Math Methods for Economics
and Microeconomic Theory
Cesar E. Tamayo
Department of Economics, Rutgers University
[email protected]
Class notes: fall 2010
Contents
I
Math Methods
4
1 Topology and analysis
1.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . .
1.2 Series, sequences and subsequences . . . . . . . . . .
1.3 Open and closed sets . . . . . . . . . . . . . . . . . .
1.4 Completeness, boundedness and compactness . . . . .
1.5 Continuity of functions . . . . . . . . . . . . . . . . .
1.6 Continuity of correspondences (set-valued mappings)
1.7 Maximum theorems . . . . . . . . . . . . . . . . . . .
1.8 Fixed point theory . . . . . . . . . . . . . . . . . . .
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5
5
6
7
9
11
14
16
18
2 Convex optimization
2.1 Convex sets . . . . . . . . . . . . . . . . . . .
2.2 Separating hyperplanes . . . . . . . . . . . . .
2.3 Convexity of functions and subgradients . . .
2.4 Support function theory . . . . . . . . . . . .
2.5 Kuhn-Tucker theory . . . . . . . . . . . . . .
2.5.1 Introduction to nonlinear programming
2.5.2 The Kuhn-Tucker conditions . . . . . .
2.5.3 Constraint quali…cations . . . . . . . .
2.5.4 Non-negativity and equality constraints
2.5.5 Su¢ ciency conditions . . . . . . . . . .
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21
21
22
24
29
32
32
32
35
36
38
II
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Microecomic Theory
40
3 Producer and consumer theory
3.1 Producer theory . . . . . . . . . . . . . .
3.1.1 Production sets and technologies
3.1.2 Cost minimization . . . . . . . .
3.1.3 Pro…t maximization . . . . . . . .
3.2 Consumer theory . . . . . . . . . . . . .
3.2.1 Utility maximization . . . . . . .
1
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41
41
41
43
46
47
47
3.2.2
Expenditure minimization . . . . . . . . . . . . . . . . . . . . . . .
49
4 Game Theory and General Equilibrium
4.1 Game theory . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Zero-sum games . . . . . . . . . . . . . . . . . . . . .
4.1.2 Non-zero-sum games and Nash Equilibrium . . . . .
4.1.3 The generalized game . . . . . . . . . . . . . . . . . .
4.2 General Equilibrium theory . . . . . . . . . . . . . . . . . .
4.2.1 Prelminaries . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 Existence of equilibrium in pure exchange economies
4.2.3 Existence of equilibrium in production economies . .
4.2.4 Welfare theorems (pure exchange economies) . . . . .
4.2.5 Relaxing the Walrasian assumptions . . . . . . . . .
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51
51
51
51
53
54
54
56
58
61
66
5 Decision making under uncertainty
5.1 Expected utility hypothesis . . . . . . . . . . . . . . . .
5.2 Risk aversion . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Application: portfolio choice . . . . . . . . . . . .
5.3 Comparative risk aversion . . . . . . . . . . . . . . . . .
5.3.1 Application: portfolio choice . . . . . . . . . . . .
5.3.2 Application: insurance . . . . . . . . . . . . . . .
5.4 First order stochastic dominance (FOSD) . . . . . . . . .
5.4.1 FOSD and precautionary savings . . . . . . . . .
5.4.2 FOSD and portfolio choice . . . . . . . . . . . . .
5.5 Likelihood ratio stochastic dominance . . . . . . . . . . .
5.6 Concave and second order stochastic dominance (SOSD)
5.6.1 Concave order SD and pro…t maximization . . . .
5.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . .
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70
70
71
72
74
76
76
79
81
83
84
85
88
89
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A Review of functions, di¤erentiation and integration
90
B Review of vectors and matrix algebra
93
2
Summary
These notes cover the …rst semester mathematics and microeconomics material of the
PhD program at Rutgers University. The range of mathematical tools presented below
is wide, including topology, real analysis, convex optimization, …xed point theory and
stochastic dominance, but attention is limited to their use in economic theory. The notes
are almost entirely based on lectures by Professor Richard P. McLean, but all errors
and omissions are my sole responsability. Ocasionally, I also refer to examples and
de…nitions found in Carter’s (2001) "Foundations of Mathematical Economics", Mas-Colell
et. al’s (1995) Microeconomic Theory and Varian’s (1992) Microeconomic Analysis.
3
Part I
Math Methods
4
Chapter 1
Topology and analysis
1.1
Metric spaces
De…nition 1 Let X be a non-? set. A metric on X is a function d : X
the following 8 x; y; z 2 X :
X ! R satisfying
1:d (x; y) 0
2:d (x; y) = d (y; x)
3:d (x; y) = 0 , x = y
4:d (x; y) d (x; z) + d (z; y)
For the case when X = Rn we have:
d1 (x; y) =
n
X
i=1
jxi
v
u n
uX
(xi
d2 (x; y) = t
yi j
yi )2
i=1
..
.
d1 (x; y) = max fjxi
i=1:::n
yi jg
De…nition 2 A metric space is a pair (X; d) where X is a non-? set and d is a metric on
X:
De…nition 3 Let (X; d) and (Y; ) be metric spaces. The Box metric on X
maxfd(x; x^); (y; y^)g
Y is ((^
x; y^); (x; y)) =
De…nition 4 The point-to-set distance d(x; A) = miny2A d(x; y)
Notation 1 The symbol B" (x) denotes the open ball of radius " centered at x:
De…nition 5 Given A; B compact (see de…nitions 25-26 below), the Hausdor¤ distance
is: Hd (A; B) = maxfmaxx2A d(x; B); maxx2B d(x; A)g: Note that each max function is well
de…ned since A; B are compact (see Theorem 25 below)
5
1.2
Series, sequences and subsequences
De…nition 6 A sequence in X is a function f : N ! X (a 1-to-1 mapping from N to X).
De…nition 7 A sequence is called convergent with limit x if, for every " > 0; 9 N s.t.
whenever n > N; xn 2 B" (x).
Lemma 1 A convergent sequence fxn g in a metric space (X; d) has exactly one limit.
Proof. Suppose that xn ! x1 and xn ! x2 . Then we must show that d (x1 ; x2 ) = 0.
To see this, note that d (x1 ; x2 ) = 0 , d (x1 ; x2 ) < " 8 " > 0. Choose " > 0. Since
limn!1 fxn g = x1 we know that 9 n
^ 1 such that n > n
^ 1 ) xn 2 B 2" (x1 ). Likewise,
since limn!1 fxn g = x2 we know that 9 n
^ 2 such that n > n
^ 2 ) xn 2 B 2" (x2 ). So
choose m = max f^
n1 ; n
^ 2 g and note that xm 2 B 2" (x1 ) and xm 2 B 2" (x2 ); to complete the
argument, use the triangle inequality to conclude:
d (x1 ; x2 )
<
=
)
)
d (x1 ; xm ) + d (x1 ; xm )
" "
+
2 2
"
d (x1 ; x2 ) < " 8" > 0:
x1 = x2
Example 2 Let X = R1 and d =Euclidean metric. The sequence xn = 1=n is convergent
with limit zero, i.e., limn!1 xn = 0. To see why, choose " > 0 and n
^ such that n
^ > 1=".
Now, n > n
^ implies:
1
1
d (0; xn ) = j0 xn j = < < "
n
n
^
Example 3 A constant sequence xn = k is trivially convergent (with limit k):
Example 4 Let X = R1 and d =Euclidean metric. The sequence xn = ( 1)n is not convergent. To see this note that any element other than 1 or 1 cannot be a candidate for
a limit since we can always …nd an open ball around it which contains no elements of the
sequence. Furthermore, 1 or 1 cannot be limits of the sequence either since @ n
^ such that
8 " > 0; n > n
^ ) d (xn ; 1) < " or d (xn ; 1) < ":
De…nition 8 fxn g1
n=1 is a Cauchy Sequence if for any " > 0; 9 N s:t: 8 i; j
".
N; d(xi ; xj ) <
1
De…nition 9 A subsequence of fxn g1
n=1 is a subset fxnm gm=1 such that n1 < n2 < n3:::
Remark 1 Every sequence is a subsequence of itself.
6
Remark 2 fxn g1
n=1 is convergent , 9 x such that every subsequence of fxn g is convergent
with limit x:
We note that if limfxn g = x and limfyn g = y, then limfxn + yn g = x + y. Likewise for
multiplication. It is also true that a sequence is convergent iif every one of its subsequence
is convergent with the same limit
In the last example, we can see that constructing two subsequences, one for n =even
and one for n =odd we obtain di¤erent limits for each one of them. Thus, we con…rm that
the sequence xn = ( 1)n is not convergent.
Lemma 2 Let fxk g be a sequence in (X; d) and x 2 X. If xk 2 B 1 (x) 8 k then fxk g is
k
convergent with limit x
Proof. Suppose that xk 2 B 1 (x) 8 k:.Pick " > 0 and choose K such that K > 1" . Then
k
k > K ) xk 2 B 1 (x). Also, k > K ) K1 > k1 so that:
k
B 1 (x)
k
B 1 (x)
K
B" (x)
) fxk g is convergent with limit x
Lemma 3 (sandwich I) If yn ! !, zn ! ! and yn
xn
zn ; 8 n then xn ! !.
De…nition 10 If xk ; x 2 Rn we say that xk ! x () xk is pairwise convergent to x in
R1 that is, if. x1k ! x1 ; x2k ! x2 :::
1.3
Open and closed sets
De…nition 11 Let (X; d) be a metric space (m:s:) and let S
X. If, for x 2 X; 9 " >
0 s:t: B" (x)
S, then x is an interior point of S. The set of all interior points of S is
intS.
