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NECESSARY AND SUFFICIENT CONDITIONS
FOR THE PERFECT FINITE HORIZON FOLK THEOREM
LONES SMITH
94-17 (93-6 Rev.)
May 1994
Massachusetts
institute of
technology
50 memorial drive
Cambridge, mass. 02139
NECESSARY AND SUFFICIENT CONDITIONS
FOR THE PERFECT FINITE HORIZON FOLK THEOREM
LONES SMITH
94-17 (93-6 Rev.)
May 1994
M.I.T.
LIBRARIES
JUL 1 2
1994
RECEIVED
Forthcoming
in
Econometrica
Necessary and Sufficient Conditions
for the Perfect Finite
Horizon Folk Theorem
by Lones Smith*
Department of Economics
Massachusetts Institute of Technology
Cambridge,
May
MA 02139
23,
1994
INTRODUCTION
1.
For a given n-player normal form game G,
where the objective function
finitely-repeated
let
G(8,T) be the T-fold repeated game
the average discounted
is
game, the perfect folk theorem
is
sum
said to hold
For this
of stage payoffs.
if
the set of subgame perfect
equilibrium (SPE) payoffs includes any feasible and strictly individually rational payoff
vector of
G for large enough T <
oo and 8
<
Benoit and Krishna (1985) [hereafter, BK]
1.
produced a perfect folk theorem for G(6, T) that obtains when
— namely, the full-dimensionality condition
conditions for the infinite-horizon folk theorem
and Maskin (1986), and
of Fudenberg
(ii)
G satisfies the sufficient
(?)
each player has distinct Nash payoffs in G.
Recently, Abreu, Dutta and Smith (1993) discovered an (essentially)
— non-equivalent
(NEU)
utilities
I
Nash
(i)
NEU, and proved
a revised general
'folk
provide a simple, easily verifiable, and necessary and sufficient condition
theorem that replaces
for the finite-horizon folk
distinct
condition for
— that neatly supplants full-dimensionality. Wen (1993)
subsequently examined games possibly violating
theorem'. Here
iff
The
(ii):
payoffs, only insists that players' behavior
intuitive condition, recursively
be
iteratively leveraged near the
end of the repeated game.
The primary
of
contribution of this note
1
BK, and thus complete
NEU
is
purely conceptual:
finish the
work
only differed from full-dimensionality by a nongeneric class of games, so too, within
Nash
I
wish to
the perfect information folk theorem program. Indeed, just as
the class of stage games with recursively distinct
I
I
payoffs for all players
later provide
also recast
a necessary and
BK
in
simple
new proof
means
in
The
Wen's
Nash
is
But conceptual
clarity is not
of the necessity of
tiered 'folk'
BK
without
sufficient condition for the finite-horizon
Wen's more encompassing framework.
logic of the
(1990/92)
zero.
is
Nash payoffs, the measure without distinct
NEU
in
Abreu
2
its
Nash
In so doing,
et al. (1993),
own reward,
I
as
folk theorem.
happen upon a
and of what
'necessity'
theorems more generally.
folk theorem, as best exemplified in
Krishna (1989) or Smith
rather simple: Late in the repeated game, because
payoffs, the behavior of
all
players have distinct
any one of them can be leveraged by threatening to
finish off
with a fixed number (say S) of plays of that player's worst Nash payoff, rather than cycle
through
all his
best
Nash
ishments work perfectly well.
effect
Away from the end of the game, the infinite-horizon punAnd because S is fixed independent of the horizon length, the
payoffs.
on the average payoff can be made arbitrarily small. This essentially
Rather than formally define
player example
*It
I
am
condition,
I first
motivate
it
their proof.
with the following three-
game G.
BK conjecture under weak conditions that only one player need have distinct
paper might in part be seen as proving a precise and rigorous statement of this conjecture.
grateful to a referee for pressing me on this point.
should be noted that
payoffs. This
2
my
is
In this game,
2,2,3
2,2,2
2,2,2
2,1,-1
0,-1,-1
0,-1,-1
2,2,2
2,2,2
2,2,2
-1,0,-1
-1,-1,-1
-1,-1,-1
2,2,2
2,2,2
-1,-1,0
-1,0,-1
-1,-1,-1
-1,-1,-1
chooses rows (actions
1
[/,
M,D),
2 chooses columns (actions
chooses matrices (actions L,R). Player 3 strictly prefers
is
weakly dominated
(pure or mixed) payoffs are
minimax payoff
the
all
convex combinations of
for all three players.
is
Because players
and 2 have unique Nash
1
payoffs of 2
and
3,
so that his behavior
only threaten to switch from (U,
large enough, 3
is
induced for players
is
(2, 1).
