A Purification Theorem for
Perfect-Information Games∗
Adam Brandenburger†
Amanda Friedenberg‡
September 2002
Revised January 2007
Kalmar [2, 1928-9] proved that Chess is strictly determined. Von Neumann-Morgenstern [5,
1944] proved the same for any finite two-person zero-sum perfect-information (PI) game. The latter
result yields a minimax theorem for (finite) non-zero-sum PI games.1 Fix a PI, and a player, Ann.
Convert this game to a two-person zero-sum game between Ann and the other players (considered
as one player), in which Ann gets the same payoffs as in the initial game. In the new game, the
min max of Ann’s payoffs is equal to the max min of Ann’s payoffs. Since this statement involves
only Ann’s payoffs, it must also hold in the initial game.
In this note we first prove a generalization of this fact: The min max and max min of Ann’s
payoffs are still equal, when they are taken over subsets of strategies, provided the subsets for the
players other than Ann are rectangular.
We then use this generalization to prove a purification result: Fix a PI game, a strategy s for
Ann, and rectangular subsets of strategies for the players other than Ann. Suppose s is weakly
dominated with respect to these subsets. That is, suppose there is a mixed strategy for Ann that
gives her at least as high a payoff for each profile of pure strategies for the other players, taken from
the given subsets, and a strictly higher payoff against at least one such profile. Then there is a pure
strategy for Ann that weakly dominates s in the same sense.
Kalmar’s [2, 1928-9] proof was forward looking. Another forward-looking argument is Kuhn’s [3,
1950], [4, 1953] well-known proof that every finite PI game has a pure-strategy Nash equilibrium.”
Our arguments, too, are forward looking, proceeding by induction on the length of the tree.
1
Perfect-Information Games
An extensive-form PI game is defined as follows: Let {a, b} be the set of players.2 A tree is given by a
finite set of nodes N , which are partially ordered by . The set N has a least element φ with respect
to . That is, φ is the initial node. The set of terminal nodes is Z = {z ∈ N : z z → z = z }.
∗ Previously titled “A Generalized Minimax Theorem for Perfect-Information Games.” We thank Elchanan Ben
Porath, Konrad Grabiszewski, and Jeroen Swinkels for helpful comments. Financial support from the Stern School
of Business and the Department of Economics, Yale University, is gratefully acknowledged. p t p i - 0 1 - 1 6 - 0 7
† Address:
Stern
School
of
Business,
New
York
University,
New
York,
NY
10012,
[email protected], www.stern.nyu.edu/∼abranden
‡ Address: Olin School of Business, Washington University, St.
Louis, MO 63130, [email protected],
www.olin.wustl.edu/faculty/friedenberg
1 Stated in, e.g., Ben Porath [1, 1997]. We are grateful to Elchanan Ben Porath for the following argument.
2 This is for notational convenience only. Our arguments immediately extend to games with three or more players—
provided the rectangularity conditions hold.
1
The set of non-terminal nodes is X = N \Z. Information sets are given by sets {x}, where x ∈ X.
Often, we will identify an information set {x} with x. The set of choices available at node x ∈ X
is the set of immediate successors of x; denote this set by C (x).3 The mapping ι : X → {a, b}
specifies the player who moves at each information set. Let X a = {x ∈ X : ι (x) = a}, and define
X b analogously. Extensive-form payoff functions are given by Πa : Z → R and Πb : Z → R.
A strategy for player a is a mapping from information sets into available choices. That is, the
strategy set for a is S a = ×x∈X a C (x). Let S = S a × S b . A strategy profile s ∈ S determines a
path through the tree. Let ζ : S → Z where ζ (s) = z if and only if s reaches the terminal node z.
Strategic-form payoff functions are given by πa = Πa ◦ ζ and πb = Πb ◦ ζ.
2
A Generalized Minimax Theorem
a
Take C (φ) = {1, .., K}. For each k, let Ska = ×x∈X a :kx C (x). If ι (φ) = a, S a = C (φ)× ×K
k=1 Sk .
a
a
a
In this case, denote by S a (k) = {k} × Ska . If ι (φ) = a, S a = ×K
k=1 Sk . Write projSka s = sk if
a
a
a
a
s = (j, s1 , ..., sk , ..., sK ).
