Supplemental Appendix to "Common Assumption of
Cautious Rationality and Iterated Admissibility"
Nicodemo De Vitoy
Emiliano Catonini
June 2016
These appendices expand on Appendix A and Appendix B in the main text. Appendix
C treats stronger notions of Assumption; Appendix D provides the proofs of some technical
claims in the main text; Appendix E treats epistemic analyses of IA under the alternative
de…nitions of Assumption; Appendix F shows the existence of a belief-complete type
structure with an empty RC CARC set; …nally, Appendix G provides the proof of Theorem
4.
Appendix C: Alternative notions of Assumption
In this Section, we put forward two stronger versions of the notion of Assumption employed
in the main body of the paper (De…nition 11). These are based on two non-monotone
"in…nitely more likely than" order relations between uncertain events, which are introduced and studied in Subsection C1. Building on these notions, we obtain (Subsection
C2) the counterparts of BFK-Assumption and PWD-Assumption in Dekel et al. ([8]); a
comparison with these two concepts is made in Subsection C3. Finally (Subsection C4),
we provide a foundation of TWD-Assumption ([8, De…nition 4.6]) in terms of our notion
of "in…nitely more likely than" (De…nition A.3), as opposed to the foundation, analogous
to the one of PWD-Assumption, given by Dekel et al. [8].
We adopt the decision-theoretic framework of Appendix A. Notation is as in the main
text, unless otherwise stated.
C1 Likelihood orders
In the Subjective Expected Utility framework an event E is deemed "in…nitely more
likely than" a disjoint event F if and only if F is Savage-null under %. By constrast, the
Lexicographic Expected Utility framework admits a richer likelihood orders on events. In
F , and we studied its main
Appendix A we introduced a partial order on events, E
properties. We now examine two closely related partial orders, denoted by S and W D ;
y
Higher School of Economics, Moscow, [email protected]
Bocconi University, [email protected]
1
the …rst one is due to Blume et.al ([4, De…nition 5.1]), while the second one is new, and
it is based on a weak-dominance concept (cf. [2]).
The following de…nition is taken from Dekel et al. ([8, De…nition 4.1]):
De…nition C.1 Fix a non-empty event E
dominates g on E, and write f PWDE g, if
1 f (!)
and f; g 2 ACT ( ). Say f P-weakly
g (!) for all ! 2 E, and
E which is non-null under % and satis…es f (!) >
2 there exists some Borel set F
g (!) for all ! 2 F .
Condition 2 in De…nition C.2 requires strict preference on a subset of E that is "significant". As Dekel et al. put it ([8, Section 4.1]), P-weak dominance employs a Preferencebased notion of signi…cance, which coincides with the usual (topological) notion of weak
dominance if is a …nite set.
Next, a formal de…nition of the relation S and W D in terms of preferences:
De…nition C.2 Fix disjoint events E; F
E WD F )
with E 6= ;. Thus E
S
F (resp.
(1) E is non-null under % , and
(2) for all f; g 2 ACT ( ), f
E
g (resp. f PWDE g) implies f
E[F
g.
Condition (2) in De…nition C.2 states that, when comparing any two acts f and g that
give the same consequences in states not belonging to E [ F , if f E g or f PWDE g, then
the consequences in F "do not matter" for the unconditional strict preference f
g.1
Note that, according to De…nition C.2, event F may, but need not, be Savage-null under
% if E S F or E W D F .
Proposition C.1 Fix = ( 1 ; :::; n ) 2 N ( ) and pairwise disjoint events E; F; G
,
with E 6= ;.
(i) If E S F , then E W D F . The reverse implication holds provided 2 L( ).
(ii) Let E1
be a non-empty event such that E1
E and E1 is non-null under
% . Thus, if E S F (resp. E W D F ), then E1 S F (resp. E1 W D F ).
(iii) Let F 6= ;. If E W D G and F W D G, then E [ F W D G.
The proof of Proposition C.1 (as well as the proofs of Propositions C.2 and C.3 below)
is postponed at the end of the current appendix section.
1
E
The de…nition of the partial order S is taken from [3, p. 95]. De…nition 5.1 in [4] states that
S F if Condition (i.2) in De…nition C.2 is replaced by the following condition:
f
E
g implies f
nF ; hF
E[F
g
0
nF ; hF
for all h; h0 2 ACT ( ). (Condition (i.1) is automatically satis…ed in [4, De…nition 5.1], since the Authors
consider a …nite state space without Savage-null events.) It is easy to check the equivalence between the
two de…nitions.
2
While the likelihood order W D is weaker than S , they are nonetheless equivalent
for LCPS’s, as part (i) of Proposition C.1 states. Example A.1 shows that, in absence of
mutual singularity, the reverse implication of part (i) of Proposition C.1 need not be true.
Furthermore, the example also illustrates that an analogous result to the one in part (iii)
does not hold for the order S .2
Part (ii) of Proposition C.1 establishes an important property of both likelihood orders,
and it will play a central role in the statement of our …rst characterization result (Theorem
C.1 below). Of course, the reverse implication of part (ii) of Proposition C.1 does not hold.
In other words, unlike
, both W D and S do not satisfy a monotonicity property,
as the following example shows.
Example C.2 Consider an abstract space = [0; 1], and take E = [0; 1), E1 = (0; 1) and
F = f1g. Consider the LPS = ( 1 ; 2 ) where 1 is uniform on [0; 1] and 2 (f0g) =
2
(f1g) = 12 . Note that 2 L( ), so by part (i) of Proposition C.1, the orders W D and
S are equivalent. It turns out that E1
S F - see Proposition C.3 below. But E
S F
does not hold. To see this, …x some z 2 (0; 1) and consider acts f; g 2ACT( ) de…ned by
!
!
!
f = (!
z f0g ; 0 f0g ) and g = ( 1 f1g ; 0 f1g ). It is easily veri…ed that f PWDE g, while
g E[F f .
We point out that an analogoue of Proposition C.1.(ii) does not hold for the order
, as Example C.3 below shows.
Example C.3 Consider an abstract space = f ; ; g, and take E = f ; g, E1 = f g
= ( 1 ; 2 ) where 1 (f g) = 1 and 2 (f g) =
and F = f g. Consider the LPS
1
2
(f g) = 2 . It turns out that E
F , but the non-null event E1 E does not satisfy
E1
F.
We now provide an LPS-based characterization of the likelihood orders S and W D .
The following notation is essential. Given = ( 1 ; :::; n ) 2 N ( ) and non-empty event
E
, let
l
IE = inf l 2 f1; :::ng
SE = sup l 2 f1; :::ng
Proposition C.2 Fix
1 E
S
= ( 1 ; :::;
n
(E) > 0 ,
l
(E) > 0 .
) 2 N ( ) and disjoint events E; F
with E 6= ;.
F if and only if the following conditions hold:
(1.1) IE < IF ;
(1.2) for all l
IF , there exists
l
1 ; :::;
l
l
IF
1
(G) =
2 RIF
IX
F 1
l
j
j
1
such that, for each Borel G
E,
(G) .
j=1
2
Recall that the analysis in [4] is carried out in a …nite space . Thus, Proposition C.1 can be viewed
as a generalization of the results in [4] for the case of arbitrary (possibly in…nite) spaces.
3
2 E
WD
F if and only if the following conditions hold:
(2.1) IE < IF ;
(2.2) for all l IF and each Borel G
that k (G) > 0.
E, if
l
(G) > 0, then there exists k < IF such
Note: for the speci…c case in which F =
E, conditions (1.1) and (1.2) are equivalent
l
to the requirement that (E) = 1 for all l I E 1. As we shall see below, the level
I E 1 of the LPS ( 1 ; :::; n ) is the level under which an event E is S-assumed andnor
WD-assumed.
is weaker than
Remark C.1 As Proposition A.2 also shows, the likelihood order
,
which
in
turn
is
weaker
than
.
Indeed,
Condition
(1.2)
implies
Condition
WD
S
F . Neverthless, as will be shown in
(2.2), and they are not needed to characterize E
Lemma C.2 below, all three likelihood orders are equivalent if both E and F are singleton
sets.
Recall that, by Proposition C.1.(i), the likelihood orders S and W D are equivalent
for LCPS’s. The following proposition records the LCPS-based representation result for
S.
Proposition C.3 Fix
Thus,
n
= ( 1 ; :::;
E
) 2 L( ) and disjoint events E; F
S
with E 6= ;.
F () SE < IF .
C2 S-Assumption and WD-Assumption
We now introduce the notions of Strong Assumption and Weak-Dominance Assumption in terms of the preference relation % . These are the analogues of the notions
of BFK-Assumption and PWD-Assumption in [8].
De…nition C.3 Fix
= ( 1 ; :::;
n
) 2 N ( ) and a non-empty event E
.
1 Relevance: Say that E is relevant under % if, for every elementary cylinder C =
fs i g T i , if E \ C 6= ; then there exist f; g 2 ACT ( ) such that f E\C g.
2 Strict Determination: Say that E strictly determines % if, for all f; g 2 ACT ( ),
f E g implies f
g.
3 PWD Determination: Say that E PWD-determines % if, for all f; g 2 ACT ( ),
f PWDE g implies f
g.
The event E
is strongly assumed ( S-assumed) under % if it is relevant
under % and strictly determines % .
