Strong Belief and Weak Assumption
Chih-Chun Yang
Institute of Economics, Academia Sinica, Taipei 115, Taiwan
April 2015
Abstract
Battigalli and Siniscalchi’s [Journal of Economic Theory 106, 356-391
(2002)] notion of “strong belief ”in conditional probability systems and Yang’s
[Journal of Economic Theory, forthcoming] notion of “weak assumption” in
lexicographic probability systems are uni…ed by the same requirements on
preferences. Our analysis, hence, reconciles the tension between Battigalli
and Siniscalchi’s characteriztion of extensive-form rationalizability and Brandenburger, Friedenberg, and Keisler’s [Econometrica 76, 307-352 (2008)] impossibility result.
Keywords: Strong Belief, weak assumption, rationality.
1
Introduction
Brandenburger, Friedenberg and Keisler (2008) (henceforth, BFK) provide an epistemic
analysis for iterative admissibility (henceforth, IA) by using the set of lexicographic
probability systems (LPS’s). They de…ne the notion of assumption, the counterpart of
knowledge, in the model constructed by LPS’s. If a player is not indi¤erent in every outcome, RCAR is empty in a complete and continuous type structure. This impossibility
result implies that RCAR is not an epistemic foundations for IA. Yang (2015) investigates IA by modifying the notion of assumption to obtain a characterization of IA. BFK
use Nontriviality and Strict Determination to characterize the notion of assumption.
By keeping Strict Determination and introducing Nontriviality , Yang (2015) de…nes
Fax:+886-2-2785 3946. E-mail: [email protected].
1
a new notion, weak assumption and shows that rationality and common weak assumption of rationality (RCWAR) is nonempty in a complete and continuous type structure.
Moreover, RCWAR and IA are outcome equivalent.
On the other hand, Battigalli and Siniscalchi (2002) (henceforth, BS) de…ne the
notion of “strong belief ” for conditional probability systems (CPS’s) and use common
strong belief of rationality to characterize extensive-form rationalizability, which is outcome equivalent to IA in generic trees.
It is shown in the literature that strong belief in a CPS and assumption in an LPS is
closely related: BFK (2008), Halpern (2010) and Tsakas (2014). The literature focus on
the connections based on probability/measure theory. However, BS’s (2002) possibility
result and BFK’s (2008) result implies that strong belief and assumption are, though
closely related, di¤erent.
In this paper, we show that strong belief and weak assumption are essentially identical: We uni…ed them by characterizing them by the same properties on preferences.
We …rst extend the notion of weak assumption for non-full support LPS’s by introducing Nontriviality . Then use this extended notion of weak assumption to characterize
strong belief.
Our preference based analysis provides additional insights into the relationships
among di¤erent notions of belief/knowledge in the literature: Savage’s notion of knowledge, BS’s (2002) strong belief, BFK’s (2008) assumption and Yang’s (2015) weak assumption. Once an LPS and a CPS represent the same preference, an event is strongly
believed (under the LPS) if and only if it is weakly assumed (under the CPS). Our
analysis reconcile the tension between BS’(2002) characterization theorem of extensive
form rationalizability and BFK’s (2008) impossibility theorem.
The rest of this paper is organized as follows. Section 2 introduces notations in a
normal-form game and the notion of weak assumption. Section 3 introduces notations in
an extensive-form game and the notion of strong belief. Section 4 proves the equivalence
between these two notions.
2
Normal-form Game, LPS and Weak Assumption.
Let A be the set of measurable functions from a state space
to [0; 1]. Let % be a
transitive and re‡exive binary relation, satisfying Independence axiom, on A. For each
x; y 2 A and event E
, de…ne
xE ; y
nE
x (!) if ! 2 E
.
y (!) if ! 2
=E
(!) =
De…ne x %E y if there is z 2 A such that xE ; z nE % yE ; z nE .
An event E
is called %-null if E is Savage-null under %: For each x; y 2 A,
x E y.
We …rst introduce the normal form game.
2
Let M ( ) be the space of Borel probability measures on a separable metric space
. Let N ( ) be the set of all …nite sequences of Borel probability measures on .
Write supp for the support of 2 M ( ). Say E
is an event if E is Borel.
Say = 0 ; :::; n 1 2 N ( ) is a lexicographic probability system (LPS) if satis…es
mutual singularity: For each i = 0; :::; n 1, there are events Ui with i (Ui ) = 1 and
n 1
2 N ( ) is a full-support sequence if = [i=0
supp i . We
i (Uj ) = 0 for i 6= j. Say
+
write L ( ) for the set of LPS’s and L ( ) for the set of full-support LPS’s.