So player
G(R)
(i.e.
2's
when
play
(r,
1
and
2.
It
player 3 plays L).
payoff of 2 in G(£, L).
the game! That
I
£,
G
I
is 1,
which
what player
I
NEU,
satisfies
And my new
Namely,
theorem'.
behavior of
The
all
if
it is
condition
for
1
3
wish just prior to this
and 2 do, a new game
By
choosing S' (and by implication S) large
many
periods, say 5", as
In particular, players 2
0,
which
new
differs
I
wish
and 3 are willing
(one-player)
from
G(r, R)
unique optimal
l's
1
game
to
near the end of
sufficient for the
is
G has recursively distinct Nash payoffs.
BK result: So long as such a recursive
all players,
then a perfect folk theorem
necessary too for the general perfect finite-horizon
any such chain of recursive reductions,
I
SPE
payoffs
is
is
also rather simple. For
'folk
cannot leverage the
either point-valued, or
his behavior
SPE
it is
any given horizon length T,
not. Evidently,
if
the behavior of
payoffs are multivalued for large enough
cannot be leveraged, then he entertains a unique equilibrium
But observe that no one player can simultaneously minimax the other two, a
on in my remarks on the necessity of NEU.
later
S
players, then a 'folk theorem' does not obtain.
set of
if
choosing
from his unique Nash payoff
differs
does, yielding a
a player can be leveraged as above, then his
T, while
need
a folk theorem now follows by the same proof as before.
intuition behind the necessity
a player's
I
have now also leveraged the behavior of player
satisfies
By
I
have now leveraged the behavior of player 2 near
procedure eventually leverages the behavior of
obtains.
and thus the standard
L) for the 5-period phase.
summarize the above procedure by saying that
game
a
G
Furthermore,
leveraged near the end of the game:
has the unique optimal payoff of
It
(2, 2, 3).
has the unique Nash equilibrium payoff vector
Nash payoff from G(R)
irrespective of
1.
and
G satisfies NEU,
,
R)
(resp. r)
G does not satisfy the BK condition
can induce player 2 to play £ m, or r for as
I
for player
(2, 2, 2)
R for as many periods, say 5', as
plays R irrespective of what players 1
just prior to this penultimate 'Nash' phase.
If
payoffs,
is
L) to (M,
£,
the end of the game! Iterate this process.
enough,
D
easy to see that the only Nash
willing to play
Nash phase. When player 3
G(R)
to R, while action
and 3
r),
theorem applies to G. Next, player 3 enjoys the distinct (extremal)
infinite-horizon folk
Nash
it is
m,
3
Nonetheless, a folk theorem does obtain! For
(ii).
in
Thus
for player 1 (resp. player 2).
L
£,
fact of
some importance
payoff for
all
T. Clearly, in the latter case,
G
instance, in
1
game has
T>
5 and
T>
all
satisfying
T,
19, respectively.
one player has distinct Nash payoffs, then the
if
two pure and generically a third mixed Nash equilibrium. Thus,
at least
games
generic
payoffs for
payoff set for
G speak to the genericity of the distinct Nash payoff requirement
theorem obtains. Indeed,
folk
SPE
only receive distinct
special payoffs in
when the
SPE
with no discounting, player 3 has a multivalued
while players 2 and
The
cannot possibly hope for anything more. For
I
my
condition,
Nash
players enjoy distinct
all
my
Section 2 focuses on the stage game, and describes
payoffs.
iterative procedure.
and necessity
sufficiency (the folk theorem) in section 3.1,
for
I
dwell on
(in all respects) in section 3.2.
THE STAGE GAME
2.
2.1 Basic Definitions
Let
G=
(Ai,
=
i
7r,-;
i's finite
set of actions,
set 7r(a)
=
(tti(o),
Simply write
.
.
,
.
G
.
.
n) be a finite normal form n-player game, where Ai
,
.
A=
and
"
x
t
=1J4,-. Let player i's utility function
n n (a)). Let Mi be player
7r,(^) for
The game
1,
utilities
genstern utility functions are equivalent,
(1993), let {lb
in the
to
i.
