Copyright (C) 2011 David K. Levine
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Period Length in LR-SR Games
payoff matrix
Player 2
L
Player 1 +1 - x ,0
R
1,1
-1 - x ,0 2,-1
player 2 plays L in every Nash equilibrium
player 1’s static Nash equilibrium payoff is - x
also the minmax payoff for player 1
1
end of each stage game, public signal z Î ¡ depends only on action
taken by player 1; probability distribution of the public signal is F (z | a1 )
F either differentiable and strictly increasing, or corresponds to a
discrete random variable
f (z | a1 ) denote the density function
monotone likelihood ratio condition f (z | a1 = + 1) / f (z | a1 = - 1) nondecreasing in z
player 2’s action is publicly observed
availability of a public randomization device
2
t is length of the period
t player 1 a long-run player with discount factor d = 1 - r t
faces infinite series of short-run opponents
most favorable perfect public equilibrium for LR characterized by the
largest value v that satisfies the incentive constraints
v = (1 - d)1 + dò w(z ) f (z | a1 = + 1)dz
v = (1 - d)2 + dò w(z ) f (z | a1 = - 1)dz
v ³ w (z ) ³ - x
or v = 0 if no solution exists
second incentive constraint must hold with equality, since otherwise it
would be possible to increase the punishment payoff w while
maintaining incentive compatibility
3
from the monotone likelihood ratio condition, iif there is a solution, w(z )
it must have cut-point form
ìï v z ³ z *
.
w(z ) = ïí
ïï w * z < z *
î
4
Proof:
Any w that is part of a solution to maximizing v subject to the incentive
constraints must be a solution, for some constant c to
max ò w(z ) f (z | a1 = + 1)dz subject to max ò w(z ) f (z | a1 = - 1)dz £ c
Let l be the Lagrange multiplier for this constraint; the first order
condition is f (z | a1 = + 1) = l f (z | a1 = - 1) , or
f (z | a1 = + 1)
= l
f (z | a1 = - 1)
The monotone likelihood ratio condition implies any solution to this has
the cutoff form.

5
Define
p=
z*
ò- ¥
f (z | a1 = + 1)dz , q =
z*
ò- ¥
f (z | a1 = - 1)dz .
enables us to characterize most favorable perfect public equilibrium for
LR as largest value v that satisfies incentive constraints
v = (1 - d)1 + d(v - p(v - w ))
v = (1 - d)2 + d(v - q(v - w ))
v ³ w³ -x
or v = 0 if no solution exists.
linear programming problem characterizing the most favorable
equilibrium has solution if and only if
(*) r £
1
t + ((1 + x ) [(q - p) / t ] - [ p / t ])-
1
in which case the solution is given by
6
(**)
v * = 1 - p / (q - p) .
7
The Poisson Case [Abreu, Milgrom and Pearce, 1991]
signal generated by underlying Poisson process in continuous time
Poisson arrival rate l p if the action taken by LR is +1 and l q > l
action taken by LR is –1
p
if the
Poisson signal is being “bad news”
the number of signals received during the previous interval of length t .
cutoff point is how many “bad signals” received before punishment
v - w is triggered
period length t is short, can ignore possibility of more than one signal,
so the cutoff is necessarily a single signal
approximation: the probability triggering punishment p = l p t , q = l q t .
8
in (**)
v* =
lp
lq - l
.
p
in (*)
r £ (1 + x )(l q - l p ) - l
p.
independent of the period length
if satisfied, best equilibrium payoff is v * , independent of period length
and worst possible punishment x .
9
The Diffusion Case [Faingold and Sanikov [2005]
signaled generated by underlying diffusion process in continuous time
drift in process given by the long-run player’s action (a1 = +1 or –1) and
the instantaneous variance of the diffusion s 2
random variable z is increment to the diffusion process during the
previous interval of length t
if the long-run player’s action is a1 signal is normally distributed with
mean a1t and variance s 2 t .
10
allow the variance of the signal z to be given by s 2 t 2a where a < 1
diffusion case corresponds to a = 1/ 2
if F is the standard normal cumulative distribution
æz * - t
p = Fç
ç
è st a
æz * + t
q = Fç
ç
è st a
ö
÷
÷
ø
ö
÷
÷
ø
11
fix x and consider best equilibrium for player 1 as t ® 0
Proposition 1: For any a < 1 there exists t > 0 such that for
0 < t < t (*) has no solution.
