This work proposes a new definition of common knowledge which extends the exist- ing definitions for situations where agents may have unusual knowledge structures. This new definition maintains the Positive Introspection property even for arbitrarily strange knowledge structures, and reduces to the usual Aumann and Lewis definitions when knowledge structures are sufficiently well-behaved. Specifically, when agents’ knowledge operators are Distributive, then Behavioral common knowledge is the same as Aumann or Lewis common knowledge.
Key to defining Behavioral Common Knowledge is the construction of the set of knowledge questions. These are questions of the form ‘Does agent 1 know that agents
2 and 3 know that the event has occurred’. This structure allows a formal definition for Behavioral Common Knowledge which guarantees that all knowledge questions will be answered by the common knowledge. A set-theoretic construction is devel- oped which allows for easy investigation of the properties of Behavioral Common Knowledge. Alternative ways of defining common knowledge for behavioral agents are discussed, but shortcomings are shown for each.
An area of future research would be to investigate the consequences of using Be- havioral Common Knowledge in games of incomplete information. This may be of particular relevance for developing the appropriate equilibrium concept when agents have poorly behaved information structures.
Another area of future research is to extend the existing definitions and results from finite state space to arbitrary state spaces. If the definitions are used as given, except that the state space is allowed to be infinite, then the results of this Chapter do not hold. In particular, Behavioral Common Knowledge ceases to have Positive Introspection. The inductive definition of the set of knowledge questions would need to be extended to allow for a degree of trans-finite induction.
3.A
Proofs
The appendix contains the remaining proofs. Proof of Proposition 3.1:
LetKj(E∩F) =KjE∩KjF for allj ∈J, E, F ⊂Ω. The everybody knows operator
fsatisfies the Distributive property by Proposition 2.11. AsCeE =
T∞
n=1fnE, fix a
positive integerm ∈ Nand considerfmCeE. As Ωis finite, the setsfnE can take on
only finitely many values asnranges overn ∈N. Therefore
fm ∞ \ n=1 fnE ! = ∞ \ n=1 fmfnE We then have fm ∞ \ n=1 fnE ! = ∞ \ n=1 fmfnE = ∞ \ n=m+1 fnE ⊃ ∞ \ n=1 fnE =CeE
That is, for each m ∈ N, fmC
eE ⊃ CeE. Therefore,
T∞
m=1fmCeE ⊃ CeE and so
To showCe=Cs, letE ∈2Ω. Then CeE = ∞ \ n=1 fnE = ∞ \ n=1 " \ j1∈J Kj1 \ j2∈J Kj2 · · · \ jn∈J KjnE ! · · · !!#
Using the Distributive property to collect the intersection signs,
CeE = ∞ \ n=1 " \ j1∈J \ j2∈J ,· · · , \ jn∈J Kj1Kj2· · ·KjnE # = ∞ \ n=1 " \ j1,...,jn∈Jn Kj1Kj2· · ·KjnE #
By the definition of the set of finite sequenceSeq(J),
CeE = \ s∈Seq(J) K(s)E =CsE as required. Proof of Lemma 3.1:
Let E, F ∈ 2Ω and E ⊂ F. For all j ∈ J,K
jE ⊂ KjF, so
T
j∈JKjE ⊂
T
j∈JKjF.
That is,fE ⊂fF. By induction
E ⊂F =⇒ fE ⊂fF =⇒ f2E ⊂
f2F =⇒ · · · =⇒ fnE ⊂
fnF
In conclusion, the operatorsfn : 2Ω →2Ωsatisfy Monotonicity for everyn∈
N.
Proof of Lemma 3.2:
LetC : 2Ω → 2Ω satisfy assumptions (i) and (ii). From assumption (i) CE ⊂ fE.
Suppose, for the purposes of induction, that CE ⊂ fnE for some n ∈
N and all
E ∈ 2Ω. By the inductive hypothesis on the eventCE, we have CCE ⊂ fnCE. As
eachKj is Monotonic, by Lemma 3.1,fn is Monotonic. Therefore,CE ⊂fE implies
fnCE ⊂fnfE. By assumption (ii),CE ⊂CCE, so overall
CE⊂CCE ⊂fnCE ⊂
fn+1E
By the principle of induction,CE ⊂fnE for alln∈
N; and any operatorCsatisfying assumptions (i) and (ii) will satisfy
CE ⊂
∞
\
n=1
fnE for allE ∈2Ω
Proof of Proposition 3.2:
LetC : 2Ω → 2Ω satisfy assumptions (i), (ii), and (iii). As eachKj is Distributive,
it is also Monotonic. AsCsatisfies assumptions (i) and (ii), by Lemma 3.2,C is more informative thanCe.