De…nition 12 x is NOT an interior point of S if B" (x) \ X n S 6= 0; 8 " > 0.
This esures that an (n 1)-dimensional subset of Rn cannot be an open set since a
n-dimesional ball around any of its points would contain points outside the subset.
De…nition 13 Let (X; d) be a metric space (m:s:):S
X.is an open set if intS = S:
De…nition 14 Suppose that (X; d) is a metric space, x 2 X and " > 0. Then an open ball
centered at x with radius " is the set:
B" (x) = fy 2 X j d (x; y) < "g
7
Example 5 Let X = R1 and d1 (x; y) above, the open ball aroun x is:
B" (x) =
=
y 2 R1 j d1 (x; y) < "
y 2 R1 j x " < y < x + "
Example 6 Let X = R2 and d2 (x; y) ; the open ball aroun x is:
B" (x) =
=
y 2 R2 j d2 (x; y) < "
q
2
y 2 R j (x1 y1 )2 + (x2
y2 )2 < "
De…nition 15 Two metrics d; d0 are equivalent if an open set in (X; d) is open in (X; d0 ),
that is, if for x 2 X, B"d (x) B"d0 (x) and vice versa.
Example 7 If S = ? then S is open. If S = X then S is open in (X; d):
Example 8 Let X = R1 and d1 (x; y). Let S = (0; 1). We can claim that S is an open set.
To see this, choose any x 2 S and " = min f1 x; xg : Then: B" (x) (0; 1). To see why
choose any y 2 B" (x) and note:
y 2 B" (x) ) jx yj < "
) " < y x < " (de…nition of abs value)
)x "<y <x+"
) x x < x " < y < x + 1 x (by def of ")
)0<y<1
) y 2 B" (x) ) y 2 S = (0; 1)
) B" (x) S = (0; 1)
) S is open
Example 9 Let X = R1 and d1 (x; y). Let S = Q =rational numbers.Then intS = ?:
Remark 3 Between any two real numbers, there are at least one rational and one irrational
number.
Proposition 1 Every open ball is an open set.
Proof. Let (X; d) be a metric space. Choose x 2 X and pick r > 0. Next, choose y 2 Br (x).
Now let " = r d (x; y) and note that " > 0 since r d (x; y) > 0: We must show that any
point in B" (y) is also a point in Br (x). To do this, choose z 2 B" (y). Then:
d (z; x)
<
=
=
)
)
)
d (z; y) + d (y; x)
" + d (y; x)
[r d (x; y)] + d (y; x)
r
d (z; x) < r
z 2 B" (y) ) z 2 Br (x)
B" (y) Br (x)
8
De…nition 16 x is a closure point of S , B" (x) \ S 6= ? 8 " > 0: In turn, x is NOT a
closure point of S , B" (x) S c 8 " > 0.
Note that it is always true that: S
clS, since x 2 S ) x 2 B" (x) \ S, thus,
B" (x) \ S 6= ;, hence, every x 2 S is a closure point of S.
De…nition 17 S is closed , clS = S, which by the remark above implies that S is closed
, clS S.
Example 10 S = X is closed in (X; d). S = ? is closed.
Example 11 Let X = R1 and S = f0; 1g. Then nothing outside [0; 1] is a closure point.
To se why, note that there will always 9 " > 0 su¢ ciently small such that B" (x)\S = ? for
x 2 ( 1; 0)[(1; 1). Likewise, nothing in (0; 1) is a closure point of S. Thus, clS = f0; 1g.
) S is closed.
De…nition 18 (sequential characterization of closure point) x is a closure point of S if
9 fxk g s:t: xk 2 S 8 k and fxk g ! x.
De…nition 19 (sequential characterization of closed set) S in Rm is a closed set if whenever fxn g is a convergent sequence completely contained in S, it follows that limfxn g 2 S.
De…nition 20 Discrete metric: TBC
Remark 4 The intersection of a …nite collection of open sets is an open set.
Remark 5 The union of any collection of open sets is an open set.
Remark 6 The intersection of any collection of closed sets is a closed set.
Remark 7 The union of a …nite collection of closed sets is a closed set.
De…nition 21 x is a boundary point of S (denoted x 2 @S) if 8 " > 0; B" (x) \ S 6= ? and
B" (x) \ S c 6= ? (i.e., if x 2 clS but x 2
= intS).
1.4
Completeness, boundedness and compactness
De…nition 22 A m.s.(X; d) is complete if any cauchy sequence in (X; d) is convergent with
limit 2 X.
9
Example 12 Let X = [0; 1) and d =Euclidean. Claim: "(X; d) is not complete". To see
k
and note that fxk g is a Cauchy sequence. This can
why consider the sequence xk = k+1
be shown by choosing " > 0 and noting that 9 N such that k > N ) d (xk ; 1) < "=2 and
m > N ) d (xm ; 1) < "=2; thus, by the triangle inequality:
d (xk ; xm )
d (xm ; 1) + d (xk ; 1)
< "=2 + "=2
= "
) xk is a Cauchy sequence
Furthermore, xk 2 X 8 k. However, limk!1 xk = 1 2
= X:
De…nition 23 A m.s. (X; d) is totally bounded if 9 …nite set fx1 :::xm g
B" (x1 ) [ ::: [ B" (xm )
De…nition 24 A set S is bounded if 9 B s:t: 8 x 2 S; kxk
X s:t: X
B.
Lemma 4 If set S is bounded above, it has a smaller upper bound, or supremum: or
K = supS.
De…nition 25 A m.s. (X; d) is compact () it is complete and totally bounded
De…nition 26 Let (X; d) be a m.s. and S
a compact m.s.
X; then S is compact in (X; d) () (S; d) is
De…nition 27 A m.s. (S; d) is compact () every sequence fxk g contains a subsequence
fxkm g1
m=1 whose limit 2 (S; d) (i.e.every sequence contains a convergent subsequence in
(S; d))
De…nition 28 Let (X; d) be a metric space. We say that S
X is compact set if the
folowing holds: whenever C is a collection of open sets in (X; d) whose union contains S,
9 a …nite subcollection from C whose union contains S:The collection C is called an open
cover. Conversely, S is not compact if 9 an open cover (or collection), C from which we
cannot extract a subcollection whose union contains S:
Remark 8 C is not a subset of X but a collection of subsets of X: That is, C is not a union
of sets.
Example 13 Let X = [0; 1), d =Euclidean metric and S = [0; 1). Claim: S is not compact. To see why we must exhibit an open cover from where no …nite subcovers containing
S can be extracted. Consider:
C=
0;
n
n+1
10
jn
1
n
is open and:
and note that each 0; n+1
1
[
n=1
0;
n
n+1
= [0; 1)
n
i.e., every point in [0; 1) belongs to at least one of the sets 0; n+1
. Thus, C is an open
cover for S. However, we cannot extract a …nite subcollection of C whose union contains
S:
Remark 9 Every closed subset of a compact set is compact.
Remark 10 A …nite subset of a metric space is compact.
Remark 11 The real numbers (Rn ) with the Euclidean metric is a complete metric space.
Theorem 14 (Heine-Borel) Let X = Rn , d = Euclidean metric. We say that S
compact , S is closed and bounded.
Rn is
Proof. ()) Suppose that S is compact. If S = ? then S is closed and bounded. If S 6= ?
then S compact ) (S; d) is compact. That it, (S; d) is complete and totally bounded. Since
(S; d) is complete, every Cauchy sequence is convergent in (S; d) ; since every convergent
sequence is a Cauchy sequence, every convergent sequences has its limit in (S; d), thus, S
is closed. Since (S; d) is totally bounded it follows that S is bounded.
(() Suppose that S is closed and bounded. If S = ? then trivially S is compact. If
S 6= ? then by the remark above, we know that X = Rn ) (X; d) is complete and S X
plus (X; d) complete, implies that (S; d) is complete. Also, S bounded implies that (S; d)
is totally bounded. Thus, (S; d) is a compact metric space. Hence, S is compact in (X; d).
Corollary 1 If S is a compact subset of the metric space (X; d), then S is closed and
bounded in (X; d) :
Proposition 2 A metric space (S; d) is compact () every sequence fxk g contains a
1
subsequence fxkm g1
m=1 whose limit 2 (S; d) (i.e.fxkm gm=1 is convergent in (S; d))
Proposition 3 Let qk = kppkk k : Then kqk k = 1 and the set fqk j kqk k = 1g is compact, i.e.
9 qkm and q s.t. kqk = 1 and qkm ! q
1.5
Continuity of functions
Notation 15 Denoting the function f : X ! Y we call X the domain and Y the co-domain.
The co-domain is a subset of the range.
11
Notation 16 Let d be a metric de…ned on X and
denoted f : (X; d) ! (Y; ) :
a metric on Y then a function is
De…nition 29 A function f : (X; d) ! (Y; ) is continuous at x if, for every " > 0, 9
> 0 such that whenever x 2 B (x) ) f (x) 2 B" (f (x)).
De…nition 30 A function f : (X; d) ! (Y; ) is continuous if it is continuous at every
x 2 X.