The event E
is weak-dominance assumed ( WD-assumed) under % if it
is relevant under % and PWD-determines % .
4
Remark C.2 Relevance is equivalent to the requirement that each relevant part of E is
non-null under % . So, it ensures that Strict Determination and PWD Determination do
not hold vacuously. Indeed, if any relevant part Es i is non-null under % , so is E.
Then it is immediate to observe that, under Relevance, Strict Determination and
PWD Determination coincide, respectively, with E
nE and E
nE (cf.
S
WD
De…nition C.2). S-Assumption and WD-Assumption can also be given an alternative
characterization using the corresponding likelihood orders as follows.
Theorem C.1 Fix
n
= ( 1 ; :::;
) 2 N ( ) and a non-empty event E
.
1 E is S-assumed under % if and only if the following conditions are satis…ed:
(1) E
S
E;
(2) for every relevant part Es
i
E, it holds that Es
2 E is WD-assumed under % if and only if Es
Es i E.
i
i
WD
S
E.
E for every relevant part
Proof: Part 1: For the su¢ ciency part: If Es i S
E for every relevant part Es i
E, then each Es i is non-null under % , so Relevance is satis…ed. Strict Determination
follows trivially from Condition (1). For the necessity part: By Remark C.2, Relevance
implies that E and each relevant part Es i are non-null under % . So Condition (1) is
equivalent to Strict Determination, and, under Condition (1), Condition (2) follows from
Proposition C.1.(ii).
Part 2: For the su¢ ciency part: for Relevance, see Part 1. PWD Determination follows
from Proposition C.1.(iii). For the necessity part: By Remark C.2, Relevance implies that
E and each relevant part Es i are non-null under % . Then, by PWD Determination,
E WD
E., and the conclusion follows from Proposition C.1.(ii).
Note that, in the …rst statement of Theorem C.1, condition (1) has to be explicitly
stated, since, if each relevant part Es i satis…es Es i
E, the same conclusion
S
need not follow for E (cf. Proposition C.1). Theorem C.1 speci…es a class of subsets of
E which must be "in…nitely more likely than"
E. Especially for WD-Assumption,
the characterization is analogous to our main notion of Assumption. However, there is
a caveat: Both S and W D have implications for the "in…nitely more likely than"
relation between Borel subsets of E and
E, whereas
does not impose any such
requirement.
The next step is to characterize the notions of S-Assumption and WD-Assumption in
terms of LPS’s representing % . We …rst need LPS-based de…nitions of S-Assumption
and WD-assumption.
De…nition C.4 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
that E
is S-assumed under at level m n if
l
(i) (E) = 1 for all l m,
5
. We say
l
1 ; :::;
(ii) for all l > m, there exists
l
(F ) =
l
m
m
X
2 Rm such that, for each Borel F
l
j
j
E,
(F ) ,
j=1
and
(iii) for every elementary cylinder C = fs i g T i , if E \ C 6= ; then
for some k n.
We say that E
is S-assumed under if it is S-assumed under
m n.
k
(E \ C) > 0
at some level
De…nition C.5 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
. We say
that E
is WD-assumed under at level m n if
l
(i) (E) = 1 for all l m,
(ii) for each l > m and each Borel set F E, if l (F ) > 0, then there exists k m
with k (F ) > 0, and
(iii) for every elementary cylinder C = fs i g T i , if E \ C 6= ; then k (E \ C) > 0
for some k n.
We say that E
is WD-assumed under if it is WD-assumed under at some
level m n.
= ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
. Thus E
Theorem C.2 Fix
is S-assumed (resp. WD-assumed) under % if and only if E is S-assumed (resp. WDassumed) under .
Proof: The proof is immediate if n E is Savage-null under % , so, in what follows, let
n E be non-null under % .
(Su¢ ciency) Suppose that E is S-assumed (resp. WD-assumed) under at level m.
By Proposition A.1, Condition (iii) in De…nition C.4 (resp. De…nition C.5) implies that
every relevant part is non-null under % . This in turn implies that also E is non-null
under % (Remark C.2). The result follows from Proposition C.2.1 (resp. Proposition
C.2.2), with m = I E 1.
(Necessity) We need to show that if a non-empty event E
is S-assumed (resp.
WD-assumed) under % , then there is some m n such that conditions (i), (ii) and (iii)
of De…nition C.4 (resp. De…nition C.5) hold.
By Proposition C.2.1 (resp. Proposition C.2.2) there exists m = I E 1 that satis…es
conditions (i) and (ii) in De…nition C.4 (resp. De…nition C.5). Suppose that such m does
not satisfy condition (iii) in De…nition C.4 (resp. De…nition C.5). We show that Relevance
is violated, hence E is not S-assumed (resp. WD-assumed) under % , as required.
Fix an elementary cylinder C = fs i g T i with E \ C 6= ; but k (E \ C) = 0 for
every k
m. Since condition (ii) in De…nition C.4 (resp. De…nition C.5) holds, then
l
(E \ C) = 0 for all l n. Hence Relevance is violated.
The following characterization result of S-Assumption for LCPS’s is an immediate
consequence of Proposition C.3 and Theorem C.2.
Corollary C.1 Fix
= ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
. Suppose
that 2 L( ). Thus E is S-assumed under at level m n if and only if Condition (i)
and Condition (iii) of De…nition C.4 are satis…ed plus the following condition:
6
(ii)’
l
(E) = 0 for all l > m.
C3 Comparison with Dekel et al. [8]
We now compare the notions of S-Assumption and WD-Assumption to the corresponding
versions in [8].
n
= ( 1 ; :::;
De…nition C.6 Fix
) 2 N ( ) and a non-empty event E
.
1 Non-Triviality: Say that E is non-trivial under % if, for every open set O
E \ O 6= ; then there exist f; g 2 ACT ( ) such that f E\O g.
, if
2 BFK-Assumption: Say that E is BFK-assumed under % if it is non-trivial under
% and strictly determines % .
3 PWD-Assumption: Say that E is PWD-assumed under % if it is non-trivial
under % and PWD-determines % .
We say that EO is a part of the event E
if EO = O \ E 6= ; for some open set
.3 Clearly, every non-empty event E can be written as union of all its parts.
Note, a relevant part of an event E is also a part of E. This is so because, technically,
every elementary cilinder is (cl)open in . Therefore, Non-Triviality implies Relevance.
This leads to the following analogue of Theorem C.1.
O
Theorem C.3 Fix
= ( 1 ; :::;
n
) 2 N ( ) and a non-empty event E
.
1 E is BFK-assumed under % if and only if the following conditions are satis…ed:
(1) E
S
E;
(2) for every part EO
E, it holds that EO
S
2 E is PWD-assumed under % if and only if EO
E.
WD
E for every part EO
E.4
The proof of Theorem C.3 is virtually identical to that of Theorem C.1, and so omitted.
We now recall the LPS-based de…nitions of BFK-Assumption and PWD-Assumption.
De…nition C.7 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
. We say
that E
is BFK-assumed (resp. PWD-assumed) under at level m
n if
conditions (i)-(ii) of De…nition C.4 (resp. De…nition C.5) are satis…ed plus the following
condition:
(iii) for each part EO of E there exists k m such that k (EO ) > 0.
We say that E
is BFK-assumed (resp. PWD-assumed) under if it is
BFK-assumed (resp. PWD-assumed) under at some level m n.
3
4
Alternatively put, a part of an event E is a non-empty, relatively open subset of E.
The symbol " " stands for strict inclusion.
7
Dekel et al. propose di¤erent, yet equivalent, de…nitions of BFK-Assumption and
PWD-Assumption (cf. [8, De…nition 3.2 and De…nition 4.5]). Note the following:
Remark C.3 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
. Suppose that
condition (i) of De…nition C.4 (or De…nition C.5) is satis…ed for some m
n. Thus
condition (iii) of De…nition C.7 holds if and only if E [l m Supp l .
Proof: If E
[l m Supp l holds, then condition (iii) of De…nition C.7 follows from
Remark 5.2 in [8]. For the reverse implication, suppose that E * [l m Supp l . Hence
EO = E \ O 6= ;, where O =
[l m Supp l is open in . It follows that k (EO )
k
(O) = 0 for all k m. Therefore, condition (iii) of De…nition C.7 does not hold.
We have the following analogue of Theorem C.2.
Theorem C.4 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
. Thus E is
BFK-assumed (resp. PWD-assumed) under % if and only if E is BFK-assumed (resp.
PWD-assumed) under .
Going through the proof of Theorem C.2, it is easy to see how the same proof, with
obvious changes, establishes the results in Theorem C.4.
C4 TWD-Assumption: an alternative foundation
We now show how the notion of TWD-Assumption, due to Dekel et al. ([8]), can be given
a characterization in terms of the "in…nitely more likely than" relation
. We …rst give
an LPS-based de…nition of TWD-Assumption which is— by Remark C.3— equivalent to
[8, De…nition 4.6].
. We say
De…nition C.8 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
that E is TWD-assumed under at level m n if the following conditions hold:
(i) l (E) = 1 for all l m;
(ii) for each part EO of E there exists k m such that k (EO ) > 0.
We say that E is TWD-assumed under if it is TWD-assumed under at some
level m n.