Let the notion L represent the lexicographic order. That is, for each (a0 ; :::; an 1 ),
(b0 ; :::; bn 1 ) 2 Rn , (a0 ; :::; an 1 ) L (b0 ; :::; bn 1 ) if and only if whenever ai < bi , there is
j < i such that aj > bj . Write margS b i for the marginal on S b of the measure i . A
strategy sa is optimal under = 0 ; :::; n 1 2 N S b T b if for each ra 2 S a ,
(
a
(sa ; margS b
n 1
i ))i=0
L
(
a
(ra ; margS b
n 1
i ))i=0
:
Note that represents the space of states of the world. Let a
S b T a , where
a
each element of T represent player a’s beliefs over player b’s behavior, and all possible
higher order beliefs.
For each
=
2 L ( ), we are now ready to de…ne the preference
0 ; :::; n 1
represented by an LPS .
De…nition 1. For each LPS = 0 ; :::; n 1 2 L ( ), say a preference % on A
satis…es LPS basis as follows: For each x; y 2 A, x % y i¤
Z
n 1
x (!) d
i
L
(!)
i=0
Z
n 1
y (!) d
i
(!)
:
i=0
Let %L be the set of LPS-based preference on A.
We consider a domain of preferences represented by LPS’s. First, the de…nition of
weak assumption under % when 2 L+ ( ) is reviewed (De…nition 2), and then an
extension for the case 2 L ( ) nL+ ( ) is de…ned (De…nition 4).
De…nition 2. Let
assumed under % if
2 L+ ( ). Yang (2015) de…nes that an event E
is weakly
(i) Nontriviality : E is nonempty and for each s 2 projS E, there are acts
x; y 2 A such that x E\[s] y;
(ii) Strict Determination: For each act x; y 2 A, x
E
y implies x
y.
Nontriviality says that for each signi…cant event [s], if E \ [s] 6= ;, then E \ [s] is
not % -null.
3
We now extend the notion of weak assumption under an LPS even when doesn’t
satisfy full support, i.e., 2 L ( ) nL+ ( ). When 2 L+ ( ), for each s 2 S, [s] is
not % -null. When is non-full support LPS, there is cylinder [s] that is % -null. We
would like to exclude these null cylinder in checking Nontriviality . Formally, we de…ne
Nontriviality under % as follows.
De…nition 3. Let % be a transitive and re‡exive binary relation, satisfying Independence axiom, on A.
An event E
satis…es Nontriviality under % if E is nonempty and for each
s 2 S, if [s] is not %-null and E \ [s] 6= ;, then neither is E \ [s], i.e., there are acts
x; y 2 A such that x E\[s] y.
Our notion of Nontriviality , de…ned in the same spirit of Nontriviality , requires
this additional quali…er for [s]. We need it because in BS (2002), rationality doesn’t
require full support condition on beliefs.
is weakly
De…nition 4. For each LPS = 0 ; :::; n 1 2 L ( ), an event E
assumed under % if E
satis…es Nontriviality and Strict Determination under
% .
For each LPS = 0 ; :::; n 1 2 L ( ), de…ne margS
margS 0 ; :::; margS n 1 .
Yang (2015) characterizes weak assumption under % when 2 L+ ( ).
Consider a full-support LPS 2 L+ ( ). Yang shows that an event E
is weakly
assumed under % if and only if there is 0 j n 1 such that
(i) for each i
j,
(ii) projS E = [i
i
j
(E) = 1 and for each i > j,
supp margS
i
(E) = 0;
i.
If is a full-support LPS and an event E
is weakly assumed under % , then
we say E
is weakly assumed under . We also have a similar characterization for
2 L ( ) as follows.
Lemma 1. For each LPS = 0 ; :::; n 1 2 L ( ), an event E
assumed under % at level j if and only if there is 0 j n 1 such that
(i) for each i
j,
(ii) projS E \ [i
i
n 1
(E) = 1 and for each i > j,
supp margS
i
= [i
j
i
is weakly
(E) = 0;
supp margS
i.
When conditions (i), (ii) in Lemma 1 are satis…ed, we say E
under at level j.
is weakly assumed
Proof of Lemma 1. For each real c, let c be the act assigning c to each state in .
