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Necessary and Sufficient Conditions for
the Perfect Finite Horizon Folk Theorem
Lones Smith
No.
93-6
May 1993
massachusetts
institute of
technology
50 memorial drive
Cambridge, mass. 02139
Necessary and Sufficient Conditions for
the Perfect Finite Horizon Folk Theorem
Lones Smith
No.
93-6
M.I.T.
LIBRARIES
May 1993
-
DEWEY
RECBVH}
Necessary and Sufficient Conditions
Horizon Folk Theorem
for the Perfect Finite
Lones Smith*
Department of Economics
Massachusetts Institute of Technology-
MA 02139
Cambridge,
May
7,
1993
Abstract
Benoit and Krishna (1985) proved a
folk
finite- horizon
n-player perfect
theorem that assumed that every player has distinct Nash payoffs
in the stage
game and
(essentially) that
infinite-horizon folk theorem. Abreu,
it
satisfy the conditions of the
Dutta and Smith (1993) have
cently provided the necessary and sufficient condition
infinite-horizon folk theorem.
We
do prove that
NEU
(NEU)
is
re-
for the
necessary for
the finite-horizon folk theorem, but more importantly, this note substitutes the distinct
Nash payoff condition requirement with a weaker
necessary and sufficient condition, that players have recursively distinct
Nash
folk
payoffs.
As a consequence, we show how the n-player
theorem might even obtain
offs in
if
finite-horizon
only one player has distinct Nash pay-
the stage game. Conversely,
when the
stage
but not our condition, the folk theorem's failure
is
game
satisfies
NEU
rather dramatic: In
particular, at least one player's limit equilibrium payoff set
is
a singleton.
'The motivation for this note, namely seeking the necessary conditions for a folk theorem,
stemmed from collaboration with Dilip Abreu and Prajit Dutta, and in particular from their
earlier work Abreu and Dutta (1991).
1.
INTRODUCTION
For a given n-player normal form game G,
where the objective function
is
G(5, T) be the T-fold repeated
the average discounted
game, the perfect
finitely-repeated
let
folk
theorem
is
perfect equilibrium payoffs includes any feasible
payoff vector of
G
enough
for large
in the finitely-repeated Prisoners'
'defect' every
period for any
T<
T<
oo and S
G(S, T) obtains
folk
theorem
(1986),
and
and
<
1.
condition for
when
G
(i)
if
the set of
subgame
strictly individually rational
It
has been long known that
Dilemma, the only subgame perfect equilibrium
oo.
This
is
is
the ineluctable consequence of a simple
and Krishna
this fact, Benoit
theorem
led to correctly conclude that a perfect folk
satisfies the sufficient
for
conditions for the infinite-horizon
— namely, the full-dimensionality condition of Fudenberg and Maskin
(ii)
Recently,
BK] were
of stage payoffs. For this
said to hold
backwards induction argument. Perhaps motivated by
(1985) [hereafter,
sum
game
each player has distinct Nash payoffs in G.
Abreu
(i),
et al. (1993) discovered
which they termed NEU,
an
(essentially) necessary
and
sufficient
for non-equivalent utilities, that neatly
supplants the full-dimensionality condition. While we do prove that
NEU
is
nec-
essary for the finite-horizon folk theorem, this note primarily provides a necessary
— recursively distinct Nash payoffs — for the finite-horizon
and
sufficient condition
folk
theorem that replaces
The
thus extending
contribution of this note
tended to
finish the
is
BK
to a complete characterization.
purely conceptual:
The
characterization
work of BK, thus completing the perfect information
rem program. For just
class of
(ii),
as
is in-
folk theo-
NEU only differed from full-dimensionality by a nongeneric
games, so too, within the class of stage games with recursively distinct Nash
payoffs, the
measure with distinct Nash payoffs
for all players is zero. In addition to
the stated goal of this note, our condition helps explain
how
finite-horizon
subgame
perfect equilibria behave as the horizon grows.