Then, a de…nition of TWD-Assumption in terms of the preference relation % :
. We say
De…nition C.9 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
that E is TWD-assumed under % if, for every part EO of E, it holds that
E.
EO
We can now state and prove the following characterization result:
Theorem C.5 Fix = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
TWD-assumed under % if and only if E is TWD-assumed under .
8
. Thus E is
Proof: The proof is immediate if n E is Savage-null under % , so, in what follows, let
n E be non-null under % .
(Necessity) We show that if a non-empty event E
is TWD-assumed under % ,
then there is some m
n such that conditions (i) and (ii) of De…nition C.8 hold. Let
m = I nE 1. Then, for every k m, k ( n E) = 0, hence k (E) = 1. Thus, m satis…es
condition (i) of De…nition C.8. Since every part EO of E satis…es EO
n E, then
Proposition A.2 yields IEO m. Thus, Condition (ii) of De…nition C.8 is satis…ed.
(Su¢ ciency) If E is TWD-assumed under at level m, then condition (i) of De…nition
C.8 implies m + 1 = I nE . By condition (ii) of De…nition C.8 each part EO satis…es
IEO < I nE , hence, by Proposition A.2, EO
n E.
We conclude this sub-section with a brief review of Dekel et al.’s ([8, De…nition 4.6])
axiomatization of TWD-Assumption. Fix a non-empty event E
and f; g 2ACT( ).
According to [8, De…nition 4.2], say f T-weakly dominates g on E, and write f TWDE g,
if
(a) f (!)
g (!) for all ! 2 E, and
(b) there exists some open set O
! 2 O \ E.
such that O \ E 6= ; and f (!) > g (!) for all
Di¤erently from P-weak dominance, this notion of T-weak dominance instead employs
a T opological notion of "signi…cance" for the subsets of E (namely, its parts) on which
act f is strictly better than g. It is also noteworthy that condition (b) does not require
EO = O \ E to be non-null under % .
[8] propose the following de…nition of TWD-Assumption in terms of the preference
relation % . A non-empty event E is TWD-assumed under % if it is non-trivial under
% and the following holds:
TWD Determination: for all f; g 2 ACT ( ), f TWDE g implies f
g.
Here, we wish to point out that the axiom of Non-Triviality is not needed for the
representation of TWD-Assumption. This is a consequence of the following
Remark C.4 TWD Determination implies Non-Triviality.
Proof: We prove this by contraposition. Suppose that Non-Triviality does not hold,
i.e., there is a part EO which is Savage-null under % . We show the existence of f; g 2
!
!
!
ACT ( ) such that f TWDE g but g % f . Let f = ( 1 EO ; 0 nEO ) and g = 0 . Clearly,
f (!) = g (!) for all ! 2 E n EO and f (!) = 1 > 0 = g (!) for all ! 2 EO , hence
f TWDE g. Using the fact that EO is Savage-null under % , we obtain
Z
Z
l
l
f d = (EO ) = 0 = gd l
for all l = 1; :::; n. Hence g
f.
Note that, under Non-Triviality, f TWDE g implies f PWDE g. Therefore, if Nontriviality holds, then PWD Determination implies TWD Determination— equivalently,
PWD-Assumption implies TWD-Assumption. If is …nite, PWD- and TWD-Assumption
coincide (see [8, Section 4.2]).
9
C5 Proofs of Propositions C.1-C.3
We …rst prove some auxiliary lemmas that will be used in the proofs that follow.
Lemma C.1 Fix = ( 1 ; :::; n ) 2 N ( ) and disjoint events E; F
with E 6= ;. Let
E1 be a non-null Borel subset of E such that k (E1 ) = 0 for all k < IF . Then there exist
acts f; g 2ACT( ) such that f PWDE g (hence f E g), but g %E[F f .
Proof: De…ne acts f; g 2ACT( ) as follows:
IF
(F ) ,
0,
! !
g = ( 0 E ; 1 nE ).
f (!) =
!2
! 2 E1 ,
n E1 ,
It is easy to observe that f PWDE g. To see that g %E[F f , note that
Z
IF
fd
IF
=
(F )
IF
(E1 ) <
IF
(F ) ,
E[F
where the strict inequality follows from
each k < IF ,
Z
fd
k
IF
=
(E) < 1 (by disjointness of E and F ); and for
IF
(F )
k
(E1 ) = 0,
E[F
where the last equality follows from IF IE . On the other hand,
Z
gd IF = IF (F ) .
E[F
Lemma C.2 Fix
then E W D F .
2 N ( ) and disjoint events E; F
with E 6= ;. Thus, if E
Proof: Let f; g 2ACT( ) such that f PWDE g . This implies f
then f E[F g. Therefore E W D F .
Lemma C.3 Fix
= ( 1 ; :::;
Thus, if E W D F (resp. E
n
E
) 2 N ( ) and disjoint events E; F
S F ), then
g. Since E
S
F,
S
F,
with E 6= ;.
IE < IF .
The reverse implication is true provided E = f!g and F = f! 0 g.5
Proof: The statement is clearly true if F is Savage-null under % , so that, by Proposition
A.1, IF = inf ; = +1. So, in what follows, let F be non-null under % . It is enough
to show the case in which E
W D F . For E
S F , the result is a consequence of
5
Note that, in their study of equilibrium re…nements, Blume, Brandenburger and Dekel adopt the
characterization result of Lemma C.1 as primary de…nition of the relation S , but of course for the
speci…c case in which events are "states", i.e., E = f!g and F = f! 0 g— see [5, De…nition 2].
10
Lemma C.2. Arguing contrapositively, suppose that IE IF . Then, either E is null, and
E W D F fails; or, m (E) > 0 for some m IF and k (E) = 0 for all k < IF . Then,
by Lemma C.1, there are acts f; g 2ACT( ) such that f PWDE g and g %E[F f , hence
E W D F fails.
For the reverse implication, let E = f!g and F = f! 0 g, and consider f; g 2ACT( )
such that f f!g g. Thus, we have f (!) IE (f!g) > g (!) IE (f!g) and j (f!g) = 0
for all j < IE . Since IE < IF , then j (f! 0 g) = 0 for all j IE . Therefore,
Z
fd
f!g[f! 0 g
j
IE 1
Z
=
gd
j
f!g[f! 0 g
j=1
IE 1
= 0,
j=1
while
Z
fd
IE
= f (!)
IE
(f!g) > g (!)
IE
(f!g) =
f!g[f! 0 g
which yields f
Z
gd
IE
,
f!g[f! 0 g
f!g[f! 0 g
g, as required.
= ( 1 ; :::; n ) 2 N ( ) and disjoint events E; F
with E 6= ;.
Lemma C.4 Fix
Thus, if E S F , then for all f; g 2ACT( ),
Z
Z
l
(f g) d = 0 for all l < IF =)
(f g) d l = 0 for all l n.
E
E
Proof:RThe proof is by contraposition. Suppose there are f; g 2ACT( ) such that
(a) E (f g) d l = 0 for all l < IF , and R
(b) there exists k 2 fIF ; :::; ng such
that E (f g) d k 6= 0.
R
De…ne k 0 = min k 2 fIF ; :::; ng E (f g) d k 6= 0 . Without loss of generality, we
R
0
can assume that E (f g) d k > 0. We now show the existence of f 0 ; g 0 2ACT( )
satisfying f 0 E g 0 and g 0 E[F f 0 , hence E S F fails.
Construct f 0 ; g 0 2ACT( ) as follows:
f 0 (!) =
g 0 (!) =
To see that f 0
IF
! 2 E,
!2
E,
(F ) f (!),
0,
IF
! 2 E,
!2
E.
(F ) g(!),
1,
R
g 0 , …rst note that, since E (f g) d j = 0 for all j < k 0 , then
Z
Z
Z
Z
0
j
IF
j
IF
j
fd =
(F ) f d =
(F ) gd =
g0d j ;
E
E
thus, using the fact that
Z
0
f 0d k =
E
E
E
E
IF
(F ) > 0, we get
Z
0
IF
(F ) f d k >
E
11
IF
(F )
Z
E
gd
k0
=
Z
E
g0d
k0
.
To see that g 0
E[F
f 0 , note that, for all l n,
Z
Z
0
l
IF
gd =
(F ) gd
E[F
and
Hence
R
E[F
0
gd
l
=
Lemma C.5 Fix
Thus,
R
0
fd
l
IF
=
E[F
l
for all l < IF and
= ( 1 ; :::;
n
Z
(F )
E[F
fd
R
l
(F ),
2 L( ),
f d l.
E
E[F
g0d
IF
>
R
E[F
f 0d
) 2 N ( ) and disjoint events E; F
SE < IF =) E
and if
+
E
Z
0
l
E
WD
S
IF
.
with E 6= ;.
F,
F =) SE < IF .
Proof: The statement is clearly true if F is Savage-null under % , so that, by Proposition
A.1, IF = inf ; = +1. So, in what follows, let F be non-null under % , and …x acts
f; g 2ACT( ) such that f E g. So, there is k 2 f1; :::ng such that
Z
Z
j
gd j , 8j < k,
fd
=
ZE
ZE
k
fd
>
gd k .
E
Such k satis…es
k
E
(E) > 0; hence, by SE < IF , we have k < IF , hence
Z
Z
j
fd =
gd j , 8j k.
