Econ 618
Rationalizable Sets, Iterated Elimination of Strictly
Dominated Strategies (IESDS)
Sunanda Roy
1
Rationality and Common Knowledge of Rationality
NE as a prediction for the outcome of a game requires that each player (1) know the equilibrium
of the game and (2) correctly expects it to be played. Then and then only, will each player have
no incentive to unilaterally deviate from his equilibrium strategy as he expects his rivals to play
their equilibrium strategies. The two assumptions are sensible if a game has been played a number
of times and a specific steady state of the game observed. Thus a NE equilibrium is a plausible
prediction of the outcome of a game if the last statement is true, but in general not if such conditions
are not satisfied (as in one shot games with players choosing actions simultaneously). Recall
from previous discussions that even if each player knows the equilibria of a game, a problem of
coordinating on one arises if there are multiple equilibria. Similarly, the issue of mistakes arise if
an equilibrium is not strict.
An area of game theory explores how far can we move towards predicting an outcome of a
game assuming only that (1) strategy spaces and payoffs are common knowledge and (2) each
player is rational and rationality of all players is common knowledge.
A belief of player i about his opponents is a probability distribution σ−i on A−i , his rivals’ pure
strategy space. A player is rational if he uses only those strategies (pure or mixed) in his strategy
space which are BR to some belief σ−i about A−i .
Define as level 1 knowledge of rationality, ”each player i thinks that each player j is rational”.
1
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If player i has level 1 knowledge of rationality, he will not hold arbitrary beliefs σ−i over A−i . He
will hold only those beliefs σ−i under which a positive prob is assigned to an action by j if the
action is j’s BR to some belief that j holds over A− j .
Define as level 2 knowledge of rationality, ”each player i thinks that each players j thinks that
each player i is rational”. If player i has level 2 knowledge of rationality, then according to i, j will
not hold arbitrary beliefs σ− j over A− j . j will hold only those beliefs σ− j , under which a positive
prob will be assigned to an action by i if the action is a BR to some belief that i holds over A−i .
We can continue as above and define level n knowledge of rationality for each playeri. Common
knowledge of rationality is knowledge of rationality at all levels.
As before Ai is the set of pure strategies of player i and ∆Ai is the mixed extension of Ai or
the set of mixed strategies over Ai . A pure strategy ai ∈ Ai is strictly dominated for player i, if
∃, σi ∈ ∆Ai such that Ui (σi , a−i ) > Ui (ai , a−i ) for all a−i ∈ A−i . A pure strategy ai ∈ Ai is weakly
dominated for player i, if ∃, σi ∈ ∆Ai such that Ui (σi , a−i ) ≥ Ui (ai , a−i ) for all a−i ∈ A−i with strict
inequality for at least one a−i .
We now describe an iterative procedure of deleting strictly dominated pure strategies for each
player i. Under this procedure each iteration assumes that each player i has one more level of
knowledge of rationalty of other players. The objective is to make the strategy space of each
player i ”smaller”. If after infinite iterations, we are left with a singleton Ai for each i, the game
has a unique outcome. Note that while the use of this procedure to predict an outcome of a game
does not require a player to know the equlibrium, it does require infinite levels of ”rationalizing”
by a player - a requirement which is no less demanding according to some behavioral theorists.
2
Iterated elimination of strictly dominated strategies (IESDS)
Define A0i = Ai and ∆A0i = ∆Ai . Define the set of pure strategies for player i, left over after the first
round of elimination as,
A1i = {ai ∈ A0i | there is no σi ∈ ∆A0i , such that Ui (σi , a−i ) > Ui (ai , a−i ) for all a−i ∈ A0−i }
and define
∆A1i = {σi ∈ ∆A0i | σi (ai ) > 0, only if ai ∈ A1i }
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That is A1i consists of only those pure strategies in A0i = Ai which are not dominated against A0−i
and consider only those mixtures in ∆A0i which assign strictly positive prob to an ai only if ai ∈ A1i .
That is, get rid of all mixtures from ∆A0i which assign strictly positive prob to strictly dominated
pure strategies in A0i . Note that this definition allows the player i to retain mixed strategies from ∆A0i
which assign zero probs to strictly dominated strategies. Further and more importantly, ∆A1i may
contain mixed strategies which assign positive probs to undominated strategies (not even weakly
dominated) but are themselves strictly dominated. We shall revisit this issue with an example later.
