REVEALED PREFERENCE FOR
MAXIMIZING CONDITIONAL EU
FATEMEH BORHANI AND EDWARD J. GREEN
PENNSYLVANIA STATE UNIVERSITY
1. Introduction
This paper concerns whether or not Bayesian decision theory has any revealed-preference implications regarding the asymptotic behavior of a rational agent. Specifically, it concerns ideas that
might be thought to follow from two features of Bayesian statistical theory.
(1) A premise of the Bayesian theory—and presumably of all theories of rational decision under
uncertainty—is that an agent does not “throw information away” after having made relevant
observations. Bayesian statistics models the agent’s prior belief, and the agent’s posterior
belief at each stage after having made a new observation, as probability measures.
(2) A theorem of Bayesian statistics is that, with respect to the prior probability measure over
possible sequences of observations, the agent’s posterior belief converges in probability, and
almost surely, to some probability-measure-valued random variable.1
In view of these features of the theory, it is intuitive to suppose two things about the asymptotic
behavior of an agent has repeated opportunities to revise a choice of an action after making new
observations, and who maximizes posterior expected utility.
(1) Eventually, the chosen action will not depend solely on the most recent observation, since the
cumulative information from past observations outweighs the incremental information from
the most recent observation. That is, for any possible value that a single observation might
take, there are two sequences of observations, both of which terminate with observations
having that value but following which the agent makes distinct action choices.
(2) Along (almost) any infinite sequence of observations, the agent’s choice of action after finite
initial sequences will eventually “settle down” to a constant action, since the agent is maximizing expected utility with respect to a posterior probability measure that is converging
to a limit.
However, despite their intuitive plausibility, it will be shown here that neither of these suggestions
is an implication of Bayesian decision theory. In fact, the theory has no implication specifically
about the agent’s asymptotic behavior. Its only restriction on that behavior is an elementary one,
immediately implied by Savage’s sure-thing principle, concerning how the chosen action can evolve
after a single incremental observation has been made.
2. A fragment of formal decision theory
2.1. The situation to be modeled. Consider an agent who, at each stage t ∈ N+ makes an
observation having value xt ∈ X.2 At some stage t ∈ N, and after having observed the values
Date: 2014.11.18 (Preliminary and incomplete draft).
1
This result is often called “convergence to truth.” However, if the observations are not completely informative
about the truth, then the limiting probability will not necessarily have one-point support. Moreover, the informal
statement of the result that we make here has implicit premises, notably that an observation can take only finitely
many possible values.
2N = {0, 1, 2 . . . } and N = {1, 2, 3 . . . }.
+
1
x1 , . . . , xt , the agent must take an action a from a set A. (The action may have to be taken at stage
0, before having made any observation.) We have in mind a situation in which the observations could
potentially influence the agent’s preference among available actions, but in which that preference
does not depend on the stage at which the action must be taken.
In the formal theory, we will assume that
X and A are finite sets.
(1)
2.2. Observation
sequences and contingent plan. Define the set of observation sequences,
S
∗
t
X = t∈N X , and define λ: X ∗ → N by λ(~x) = t ⇐⇒ ~x ∈ X t . Let ∅ denote the null sequence
(defining λ(∅) = 0) and let ~x ~y denote that ~y is an extension of ~x and let ~x ≺ ~y denote that ~y is
a proper extension of ~x. If ~x = hx1 , . . . , xn i, then for 1 ≤ t ≤ n, ~xt denotes xt . The fact that X ∗ is
a countable set will be invoked, without explicit mention, in proofs below.
The immediate extensions of ~x ∈ X ∗ consist of ~x followed by a single observation. Let ~x ∗ y
denote the immediate extension of ~x by y.3
The usual representation of observations in terms of measurable functions is adopted here. A
measurable space (Ω, B) is constructed from X by taking Ω = X N+ and Xt (ω) = ωt , and by
taking B to be the smallest σ-algebra with which all of the projection functions Xt are measurable.
