizable strategy pro…le (hence, the
unique NE) is for everyone to choose 0. For example, suppose p = 2=3.
Then, any number greater than 200=3 will never be a best response.
If a person is rational and knows that every one is rational, then it is
not a best response for this person to choose any number greater than
400/9. And so on.
4. There is a great number of paper that conduct this game experimentally, and the results consistently show that 1) a greater majority of
players do not choose zero, and 2) zero is actually a poor choice empirically.
5. Instead, the results points to the prevalence of the iterated-best-response
rule— there is a spike in the number of people choosing 50p and 50p2 .
Empirically, one tends to do well by thinking two steps ahead or slightly
more in this game.
6. Two observations: 1. Common knowledge of rationality is too strong.
Most likely players do not think in terms of higher order beliefs. 2.
Not all best responses are equally likely. 50p is much more likely than
0 even though both are possible best response.
7. Vince Crawford and others have developed a level-k model and applies
to a variety of settings.
13
8. Level 0 player plays non-strategically. Depending on situations, they
may be choosing randomly (in the p-beauty contest) or non-weaklydominated strategies (like bidding a randomly chosen amount that is
less than the value of the good in an auction). Level one player chooses a
best response against level zero players. More generally, level n players
best respond against level n-1 players.
9. Unlike NE, a level-k model is a non-equilibrium theory.
10. Crawford and his collaborators argue that this model …ts the experimental data of many non-repeated one-shot games better than other
alternatives. My view is mixed (though I haven’t read the literature
carefully. The reason is that the level zero behavior is arbitrary, so
there is scope to rationalize any data by choosing the level zero behavior judiciously.
11. Colin Camerer has a related model in which a higher level chooses a
best respond against a mixture of lower level players.
12. References: Bosch-Domenech et. al., 2002, “One, Two, (Three), In…nity,...: Newspaper and Lab Beauty-Contest Experiments”AER Vol.92.
13. Camerer et. al., 2004, “A Cognitive Hierarchy Model of Games,”QJE
August.
14. Crawford’s paper can be found on his personal webpage.
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