F
F
It follows that
Z
Z E[F
E[F
fd
fd
j
k
=
>
Z
ZE[F
gd j , 8j < k,
gd
k
,
E[F
i.e., f E[F g. Since f and g are arbitrary, this shows that E S F .
Next, let 2 L( ). Suppose by contraposition that SE IF . By mutual singularity,
there are Borel sets (Ul )l n in such that l (Ul ) = 1 for every l n, and k (Ul ) = 0 for
k 6= l. Moreover, SE (E) > 0. Then we have:
SE
k
(E \ USE ) > 0;
k
(E \ USE )
(USE ) = 0; 8k < IF
SE :
Hence, by Lemma C.1 there exist f; g 2ACT( ) such that f PWDE g and g %E[F f , hence
E W D F fails.
Proof of Proposition C.1: Part (i): The …rst statement follows from Lemma C.2.
12
For the reverse implication, let 2 L( ). Then, according to the second statement of
Lemma C.5, we get SE < IF . By the …rst statement of Lemma C.5, this implies E S F .
Part (ii): Suppose that E1 S F fails. So there are f; g 2ACT( ) such that f E1 g
and g %E1 [F f . We show the existence of f 0 ; g 0 2ACT( ) such that f 0 E g 0 and
g 0 %E[F f 0 ; this will violate E S F .
De…ne f 0 ; g 0 2ACT( ) as follows:
!
f 0 = (fE1 [F ; 1
!
g 0 = (gE1 [F ; 1
To see that f 0
f
0
g 0 , observe that
Z
0
f 0d
E g ()
E
Z
Z
l
()
fd +
(E1 [F ) ),
(E1 [F ) ).
E
E1
()
Z
E
l
E1
() f
l=1
n
d
E1
>L
>L
l=1
l
gd
+
n
Z
n
d
E
E1
fd
l
l
l=1
l
gd
E1
g.
l
E1
l=1
Z
l=1
E1
l
n
g0d
E
Z
>L
n
fd
Z
n
l
l=1
Next note that
g
0
%
()
Z
0
E[F f ()
Z
E1 [F
gd
gd
E1 [F
n
l
L
l
+
Z
Z
n
d
E
E1
() g %E1 [F f ,
L
l=1
n
fd
Z
E1 [F
+
Z
E
n
d
E1
l
l=1
l
E1 [F
l=1
l
l=1
hence g 0 %E1 [F f 0 , as required.
Suppose now that E1 W D F fails. So there are f; g 2ACT( ) such that f PWDE1 g
and g %E1 [F f . De…ne f 0 ; g 0 2ACT( ) as above. Then it is easily seen that f 0 PWDE g 0
and g 0 %E[F f 0 , hence E W D F fails.
Part (iii): Let f; g 2 ACT ( ) such that f PWDE[F g. We need to show that f E[F [G
g. To this end, note that f (!)
g (!) for all ! 2 E [ F , and there is a Borel set
B
E [ F which is non-null under % and satis…es f (!) > g (!) for all ! 2 B. Note
that B = (E \ B) [ (F \ B), where E \ B and F \ B are disjoint. Since B is non-null
under % , one of the sets E \ B and F \ B is non-null under % , say E \ B. We thus
have f PWDE g, which implies f E[G g; that is,
Z
Z
n
n
l
l
fd
>L
gd
.
E[G
E[G
l=1
13
l=1
Since f (!)
g (!) for all ! 2 F , it follows that
Z
Z
Z
n
n
l
l
fd
=
fd
+
fd l
E[F [G
E[G
F
l=1
l=1
Z
Z
n
l
> L
+
gd
gd
E[G
F
l=1
Z
n
=
gd l
,
E[F [G
which shows that f
E[F [G
n
l=1
n
l
l=1
l=1
g, as required.
Proof of Proposition C.2: Both statements hold vacuously if F is Savage-null under
% ; so, in what follows, let F be non-null under % .
Part 1: (Su¢ ciency) If Condition (1.1) holds, then IE < +1, hence E is non-null
under % . So Condition (1) of De…nition C.2 holds true. Fix f; g 2ACT( ) for which
f E g. So, there is k n such that
R
R
(a) E f d l = E gd l for each l < k,
R
R
(b) E f d k > E gd k .
We claim that k < IF . To see this, suppose instead that k
(1.2), there exists k1 ; :::; kIF 1 2 RIF 1 such that
Z
(f
g) d
k
=
E
IX
F 1
k
l
l=1
Z
(f
g) d
l
IF . Then, by Condition
= 0,
E
where the second equality follows from (a). This contradicts (b). Therefore k < IF .
Then, using the fact that l (F ) = 0 for all l < IF , we obtain f E[F g.
(Necessity) Condition (1.1) follows from Lemma C.3. By [8, Lemma 5.4], if IF satis…es
the conclusion of Lemma C.4, then, for all l
IF , there exist scalars l1 ; :::; lIF 1 2 R
such that, for each Borel G E,
l
(G) =
IX
F 1
l
j
j
(G) .
j=1
This shows that Condition (1.2) holds, and it concludes the proof of the necessity part.
Part 2: (Su¢ ciency) If Condition (2.1) holds, then IE < +1, hence E is non-null
under % . So Condition (1) of De…nition C.2 holds true. Fix f; g 2ACT( ) for which
f PWDE g. That is, f (!) g (!) for all ! 2 E, and there exists some Borel set G E
which is non-null under % and satis…es f (!) > g (!) for all ! 2 G. For each l < IF , it
holds that
Z
Z
Z
Z
l
l
l
fd =
fd
gd =
gd l .
E[F
E
E
E[F
Since G is non-null under % , then
(G) > 0 for some k
can take k < IF . Hence
Z
Z
Z
Z
k
k
k
fd =
fd >
gd =
k
E[F
E
E
14
E[F
n. By condition (2.2), we
gd
k
,
and this shows f E[F g, as required.
(Necessity) We argue by contraposition. That is, if at least one of conditions (2.1) and
(2.2) fails, then E W D F does not hold. To this end, it is enough to consider two cases:
(a) IE IF ;
(b) Condition (2.1) holds, and there is some Borel set G E such that l (G) = 0 for
all l < IF and k (G) > 0 for some k IF .
In case (a), it follows from Lemma C.3 that E W D F fails. In case (b), by Lemma
C.1 there exist f; g 2ACT( ) satisfying f PWDE g and g E[F f , hence E W D F fails.
Proof of Proposition C.3: Immediate from Lemma C.5 and Proposition C.1.(i).
15
Appendix D: Proof of measurability of the relevant
sets
The aim of this Section is to show that, for a given type structure T = hSi ; Ti ; i ii2I , the
sets Rim , m > 1, as de…ned in the main text, are Borel subsets of Si Ti . We do this by
…rst showing that Ai (E) Si Ti is Borel for every event E S i T i . Lemma D.1
below additionally shows that an analogous conclusion holds for the set of all LPS’s under
which an event is S-assumed or WD-assumed (we will use this fact in Appendix E).
Lemma D.1 Fix a type structure T = hSi ; Ti ; i ii2I and non-empty event E S i T i .
2 N (S i T i ) under which E is assumed (S-assumed, WDThus, the set of all
assumed) is Borel in N (S i T i ).
Proof: By [11, Theorem 17.24] it follows that, for a given event E
S i T i , the
set of probability measures satisfying (E) = p for p 2 Q \ [0; 1] is measurable in
M(S i T i ). So the sets of all 2 M(S i T i ) satisfying (E) = 1 or (E) = 0 are
Borel in M(S i T i ). Now, …x n and m < n. By the above argument and by de…nition
of Nn (S i T i ), it turns out that the set
A1n;m = f 2 Nn (S
= \l
T i) j
i
2 Nn (S
m
i
m
(E) = 1; 8l
T i)
l
(E) = 1
mg
is Borel in Nn (S i T i ). Note that A1n;m is the set of all 2 Nn (S i T i ) for which
Condition (i) in the de…nition of Assumption (S-Assumption, WD-Assumption) holds for
level m.
By the same argument, it follows that, for every s i 2 ProjS i (E), the set
Asn:mi =
= \l
is Borel in Nn (S
i
2 Nn (S
m
i
l
T i)
2 Nn (S
T i)
i
T i ). Note that the set
T
A3n;m =
s
(fs i g
i 2ProjS i (E)
l
T i ) = 0; 8l
(fs i g
m
T i) = 0
Nn ( ) Asn;mi
is the (measurable) set of all 2 Nn (S i T i ) satisfying Condition (ii) (resp. Condition
(iii)) in the de…nition of Assumption (resp. S-Assumption and WD-Assumption) for level
m. De…ne An;m = A1n;m \ A3n;m ; clearly, An;m is a Borel subset of Nn (S i T i ). So the
set of all 2 N (S i T i ) under which E is assumed is given by [n2N [m2N An;m , so
Borel in N (S i T i ).
By [8, Lemma A.10] shows that the set AS;2
2 Nn (S i T i ) satisfying
n;m of all
3
Condition (ii) of De…nition C.4 for level m is also Borel. Hence ASn;m = A1n;m \AS;2
n;m \An;m
is a Borel subset of Nn (S i T i ). Thus, the set of all 2 N (S i T i ) under which E
is S-assumed is given by [n2N [m2N ASn;m , so Borel in N (S i T i ).
