NECESSARY AND SUFFICIENT CONDITIONS FOR
EXPECTED UTILITY MAXIMIZATION:
THE FINITE CASE, WITH A PARTIAL ORDER
by
Leonard Shapiro
Discussion Paper No. 77-80,
February 1977
Center for Economic Research
Depa rtment of Economics
University of Minnesota
Minneapolis, Minnesota 55455
NECESSARY AND SUFFICIENT CONDITIONS FOR
EXPECTED UTILITY MAXIMIZATION:
THE FINITE CASE, WITH A PARTIAL ORDER
'f,-
Leonard Shapiro'
*Supported by NSF Grant MCS3l276X
The author wishes to thank the Center for Applications of Mathematics to Economics and Management Science, Northwestern University, for its hospitality during part of the preparation of
this manuscript. Professor M. K. Richter's many suggestions and
criticisms, some but not all of which we have been able to answer,
have been most valuable. This paper arose out of joint work with
Richter on a related question (cf. Richter-Shapiro, 1977).
Thanks
are due to Professor R.D. Luce for pointing out errors in a related
theorem, which does not appear in this paper.
i
In troduc ti on
An individual is to choose among various possible decisions,
each being a mapping from a set of possible states of nature
a set of consequences.
If we fix a utility
consequences and a probability measure
~
states, the expected utility of a decision
u
on the set of
on the set
f
to
is
E
of
Ju(f(e)}:1~(e)
E
The "expected utility hypothesis"(EUH) is that there exists a
utility and a measure such that the choices of the individual are
consistent with expected utility maximization.
This hypothesis is a very useful one in modeling choice under
uncertainty, and many authors have attempted to justify it by describing reasonable conditions (or axioms) on behavior, which imply
EUH.
Certain hypotheses are common in these treatments: an in-
finite, atomless space of states [Savage (1954)) or, in the case
of a finite set of events, a complete order
(i.e., every pair of
events comparable) on the set of all decisions <tonsidered
and Krantz, 1971).
which
a~
(Luce
It is also common to present axioms
sufficient, but not necessary, to establish EUH.
Here we assume the space of states and of consequences to be
finite and do not assume the order to
~e ~omplete,
and our conditions
are equivalent to EUH, modulo the finiteness assumptions on states
and consequences.
ii
Of course this comes at some cost:
easily stated.
Our conditions are not
They include statements about repetitions of the
states, which corresponds to considering states with smaller probabilities than those of the original finite set - however the
number of repetitions is finite (it is bounded by the number of
consequences), so we do not get into an atomless measure space.
Finally, they include a hypothesis that the individual is able
to make certain other simple choices, not in the given set of
choices, in a consistent way.
This feature is common to other
treatments and it does not imply an extension of the original
partial order to a complete order.
In fact these other choices
do not involve utility at all.
The key tool in the proof that our conditions are necessary
and sufficient is a s light generalization of the "FourierMotzkin Elimination Procedure" (Kuhn, 1956).
Since the set of states and consequences are both assumed
finite and we assume strict preferences, if a (necessarily finite)
set of decisions satisfies EUH for a fixed
ul
,VI
sufficiently close to
u
and
v
u
and
v
then any
will also work.
the utilities we describe cannot be cardinal.
Thus
1.1
1.
Let
Definitions, Outline of the Paper
X denote a finite set of outcomes, or prizes. *
assume the set of states, or events
E
We as
to be finite; for simpli-
city we will further assume it contains two elements, say
E
=
fH, T}.
ticket.
An ordered pair
(x,y)
E
X x X is a decision, or
It is interpreted to yield its owner the prize
H occurs and the prize
y
if
T
x
if
occurs.
We imagine a single subject who cannot influence the events
Hand
T.
He may have some information about each event, e.g.
from having looked carefully at a coin or (if
sun) reading a barometer, etc.
H
= rain,
note the fact that he has chosen
For example, if
X
=
=
We are given the information
that our subject has made some choices between tickets.
(z,w).
T
(x,y)
over
(z,w)
by
We de(x,y) >
[x,y,z} , the subject may have made
the choices
(x,z) > (y,z)
(w,z) > (y,z)
(x,y) :> (y,x)
(y,w) > (w,y) •
Note that the subject can make only finitely many choices (since
X is finite) and that he need not make a choice between every
'1(
We do not assume that the elements of
parable.
Thus if
"same as"(%)x •
X are divisible or com-
x E X we cannot require some
y E X to be the
1.2
available pair of tickets.
These choices may have arisen as a
result of our subject's market behavior (involving a price structure, as in the theory of revealed preference:Richter, 1966) but
we adopt a more general approach and ask not whence the information of these choices came.
There are several methods which can be used to make such
choices, for example maximizing minimum uti1ity,or expected
utility.
and
q
(1.1)
(1.2)
Given a utility
(of
T)
u
and
probabilities
(of
the expected utility of the ticket
H)
(x,y)
is
E(x,y) = pu(x) + qu(y) •
Definition: Let
denote a set of choices
g
1, •.. ,n}
(1.3)
where
with
p
x .. E X for each
i
J1-
and
j.
We say
g
is compatible
EUH (expected utility hypothesis) if there exist
positive and a function
u
from
X
to the rea1s R
p,q
such that
(1.4)
Notice that we have required
than just nonnegative.
p
and
q
to be positive, rather
This assumption is merely for convenience,
for if the equations (1.4) were solvable with, for example
p=O,
then, since the inequalities involved are strict, they would have
a solution for sufficiently small positive values of
p •
1.3
(1.5)
Notation.
For simplicity we may delete the
mu1as such as (1.4) so
u
in for-
x E X stands interchangeably for a prize
and for the utility of that prize.
Thus instead of asking for a
utility function
u
X ~ R satisfying (1.4), we will ask for
real numbers
' 1
~
x
ji
j
~
4, 1
~
~
i
n , such that
for all
where we of course intend that if
x ..
i
and
J1.
refer to the
same prize, then they should be equal when we are referring to
utilities.