Further note that A1i ⊆ A0i and ∆A1i ⊆ ∆A0i .
Define the set of pure strategies for player i that are left over after the second round of elimination as,
A2i = {ai ∈ A1i | there is no σi ∈ ∆A1i , such that Ui (σi , a−i ) > Ui (ai , a−i ) for all a−i ∈ A1−i }
and
∆A2i = {σi ∈ ∆A0i | σi (ai ) > 0, only if ai ∈ A2i }
Thus after the second round of elimination, the surviving set of pure strategies A2i ⊆ A1i , contains only those pure strategies in A1i which are not dominated against A1−i . That is in the second
round of elimination we are dropping those pure strategies from A1i which were only best responses
to strategies in A0−i which were dropped in the first round of elimination, but are otherwise dominated. The mixed strategy set ∆A2i is defined accordingly.
Recursively define the set of pure strategies for player i that are left over after the nth round of
elimination as,
Ani = {ai ∈ An−1
| there is no σi ∈ ∆An−1
, such that Ui (σi , a−i ) > Ui (ai , a−i ) for all a−i ∈ An−1
i
i
−i }
and
∆Ani = {σi ∈ ∆A0i | σi (ai ) > 0, only if ai ∈ Ani }
Define the set A∞
i =
n
n=0 Ai
T∞
∞
and the set A∞ = ×i∈N A∞
i . The set Ai represents the set of pure
strategies that survive iterated elimination of strictly dominated strategies (IESDS). Note that A∞
i
may or may not be a singleton. In fact, as one example below shows it may be an infinite set and
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0
further A∞
i = Ai = Ai is a possibility. That is to say, we may not be able to drop a single pure
strategy from the initial set.
Noting that a pure strategy is a degenerate mixed strategy, we now ask ourselves, what is the
set of player i’s mixed strategies which survive iterated elimination of strictly dominated (all kinds
of) strategies? Contrary to the natural guess, this is not the set of mixtures on A∞
i . This set is given
by,
0
0
∞
∆A∞
i = {σi ∈ ∆Ai | there is no σi ∈ ∆Ai , such that Ui (σi , a−i ) > Ui (σi , a−i ) for all a−i ∈ A−i }
That is ∆A∞
i contain those mixtures from the initial set ∆Ai which are not dominated against
A∞
−i . Several remarks are in order.
Remark 1: We are guilty of a slight abuse of notation here as ∆A∞
i is not defined to be the set of
∞
mixtures on A∞
i , to maintain consistency with the previous notation ∆Ai . In fact, ∆Ai is contained
in the set of all mixtures over A∞
i and could be a strictly smaller set, as the following example
shows.
Example 1:
L
R
U (1,3) (-2,0)
M (-2,0) (1,3)
D (0,1) (0,1)
0
∞
0
Note that A∞
1 = A1 = A1 = {U, M, D} and A2 = A2 = A2 = {L, R}. That is no pure strategy
could be deleted from the initial set and as a matter of fact, each pure strategy of each player
is undominated (not even weakly dominated). Consider the mixture for player 1, {1/2, 1/2, 0}.
Assuming that player 2 plays the mixture {p, 1 − p}, the expected payoff for player 1 is, (1/2)p +
(1/2)(1 − p)(−2) + (1/2)p(−2) + (1/2)(1 − p) = −(1/2). Thus the mixture {1/2, 1/2, 0} is
dominated by D against player 2’s mixed strategy set. The essence of this example is contained in
the following statement: A mixed strategy that assigns positive prob to a dominated pure strategy
is dominated. However a mixed strategy that assigns positive prob only to pure strategies that are
undominated (not even weakly dominated) may be strictly dominated.
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∞
∞
Remark 2: It is natural to ask whether the limit set A∞
i or A = ×i∈N Ai depends on the order
of deletion. That is, we have assumed that at each iteration or round of elimination, the strictly
dominated pure strategies of all players are deleted simultaneously. Suppose some other order
of deletion is followed. For example, suppose that in the first round we delete only player 1’s
strategies from A01 , followed by player 2’s strategies from A02 in round two, followed by player
3’s strategies from A03 in round three and so on and come back to player 1 again to proceed with
deletion from A11 . That is, what if we followed sequential deletion instead of simultaneous deletion.
Do we get a different A∞
i ? The answer is ”No”, if we are deleting strictly dominated strategies.