D;2
By [8, Lemma A.11] shows that the set AW
of all 2 Nn (S i T i ) satisfying
n;m
D
1
W D;2
3
Condition (ii) of De…nition C.5 for level m is Borel. Hence AW
n;m = An;m \ An;m \ An;m
is a Borel subset of Nn (S i T i ). So, the set of all 2 N (S i T i ) under which E is
D
WD-assumed is [n2N [m2N AW
T i ).
n;m , which is Borel in N (S i
16
By the measurability of each belief map in a lexicographic type structures, it follows
that
Corollary D.1 Fix a type structure T = hSi ; Ti ; i ii2I . Thus, for every i 2 I, if E
S i T i is a non-empty event, then Ai (E) is a Borel subset of Si Ti .
The next step is to show that, for a given type structure T = hSi ; Ti ; i ii2I , the set Ric
(De…nition 10) is a Borel subset of Si Ti . We …rst report some auxiliary technical facts
we shall be using in the proofs that follow.
Fact D.1 Fix two countable collections of pairwise disjoint topological spaces (Xn )n2N
and (Yn )n2N . Let X = [n2N Xn and Y = [n2N Yn . Let (fn )n2N be a countable family of
mappings fn : Xn ! Yn . Thus, if each map fn is continuous (resp. Borel measurable),
then [n2N fn : X ! Y is continuous (resp. Borel measurable).
Proof: Let O be open in Y . By de…nition of direct sum topology, the set O can be
written as O = [n2N On , where each On = O \ Yn is open in Yn (see [9, p. 74]). Thus
([n2N fn )
1
(O) = [n2N fn 1 (O) = [n2N fn 1 (On ) .
Therefore, if each fn is continuous (resp. Borel measurable), then each fn 1 (On ) is open
(resp. Borel), which in turn implies that ([n2N fn ) 1 (O) is open (resp. Borel).
Fact D.2 Let X and Y be Lusin spaces, and …x a map f : X ! Y . Thus, if f is
continuous (resp. Borel measurable), then fb : N (X) ! N (Y ) is continuous (resp. Borel
measurable).
Proof: Since fb is the combination of the functions fb(n)
n2N
, where fb(n) : Nn (X) !
Nn (Y ), by Fact D.1 it is enough to show that, for each n 2 N, fb(n) is continuous or Borel
measurable. By [1, Theorem 15.14], the image measure map fe is continuous, provided
f is continuous. If f is assumed to be only Borel measurable, we conclude that fe is
Borel measurable by using two mathematical facts. First, the Borel -…eld on M (X) is
generated by sets of the form f 2 M (X) : (E) pg, where E 2 X and p 2 Q\ [0; 1]
(use [11, Theorem 17.24]). Second, each set fe 1 (f 2 M (Y ) : (E) pg), E 2 Y , can
be written as f 2 M (X) : (f 1 (E)) pg. Since each space Nn (X) is endowed with
the product topology, it follows from [9, Proposition 2.3.6] that the map fb(n) is continuous
provided f is continuous. The conclusion that fb(n) is Borel measurable follows from [1,
Lemma 4.49].
2
Lemma D.2 Fix a type structure T = hSi ; Ti ; i ii2I and si 2 Si . The set of all
N (S i T i ) such that si is a lexicographic best reply to margS i is Borel in N (S i T i ).
To prove Lemma D.2, we need the following auxiliary result:
Lemma D.3 Fix si ; s0i 2 Si . Let OsWi ;s0 and OsSi ;s0 be subsets of M(S
i
i
follows:
OsWi ;s0i =
OsSi ;s0i
=
2 M(S
i
2 M(S
i
T i)
T i)
i (si ; margS
i (si ; margS
17
i
i
T i ) de…ned as
i
)
0
i (si ; margS
i
) ,
)>
0
i (si ; margS
i
) .
Thus OsWi ;s0 and OsSi ;s0 are closed and open in M(S
i
gS
Proof: First recall that the map Proj
7! margS
i
: M(S
,
i
T i ), respectively.
i
i
T i ) ! M (S i ) de…ned by
i
2 M(S
i
T i ),
is continuous. Moreover, for each sei 2 Si , the function i (e
si ; :) : M (S i ) ! R is also
continuous. De…ne the real valued map fsi ;s0i : M (S i ) ! R as
fsi ;s0i margS
=
i
i (si ; margS
i
0
i (si ; margS
)
),
i
2 M( ).
The map fsi ;s0i is clearly continuous, and the set OsWi ;s0 can be written as
i
=
gS
Proj
=
fsi ;s0i
OsWi ;s0i =
gS
Proj
1
margS
i
1
i
2 M (S i ) fsi ;s0i margS
i
i
0
fsi ;s1 0 ([0; +1))
i
gS
Proj
1
([0; +1)) ,
i
i.e., OsWi ;s0 is the inverse image of the set [0; +1), closed in R, under the continuous map
i
g S , hence OW 0 is closed in M( ). An analogous argument shows that set
fsi ;s0i Proj
si ;si
i
OsSi ;s0 can be written as
i
OsSi ;s0i = fsi ;s0i
hence OsSi ;s0 is open in M(S
i
i
T i ).
gS
Proj
1
((0; +1)) ,
i
Proof of Lemma D.2:6 Let Unsi be the set of all 2 Nn (S i T i ) for which si is
a lexicographic best reply to margS i . By Lemma D.3, the sets OsWi ;s0 and OsSi ;s0 are,
i
i
respectively, closed and open in M( ), hence the set OsEi ;s0 = OsWi ;s0 OsSi ;s0 is closed in
i
i
i
M(S i T i ). The set Unsi can be expressed as
1
0
S
E
S
N
(
)
[
O
O
N
(
)
[
O
n 2
T
n 1
si ;s0i
si ;s0i
si ;s0i
A;
@
Unsi =
E
E
W
0
::: [ Osi ;s0 Osi ;s0 ::: Osi ;s0
si 6=si
i
i
i
and this shows that Unsi is Borel in N (S i T i ). The set of all 2 N (S i T i ) for
which si is a lexicographic best reply to margS i can be written as [n2N Unsi , hence it is
Borel.
Given a type structure T = hSi ; Ti ; i ii2I , we denote by A0 the set of all
T i ) such that margS i 2 N + (S i ); that is,
A0 =
6
2 N (S
i
T i ) margS
i
2 N + (S i ) .
The proof closely follows the lines of the proof of [8, Lemma A.6].
18
2 N (S
i
Lemma D.4 Fix a type structure T = hSi ; Ti ;
Proof: Note that
dS
A0 = Proj
i ii2I .
1
i
The set A0 is Borel in N (S
i ii2I .
1
i
i
: N (S
i
Thus, for every i 2 I, the set Ric
Proof: It follows from Lemma D.4 and from the measurability of
1
0
Ti . De…ne the set Ri as
i (A ) is Borel in Si
Ri = [si 2Si fsi g
T i ).
N + (S i ) .
dS
Since N + (S i ) is Borel in N (S i ) ([6, Corollary C.1]) and the map Proj
0
T i ) ! N (S i ) is continuous (Fact D.2), then A is Borel in N (S i T i ).
Corollary D.2 Fix a type structure T = hSi ; Ti ;
is Borel in Si Ti .
i
i
that the set Si
(B si ) ,
where B si stands for the set of all i 2 N (S i T i ) for which si is a lexicographic best
reply to margS i i . By Lemma D.2 and measurability of i , it follows that Ri is Borel in
1
0
Si Ti . Since Ric = Ri \ Si
i (A ) , the conclusion follows.
We can now state and prove the desired result:
Lemma D.5 Fix a type structure T = hSi ; Ti ; i ii2I . Thus, for each i 2 I and each
m > 1,
Rim+1 = Ri1 \ \l m Ai Rl i ,
and Rim is Borel in Si
Ti .
Proof: The equality Rim+1 = Ri1 \ \l m Ai Rl i is obvious. By Corollary D.2, it
follows that, for each i 2 I, the set Ri1 = Ric is Borel in Si Ti . By Corollary D.1, the set
Ai R1 i is Borel in Si Ti . The conclusion follows from an easy induction on m.
19
Appendix E: Epistemic Analysis of IA under alternative notions of Assumption
This Section illustrates how the results and arguments of the previous sections (Appendices C and D) apply to the setting of our main paper. Speci…cally, we show that an
analogue of Theorem 2 holds under the notions of S-Assumption and WD-Assumption.
There are two issues. First, we modify Lemma 3 in the main text (see Lemma E.2 below)
to show that an analogous invariance property under type morphisms also holds under
those alternative notions of Assumption. Second, we show how the …nite type structure
constructed in Section 6.1 for the proof of our main result (Theorem 2.(ii)) satis…es the
premises for the application of Lemma E.2.
For later use, we …rst state a version of Lemma B.1 and Lemma B.2 for the cases of
S-Assumption and WD-Assumption (the proofs are identical and so omitted).
. Thus
Lemma B.1* Fix an LPS = ( 1 ; :::; n ) 2 N ( ) and a non-empty event E
E is S-assumed (resp.WD-assumed) under if and only if there exists m n such that
satis…es Condition (i) and Condition (ii) of De…nition C.4 (resp. De…nition C.5) plus
the following condition:
(iii)’ E
[l m SuppmargS i l
T i.