(In some sense we are viewing the elements of
variables.)
for
T
Also we denote
over
H".
q/p
by
Dividing (1.4) by
if and only if there are real numbers
p
p
p
as
is the "odds
and using our variab Ie
~
notation, definition (1.1) becomes:
so
0,
X
is compatible with
x
and a positive
ji
EUH
numb~
such that:
for all
(1.6)
Since
p
i
1, ... ,n .
will always denote these odds and we require
it will simplify matters to make the convention: the variable
p> 0
p
is assumed to take on only positive values, for the remainder of
the paper.
We pause for
(1. 7)
g:
an
example.
If
g
is given by
(1)
(X,z) > (y,z)
(3)
(w,z) > (y,z)
(2)
(X,y) > (y,x)
(4)
(y,w) > (w,y)
1.4
then
g
can't be compatible with EUH.
says
x
is worth more than
likely than
Intuitively, choice (1)
y, so (2) says
H
T • Similarly, (3) and (4) say
T
H , a contradiction of our intuition.
than
is more
is more likely
This intuitive
reasoning could be made rigorous, as we do for equations (1) and (2):
This
g
and
p
(1 )
(x,y) > (y,y)
(2 )
(x,y) > (y,x)
satisfies
EUH iff we can find real numbers
x
and
y,
> 0 , such that
(1' )
x + Py > y + py
(2' )
x
+
py
> y + px
These equations are equivalent to
(1" )
(2" )
(1 -
p)(x - y) > 0 ,
and these can be solved if and only if
to
q
p < 1
9
which is equivalent
< p ,i.e. H more likely than T •
We finish this section by
outlinin~
the rest of the paper.
In section 2 we discuss other approaches to the problem of finding
necessary and sufficiept conditions for EUH, and describe criteria
we would like such conditions to meet.
None of the approaches
1.5
we survey, nor even our own, fully meet these criteria.
Section 3 consists of a detailed example which illustrates the
Fourier-Motzin Elimination Procedure and its extension Which is
at the heart of our proofs of necessity and sufficiency.
As
mentioned above, our conditions for EUH entail repetitions of
the H - T
experiment, and they involve more complicated tickets,
which we call compound tickets. These and related definitions are
given in section 4, where we also extend the definition of EUH to
cover choices between compound tickets, so that ordinary tickets
and the EUH defined above are special cases.
We will find it
more convenient to give necessary and sufficient conditions for
EUH in this more general setting.
Since our conditions are not
easily stated we devote section 5 to carefully explaining and
motivating them.
Section 6 contains the detailed statements of
the conditions, along with some more motivation, and section 7
is the proof of necessity and sufficiency.
In section 8 we sur-
vey the success of our necessary and sufficient conditions in
meeting the criteria discussed in section 2.
2.1
2.
Other Approaches
Given our restatement of EUH as the solvability of the quadratic inequalities (1.6), it seems natural to use duality theory.
p, the equations (1.6) are linear.
For fixed
be written in the form
p > 0
each
prizes.
and
x
A(p)x >
y(p»> 0
compatible with
for every
y»
where
A(o)
is a matrix for
is a vector with as many components as
By (Gale, 1960),
is no vector
a
In fact they can
EUH iff
0 , yA(p)
A(p)x > 0
such that
has a solution iff
y(p)A(p)
0
there exists some
~
p > 0
Thus
iliere
3
is
such that
0 •
This first approach, using duality, is deficient on at
least two grounds.
(2.1a)
It is not constructive.
Given a specific
not tell us how to go about finding whether or not
patible with
EUH.
whether there is a
Of course for a single
y(p»> 0
with
p
y(p)A(p)
3, it does
g
is com-
one can test
= 0 (Kuhn, 1956),
but we might have to carry our this procedure for infinitely
many
(2.lb)
p > 0
before possibly finding one that worked.
The conditions do not intuitively justify EUH; in
fact they are almost a restatement of EUH.
We would like to
find conditions which are reasonable in the sense that they seem
to impose less restrictions on a subject than does EUH.
A proto-
type of such conditions is the strong axiom of revealed preference.
This requires, in the context of choice under certainty, only
that a subject not express a cycle of choices (e.g.
xl > x
2
2.2
> x3 > xl) , and eminently reasonable requirement, but (under
certain conditions) it is equivalent to the utllity hypothesis,
that a subject acts to maximize the value of a utility function
(Richter, 1966).
A second approach to this problem can be found in the work
of Tarski.
Equations such as (1.6) are polynomial equations in
their unknowns which are
p and the elements of
X. The state-
ment that they have a solution is therefore a theorem of the firstorder theory of algebra ana by
Ta~ski's
theorem on the completeness
of the theory of real-closed fields (Tarski, 1951) its truth or
falsity can be verified in a constructive way.
This solution
does satisfy our requirement (2.la) above, of constructivity, although the methods Tarski uses are not at all simple to apply.
But it does not satisfy our requirement (2.lb) as far as we know:
we have been unable to interpret Tarski's work to yield what we
would call reasonable conditions for a
s~bject's
behavior under
uncertainty.
The comments we have made about Tarski's work also apply to
the results in Richter (1975).
There another constructive method
does not supply reasonable conditions.
We have previously mentioned the axiom systems of Savage
(1954) and Luce and Krantz (1971).
We pointed out that Savage's
axioms presume a nonatomic event space.
Luce and Krantz' require
.(Axiom 2, p. 265) a complete orner on the set of tickets considered.
Actually their theory refers to "decisions", a concept more general
than our "tickets", and their conditions do not guarantee the
existence of a utility function on prizes.
3.1
3.
Illustration of the Elimination Procedure
We have mentioned that a generalization of the Fourier-Motzkin
Elimination Procedure is central to our results.
That procedure
is used to solve linear inequalities by elimination of variables.
To illustrate its use here, we consider another example:
(x,y) > (v,x)
(3.1)
(v,y) > (y,x)
(x,y) > (y,x)
First we rewrite the system in inequality form and collect terms
to get:
x - v
+ p(y -
(2)
v - y
+ p(y - x) >
(3)
x - y + p(y - x) > 0
x) > 0
0
Adding equations (1) and (2) cancels the
(1 + 2)
gl:
(3)
Clearly
go
go
~
gl
v
and we get
x - y + 2p(y - x) > 0
x - y + p(Y - x) > 0
in the sense that if
then the same values of
x,y
and
x,y,v
p
and
satisfy
r
gl
satisfy
.