Bt ⊆ B is the σ-algebra (a finite Boolean algebra, given the assumption that X is finite) generated
by X1 , . . . , Xt . In particular, every set
B ∗ (~x) = {ω|∀t ≤ λ(~x) Xt (ω) = ~xt }
(2)
is measurable with respect to B and to Bt for λ(~x) ≤ t. If ∀t ≤ λ(~x) Xt (ω) = ~xt , then B ∗ (~x) =
B(ω, λ(~x)). Define φ: Ω × N → X ∗ and α∗ : Ω × N → A by
ξ(ω, 0) = ∅ and, for t > 0, ξ(ω, t) = hX1 (ω), . . . , Xt (ω)i .
(3)
∗
It will be convenient to enlarge the domain of ξ(,t)o Ω ∪ X by defining
ξ(~x, 0) = ∅ and, for t > 0, ξ(~x, t) = hX1 (~x), . . . , Xmin{t,λ(~x)} (~x)i .
(4)
Define
~x ≺ ω ⇐⇒ ξ(ω, λ(~x)) = ~x.
A (single-valued) contingent plan is a correspondence α: X ∗ → A. Define
α∗ (ω, t) = α(ξ(ω, t)) .
(5)
(6)
2.3. Bayesian rationality. The Bayesian concept of rationality interprets the agent as having
beliefs represented by a probability space (Ω, B, P ) and a bounded, state-contingent utility function
u: A × Ω → <, and interprets the observations xt as the values of random variables Xt : Ω → X. P
is assumed to be countably additive.
A Bayesian-rational agent almost surely (that is, after any sequence of observations that has
positive probability according to the agent’s prior beliefs) chooses an action that maximizes the
posterior expected utility
R
R
B ∗ (ξ(ω,t)) u(a, ψ) dP (ψ)
B(ω,t) u(a, ψ) dP (ψ)
=
,
(7)
U (a, ω, t) =
P (B(ω, t))
P (B ∗ (ξ(ω, t)))
where
B(ω, t) = {ψ ∈ Ω|∀s ≤ t Xs (ψ) = Xs (ω)} = B ∗ (ξ(ω, t)) .
(8)
Define
R
B ∗ (~
x) u(a, ψ) dP (ψ)
U ∗ (a, ~x) =
(9)
P (B ∗ (~x))
3In general, let ~
x∗~
y be the concatenation of sequences ~
x and ~
y . Then ~
x ∗ y is shorthand for ~
x ∗ hyi.
2
so that, if P (ξ(ω, t)) > 0, then
U (a, ω, t) = U ∗ (a, ξ(ω, t)) .
(10)
Intuitively, a Bayes contingent plan is one that might be followed by an agent who maximizes
posterior objective utility, conditional on any sequence of observations. Formally, a contingent plan
α is Bayes for hΩ, B, P, {Xt }t∈N+ i (or simply Bayes for P ) if α if it satisfies, for all ~x ∈ X ∗ ,
h
i
∗ 0
P (B ∗ (~x)) > 0 =⇒ ∀a ∈ A a = α(~x) ⇐⇒ U ∗ (a, ~x) = max
U
(a
,
~
x
)
(11)
0
a ∈A
with respect to some von Neumann-Morgenstern utility function and to that stochastic process.
Call α Bayes if it is Bayes for some P , and Bayes for u and P if u induces U , which satisfies (11)
for α and P .
Stochastic processes for which every observation sequence occurs with positive probability will
be of special interest. That is, statements of some results will involve the condition that
∀~x ∈ X ∗ P (B ∗ (~x)) > 0 .
(12)
Conceptually, condition (12) ensures that Bayes rule is always applicable.4
3. Nearly sound conjectures
3.1. The conjectures are not vacuous. At the end of section 1, it was conjectured that Bayes
contingent plans might have two properties.
Conjecture 1. A Bayes contingent plan depends non-trivially on past observations, as well as on
the current observation, after some observation sequences.
Conjecture 2. Almost surely, a Bayes contingent plan is asymptotically constant (if, as is assumed
in this paper, there are only finitely many feasible actions).
It is obvious that there are contingent plans that violate each of those conjectures. For example,
let X = {1, 2} and A = {0, 1, 2}, and define
α1 (∅) = α2 (∅) = 0 and
λ(~
x)
α1 (~x) = ~xλ(~x) and α2 (~x) ≡
X
xt
(mod 3).