The order of deletion however matters if we are instead deleting weakly dominated strategies.
The following example illustrates why.
Example 2:
U
D
L
R
(1,0) (0,1)
(0,0) (0,2)
(In this game, the two PSNEs are (U, R) and (D, R).) Suppose in the first round of elimination
we delete only player 1’s weakly dominated strategies and follow up by deleting player 2’s weakly
1
∞
∞
dominated strategies in the second round. Then A11 = {U} = A∞
1 and A2 = {R} = A2 . And A =
{U} × {R} = (U, R) is the unique outcome of the game. If however, in the first round we delete
only player 2’s weakly dominated strategies and follow up by deleting player 1’s weakly dominated
1
∞
strategies in the second round, we are left with A12 = {R} = A∞
2 and A1 = {U, D} = A1 . The set of
outcomes are given by, A∞ = {U, D} × {R} = {(U, R), (D, R)}.
The intuitive explanation as to why the order of deletion matters for weakly dominated strate(1)
gies is as follows: Consider a two player game. For player 1 a1 is a weakly dominated strategy
(1)
if ∃ a σ1 such that U1 (σ1 , a2 ) ≥ U1 (a1 , a2 ) for all a2 ∈ A2 with strict inequality for at least one
(1)
a2 . Suppose that the strict inequality holds for only one such a2 = a2 . If we begin the process of
(1)
deleting weakly dominated strategies with player 2 first and a2 gets eliminated, then in the second
(1)
(1)
and all successive rounds a1 stays in A21 and A∞
1 . If however, we start with player 1 first, a1 gets
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knocked out. Note that such a problem never happens with strictly dominated strategies.
Remark 3: In the above procedure, we first eliminated the dominated pure strategies in successive rounds and eliminated the dominated mixed strategies at the very end. It is natural to ask if we
∞
∞
get the same ∆A∞
i and the same ∆A = ×i∈N ∆Ai if we eliminated all the dominated strategies -
pure and mixed - at each iteration. Note that ai is dominated against all pure a−i iff it is dominated
against all mixed σ−i . Hence if we keep a σi which is dominated in player i’s strategy set at some
step, it does not alter which strategies of j are getting deletd at the next step or round. Thus once
again, the order of the procedure does not matter.
∞
A game is dominance solvable if for each player i, A∞
i and hence A is a singleton. Under the
situation a unique joint strategy profile emerges as the outcome of the game. This strategy profile
is a PSNE and in fact the unique PSNE of the game (the proof is easy - see F-T). The converse
however is not in general true except in two player games. That is, a game may have a unique PSNE
but may not be dominance solvable. We shall revisit this issue in the next unit when we discuss
games of strategic complements and substitutes. However dominance solvability and uniqueness
of PSNE are equivalent in a Cournot duopoly, which we study as an illustrative example below.
3
Rationalizability
An action of player i is ”never a best response (NBR)” if it is not a BR to any (at least one) belief
of player i about his rivals.
For any number of players, a strictly dominated strategy is never a best response - the strategy
which dominates is always strictly a better response. The converse, ”a strategy is never a best
response if it is strictly dominated”, is not in general true except in two player games. That is,
in more than two player games, a strategy may be undominated but never a best response, as the
following example shows.
Example: (pending)
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Under IESDS, we iteratively eliminated strategies that were strictly dominated, under the
premise that rational players do not choose strategies which are strictly dominated. Under the
concept of ”rationalizability” our premise is the question, ”What strategies do rational players
play?” Our answer is, only those which are best responses to at least one belief about other players. Working under this premise, we shall iteratively eliminate strategies which are never best
responses because according to our answer, such strategies will never be used by rational players.
˜ 0 = ∆Ai , the latter being the set of mixed strategies on the initial strategy set Ai , for
Define ∆A
i
player i.
˜ 0 | ∃ σ−i ∈ × j∈N convex hull(∆A
˜ 0 ) such that σi is a best response to σ−i }.
˜ 1 = {σi ∈ ∆A
Define ∆A
i
i
j
˜ 0 ).
Thus at step 1, we drop those strategies in ∆Ai which are never best responses to × j∈N convex hull(∆A
j
Working iteratively, at step n,
˜ n = {σi ∈ ∆A
˜ n−1 | ∃ σ−i ∈ × j∈N convex hull(∆A
˜ n−1 ) such that σi is a best response to σ−i }.