Notice
3.2
that also
(1 + 2)
gl
~
gO.
For if
x,y
and
p
satisfy
gl
then by
we have
x + p(y - x) > y - p(y - x)
so we can choose
v
between them,
i.e.,
x + P (y - x) > v > y - p (y - x)
(i<)
The first inequality in (*) shows that
the second inequality shows (2).
gO.
The equivalence
go
~
gl
x,y,v
Thus
and
x,y,v
p
satisfy (1) and
and
p
satisfy
is a special case of the Fourier-
Motzkin Elimination Procedure, which can be applied since the
variable we eliminated,
system
gl
v,
ap~ars
linearly.
we introduce the variable
z
=x
Now to solve the
- y .
Then
is
equivalent to
g2:
(4)
z(l - 2p) > 0
(5)
z(l - p) > 0
We could solve this system in a few simple steps, but it will be
instructive to continue an "elimination procedure".
eliminate
and
z
from
if we knew that its coefficients,
1 - P , differed in sign.
1 - p > 0
1 - 20 > 0
coefficients
(1 - p) > 0
add them together and get
and
1 - p
1 - p < 0
or
1 - 2p
then
0 > 0
0 > O.
equals
1 - 2p
1 - 2p < 0
For example, if
then we could multiply (4) by
-(1 - 2p) > 0,
if
g2
We could
and
and (5) by
Similarly,
If either of the
0, then clearly
0 > 0 •
3.3
Summarizing,
~
g2
(7)
g3
where
g3
is:
if
1 - 2p
~
0
and
1 - p
~
0
then
0 > 0
if
1 - 2p
~
0
and
1 - p
~
0
then
0 > 0
As is proved in general below in section 7,
g2.
Of course
1 - 20 > 0
that either
1 - 20 < 0
p
0 > 0
is equivalent to
is impossible, so (6) amounts to requiring
or
1 - p > O.
or
g3
1 - P < 0 •
Similarly, (7) says
Summarizing,
g3
is equivalent to:
is in the set
[[0: 1 - 2p > O} U [p: -(1 - p) > O}J n r(p: -(1 - 2p) > O}
U
[p:
1-
p >
O}J
(The first bracketed term is from equation (6), the other from
equation (7».
(O,~)
U
(1,~)
It is easily checked that this last set is
, which is certainly nonempty, so
Therefore so does
E~.
go' so the
g
in (3.1) is
g3
has solutions.
compatible
with
4.1
4.
Compound Lottery Tickets
Ideally, a set of necessary and sufficient conditions for
sets
g
g
of choices to satisfy
EUH should involve only the sets
themselves, e.g. requiring that no cycles (appropriately de-
fined) exist.
We are unable to find such donditions which are
also constructive and reasonable, as defined in (2.1).
Our con-
ditions, though constructive and (we think) reasonable, require
that some "choices", other than those in
g, be made.
These
choices, finite in number, concern tickets involving a repeatins
of the experiment with outcomes
Hand
T.
Thus these results
are not strictly in the spirit of the subjectivist school, which
claims expected utility maximization even in the case of unrepeatable experiments. See section 8 for more information on this point.
The main result of Richter-Shapiro (1976) is that the set
of
pIS
which appear in solutions to (1.6) can be an arbitrary
polynomial set, i.e., polynomials of arbitrarily high order are
needed to describe such sets.
This result suggests that repe-
titions of events will be needed in any necessary and sufficient
conditions.
In describing our necessary and sufficient conditions it will
be helpful to use tickets which involve positive integer multiples
of the prizes in
ment.
X , as well as repetitions of the H-T
experi-
(We can use a fair coin to interpret these multiples, as
suggested by Ramsey and described below in (4.6)).
be convenient to use notation such as
It will also
4.2
(x - z, y - w) > 0
(4.1)
instead of
(x,y) > (z,w)
(4.2)
Even with more complicated tickets one can always go from the
form (4.1) to (4.2) by merely putting all the "negative" terms
on the right.
One might interpret (4.1) in its own right, not just as a
notation for (4.2), by viewing it as a ticket which yields
x - z
(i.e., the subject gets
H ,and
y - w
in case of
quite different than (4.2).
T.
x
but must pay
z)
in case of
But this interpretation is
For example, a subject who did not
have a "z" might make the choice (4.2) but not want to risk
having to pay "z", as in this interpretation of (4.1).
Even though
this interpretation differs from the one we have in mind, it may
be the appropriate one in some circumstances.
We now give a rigorous description of the compound tickets
we have been mentioning, which involve repetitions of the H-T
experiment
and positive integer multiples of the prizes in
For fixed
K
~
1,
define
to be a certain set of outcomes of the
K times, namely
first
k
trials,
X.
H-T
denotes the outcome of
on the other
K-k
experiment repeated
H occuring on the
trials.
Notice that
4.3
e
k
EK
E
differs from
e
E
k
E ,
K
if
K';' K' ,
meaning will be clear from the context.
of the
H-T
The set
EK
H"
set of prizes to
1
~
T.
"T
then
E2 •
X as
x1, ••• ,x
and we expand the
L
P. ~ L ,
G consists of expressions of the form
Q' ~
is an integer (possibly negative or
Note that the zero prize (all
zero) •
is in
to
G, the free abelian group on the origiqal prize
The group
where for
H and
is not the set of all outcomes; for example
We number the elements of
X.
is just the outcomes
corresponds to
experiment:
is not an outcome in
set
EI
but the appropriate
G, and one adds or subtracts elements in
G
Of
J.,
= 0)
in the ob-
vious way.
A compound lottery ticket is a function from EK
i.e., it associates with each "compound event" in
pound prize" in
denote
G.
If
h by its values:
(x,2y - z, 0)
the outcome
the outcome
h
EK
a "com-
is a compound lottery ticket we can
(h(eO),h(el), ••• ,h(e
».