4
If part. Consider 2 L ( ) and an event E
satisfying conditions (i) and (ii).
(Nontriviality ) Let s 2 S such that [s] is not % -null and E \ [s] 6= ;. Therefore,
there is i n 1 such that i ([s]) > 0 and s 2 supp margS .
By (ii) s 2 projS E \ [i n 1 supp margS i = [i j supp margS i , there is i0 j such
that i0 ([s]) > 0. By (i), i0 (E) = 1 and hence, i0 (E \ [s]) = i0 ([s]) > 0. Therefore,
1 E\[s] 0.
(Strict Determination.) The proof is identical to that for Lemma 1 in Yang (2015)
and hence, omitted. Consider acts x; y 2 A such that x E y. Let x0
xE ; 0 nE and
0
y
xE ; 0 nE . By (i),
R 0
R
R
n 1
x (!) d i (!) i=0 =
x (!) d i (!) ; :::; x (!) d j (!) ; 0; :::; 0 and
R
R
R
x (!) d i (!) ; :::; x (!) d j (!) ; 0; :::; 0 .
R
R
j
j
Since x E y, x0
y 0 . Therefore,
x (!) d i (!) i=0 >L
y (!) d i (!) i=0 and
R
R
n 1
n 1
hence,
x (!) d i (!) i=0 >L
y (!) d i (!) i=0 , i.e., x
y.
Only If part. Let event E
be weakly assumed under % and 2 L ( ).
Suppose, in negation, that (i) is violated: Either (a) there are i; j such that i < j,
i (E) = 0 and j (E) = 1 or (b) there is j such that j (E) 2 (0; 1). If case (a), let
x0 (!) d
i
(!)
n 1
i=0
=
1
(E)
j
. By mutual singularity, there are events
c 1. Otherwise, let c 2 (0; 1) \ 0; (E)
j
U0 ; :::; Un 1 such that for each i = 0; :::; n 1, i (Ui ) = 1 and for each j 6= i, i (Uj ) = 0.
Now, let x
cUj \E 0 n(Uj \E) and y
0E 1 nE . However, we have x E y and
y
x, contradicting Strict Determination.
Suppose that there is s 2 (projS E \ [i n 1 supp margS i ) n [i j supp margS i : s 2
projS E and there is i > j such that i ([s]) > 0 and for each k j, k ([s]) = 0. Since
i ([s]) > 0, 1
[s] 0 and hence, [s] is not % -null. By hypothesis, s 2 projS E and for
each k j, k ([s]) = k (E \ [s]) = 0. For each k > j, by (i), k (E \ [s]) = k (E) = 0.
Therefore, [s] is not % -null and E \ [s] 6= ;, but for each k 0, k (E \ [s]) = 0. That
is, E \ [s] is % -null, contradicting Nontriviality .
Remark. When 2 L+ ( ), Nontriviality
tion 4 is an extension of De…nition 2.
3
implies Nontriviality . That is, De…ni-
Extensive-form Game, CPS and Strong Belief
We consider a two-player …nite extensive-form game as follows: Let fa; bg be the set of
players. Let H be a …nite collection of nonterminal histories, sequences of consecutive
action pro…les. To avoid the confusion between the empty history in H and empty set
(;), we denote the former by ?.
The set of feasible actions for a player i may depend on previous history, and it is
denoted Ai (h). Let Z be a …nite collection of terminal histories. Let ui : Z ! R be the
payo¤ function of player i.
5
We say player i is active at h 2 H if jAi (h)j 2. If at h 2 H at least two players
are active, then there are simultaneous moves at h. If there is no simultaneous move at
each h 2 H, we say that the game has perfect information. For each h 2 H [ Z, let
S (h) denote the set of strategy pro…les inducing h. The projection of S (h) on S a (S b )
is S a (h) (S b (h)).
A strategy of player a is a function sa : H ! [h2H Aa (h) such that sa (h) 2 Aa (h)
for each h 2 H. Let S a denote the set of strategies for player a. De…ne a strategic-form
payo¤ function a : S a S b ! R in the usual way: For all z 2 Z, if sa ; sb 2 S (z), then
let a sa ; sb = ua (z). For every strategy sa 2 S a , let H (sa ) fh 2 H : sa 2 S a (h)g,
i.e., the set of histories consistent with sa . De…ne S b , b , S b (h) and H sb in a similar
way.