C
Denote the
nj(-).
X, b
=
1,
.
.
.
=
=
>
0, for all i}
F*
^
0. Note that under
Wen
Kj(a)
and
NEU, w
= max a
<
j
for all
i
and
j, 7r,(-) is
1 =
=
denote by X{i)
ttj(-)
7r,(aj,
€
for all j
all
a_j) of player
1(i).
l
To
it
i
is
co{7r(//)
:
/j,
reduces to the standard
€
I.
(/ii,
l
j€I
.
//„)
€ M.
Following
I
Wen
such that any two players
The
effective
minimax payoff
the best payoff that anyone in
all
4
players in I(i)-
e M}, and
set.
To
call
F*
:
sidestep trivialities, let
minimax strategy
payoff
= {w G F
for player
z,
set.
it is true.
Indeed, w' is certainly a minimax profile
weakly exceeds any minimax payoff for all j £ Z{i), let
)
Analogous to the fact that the minimax
= minmax7Tj(a) > maxmin Jrj(a) = max
a
.
not a positive affine
> maximin
games,
Tr^u;')
.
M,-.
as such, but
see that iri(w
nj(a'j,a.-j) for j
=
)-
:
M = x"=1
with
{1,2, ... ,n}.
minimax payoff over
F=
fi
,
players with equivalent utilities
1{i).
(strictly) individually rational
(1993) does not describe
€
set,
ir,
player
no two players' von Neumann Mor-
the coarsest partition of
Define
i.
if
the feasible and (strictly) rational payoff
feasible
some player
i.e.
the greatest
i.e.
for all
Wi
and F* the
7r,-(-)
(NEU)
set of players as
utilities;
min a maXj 6 i(,) max„
Normalize n^w')
for
B} be
Normalize payoffs so that
X(i) can guarantee himself,
4
,
same X& have equivalent
level 7r,(w')
mixed strategy
Vs expected payoff under the mixed strategy
has non-equivalent
transformation of
i's
be
is
A — M, and
j€/
a
j€/
minmax7r,(a,-,a_,) )
\ a -j a i
J
[
.
in constant
sum
2.2 Recursively Distinct
Nash Payoffs
Given a subset of players 5 J'
=
a j,
let
=
{ji, j%,
(ah ,ah ,..., a jm )
G(aj>) be the induced (n
.
.
,
C X and their
jm }
Mh x Mh x
€
— m)-player game
•
•
x
•
(possibly mixed) actions
M
jm
X\J'
for players
=
Mj>,
(1)
obtained from
G
when
the actions of players J' are fixed to aj>
G as an increasing sequence of h >
Define a Nash decomposition of
{0
=
y{ e Jg -i) of
1,
nonempty subsets
namely
of players from I,
so that for g
1
.
.
h, actions
,
.
= JoC JiC J2 C---C JhQX},
G(e^_,) and y{fjg _
ejg _
x
)
x
of
,
fjg _ G
y
G{fJg _
Mj _
g
a pair of Nash payoff vectors
different exacta/ for players in
)
1
exist with
Y
(2)
Jg \ Jg _ u
Viej^i * V(fs^)i
i.e.
(3)
E Jg \ Jg -\- For instance, players in J\ have distinct Nash payoffs in G.
The game G has recursively distinct Nash payoffs if there is a Nash decomposition with
for all
=
Jh
i
X. So
holds.
if,
BK,
as in
The converse
is
all
players have distinct
not true, as illustrated
not have recursively distinct Nash payoffs,
then
arise, since
union of
distinct
There
Nash
t
It is
An
G
that do
ambiguity
may
not inconceivable that the
player. In fact, this cannot occur.
J* C X who have
6
recursively
FINITELY-REPEATED GAMES
Nash
games with
(a,i
,
.
.
.
,
oj.t)
all
players.
be a behavior strategy for player
with the strategy profile a. Player
pected discounted
Payoffs
perfect monitoring, allowing each player to
condition his current actions on the past actions of
in period
impossible.
games
payoffs.
shall analyze finitely-repeated
=
is
a well-defined maximal set of players
is
3.1 Sufficiency of Recursively Distinct
Let a,
later consider
I
=X
Jh
payoffs in G, then this condition
Nash decompositions need not be unique.