K
For example,
is a compound lottery ticket which awards
eo '
2y - z
x
for
and nothing for
for the outcome
(See (4.6) for the interpretation of
A linear utility on
G,
into
G is a real-valued function
2y - z .)
u
on
satisfying
u(x + y)
= u(x) + u(y)
and integers
~.
and
u(ax)
Q'u(x)
for
x, y E G
G
4.4
For a fixed linear utility
q
u
on
G
and posttive numbers
we define the expected utility function
E
p
and
for compound lottery
tickets by
(4.3)
K
K-k k
P
q u(h(e )
~
E(h)
E(h) > 0
h
by
r. p
Note that if
h
p.
Thus
> 0
u(h(e k )
a simple lottery ticket if
X.
yare in
q/p
K k
iff
k=O
We call
on the event
set
As in section 2, we will denote
(4.4 )
h
for tickets
k
k=O
h
=
(x,y)
where
x
and
is a simple lottery ticket then
(4.3) reduces to the definition of expected utility given in (1.1).
We use the phrase "h is desirable" to mean
h > 0 , as in (4.1).
In section 1, EUH meant that there existed a utility
probabilities
implied
u , p
p,q
E(x,y) >
q.
and
such that a preference for
E(z,w) ,where
E
Here it mean$ that
desirability of h, h > 0, implies
(x,y)
over
u
and
(z,w)
was defined (1.1) in terms of
u, p, q
exist such that the
E(h) > 0, where now E is defined by
(4.3).
(4.5)
Definition:
;r
where each
= [h. > 0:
J
h.
J
J
Let
j
denote a set of the form
1,2, ... ,J}
is a compound lottery ticket.
Then
patible with EUH if there exist a linear utility
p,q > 0
such that, for the
E
defined by (4.3),
u
;r is comon
G
and
4.5
E (h.) >
J
°
for
j
1,2, ••. , J .
Since (4.3) is a special case of (1.1), as remarked above, this
definition is a generalization of (1.2).
Thus, our goal of
finding necessary and sufficient conditions for EUH will be
reached if we use definition 4.5.
(4.6)
To interpret multiples and sums of a prize
x
example, from which it should be easy to generalize.
we give an
Consider
the statement
(x + w, 3y - z) > 0,
(4.7)
or
(x + w, 3y) > (O,z) ,
and consider a fair 3-sided "coin" with sides marked
a, band
c, so each side is expected to come up with equal probability
equal to
(4.8)
Define the ticket
l
3" •
(x,y)
as follows:
~
(w,y) ffi (O,y)
The fair coin is flipped.
ject gets the ticket
(O,y)
(4.9)
in case
c.
(x,y), he gets
If
a
(w,y)
comes up the subin case
b, and
The ticket
(O,z) ffi (0,0) $ (0,0)
is interpreted similarly.
E(x + w, 3y - z) >
° iff
greater than that of (4.9).
Then it is easy to check that
the expected utility of (4.8) is
Thus we can interpret (4.7) as the
statement that (4.8) is preferred over (4.9).
4.6
If one objects to the zero prize, it is possible to go further and write the choice of (4.8) over (4.9) as a choice between
lottery tickets not involving the zero prize.
We omit the details.
5.1
5.
Motivat ion
If any conditions claim to justify EUH (cf.(2.l) above) they
must in particular explain why the subject who made the choices
in example (1.7), repeated below, is unreasonable.
at that example carefully.
(x-y, y-x) > 0
more likely than
and
T).
There the subject made the choices
(x-y, 0) > 0 , which implied
Call these two choices
other choices, which we'll call
g
gl U g2
Let us look
p > q
gl
q > p,
so
But what does the term
"implied" mean in this context?
On the mathematical level it
means that, for example, if
were compatible with EUH then
the
p
and
q
gl
p > q.
involved would satisfy
is
He made two
g2 , which implied
was incompatible with EUH.
(H
But our conditions
should justify, as well as be mathematically equivalent to, EUH.
So there is another sense in which we might intend the word
"implied" -- that a subject who made the choices in
believe that
H
is more likely than
T.
~l
should
We use the word
"re-
veal" in place of "imply" for this latter, subjective, meaning.
g
Even if we agree that
likely than
T ,and
g2
reveals
T
1
reveals
H
is more
is more likely than
H, we
must also agree that these revelations are inconsistent, and not
the behavior we expect of a "reasonable" subject.
5.2
That is, our conditions for EUH should have two parts:
(i)
Showing how ticket-choices reveal information about
the relative likelihood of
(ii)
Hand
T ;
ensuring that this information is consistent.
Let us look at (i) more carefully, in the case
gl
[(x-y, y-x) > 0,
The choice
p > q
(x-y, y-x) > 0
or
p < q,
potentially reveals that either
for if the two events
likely then the ticket
(y-x, x-y)
the sum of the two tickets,
tradiction.
(x-y, 0) > O}
y
would also be desirable, and
p < q
If we suppose
is more useful than
(H
less likely than
(x-y, y-x) > 0
x, so
(y-x,O)
to get that
So we reject
p < q
(0,0)
T)
to
also reveals
should be desirable.
Now add this latter ticket to the other ticket in
(x-y, 0)
T were equally
(0,0), would be desirable -- a con-
be what is actually revealed then
that
Hand
gl , namely
is desirable -- again a contradicti on.
and conclude that
p > q
is actually revealed.
Generalizing from this example, we can split (i) into two
parts, first that choices (like (x-y, y-x) > 0) potentially reveal some things about
p
and
q
(like
p < q
or
p > q),
and
second that these potential revelations are actual when they do
not allow us to derive a contradiction (like
(0,0) > 0) •
In summary, we would like our conditions for EUH to involve
three steps:
5.3
S1
Show how choices reveal (potential) information about
the relative likelihood of
Hand
T •
S2 Ensure that the information is consistent.
S3 Show how this information can be used to derive new
choices and, pOSSibly, a contradiction.
Notice that (i) has become Sl
and
In general, new choices found in
S3.
S3 must be fed into Sl
to reveal new potential information, but this process stops
after
L (= the number of prizes in
The condition
82
X) stages.
is analogous to the axiom of revealed
preference (Richter, 1965).
There the "new" choices are derived
by applying transitivity, and a contradiction is a choice of the
x > x , analogous to our
form
(0,0) >
°.
We need a few definitions in order to clarify Sl·.
on we number the prizes in
Let
X :
be a compound lottery ticket on
h
From here
EK ' then
is given
h
by
k
L
I:~=1 ex px 0
h(e )
k
for some integers
k
x
P.