The
logic of the
Smith (1990/92)
is
BK
folk theorem, as best exemplified in
Krishna (1989) or
rather simple: Late in the repeated game, because
all
players
have distinct Nash payoffs, the behaviour of 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
the players' best Nash payoffs.
from the end of the game, the infinite-horizon punishments work perfectly
And
because
S
is
fixed independent of the horizon length, the effect
Away
well.
on the average
made
payoff can be
arbitrarily small. This essentially
Rather than formally define our condition, we
three-player example
2,1,-1
-1,-1,-1
1,-1,-1
2,2,2
2,2,2
2,2,2
-1,-2,-1
-2,-1,-1
1,-2,-1
2,2,3
2,2,2
-1,-1,0
-1,1,-1
-2,1,-1
0,0,-1
chooses rows (actions U,
1
weakly dominated
M,D),
2 chooses columns (actions
strictly prefers
for player 1 (resp. player 2).
Nash (pure or mixed)
payoffs are
1
and 2 have unique Nash
payoffs,
m,
G
Thus
it is
easy to see
(2, 2, 2)
satisfies
1
BK
does not satisfy the
G
r),
to R, while action
for all three players.
is
Nonetheless, a folk theorem does obtain! For
(ii).
L
£,
convex combinations of
all
Furthermore, the mutual minimax payoff
Because players
condition
with the following
2,2,2
that the only
(2, 2, 3).
it
2,2,2
(resp. r) is
and
motivate
game G.
and 3 chooses matrices (actions L, R). Player 3
D
their proof.
2,2,3
game,
In this
first
is
NEU, and
thus the infinite-horizon folk theorem applies to G. Next, player 3 enjoys the distinct
Nash
(extremal)
of the game:
We
so that his behaviour
is
need only threaten to switch from (U,
£,
By choosing S
S-period phase.
periods, say S', as
irrespective of
and
2.
It
and
payoffs of 2
we wish
3,
large enough, 3
just prior to this
what players
G(R)
player 3 plays L).
which
is 1,
We
differs
from
can induce player 2 to play
£,
m, or
r for as
G(r,R)
for player
1.
It
in G(£, L).
We
near the end of the game! That
same proof
(2, 1).
R
player 3 plays
1
2's
Nash
(i.e.
when
So player
S' (and hence S) large enough,
many
periods, say S", as
1
does, yielding a
has the unique optimal payoff of
unique optimal payoff of 2
the
many
induced for players
In particular, players 2
R) irrespective of what player
(r,
L) for the
for as
unique Nash payoff in G(R)
By choosing
prior to this penultimate 'Nash' phase.
1
his
is
R
£,
have now leveraged the behaviour of player 2 near the end
of the game! Iterate this process.
to play
When
Nash phase.
and 2 do, a new game G(R)
1
L) to (M,
willing to play
has the unique Nash equilibrium payoff vector
payoff from
player
is
leveraged near the end
0,
new
we wish
we
just
and 3 are willing
(one-player)
which
differs
game
from
l's
have now also leveraged the behaviour of
G
satisfies
a folk theorem
now
follows by
as before.
*But observe that no one player can simultaneously minimax the other two, a fact of some
importance later on in our proof of necessity.
We
Nash
summarize the above procedure by saying that
payoffs. If
G
satisfies
NEU,
G
has recursively distinct
a sufficient condition
it is
for the
BK
long as such a recursive procedure eventually leverages the behaviour of
then a perfect folk theorem obtains. Moreover,
all
So
players,
G satisfies NEU, our new condition
if
necessary too for the perfect finite-horizon folk theorem. Namely,
is
result:
chain of recursive reductions, we cannot leverage the behaviour of
if
for
any such
all players,
then
a folk theorem does not obtain.