∆A
i
i
j
The rationalizable set of strategies for player i is the set Ri =
˜ n
n=0 ∆Ai .
T∞
˜ n−1 ) is the smallest convex set containing ∆A
˜ n−1 . It is possible
Remark 4: The convex hull(∆A
j
j
˜ n−1 (because both are BR to some
that after (n − 1) iterations, strategies σ0j and σ” j are in ∆A
j
belief of j about his rivals) but a convex combination, say ((1/2)σ0j , (1/2)σ” j ) is not. However
((1/2)σ0j , (1/2)σ” j ) is an admissible belief on the part of player i about player j at step n, since
˜ n−1 .
both σ0j and σ” j are in ∆A
j
Theorem 1 (Bernheim 1984; Pearce 1984) Ri is non-empty and contains at least one pure strategy.
Further each σi ∈ Ri is in ∆Ri , a best response to an element of × j∈N convex hull(R j ).
More importantly from an applied point of view and in light of our previous comment, the
following result holds.
Theorem 2 (Pearce 1984) The rationalizable set Ri and the set which survives IESDS, ∆A∞
i , are
identical in two player games.
Our previous comment implies that the set Ri is contained in the set ∆A∞
i in general. Pearce’s
theorem implies that both are equal to each other in two player games. The practical usefulness of
this theorem is that in two player games if it is (from a technical point of view) easier to identify
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the rationalizable set than it is to identify the set which survives IESDS or vice versa, we may
invoke this theorem to draw conclusions about both. The following example serves to illustrate.
3.1
The Cournot duopoly example
Consider first Fig 1 which represents the case of a unique PSNE. The BR functions are depicted as
linear (basically so that I could use a straight edged ruler to draw them!) but they need not be. All
we need is a unique intersection. Note that since by definition, the BR of firm 1 must be defined
for all outputs of firm 2 and the BR of firm 2 must be defined for all outputs of firm 1, a unique
intersection means that the firm 1’s BR must intersect firm 2’s BR from above - that is the BR of
firm 1 must be locally steeper than firm 2’s BR. Let the unique PSNE be (x1∗ , x2∗ ).
The iterated deletion of firm 1’s strictly dominated strategies takes the following form:
A01 = (0, x1max );
A11 = (0, a), because (a, x1max ) is never BR for firm 1 to any choice by firm 2;
A21 = (b, a), because firm 2 will delete (a0 , x2max ) at the previous step;
A31 = (b, c), because firm 2 will delete (0, b0 ) at the previous step ... and so on.
Note that A01 ⊆ A11 ⊆ A21 . . .. Assume that at the nth step of the iteration An1 = (xn1 , x̄1n ) and the
∗
n
∞
∗
IESDS process implies that limn→∞ xn1 = x∞
1 ≤ x1 and limn→∞ x̄1 = x̄1 ≥ x1 . Note that the deletion
process form below and above cannot cross x1∗ , because x1∗ cannot be deleted if x2∗ remains and
∞
∗
∞
vice versa. We want to argue that x∞
1 = x̄1 = x1 and hence A1 is a singleton. Here is the simple
geometric argument.
If the IESDS process stops somewhere from below and above for both firms, by definition of
∞
∞
∞
∞
∞
∞
∞
these limits, x∞
1 = BR1 (x̄2 ), x̄1 = BR1 (x2 ), x2 = BR2 (x̄1 ) and x̄2 = BR2 (x1 ).
∞
∞
From the above relations, x̄1∞ = BR1 (x∞
2 ) = BR1 (BR2 (x̄1 )). Thus x̄1 must lie at a point of
∗
∗ ∗
intersection of the two functions. Hence it cannot be true that x∞
1 < x1 because (x1 , x2 ) is the
unique point of intersection.
Fig 2 depicts the case of non-linear BR functions with multiple intersections. The IESDS
∞ ∞
∞
∞ ∞
process leads to the sets A∞
1 = (x1 , x̄1 ) and A2 = (x2 , x̄2 ) as shown in the figure. This game is not
dominance solvable (and in fact the sets which survive IESDS is infinite).
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Concluding remarks: A NE strategy belongs to both sets Ri and ∆A∞
i . But in general, there
are more strategies in both sets in addition to the NE strategies (strategies which form part of a
correlated equilibrium also belong to these sets) - a feature that limits their usefulness in predicting
the outcome of a game.