Let a
S b T b and b
S a T a . Let a be a topological space, and a is
a
the Borel sigma-algebra on
. For a given ( a ; a ), consider a nonempty collection
a
Fa
of events such that ; 2
= F a.
Following BS (2002), we require that a player’s belief at an information set is over
strategies and types to analyze players epistemic status/conditions in a game.
We take the collection of relevant conditioning events to be the histories. In perfect
information setting.
In this section, following BS (2002), we also require F a to be clopen in a and
consider the “relevant hypotheses” corresponding to the event that a certain partial
history has occurred and hence, each element of F a to is clopen in a . For each h 2 H, let
a
a
hhia
E
: E = S b (h) T b and formally, we de…ne F a fhhia
: h 2 Hg.
For each CPS p and h 2 H, we, slightly abusing the notations, write p ( jh) for p ( j hhi) 2
M ( a ).
Each conditional probability system (CPS) on ( a ; a ; F a ) represents player a’s belief
over his opponent’s strategies and higher order beliefs, conditioning on each observable
event in F a .
De…nition 5. Conditional Probability Systems (CPS’s) of player a. (See Rênyi
(1955), Myerson (1987), and BS (2002).) A conditional probability system (CPS) on
( a ; a ; F a ) is a mapping p ( j ) : a F a ! [0; 1] such that, for all B; C 2 F a and
A 2 a,
1. p (BjB) = 1;
2. p ( jB) = 1 is a probability measure on (
3. A
B
a
;
a
);
C implies p (AjB) p (BjC) = p (AjC).
The set of conditional probability systems on ( a ; a ; F a ) is a subset of [M ( )]F
a
a
and is denoted by MF ( a ). We endow MF ( a ) with the product topology.
6
a
De…nition 6. The conditional belief operator for player a given history h 2 H is a
map B a;h : a ! b de…ned as follows. For each E 2 a ,
B a;h (E)
De…ne hhi
S0
(sa ; ta ) 2
T b : S 0 = S (h)
b
:
a;h
(ta ) (E) = 1 .
S a , i.e., hhi is the event history h occurs.
De…nition 7. The strong belief operator for player a is a map SB a :
de…ned as follows. Let SB a (;) = ; and for each E 2 a n;,
T
SB a (E)
B a;h (E) .
a
!
b
,
h2H:E\hhi6=;
Let the notion L represent the lexicographic order. That is, for each (r0 ; :::; rn 1 ),
(s0 ; :::; sn 1 ) 2 Rn , (r0 ; :::; rn 1 ) L (s0 ; :::; sn 1 ) if and only if whenever ri < si , there
is j < i such that rj > sj . Let p be a CPS on ( a ; a ; F a ). Write margS b p for the
marginal on S b of the CPS p.
For each ( a ; a ; F a ), we are now ready to de…ne the preference represented by a
CPS p ( j ). Note F 2 F a if and only if there isRh 2 H such that hhi = F 2 F. For each
2 M ( ), x 2 A and hhi 2 F a , let E (x)
x (!) d (!).
a
a
a
For each ( ; ; F ), we are now ready to de…ne the preference represented by a
CPS p ( j ). Let P ( a ) be the set of complete, transitive, and re‡exive binary relations
on a , satisfying Independence axiom.
De…nition 8. CPS-based preferences (P cps ). Let p be a CPS on ( a ; a ; F a ).
We say a preference %p 2 P ( a ) is p-based if it satis…es re‡exivity, transitivity, Relevance
and Responsiveness:
(i) Relevance: For each x; y 2 A, if for each hhi 2 F a ; Ep( jh) (x) = Ep( jh) (y),
then x p y;
(ii) Responsiveness: For each x; y 2 A, if for each hhi 2 F a , Ep( jh) (x)
Ep( jh) (y) and for some hhi 2 F a , Ep( jh) (x) > Ep( jh) (y), then x p y.
Relevance says that in only the conditional expectations Epjh of acts matter in determining a preference. Responsiveness says that if the conditional expectations on all
hhi of one act x is weakly higher than that of another act y, and on some hhi strict
inequality holds, then x is strictly preferred to y.
We denote the set of all p-based preferences in P ( ) by P cps ( ), called CPS-based
preferences. Let Ha H be the set of histories where a is active.
Remark. Siniscalchi (2012) provides a revealed preference axiomization for CPS. Our
de…nition of CPS-based preferences is consistent with Siniscalchi’s (2012) axioms.