The
intuition behind the necessity
length T, a player's set of
or
it is
not.
subgame
evident that
It is
if
is
perfect equilibrium payoffs
enough T, while
leveraged, then he entertains a unique payoff for
all
SPE
for instance, player 3 has a multivalued
The
SPE
special payoffs in
payoffs for
G
T. Clearly in the latter case,
G
game has
our condition,
The
of our condition
Indeed,
fail.
two Nash
we focus on the
folk
theorem
and of
NEU
2.
4 and
if
T>
T, while players 2 and
all
8, respectively.
Nash
Thus, for generic games satisfying
payoffs.
stage game, and precisely define our iterative
We
established in section 3.1.
is
which distinct
in
one player has distinct Nash payoffs,
equilibria.
players enjoy distinct
all
In section 2,
procedure.
at least
payoff set for
we
with no discounting,
might suggest the nongeneric way
Nash payoff requirement can
then the
T>
behaviour cannot be
his
if
cannot possibly hope for anything more. In the above game
only receive distinct
either point- valued,
is
the behaviour of a player can be leveraged, then
his payoffs are multivalued for large
1
any given horizon
also rather simple. For
prove the necessity
in section 3.2.
THE STAGE GAME
2.1 Basic Definitions
Consider a
is
player
A—
\
write
finite
set of actions,
i's finite
E. Let
7Tj(/i)
Mi be
—
best response for
i
payoffs
is
F'
:
/j,
i's
A =
G=
II" A{
mixed
.
(A,-,
Player
strategies
i's
and
l)-profile of
mixed
strategies
= {w G F
1,
.
.
.
,
n),
where A{
utility function is
let
\i
M = II"M,.
=
(/ij,
.
which minimax player
when being minimaxed. Normalize
€ M},
=
i
7r,-;
expected payoff under the mixed strategy
for i's
co{n(fi)
and
the set of player
Let ml,- be an (n
F=
normal form n-player game
7r,(m')
=
.
.
i,
for all
7r,
:
Simply
/lx„)
6 M.
and m\ a
i.
Define
so that the set of feasible and strictly individually rational
:
m,
>
0, for all i}
^
0, to sidestep
trivialities.
.
Denote the
=
X'
set of players
,i
m}
av
=
{ii,i2,
when
2.2
(a, 1 ,a,- 2 ,.
.
,a im
—
€
)
MM x M
•
•
G
The game
Mx
=
>,
X \ X'
obtained from
utilities (or satisfies
NEU)
if
game
G
no two players'
utility functions are equivalent, i.e. for all
i
and
j, ni(-)
7Tj(-).
has recursively distinct Nash payoffs
for
if
some h
>
1,
there
an increasing sequence of h nonempty subsets of players from I, namely
{0=Zo£Z £I2 £---QIh
1
Mj
and actions a%g E
and
£
bj
Mj
for each
g
=
=Z},
1, 2,
Nash payoff vectors y{a Ig ofG(ai g ) and y(bIg o(G(bIg
)
lg \Ig -i,
)
i
)
(1)
..
.
,h,
with some pair of
different exactly for players
i.e.
y(ai.)i
for all
im
Games
has non- equivalent
G
The game
2.
in
M
x
•
for players
not a positive affine transformation of
exists
x
l2
m)-player
Properties of Stage
von Neumann Morgenstern
is
.
actions
the actions of players X' are fixed to aji
Two
1.
{l,2,...n}. For a given strict subset of players
QX and corresponding choice of (possibly mixed)
G(aj') be the induced (n
let
asJ =
G Xg \Z
ff
-i.
Even
if
Ir Q
*
y(bx,)i
(2)
I, call such a sequence
{Ig } a Nash decomposition
of G.
Observe that
if all
players have distinct
Nash
payoffs as
Krishna (1985), then our condition holds. The converse
is
assumed
in
Benoit and
not true, as illustrated
earlier.
It
later proves fruitful to consider the case
distinct
Nash
Nash equilibrium
Nonetheless,
it is
payoffs y(aj
G
does not have recursively
Nash decomposition and correspond-
payoffs. In that case, for every
ing choice of
where
and y(bj
)
conceivable that the set of
all
),
J/,
£ X
always obtains.
such Nash decompositions could
include every player. In fact this cannot occur.