Definition 2.1 A subset Y a of S a is rectangular if it is a product set, i.e., if Y a = ×x∈X a Y a (x)
where, for each x ∈ X a , Y a (x) ⊆ C (x) with Y a (x) = ∅.
Notice that if both Y a and Y b are rectangular subsets, then there is another PI game Γ where
Y and Y b can be identified with strategy sets of Γ . But, if rectangularity of one of these sets fail,
there may be no such tree.4 Thus, the following result is a generalization of the Minimax Theorem:
a
Theorem 2.1 Fix a PI game. Let Y a be a subset of S a and Y b be a rectangular subset of S b .
Then,
a b
a b
a
a
min max
π
s
,
s
=
max
min
π
s ,s .
a
a
a
a
sb ∈Y b s ∈Y
s ∈Y
sb ∈Y b
Proof. By induction on the length of the tree.
Length = 1: If ι (φ) = a then S b = {∅}. So, certainly
πa sa , sb = max
π a (sa , ∅) = max
min π a sa , sb .
min max
a
a
a
a
a
a
sb ∈Y b s ∈Y
s ∈Y
s ∈Y
sb ∈Y b
If ι (φ) = b then S a = {∅}. We then have
min
maxa πa sa , sb = min
πa ∅, sb = max
minb πa sa , sb ,
a
a b
b
b a
b
b
s ∈Y s ∈Y
s ∈Y
s ∈Y
s ∈Y
as desired.
Length ≥ 2: Assume the result is true for any tree of length λ or less. Fix a tree of length
λ + 1.
b
b
First suppose ι (φ) = a. Since Y b is rectangular, we can write Y b = ×K
k=1 Yk , where each Yk is
b
a rectangular
subset
of
S
.
k
a b
a
Let sak , sbk ∈ arg a max
min
π
s , sk . Consider sb ∈ S b such that projSkb sb = sbk for
s ∈Y a ∩S a (k) sbk ∈Ykb
each k. Since Y b is rectangular, sb ∈ Y b . Let sa1 , sb1 ∈ Y a × Y1b be such that πa sa1 , sb1 ≥
πa sak , sbk for each k with Y a ∩ S a (k) = ∅.
3 An
4 We
immediate successor of x is a node y ∈ N with x y and x y → y y .
thank Jeroen Swinkels for this point.
2
It follows that sa1 , sb ∈ arg max
min πa sa , sb : For any sa ∈ Y a , sb ∈ arg min π a sa , sb
a
a
b
b
b
b
s ∈Y s ∈Y
s ∈Y
if and only if projSkb sb ∈ arg min πa sa , sbk for k such that sa ∈ S a (k). So, certainly, sb ∈
sbk ∈Ykb
a b
a
arg min π s , s for some sa ∈ Y a . Suppose there exists sb ∈ arg min πa sa , sb such that,
b
b
b
b
s ∈Y
s ∈Y
for some sa ∈ Y a , πa (
sa , sb ) > πa sa1 , sb . Without loss of generality, take (
sa , projSkb sb ) ∈
a b
a
a
a
arg a max
min π s , sk for k with s ∈ S (k). This contradicts the choice of sa1 , sb1 .
a
a
b
b
s ∈Y ∩S (k) sk ∈Yk
πa sa , sbk .
Similarly, for each k, let (sak , sbk ) ∈ arg min a max
Let πa sa , sb :=
a
a
b
b
sk ∈Yk s ∈Y ∩S (k)
a b
a
π
s
,
s
.
Define
min max
a
a
sb ∈Y b s ∈Y
π a ·, sb for some sb ∈ Y b }.
MAX (Y a ) = {sa ∈ Y a : sa ∈ arg max
a
Y
It will be shown that, for some j with Y a ∩ S a (j) = ∅, πa (saj , sbj ) ≥ πa sa , sb . To do so, it
suffices to show that for some such j with Y a ∩ S a (j) = ∅, saj ∈ MAX (Y a ). Then
max
πa sa , sb ≤ πa (saj , sbj ),
a
Y
for all
sbj
∈
Yjb .