7
4
Main Result
We now provide the preference basis for the notion of strong assumption in a CPS.
Theorem 1. Fix a CPS p on ( a ; a ; F a ) and …nitely many events ; =
6 E1 ( E2 (
k
a
::: ( Ek
. (1) If each of fEi gi=1 is strongly believed under p, then there is %p 2
P cps ( a ) such that each of fEi gki=1 satis…es Nontriviality and Strict Determination
under %p . (2) For each %p 2 P cps ( a ), if each of fEi gki=1 satis…es Nontriviality and
Strict Determination under %p , then each of fEi gki=1 is strongly believed under p.
Remark. We can’t replace (1) by the following statement: There is %p 2 P cps ( a )
a
such that whenever an event E
is strongly believed under p, E satis…es Nontriviality
and Strict Determination under %p . That is, we can’t have a …xed %p such that every
strong believed event E satis…es Nontriviality and Strict Determination under %p .
A CPS is coarser in the sense that the corresponding preference is not complete, while
an LPS always represents a complete preference. There may multiple LPS’s having the
same CPS representation as illustrated in the example discussed above. Therefore, in
general, we can’t …nd a uniform preference satisfying Nontriviality and Strict Determination for all strongly believed events.
Proof of Theorem1. For each real c, let c be the act assigning c to each state in
.
a
is strongly believed
Sketch of Proof for Part (i). Suppose that each of fEi gki=1
a
a
a
under a CPS p on ( ; ; F ). We want to construct a complete preference %p 2
P cps ( a ) such that each of fEi gki=1 satis…es Nontriviality and Strict Determination
under %p . We show it in three steps.
1. We de…ne %p % by constructing
2. We verify that %p % 2 P cps (
vance and Responsiveness.
a
2 L(
a
) based on p.
). That is, %p 2 P (
3. We verify that each of fEi gki=1 satis…es Nontriviality
termination under %p .
a
) satis…es Releand Strict De-
Step 1. De…ne H 0
f?g and for each n 0, inductively de…ne H n+1 by the set
of histories immediately following an element h in H n . Let M + 1 be the length of
the longest non-terminal history. De…ne E0 = ; and inductively, for each i = 1; :::; k,
de…ne Di
Ei nEi 1 . For each i = 1; :::; k, de…ne Xi
fh 2 H : p (Di jh) > 0g. Since
E1 \ h?i = D1 = E1 6= ;, by strong belief of E1 , p (E1 j?) = 1 > 0 and hence,
? 2 X1 6= ;.
De…ne Y0
f?g
X1 , V0
supp p ( j?) and W0
fh 2 X1 : p (V0 jh) > 0g. For
each 0 i k 1, iM +1 n iM +M , inductively de…ne Yn
Xi+1 \ H n iM n[n0 n
Wn0 , Vn [h2Yn supp p ( jh) and Wn fh 2 Xi+1 : p (Vn jh) > 0g.
8
1
0
De…ne 0
Y0 . Suppose that n0 2 fYn gkM
n and in
n=0 is de…ned for each n
0
particular, n
Ym . Inductively, de…ne n+1
Ym0 if m > m, Ym0 6= ; and for each
m0 > m00 > m, Ym00 = ;. Let f m gnm=0 be the set of m de…ned as above, where n kM .
For each m = 0; :::; n, let m be the mean of fp ( jh) 2 M ( a ) : h 2 m g and Um
a
[h2 m supp p ( jh). Let
( 0 ; :::; n ). We …rst verify that
P 2 L ( ).
1
By the de…nition of m and Um , each m (Um ) = j m j h2 m p (Um jh) = 1. Consider
0 m0 < m00 n. To show mutual sigularity of , it remains to show that m0 (Um00 ) = 0
and m00 (Um0 ) = 0.
Suppose that h0 2 m0 = Yn0
Xi0 and h00 2 m00 = Yn00
Xi00 , where n0 m0 and
n00
m00 . Since m0 < m00 and h00 2 m00 , by the construction of m00 , p (Um0 jh00 ) = 0.
P
Hence, m00 (Um0 ) = j 1 00 j h00 2 m00 p (Um0 jh00 ) = 0.
m
On the other hand, we show that m0 (Um00 ) = 0 by showing p (supp p ( jh00 ) jh0 ) = 0.