Lemma
1
There
sively distinct
Nash
is
a well-defined
payoffs.
maximal
set of players
J
C X who
have recur-
.
The proof
is
in the appendix.
FINITELY-REPEATED GAMES
3.
NEU
3.1 Sufficiency of
We
and Recursively Distinct Nash Payoffs
games with
will analyze finitely-repeated
perfect monitoring, allowing each
player to condition his current actions on the past actions of
simplicity, public randomization
is
players. Also, for
allowed: In every period players publicly observe
the realization of an exogenous continuous
its
all
random
variable and can condition on
outcome.
Denote by
a,
=
{an,
.
expected payoff in period
function in G(5, T)
The
set of
is
subgame
q,t) a behavior strategy for player
,
.
t
given the strategy profile a.
the expected discounted
Beyond the replacement
differs
and by
Each player
sum of his payoffs:
perfect equilibrium payoffs
i
jz$r
^
7r i«(
a )-
V(5, T).
is
from Theorem 3.7 of Benoit and Krishna (1985) on a few
it is
So
objective
_1
Nash payoff requirement, Theorem
of the distinct
admits payoff discounting. As such,
i's
7r,t(a) his
fronts.
1
First, it
a uniform folk theorem, meaning that the
discount factor and horizon length can vary independently over the relevant range.
Second,
NEU
replaces full-dimensionality, as already noted. Third, unlike the (more
intuitive) use of long deterministic cycles in Benoit
rely
and Krishna (1985), we simply
upon correlated outcomes. 2
Theorem
1
(The Folk Theorem)
and has recursively
distinct
VueF* and Ve >
0,
3T
Nash payoffs. Then for
<
G satisfies NEU
Suppose that the stage game
oo and S
<
1
the finitely -repeated
so that
T
>T
and
8
<E
[S
game
,
1]
G(8, T),
=>
3v G
V{6,T) with \\v-u\\ <e.
Proof:
We
suppose
modify the argument
first
that mixed strategies are observable, and discuss
how
to
later.
Fix a Nash decomposition
(1) of players for
which inequality
(2) obtains,
and
define
c
9
2
=
mn
CT xV
\\y^)i
-
y(6r,)«ll
>
°
It is also possible to take advantage of the correlating device to produce an exact, rather than
an approximate, folk theorem. We omit such a laborious exercise.
for
=
g
means
1, 2,
.
.
>
Let the p
h.
,
.
be the largest payoff range for any player in G. By
of public randomizations, let y9 (resp. z 9 ) be a payoff vector according equal
weight to the preferred (resp.
and y(bx
Xg
Xg -\
\
for players in
)
is
Xg \ Xg -\.
Then, without payoff discounting, any player
willing to suffer through k periods of any action profile
is strictly
alternative
Nash payoff vectors amongst y(aj
less preferred)
that at least fg {k)
=
+1
[knp/cg \
if
periods of y9 are switched to z 9
-kp + fg (k)yf>fg (k)zf.
For any
m>
0,
-
s g (m)
for
g
g
=
=
h
1, 2,
.
—
.
.
— 2,
l,h
,
.
.
,
.
+
fg (m
=
where each y9 phase
vectors x ,x
.
,
rationality), x\
and
t
s
(m)
.
.
<
,x
n
h
periods,
have shown that
et al. (1993)
2
+ r)
such that for
x\ (payoff
These inequalities are
all
and u E F*
NEU
i
^
i.e.
h s h (m))
-\
...,u;y ,..., y ;...;y\...,y
lasts s g (q
the
(3)
=
S\(m)
The T-period equilibrium outcome sequence
h.
h
l
0,
in
fh(m) and
s g+ i(m)
Define to(m)
1.
it,
Abreu
=
recursively define Sh(m)
,
)
is
+
•
•
+
s g (m) for
is
l
played
T — t^q + r)
times.
implies the existence of feasible payoff
we have
j,
asymmetry), and
x)
<
u,-
3>
x'
(strict individual
(target payoff domination).
later essential.