Proposition 4 (sequential characterization of continuity) A function f : (X; d) ! (Y; )
is continuous at x ,, whenever xk ! x it follows that f (xk ) ! f (x)
Proof. ()) Suppose that f is continuous at x 2 X. Choose fxn g such that xn ! x:
We must show that f (xn ) is convergent in (Y; ) with limit f (x) : Equivalently, we must
show that 8 " > 0 9 > 0 such that k > k^ ) f (xn ) 2 B" (f (x)). To see this choose
" > 0. Since f is continuous, at x it follows that 9 > 0 such that x 2 B (x) )
f (x) 2 B" (f (x)) : Next, since xk ! x we know that 9 k^ such that k > k^ ) xn 2 B (x).
) k > k^ ) xn 2 B (x) ) f (xn ) 2 B" (f (x)) :
(() (contrapositive) Now suppose that f is not continous at x. Since f is not continous
9 some " > 0 such that 8 > 0 we can …nd x 2 B (x) with the property that f (xn ) 2
=
B" (f (x)). Therefore, 9 " > 0 such that 8 n we can …nd xn 2 B1=n (x) but f (xn ) 2
=
B" (f (x)). Thus from lemma 34, we conclude that xn ! x but f (xn ) 2
= B" (f (x)). Thus
f (xn ) is not convergent with limit f (x) :
De…nition 31 f is uniformly continuous if for every "; 9 a that "works" for any arbitrary point x (i.e., does not depend upon x as in mere continuity)
We note that a continuous function on a compact domain is uniformly continuous.
Also, a linear combination of continuous functions is continuous. Finally, the composition
of continuous functions is continuous.
Proposition 5 Every uniformly continuous function is continuous.
Proof. Suppose that f : (X; d) ! (Y; ) is uniformly continuous. Choose " > 0: Then
9 some > 0 such that (f (x) ; f (y)) < " whenever x 2 X, y 2 Y and d (x; y) < : In
particular, (f (x) ; f (y)) < " whenever d (x; y) < so that f is continuous at x. Since f is
continuous at some arbitrary x it is continuous at every point and therefore is continuous.
Exercise 17 (A.11, McLean) Again let f : (X; d) ! (Y; ) and suppose that X =
Y = (0; 1) and d =
= Eclidean. Show that the function f (x) = 1=x is continuous
but not uniformly continuous.
12
Proof. To see that f is continuous, choose " > 0 and x > 0. Next, choose:
0<
so that:
jx
yj <
) jx
jx
yj <
)
x2 " x
;
2 2
< min
yj <
x
)x
2
y<
x
x
) <y
2
2
consequently:
1
jx yj
=
y
xy
1
x
xy
<
2
<"
x2
However, f is not uniformly continuous. To see why, choose " = 12 : For each integer m > 0
1
: Then:
choose xm = m1 and ym = 2m
jxm
but:
If
> 0, then choose m so that
1
m
1
2m
ym j =
1
2m
1
=m
2m
< . Then jxm
ym j <
but
1
m
1
2m
= m > 12 :
De…nition 32 f : (X; d) ! (Y; ) is uniformly continuous if, whenever xk is a Cauchy
sequence in (X; d) ; it follows that f (xk ) is Cauchy sequence in (Y; ) :
Example 18 (2.82, Carter) Let f : [0; 1) ! R be de…ned by:
f (x) =
x
1
x
then f is continuous but not uniformly continuous. To see why, note that the sequence
xk = 1 k1 is a Cauchy sequence in X but f (xk ) is not a Cauchy sequence in Y:
Example 19 If xk ! x then jxk j ! jxj i.e., the absolute value function is continuous.
Lemma 5 (sandiwch II) If g(x) ! l; h(x) ! l as x ! x and g(x)
f (x) ! l as x ! x.
De…nition 33 The function f : Rn ! Rm de…ned as:
0
1
f1 (x)
B f2 (x) C
B
C
f (x) : B .. C
@ . A
fm (x)
f (x)
is continuous if and only if fi : Rn ! R1 is continuous for each i = 1; :::m
13
h(x), then
Remark 12 The function f (x) = 1=x is unbounded from above but is bounded from below.
De…nition 34 The mapping f : (X; d) ! (Y; ) is upper semi continuous if the set
fx 2 X j f (x)
g is closed in (X; d) :
De…nition 35 The mapping f : (X; d) ! (Y; ) is lower semi continuous if the set
fx 2 X j f (x)
g is closed in (X; d) :
Example 20 Consider the function:
f (x) =
1
if x 0
1 if x < 0
we can claim that f is USC but not LSC. To se why, note that if > 1 the set fx 2 X j f (x)
g=
? which is closed. Next, if 1 <
1 the set fx 2 X j f (x)
g = fx 2 X j x 0g
which is also closed (the set [0; 1) is closed). Finally, if < 1 the set fx 2 X j f (x)
g=
R, also a closed set. Thus, for any value of , the set fx 2 X j f (x)
g is closed. On
the other hand, if e.g., 0 < < 1 then the set fx 2 X j f (x)
g = ( 1; 0) which is not
a closed set.
Remark 13 The distribution function of a random variable is USC.
Remark 14 A function is continuous if and only if it is USC and LSC.
Proposition 6 Suppose that f : (X; d) ! (Y; ) is continuous. If S is compact in (X; d),
then f (S) is compact in (Y; ) :
Proof. Choose some sequence fyk g such that yk 2 f (S) 8 k: This in turn implies that for
each k, 9 xk 2 S such that yk = f (xk ). Since fxk g is a sequence in S and S is compact,
9 a subsequence fxkm g and x such that xkm ! x. Moreover f continuous implies that
f (xkm ) ! f (x). But f (xkm ) = ykm so that ykm ! f (x) :Thus, fykm g is a subsequence of
fyk g convergent with limit in f (S). Hence, f (S) is compact.
1.6
Continuity of correspondences (set-valued mappings)
De…nition 36 Suppose that X 6= ?. A relation in X is a "rule" that associates with each
x 2 X a non-? subset of X, denoted (x) :
De…nition 37 (preference relation) Let : X
X be a relation.Then: i) is re‡exive
if x 2 (x) ; ii) is transitive if y 2 (x) and z 2 (y) ) z 2 (x) and iii) is complete
if x 6= y ) x 2 (y) and/or y 2 (x) : We say that is a preference relation if it sats…es
i) iii):
14
De…nition 38 Let : X
X be a relation. A function u : X ! R is a representation for
if u (x) u (y) , x 2 (y) :
De…nition 39 A correspondence or set-valued mapping, : (X; d)
assigns to every element x 2 X, a non-? subset (x) of Y.
(Y; ); is a rule that
Proposition 7 Let (X; d) is a m.s. and if: i)
: X
X is re‡exive, transitive and
complete, ii) (x) is closed in (X; d), then (x) has an USC representation.
De…nition 40 : (X; d)
(Y; ) is upper hemi continuous (UHC) at x if for every open
set U (Y; ) s.t. (x) U 9 > 0 s.t. whenever x 2 B (x) ) (x) U
De…nition 41
is lower hemi continuous (LHC) at x if for every open set U
s.t. (x) \ U 6= ?, 9 > 0 s.t. whenever x 2 B (x) ) (x) \ U 6= ?:
De…nition 42 The graph of
: (X; d)
(Y; ) is de…ned as: Gr( ) = f(x; y) 2 X
(Y; )
Yjy2
De…nition 43 : X
Y is closed graph (w.r.t. box metric) at x if i) xk ! x ii) yk ! y
and iii) yk 2 (xk ) 8 k ) y 2 (x):
De…nition 44 Alternatively is closed graph at x if Gr( ) is a closed set, i.e., whenever
(xk ; yk ) ! (x; y) and (xk ; yk ) 2 Gr( ) 8 k then it follows that (x; y) 2 Gr( ).
Lemma 6
is continuous , it is UHC and LHC.
Remark 15 A constant correspondence
and LHC).
Lemma 7 If
is closed graph )
(x) = K 8 k is trivially continuous (i.e. UHC
is closed-valued
Example 21 (2.88, Carter) The correspondence
(x) =
1
x
f0g
: R+
R de…ned by:
if x > 0
if x = 0
has closed graph but is not UHC. To see why, note that Gr( ) = f(x; 1=x) 2 R2 jx > 0g [
(0; 0) which is a closed set in R2 ; however, for every sequence fxk g such that x ! 0; the
sequence yk 2 (xk ) does not converge.
Example 22 The constant correspondence (x) = (0; 1) is UHC but not closed-valued
and therefore does not have closed graph (closed graph)closed valued, so qclosed valued)
qclosed graph).
Proposition 8 If 1 :(X; d)
(Y; ) is closed graph and
compact-valued, then 1 \ 2 is compact-valued.
15
2 :(X; d)
(Y; ) is UHC and
(x)g
Proposition 9 The product of UHC correspondences is UHC. Thus, the product of UHC
and compact correspondences is UHC and compact.
Proposition 10 If
compact set.
: (X; d)
(Y; ) is closed graph, then
is UHC whenever Y is a
Example 23 (2.89, Carter) Let P denote the domain of the budget correspondence, that
is, the set of all prices and incomes pairs for which some consumption is feasible:
)
(
n
X
pi xi m
P = (p; m) 2 Rn Rj min
x2X
i=1
where X is the consumption set, the graph of the budget correspondence X (p; m) is given
by :
(
)
n
X
Gr (X (p; m)) = (p; m; x) 2 P X j
pi xi m
i=1
which is a closed set in P X. To see why, let fxk g be a sequence of consumption bundles
X (p; m). Since X (p; m) is bounded, xk ! x for some x 2 X: Thus,
p1 x1k + ::: + pn xnk
m
p1 x1 + ::: + pn xn
m
so that:
so that (p; m; xk ) ! (p; m; x) and (p; m; x) 2 Gr (X (p; m)). Therefore, Gr (X (p; m))
is closed. Consequently, if the consumption set X is compact, the budget correspondence
X (p; m) is UHC.
