Preference and Utility
Econ 2100
Fall 2015
Lecture 3, 9 September
Problem Set 1 is due now
Outline
1
Utility Functions and Preference Representation
2
Continuous Preferences
3
Existence of Utility Functions
Preferences and Utility Functions
Last week, we discussed what rationality implies for choices: given a preference
relation %, the induced choice rule C% is de…ned by
C% (A) = fx 2 A : x % y for all y 2 Ag:
On the other hand, from initermediate micro, we are used to modeling
consumers’behavior using a utility function: the consumer chooses, among all
a¤ordable bundles, one that maximizes utility:
Cutility (A) = fx 2 A : u(x) u(y ) for all y 2 BudgetSetg:
Next we study when these two ways of looking at choices are equivalent.
When is it possible for a preference relation to be described by a function?
What does this mean?
A utility function maps consumption bundles to real numbers.
It represents preferences if the function’s values rank any pair of consumption
bundles in the same way the preferences would rank them.
If it exists, a utility function provides a tremendous simpli…cation: instead of
comparing two highly dimensional objects using preferences, just look at which
of two real numbers is largest.
Question
Under what assumptions can a preference relation be represented by a utility
function?
Preference Representation
De…nition
A utility function on X is a function u : X ! R.
De…nition
The utility function u : X ! R represents the binary relation % on X if
x % y , u(x)
u(y):
Question 1, Problem Set 2 (due Monday September 14).
Let % be a preference relation. Prove that u : X ! R represents % if and only if:
x % y ) u(x)
u(y);
and
x
y ) u(x) > u(y):
This result provides an alternative way to show that that a preference is
represented by some function (very useful).
Existence of a Utility Function and Axioms
The following result says that completeness and transitivity are necessary for
existence of a utility function representing some preference relation.
Proposition
If there exists a utility function that represents %, then % is complete and transitive.
Proof.
Let u ( ) be a utility function that represents %.
Transitivity.
Suppose x % y and y % z.
Since u ( ) represents %, we have u(x)
Therefore, u(x)
u(y) and u(y)
u(z).
u(z).
Since u ( ) represents %, we conclude x % z. Hence, % is transitive.
Completeness.
Since u : X ! R, for any x; y 2 X , either u(x)
Since u ( ) represents %, we have x % y or y % x.
Therefore, % is complete.
u(y) or u(y)
u(x).
Preferences, Utility Functions, and Ordinality
Proposition
Suppose u : X ! R represents %. Then v : X ! R represents % if and only if
there exists a strictly increasing function h : u(X ) ! R such that v = h u.
Proof.
Question 2, Problem Set 2; due Monday September 14.
Note: One direction is easy while the other is not.
Exercise
De…ne the binary relation on RX by u v if there exists a strictly increasing
h : u(X ) ! R such that v = h u. Show that is an equivalence relation on RX .
These results are uniqueness statements of sorts.
not literally because there are many utility functions, rather than a single utility
function, in RX that represent %.
There is a single equivalence class, under , of utility functions that represents
%. The representation is therefore unique in RX = .
Existence of a Utility Function When X is Finite
Theorem
Suppose X is …nite. Then % is a preference relation if and only if there exists some
utility function u : X ! R that represents %.
When the consumption set is …nite the problem of …nding a utility function
that represents some given preferences is easy, almost trivial.
The proof is constructive: a function that works is the one that counts the
number of elements that are not as good as the bundle in question.
This function is well de…ned since the are only …nitely many bundles that can be
worse than something.
In other words, the utility function is
u(x) = j- (x)j
where - (x) = fy 2 X : x % yg is the lower contour set of x.
Notice that we only need to prove that this function represents - (x) because
from a previous result we know that if that is the case then - is a preference
relation.
Existence of a Utility Function When X is Finite
If X is …nite and % is a preference relation ) 9 u : X ! R that represents %.
Proof.
Let u(x) = j- (x)j. Since X is …nite, u(x) is …nite and therefore well de…ned.
Suppose x % y. I claim this implies u(x)
u(y).
Let z 2- (y), i.e. y % z.
By transitivity, x % z, i.e. z 2- (x). Thus - (y) - (x).
Therefore j- (y)j j- (x)j. By de…nition, this means u(y)
Now suppose x
u(x).
y. I claim this implies u(x) > u(y).
x y implies x % y, so the argument above implies - (y) - (x).
x % x by completeness, so x 2- (x). Also, x y implies x 2= (y).
Hence - (y) and fxg are disjoint, and both subsets of - (x). Then
- (y) [ fxg
- (x)
j- (y) [ fxgj
j- (x)j
j- (y)j + jfxgj
j- (x)j
u (y) + 1
u (y)
This proves
u (x)
<
u (x)
x % y ) u(x) u(y)
and thus we are done.
x y ) u(x) > u(y)
Utility Function for a Countable Space
Existence of a utility function can also be proven for a countable space.