We now explicitly describe the players' strategies which support this equilibrium. 3
For ease of exposition, late deviations are those occurring during the final q + r +
th(q + r) periods of the repeated game; all others are called early deviations.
1.
if
2.
some player
For g
[If
3.
3
to
T — th(q + r) —
Play u until period
=
in
Xg
in
Xg
[If
deviates late, start
>
h,h-l,...,l: Play y9
some player
1.
>
[If
in periods
player j
^
i
deviates early, start 3;
T-t g (q + r), ... ,T-t g _i{q + r) - 1.
g'
<
g, start 5.]
deviates, start 4.]
Then
set j <—
i.
i and j denote arbitrary players.
Also, for clarity, we use the simple notation j <— i
"assign j the value i." Also, steps always follow sequentially, unless otherwise indicated.
Below,
mean
i
5.]
deviates late, where
Play m' for q periods.
any player
Bracketed remarks refer to off-path play,
i.e.
following deviations.
Play xi for
4.
r periods. [If
any player
Then
in Xg> deviates late, start 5.]
'
<
g"
The
g',
return to step
T — t g >-i(q + r) — 1.
Play z 9 until period
5.
deviates early, restart 3; if some player
i
and
set g' <— g"
restart 5.]
[If some
Then go
verification that these strategies constitute a
suitable choices of
Remark:
If
the
q, r,
and To
5
found
is
minimax strategy m'
in the
strategies, then unobservable deviations
of m'j
must somehow be deterred.
show how, 4
(1993)
al.
player in
lg »
deviates,
where
to step 2.
subgame
perfect equilibrium for
appendix.
<0>
some punishers playing completely
entails
mixed
2.
by
all
players j within the support
Fudenberg and Maskin (1986) and Abreu
et
an infinite-horizon context, to make players indifferent
in
across their pure strategies: After playing
probabilistically replaced by one of
n
—
m
!
play proceeds to step
,
but with
x'
possible substitute reward phases, with
1
make each minimaxing
probabilities chosen as to
4,
player j
^
indifferent over all
i
actions in the support of m'-. This technique applies here equally well.
In our finitely repeated context,
we must
from any mixed strategies required
to maintain the incentives of the
recursively distinct
Nash
payoff vector y 9
=
than y9 and yet
still
,
for playing
may
a
9
y /2
z 9 /2 that
NEU.
strictly
is
strictly preferred to z 9
5
.
First replace y 9 with the
Since
g'
>
g,
Second, to reward player j
the support of a%
fashion analogous to Abreu et
that will maintain
3.2 Necessity of
Abreu
all
incentives.
NEU
,
6
Mj
et al. (1993)
first
where
Nash outcome among
>
\
I9'_i.
Finally,
it is
affect the
possible to define in a
(1993) a proper probabilistic transition function
We
omit the tedious algebraic
verification.
have already established the (near) necessity of
That proof
NEU
for the
easily extends to our finite-horizon case.
authors discovered the idea, while the latter modified their argument to handle stage
games that do not
The
,
,
and Recursively Distinct Nash Payoffs
infinite-horizon folk theorem.
The
al.
Xg \ Ig _i
G Tg \Ig -\
g' > g, we
any such substitution cannot possibly
I5
new
worse for every player j E
probabilistically replace y 9 in step 2 with player fs best
incentives of the original deviant in
5
For this we provide another way
minimaxing players that takes advantage of the
less preferred action in
y{a Ig ) and y{bxg )-
4
in step 2.
payoffs and not of
+
also deter late unobserved deviations
but only the weaker NEU assumption.
be modified to fg (k) = [2knp/cg \ + 1.
satisfy full-dimensionality,
definition of fg {k) will have to
For consider the infinitely-repeated game with stage game
G(<5,
T).