3.
I
earlier.
Nash decompositions could include every
all
Lemma
i.e.
Nash
sum of his payoffs: -—^ Y.J
5
Throughout,
ACB
6
The proof of
this result is omitted; see
means that
A
is
a
strict
my
i's
and nu (a)
his expected payoff
objective function in G(S, T)
(-1
<5
i,
7r, (
(a).
subset of B,
i.e.
The set
of SPE payoffs
AC B and A / B.
(1994) working paper.
is
is
the ex-
V(6, T).
Beyond replacing the
Theorem
Nash payoff requirement, Theorem
distinct
3.7 in three ways. First,
it
admits payoff discounting, where 5 and
Second,
independently over the relevant range.
Wen
pursued by
BK's
(1993), which will imply
period, players can condition on the
random
variable.
7
BK,
Just as in
with
NEU.
T
can vary
theorem', as
'folk
Third, unlike the (more
public randomization
is
used: In every
outcome of a publicly observed exogenous continuous
assume that players can observe deviations within
shall
I
result
BK,
intuitive) use of long deterministic cycles in
a more general
it is
from BK's
1 differs
the support of strictly mixed strategies. Alternatively, just assume that the folk theorem
pure strategy minimax payoff
refers to the
level.
8
G has recursively distinct Nash payoffs. Then for the finitely-repeated game G(5,T), V« € F* and Ve > 0,
3v G V(S, T) with \\v - u\\ < e.
3 T < oo and S < 1 so that T>T and S G [5 1]
Theorem
(The 'Folk Theorem')
1
Suppose that the stage game
=
,
Fix a Nash decomposition
Proof:
c9
g
for
=
1, 2,
.
.
.
,
= .*W£
*€ Jg\ J 9-1
Let the p
h.
(2) for
>
\\y(
which inequality
e J.-i)i
~ y(fjg -i)i\\ >
be the largest payoff range
best and worst payoffs) for any player in G, and
2kp/cg
ones.
Saving on notation,
.
let
y
9
Further, let z 9A be the less
y(fjg -i) for player
i
€
Jg \ Jg -\.
ip
g
and define
°
(i.e.
the difference between
the least even
(k)
number above
and y{fjg _,
preferred Nash payoff vector amongst y(ej
denote y{ejg _^
Since for
-kp +
it
(3) obtains,
all
i
)
in even periods
?>i>g (k)z?>
follows that without payoff discounting, any player
m > 0,
For any
for
7
g
=
h
—
One may
l,h
—
2,
.
.
.
,
1.
=
ijj
g
(m
Define
=
ip h
and
i
(4)
,
has a strict incentive to conform
,
if
deviations
+
s g (m) for
(m) and
+ s g +i(m)
£o(
)
.
recursively define Sh(m)
s g (m)
_,
odd
i
to k consecutive periods of any action profile followed by ipg{k) periods of y9
are punished by switching each y9 to z 9,t
in
Jg \ Jg -u
e
g (k) y
ii>
)
m =
)
0>
H
and
also take advantage of the correlating device to
1-
t
g
s h (m))
(m)
=
si(ra)
+
•••
produce a nearly exact, rather than an
one can use the correlation device to do away with the ^approximation
for all payoff vectors not on the boundary of F. I omit such a laborious exercise.
This assumption is not necessary to detect deviations from the minimax phase (step 3, below), as
noted in my (1994) working paper. But for the two recursively Nash phases, no such workaround exists.
It would thus be interesting to know whether Theorem 2 goes through when J* is defined given pure
actions (1), in which case this would not be a concern.
approximate, folk theorem. That
is,
=
g
1,
.
.
,
.
The T-period equilibrium outcome sequence
h.
is
h
h
a,...,a;y ,...,y ;...;y\...,y
where a
If
v
is
I
9
and y9
^ j,
Ui (target payoff
now
x
x 3>
(strict IR), x\
<
\\v
—
<
u\\
e for big enough 5
£
x? for all j
periods of the repeated game;
some player
2.
Jg
<
.
Reward Phase.
Play xJ for r periods.
Then
Bad Recursive Nash Phase.
deviates, where g" < g' set g' «—
,
straightforward (or see
SPE
for
a
game
rather stark.
9
10
[If
some
my
q, r
G
[If
£
any
i
+ r + th(q +
r)
i
deviates early, start 3; if
i -f-
j
g (q
and
+
5.]
i.
some
or 2.