So, certainly
πa sa , sb ≤ πa (saj , sbj ).
max
Ya
Pick j such that π a (saj , sbj ) ≥ πa sak , sbk for each k with Y a ∩ S a (k) = ∅. Suppose for each k
with Y a ∩ S a (k) = ∅, there exists sbk such that π a (saj , sbj ) ≥ maxY a ∩S a (k) π a sa , sbk = πa sak , sbk .
Pick sb ∈ S b so that projSjb sb = sbj , projSkb sb = sbk for k with Y a ∩ S a (k) = ∅, and projSkb sb ∈ Yk
otherwise. Since Y b is rectangular, sb ∈ Y b . Then saj ∈ MAX (Y a ). So, if saj ∈
/ MAX (Y a ), there
exists k with Y a ∩ S a (k) = ∅ such that π a sak , sbk > π a (saj , sbj ), for all sbk ∈ Ykb . Certainly, for this
k, πa sak , sbk > πa (saj , sbj ), a contradiction.
With this πa sa1 , sb = πa sa , sb . Indeed, suppose not. Then
πa sa1 , sb < π a sa , sb
≤ π a (saj , sbj ) = πa saj , sbj
≤ π a sa1 , sb1 = π a sa1 , sb ,
where the first line comes from max min π a (·, ·) ≤ min max π a (·, ·), the second line comes from
the
choice of (saj , sbj ) and the induction hypothesis, and the third line comes from the choice of sa1 , sb1 .
But this is a contradiction.
min
πa sa , sbk . Further, let
Now suppose ι (φ) = a. For each k, let πa rak , rbk = max
a
a
sk ∈Yk sb ∈Y b ∩S b (k)
a b
a
πa ra , rb = max
min
π
s
,
s
.
Define
a
a b
b
s ∈Y
s ∈Y
MIN Y b = {sb ∈ Y b : sb ∈ arg min
πa (sa , ·) for some sa ∈ Y a }.
b
Y
It will be shown that, for some 1 with Y b ∩ S b (1) = ∅, πa ra1 , rb1 ≤ π a ra , rb . To do so, it
suffices to show that for some such 1 with Y b ∩ S b (1) = ∅, rb1 ∈ MIN Y b . Then
min π a ra , sb ≥ πa r1a , rb1 ,
Yb
3
for all r1a ∈ Y1a . So, certainly
min π a ra , sb ≥ πa ra1 , rb1 .
b
Y
a b
a
a
Pick 1 such that π r1 , r1 ≤ π rak , rbk for each k : Y b ∩ S b (k) = ∅. Suppose for each k with
Y b ∩ S b (k) = ∅, there exists rka such that π a ra1 , rb1 ≤ minY b ∩S b (k) πa rka , sb = π a rka , rbk . Pick
ra ∈ S a so that projS1a ra = ra1 , projSka ra = rka for k with Y b ∩ S b (k) = ∅, and projSka ra ∈ Yka
otherwise. Since Y a is rectangular, ra ∈ Y . Then rb1 ∈ MIN Y b . So, if rb1 ∈
/ MIN Y b , there
∅ such that πa rka , rbk < πa ra1 , rb1 , for all rka ∈ Yka . Certainly, for this
existsk with Y b ∩ S b (k) =
k, πa rak , rbk < πa ra1 , rb1 , a contradiction.
Let πa rak , rbk = minsb ∈Y b ∩S b (k) maxsak ∈Yka π a sak , sb . Define ra ∈ S a so that projSka ra = rak ;
note that ra ∈ Y , as Y is rectangular. Pick raj , rbj so that π a raj , rbj ≤ πa rak , rbk for each k with
Y b ∩ S b (k) = ∅.
Note, (ra , rbj ) ∈ arg min max
π a sa , sb : For each sb ∈ Y b ∩ S b (k), sa ∈ arg maxY a π a sa , sb
a ∈Y a
b
b
s
s ∈Y
if and only if projSka sa ∈ arg maxYka π a sak , sb . So, certainly, ra ∈ arg maxY a πa sa , sb for
some sb ∈ Y b .