(13)
t=1
Contingent plan α1 clearly violates conjecture 1. Also α2 violates conjecture 2. If ~x ∈ Ω and y ∈ X,
then α2 (~x ∗ y) ≡ α2 (~x) + y (mod 3) so α2 (~x ∗ y) 6= α2 (~x).
3.2. Qualified versions of the conjectures are sound. When qualified by some hypotheses,
the two conjectures become sound propositions.
Proposition 1. Suppose that f : X → A and contingent plan α satisfy
∀~x ∀y α(~x ∗ y) = f (y)
(14)
and that, for some observation value y such that f (y) 6= α(∅),
P (∃t Xt = y) = 1.
(15)
Then α is not Bayes for P
4Mathematically, this condition states that P has full support with respect to the product topology of X N+ , since
the sets B ∗ (~x) are a sub-base of that topology.
3
Proof. Define Y = {~x | P (B ∗ (~x)) > 0 and ~xλ(~x) = y and
∀t < λ(~x) ~xt 6= y}. RDefine Y ∗ =
R
S
∗ x). Then, by (15), P (Y ∗ ) = 1. By condition (11),
~
x∈Y B (~
B ∗ (~
x) u(f (y), ω) dP (ω) > B ∗ (~
x) u(α(∅), ω) dP (ω)
∗
∗
for every ~x ∈ Y . Note that, if w
~ 6= ~x and w
~ ∈ Y and ~x ∈ Y , then B (w)
~ ∩ B (~x) = ∅.
Therefore, since integrals of the L
functions
u(f
(y),
ω)
and
u(α(∅),
ω)
with
respect to P deR1
R
∗
fine
Y ∗ u(f (y), ω) dP (ω) > Y ∗ u(α(∅), ω) dP (ω). Since P (Y ) = 1,
R countably additive measures,
∗
(f (y), ∅) and
RY ∗ u(f (y), ω) dP (ω) = U
∗
∗
∗
Y ∗ α(∅), ω) dP (ω) = U (α(∅), ∅). Thus U (f (y), ∅) > U (α(∅), ∅), contradicting condition (11).
Proposition 2. Let B ∈ B and C ∈ B be defined, for some a∗ ∈ A, by
B = {ω | ∀a 6= a∗ [u(a∗ , ω) > u(a, ω)]} and C = {ω | ∀s ∈ N ∃t > s [α∗ (ω, s) 6= a∗ ]} .
(16)
Suppose that
P (B ∩ C) > 0.
(17)
Then α is not Bayes for P .
This proposition will be proved in section 8
3.3. Discussion. Proposition 1 and proposition 2 qualify conjecture 1 and conjecture 2, respectively, in ways that make them sound. Condition (15) of proposition 1 shows that if observations
are dichotomous, for example, then the agent would have to attribute positive prior probability to
the event of making identical observations forever, in order for contingent plan α1 (which satisfies
condition (14)) to be Bayes with respect to prior beliefs. Condition (17) of proposition 2 shows
that the agent would have to believe that almost surely several actions will be tied for being the
best, in order for contingent plan α2 to be Bayes with respect to prior beliefs.
The hypotheses of both of these results apply straightforwardly to canonical examples of decision
making according to posterior-expected-utility maximization, such as the exchangeable-distribution
model of a risk-averse agent’s decision whether to bet on Heads, on Tails, or not at all, on the toss
of a coin of unknown bias, after a finite sample of tosses has been observed.
Nevertheless, the hypotheses of these results that rule out divergent sequences of posteriorEU-maximizing actions are considerably more restrictive than those of the corresponding results
that rule out divergent sequences of posterior beliefs for a Bayesian agent. At best, the results
presented in section 3.1 only suggest asymptotic properties that Bayes contingent plans will possess
under conditions that are ad hoc, albeit typically satisfied. These results do not identify intrinsic
properties that are dictated by the Bayesian decision criterion itself. In fact, the remainder of this
paper is devoted to showing that there are no such intrinsic asymptotic properties.