1
Play z 9 '^ until period
g" and
t
T - V-i(? + r
restart 5.]
Then go
)-
pfj e
to step
<?g"
2.
(1994) working paper) to verify that these strategies
and big enough
5q
Nash
and
To.
11
Payoffs
does not have recursively distinct Nash payoffs, the consequences are
The
following result has a flavour of the "zero-one" laws of probability
Note that q and r are implicitly defined in steps 3 and 4 below.
Below, i and j denote arbitrary players, and g, g' and g" arbitrary
shall use the simple notation j <—
i
to
mean
"assign j the value
i."
indices in {1, 2,
.
.
.
,
h). For clarity,
Also, steps always follow sequentially,
unless otherwise indicated. Bracketed remarks refer to off-path play,
11
any
deviates early, restart 3; if
,
I
n
For
X(i) deviates, start 4.] Set j 4-
return to step
3.2 Necessity of Recursively Distinct
If
,x
10
this equilibrium.
>
w' for q periods. [If j
Jg> deviates late, start 5.]
constitute an
.
5.]
i
4.
is
.
and
which support
T — th(q + r).
.
Minimax Phase. Play
G
.
others are called early deviations.
deviates late, start
3.
i
It
,
Good Recursive Nash Phase. For g = h,...,l\ Play y 9 in periods T r) + 1,
T — t g -i(q + r). [If some G Jg deviates late, where g' < g, start
.
5.
all
Play a until period
in
and T.
1
domination).
explicitly describe the players' strategies
Main Path.
0.
Z(i) (payoff asymmetry),
ease of exposition, late deviations are those occurring during the final q
1.
>
played for s g (q+r) periods. Fix £
is
have established the existence of feasible payoff vectors x
et al. (1993)
such that for alii
<
periods,
the (5-discounted average payoff vector, then
is
Abreu
x\
T—th(q+r)
played for
1
i.e.
following deviations.
Absent public randomization, the simple direct proof presented above would require two modifications:
First, the target outcome a would have to be replaced by an approximating finite outcome profile. Second,
ditto for each x' vector. Finally, to my knowledge, public randomization is essential if one wishes to work
with the mixed minimax strategy, which I noted earlier was possible. See for instance, Abreu et al. (1993).
T —
Namely, as
theory.
payoff set F*, or
some
oo, either
>
V(S,T) tends to the
players receive a unique
SPE
payoff.
strictly individually rational
There
is
no middle ground.
Theorem 2 For any T < oo and any S £ (0, 1], players inX\J* receive
SPE of G(S,T), equal to their unique Nash equilibrium payoff of G. 12
X \ J* has a
in I \ J* has
Proof:
Every player in
unique Nash and hence
Assume
that everyone
a unique
payoff in G, for
T=
1,
.
.
.
,T
Then
.
SPE
SPE
a payoff in any
payoff in G(S,
=
1)
Nash
payoff in G(8, T) equal to his
in the first period of
+
G(S,T
1),
players in
G.
X \ J*
must play a Nash equilibrium of G(ej.),
for
any ej- € Mj., because they have a unique
SPE
By
induction, the result obtains for
continuation payoff by assumption.
all
T.
Remarks:
1.
This stark necessity result contrasts markedly with the partial failure of NEU, where
we need only
Let F*
2.
=
{Xf,}.
profile separately holding
,
13
all
with the effective minimax payoff
If (+)
7t i(
aji a -j) V«}, where the partition
for no such partition {lb} does there exist any strategy
SPE
> mina max
€ f,^
max^
7r,-(oy,o_j-), for
By
corollary, the necessity result of
(a particularization of (*)),
X(i)
payoffs Pareto dominate the effective
This yields 'necessity' in another respect: The folk theorem
F*.
level.
two or more players in any Xb to their worst feasible payoff in
necessarily 7r,(w')
(1993) proves that
level
> mina maxj€ x(t) maxaj
{v € F*\v{
{lb} strictly refines
F* then
minimax payoff
replace the
if
NEU
fails,
Abreu
et
al.
fails
X(i).
But
Wen
minimax outcome. 14
when F*
(1993) follows!
then {It} admits a
C
is
replaced by
For under
strict refinement,
NSM
and the
(standard) finite- or infinite-horizon folk theorem cannot obtain!