Theorem 24 Suppose that :(X; d)
(Y; ) is UHC and compact valued. If K is compact
in (X; d) then (K) is compact in (Y; ).
Proof. Choose yk 2 (K). We must show that 9 a subsequence ykm ! y and y 2 (K).
To see this, choose xk 2 K such that yk 2 (xk ) : Now, K compact implies that even
if xk is not convergent, it must contain a subsequence fxkm g with xkm ! x and x 2 K:
Summarizing: 1) xkm ! x, 2) yk 2 (xk ) so ykm 2 (xkm ). Now, UHC and compact
1
valued implies that 9 ykmt t=0 and y such that ykmt ! y. Finally, x 2 K so that
(x)
(K). Therefore, y 2 (K), so we have constructed a subsequence of yk which is
convergent and whose limit belongs to (K) :
1.7
Maximum theorems
First we introduce the main theorem that relates topological assumptions of a maximization
problem with the existence of optima.
16
Theorem 25 (Weierstrass) Suppose that (X; d) is a metric space and f : X ! R is a
continuous function. If S
X is non-? and compact, then f attains a maximum and a
minimum in S:
Proof. First, note that f (S) is non-? and compact in R. Hence, by the Heine-Borel
theorem f (S) is bounded and in particular, bounded from above. Next, we notice that
any set in R has a least upper bound or sup. Let sup f (S) = . Now, let Ak = f (S) \
1
1
. Note Ak 6= ? 8 k since there must be something between
and
y2Rjy
k
k
which belongs to f (S) or else would not be the least upper bound. So, choose fyk g
1
1
which implies that
yk
such that yk 2 f (S) \ y 2 R j y
. Thus,
k
k
yk ! :. But f (S) is compact and therefore closed so it must be that 2 f (S) : Now if
2 f (S), then 9 x 2 S such that f (x) = . Therefore, f (x)
= f (x) for every x 2 S.
Thus x is a maximizer for f in S. An identical argument can be applied to show that f
attains a minimum in S:
Next, we study the properties of the maximizer correspondence and the optimal value
function of maximization prtoblems. Consider the general constrained maximization problem:
max f (x; y)
s.t. y 2 (x)
Theorem 26 (The continuous maximum) If f : X Y ! R is continuous (w.r.t. box
metric) and : X
Y is non-?, UHC, LHC, compact valued, then: 1)
: X
Y
where (x) = arg maxy2 (x) f (x; y) is non-?; UHC and compact valued (hence UseqC)
correspondence and 2) V : X ! R where V (x) = maxff (x; y)j y 2 (x)g is a continuous
function of x.
Proof. First note that f continuous w.r.t. box metric) y 7! f (x; y) is continuous. Since
(x) is compact valued, the Weierstrass theorem implies that (x) 6= ?. Next, to se that
closed-valued...
Theorem 27 (The Concave Maximum) If f : X Y ! R is continuous and quasiconcave and : Y
X is convex valued, then: 1) : X
Y where (y) = arg maxx2 (y) f (x; y)
is convex-valued and 2) if f is strictly concave, then V : X ! R where V (y) = maxff (x; y)j
x 2 (y)g is concave in x.
Proof. To see that is convex-valued, choose x1 ; x2 2 (y). Naturally, these two maximizers yield the same maximum value. So let = f (x1 ; y) = f (x2 ; y). Choose t 2 [0; 1] and
note that f quasiconcave implies that the set tx1 + (1 t) x2 2 fx 2 (y)jf (x1 ; y)
g.
Thus, f (tx1 + (1 t) x2 )
. But (y) is a convex set and x1 ; x2 2 (y)
(y). Hence,
tx1 + (1 t) x2 2 (y) which implies that f (tx1 + (1 t) x2 )
. Therefore, we conclude
that f (tx1 + (1 t) x2 ) = and tx1 + (1 t) x2 2 (y):
17
1.8
Fixed point theory
We survey two main classes of …xed point theorems (FPT). The …rst class of FPT allows
for ‡exible domain sets, but requires heavy structure on the objectve function. The second
class of FPT requires a highly structured domain sets but requires only mild assumptions
about the objective function.
De…nition 45 f : (X; d) ! (X; d) is a contraction mapping if 9
d(f (x); f (y))
d(x; y) 8 x; y 2 X:
2 [0; 1[ s.t.
Remark 16 A contraction is always a continuous function.
Theorem 28 (Banach Fixed Point) Let f : (X; d) ! (X; d) be a contraction mapping
and (X; d) be a complete metric space, then 9 a unique …xed point f (x) = x and if xk is a
sequence satisfying xk+1 = f (xk ) for each k 0, then, xk is convergent with limit x:
Corollary 2 Every contraction mapping f : (X; d) ! (X; d) on a complete m.s. has a
unique …xed point.
The Banach FPT guarantees the existence of a stationary point x under the appropriate
contracion and completeness assumptions. However, stability around such stationary point
is another matter. A …xed point is globally stable if, starting from any point in the domain
of f , we can generate a sequence that converges to x. On the other hand, a …xed point
is called locally stable, if, starting from a neighborhood of x (say B" (x)) we can generate
a sequence converging towards x: The following proposition outlines the main conditions
under which local stability can be established.
Proposition 11 Suppose that f : Rn ! Rn is a contraction de…ned in a complete metric
space (i.e., it has a …xed point x). Then if Jf (x) ; the Jacobian or matrix of partial derivatives at x has all its eigenvalues lying inside the unit circle the stationary point x is locally
(asymptotically) stable.
Remark 17 In one dimension this condition trivially reduces to jf 0 (x)j < 1:
Theorem 29 (Brower’s Fixed Point) Let f : C ! C be a continuous function. If
C Rn is non-?, compact and convex, then f has a …xed point x = f (x).
Remark 18 Note this is a su¢ ciency theorem, so one can make examples that violate
these assumptions and have a …xed point.
Exercise 30 (Excess
demand theorem 2.6.1 Carter) De…ne the relative price set n
P
n
n 1
fp 2 Rn jpi 0 and
! Rn to be a continuous function
i=1 pi = 1g and de…ne z :
satisfying p z (p) = 0 8 p 2 n 1 . Then there exists p 2 n 1 such that z (p ) 0.
18
1
=
n 1
Proof. First de…ne the following "adjustment function" gi :
gi (p) =
n 1
and the corresponding g :
:
pi + max f0; zi (p)g
P
1 + nj=1 max f0; zj (p)g
n 1
!
!
as:
g (p) = g1 (p) g2 (p) ::: gn (p)
0
then p 7!gi (p) is continuous for each p and therefore p 7!g (p) is a continuous function.
Since n 1 is convex, compact non-? and g ( ) is continuous, we can apply Brower’s theorem and justify the existence of p such that g (p ) = p . That is, for each i :
pi =
or:
pi
pi + max f0; zi (p )g
P
1 + nj=1 max f0; zj (p )g
n
X
max f0; zj (p )g = max f0; zi (p )g
n
X
j=1
max f0; zj (p )g = zi (p ) max f0; zi (p )g
n
X
n
X
j=1
next, multiply both sides by zi (p ) :
zi (p ) pi
and summ over i :
n
X
zi (p ) pi
i=1
now, since by assumption
j=1
Pn
max f0; zj (p )g =
i=1 zi
n
X
i=1
i=1
zi (p ) max f0; zi (p )g
(p) pi = 0, the LHS is zero so that:
zi (p ) max f0; zi (p )g = 0
now, every term of this sum is nonegative since it is either 0 or [zi (p )]2 . Thus, for a sum of
nonnegative numbers to be equal to zero, it must be the case that all the terms of the sum
are zero so that max f0; zi (p )g = 0 and zi (p ) 0 for each i implying that z (p ) 0:
Theorem 31 (Kakutani’s Fixed Point) Let : C
C be UHC, non-?; convex and
compact-valued. If C Rn is non-?, compact and convex, then has a …xed point x 2
(x):
Exercise 32 (A.52, McLean) If f : C ! Rn is continuous, the Variational Inequality
Problem for f; Rn+ is the following: …nd x 2Rn+ such that f (x) (x x) 0 8 x 2Rn+ .
Suppose that C
Rn is compact, convex, non-?. Show that the VIP for f; Rn+ has a
solution.
19
Proof. De…ne the constant correspondence (C) = C. And note that
is compact,
non-?, UHC and LHC (see remark (15) and theorem (24)). Next, de…ne the problem
minz2 (C) z f (x) and note that z 7! z f (x) is continuous and quasiconcave for each
x. Therefore, the maximum theorems imply that (x) = minz2 (C) z f (x) is non-?
(Weierstrass), compact, UHC (Continuous Max) and convex (Concave Max): Next note
that (x) C for each x: That is, : C
C satis…es all the assumptions required for
Kakutani’s theorem to apply and we conclude that has a …xed point; i.e., 9 x 2 (x) =
minz2C z f (x) so that f (x) x f (x) x 8 x 2Rn+ and we conclude that x solves the VIP
for f; Rn+ :
Theorem (28) establishes that, under the required topological conditions, the existence
of a unique …xed point is ensured. However, theorems (29) and (31) are merely existence
theorems. Uniqueness of …xed points, equilibria and solutions to system of equations require
"monotonicity" conditions that generalize to multidimensional problems the notion of a
strictly increasing function of one variable. For de…nition and examples of monotonicity
see section (2.3).