Suppose there exists rb ∈ Y b ∩ S b (k) such that πb (
ra , rb ) < πb (ra , rbj ) for
a b
a
a
a
some r ∈ arg maxY a π s , s . Without loss of generality, take (
πa (
ra , rb ) =
r , rb ) so that
minrb ∈Y b ∩S b (k) maxsak ∈Yka πa sak , sb . But this contradicts πa (raj , rbj ) ≤ πa rak , rbk for each k with
Y b ∩ S b (k) = ∅.
It follows that πa ra1 , rb = πa (raj , rb ), since
πa (ra , rbj ) = π a (raj , rbj )
≤ π a ra1 , rb1
= π a ra1 , rb1
≤ π a ra , rb ≤ π a (ra , rbj ),
where the second line comes from the choice of (raj , rbj ), the third line comes from the induction
hypothesis, and the fourth line comes from the choice of ra1 , rb1 and the fact that max min π a (·, ·) ≤
min max πa (·, ·).
3
A Purification Theorem
We need a preliminary result: If a strategy for Ann is inadmissible with respect to a rectangular
subset of strategies for Bob, then Ann’s strategy must be weakly dominated by a mixture that
reaches a single subtree k.
Some notation: For a finite set X, let M(X) denote set of all probability measures on X. Then:
Lemma 3.1 Fix a PI game with ι (φ) = a. If sa is inadmissible with respect to a rectangular subset
Y b of S b , then there exists k and σa ∈ M (S a ), with σ a (S a (k)) = 1, such that
πa σa , sb ≥ πa sa , sb for all sb ∈ Y b ,
πa σa , sb > πa sa , sb
for some sb ∈ Y b .
Proof. Let σ
a ∈ M (S a ) be such that
a b
πa σ
, s ≥ π a sa , sb πa σ
a , sb > π a sa , sb
4
for all sb ∈ Y b ,
for some sb ∈ Y b .
Let k = 1, .., K ≤ K be such that σ
a (ra ) > 0 for some ra ∈ S a (k). For each k = 1, .., K, define
σ
ak (ra ) =
σ
a (ra )
.
σ
a (q a )
qa ∈S a (k)
It is readily verified that σ
ak (S a (k)) = 1. Also, for each sb ∈ Y b ,
K
a b ,s =
πa σ
[(
k=1 q a ∈S a (k)
a b
k , s ].
σ
a (q a )) × πa σ
Suppose, for each k = 1,
.., K, σ
ak does
sa with respect to Y b . Then, for
anotb weakly dominate
a
a
b
a
b
b
each such k, either: (i) π σ
k , s = π s , s for each s ∈ Y ; or (ii) there exists rb ∈ Y b with
a b
a b
a
a
k , r . Notice that if (i) holds for all k = 1, .., K, then for all sb ∈ Y b ,
π s ,r > π σ
K
a b πa σ
,s =
[(
k=1 qa ∈S a (k)
σ
a (q a )) × πa sa , sb ] = π a sa , sb ,
a contradiction. So, (ii) must hold for some such k.
For each k = 1, ..., K satisfying (ii), let rkb = projS b (k) rb and, if sa ∈ S a (j) for j = k, let
b
b
rj = projS b (j) rb . Set rb = r1b , ..., rK
where rkb is as above, if defined, and is otherwise an arbitrary
element of Yk . Since Y b is rectangular, rb ∈ Y b . But
K
a b [(
πa σ
,r =
k=1 q a ∈S a (k)
a b
σ
a (q a )) × πa σ
k , r ] < πa sa , rb ,
where the inequality comes from the fact that (ii) holds for some k = 1, ..., K.
We now come to the purification result:
Theorem 3.1 Fix a PI game. If sa is inadmissible with respect to a rectangular subset Y b of S b ,
then there exists ra ∈ S a such that
πa ra , sb ≥ π a sa , sb for all sb ∈ Y b ,
πa ra , sb > π a sa , sb
for some sb ∈ Y b .
Proof. By induction on the length of the tree.
Length = 1: Without loss of generality, let ι (φ) = a. Here S b = {∅}, so each sb is admissible.