SinceCis the maximal operator satisfying (i) and (ii), ifCeE =T
∞
n=1f
nEsatisfies (i)
and (ii) thenC =Ce. By Lemma 2.1,Cesatisfies (i), and by Proposition 3.1Cesatisfies
Positive Introspection andCe=Cs. ThereforeC =Ce =Cs, as required.
Proof of Proposition 3.3:
As there is some agentjE ∈J whereKjEE ⊂E, thenfE ⊂Efor all eventsE ∈2Ω.
The infinite sequence(E, fE, f fE, f f fE, . . .)is non-increasing. AsΩis finite, and thus 2Ω is finite, this sequence eventually reaches a fixed point. Therefore for
sufficiently largeM, the following are equal
CeE = ∞ \ n=1 fnE =fME = ∞ \ n=1 fnfM E = ∞ \ n=1 fn ∞ \ m=1 fmE ! =CeCeE Hence,CeE =CeCeE.
By contrast, let Ω = {a, b, c}, J = {1,2} and K1, K2 be given by K1Ω = {a, b},
K1{a} = ∅,K2Ω = {a, c}, and KjE = E otherwise. Then CsΩ = {a, b} ∩ {a, c} ={a},
whileCs{a} = ∅. ThereforeCsΩ 6⊂ CsCsΩ. In particular,Cs does not satisfy Positive
Introspection.
Proof of Proposition 3.4:
LetΩ ={a, b, c, d, e},J ={1,2}. Define knowledge operatorsK1 andK2 as follows:
K1{a, b}={a, c, d} K2{a, b}={a, c, e}
K1{a, c}={a, b, d} K2{a, c}={a, b, e}
K1{a}=K2{a}=∅
K1E =K2E =E, otherwise.
By the definition of everybody knows,f:
f{a, b}={a, c}, f{a, c}={a, b}, f{a}=∅, f∅=∅
Then, Ce{a, b} ={a}, and Ce{a} = ∅. Hence,Ce{a, b} 6⊂ CeCe{a, b}. Therefore,Ce
For any sequence of agentsKj1,...,jn,1{a, b}={a, c, d}, andKj1,...,jn,2{a, b}={a, c, e}, soCs{a, b} ={a, c}. HoweverKj1,...,jn,1{a, c} ={a, b, d}, andKj1,...,jn,2{a, c} ={a, b, e}, soCs{a, c} = {a, b}. HenceCs{a, b} 6⊂ CsCs{a, b}. Therefore,Cs does not satisfy Posi-
tive Introspection.
Proof of Proposition 3.5:
It is sufficient to provide an example. Let Ω = {a, b, c, d}, and J = {1,2,3}. Let
K1Ω = {a, b, c}, K2Ω = {a, b, d},K3Ω = {a, c, d}, K1{a, d} = ∅, and KjE = E for all
other eventsE ∈2Ωand agentsj ∈J.
Then,fΩ = {a, b, c}∩{a, b, d}∩{a, c, d}={a}, andffΩ = f{a}={a}. Continuing in this fashion givesCeΩ ={a}.
For any sequences= (j1, . . . , jn)∈Seq(J), asKjKj1Ω =Kj1Ωfor allj ∈J, then
KsΩ =Kjn· · ·Kj1Ω =Kj1Ω Hence,CsΩis CsΩ = \ s∈Seq(J) KsΩ =K1Ω∩K2Ω∩K3Ω = {a} However, forK1(K2Ω∩K3Ω) K1(K2Ω∩K3Ω) =K1({a, b, d} ∩ {a, c, d}) = K1{a, d}=∅ Therefore,CeΩ6⊂K1(K2Ω∩K3Ω)andCsΩ6⊂K1(K2Ω∩K3Ω). Proof of Proposition 3.6:
Extend the knowledge model from the proof of Proposition 3.4 by adding elements to the state space untilΩ = {a, b, c, d, e,ea,eb,
e
c}, and J = {1,2}. Define the knowledge operatorsK1 andK2 as K1{a, b}={a, c, d} K2{a, b} ={a, c, e} K1{a, c}={a, b, d} K2{a, c} ={a, b, e} K1{a}=K2{a}=∅ K1{ea,eb,ec}={ea,eb} K2{ea,eb,ec} ={ea,ec} K1{ea}=K2{ea}={ea} K1E =K2E =∅, otherwise.