20
Chapter 2
Convex optimization
2.1
Convex sets
De…nition 46 S
R is a convex set if x+(1
)y 2 S whenever x; y 2 S and
Remark 19 Note that the convex combination x + (1
nation.
2 [0; 1].
)y is a particular linear combi-
De…nition 47 The convex hull of S, denoted convS is the intersection of all convex supersets of S :
)
(
m
X
n
f orxi 2 S
convS = y 2 R j y =
i xi
i=1
and
i
> 0;
m
X
i
=1
i=1
Remark 20 The intersection of any collection of convex sets is convex so that, trivially,
convS is a convex set.
Remark 21 S is convex () S = convS
Exercise 33 (1.232, Carter) The consumer’s budget set (correspondence) is convex.
Remark 22 The closure of a convex set is convex.
Lemma 8 The Minkowski’s sum of two convex sets A
is a convex set
B = fa + b j a 2 A and b 2 Bg
Lemma 9 Also, if A; B are convex, then the cartesian product A
21
B is a convex set.
2.2
Separating hyperplanes
Remark 23 Recall that the equation of the tangent of a curve at x is y = f (x)+f 0 (x)(x x)
De…nition 48 If p is a non-0 vector, the vector p x = 0 is orthogonal to p and p is
normal to p x = 0
Remark 24 p "points" in the direction where p x > 0
Theorem 34 (Separating Hyperplane) (aka Minkowski’s theorem): Let S be non-?,
closed and convex, and let x 2
= S, then 9 p 6= 0 and x0 2 S s.t.: p x > p x0 p x; 8
x2S
Proof. (step 1) De…ne f (x) = (x x) (x x) and claim 9 x0 2 S such that f (x) f (x0 )
8 x 2S: To see why this is true note that f as de…ned above is continuous. Morover, choose
r > 0 and the closed ball Br (x) so that Br (x) \ S is a compact set (recall S is closed and
Br (x) is compact). Thus, Weierstrass theorem ensures the existence of the minimizer x0
on Br (x) \ S. This in turn implies that x0 is actually a minimizer for f on S (if x
^2
= Br (x)
the function f (^
x) is even greater than f (x0 )):
(step 2) Next, let p = x x0 and note that p 6= 0 since by assumption x 2S
= while x0 2S.
Therefore p p > 0 implies:
p p =
=
=
>
)
(x x0 ) (x x0 )
p (x x0 )
p x p x0
0
p x > p x0
(step 3) Finally we show that p x p x0 8 x 2S: To do so, choose x 2S and t 2 (0; 1).
Let xt = tx+ (1 t) x0 and note that so that xt 2S since S is convex by assumption. Now:
xt
x
xt
x = t2 (x
x
xt
x
x0 ) (x
x0 ) + 2t (x
x0 ) (x0
x) + (x0
x) (x0
x)
x0 ) (x0
x)
or:
xt
now, since (xt
(x0
x) (xt
x) (x0
x)
(x0
x) = t2 (x
x) (x0
x0 ) (x
x0 ) + 2 (x
x0 ) (x0
x0 ) (x0
x0 ) (x0
(x x0 ) (p) 0
) p x p x0
22
x0 ) + 2t (x
0 it follows that:
x)
t2 (x x0 ) (x x0 ) + 2t (x
t (x x0 ) (x x0 ) + 2 (x
and since limt!0 t (x
that:
x0 ) (x
x)
x)
x) = 2 (x
0
0
x0 ) ( p) we conclude
Corollary 3 Under the same assumptions it is also true that 9 q 6= 0 and x0 2 S s.t.:
q x > q x0 q x; 8 x 2 S:
n
o
P
De…nition 49 If A is a m n matrix, posA = y 2 Rm j y = nj=1 aj xj and xj 0 :
Remark 25 Note that aj are the columns of A so posA is the cone formed by the columnvectors of A
Theorem 35 (Farka’s lemma) Let A be n
AT y 0 imply bT y 0
m matrix; then b 2 posA , y 2 Rm and
Pn
Proof. ()) Suppose b 2 posA then 9 x 2 Rm
+ such that b =
j=1 aj xj = Ax. Now, if
T
T
T
T
T
A y 0 then 0 x A y = (Ax) y = b y:
(() (contrapositive and separating hyperplane theorem) Suppose that b 2
= posA. Since
posA is closed convex and non-? there exists p 6= 0 and y0 2 posA such that p b < p y0
p y 8 y 2 posA. Now since posA is a cone, it follows that 0 2 posA and 2y0 2 posA. Thus
p y0 p 0 = 0 and p y0 p 2y0 so that p y0 0 and we conclude that p y0 = 0
which in turn implies that p b < 0:Therefore, we conclude that AT y 0 so AT p 0 but
pT b < 0. Summarizing, b 2
= posA ) y 2 Rm and AT y 0 but bT y < 0:
Proposition 12 (supporting hyperplane) Let S be non-? and convex, and suppose
that xk 2
= clS 8 k with xk ! x. Then 9 p 6=0 such that p x p x 8 x 2 S.
Proof. Recall that S convex ) clS convex. Thus all the conditions required to apply
the separating hyperplane theorem are satis…ed. However, note that each xk requires a
di¤erent qk . So we conclude that for each xk , 9 qk 6=0 such that qk xk qk x 8 x 2 clS
and since S cls it follows that qk xk qk x 8 x 2 S. Now, since qk 6=0 then kqk k =
6 0 so
that:
qk
qk
xk
x 8x 2 S
kqk k
kqk k
and since
qk
kqk k
= 1 it follows that:
qk
2 A = fy 2Rn j kyk = 1g
kqk k
note that A is the pre-image of f (y) = kyk = 1. Now this function is trivially continuous
so that the set A is closed (apply sequential characterization of closedness and continuity).
Moreover, A is bounded so the Heine-Borel theorem ensures that A is compact. Thuse, we
can extract a convergent subsequence:
qk m
qkm
such that
! p and p 2A
kqkm k
kqkm k
Next, note that:
qk m
qkm
qkm
xk
x 8x 2 S )
x
kqkm k
kqkm k
kqkm k
23
qkm
x 8x 2 S ) p x
kqkm k
p x
Corollary 4 Let S be non-?; closed and convex, and suppose that x 2 @S then 9 p 6=0
such that p x p x:
2.3
Convexity of functions and subgradients
De…nition 50 f is a convex function if: f (x) + (1
x; y 2 S and 2 [0; 1].( for concavity)
)f (y)
f ( x + (1
)y) whenever
De…nition 51 Let f : S ! R and S
Rn . The epigraph of f is epi (f ) = f(x; y) 2 Rn+1 jx 2S and y
De…nition 52 Let f : S ! R and S
Rn . The hypograph of f is hyp (f ) = f(x; y) 2 Rn+1 jx 2S and y
Lemma 10 f is convex , epi (f ) is a convex set.
Lemma 11 f is concave, hyp (f ) is a convex set.
Remark 26 Note that Grf
epif and that (x; f (x)) 2 @epif
Proposition 13 If S
Rn is closed and convex, and is f : S ! R is continuous and
convex on S then epif is closed and convex.
Proof. That epif is convex follows from the assumption that f is convex. Next, to see that
epif is closed note that epif Rn+1 so choose (xk ; yk ) 2epif such that (xk ; yk ) ! (x ; y ) :
If (xk ; yk ) 2epif 8 k then yk
f (xk ) which, given that f is continuous, implies that
y
f (x ). Summarizing, (xk ; yk ) 2epif; (xk ; yk ) ! (x ; y ) and (x ; y ) 2epif so we
conclude that epif is a closed set.
De…nition 53 Let f : S ! R , then p is a subgradient for f at x 2 S if f (x)
f (x) + p (x x) 8 x 2 S:
De…nition 54 The subdi¤erential @f (x) is the set of all the subgradients of f at x
Proposition 14 Let f : S ! R with S a convex subset of Rn . If @f (x) 6= ? for each
x 2S then f is convex.
Proof. Choose x1 ; x2 2 S and 2 [0; 1]. Let xt = tx1 + (1
?. Next choose p 2 @f (xt ) and note:
f (x1 )
f (x2 )
f (xt )
f (xt )
p x1
p x2
t) x2 and note that @f (xt ) 6=
xt
xt
…nally multimply the …rst of these weak inequalities by t and the second one by 1
them together and obtain:
tf (x1 ) + (1
t) f (x2 )
so we conclude that f is convex.
24
f (tx1 + (1
t) x2 )
t; add
f
Proposition 15 Let g : Rn ! R and g(x)
g is di¤erentiable. at x, then rg(x) = 0.
Corollary 5 Now let g(x) = q x if q x
g(x) whenever x 2 B" (x) for some " > 0; if
q x 8 x 2 B" (x) then q = 0 = rg(x):
Proposition 16 Let S be non-?; and convex and let f : S ! R be convex and continuous
at x, if x 2 intS then @f (x) 6= ? (i.e., there exists at least one subgradient).