4. Consistent contingent plans
Define α to be consistent at ~x iff
∀a ∈ A \ {α(~x)}
∃y ∈ X
α(~x ∗ y) 6= a .
(18)
Consistency is an obvious revealed-preference implication of the sure-thing principle formulated by
Savage-1954.5 Therefore, a Bayesian agent must consistent.
If there is a finite bound on the length of observation sequences, then the converse is true: for
every P , a finite contingent plan that is consistent at every node must be Bayes for P and some
u. GreenPark-1996 prove this fact, in the course of proving theorem 1 in that paper. (Antecedent
investigations of the relationship between being consistent and being Bayes include Weller-1976,
5In the formal theory introduced by Savage, which has been adopted by most subsequent researchers, acts are
defined entities. Despite that difference from the present theory (in which A is a primitive of the theory), clear
parallels can be drawn.
4
Hammond-1988, and EpsteinLeBreton-1993.) However, they asserted a result that encompasses
both finite- and infinite-horizon plans, and the assertion is false as it applies to infinite-horizon
plans. (Green and Park’s proof makes the unsound assumption that an a.s. convergent martingale
must always converge also in L1 .)
Proposition 6 in the present paper, to be stated and proved below, is a sound weakening of
Green and Park’s theorem 1. The lemmas following the next definition will be used in the proof of
proposition 6.
Define
C(ω, t) = {α∗ (ω, λ(~x) + n) | n ∈ N}
D(a, ~x) ={ω | ~x ≺ ω and ∀n ∈ N α∗ (ω, λ(~x) + n) 6= a}
(19)
={ω | ~x ≺ ω and a ∈
/ C(ω, λ(~x))}
Lemma 3. If α is consistent and a 6= α(~x), then D(a, ~x) 6= ∅.
Proof. A recursive construction, using (5) at each stage n > 0, establishes that there is a sequence
~x0 , ~x1 , . . . such that ~x0 = ~x and, for every n ∈ N, λ(~xn ) = λ(~x) + n and ~xn ≺ ~xn+1 and a 6= α(~xn ).
Then, defining ωt = x ⇐⇒ ∃n ~xnt = x produces an element of Ω that satisfies the lemma.
The next lemma follows immediately from lemma 3.6
Lemma 4. If α is consistent and there exists a function d: A × X ∗ → Ω ∪ {∅} such that, for every
~x,
d(~x, α(~x)) = ∅ and ∀a 6= α(~x) d(a, ~x) ∈ D(a, ~x) .
(20)
Ψ = {d(a, ~x) | a 6= α(~x)} .
(21)
Define Ψ ⊆ Ω by
Ψ is countable, since it is a subset of the range of a function with a countable domain. Therefore
Lemma 5. There is a countably additive probability measure P on Ω that satisfies
∀ψ ∈ Ψ P ({ψ}) > 0 .
(22)
Condition (44) entails the full-support condition (12).
5. The main result
From lemmas 3–5, we will prove the following result.
Proposition 6. If α is consistent, then there are a bounded utility function u and a countably
additive probability measure P such that α is Bayes for u and P , and every finite sequence of
observations receives positive probability according to P .
6It can be proved that, for each ~
x, at most one of the sets D(a, ~
x) (for a 6= α(~
x)) can be finite. It would be
straightforward to define d in lemma 4 so that
If ~
y≺~
x ≺ d(a, ~
y ) and a 6= α(~
x), then d(a, ~
x) = d(a, ~
y)
would also be satisfied.
5
1
2
1
1
1
2
1
2
1
2
Figure 1. ω 8 = (1, 2, 2, 1, 1, 1, 2, 1)
6. Leading example:
Set of signals: X = {1, 2}
Action set: A = {0, 1, 2}
A typical element in Ω = {1, 2}N is shown by ω = (ω1 , ω2 , ...)
ω t = (ω1 , ω2 , ..., ωt ) is the initial segment of ω
At each t, the agent has observed (and recalls) ω t .