3.
Recursively distinct Nash payoffs
the finite horizon
12
13
14
Note that
That is,
this
Nash
folk theorem,
is
also the necessary
for all
i,
and
sufficient condition for
due to Benoit and Krishna (1987). 15
theorem says nothing about the payoff
£ l(i) C l(i)
and
X(i)
C I(»)
for
set of players in
some
J*.
i.
Unlike his folk theorem, this result of his actually doesn't rely on observable mixed strategies.
And
it
holds for finitely-repeated games too.
15
I am grateful to a refereefor suggesting this. For a proof of sufficiency, consider the following strategies:
Let the equilibrium path follow steps 1 and 2. Off-path, retain the minimax phase, step 3, but allow it to
last until the end of the game, and omit any reference to play after deviations, including step 4 (as it is
not needed in a Nash equilibrium); finally, retain step 5, but omit the references to off-path play. To see
necessity, just observe that
Theorem
2
still
obtains
if
SPE
is
replaced by
Nash
equilibrium.
References
Abreu, Dilip, and Prajit Dutta (1991) 'Folk Theorems for Discounted Repeated Games:
a
New
Condition.' University of Rochester
mimeo
Abreu, Dilip, Prajit Dutta, and Lones Smith (1993) 'The Folk Theorem for Repeated
Games: a
NEU
Condition.' Forthcoming in Econometrica
Benoit, Jean-Pierre, and Vijay Krishna (1985) 'Finitely Repeated Games.' Econometrica
53,
—
905-922
(1987) 'Nash Equilibria of Finitely Repeated Games.' International Journal of
Theory
16,
Game
197-204
Fudenberg, Drew, and Eric Maskin (1986) 'The Folk Theorem in Repeated
Games with
Discounting or Incomplete Information.' Econometrica 54, 533-554
Krishna, Vijay (1989) 'The Folk Theorems for Repeated Games.'
mimeo
89-003, Harvard
Business School
Smith, Lones (1990/92) 'Folk Theorems: Two-Dimensionality
mimeo
Wen, Quan
(1993) 'The 'Folk Theorem' for Repeated
Information.' Forthcoming in Econometrica
is
(Almost) Enough.'
Games with Complete
MIT
APPENDIX: OMITTED MATERIAL
4.
Proof of
4.1
If for
Lemma
some Nash decomposition
I
= X,
have Jh
then unambiguously
J* =
2, and
I
am
done.
Otherwise,
let
£,(G) be the set of
Nash
all
the alternative unique nested sequence of
{0
=J C
h!
C
Ji
payoffs for player
nonempty
J2 C
•
• •
X\J for whom {£i(G(aj aj
players Jg+ \ J exist, set h! = g.
€
G
)),
g
\
Note that
Mj
{Jg }
in the
game G. Consider
satisfying
C Jh C J},
>
given by the following (well-defined) recursion: Given
i
sets
i
Jg
,
the set
Jg +i \ Jg
is all
does not consist of a singleton.
}
If
players
no such
g
it
may not be possible to find
only two Nash equilibria providing distinct pay-
Jg +\\Jg Still, can show that Jh = Jh for all Nash
Moreover, for any such nested sequence
decompositions (1). For
is clear that Jh C Jh
For
{£i(G(a,j,)),aj, €
(1), all players in any Jg are eventually included in some Jg
simultaneously for
offs
players
all
i
G
I
-
it
<.
>
Mj,} does not
if
.
consist of a singleton, then neither does {£i(G(aj„)),aj„
G Mj,,}
if
J'CJ".
Omitted Parts of Proof of Theorem
4.2
Step
Observable Mixed Strategies
1:
proceed recursively as
I
I
ensure subgame perfection.
some
that all deterrents are strict by
continuity of discounted
S
G
1
[5o,l], for
some
5q
sums
<
1.
positive margin, say
in 5, they will thus
I
remain
shall check that after
First note that J only check
1,
without discounting. Given
strict for
any
level of discounting
no history consisting of isolated
deviations does a one-shot deviation yield a higher payoff than the strategy profile; by the
"unimprovability" criterion, that profile will be a
I
now
select q
and
r.
In light of x\
w\
for all
x\
<
16
i,
where
U{ too, (4)
The
/?'
+
>
0, let
qx)
>
subgame
q satisfy
PI
+
perfect equilibrium.