Clearly,
u
if
G as a sequence of games
a subgame perfect discounted average payoff for G(8,T),
is
then by concatenation of the equilibria,
it is
also
one for the
infinitely repeated
game
with stage game G. This establishes
Theorem
2 (Necessity of
NEU)
Suppose that no two players can be simulta-
neously held to their lowest feasible payoff in G.
Then
NEU
necessary for the
is
conclusion of the folk theorem.
If
a
game G, however, does not have
recursively distinct
Nash
payoffs, the conse-
quences are rather stark. The following result has the flavour of the "zero-one" laws
of probability theory. Namely, in a T-period repeated
subgame
,
F*
of G, or
some
players receive a unique equilibrium payoff.
no middle ground.
is
Theorem
game
G(5, T) either the set of
perfect equilibrium payoffs of all players tends to the strictly individually
rational payoff set
There
game
3 (Necessity of Recursively Distinct
perfect equilibrium of G(5,T), for
Nash
In any sub-
Payoffs)
any T, players in
X
J
\
receive a payoff
Nash equilibrium payoff of G.
equal to their unique
payoff of G(S,T)
X\J has a unique subgame perfect equilibrium
equal to his Nash payoff in G, for T = 1,2,
,T (Since J\ J ^ 0,
we know
be true for To
Assume
Proof:
in
1\
J
that every player in
.
this to
=
1.)
Then
.
.
.
in the first period of G(5, Tq
must play a Nash equilibrium of G(aj),
for
any aj E
Mj,
As a simple
set
V(5,T) as
grows.
to say, but otherwise,
Corollary
Tg -i =
i
g
for
proji g (V(5,T))
g
is
=
induction,
<)
we can
If
many
offer
some
possible
insight into the behaviour of the payoff
Nash decompositions
we can sometimes make
1,2,
.
.
.
exist,
,
h, then there exist
single valued for all periods
we have
little
rather sharp predictions.
unique Nash decomposition of G, and
7/ there exist a
all
players
T.
all
corollary,
T
1),
because they
By
have a unique (subgame perfect) continuation payoff by assumption.
the result obtains for
+
times T\
T < Tg
,
< T2 <
•
•
it
•
<
satisfies
Ig
\
Th such that
and otherwise multi-valued.
CONCLUDING REMARKS
4.
This note has illustrated an easily- verifiable, and yet necessary and
sufficient
replacement of the distinct Nash payoff requirement of Benoit and Krishna (1985).
Although
its
theoretical advantage
is
nongeneric,
theorems only imply that players' behaviour
is
hopefully underscores that folk
it
iteratively leveraged near the
end of
the repeated game. This conceptual contribution might have generic application in
other circumstances.
The
analysis has also offered insight into the behaviour of
fulfilling
our condition:
are singletons.
When
not
all
effective
minimax payoff
to their
level.
Proof of
If for
Lemma
(1993),
who has
pointed
might not be so dramatic:
minimax payoff level, but
their (newly defined)
6
APPENDIX
5.
5.1
Wen
NEU
out that the consequences of the partial failure of
down not
partially
players can be leveraged, their payoff sets
This contrasts with recent work by
Players can be held
games only
1
=
some Nash decomposition we have Xh
X, then unambiguously
J
=
X,
and we are done.
Otherwise,
let
£,(G) be the set of
all
Nash
payoffs for player
Consider the alternative unique nested sequence of
{0 = Jo £ Jx £
J2 £
•
•
£
h'
Jh>
nonempty
If
i
£
no such
T\Jg for whom {£,(G(axJ),aj G MXg
players Jg+ \ Jg exist, set h' = g.
Note that
9
it
may
Nash decompositions {Xg }. For
This equals
i.
{Xg }
satisfying
sets
Jg
,
the set
Jg+i \ Jg
is all
does not consist of a singleton.
not be possible to find only two Nash equilibria providing
i
not coincide with any Nash decomposition {Tg }.
utilities to
game G.
i
distinct payoffs simultaneously for all players
for all
)
in the
= J £ X},
given by the following (well-defined) recursion: Given
players
i
mm a _
.