ω t ∗ i = (ω1 , ω2 , . . . , ωt , i)
A contingent plan assigns P
an action to each finite sequence ω t
t
Contingent plan: α(ω ) = t0 ≤t ωt0 (mod 3) 7
0
1
2
2
1
0
2
1
0
2
1
0
0
2
1
2
1
0
1
1
1
2
2
2
0
0
0
1
1
2
1
0
2
0
1
2
0
1
2
2
1
2
0
1
2
0
1
0
2
1
2
0
1
2
0
1
1
0
1
2
0
1
2
0
2
1
0
1
2
0
1
2
0
0
2
0
1
2
0
1
2
1
0
2
0
1
2
0
2
0
2
1
0
2
1
0
2
1
0
2
1
0
Figure 2. Suggested actions by α at ∅, ω 1 , ω 2 , . . .
7This is the same contingent plan that was denoted by α in equation (13).
2
6
α(ω t ), α(ω t ∗ 1) and α(ω t ∗ 2) are distinct values from one another.
Therefore for any ω chosen actions never settle down.
But the beliefs of a Bayesian agent typically converge.
So, isn’t behavior according to α evidence against the agent being Bayesian rational?
Contingent plan α is consistent at ~x iff
∀a ∈ A \ { α(~x)}
∃y ∈ X
α(~x ∗ y) 6= a
α is consistent.
By proposition 6, there is a probability/utility framework that rationalizes α.
The key to rationalizing α is to find a set of states Ψ which satisfies three properties. Subsequently
we will assign positive probabilities, summing to one, to the elements of Ψ.
• Ψ is countable.
• For any finite initial sequence of observation, there exists some ψ ∈ Ψ which is not in
contradiction with observations.
• For any ω ∈ Ψ at least one action is chosen by α only finitely many times along ω.
– We construct ω by recursively applying consistency.
– τ (ω) indicates the first time that some action is never chosen at it or later along ω.
Define Ψ to be the set of states that, after some finite time, oscillate with period 2 (zigzag forever
between 1 and 2).
Ψ = {ω ∈ Ω|∃t ∈ N ∀t0 ≥ t ωt0 6= ωt0 +1 }
(23)
Let τ : Ω → N specify the date at which each state in Ψ becomes periodic.
τ (ω) = min{t|∀t0 > t
ωt0 6= ωt0 +1 }.
(24)
Once a state begins to oscillate, α also begins to oscillate between only 2 of the 3 actions in A.
The third action will never be picked by α and this makes this definition of τ the same as previous
slide.
Let set C(ψ) be such that for ψ ∈ Ψ, it consists of the two actions that are picked by α at time
τ (ψ) and later:
X
X
C(ψ) = {
ωt (mod 3),
ωt (mod 3)}.
(25)
t≤τ (ω)
t≤τ (ω)+1
For every finite signal sequence, to define the set of states in Ψ which begin with that sequence,
and oscillate immediately thereafter.
We are going to assign a high probability to every state in this set, conditional on the sequence
being observed.
Let B ∗ (ω t ) be the cylinder set defined as
B ∗ (ω t ) = {ζ ∈ Ω|ζ t = ω t }.
(26)
Define Ψ̂(ω t ) to be the set of states in Ψ such that their initial part be equal to ω t and begin to
zigzag at t or before it.
Ψ̂(ω t ) = {ψ ∈ Ψ ∩ B ∗ (ω t )|τ (ψ) ≤ t}.
For any
ωt
the set
Ψ̂(ω t )
contains only two states and
\
C(ζ) = {α(ω t )}.
(27)
(28)
ζ∈Ψ̂(ω t )
Pick δ ∈ (0, 21 ) and consider additive probability function and state dependent utility function
which satisfy
7
0
1
2
2
1
0
2
1
0
2
1
0
1
0
2
1
2
1
1
0
0
0
1
0
1
1
2
0
0
2
ω
1
0
2
1
2
1
0
0
1
2
2
1
2
0
2
0
0
1
0
2
1
2
1
2
2
0
2
1
0
1
0
1
2
2
1
2
2
1
0
2
0
2
1
0
1
0
2
1
2
0
2
0
2
0
1
1
0
2
0
2
1
0
2
1
ω0
0
2
1
0
2
1
1
0
Figure 3. τ (ω) = 7, C(ω) = {0, 1}, τ (ω 0 ) = 8, C(ω 0 ) = {0, 2}
P ({ω}) = κδ τ (ω) 1Ψ (ω)
(29)
u(a, ψ) = 1C(ψ) (a).