As noted in the proof of Proposition 3.4,Ce{a, b}={a}, butCs{a, b}={a, c}. Thus,
there exists an eventE such thatCeE (CsE.
Similarly, as K1K1{ea,eb,ec} = K1{a,e eb} = ∅, we have Cs{ea,eb,ec} = ∅. However
Ce{ea,eb,ec}={ea}. Thus, there exists an eventEesuch thatCsEe (CeEe.
Proof of Proposition 3.7:
LetQnbe given as in Definition 3.6, andQe = S∞
n=1Qn.
ClearlyQ1 ⊂Qas for allj ∈ J,(j,∅)∈ Q. Suppose, for the purposes of induction,
thatQn ⊂Q. Letq∈Qn+1. Eitherq∈Qn, orq= (j,{q1, q2, . . . , qn})where eachqi ∈Qn.
Ifq∈Qnthen by the inductive hypothesisq∈Q. Ifq= (j,{q1, q2, . . . , qn})then by the
inductive hypothesis eachqi ∈ Q, so by assumption Q.2,q ∈ Q. ThereforeQn+1 ⊂ Q.
By the induction principle, eachQn⊂Q, so
e Q= ∞ [ n=1 Qn⊂Q Now we showQe = S∞
n=1Qnsatisfies requirements Q.1 and Q.2. AsQ1 =J× {∅}, for
allj ∈J then(j,∅)∈ S∞
n=1Qn. Letj ∈J, andq1, . . . , qm ∈
S∞
n=1Qn. Eachqi is in some
Qi, so writeqi ∈ Qli. As the sequence(Qn)n∈N is non-decreasing,qi ∈ Qmaxklk for all i.
Therefore,q= (j,{q1, q2, . . . , qn})∈Q1+maxklk, and in particular,q ∈
S∞
n=1Qi.
As S∞
n=1Qn ⊂ Q, and
S∞
n=1Qn satisfies the requirements for the set of knowl-
edge questions, it must be the minimal set which satisfies these properties. Therefore
Q=Qe.
Proof of Lemma 3.3:
Let F ∈ Sn−1[E], then F = F ∩F ∈ Sn[E]; so Sn−1[E] ⊂ Sn[E]. Therefore the
sequence(Sn[E])n∈Nis non-decreasing.
Proof of Proposition 3.8:
First we showCEe ⊂CBEfor allE ∈2Ω. Define the length of knowledge question
q, denoted|q|, by|(j,∅)|= 1, and forq= (j,{q1, . . . , qn}), by
|q|= 1 +
n
X
i=1
We proveCEe ⊂CBEby induction over the length of knowledge questions. Suppose
|q|= 1. Then,q = (j,∅)for somej ∈J, andKqE =KjE ∈S1[E].
Now suppose, for the purposes of induction, that for all questions r of length m
or less, KrE is in Sk[E] for some k. Let |q| = m + 1. As q = (j,{q1, q2, . . . , qn}) for
j ∈ J andqi ∈ Q, and |q| = 1 +Pi|qi|, then |qi| ≤ m for each i. Therefore, by the
inductive hypothesis,KqiE ∈Ski[E]for someki. From Lemma 3.3,(Sk[E])k∈Nis a non-
decreasing sequence, soKqiE ∈SM0[E]for alli∈ {1, . . . , n}, whereM0 = maxiki. Then
T
iKqiE ∈ SM0+n−1[E]as SM0+n−1[E] contains all intersections of n events in SM0[E].
LetM =M0+n−1. Therefore:
KqE =Kj(inf(Kq1E, . . . , KqnE)∈SM[E]
By the principle of induction, for every knowledge questionq,q∈SM[E]for someM.