If supp p ( jh0 ) \ supp p ( jh00 ) = ;, then
p (supp p ( jh00 ) jh0 ) = p ((supp p ( jh0 ) \ supp p ( jh00 )) jh0 )
= 0:
Suppose that supp p ( jh0 ) \ supp p ( jh00 ) 6= ;. Note that . supp p ( jh0 )
hh0 i and
00
00
0
00
supp p ( jh )
hh i. Hence, hh i \ hh i 6= ;. Since the game is perfect information,
we have either h0
h00 or h00
h0 . We …rst show that it not true that h00
h0 . If
so, then hh0 i hh00 i. Since supp p ( jh0 ) \ supp p ( jh00 ) 6= ;, S 0 supp margS b p ( jh0 ) \
supp margS b p ( jh00 ) 6= ;. By the de…nition of support, margS b p (hS 0 i jh00 ) > 0 and
hence, p (hS 0 i jh00 ) > 0. Since hh0 i
hS 0 i, p (hh0 i jh00 )
p (hS 0 i jh00 ) > 0. Note that
p (supp p ( jh0 ) \ hh0 i jh0 ) = 1. Since hh00 i ; hh0 i 2 F a and supp p ( jh0 )\hh0 i hh0 i hh00 i,
by Bayes’rule (CPS’s (iii)),
0 < p (hh0 i jh00 )
= p (supp p ( jh0 ) \ hh0 i jh0 ) p (hh0 i jh00 )
= p (supp p ( jh0 ) \ hh0 i jh00 )
p (Um0 jh00 ) .
However, p (Um0 jh00 ) = 0, a contradiction.
Hence, it must be the case that h0 h00 , i.e., hh00 i hh0 i. By the construction of m00
, p (supp p ( jh0 ) jh00 ) = 0 and hence, p (supp p ( jh0 ) \ supp p ( jh00 ) jh00 ) = 0. Therefore,
since hh00 i ; hh0 i 2 F a and supp p ( jh0 )\supp p ( jh00 ) hh00 i hh0 i, by Bayes’rule (CPS’s
(iii)),
p (supp p ( jh00 ) jh0 ) = p (supp p ( jh0 ) \ supp p ( jh00 ) jh0 )
= p (supp p ( jh0 ) \ supp p ( jh00 ) jh00 ) p (hh00 i jh0 ) = 0.
Since for each h00 2
m00 ,
p (supp p ( jh00 ) jh0 ) = 0, we have
p (Um00 jh0 ) = p [h00 2
= 0:
9
m00
supp p ( jh00 ) jh0
P
Therefore, m0 (Um00 ) = j 1 0 j h0 2 m0 p (Um00 jh0 ) = 0.
m
Step 2. We verify that %p 2 P ( a ) satis…es Relevance and Responsiveness. It is
easy to see that % is a complete, transitive, and re‡exive binary relation, satisfying
Independence axiom, on A.
Relevance. Let x; y 2 A such that for each hhi 2 F a , Epjh (x) = Epjh (y). By the
construction of , for each m = 0; :::; n,
Z
1 P
x (!) d m (!) =
Epjh (x)
j m j h2 m
1 P
Epjh (y)
=
j m j h2 m
Z
=
y (!) d i (!) .
Therefore, by the de…nition of %p % , x p y and x p y.
Responsiveness. Let x; y 2 A such that for each hhi 2 F a , Epjh (x) Epjh (y) and
there is hh0 i 2 F a such that Epjh0 (x) > Epjh0 (y). Hence, by the construction of , for
each m = 0; :::; n,
Z
1 P
x (!) d m (!) =
Epjh (x)
j m j h2 m
1 P
Epjh (y)
j m j h2 m
Z
=
y (!) d i (!) .
Let E
f! 2 a : x (!) > y (!)g and U
[nm=0 Um . Since Epjh0 (x) > Epjh0 (y),
p (U \ Ejh0 ) = p (Ejh0 ) > 0.
R
n
0
,
there
is
0
m
n
such
that
By
the
construction
of
f
g
m
m=0
R
R
Rx (!) d m0 (!) >
y (!) d m0 (!). Otherwise, for each m = 0; :::; n,
x (!) d m (!) = y (!) d m (!).
Hence, by the construction of , for each m 2 f m gnm=0 and h 2 m , Epjh (x) = Epjh (y).
Z
1 P
x (!) d m (!) =
Epjh (x)
j m j h2 m
1 P
Epjh (y)
j m j h2 m
Z
=
y (!) d i (!) .