(30)
Every state in Ψ has positive probability.
Therefore every finite initial sequence of observations is observed with positive probability.
For every ω t , p(Ψ̂(ω t )) > p(B ∗ (ω t ))/2.
After having observed ω t , agent assigns a very high probability to each of the two states in ψ̂(ω t ).
α(ω t ) is the only action with highest possible utility in both states in ψ̂(ω t ).
Probability assigned to any state in ψ̂(ω t ), is higher than the summation of probabilities assigned
to all the still possible states outside this set.
The maximum utility attainable in any state is one.
Therefore alpha is the unique expected utility maximizer.
So (Ω, B∗, P ) and u rationalize α as a Bayes contingent plan.
7. Sketch of Proof of Proposition 6
The proof generalizes the preceding example. Set Ω = X N+ .
First we define Ψ a subset of Ω which has three properties:
• It is countable.
• For any finite initial sequence of observation, there exists some ψ ∈ Ψ which is not in
contradiction with observations.
• For any ω ∈ Ψ at least one action is chosen only finitely many times along ω.
– τ (ω) indicates the first time that an action is never chosen at it or later along ω.
(These are the features of Ψ that we used to construct a Bayes representation for the example.)
Lemma 3 to lemma 10 in the paper are used in defining Ψ and proving these properties.
Given the first two properties of Ψ, we define
• a probability measure such that every finite initial sequence has positive probability of being
observed
• a bounded state dependent utility function
8
Probability measure
• P (Ψ) = 1.
• Any state in Ψ has positive probability.
• Every finite initial sequence of observations is observed with positive probability.
• P assigns probability to states according to their index τ (ω).
• States with smaller index are assigned a much higher probability.
State dependent utility function
1 if a ∈ C(ω) and ω ∈ Ψ
u(a, ω) =
0 otherwise.
(31)
Set C(ω) consists of all actions chosen along ω but the action(s) which is abandoned first.
We will show that α is Bayes for P and u.
7.1. Step 1: defining Ψ ⊆ Ω. Define
C(ω, t) = { α(ω t+n )|n ∈ N+ }
(32)
D(a, ~x) = {ω|~x ≺ ω, a 6= C(ω, t)}
(33)
Lemma 7. If α is consistent and a 6= α(ω, t), then D(a, ~x) 6= ∅.
Proof. A recursive construction, using consistency at each stage n > 0, establishes that there is a
sequence ~x0 , ~x1 , . . . such that ~x0 = ~x and, for
every n ∈ N, λ(~xn ) = λ(~x) + n and ~xn ≺ ~xn+1 and a 6= α(~xn ). Then, defining ωt = x ⇐⇒
∃n ~xnt = x produces an element of Ω that satisfies the lemma.
Lemma 8. If α is consistent and there exists a function d: A × X ∗ → Ω ∪ {∅} such that, for every
~x,
d(α(~x), ~x) = ∅ and ∀a 6= α(~x) d(a, ~x) ∈ D(a, ~x) .
(34)
Define Ψ ⊆ Ω by
Ψ = {d(α(~x), ~x) | a 6= α(~x)}
(35)
τ (ψ) = min{ t ∈ N|for some a 6= α(ψ t ), a ∈
/ C(ψ t ).}
(36)
Define τ : Ψ → N that satisfies
Define Ψ̂(ω t ) ⊆ B ∗ (ω t ) ∩ Ψ by
Ψ̂(ω t ) = {ψ ∈ Ψ ∩ B ∗ (ω t )|τ (ψ) ≤ t}.
Note that equivalently we can restate Ψ̂(ω t ) as
[ [
Ψ̂(ω t ) =
0
{ d(a, ω t )}.