Proof. Let A =epif . By proposition (13) we know that A is convex. Next we can claim
that (x;f (x)) 2 @A: To see why, notice that (x;f (x)) 2 A since Gr(f ) epif = A. Thus,
B" (x;f (x)) \ A 6= ?. However, (x;f (x) 1=k) 2
= A so that B" (x;f (x)) \ Ac 6= ?. Thus
x;f (x) 2 A but x;f (x) 2 intA so x;f (x) is a boundary point and the claim is true. Now,
using corollary (4) it follows that (x;f (x)) 2 @A ) 9 (q; r) 6= 0 such that:
q x+rf (x)
for each (x;y) 2 A; but since Gr(f )
q x+ry
epif = A it is also true that:
q x+rf (x)
q x+rf (x)
(2.1)
for each x 2S. These two facts can only mean that r
0. In fact we can deduce that
r < 0 since r = 0 implies q x q x, which using corollary (5) implies that q =0 which
contradicts the hypothesis that (q; r) 6= 0. Now, since r < 0 we can divide (2.1) by r to
obtain:
1
q (x x)
f (x) f (x) +
r
| {z }
subgradient
thus,
1
q
r
2 @f (x) and therefore we conclude that @f (x) 6= ?:
Lemma 12 (from subgradient to gradient) Suppose that S
Rn is convex and x 2
intS. If @f (x) 6= ? and if f : S ! R is di¤erentiable at x, then @f (x) = frf (x)g
Proof. Since @f (x) 6= ? choose q 2@f (x). Then:
f (x)
f (x) + q (x
Now, since x 2 intS then 9" > 0 such that B" (x)
f (x)
q x f (x)
| {z }
g(x)
x) 8 x 2S
S. hence:
f (x) + q (x x) 8 x 2B" (x)
q x f (x) 8 x 2B" (x)
|
{z
}
g(x)
so that g (x)
g (x) 8 x 2B" (x). Finally, note that g is di¤erentiable at x since f is
di¤erentiable at x so that applying corollary (5) we conclude that rg (x) = 0 or:
0 = rg (x) = q rf (x)
) q =rf (x)
25
De…nition 55 (strong monotonicity) A function f : Rn ! Rn+ is said to be (strongly)
monotone if [f (x) f (y)] (x y) 0 (> for strong)
Example 36 If f : Rn ! Rn+ is strongly monotone, the system f (x) = 0 has at most one
solution.
Proof. (by contradiction) Suppose that x2 and x2 solve the system f (x) = 0 and x2 6=
x2 . Then strong monotonicity and x2 6= x2 imply:
[f (x1 )
f (x2 )] (x1 x2 ) > 0
but by hypothesis f (x1 ) = 0 =f (x2 ) since x2 and x2 solve the system. Therefore:
0 (x1 x2 ) > 0 ) 0 > 0
a contradiction.
Lemma 13 (homothetic function) Let u : Rm ! R be a strictly monotone increasing
function. Then u is homothetic if and only if, 8 x; x0 2 Rm and 8 t > 0:
u (x)
u (x0 ) () u (tx)
u (tx0 )
Lemma 14 (homothetic function II) : Suppose f : Rm ! R is homothetic and C 1 .
Then for any x 2 R and > 0, there is a k > 0 such that rf (x)=rf ( x) = k
Proposition 17 (monotonicity of subgradients) Let f : S ! R: If p1 2@f (x1 ) and
p2 2@f (x2 ) then (p1 p2 ) (x1 x2 ) 0:
Proof. p1 2@f (x1 ) ) f (x2 ) f (x1 ) + p1 (x2 x1 ) while p2 2@f (x2 ) ) f (x1 )
p2 (x1 x2 ). Adding these two inequalities together we obtain (p1 p2 ) (x1
f (x2 ) +
x2 ) 0:
Corollary 6 (monotonicity of gradients) Let f : S ! R and S
Rn open, convex
and non-?: If f is di¤erentyiable and convex, then (rf (x1 ) rf (x2 )) (x1 x2 ) 0 for
each x1 ; x2 2 S:
Remark 27 Note that if n = 1 then (rf (x1 ) rf (x2 )) (x1 x2 ) 0 becomes (f 0 (x1 )
(x1 x2 ) 0 so that x1 x2 ) f 0 (x1 ) f 0 (x2 ), i.e., the derivative is nondecreasing.
De…nition 56 (single crossing) A function f : X
crossing if s < t and x < y imply:
f (y; s)
f (x; s) < f (y; t)
Remark 28 Supermodularity implies single crossing.
26
Y ! R is said to satisfy single
f (x; t)
f 0 (x2 ))
Proposition 18 (A.45, McLean) If f is di¤erentiable, the supermodularity property is
@2f
equivalent to @x@t
> 0:
Proof. Suppose that f : [a; b] [0; 1] ! R is continuously di¤erentiable. Choose x < y
@2f
and s < t: Then for each z 2 [a; b], @x@t
> 0 implies:
Z t 2
@f
@f
@ f
(z; w) dw =
(z; t)
(z; s)
0<
@x
@x
s @x@t
and:
0<
Z
b
a
@f
(z; t)
@x
@f
(z; s) dz = f (y; t)
@x
f (x; t)
f (y; s) + f (x; s)
De…nition 57 A function f : Rn ! Rm is quasiconvex if the set fx 2 Rn j f (x)
a convex.set.
De…nition 58 A function f : S ! R is quasiconcave if fx 2 S j f (x)
set
g is a convex
De…nition 59 A function f : Rn ! Rm is strictly quasiconcave if f (x)
f ( x + (1
)y) > f (y)
Proposition 19 If f is di¤erentiable on the open-convex set S
g is
f (y) )
Rn , then:
f is strictly convex , f (x2 ) > f (x1 ) + rf (x1 ) (x2 x1 )
f is convex , f (x2 ) f (x1 ) + rf (x1 ) (x2 x1 )
Proof. ()) Suppose f is convex. Choose x1 ; x2 2 S and
f ( x2 + (1
f [ (x2 x1 )] + x1
f [ (x2 x1 )] + x1
) x1 )
f (x1 )
f (x1 )
2 [0; 1]. then:
f (x2 ) + (1
) f (x1 )
[f (x2 ) f (x1 )]
f (x2 )
f (x1 )
if > 0 this is a Newton quotient de…ning the directional derivative of f in the direction
of (x2 x1 ) so that:
rf (x1 ) (x2 x1 ) f (x2 ) f (x1 )
(() Now choose x2 ; x1 2 S and suppose that f (x2 ) f (x1 )+rf (x1 ) (x2 x1 ) 8x2 ; x1 2 S.
Pick 2 [0; 1], denote x = x1 + (1
) x2 and note that S convex) x 2 S. Thus:
f (x2 )
f (x
= f (x
f (x1 )
f (x
= f (x
) + rf (x
) + rf (x
) + rf (x
) + rf (x
27
)
)
)
)
(x2 x
((1
(x1 x
( (x1
)
) (x2 x1 ))
)
x2 ))
multiply the …rst inequality by , the second by (1
f ( x2 + (1
) x1 )
) and add them together to get:
f (x2 ) + (1
) f (x1 )
so that f is convex.
Proposition 20 If S Rn open and convex, and if f : S ! Rn is di¤erentiable, then f
is convex if and only if its gradient is monotone i.e.:
f is strictly convex , [rf (x2 )
f is convex , [rf (x2 )
rf (x1 )] (x2
rf (x1 )] (x2
x1 ) > 0
x1 ) 0
Remark 29 In general, the product of two convex functions is not convex. However, if
f; g are both convex, monotone and positive, f g is convex.
n. We say that A is PSD if xT Ax
De…nition 60 Let A be n
Remark 30 In R1 we say that f ( ) is non-decreasing , f 0 ( )
non-decreasing by monotonicity, f 0 ( ) by the Jacobian, Jf ( ); and
de…niteness (PSD).
Proposition 21 g : Rn
0 for all x 6= 0
0: In Rn we replace
0 with positive-semi-
S ! Rn is monotone ,its Jacobian (Jg ) is PSD.
Proof. (() Assume that Jg is PSD. Choose x2 ; x1 2 S and for each t 2 [0; 1] de…ne the
function (t) = g ((1 t) x1 + tx2 ) (x2 x1 ). Since (t) is a function of t and a scalar
for each t, we can invoke the Mean Value Theorem (MVT) to deduce that 9 t 2 (0; 1)
such that:
(1)
(0)
= 0 (t )
1 0
0
writing out this expression for (t ) we get:
[g (x2 )
g (x1 )] (x2
x1 ) = (x2
|
x1 )T Jg ((1
0 (t
and since Jg is PSD we conlcude that:
[g (x2 )
t ) x1 + t x2 ) (x2
{z
g (x1 )] (x2
x1 )
)
x1 )
}
0
()) Now suppose that g is monotone. Choose x 2 S. Next, choose such that x +
y 2 B" (x). Not that if 0 < < then x+ y 2S. Now de…ne (t) = g ((1 t) (x+ y) + tx)
(x+ y x). Again using the MVT we obtain:
[g (x+ y)
|
g (x)] (x+ y
{z
x1 ) = ( y)T Jg (z ) ( y)
}
0 since g is assumed to be monotone
where z = (1 t ) (x+ y) + t x. We want to show that ( y)T Jg (x) ( y) 0 or equivalently yT Jg (x) y 0. So let ! 0 so that z ! x.and we arrive at the conclusion that
yT Jg (x) y 0 so that Jg is PSD.
28
Proposition 22 f is convex , the Jacobian (Jf ) of the gradient is PSD.
Corollary 7 Equivalently, f is convex ,the Hessian (Hf ) of the original function is PSD.
Corollary 8 Equivalently, if C Rn is convex and Q : C ! Rn with Q (x) = xT Ax + bx
then Q (x) is convex whenever A is PSD (NSD for concavity)
2.4
Support function theory
Many optimization problems can be cast in the form:
max q x
subject to x 2 S
which is the cannonical form of a type of convex optimization problem; the support function
problem.
De…nition 61 K
Rn is a pointed cone if x 2 K whenever x 2 K and
0
Remark 31 If K is a cone, 0 2 K.