Theorem
Suppose X is countable. Then % is a preference relation if and only if there exists
some utility function u : X ! R that represents %.
Proof.
Exercise
Notice that the previous construction cannot be applied directly here because
the cardinality of the lower contour sets can be in…nite.
Lexicographic Preferences
We want conditions on a preference relation which guarantee the existence of
a utility function representing those preferences even if X is not countable.
Typically, we assume that X lives in Rn , but ideally we would want as few
restrictions on X as we can get away with.
We know that without completeness and transitivity a utility function does not
exist, so we cannot dispense of those.
What else is needed?
Next, a counterexample of the existence of a representation.
Question 3, Problem Set 2 (due Monday September 14).
Show that the lexicographic ordering on R2 de…ned by
8
x1 > y1
<
or
(x1 ; x2 ) % (y1 ; y2 ) ,
:
x1 = y1 and x2
y2
is complete, transitive, and antisymmetric (i.e. if x % y and y % x, then x = y ).
The lexicographic ordering is a preference relation (it is complete and
transitive), yet admits no utility representation.
The Lexicographic Ordering in R 2 admits no utility representation
8
x1 > y1
<
or
Let X = R2 and de…ne % by (x1 ; x2 ) % (y1 ; y2 ) ,
:
x1 = y1 and x2
y2
Suppose there exists a utility function u representing % (…nd a contradiction).
For any x1 2 R, (x1 ; 1)
represents %.
(x1 ; 0) and thus u(x1 ; 1) > u(x1 ; 0) since u
The rational numbers are dense in the real line (for any open interval (a; b),
there exists a rational number r such that r 2 (a; b)); hence, there exists a
rational number r (x1 ) 2 R such that:
u(x1 ; 1) > r (x1 ) > u(x1 ; 0)
(A)
De…ne r : R ! R by selecting r (x1 ) to satisfy u(x1 ; 1) > r (x1 ) > u(x1 ; 0).
CLAIM: r ( ) is a one-to-one function.
Suppose x1 6= x10 ; and without loss of generality assume x1 > x10 .
since (x1 ;0) (x10 ;1)
by (A)
by (A)
Then:
z}|{
z}|{
z}|{
0
r (x1 ) > u(x1 ; 0)
>
u(x1 ; 1) > r (x10 )
Thus r ( ) is an a one-to-one function from the real numbers to the rational
numbers, which contradicts the fact the real line is uncountable.
Continuous Preferences
Continuity will get rid of this example.
De…nition
A binary relation % on the metric space X is continuous if, for all x 2 X , the upper
and lower contour sets, fy 2 X : y % xg and fy 2 X : x % yg, are closed.
Examples
De…ne % by (x1 ; x2 ) % (y1 ; y2 ) ,
8
<
x1 > y1
or
:
x1 = y1 and x2
y2
2
Draw the upper contour set of (1; 1); this preference on R is not continuous.
8
< x1 y1
and
De…ne % by (x1 ; x2 ) % (y1 ; y2 ) ,
:
x2 y2
Draw the upper contour set of (1; 1); this preference on R2 is continuous.
Continuous Preferences
De…nition
A binary relation % on the metric space X is continuous if, for all x 2 X , the upper
and lower contour sets, fy 2 X : y % xg and fy 2 X : x % yg, are closed.
Proposition
A binary relation % is continuous if and only if:
1
2
If xn % y for all n and xn ! x, then x % y; and
If x % yn for all n and yn ! y, then x % y.
Proof.
This follows from the fact that a set is closed if and only if it contains all of its
limit points.
Debreu’s Representation Theorem
The main result of today is due to Gerard Debreu, and it provides necessary and
su¢ cient conditions for the existence of a continuous utility function.
Theorem (Debreu)
Suppose X Rn . The binary relation % on X is complete, transitive, and
continuous if and only if there exists a continuous utility representation u : X ! R.
We will prove su¢ ciency next (under a couple of simplifying assumptions, but
you should ckeck what happens without those).
You will do necessity as homework.
Question 4, Problem Set 2.
Suppose X Rn . Prove that if u : X ! R is a continuous utility function
representing %, then % is a complete, transitive, and continuous preference relation.
Debreu’s Representation Theorem
Theorem (Debreu)
Suppose X Rn . The binary relation % on X is complete, transitive, and
continuous if and only if there exists a continuous utility representation u : X ! R.
Two simplifying assumptions:
X = Rn ; and
% is strictly monotone, (i.e. if xi
yi for all i and x 6= y, then x
y).
When strict monotonicity holds we have
, e % e:
(mon)
where e = (1; 1; : : : ; 1) (make sure you check this).
To prove:
if % is a continuous and strictly monotone preference relation on Rn , then
there exists a continuous utility representation of %.
How do we …nd a utility function? Look at the point on the 45o line that is
indi¤erent:
u(x) =
(x) where
The proof is in 3 steps.
(x) is the real number
such that
e
x
Step 1: There exists a unique
(x) such that
e
x.