(37)
(38)
t0 ≤t a6=α(ω t0 )
therefore Ψ̂(ω t ) is finite and
(|A|−1) ≤ |Ψ̂(ω t )|≤ (1 + t) × (|A|−1).
(39)
Lemma 9. Ψ is countable.
Proof. Ψ is a subset of the range of a function with a countable domain therefore it is countable. 9
Lemma 10.
\
C(ψ τ (ψ) ) = α(ω t )
(40)
ψ∈Ψ̂(ω t )
Proof. Please note that
∀ψ ∈ Ψ̂(ω t )
\
α(ω t ) ∈ C(ψ τ (ψ) ),
(41)
C(ψ τ (ψ) ) = α(ω t ).
(42)
{ψ∈Ψ̂(ω t )|τ (ψ)=t}
Result immediately follows from the above two equations.
7.2. Step 2: probability measure and utility function. Since Ψ is countable and every Ψ̂(ω t )
is nonempty and finite, and all information partitions are countable, we can specify a countably
additive probability measure P such that
P (Ψ) = 1,
(43)
∀ψ ∈ Ψ P ({ψ}) > 0
(44)
and
X
P ({ζ}) = δ × min {P ({ω})} > 0
(45)
Ψ̂(ω t )
ζ∈Ψ∩B ∗ (ω t )\Ψ̂(ω t )
for δ ∈ (0, 1).
Note that condition (44) entails that every finite sequence of initial observations has positive
probability.
Define bounded utility function
1 if a ∈ C(ψ τ (ψ) ) and ω ∈ Ψ
u(a, ω) =
(46)
0 otherwise.
Next we show that the consistent contingent plan, α, is Bayes with with respect to P and u so
defined.
7.3. Step 3: rationalization.
Lemma 11. If P and u satisfy conditions (44) and (46), and a 6= α(ω, t), then
U (a, ω, t) < U (α(ω t ), ω, t) .
(47)
where U is derived from uδ according to
R
U (a, ω, t) =
B ∗ (ω t ) u(a, ψ) dP (ψ)
B ∗ (ω t )
Proof. Maximum gain of deviating from α(ω t ):
X
P ({ζ}) × 1 = δ × min {P ({ω})}
(48)
(49)
Ψ̂(ω t )
ζ∈Ψ∩B ∗ (ω t )\Ψ̂(ω t )
Minimum loss of deviating from α(ω t ):
min {P ({ζ})} × 1
(50)
ζ∈Ψ̂(ω t )
By comparing the above two equations we can see that the maximum gain is strictly smaller
than the minimum loss.
Proposition 6 follows immediately from lemma 11.
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8. Proof of Proposition 2
Here we prove proposition 2:
Let B ∈ B and C ∈ B be defined, for some a∗ ∈ A, by
B = {ω | ∀a 6= a∗ [u(a∗ , ω) > u(a, ω)]} and C = {ω | ∀s ∈ N ∃t > s [α∗ (ω, s) 6= a∗ ]} .
Proof. Suppose that α is Bayes for P . A contradiction will be derived.
R: the smallest ring containing {B ∗ (~x) | ~x ∈ X ∗ }
S(R): the smallest σ-ring containing R.
En = {ω ∈ B(a∗ ) ∩ C(a∗ ) | u(a∗ , ω) > n1 + maxa6=a∗ u(a, ω)} ∈ S(R)
By countable additivity, there is some n large enough so that P (En ) > 0.
By a theorem in Halmos-19708 for any > 0 there exists a set E ∈ R such that P (En 4E ) < .
S
Since E ∈ R there exists a finite set Ô ⊂ X ∗ such that E = ~x∈Ô B ∗ (~x). Let Ô be such set
with minimum size. Then for any ~x, ~y ∈ Ô if ~x 6= ~y
B ∗ (~x) ∩ B ∗ (~y ) = ∅.
(51)
S
∗
∗
0
set Q~x of extensions of ~x such that P (B (~x) \ ~y∈Q~x B (~y )) = 0 and for any ~y , ~y ∈ Q~x , α(~y ) 6= a∗
and B ∗ (~y ) ∩ B ∗ (~y 0 ) = ∅ if ~y 6= ~y 0 .