By the construction of f m gnm=0 , there are h00 2 H and m0 2 f m gnm=0 such that
hh00 i = m0 and p (h0 jh00 ) > 0. Note that by Bayes’rule, p (Ejh00 ) = p (Ejh0 ) p (hh0 i jh00 ).
10
Epjh00 (x) =
Z
x (!) dp ( jh00 ) (!)
P
00
Z
p (hhi jh ) x (!) dp ( jh) (!)
P
= p (hh0 i jh00 ) J?Epjh0 (x) + h h00 ;hfollowsh00 ;h6=h0 p (hhi jh00 ) J??Epjh (x)
P
> p (hh0 i jh00 ) Epjh0 (y) + h h00 ;hfollowsh00 ;h6=h0 p (hhi jh00 ) J??Epjh (y)
= Epjh00 (y) .
=
h h00 ;hfollowsh00
Therefore, x y.
Step 3. Let Ei 2 fEi gki=1 , sa 2 projS a E such that [s] is not %-null and let h 2
H a (s). We want to show that there are acts x; y 2 A such that x E\[s]\hhi y.
Since %p % and 2 L ( a ), by Lemma 1, it remains to show that for each Ei
there is 0
mi
n such that Ei
is weakly assumed under at level mi . That
is, (i) for each m mi , m (Ei ) = 1 and for each m > mi , m (Ei ) = 0; (ii) projS Ei \
supp margS = [m mi supp margS m .
Note that for each h 2 H and Ei , if p (Ei jh) > 0, then Ei \ hhi 6= ; and hence, by
strong belief of Ei , p (Ei jh) = 1. Therefore, p (Ei jh) > 0 i¤ p (Ei jh) = 1.
Let h 2 [ii0 =1 Xi0 , i0 > i and h0 2 Xi0 . Since p (Ei jh) > p (Di jh) > 0, p (Ei jh) = 1.
Since p (Ei jh0 )
1 p (Di jh0 ) < 1, p (Ei jh0 ) = 0 and hence, p (Di jh0 ) = 0. Thus, for
each i0 6= i, Xi \ Xi0 = ;. By the construction of fDi gki=1 , each Ei = [ii0 =1 Di0 . Therefore,
fh 2 H : p (Ei jh) > 0g = fh 2 H : p (Ei jh) = 1g = [ii0 =1 Xi0 .
De…ne mi
jf m 2 f m gnm=0 : m 2 [ii0 =1 Xi0 gj. By the construction of f m gnm=0 ,
for each m
mi , m 2 [ii0 =1 Xi0 . Hence, for each m
mi and h 2 m
[ii0 =1 Xi0 ,
p (Ei jh) = 1. For each m > mi P
and h 2 m
[ni0 i+1 Xi0 , p (Ei jh) = 0. Therefore,
for each m mi , m (Ei ) = j 1m j h2 m p (Ei jh) = 1 and for each m > mi , m (Ei ) =
P
1
h2 m p (Ei jh) = 0.
j mj
It remains to show that projS Ei \ supp margS = [m mi supp margS m . We prove
it by contradiction: Suppose not and consider two cases as follows.
Case 1. There is s 2 [m mi supp margS m n (projS Ei \ supp margS ). That is,
there is m
mi such that m ([s]) > 0 and [s] \ Ei = ;. However, for each m
mi ,
m (Ei ) = 1 and hence, m ([s]) = m ([s] \ Ei ) = 0, a contradiction.
Case 2. There is s 2 (projS Ei \ supp margS ) n [m mi supp margS m . That is, we
have s 2 projS Ei and there is m0 > mi such that m0 ([s]) > 0 and for each m mi ,
0
0
m0 such that p ([s] jh ) > 0 and
m ([s]) = 0. By the de…nition of m0 , there is h 2
hence, [s] hh0 i. Note that ; 6= [s] \ Ei hh0 i \ Ei and hence, by strong belief of Ei ,
p (Ei jh0 ) = 1. However, h0 2 m0 [ni0 i+1 Xi0 and hence, p (Ei jh0 ) = 0, a contradiction.
a
Part (ii). Let %p 2 P cps ( a ) and Ei
satisfy Nontriviality and Strict Deterp
mination under % . We show that Ei is strongly believed under p. We prove this by
6 ; and p (Ei jh0 ) < 1.
contradiction. Suppose not, i.e., there is h0 2 H such that Ei \hh0 i =
The proof is similar to the “only if”part of the proof for Lemma 1 and hence, omitted.
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