,
denoted
hlx
1
~
k
~
K
We define the restriction of
Q'
P.
prize
for
p.
h
, to be the ticket defined by
to a
5.4
Thus
hlx
~
is the game
h
with all prizes ignored except
Given positive probabilities
I
the expected value of
K
= g/p
0
(5.1)
( hIx )> 0
E
K-k k k
q, and a linear utility u ,
q
Ci U
~.
(x )
.P,
,we obtain
iff u ( x )
t
Kk
k
lk=o
01 P
t
t
> 0
Kkk
P is called the characteristic polynomial
The polynomial
of
t
is clearly
h xl,
= lk=o P
Setting
p and
x
~=o CI
fJ
prhl x
h, and it is denoted
the coefficient of
u(x)
J •
Note that
in the formula for
1,
prhlx~J
tive meaning of
p,
is this:
is just
E(h).
it is positive
The intui-
iff
x
n
more likely to be gained than lost, from using the ticket
is
h.
Given the previous definitions, we can explain S 1 :
each ticket-choice
X EX,
that
p[hlxJ
h > 0
potentially reveals, for each
is positive, zero, or negative.
Next we explain the term "consistent" in
(5.2)
Definition.
polynomials
=, 0)
A set of choices of signs of characteristic
prhjlXtJ
for various
tive value of
j
p, say
(i.e., choices of their being
and
0,
p,
j
and
>, <,
or
is consistent if there is a posiwhich gives all those polynomials
their chosen sign, e.g. if we have chosen
positive for some
82 •
1
prh Ix
j
then we must have
t
J to be
P[hJ.lx )(p) > 0
I,
5.5
for that
j
and
p.
8tep 81 , deriving new choices, is complicated and we postpone it for now.
But a contradiction is as hinted above
the choice (as desirable) of a ticket
to zero,
i.e.,
h(e k ) -
In general the steps
° for all
h
k
it is
having all prizes equal
for example
(0,0,0) >
81- 83 get quite involved
°
there
are potential revelations for each ticket and each prize in
X,
and these lead to new tickets, then new potential revelations, and
so on ad nauseam (but not ad infinitum -- everything is finite).
Meanwhile we must keep track of which "potential" revelations lead
to contradictions, so they can be discarded (the remaining revelations are "actual").
It will simplify matters greatly if we state our conditions
contrapositively -- that is, we require a set of "actual" revealed
information (revelations which do not lead to a contradiction) to
exist.
(1)
We emphasize three points:
This revealed information involves whether
positive, negative, or zero, where
x
E
prhlxJ
X and
h
is
is
either a ticket from the original set or a derived
new ticket (see 83) ,'.
(2)
The number of "new" tickets created is finite, so in
deciding whether a set of revelations is actual -- i.e.,
does not lead to a contradiction -- one needs go through
only a finite number of steps.
5.6
(3)
It is possible to express the decision as to whether
Prh Ix]
tickets
is
>, <,
or = 0
in terms of choices between
we explain this at the end of section 6.
6.1
6.
Statement of Necessary and Sufficient Conditions
- The Main Theorem
(6.0)
Theorem:
Let
g
be a set of simple ticket-choices,
1,2, ••• ,J}
j
Define
h.
J
j = 1, ...
{h. > 0
J
Then
iff
g
is compatible with EUH
characteristic polynomials
cl
P[h\x~] ,
is compatible with EUH,
l
h
for
€
cl p , such that
The choices are consistent i.e. there exists
such that each number
h
b)
iff
t = 1.2, •••• L it is possible to choose signs of
for each
a)
,J}
€
cl
I
,
1,
,
for
1 :S
has a sign equal to that chosen for
cl
The sets
P[h\x ](p)
p
> 0
0
1,
:S L
and
P[h\x.t]
.
defined (inductively - see below) by
these choices do not contain a contradiction. i.e •• a
member
h > 0
h(e )
k
where
= 0 for all k.
Remarks:
(i)
The sets
cl
1,
,P = 1,2, ••. ,L+l , will be defined in the
remainder of this section.
structed from the tickets in
restricted polynomials
For
clj.
P[h\xp,]
1:S t
~
L,
clf,+l
will be con-
and the choices of signs of the
as
h
ranges over
clf, •
6.2
(ii)
A similar theorem could be stated where
g
contains
arbitrary compound tickets, of arbitrary degree, and the proof
would differ only in notation.
(iii)
That
is compatible with EUH
g
iff
J
1
is,
was
shown in the comments following (4.5); the rest of the proof of
is in section 7.
6.0
We will now define the sets
in the Theorem.
K
=
2
),-1
Each
J
Fix
x
. it consists of the "new" ticket-
S3.
it consists of tickets on
Jl,
Assume
E , K
K
involves only the prizes
J),
by
t '
1 s p S L.
)"
EK '
will be built from
"eliminating" the prize
choices mentioned in
, t = 2, ••• ,L+1 , mentioned
?
will consist of tickets on
j,
Intuitively,
.
J
hypotheses are satisfied for
has been defined So that
= 2),-1 , and each ticket in
x£,x + ' ••• ,x •
L
t 1
f
1
=
(Clearly these
by the Theorem's
J
1
.)
As in the Theorem, assume we are given signs of the characteristic
I
polynomials
to define
satisfying
Nl )
h
O
for
€
J
)'+1
K < k
P [h x£]
J £+1
NI, N2
Snppose
where
--S
2K •
for all
The set
and
h
0
€
N3
;r
x
J
h
}.+1
Jp, ; these signs will be used
€
is the smallest set of tickets
helc)w:
and
h (e ) = h(e )
k
k
for
1
for all
0
hlxj:(e k )
S
k
S
K and
.
k
0
Then
h (e ) = 0
k
6.3
N2)
Suppose h
h
E!
if
hlx (e ) # 0 for some k. If
/., k
where hiF is defined as follows: If
and
3'
;,
e 3'}, +1
L
h(ek ) =
o :;
for
r, t"VkrtR.,
m=J,
k ::;; K ,
then
L
hiF(e ) =
k
l,
(j/
:x
o :;
for
m=;,+l km;,
k ::;; K
and
hiF (e ) = 0
for
Conditions
N1
k
x
P,
is irrelevant in
3' +1 ."