Lemma B.2* Let ti 2 Ti S-assume (WD-assume) the event E
where i (ti ) = ( 1i ; :::; ni ). Thus
[l
m SuppmargS
i
l
i
S
T
i
i
at level m,
= ProjS i (E) .
Fix a type structure T = hSi ; Ti ; i ii2I . For each player i 2 I, let ASi :
WD
: S i T i ! Si Ti be the operator de…ned by
Si Ti and Ai
ASi (E i ) = f(si ; ti ) 2 Si
D
AW
(E i ) = f(si ; ti ) 2 Si
i
Ti jti S-assumes E i g , E i 2 S
Ti jti WD-assumes E i g , E i 2
T
i
S
i
S
i
i
T
i
!
.
T
i
.
By Lemma D.1 and the measurability of each belief map i , it follows that the sets
D
(E i ) are Borel in Si Ti for every event E i
S i T i ; so the
ASi (E i ) and AW
i
S
WD
operators Ai and Ai are well-de…ned.
In Lemma E.1 below, we denote by E the complement of a set E.
Lemma E.1 Let T = hSi ; Ti ; i ii2I and T = hSi ; Ti ; i ii2I be lexicographic type structures, such that there exists a type morphism ('i )i2I : T ! T from T to T . Let
E i
S i T i and E i
S i T i be non-empty events satisfying the following
conditions:
1) IdS i ; ' i (E i ) E i ;
2) IdS i ; ' i ( E i )
E i;
3) ProjS i (E i ) = ProjS i E i .
D
D
Thus (IdSi ; 'i ) ASi (E i )
ASi E i and (IdSi ; 'i ) AW
(E i )
AW
E i .
i
i
D
Proof: Let (si ; ti ) 2 ASi (E i ) (resp. (si ; ti ) 2 AW
(E i )), and set 'i (ti ) = ti . We
i
show that event E i is S-assumed (resp. WD-assumed) under i (ti ), that is, conditions
(i)-(iii) of De…nition C.4 (resp. De…nition C.5) are satis…ed.
20
Since event E i is S-assumed (resp. WD-assumed) under i (ti ) = 1i (ti ) ; :::;
then there exists m n such that li (ti ) (E i ) = 1 for all l m. Next note that
E
i
1
IdS i ; '
IdS i ; '
i
(E i )
i
1
IdS i ; '
i
E
i
n
i
(ti ) ,
,
where the …rst set containment is obvious, while the second one follows from condition
1). Hence, by de…nition of type morphism, it follows that, for all l n,
l
i
l
i
(ti )
=
(
l
i)
('i (ti ))
=
(
l
i)
(ti )
(ti ) (E i )
IdS i ; '
1
E
E
i
E
i
i
i
,
which implies ( i )l (ti ) E i = 1 for all l m, hence condition (i) of De…nition C.4
(resp. De…nition C.5) is satis…ed.
To show that Conditions (ii) in De…nition C.4 and De…nition C.5 are also satis…ed, we
need the following
Claim E.1 For every Borel set F
i
E i,
IdS i ; '
1
i
E i.
(F i )
Proof of Claim E.1 : Observe that, by condition 2) of the Lemma,
(
E i)
1
IdS i ; '
E
i
.
i
Hence,
IdS i ; '
1
i
(F i )
IdS i ; '
=
1
IdS i ; '
(
E
i
1
i
i
E
i
E i)
= E i,
where the …rst set containment is due to F
from the observation above.
i
E i , while the second containment follows
1
Fix an arbitrary Borel set F i
E i . By Claim E.1, IdS i ; ' i
(F i ) is a Borel
subset of E i . If E i is S-assumed under i (ti ) at level m
n, then there are scalars
l
l
1 ; :::; m 2 R such that, for all l > m,
l
i
(ti )
IdS i ; '
1
i
(F i ) =
m
X
l
j
j
i
(ti )
IdS i ; '
1
i
(F i ) .
j=1
If instead E i is WD-assumed under i (ti ) at level m
n, then there exists k
m
1
1
k
l
such that i (ti ) IdS i ; ' i
(F i ) > 0 provided i (ti ) IdS i ; ' i
(F i ) > 0
21
1
(F i ) = ( i )l (ti ) (F i ) for all l n, and F i is
for l > m. Since li (ti ) IdS i ; ' i
arbitrary, it turns out that in both cases Conditions (ii) of De…nition C.4 and De…nition
C.5 are satis…ed.
To show that Condition (iii) of De…nition C.4 (resp. De…nition C.5) is also satis…ed,
we proceed as follows. Consider an elementary cylinder C = fs i g T i satisfying
E i \ C 6= ;. It turns out that
IdS i ; '
1
i
C \E
i
=
=
1
IdS i ; '
fs i g
i
'
1
i
(C) \
IdS i ; '
\
IdS i ; '
T
i
1
= (fs i g
T i) \
IdS i ; '
i
(fs i g
T i) \
IdS i ; '
i
(fs i g
T i) \ E i,
1
i
1
i
E
i
IdS i ; '
i
i
E
1
E
i
(E i )
where the fourth line follows from condition 1) of the Lemma. Since (fs i g T i )\E i 6=
; (by condition 3) of the Lemma) and E i is S-assumed (resp. WD-assumed) under i (ti )
at level m n, then there exists l m such that li (ti ) ((fs i g T i ) \ E i ) > 0. But
l
i
(ti ) ((fs i g
l
i
(ti )
(
l
i)
T i) \ E i)
=
IdS i ; '
(ti )
1
i
C \E
i
C \E
i
,
and since C is an arbitrary elementary cilynder, this shows that Condition (iii) of De…nition C.4 (resp. De…nition C.5) is satis…ed, thus concluding the proof.
In what follows, …x a type structure T = hSi ; Ti ; i ii2I and, for each player i 2 I, let
be the set of all cautiously rational strategy-type pairs (si ; ti ) 2 Si Ti . As in the
main text, for each m > 1, rede…ne Rim inductively by
Ri1
Rim+1 = Rim \ ASi Rmi .
Q
m+1
0
1
m
We
write
R
=
S
T
and
R
=
\
R
for
each
i
2
I.
Thus
(resp.
i
i
m2N
i
i
i
i2I Ri
Q
1
R
)
is
the
set
of
all
states
at
which
there
is
cautious
rationality
and
mth-order
i
i2I
(resp. common) S-Assumption of cautious rationality. Each set Rim is Borel in
Si Ti (this follows from the fact that the operator ASi is well-de…ned and from arguments
analogous to those in Lemma D.5). Likewise,
we can rede…ne the
sets Rim , m > 1, with
Q
Q
D
1
the operator ASi replaced by AW
, so that i2I Rim+1 (resp.
i
i2I Ri ) is the set of all
states at which there is cautious rationality and mth-order (resp. common) WDAssumption of cautious rationality. We now show that an analogue of Theorem 2
holds when each set Rim is de…ned according to these alternative notions of Assumption.
First note that the proof of Part (i) of Theorem 2 relies, for the case where m 2, on
Lemma B.2, which concerns only Condition (ii) of Assumption (De…nition 11), and this
in turn is equivalent to Condition (iii) in De…nition C.4 and De…nition C.5 (cf. Lemma
B.2*). It is also noteworthy that the inductive step actually shows that Condition (ii)
l
l
in De…nition C.4 is satis…ed with
= (0; :::; 0) (cf. Footnote 25). So, the
1 ; :::; m
22
very same proof of Part (i) of Theorem 2 shows that an analogous result holds under
S-Assumption or WD-Assumption.
To show that the analogue of Part (ii) of Theorem 2 holds under the notion of SAssumption, we need to establish an analogue of Lemma 5. This amounts to show that the
type structure T constructed in Section 5.1 of the paper satis…es the required properties.
Moreover, since T is an LCPS type structure (Claim 2), this will show that Part (ii)
of Theorem 2 will continue to hold under the notion of WD-Assumption, by virtue of
Corollary C.1. To this end, we report here the required modi…cations in the proofs of
Claim 5 and Claim 3, respectively. The proofs of Claim 4 and Lemma 5 apply here
verbatim.
Modi…ed Proof of Claim 3: We only need to prove the inductive step of part (2).
Speci…cally, we only have to show that (si ; (si ; k + 1)) satis…es two requirements, namely
(a) (si ; (si ; k + 1)) 2 \l k ASi R ;li , and
.
(b) (si ; (si ; k + 1)) 2
= ASi R ;k+1
i
The proof of property (b) is the same as that in the main text. To show that property
(a) holds, …rst note that, by the induction hypothesis, for each l k 1, R ;li is S-assumed
under i ((si ; k)) = ksi ; ksi 1 ; :::; 1si at some level m. This implies that
;l
j
si (R i )
= 1, 8j
k+1
(0.1)
m,
and, since T is an LCPS type structure, then Corollary C.1 yields
;l
j
si (R i )
= 0, 8j < k + 1
(0.2)
m.