Claim For any ~x ∈ Ô there exists a countable set Q~x of extensions of ~x such that
α(~y ) 6= a∗ for all ~y ∈ Q~x
(52)
B ∗ (~y ) ∩ B ∗ (~z) = ∅ if ~y 6= ~z for all ~z, ~y ∈ Q~x
(53)
and
and
P (E \
[ [
B ∗ (~y )) < .
(54)
y ∈Q~x
~
x∈Ô ~
Proof. If we can find { Q~x } ~x∈Ô that satisfies (52) and (54) we can refine it such that (53) also be
satisfied. We do this by deleting ~y from Q~x if it is an extension of some other member of Q~x . Thus
we just need to show for some { Q~x } ~x∈Ô both (52) and (54) are satisfied.
Suppose for every choice of { Q~x } ~x∈Ô that satisfies (52), (54) is not satisfied. A contradiction
will be drived.
Consider Q~x S
= { ~y ~x|α(~y ) 6= a∗S
}.
S
S
Define O = ~x∈Ô Q~x and F = ~x∈Ô ~y∈Q~x B ∗ (~y ). Then F = ~y∈O B ∗ (~y ).
For any ~x, ~y ∈ X ∗ the following is satisfied:


if ~x ~y ,
∅S
∗
∗
∗
B (~x) \ B (~y ) =
(55)
z ) if ~y ~x,
{ ~
z ~
x|~
z 6=~
y ,λ(~
z )=λ(~
y )} B (~

 ∗
B (~x)
otherwise.
Therefore
E \ F = [
[
B ∗ (~x)] \ [
~
x∈Ô
[
B ∗ (~y )] =
~
y ∈O
[
[B ∗ (~x) \ [
~
z ∈{
~
y ~
x|~
y 0 ~
y
8Page 56, Theorem D.
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for
B ∗ (~y )]].
(56)
~
y ∈Q~x
~
x∈Ô
Second equality in the above equation comes from (51).
Considering that Q~x is a subset of extensions of ~x we have
[
[
B ∗ (~x) \ [
B ∗ (~y )] =
~
y ∈Q~x
[
any~
y 0 ∈Q~x }
B ∗ (~z).
(57)
Thus
E \ F =
[
z ∈{
~
x∈Ô ~
~
y ~
x|~
y 0 ~
y
for
B ∗ (~z).
(58)
any~
y 0 ∈Q~x }
S
Define W = ~x∈Ô { ~y ~y for any ∈ Q~x } then E \F = ~z∈W B ∗ (~z). For any ~z ∈ W ,
for any ~y ~z, α(~y ) = a∗ . Thus B ∗ (~z)∩En = ∅ and as a result E \F has an empty intersection with
En and therefore [E \ F ] \ En = E \ F . Therefore P (E \ En ) ≥ P ([E \ F ] \ En ) = P (E \ F ) ≥ .
This contradicts P (En ∆E ) < S
~x|~y 0
[
~y 0
Please note that since P (E \ F ) < and P (En ∆E ) < we have P (Fn ∆E ) < 2.
For any ω ∈ E let ~xω ∈ O be defined such that ω ∈ B ∗ (~x). ~xω is unique.
Consider an alternative contingent plan α̂ defined below.
(
a∗
if ~x ∈ O ,
α̂(~x) =
α(~x) otherwise.
By assumption P (B ∗ (~x)) > 0 for any ~x therefore if α is Bayes then
X
[U (α(~x), ~x) − U (α̂(~x), ~x)]P (B ∗ (~x)) > 0 .
(59)
(60)
~
x∈O
Left hand side in turn is equal to
Z
Z
∗
[u(α(~xω ), ω) − u(a , ω)]dp(ω) +
ω∈F \En
[u(α(~xω ), ω) − u(a∗ , ω)]dp(ω) .
(61)
ω∈F ∩En
Which is less than 2M.2 − n1 (P (En ) − 2.). But for small enough this expression is negative and
this is a contradiction to (60).
E-mail address: fzb5027@psu.edu
E-mail address: eug2@psu.edu
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