~,
K
<
and
h
k ::;; 2K
N2
say essentially "if
then put
h E 3'
;,
and
h , properly extended, in
The meaning of "properly extended" is that, regardless
of the value of
p,
q
and the utility
the sign of the expected utility of
u, we do not change
h.
The following lemma,
whose proof is a simple computation, is a mathematical version
of the relevant part of this paragraph.
(6.1) Lemma.
Fix
p. q
and a utility
u
and let
E
be the
expected utility defined by them.
a)
hlx.e(e k ) = 0
If
for all
k then
E (h) > 0
iff
0
E{h~>O.
b)
If
E(hlx ) = 0 then
.e
E{h~
> 0
iff
E {hiF~ > 0 •
6.4
The third method of elimination (N3) will assume there are
h* h'
in
h ,
It is analogous to the following elimination
"
method for inequalities:
Suppose we are to eliminate
x
from the
inequalities
(6.2a)
(6.2b)
where the Ri
Ql < 0
and
Qi
are real numbers.
we can multiply (6.2a) by -Ql
If
Rl > 0
and (6.2b) by
and
Rl
to
get
(6.3a)
(6.3b)
Adding these two inequalities, the
x
terms cancel:
(6.4)
Note that if the coefficients R., Q.
1.
degree
K
1.
in (6.2) are polynomials of
then those in (6.3) are of degree
contains tickets on
The conditions
analogues of
prh\x
EK '
Rl > 0
,J > 0
and
and
J +1
j,
2K.
(That's why
contains tickets on
Qi < 0 will be
P[h'\x
~
J<
0
We now define the analogues of (6.3a) and (6 .3b), hi' and h' ,',: If
(6.5)
and
6.S
then
= -
I
m+n=k
Sm h(e)
,
n
o~
k
~
2K
o~
k
~
2K •
(6.6)
Then, as (6.4) is the sum of (6.3a) and (6.3b), we define
h* h'
;.
to be the sum of
h*
h ,,~ :
and
(6.7)
N3)
If
hand
PCh'lxtJ < 0
h'
then
are in
h*ph'
E
J
J
;,
and
p[hlx
t
in the next lemma, especially if one sees
p[h'lx
t
J
and
0
+ .
t l
More motivation for the definition of
and
J>
h* h'
t
can be found
E(h), E(h'), p[hlx
as analogues of(6.2a), (6.2b),
Rl
and
t
d
Ql
respectively.
(6.8)
Lemma:
utility
u
Fix positive probabilities
on
G , and set
and
q
and a linear
D = q/p •
a)
·
d f·1n1t10ns
..
of
the e
Us1ng
b)
If
h'~._
E(h), E(h'), P[hlx,~J(p)
positive, so is
p
E (h* h') •
i
h' *
and
in (6.6), we have
-P[h'IXtJ(p)
are
6.6
c)
The ticket
h* hI
does not include any non-zero multiple
1,
of
x
in any of its prizes.
/-
Proof:a)From the various definitions involved we get
K K
n
E(h) = p !:n=O u (h(en»p
The formula for multiplication of polynomials gives
P[h I Ix
/-
J (O)E (h)
Now use the fact that
P[h I Ix
J.,
u
is linear:
J(p)E(h)
-p~(h*) • The proof of the other
The right-hand side is clearly
half of (a) is similar.
b)
Since
and
p > 0 , the hypotheses and (a) imply that
E(h ' *)
E(h*)
E(h* hI) = E(h*) + E(h ' *)
are positive, so
/,
is positive.
c)
The terms involving
~
m+n=k
Sm~nx
and
f.
-r
x
t
in
h*
~m~nx~
m+n=k
clearly add up to zero.
!CJ
and
h '*
are
, respectively, and these
6.7
Lemma 6.8 also provides the key to an intuitive explanation
of N3:
Let us say you are presented with
h
and
hI , both
desirable tickets, both involving a certain prize
you feel you are likely to get
you are likely to lose
x
x
that for some reason the prize
x
be willing to exchange
hI
a combination of
h~'~
The ticket
hI
h
h
and
and
P[hlxJ > 0), in
(i.e.,
(i.e.,
Now suppose
is unavailable.
You should
for some new ticket which is
hI , but which does not involve
is just such
Stating the choice
h
hI
p[hllxJ < 0).
C1
t: icket,
I
is
if
x
V
(6.10)
x, and in
"Prh xJ
=
x
>, < , or
x
e
0"
as a ticket
choice.
The key to this is (5.1).
and
u(x)
is positive.
Suppose for the moment that
x
E
Then (5.1) becomes
(6.11)
Thus, instead of asking
whether
prhlx]
were positive, we could
ask a subject whether the restricted ticket
By (6.11), desirability of
P[hlxJ
hlx
were desirable.
corresponds to positivity of
Similarly, deSirability of -hlx
P[hlxJ <0.
hlx
corresponds to
As in (4.1, 4.2) desirability of tickets can be
stated in terms of choices between tickets not involving "negative" prizes.
Our assumption
u(x) > 0
is not justifiable, but
one can get around it by creating a new prize
~,which
ject agrees is itself deSirable, then replacing
hlx,
x
by ~
the subin
and asking whether the resulting ticket is desirable.
X
7.1
7.
Proofs of Necessity and Sufficiency
in the Main Theorem
Proof of Necessity in (6.0)
p, u(x ), ••• ,u(x)
1
1,
let
Assume that
For this proof,
P[h\xJ
chooses
iff
prh\xJ > 0
We claim that if
in some
then
g
(6.1a),
J,1, ~
1,
which
such that
The
can be made consistently if one
E(h\x) > 0 , etc., proving the condition
g
E
J
for any
"
This will imply that some outcome of
that 9.9b holds.
E
u
is to be computed using those values.
E
choices of signs of
6.9a.
p and
E(h) > 0
iff
1
is compatible with EUH and
denote the values of
define the expected utility function
h > 0 E J
J1
g
By induction, assume
then
E(g) > 0
.
is nonzero, proving
E(h) > 0
g E I n+ •
1
L , and let
1,
If
g
for every
h
arises from (N1)
J(;
= h O for some h
o
and
E(h ) > 0 ,
E
J
1,
and
E(h) > 0 .