Finally, as CBE = inf q∈QKqE and CEe = ∞ \ M=1 \ F∈SM[E] F Therefore,CEe ⊂CBE.
Conversely, we show CBE ⊂ CEe for all E ∈ 2Ω, again using induction over the
sequence(Si[E])i∈N. LetF ∈S1[E]. ThenF =KjEfor somej ∈J, soF can be written
as the infimum of a set of knowledge questions, in this case the singleton set{(j,∅)}. Suppose, for the purposes of induction, that for allG∈ Sm[E], there exists a set of
knowledge questions{q1, . . . , qn}such thatG= inf(Kq1E, . . . , KqnE). LetF ∈Sm+1[E]. There are two cases: (i)F =KjF0 for somej ∈ J andF0 ∈ Sm[E], or (ii)F = F1∩F2
forF1, F2 ∈Sm[E]. In case (i), asF0 ∈Sm[E], by the inductive hypothesis, there exists a
set of knowledge questions such thatF0 = inf(Kq1E, . . . , KqnE). Therefore
F =KjF0 =Kj(inf(Kq1E, . . . , KqnE)) = KqE
where q = (j,{q1, q2, . . . , qn}). The event F is the infimum over a set of knowledge
questions, in this case just the singleton set {q}. For case (ii), as F1, F2 ∈ Sm[E], by
the inductive hypothesis, F1 = inf(Kq1E, . . . , KqnE) and F2 = inf(Kr1E, . . . , Krn0E).
Therefore:
EventF can be written as the infimum of a set of knowledge questions. By the prin- ciple of induction, for allmand for every elementF ∈Sm[E],F can be written as the
infimum of a set of knowledge questions. Therefore, the intersection over all suchF
will be larger than the infimum over all knowledge questions, which isCBE ⊂CEe .
Proof of Proposition 3.9:
Chooseq∗ ∈ Q\Qb. For ease of notation, letCb = infq∈
b
QKq. To showCb 6= CB, it
is sufficient to construct a knowledge model (Ω, J,{Kj}j∈J) such that CΩb 6= ∅, and
Kq∗Ω = ∅.
LetQn be given as in Definition 3.6, and define the depth ofq ∈ Qas the unique
value n ∈ N such that q ∈ Qn \ Qn−1. Let m = depth(q∗), M = |Qm|, and Ω =
{1, . . . , M + 1}. Letσ : Qm → {1, . . . , M}be bijective such that depth(q) < depth(q0)
impliesσ(q)< σ(q0), andσ(q∗) =M.
We want to define a collection of operators Kj, and thereby operatorsKq, so that
M+ 1∈KqΩfor allq6=q∗, andKq∗Ω = ∅. OperatorsKqare first defined recursively for
questions of depth no greater thanm, then defined recursively for operators of depth greater thanm.
First, for eachq∈Qm\ {q∗}, writeq= (j,{q1, . . . , qn}). If{q1, . . . , qn}=∅, then let
KqΩ =KjΩ = Ω\ {σ((j,∅))}= Ω\ {σ(q)}
Otherwise, let
Kj(Ω\σ({q1, . . . , qn})) ={1, . . . , σ(q)−1, σ(q) + 1, . . . , M + 1}= Ω\ {σ(q)}
Running through the recursion gives
KqΩ ={1, . . . , σ(q)−1, σ(q) + 1, . . . , M + 1}= Ω\ {σ(q)}
Also let
Kq∗Ω = ∅, and Kj∅={1, . . . , M −1, M + 1} ∀j ∈J
Now, for eachq /∈ Qm, writeq = (j,{q1· · ·qn}). If qi =q∗ for somei, thenKqΩ =Kj∅
so is already defined. Otherwise, recursively defineKqΩsuch that|KqΩ|=M, and the
element ofΩnot inKqΩis the largest number not in one ofKq1Ω, . . . , KqnΩ. That is
KqΩ =Kj(Kq1Ω∩ · · · ∩KqnΩ) = Ω\ max
For everyq∈Q\ {q∗}, we haveM + 1∈KqΩ. So
b
CΩ = \
q∈Qb
KqΩ⊃ {M + 1}
However,Kq∗Ω =∅so thatCBΩ =∅. Therefore,CB 6=Cbas required.