Moreover, if there exists σ a ∈ M (S a ) such that πa (σa , ∅) > π a (sa , ∅) then certainly there exists
ra ∈ Supp σa such that πa (ra , ∅) > πa (sa , ∅), as required.
Length ≥ 2: Assume the result is true for any tree of length λ or less. Fix a tree of length
λ + 1.
First, suppose ι (φ) = a. Fix sa ∈ S a (1) and suppose sa is inadmissible with respect to a
b
a
a
b
b
rectangular
Y . Then,by Lemma 3.1, there exists k with σ (S (k)) = 1 such that, for all s ∈ Y ,
πa σ a , sb ≥ πa sa , sb , with strict inequality for some sb ∈ Y b . Without
loss
of
generality,
pick
σa so that ra ∈ Supp σa implies that there exists sb ∈ Y b with πa sa , sb = πa ra , sb .
Suppose first that Supp σ a ⊆ S a (1). Then by the induction hypothesis, there exists ra ∈ S a (1)
such that
πa ra , sb1 ≥ πa sa , sb1 ∀sb1 ∈ Y1b
;
πa ra , sb1 > πa sa , sb1
for some sb1 ∈ Y1b
5
the result is then immediate. So, let Supp σa ⊆ S a (k), for k = 1.
Pick r1b ∈ arg maxsb1 ∈Y1 π a sa , sb1 and let Y0b = {sb ∈ Y b : projY1b sb = r1b }. Let Y0a := Supp σ.
Let πa sa , sb = maxY0a minY0b πa (·, ·) and πa sa , sb = minY0b maxY0a πa (·, ·). Then, for all sb ∈
Y b,
πa sa , sb ≤ πa sa , sb
≤ πa sa , sb
= πa sa , sb
≤ πa sa , sb ,
where the first line comes from the choice of r1b , the second line from the definition of min max, the
third line from Theorem 2.1, and the
from
the definition of max min. By choice of σ a ,
last
line
a b
b
b
a
a
a b
there exists s ∈ Y , with π s , s > π s , s , establishing the desired result.
Now suppose ι (φ) = a. Let σa ∈ M (S a ) be such that, for all sb ∈ Y b , π a σa , sb ≥ π a sa , sb ,
with strict inequality for some sb ∈ Y . Write sa = (sa1 , .., saK ). For each k with Y b ∩ S b (k) = ∅,
define
σa (ra ) .
σak (rka ) =
a
r a ∈S a :projS a r a =rk
k
It is readily
verified that
(S (k)) = 1. For each k with Y b ∩ S b (k) = ∅, πa σ ak , sb ≥
πa sak , sb for all sb ∈ Y b ∩ S (k). Moreover, there exists k, with sb ∈ Y b ∩ S b (k), such that
πa σ ak , sb > πa sak , sb .
By the
induction hypothesis, for each such k, there exists rka ∈ S a (k) such that πa rka , sb ≥
πa sak , sb for all sb ∈ Y b ∩ S b (k), with strict inequality for some sb ∈ Y b ∩ S b (k). For all other k,
a
set rka = sak . Set ra = (r1a , ..., rK
). It is readily verified that ra satisfies the desired properties.
σak
b
a
6
References
[1] Ben Porath, E., “Rationality, Nash Equilibrium and Backward Induction in Perfect-Information
Games” Review of Economic Studies, 64, 1997, 23-46.
[2] Kalmar, L., “Uber Eine Shlussweise aus dem Endlichen ins Unendliche,” Acta Universitatis
Szegediensis/Sectio Scientiarum Mathematicarum, 3, 1928-9, 121-130.
[3] Kuhn, H., “Extensive Games,” Proceedings of the National Academy of Sciences, 36, 1950, 570576.
[4] Kuhn, H., “Extensive Games and the Problem of Information,” in Kuhn, H., and A. Tucker
(eds.), Contributions to the Theory of Games, Vol. II, Annals of Mathematics Study, 28, 1953,
Princeton University Press, 193-216.
[5] von Neumann, J., and O. Morgenstern, Theory of Games and Economic Behavior, Princeton
University Press, 1944 (Sixtieth Anniversary Edition, 2004).
7