E(g) > O.
Thus by lemma
A similar argument is
used in cases (N2) and (N3), appealing to Lemmas (6.1b) and
(6.8b), respectively.
Proof of Sufficiency in (6.0)
We assume the choices made of signs of
sis tent and we let
are, as in (5.2).
p
be a value of
p
P[h\x
1,
J
are con-
which shows that they
We will isolate the main step of the proof in
a definition and a 1ennna.
7.la
Definition:
p(J)
Let
J
is the set of
u(x.) , i
= 1, ••• ,L
by
and
~
p, q
Clearly,
be a set of compound lottery tickets.
q/p
(u(x )}
i
J
Then
such that for some real numbers
the expected utility function
satisfies
E(h) > 0
is compatible with EUH
iff
if
E
h E J .
p(J) # ¢ .
defined
7.2
7.1
Lemma.
Suppose
p
is a value of
Q which satisfies all
P[hlx J for all h E J
J.,
---f
p(J.e+ l ) for some J"
1 s
the choices of signs of
1
S
J.,
S
L,
and
PE
and all
p
J., ,
s L •
Then
Proof:
Let
[h. > 0
J
j = 1,2, •.•
1,2, ••. ,I}
i
By construction, the tickets in
and those in
,J}
J
i,
x,x +l' ••• '~
involve prizes
involve
(If
J.,
J,
J,
=L
then
J p+1
is the empty set or has one ticket, all of whose outcomes are zero.)
pE
That
p
'u(~)
u(xJ.,+l)' •••
function
p(J +1)
E
means there exist real numbers
such that the resulting expected utility
satisfies
is to find a value for
E(g) > 0
u(x)
J.,
for all
g
so that the
E
p,
u(x ), u(x.e+l)'···'u(~) , which we call
J.,
E' (h) > 0 for all h E J • Each of the J
J,
will impose a condition on
u(x )
j,
be satisfied.
E' (h) > 0
We denote
imposes.
h.
J
Now use the linearity of
u
J +1 .
J,
Our task
defined by
E' ,
s at is fies
conditions
EI(h.) > 0
J
and we must show they all can
by just
As in (4.4),
E
h
and see what conditions
E' (h) > 0
iff
to rearrange the left side:
7.3
L
(7.2)
E' (h) =
Pm (p)u(xm) > 0
L:
m=1,
where each
Pm(p)
is an integer polynomial in
Notice that
= P (p)u(x) and P[hlx J = P •
E'(hlx)
t
t
1,
Case 1.
1,
P (p)
x
u(x)
on
u(x) •
is in
lennna 6.1a
E'(h) > 0
So in this case
1,
hO
imposes no restriction
1,
Case 2.
=P
E'(hlx)
1,
P
1,
R
not identically zero, but
in this case
(p)u(x ) ,
£
of the value of
u(x) •
Since
£
all the choices of signs of
P[hlx..e J > 0
(N2), and
h4F
or
P[hlx
E
J +l •
t
£
J
Case 3.
P (p) :f
£
p[hlx
Again
o•
Since
E'(hlx)
f.
=0
is a value of
p
< 0 •
regardless of the value of
£
J
,regardless
p
which makes
consistent, we cannot have
Thus this case corresponds to
E (h4F) > 0 , so
u(x )
.t
,
E I (h) > 0 ,
and no restriction is imposed.
Since we have eliminated all other
h.
J
cases, we can without loss of generality assume that all the
in
J
fall into this case.
£
For each
h.
J
we rewrite
J
L
(7.3)
E' (h.)
J
l,
m=;,
pl(p)u(xm) > 0 •
By renumbering, we can assume that
for
1 s: j s: JI
·1
P~ (p) < 0
1 < JI < J
for
since if,
J' < jl s: J .
for example,
We can also assume
IS
E'(h.) > 0
as in (7.2):
and
h
E'(h) > 0 , regardless of the value
implies
of
Then the ticket
, so by (Nl) the ticket
1,
o
E(h ) > 0 , and
Thus
1,
is the zero polynomial.
P.
doesn't involve the prize
Jf,+l.
p.
7.4
p~(p) >
0
for all
by just making
u(x)
~
1
j,
j
~
J,
then we can achieve (7.3)
sufficiently positive.
p,
We can rearrange
(7.3) as:
1 s: j
~
JI
(7.4)
.IlL
~
.I
_(pJ (p))-
)(;~
m=p,+l
It is clear that, given
exists a real number
j
and
jI
wi th 1
~
pJ (p)u(x ) > u(x ) ,
m
m
~
JI < jl ~ J •
y.
u(xm)
for
m
~ 1,
+ 1 and
satisfying (7.4)
u(x)
1..
j s: J I
and
JI < jI
~
iff
p , there
for every
J
(7.5)
for every
Thus the lemma will be proved if we can show:
h
j
• I
P~
h.,
and
J
(p)
< 0
(7.6)
> 0 ,
Recall that
pjl () < 0
p,
p
imply
prh·lxJ>O,
. J f
and the pair h., h. I
]
falls into case (N3).
From now on we leave off the "j"s
formulas, and denote
h.,
J
by
hI.
By N3 the ticket
in our
g
= h*h '
.e
]
7.5
is in
~t+l
in (6.7), so
E(g) > 0
g = h* + h ' * , as defined
Recall
E(g) > 0 •
,so
implies
E I (h lic) > - E I (h*)
regardless of the value of
u(x).
p.
By lemma 6.8a
P[hlx J(p)E(h ' ) > P[h ' Ix J(p)E(h)
£
Since
P[hIX
£
J
t
= P
and similarly for
£
hI
and
pI , this implies
£
P (p)E(h ' ) > P'(p)E(h)
1,
£
Now expanding
E(h')
and
E(h) , we get
L
L
P (p)[PI(p)U(X ) + ~ PI(p)u(x)J > pI(p)rp (p)u(x ) +
~ Pm(p)u(xm)J •
p,
£
£
m= j+1 m
m
1,
£
£
m= £+I
cancel.