Proof.
Let B = f 2 R : e % xg
R and de…ne
= inf f 2 R : e % xg :
|
{z
}
B
Oviously, (maxi xi )e x, so by strict monotonicity (maxi xi )e % x which
implies B is nonempty and
is well-de…ned.
Now show that
e % x and
e - x, so that
e
x.
Since
is the in…mum of B , there exists a sequence n 2 B s.t. n !
Then n e ! e (in Rn ) and n e % x because 2 B .
By continuity, one gets:
e % x as desired.
Since
is a lower bound of B , if 2 B )
.
Using the contrapositive:
<
) e x:
1
Let n =
.
By
(A),
e
x,
so
n
n e - x.
n
Also, n e ! e (in Rn ). Hence, by continuity, e - x as desired.
To prove uniqueness, suppose ^ e
By transitivity, ^ e
x.
e. Then, by (mon), ^ =
.
.
(A)
Step 2: The u(x) that represents % is de…ned as follows:
u(x) =
(x)
where
(x) is the unique
such that
e
Proof.
Show that u(x) represents %.
Suppose x % y.
By construction of
, we have:
x
(x)e
and
By transitivity, we have:
x
(x)e %
By (mon) this is equivalent to
(x)
(y )e
(y)e
y
y
(y)
Repeat the same argument to show that x y implies (x) >
x % y ) u(x) u(y)
Since we have shown that
we are done.
x y ) u(x) > u(y)
(y).
x:
Step 3: The de…ned u(x) is continuous.
(x) where (x) is the unique
such that
u(x) =
e
x
Proof.
To prove that a function f : Rn ! R is continuous, it su¢ ces to show that
f 1 ((a; b)) is open for all a; b 2 R.
Since ae
ae, we have u(ae) =
Notice that
u 1 ((a; b)) = u
1
(ae) = a for any a 2 R.
((a; 1) \ ( 1; b)) = u
1
((a; 1)) \ u
1
(( 1; b)):
But u(ae) = a, so
u
1
((a; 1)) = u
1
((u(ae); 1)) = fx 2 Rn : x
aeg;
this is open because the strict upper contour set of ae is open whenever % is
complete and continuous (why?).
An entirely symmetric argument proves that u 1 (( 1; b)) is the strict lower
contour set of be, hence it is also an open set.
Since u
1
((a; b)) is the intersection of two open sets, it is open.
Debreu’s Representation Theorem
Back to the theorem
Theorem (Debreu)
Suppose X Rn . The binary relation % on X is complete, transitive, and
continuous if and only if there exists a continuous utility representation u : X ! R.
Conclusion
When her preference relation is complete, transitive, and continuous a
consumer’s behavior is entirely described by some continuous utility function
that represents that preference relation.
The theorem asserts that one of the utility representations for % must be
continuous, not that all of the utility representations must be continuous.
Continuity is a cardinal feature of the utility function, not an ordinal feature,
since it is not robust to strictly increasing transformations.
Exercise
Construct a preference relation on R that is not continuous, but admits a utility
representation.
Choice Rules and Utility Functions
The induced choice rule for % is C% (A) = fx 2 A : x % y for all y 2 Ag
Proposition
If % is a continuous preference relation and A
then C% (A) is nonempty and compact.
Rn is nonempty and compact,
Proof.
Suppose % is continuous.
By Debreu’s Theorem, there exists some continuous function u representing
the preferences.
One can show (do it as exercise) that
C% (A) = arg max u(x):
x2A
Nonemptiness then follows from continuity of u and the Extreme Value
Theorem.
Compactness follows from the fact u 1 ( ) is bounded (it is a subset of the
bounded set A), and closed (since the inverse image of a closed set under a
continuous function is closed).
Econ 2100
Fall 2015
Problem Set 2
Due 14 September, Monday, at the beginning of class
1. Let % be a preference relation. Prove that u : X ! R represents % if and only if:
x % y ) u(x)
and
u(y);
x
y ) u(x) > u(y):
2. Suppose u : X ! R represents %. Then v : X ! R represents % if and only if there exists a strictly
increasing function h : u(X) ! R such that v = h u.1
3. Show that the lexicographic ordering on R2 de…ned by
8
x1 > y1
<
or
(x1 ; x2 ) % (y1 ; y2 ) ,
:
x1 = y1 and x2
y2
is complete, transitive, and antisymmetric (i.e. if x % y and y % x, then x = y).
4. Prove that if u : X ! R is a continuous utility function representing % then % is a continuous
preference relation.
5. Prove that if % is a re‡exive (but not necessarily complete) and transitive binary relation on a …nite
set X, then there exists a function f : X ! R such that x % y implies f (x)
f (y). Does this
function represent %? Explain.
1
Suppose Z
R. The function h : Z ! R is strictly increasing if h(z) > h(z 0 ) for all z; z 0 2 Z such that z > z 0 .
1