Proof of Proposition 3.10:
Fix an eventE ∈2Ω. The sequence(Sn[E])n∈Nis a non-decreasing sequence, from
Lemma 3.3. As Ωis finite, this sequence will eventually reach a fixed point. That is, there existsM ∈Nsuch thatSN[E] =SM[E]for allN ≥M. Therefore
CBE = ∞ \ n=1 \ F∈Sn[E] F = M \ n=1 \ F∈Sn[E] F = \ F∈SM[E] F
AsSM[E] =SM+1[E], the setSM[E]is closed under intersections. Therefore
CBE = inf(SM[E])∈SM[E]
The sequence(Sn[CBE])n∈Nis also non-decreasing, so there existsM
0 ∈
Nsuch that
SN0[CBE] =SM0[CBE]for allN0 ≥M0. ThereforeCBCBE = inf(SM0[CBE]).
AsSM0[CBE]is formed by taking repeated applications of intersections and opera-
torsKj, and as{CBE} ⊂SM[E], thenSM0[CBE]⊂SM+M0[E] =SM[E]. AsSM0[CBE]⊂
SM[E], then
CBCBE = inf(SM0[CBE])⊃inf(SM[E]) =CBE
Hence,CBE ⊂CBCBE, as required.
Proof of Proposition 3.11:
Part (i): First we show CBE ⊂ CeE. By construction, KjE ∈ S1[E], for allj ∈ J.
Therefore,Kj1E∩Kj2E ∈S2[E], and so on up tofE =
T
j∈JKjE ∈ S|J|[E]. Similarly,
fnE ∈S
n|J|[E]for alln∈N. Therefore
CBE = ∞ \ n=1 \ F∈Sn[E] F ⊂ ∞ \ n=1 fnE =CeE
Next we show CBE ⊂ CsE. Let s = j1, j2, . . . , j|s| ∈ Seq(J). As Kj|s|E ∈ S1[E], and
Kj|s|−1Kj|s|E ∈S2[E], and so on, up toKj1· · ·Kj|s|E ∈S|s|[E], therefore
CBE = ∞ \ n=1 \ F∈Sn[E] F ⊂ \ s∈Seq(J) K(s)E =CsE
as required.
Part (ii): We want to show that if Kj(F ∩G) = KjF ∩KjG for all agents j ∈ J
and events F, G ∈ 2Ω, then CBE = CeE = CsE. By Proposition 3.1, it is enough
to show CBE = CsE. We already have CBE ⊂ CsE, so only need CsE ⊂ CBE.
Consider a knowledge questionq = (j,{q1, q2, . . . , qn}). As each individual knowledge
operator Kj is Distributive, KqE = KjKq1E ∩ · · · ∩KjKqnE. Define the length of q by |q| = 1 +Pn
i=1|qi| as in the proof of Proposition 3.8. Suppose, for the purposes
of induction, that for each knowledge question r of length at most m, there exists some A ⊂ Seq(J) such thatKrE = Ts∈AK(s)E. Let |q| = m + 1. Then there exists
A1, . . . , An⊂Seq(J)such that
KqE =KjKq1E∩ · · · ∩KjKqnE =Kj \ s∈A1 K(s)E ! ∩ · · · ∩Kj \ s∈An K(s)E ! = \ s∈A1 KjK(s)E ! ∩ · · · ∩ \ s∈An KjK(s)E ! = \ s∈(A1∪···∪An) KjK(s)E
There existsA=A1∪ · · · ∪An⊂Seq(J)such thatKqE =
T s∈AK(s)E. Therefore KqE = \ s∈A K(s)E ⊃ \ s∈Seq(J) K(s)E =CsE
for each knowledge questionq, and thus
CBE =
\
q∈Q
KqE ⊃CsE
Therefore,CBE =CsE =CeE for any eventE ∈2Ω, as required.
Part (iii): It is sufficient to provide an example. LetΩ ={a, b, c},J ={1,2}. Define knowledge operator K1 by K1{a, b} = {a, c}, K1{a, c} = {b, c}, K1{a} = K1{b} =
K1{c}=∅, andK1E =Efor all other eventsE ∈2Ω. LetK2 =K1.