16
1
(4)
(resp. w') is the best (resp. worst) payoff vector for player
i
in G. Since
simultaneously renders steps 3 and 4 a strict deterrent to early deviations
inequality (4)
— in particular, why the left-hand side
is
not (q
+
l)x\
obeying the random correlating device generating x' might sometimes require
outcome.
i
— reflects the fact that
to play his worst possible
from step
for any
1,
Next, step 4 deters deviations from step 3
r.
+ rxj > Pj+ rx3] + (q-
qw'j
for all j 7^
Next,
This
i.
possible because Xj
is
[q
i
lg
G
i>; (s 9 (Q
>
s g (q
1
=
.
.
+
r)
•
+
•
s h {q
T > T
T < oo
for
—
\\u
Step
v\\
<
,
.
,
1
+
1
h.
each deterrent remains
<
and 5
j
i
for
whom
an adaptation of Abreu
is
k,
F
,
or x\
c !J to
=
c\
xJ or
,
3
l
cj
\
(b)
>
onto the
i-j
j
arj:
is
6 F* such
c'-'
€ F*
— to which the reader
coordinate plane has
l.
satisfies
is
referred
After step
that: (a)
payoff x' in step 5
4,
minimaxing players j
^
full
dimension; for
17
not indifferent between c,J and x
(via the public randomization device)
incentives of
et al. (1993)
play proceeds just as in step
There must exist payoff vectors
c
f° r
[<$o,l],
To
need only (below) consider deviations from m' by those players
I
the projection of
any other player
IJ
> q + r + th{q + r).
some 8 < 1. Finally,
if
r are fixed.
a simple argument that
7^
S €
feasible
Unobservable Mixed Strategies
2:
The argument
for
strict for all
is
large enough, the resulting average payoff vector v
and
since q
e,
1
an immediate
is
+ r)](-p) + s g (q + r)yf
Given q and r as defined above, the above program
For
Indeed, this
+ r))(-p) + s g (q + r)y?
+ r)zf'
1, 2,
1
which upon substitution yields
+ r + s g+l (q +
=
and g
,
(3),
enough that
x^.
verify that step 5 deters all late deviations.
I
consequence of inequality
for all
>
+
1)uj
r is large
if
i is
1
,
or
indifferent
c-
^
between
xj\ and
l
(c)
x'
j prefers
probabilistically replaced
is
and
l
by c i
where the probabilities are chosen to maintain the
,
i,
and player
i
is
unaffected by the transitions given
(a).
Let Pt'(aj) satisfy for
all aj,a'j in
the support of m}, and
Kj {a, raLj )
,
*-h-i
That such
real
numbers
(1
ttj (a'j
,
<
q,
wLj =
)
+S + -..+ S^Mia'j) -p^ajW/ - x)).
3
p\
-
t
{-)
exist is a
consequence of
(b).
That they can be chosen
be (non-unique) probabilities follows from this construction, for instance, when cj
3
(6):
17
Let p\ {aj)
Below,
if
=
such k
0,
where action
a,-
exist, the expression
yields the highest
u
j
^
*"
>
to
x^ in
myopic payoff to player j (when the
actually denotes j
10
(5)
^
i,
k.
other players play
m*
3
substituting p\ {aj)
8
<
1
and
=
in (5),
r.
m
1
a'j
will follow that pJ (Oj)
<
the realized sequence of action
—
1) for
each j
^
l/(n
Also observe cj
-.
p
i,
x
i{a q
)
myopic payoff to player
yields the lowest
J
it
Furthermore, p t{a")
step 4 with probability
l/(n
action
l
a" in the support of
If
If
•).
—
>
<
l/(n
\)q for any larger 8
x^ implies that
profiles is a q
= £? p^ (ajt)
=
(a,i,a,2,
for all j
^
i.
p
.
tj
.
By
and
(a")
,
—
then after
l)q for large enough
r,
>
j,
or any other action
for all
a g ), then
7
c'-
a'-.
replaces
construction,
<
yielding a well-defined probabilistic transition function.
independent of actions by players j
^
i
and of j
's
own
^t,
actions in periods s
(5)
a;*
in
ij
p (-) <
As p? is
ensures
that players are indifferent across the support of their minimaxing strategies. Finally, in
light of (c), the
punishments that previously supported step 5
11
still
obtain.
uou
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