De…nition 62 Remark 32 0 is a cone and Rn is a cone.
Exercise 37 (A.52, McLean) If f : Rn ! Rn is continuous, the Variational Inequality
Problem (see example ()) for f; Rn+ is the following: …nd x 2Rn+ such that f (x) (x x)
0 8 x 2Rn+ . Show that x solves the VIP for f; Rn+ , x 2Rn+ , f (x) 2Rn+ and f (x) x = 0.
Proof. ()) Suppose that x solves the VIP for f; Rn+ . Then f (x) (x x) 0 8 x 2Rn+ .
0 ) f (x) x
0. But Rn+ is a cone so
Since 0 2 Rn+ it follows that f (x) (0 x)
n
n
x 2R+ ) 2x 2R+ . Therefore, f (x) (2x x) 0 ) f (x) x 0 from which we conclude
that f (x) x =0. Finally, to see that f (x) 2Rn+ note that x solves the VIP for f; Rn+
implies f (x) x f (x) x = 0 so that f (x) x 0 and since x 2Rn+ we conclude that
f (x) 2Rn+ :
(() Suppose that x 2Rn+ , f (x) 2Rn+ and f (x) x = 0. Then f (x) (x x) = f (x)
x f (x) x =f (x) x 0 since f (x) 2Rn+ and x 2Rn+ . Moreover, since x 2Rn+ we conclude
that x solves the VIP for f; Rn+ :
De…nition 63 (barrier cone) Let S
Rn be non-?. The barrier cone of S, b (S) is
de…ned as follows: p 2 b(S) if and only if the problem maxx p x subject to x 2 S has a
solution.
Remark 33 b (S)
Rn and 0 2 b (S) since max q x = 0 x makes any x a maximizer.
Example 38 Suppose that S = fx 2R2 j kxk < 1g, then b (S) = f0g.
29
Example 39 On the other hand, if S = fx 2R2 j kxk
1g, then b (S) = R2 .
Example 40 Suppose that S = fx 2R2 j x1 0 ; x2 0g (i.e., the third quadrant). To …nd
b (S) in this case, …rst note that if p has a negative coordinate, @ a maximizer. Thus, p
must have both non-negative coordinates, i.e., b (S) = fp 2R2 j p1 0 ; p2 0g. Naturally,
if pi > 0 and pj = 0 the problem would have in…nitely many maximizers. On the other
hand, if p1 > 0 and p2 > 0 the problem has a unique maximizer. The support function is
found by plugging any of the maximizers in the objective function.
Lemma 15 b (S) is a cone in Rn
Proof. Suppose that p 2b (S) and choose
0. Then we know that p 2b (S) , 9 x such
that p x p x 8 x 2S. Therefore,
0 ) (p x)
(p x) or ( p) x ( p) x so
that q = p 2b (S) so we conclude that b (S) is a cone.
De…nition 64 (support function) If p 2b (S) then the function s (p) = max [p x j x 2S]
is well de…ned. The function s : b (S) ! R is called the support function of S:
Remark 34 The cost and pro…t functions are support functions.
Remark 35
s
is homogeneous of degree one, i.e., if p 2b (S) then
Remark 36
s
is a convex function, whenever b (S) is a convex set.
(tp) = t
s
(p) :
Rn convex and p 2 b(S) then x 2 arg
Theorem 41 (support function) Let S
maxx2S p x , x 2 @ s (p):
Proof. ()) Suppose that x 2 arg maxx2S p x. Then
and notice that
s (q)
s
s (p)
= p x. Next choose q 2 b(S)
q x
=0
z
}|
{
= q x+( s (p) p x)
= s (p) + x (q p)
) x 2@ s (p)
(() Suppose that x 2@
subgradient:
s (p).
s (p)
This implies that p 2 b(S): Now, by the de…nition of
s (p)
+ x (p
p) 8 p 2 b(S)
choosing p = 0 and p =2p we note that 2p 2 b(S) since p 2 b(S) and b(S) is a cone.
Thus:
s (0)
= 0
s (p)
+ x (0
) x p
s (p)
30
p)
while:
s (2p)
= 2 s (p) (by homogeneity)
p)
s (p) + x (2p
= s (p) + x p
) x p
s (p)
therefore, x p = s (p) and hence, x p x p 8 x 2 S. To complete the argument we
must show that x 2S. We do this by contradiction and using the separating hyperplane
theorem. Suppose that x 2S
= then 9 w 6= 0 such that:
w x > w x0
w x;8x 2S
which in turn implies that x0 2 arg maxx2S w x. But w x > w x0 so we would have
that:
s (w)
< w x
=0
z
}|
{
= w x+( s (p) p x)
= s (p) + x (w p)
a contradiction since x 2@
s (p):
Corollary 9 Let S Rn convex and suppose b (S) is a convex set. If p 2intb (S) and s
is di¤erentiable at p then arg maxx2S x p = fr s (p)g = x is the unique solution to the
maximization problem.
Example 42 Let S = x 2R2+ j x1 x2 1 then b (S) = fp 2R2 jp1 < 0,p2 < 0g [ f0g and
p
intb (S) = fp 2R2 jp1 < 0, p2 < 0g. Also, s (p1 ; p2 ) = 2 p1 ; p2 . Notice that s (p1 ; p2 ) is
di¤erentiable at intb (S) so that:
r
@ s (p1 ; p2 )
p2
x1 =
=
@p1
p1
r
@ s (p1 ; p2 )
p1
x2 =
=
@p2
p2
q q
p2
and (x1 ; x2 ) =
; pp12 is the solution to the max problem.
p1
So far we have de…ned all these objects in terms of a maximization problem. There is
a natural equivalence with the minimization problem:
De…nition 65 (supergradient) Let f : S ! R , then p is a supergradient for f at x 2 S
if f (x) f (x) + p (x x) 8 x 2 S:
31
De…nition 66 (barrier cone min) Let S Rn be non-?. The barrier cone of S, ~b (S)
is de…ned as follows: p 2 ~b(S) if and only if the problem minx2S p x subject to x 2 S has
a solution.
De…nition 67 (support function min) If p 2~b (S) then the function ~ s (p) = min [p x j x 2S]
is well de…ned. The function ~ s : ~b (S) ! R is called the support function of S:
Remark 37 ~ s is also homogeneous of degree one, is a concave function, whenever ~b (S)
is a convex set.
2.5
2.5.1
Kuhn-Tucker theory
Introduction to nonlinear programming
Suppose that we face the NLP problem with equality constraint:
min f (x1 ; x2 )
s:t: g (x1; x2 ) = 0
with f; g di¤erentiable. To …nd optima, we set-up the Lagrangian:
L = f (x1 ; x2 )
g (x1; x2 )
obtain FOC:
rf (x1 ; x2 ) =
"
@f (x1 ;x2 )
@ x1
@f (x1 ;x2 )
@ x2
#
=
"
@f (x1 ;x2 )
@ x1
@f (x1 ;x2 )
@ x2
#
= rg (x1 x2 )
which, says that, at the optimum, the gradient of the objective is colinear with the gradient
of the restriction. After eliminating this method yields the familiar result:
@f (x1 ; x2 ) =@ x1
@g (x1 ; x2 ) =@ x1
=
@f (x1 ; x2 ) =@ x2
@g (x1 ; x2 ) =@ x2
2.5.2
The Kuhn-Tucker conditions
Next introduce inequality constraints. Suppose that f : Rn ! R; g : Rn ! R with f; g
di¤erentiable. The cannonical NLP problem is:
min f (x)
s:t: g1 (x) 0
..
.
gm (x) 0
32
(2.2)
2 Rn
A Kuhn-Tucker pair for this problem is a pair x;
Rn satisfying:
1) gi (x) 0 8 i = 1; :::; m
2) i 0 8 i = 1; :::; m
X
3) rf (x) =
i rgi (x)
(2.3)
(2.4)
(2.5)
i
4)
gi (x) = 0 8 i = 1; :::; m
(complementary slackness)
(2.6)
i
Example 43 (linear programming) Let the objective function be f (x1 ; x2 ) = c1 x1 +
c2 x2 and the constraints be g1 (x1 ; x2 ) = a11 x1 +a12 x2 b1 and g2 (x1 ; x2 ) = a21 x1 +a22 x2 b2 .
Then the LP problem is:
min c x
s:t: aT1 x
aT2 x
a KT pair for this problem would be x;
b1
b2
2 R2
c = 1 a1 + 2 a2 ;
aT2 x b2 0;
aT2 x b2 0;
1
2
R2 satisfying:
2R2+
aT2 x b2 = 0
aT2 x b2 = 0
Theorem 44 (Kuhn-Tucker for LP) Let c 2Rn ; A be m n and b 2Rm : Then x solves
the LP problem:
min c x
s:t: Ax b
if and only if, 9
2Rm
+ such that:
Proof. (() Suppose that x;
that any other x such that Ax
0
c =AT
Ax b 0
[Ax b] = 0
(2.7)
(2.8)
(2.9)
2 Rn Rm satis…es the KT conditions. We must show
b 0 ) c x c x: Now, from (2.7):
c =AT
so that:
T
T
) c x = AT
cT x =
AT
T
=
AT
T
= (Ax
bT
= AT
= cT x
x
x
x + bT
b)T
33
+b
b
T
T
()) Now suppose that x solves the problem. To begin, de…ne the set of active constraints:
I (x) = ijaTi x b = 0 . If I (x) = ? then there are no bonding constraints so to satisfy rcT x = 0 ) c =AT we simply need = 0:Now, if I (x) 6= ?, suppose that y 2Rn
satis…es aTi y 0. Then we can claim that cT y 0; to see this, choose t > 0 such that
aTi x+taTi y >bi if i 2
= I (x) and if i 2 I (x) it follows that:
aTi (x+ty) = aTi x+taTi y
= bi + taTi y
bi
Summarizing, A (x+ty)
b ) cT (x+ty)
cT x ) cT y 0. Now, applying Farka’s
lemma (proposition (35)) we conclude that for each i 2 I (x) ; 9 i 0 such that c 2posA
that is:
X
c=
ai i
i2I(x)
or c =AT : To complete the argument let
i
=08i2
= I (x) :
Theorem 45 (duality of linear programming) Let A be m
The problem:
min c x
s:t: Ax b
n; b 2Rm and c 2Rn :
(2.10)
0
has a solution if and only if the problem:
max b y
(2.11)
s:t: AT x = c
yi 0
has a solution.