Taking into account that
k+1
si
R
;l
i
k+1
si
=
Sk i
S l i , and
Supp
k+1
si
we get
k+1
si
S l i (by de…nition of
k+1
si
and the induction hypothesis),
= S k i,
R
;l
i
= 1. It follows from (0.1) and (0.2) that conditions (i) and (ii) of
S-Assumption (De…nition C.4) under i ((si ; k + 1)) = k+1
si ; i ((si ; k)) are satis…ed for
0
k+1 m
l
m = m + 1. Finally, since SuppmargS i si
= S i , it readily follows that condition
(iii)’Lemma B.2* is also satis…ed. This shows that (si ; (si ; k + 1)) 2 ASi R
l k 1.
;l
i
for every
The modi…ed proof of Claim 5 requires the following Lemma.7
Lemma E.3 Fix the type structure T and the canonical type structure Tu . Thus, for
each player i 2 I,
(IdSi ; di )
Rik , 8k 1.
Ri ;k
7
As in Lemma D.2, we denote by
E the complement of a set E.
23
Proof: By induction on k.
(k = 1) If (si ; ti ) 2
= Ri ;1 , then (si ; di (ti )) 2
= Ri1 by Lemma 1.
(k
2) Suppose that the statement is true for k. Let (si ; ti ) 2
= Ri ;k ; we show that
(si ; di (ti )) 2
= Rik . Recall that Ri ;k = Ri ;k 1 \ ASi R ;ki 1 . First note that if k > M + 1,
then (si ; ti ) 2
= Ri ;k 1 by the version of Claim 3.4 for S-Assumption, hence the conclusion
follows from the induction hypothesis. So let k
M + 1. We consider separately two
;k 1
S
cases: (a) (si ; ti ) 2
= Ri
, and (b) (si ; ti ) 2
= Ai R ;ki 1 . In case (a), the conclusion
follows again from the induction hypothesis.
For case (b), let (si ; ti ) 2 Ri ;k 1 but (si ; ti ) 2
= ASi R ;ki , i.e., (si ; ti ) 2 Ri ;k 1 Ri ;k .
The induction hypothesis states that IdS i ; d
IdS i ; d
1
i
Rk i 1
R
;k 1
i
R
i
;k 1
i
Rk i 1 , and this implies
(cf. Claim E.1 in the proof of Lemma E.1). It follows that
f i (di (ti )) Rk i 1
=
i
(ti )
IdS i ; d
i
(ti ) R
1
i
Rk i 1
;k 1
i
= 0,
where the last equality follows from the version of Claim 3.1 for S-Assumption. So
Rk i 1 cannot be S-assumed under f i (di (ti )) in the canonical type structure Tu , hence
(si ; di (ti )) 2
= Rik .
Modi…ed Proof of Claim 5: By induction on k.
(k = 1) If (si ; ti ) 2 Ri ;1 , then (si ; di (ti )) 2 Ri1 by Lemma 1.
(k 2) Suppose that the statement is true for k 1. Let (si ; ti ) 2 Ri ;k . So (si ; ti ) 2
Ri ;k 1 and, by the induction hypothesis, (si ; di (ti )) 2 Rik 1 . So we need to show that
(si ; di (ti )) 2 ASi Rk i 1 ; this will imply (si ; di (ti )) 2 Rik , as required.
Now note that
(a) IdS i ; d i R ;ki 1
Rk i 1 ,
(b) IdS i ; d
i
R
;k 1
i
Rk i 1 ,
(c) ProjS i R ;ki 1 = ProjS i Rk i 1 .
Part (a) is the induction hypothesis, while part (b) follows from Lemma E.3; part
(c) follows from the versions for S-Assumption of Claim 4.(1) and part (i) of Theorem 2.
Since (si ; ti ) 2 ASi R ;ki 1 , then Lemma E.1 yields (si ; di (ti )) 2 ASi Rk i 1 .
24
Appendix F: A negative result for belief-complete type
structures
The aim of this section is to show that, for a …nite game G = hI; (Si ; ui )i2I i with jSi j
2, there exist associated belief-complete type structures
T = hSi ; Ti ; i ii2I such that
Q
C
C
R CAR is not possible at any state (si ; ti )i2I 2 i2I (Si Ti ) (cf. Section 6.3 in the
main text).
Theorem F.1 Fix a game G = hI; (Si ; ui )i2I i with jSi j 2 for each i 2 I. There exists
a continuous, belief-complete type structure T = hSi ; Ti ; i ii2I such that
Q
1
i2I Ri = ;.
Note that, as already remarked (Section 3.3), the type structure T in the statement
of Theorem F.1 cannot be simultaneously compact, belief-complete and continuous. It is
also noteworthy that the result of Theorem F.1 does not hinge on the continuity of belief
maps. Using arguments as those in [10], we now show that, as a consequence of Theorem
F.1, there indeed exists a (compact) belief-complete type structure with discontinuous
belief maps so that the RC CARC set is empty.
Corollary F.1 Fix a game G = hI; (Si ; ui )i2I i with jSi j 2 for each i 2 I. There exists
a belief-complete and compact type structure T = hSi ; Ti ; i ii2I such that
Q
1;
= ;.
i2I Ri
Proof: For each i 2 I, let Ti be the Cantor space f0; 1gN , which is an uncountable, compact metrizable (so Polish) space. Being Polish, each Ti has the cardinality of continuum.
The type structure T = hSi ; Ti ; i ii2I of Theorem F.1 has the following property: Each
type space Ti is Polish, and, by belief-completeness, it has the cardinality of continuum.
So, by the Borel Isomorphism Theorem ([7, Theorem 8.3.6]), there exists, for each i 2 I,
a Borel isomorphism (but not a homeomorphism) 'i : Ti ! Ti . It follows that, for each
\
i 2 I, the map (Id
Ti ) ! N (Si Ti ) is also a Borel isomorphism. So,
Si ; 'i ) : N (Si
for each i 2 I, de…ne the Borel map i : Ti ! N S i T i by
i
\
= (Id
S i ; 'i )
1
i
'i .
It is straightforward to check that each i is a well de…ned surjective belief map. So,
T = hSiQ
; Ti ; i ii2I isQa compact, belief-complete type structure. Moreover, the map
('i )i2I : i2I Ti ! i2I Ti is a type isomorphism betweenQT and T . Therefore, by
Theorem F.1 and Remark 2 in the main text, it follows that i2I Ri1; = ; holds for T .
Fix a Standard Borel space Y .8 Following the terminology in [12], we say that Y is
a Borel re…nement of a Standard Borel space X if Y has the same points as X, and
8
We recall that every Lusin space is Standard Borel, when considered as a measurable space— see [7,
Proposition 8.6.13].
25
every open set in X is open in Y . A Borel re…nement Y exists for every Standard Borel
X ([11, Section 5.2]), and Y generates the same Borel -…eld as X.
Fix type structures T = hSi ; Ti ; i ii2I and T 0 = hSi ; Ti0 ; i ii2I . We say that T 0 is a
Borel re…nement of T if T 0 and T have the same strategy sets, types and belief maps, and,
for each i 2 I, Ti0 is a Borel re…nement of Ti . Keisler and Lee ([12, Theorem 5.10]) show
that every type structure T admits a continuous Borel re…nement T 0 , i.e., a continuous
type structure which is a Borel re…nement of T .
If T 0 is a
Remark F.1 Fix type structures T = hSi ; Ti ; i ii2I and T 0 = hSi ; Ti0 ; i ii2I . Q
0
Borel
re…nement of T , then T is type-isomorphic to T . The map ('i )i2I : i2I Ti !
Q
0
0
i2I Ti , where each 'i is the identity map, is a type isomorphism between T and T .
The following result is an immediate implication of the above observation.
Corollary F.2 Fix a game G = hI; (Si ; ui )i2I i with jSi j Q
2 for each i 2 I. Let T =
1
hSi ; Ti ; i ii2I be a belief-complete type structure such that
= ;. Thus, there
i2I Ri
i
which
is
a
continuous Borel
exists a belief-complete type structure
T
=
hS
;
T
;
i
i i2I
i
Q
1;
re…nement of T and for which i2I Ri = ;.
Corollary F.2 provides an alternative route to show the existence of a continuous,
belief-complete type structure with empty RC CARC set from a type-isomorphic type
structure with discontinuous belief maps. Note indeed that the type structure T in
the statement of Corollary F.1 admits a continous Borel re…nement T 0 . Of course, by
belief-completeness, T 0 cannot be compact.
Proof of Theorem F.1:9 For each i 2 I, let Ti be the Baire space NN0 0 ,10 so that each
ti 2 Ti is an in…nite sequence of non-negative integers. The set N0 is endowed with the
discrete topology, and NN0 0 is endowed with the product topology. The basic open sets
are sets of the form
Ok = (n1 ; n2 ; :::) 2 NN0 0 j(n1 ; :::; nk ) = (o1 ; :::; ok )
for each k 2 N0 and o1 ; :::; ok 2 N0 . With this topology, a basic open set is also closed, so
sets of the form Ok constitute a clopen basis. The space NN0 0 is Polish and uncountable,
but not compact.
For each i 2 I, we partition Ti into a countable family of non-empty Borel subsets.
This is done in the following way: For each k 0, let
Tki = (n1 ; n2 ; :::) 2 NN0 0 jn1 = k .
Each Tki is a subbasic clopen subset of Ti ; moreover, each Tki is homeomorphic to the
Baire space. It is clear that Ti = [k 0 Tki , and all the Tki ’s are pairwise disjoint.