£
£ £
so we can divide through by
If we multiply out, the first terms
P£(p) > 0,
recall
< 0
(P (p)PI(p»
/.,
£
P~(p) < 0
P (p)P I Cp)u(x )
to get
which is exactly (7.6),
QED.
Now for the proof of sufficiency:
6.9b.
Consider
~L+I.
Assume conditions 6.9a and
Since all prizes have been eliminated at
that stage (see 6.8c) the only possible ticket in
zero ticket.
Then
This is excluded by assumption 6.9b.
~L+I
is the
Therefore
J L+ I ~ ¢
and,
by default,
p be a positive value of
such a
lennna
get
p satisfying all the sign-choices
exists by 6.9a.
p
= (p: p > O}. Now let
P(JL+I )
p E P(J + ) , by the previous
L I
Since
P EO(JL_I),etc.
p Ep(JL).For similar reasons
p E p(J ) .
I
Since
~
p(J I )
¢' J I
until we
is compatible with
EUH •
This proof may seem a bit too slick, although it is valid.
A more complex and perhaps more believable argument can be made
instead of
been eliminated from
~
Since all prizes but
J
L
,each
h
j
EJ
L
have
has expected utility
eq ua.l to
for some polynomial
(in fact
P.
J
Thus
iff
P.(p)u(x ) > 0
(7.7)
P
J
If the numbers
u(x~)
Pj(p)
for all
j.
are all of the same sign, one picks
to be of that sign to achieve (7.7), and
then induct using the lemma as above to prove
with EUH.
P
J
I
E
p(J ) ,
L
compatible
But this must be the case since otherwise for some
P.,(p) < 0 , and one can show that
J
then
h.*Lh.,
J
J
has all outcomes equal to zero, violating 6.9b.
8.1
8.
Criticisms
We have mentioned certain criteria that a theory of expected
utility might satisfy:
(A)
It should apply to a partial order on the set of de-
cisions and to finite sets of events and consequences.
(B)
The axioms, or conditions, should be constructive and
reasonable (cf. 2.1), and should not assume repetitions of events.
Before discussing the extent to which our theory has met
each of these criteria, we must admit that if our theory is to be
judged by this yardstick then it is surely a failure.
But we
would like to view it as an attempt to answer the question:
any theory satisfying (A) also satisfy (B)?
Can
Note that (A) con-
tains some of the most fundamental criteria for an adequate
theory put forth by the objectivist school, while (B) includes
claims made by the subjectivist school in favor of its theories.
Since some of the criteria in (B) cannot be rigorously stated,
we cannot hope to answer this question yes or no.
But, since
our conditions are necessary (as well as sufficient) it gives
evidence that any theory satisfying (A) cannot satisfy (B) any
better than does ours.
Of course this is not conclusive evidence
someone might design totally different conditions which, although
they would have to imply ours, WDu1d better meet the criteria of
(B).
But it seems to be the best available way to attack the
question.
If one is an objectivist, it is probably the closest
8.2
one can come to answering it in the negative, i.e., to showing
that any theory satisfying
(A)
cannot satisfy
(B).
Now we turn to the question of whether our theory does meet
the criteria
(A)
and
(B).
Partial order on decisions:
We meet this criteria unless
"
one interprets the statement "there exist choices of signs
in 6.9 as requiring the subject to make other choices than those
in the original partial order (cf. 6.10).
Even then, these other
choices do not require an extension of the partial order to a
complete one, or even an extension to choices between constant
ti('kf'tR.
I,'ICP
and Krantz (]97J,
requiring such other
Pc!.
2')8) make
thj~
ClllTIll1t:'nt
abL'lIt.
choice~:
"The measurer must be prepared to present for serious
consideration by the decision maker Some rather artificial
alternatives, and the decision maker must be induced to
make realistic decisions among them."
Finite set of events:
We have assumed that the set of
events contains only two elements, but this was merely for simp1icity; our methods could equally well apply to any finite set.
They do not, however, apply directly to an infinite event set.
Finite set of consequences:
this.
We have explicitly assumed
Again, our methods do not apply to an infinite set of
consequences.
8.3
Constructive:
Our method seems on the surface to be con-
structive, since everything is finite.
There are finitely many
choices of signs of characteristic polynomials
P[h\x
J
each
given
J
Each
J
thus finitely many possible
t
J£+l's
t
for
£
is finite so it is easy to check if it contains a
J£
zero ticket.
But checking whether a choice of signs is consistent
is not so simple.
In general it involves verifying that a certain
set of polynomials in one variable (the characteristic polynomials) have a common point of positive value, i.e.,
for all
P
in that set.
pep)
> 0
This problem is decidable in the mathe-
matica1 sense, since it is a special case of the solvability of
polynomial inequalities discussed in section 2, but so is the
original problem of solving the inequalities (1.6).
So by some
simple steps we have reduced the problem of deciding solvability
of the polynomial inequalities (1.6) to the solvability of some
polynomial inequalities in
~
unknown
p.
The main result of
J
Richter-Shapiro (1977) is that given any set
inequalities in one unknown, there is a set
as in section 1) ticket-choices such that
p/q
0
iff
p
is a solution to
g
g
of polynomial
of simple (i.e.,
has a solution with
J
Reasonability is a purely subjective judgement and our method
does require repetitions of events.
BIBLIOGRAPHY
1.
D. Gale (1960), The Theory of Linear Economic Models, New
York: McGraw-Hill.
2.
H. W. Kuhn (1956), "Solvability and Consistency of Linear
Equations and Inequalities", Amer. Math. Monthly,
63, pp. 217-232.
3.
R. D. Luce and D. H. Krantz (1971), "Conditional Expected
Utility", Econometrica, 39, pp. 253-271.
4.
M. K. Richter (1966), "Revealed Preference Theory",
Econometrica, 34, pp. 635-645.
5.
M. K. Richter (1975), "Rational Choice and Polynomial
Measurement Theory", Journal of Math. Psychology,
12, pp. 99-113.
6.
M. K. Richter and L. Shapiro (1977), "Revelations of a
Gambler", in preparation.
7.
L. J. Savage (1954), The Foundations of Statistics, New York:
Wiley.
8.
A. Tarski (1951), A decision method for elementary algebra
and geometry, 2nd ed., rev., Berkeley: University
of California Press.