/(
Jg +\ \ Jg
Still,
clear that
.
That
we can show
X/,
C
maa; j€ /(,)mai , j.7ri(aj,a_ J ), where I(i) axe
(
)
it is
E
Jh
all
,.
is,
{Jg }
that Jh<
need
=
Xh
Moreover, for any
players with equivalent
.
such nested sequence
Xg
For
.
{£i(G{ax,.)),ai,, e
Mr
Mf}
€
{Si{G(ax')),ai'
if
Theorem
1
proceed recursively as we ensure subgame perfection.
deterrents are strict by
some
discounting. Given continuity of discounted
sums
only check that
any
for
all
G
level of discounting 8
[5
By
yield a higher payoff.
for
1],
and
select q
In light of x\
r.
+
l
where
^
2, for
any
>
0, let
>/?«'
It will
1.
q satisfy
strict
follow that for this
7
+1
(4)
the best payoff vector for player
is
simultaneously renders steps 3 and 4 a
large
without
they will thus remain
in 6,
some 5 <
1,
the "unimprovability" criterion, the profile will be a
w\
and
positive margin, say
we
perfect equilibrium.
We now
for all j,
,
First note that
no history consisting of isolated deviations does a one-shot deviation
profile, after
subgame
some
does not consist of a singleton, then neither does
5.2 Verification of Strategies for
We
are eventually included in
>
£ 1"
I'
>) if
Jg
players in any
(1), all
<
G. Since x\
in
i
u,-
too, (4)
deviations from steps
strict deterrent to
Next, step 4 deters deviations by the punishers from step 3
r.
if
r
1
is
enough that
+
qwj
for all j
^
Next,
This
i.
we
for all
i
+ r + t h (q +
=
£ lg and g
,
if
To
j3]
+ rxij +
1
-
(q
-
l)iij
+
1
> x3j.
verify that step 5 deters all late deviations. Indeed, this
Given q and
context
>
possible because x
is
consequence of inequality
[q
rx)
'The inequality
-
t
(4)
q
g (q
+
r)]{-p)
+
s g (q
+
r)y?
>
s g {q
+ r)z° +
1
l,2,...,h.
+
t^(p
+ q)-
For
above program
T>
is
feasible for
To, each deterrent
— in particular, why the left-hand side
random
an immediate
which upon substitution yields
r as defined above, the
> p+
that obeying the
r)
(3),
is
is
not
any repeated
remains
(q+
l)x\
strict for all
— reflects the fact
correlating device generating x' might sometimes require
worst possible outcome.
10
i
to play his
S
G
for
[Sq, 1],
some
S
<
1.
Finally, for
resulting average payoff vector
is
T <
v G F*,
oo and 5o
we have
||u
—
<
v\\
1
<
large enough,
e,
since
if
p and
the
q are
fixed.
References
Abreu, Dilip and Prajit Dutta,
Games:
A New
Condition," 1991. University of Rochester mimeo.
— — and Lones
,
"Folk Theorems for Discounted Repeated
,
Condition," 1993.
Smith, "The Folk Theorem
for
Repeated Games:
A NEU
MIT mimeo.
Benoit, Jean-Pierre and Vijay Krishna, "Finitely Repeated Games,"
Econometrica, 1985, 53, 905-922.
Fudenberg, Drew and Eric Maskin, "The Folk Theorem
in
Repeated Games
with Discounting or Incomplete Information," Econometrica, 1986,
Krishna, Vijay, "The Folk Theorems
for
54,
533-554.
Repeated Games," 1989. mimeo 89-003,
Harvard Business School.
Smith, Lones, "Folk Theorems: Two-Dimensionality
1990/92.
is
(Almost) Enough,"
MIT mimeo.
Wen, Quan,
"Characterization of Perfect Equilibria in Repeated Games," 1993.
University of Windsor mimeo.
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