The next step is to construct the belief maps in such a way that, for each k
0,
ti 2 Tki and si 2 Si , (si ; ti ) does not belong to Rik+1 . For each i 2 I, we de…ne a countable
9
The construction of the required type structure follows some of the ideas in [10].
Here N0 denotes the set f0; 1; 2; :::g, i.e., N0 = N [ f0g. The Baire space is sometimes de…ned as
the set NN of all in…nite sequences of natural numbers (i.e., not including the zero; see, for instance, [1,
Section 3.14]). This di¤erence is immaterial for all the relevant topological properties we are going to use
in this section.
10
26
partition of N (S i
For each i 2 I, let
T i ) that "mirrors" the above partition of Ti . This is done as follows:
0
i
= N (S
T i ) A0i ;
i
= N + (S i ).11 Since jSi j
that is, 0i is the set of all LPS’s such that margS i i 2
0
each i 2 I, then i 6= ;.
Next, let
0
1
1
T0 i ) > 0 ,
i 2 Ai
i (S i
i =
and, for each k
k
i
= \m2f1;:::;k
2 for
2,
1g
i
2 A0i
Tmi 1 ) = 0 \
1
i (S i
i
2 A0i
1
i (S i
Tk i 1 ) > 0 .
In words: 1i is the set of all LPS’s on S i T i for which their marginals on S i are
of full-support and the …rst component measure assigns strictly positive probability to
S i T0 i ; 2i is the set of all LPS’s on S i T i for which their marginals on S i are
of full-support and the …rst component measure assigns probability 0 to S i T0 i , and
strictly positive probability to S i T1 i ; and so on.
It is immediate to check that N (S i T i ) = [k 0 ki and all the ki ’s are non-empty,
pairwise disjoint sets;12 so the countable family of all ki ’s is a partition of N (S i T i ).
Claim F.1 For each k
k
i
0,
is a Borel subset of N (S
0
i.
Proof: Since A0i is Borel (Lemma D.4), so is
Pki =
i
2 A0i
1
i (S i
T i ).
i
For each k
1, let
Tk i 1 ) > 0 .
Note that 1i = P1i , and for each k 2, ki is the intersection of Pki with the complements
of P1i ; :::; Pki 1 . Thus, to show that each ki is Borel in N (S i T i ), it is su¢ cient to
show that each Pki is Borel in N (S i T i ). Let
Mki =
2 M (S
i
T i)
(S
Tk i 1 ) > 0 .
i
Recall that, if X is a Lusin space, then the Borel -…eld on M (X) is generated by sets
of the form f 2 M (X) : (E) pg, where E 2 X and p 2 Q\ [0; 1] ([11, Theorem
17.24]). Thus, Mki is Borel in M (S i T i ). Since the canonical projection map
Proj1;n
n
i
1
i ; :::;
:
7!
Nn (S
T i ) ! M (S
i
1
i,
T i) ,
i
is continuous for each n 2 N, then Proj1;n1 Mki is Borel in Nn (S
follows from the observation that Pki can be written as
Pki = [n2N Proj1;n1 Mki
i
T i ). The conclusion
\ A0i .
11
Recall that A0i stands for the set of all i 2 N (S i T i ) such that margS i i 2 N + (S i ) (see
Section D).
12
To see that ki (k 1) is non-empty, pick any 2 M (S i ) and 2 M (T i ) such that
Supp = S i , and
(Tk i 1 ) > 0 and (Tmi 1 ) = 0 for all m = 1; :::; k 1. (For k = 1, simply require that (T0 i ) > 0.)
Thus, the product measure
i
=
is a length-1 LPS on S
27
i
T
i
which belongs to
k
i.
Recall now that every Borel subset of a Lusin space is Lusin, when endowed with the
relative topology. Moreover, every Lusin space is also Standard Borel, and so analytic
([7, Proposition 8.6.13]). Thus, by Claim F.1, each ki is analytic. Since each Tki is
homeomorphic to the Baire space, then it follows from [7, Corollary 8.2.8] (see also [11,
p. 85]) that there exist surjective continuous maps
[k]
i
: Tki !
k
i,
8k
0.
[k]
[k]
Hence, for each i 2 I, let i be the combination of the i ’s, i.e., i = [k 0 i : Ti !
N (S i T i ). By Fact D.1, i is a continuous (and so Borel) surjective map.
We now show that the type structure T = hSi ; Ti ; i ii2I satis…es the required properties.
Claim F.2 For each i 2 I,
(Si
Tki ) \ Rik+1 = ;, 8k
0.
Proof: By induction on k 0.
(Base step: k = 0) Fix i 2 I and (si ; ti ) 2 Si Ti , where ti 2 T0i . We clearly have that
(si ; ti ) 62 Ri1 because i (ti ) 2 0i , hence ti is not cautious. Thus (Si T0i ) \ Ri1 = ;.
(Inductive step: k 1) Suppose we have already shown that (Si Tki 1 ) \ Rik = ; for
each i 2 I. Fix i 2 I and (si ; ti ) 2 Si Ti , where ti 2 Tki . Thus i (ti ) = ( 1i ; :::; ni ) 2 ki ,
hence 1i (S i Tk i 1 ) > 0. Since, by the induction hypothesis, (S i Tk i 1 )\Rk i = ;, then
it must be the case that 1i (Rk i ) + 1i (S i Tk i 1 ) 1, and so 1i (Rk i ) < 1. Therefore Rk i
is not weakly believed (and so not assumed) under i (ti ); this implies (si ; ti ) 62 Ai (Rk i ).
Thus, (si ; ti ) 62 Rik+1 .
To conclude the proof, pick any (si ; ti ) 2 Si Ti . Thus there exists k 0 such that
ti 2 Tik . By Claim F.2, it follows that (si ; ti ) 62 Rik+1 . Since Ri1 = \m 0 Rim , this shows
that Ri1 = ;, as required.
28
Appendix G: Self-admissible sets
As we mentioned in the main text (Section 6.6), the notion of Rc CARc in arbitrary type
structures can be characterized in terms of SAS’s as follows.
Theorem G.1 Fix a …nite gamehI; (Si ; ui )i2I i.
Q
(1) For each type structure T = hSi ; Ti ; i ii2I , i2I ProjSi (Ri1 ) is an SAS.
Q
(2) For each SAS Q = i2I Qi S, there exists a …nite type structure T = hSi ; Ti ;
such that
Y
Y
ProjSi (Ri1 ) =
Qi .
i2I
i ii2I
i2I
As mentioned, Theorem G.1 can be proven using arguments which are similar to those
in BFK or Dekel et al. [8]. We now discuss the required modi…cations.
Part (1) is an analogue of Theorem 6.1(i) in Dekel et al. [8]. As in [8], the structure
of the argument follows the proof of Theorem 8.1(1) in BFK. Speci…cally, it is enough
to modify the proof in BFK as follows. First, Lemma 7.2 in BFK must be replaced by
Proposition 4. Then, one should use Lemma 2 and Lemma B.2 in place of Property 6.3
and Lemma D.1 in BFK, respectively. Finally, note that Lemma D.2 in BFK also holds
in our setting, and the remainder of the proof is unaltered.
Part (2) of is an analogue of Theorem 8.1(ii) in BFK. They show the result by constructing an LCPS type structure in which, for every player i, Ti = Qi [ fti g and ti is
an arbitrary label. Di¤erently from BKS’s construction, we need to require that each
ti be a non-cautious type. With this modi…cation, the remainder of the proof in BFK
applies here almost verbatim - the m-level "rationality" sets Rim in BFK must be given
our interpretation, and their version of Assumption implies the one we adopted in the
main text.
29
References
[1] Aliprantis, C.D., and K.C. Border (1999): In…nite Dimensional Analysis.
Berlin: Springer Verlag.
[2] Asheim, G., and M. Dufwenberg (2003): “Admissibility and Common Belief,”
Games and Economic Behavior, 42, 208-234.
[3] Asheim, G., and Y. Sovik (2005): “Preference-based Belief Operators,” Mathematical Social Sciences, 50, 61-82.
[4] Blume, L., A. Brandenburger, and E. Dekel (1991): “Lexicographic Probabilities and Choice Under Uncertainty,”Econometrica, 59, 61-79.
[5] Blume, L., A. Brandenburger, and E. Dekel (1991): “Lexicographic Probabilities and Equilibrium Re…nements,”Econometrica, 59, 81-98.
[6] Brandenburger, A., A. Friedenberg, and H.J. Keisler (2008): “Admissibility in Games,”Econometrica, 76, 307-352.
[7] Cohn, D.L. (2013): Measure Theory, 2nd edition. Boston: Birkhauser.
[8] Dekel, E., A. Friedenberg, and M. Siniscalchi (2016): “Lexicographic Beliefs
and Assumption,”Journal of Economic Theory, 163, 955-985.
[9] Engelking, R. (1989): General Topology. Berlin: Heldermann.
[10] Friedenberg, A., and H.J. Keisler (2011): “Iterated Dominance Revisited,”
working paper.
[11] Kechris, A. (1995): Classical Descriptive Set Theory. Berlin: Springer Verlag.
[12] Keisler, H.J., and B.S. Lee (2015): “Common Assumption of Rationality,”working paper, University of Toronto.
30

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