12
Proof:
z * ( t ) cutoff when period length is t
work with normalized cutoff
z( t ) =
t - z * (t )
st a
So:
p = F (- z( t ) )
æ2t 1q = Fç
ç
ç
è s
a
ö.
÷
- z( t ) ÷
÷
ø
if not (*) not violate then (q - p) / t and p / t both remain bounded
away from zero as t ® 0
show that if (q - p) / t remains bounded away from zero, then
necessarily p / t ® ¥
13
compute (q - p) / t using the mean value theorem to observe that for
each t there is a number f (t ) , 0 £ f ( t ) £ 1 , such that
q- p
=
t
=
æ2t 1Fç
ç
ç
è s
1
st a
( )
ö
- z( t ) ÷
- F (- z( t ))
÷
÷
ø
t
1- a ö
æ
2
t
÷
f ç
- z( t ) + f ( t )
÷
ç
÷
ç
s ø
è
a
Let c(t ) = (q - p) / t , we can invert this relationship to find
14
1æ
2
t
f ç
- z( t ) + f ( t )
ç
ç
s
è
(***)
a
ö
÷
= c( t )st a
÷
÷
ø
11
1æ
2
t
exp(- ç
- z( t ) + f ( t )
ç
2ç
s
2p
è
1æ
2
t
ç
- z( t ) + f ( t )
ç
ç
s
è
z( t ) = -
a ö2
÷
) = c( t )st a
÷
÷
ø
a ö2
÷
= - 2 log( 2pc( t )s ) - 2a log( t )
÷
÷
ø
2t 1- 2 log( 2pc( t )s ) - 2a log( t ) + f ( t )
s
a
15
want to show p / t ® ¥
c( t ) is bounded away from zero so
z(t ) £
- log b - 2a log t + a t 1-
a
and
p / t ³ F(-
- log b - 2a log t - at 1- a ) / t
by mean value theorem for some t ' Î [0, t ]
16
F (-
- log b - 2a log t - a t 1- a ) / t
æ
ö
a
- a÷
1- a
= ç
a
(1
a
)(
t
')
f
(
log
b
2
a
log
t
'
a
(
t
')
)
÷
ç
÷
ç
èt ' - log b - 2a log t ')
ø
ö
1 æ
a
- a÷
ç
=
a
(1
a
)(
t
')
÷
÷
ç
2p ç
èt ' - log b - 2a log t ')
ø
´ exp(- 1/ 2) (- log b - 2a log t '+ 2a( t ')1- a - log b - 2a log t ' + a 2( t ')2- 2a )
ö
b1/ 2 æ
a
- a÷
ç
=
a
(1
a
)(
t
')
÷
÷
ç
2p ç
èt ' - log b - 2a log t ')
ø
´ ( t ')a exp(- a( t ')1- a - log b - 2a log t ') exp(- 1/ 2) (a 2( t ')2- 2a )
a- 1
ö
ab1/ 2 æ
(
t
')
- 2a log t '
÷
ç
®
exp(
a
)
÷
ç
a- 1
÷
2p ç
( t ')
è - 2a log t ') ø
17
apply L’Hopital’s rule to show that
t a- 1
t a- 2
® (1 - a )
= 1 - a )t a - 1 ® ¥
1/ t
- log t
.

18
recall that v * = 1 - p / (q - p)
consider for fixed t taking very small cutoff z * ® - ¥
as in Mirlees [????] causes likelihood ratio q / p ® ¥ so
p / (q - p) ® 0 ; that is v * ® 1, the pure commitment value, and the
best PPE payoff for the most the long-run player can get when there is
no noise in the signal
this solution has p, q ® 0 , which means that for fixed x and t and z *
sufficiently negative, (*) must be violated
however, for any choice of z *, r , t , there is always an x sufficiently
large that (*) holds
the diffusion case the worst punishment determines the best
equilibrium
going far enough into the tail of the normal, arbitrarily reliable
information is available and can be used to create incentives, provided
sufficiently harsh punishment is available
19