Proof. It su¢ ces to show that (x; y) is a KT pair for the problem in (2.10) if and only if
(y; x) is a reduced KT pair for the problem in (2.11). Suppose that (x; y) is a KT pair for
the problem in (2.10). Then:
c =AT y, Ax b 0
y 0 and y [Ax b] = 0
therefore, since max f (x) is equivalent to min f (x) ; it follows that (y; x) is a reduced
KT pair for the problem:
min b y
T
s:t: A x = c
yi 0
34
(2.12)
where the complementary slackness condition for the problem in (2.10) becomes the colinearity of gradients condition for the problem in (2.12), the colinearity of gradients condition
for the problem in (2.10) guarantees that the equality restriction in (2.12) is satis…ed and
the condition that the multiplier is nonnegative in (2.10) guarantees that the nonnegativity
restriction in (2.12) are satis…ed.
2.5.3
Constraint quali…cations
De…nition 68 (general constraint quali…cation) Suppose that x solves the problem
min f (x)
s:t: g1 (x) 0:::gm (x)
0
Then x satis…es the general constraint quali…cation (GCQ) if I (x) 6= ? ) x solves the
problem:
min rf (x) x
s:t: rgi (x) (x x) 0 8i 2 I (x)
that is, if x solves the local-linearized problem.
Theorem 46 (Kuhn, Tucker; 1951) Supose that x solves the problem
min f (x)
s:t: g1 (x) 0:::gm (x)
If x satis…es the GCQ then 9 2Rm such that x;
the KT conditions (2.3)-(2.6).
0
is a KT pair; that is, x;
satisfy
Example 47 (GCQ is not satis…ed) Let the problem be:
s:t: x2
min x1
0; x31
x2
0
the unique solution to this problem is x = (0; 0) and:
rf (x) =
1
,
0
rg1 (x) =
0
,
1
rg2 (x) =
3x21
=
1
0
1
Then it is not possible to express rf (x) as a linear combination of rg1 (x) and rg2 (x).
The problem here is that x = (0; 0) does not solve the local-linearized problem, i.e., x = (0; 0)
does not satisfy the GCQ.
Condition 48 (Cottle CQ) A NLP satis…es the Cottle CQ if I(x) 6= ? 9 z 2 Rn s.t.
rgi (x) z > 0 8 i 2 I(x)
35
Condition 49 (linear independence CQ) A NLP satis…es linear independence if I(x) 6=
? then rgi (x) and rgj (x) are linearly independent whenever i; j 2 I(x):
Condition 50 (Slater CQ) A NLP satis…es the Slater CQ if each gi ( ) is concave and 9
x0 2 Rn s.t. gi (x0 ) > 0 8 i (all the constraint sets are convex)
Remark 38 In the last example, none of these three constraint quali…cations is satis…ed.
2.5.4
Non-negativity and equality constraints
Non-negativity constraints
If we modify the NLP problem to include non-negativity constraints we look at:
min f (x)
s:t: g1 (x) 0:::gm (x)
xi 0 8 i
0
(2.13)
In order to ignore the multipliers for these restrictions, use the (reduced) K-T system:
rf (x)
Pm
i=1
i
0 (2)
xj
0 (4)
h
@f
(x)
xJ
Example 51 Let the problem be:
i rgi (x)
gi (x)
0 (3)
i gi (x)
Pm
i=1
0 (1)
i
@g
i xJ (x)
= 0 (5)
xj = 0 (6)
min x1 + x2
s:t: x21 + 4x22 4
xi 0 8 i
the KT conditions are:
xi 0 8 i
+ 4x22 4) = 0
1 8 x2 x2 = 0
2x1
0
8x2
0,
x21
1
+
4x22
(x21
4 0;
2 x1 x1 = 0;
1
1
36
There are three cases to consider:
Case 1: x2 = 0, x1 > 0. In this case, we replace the last condition with x2 = 0 and solve:
2 x1 = 0 ) x1 =
1
1
2
1
2
2
4=0
so that:
1
and x1 = 2
4
Case 2: x1 = 0, x2 > 0. Now we replace the second-to-last condition with x1 = 0 and solve:
=
8 x2 = 0 ) x2 =
1
1
8
4
1
8
2
4=0
so that:
1
and x2 = 1
8
Case 3: Finally we look for an interior solution (i.e. x1 > 0, x2 > 0) so we solve:
=
1
2
1
1 8 x2 = 0 ) x2 =
8
2
2
1
1
+4
=4
2
8
2 x1 = 0 ) x1 =
1
so that:
p
5
4
1
, x2 = p , x2 = p
8
5
5
However, only the KT pair of the second case is a solution to the problem. To see why,
replace this candidates for solutions in the objective function and note that:
=
4
1
1<2< p +p
5
5
Remark 39 In the last example we found 3 KT pairs but only one of them is a solution
to the min problem. This issue arises because the constraint is not concave.
37
Equality constraints
Again, if we extend the NLP problem to include equality constraints we look at:
(2.14)
(2.15)
(2.16)
min f (x)
s:t: g1 (x) 0:::gm (x) 0
h1 (x) = 0:::hr (x) = 0
then the KT pair is x; ;
conditions for this problem are:
Xm
Xr
1) rf (x) =
rg
(x)
+
i
i
i rhi (x)
i=1
i=1
2) i 0 8 i = 1; :::; m
3) gi (x) 0 8 i = 1; :::; m
4) i gi (x) = 0 8 i = 1; :::; m
4) hi (x) = 0 8 i = 1; :::; r
(2.17)
(2.18)
(2.19)
(2.20)
(2.21)
Remark 40 Note that the problem min f (x) s.t. h(x) = 0 is equivalent to the problem
min f (x) s.t. h(x) 0 and h(x) 0. Therefore, if we have a problem like that in (2.14)
but with nonnegativity constraints, we can always transform the problem into one with
inequality constraints and apply the methods used fot (2.13).
2.5.5
Su¢ ciency conditions
The Kuhn-Tucker theorem (46) of the last section helps us obtain a KT pair from a solution
to the problem. Additionally, we can …nd conditions under which a KT pair is a solution
to the optimization problem (as is always the case in linear programming problems).
Theorem 52 (KT-su¢ ciency) If f : Rn ! R is convex and each gi ( ) is concave, whenever (x; ) is a K-T pair for the NLP (2.13) ) x solves the NLP (2.13)
Proof. Suppose that (x; ) is a KT pair for the NLP. Then from the KT conditions we
have:
X
rf (x) =
i rgi (x)
i
Now choose x satisfying gi (x)
0 for each i. Then:
X
rf (x) (x x) =
i rgi (x) (x
i
x)
but since gi ( ) is concave, it follows that rgi (x) (x x) gi (x) gi (x) thus:
X
rf (x) (x x)
gi (x)]
i [gi (x)
X
Xi
=
i gi (x)
i gi (x)
} | i {z
}
| i {z
0
0
38
=0 by KT cond.
thus, we have rf (x) (x
x)
f (x)
0. But f is convex so that:
f (x)
rf (x) (x x)
0
) f (x) f (x)
Theorem 53 (generalized su¢ ciency I) If f : Rn ! R is pseudo-convex and each gi ( )
is quasi-concave, whenever (x; ) is a K-T pair for the NLP (2.13) ) x solves the NLP
(2.13)
Theorem 54 (generalized su¢ ciency II) if f : Rn ! R is pseudo-convex, each gi ( )
is quasi-concave, and the equality constraint h( ) is quasiconcave (w/ multiplier ), then
whenever (x; ; ) is a K-T pair for the NLP (2.13) ) x solves the NLP (2.13)
39
Part II
Microecomic Theory
40
Chapter 3
Producer and consumer theory
In this chapter we apply the main concepts and results from convex analysis to the practical
issues of proudction, cost and expenditure minimization and pro…t and utility maximization.
3.1
3.1.1
Producer theory
Production sets and technologies
In classical producer theory, a …rm produces a single output with n di¤erent inputs. If we
let y denote the quantity of output and x the vector of quantities of the various inputs,
we can represent a production plan as the pair (y; x) where x 2Rl . The production
possibility set is the set:
Y = (y; x) 2 R
Rl j (y; x) is technologically feassible
For some production function g (x) this can be written as:
Y = (y; x) 2 Rl+1 jg (x)
y
In some applications it is convenient to work with the input requirement set:
V (y) = x 2 Rl+ j (y; x) 2 Y
which is the set of all input bundles that will produce at least y units of output. It is
standard practice to assume that V (y) is a convex set 8 y so that technology can be
said to be convex. A less restrictive assumption would be to assume that the production
possibility set is convex.
Remark 41 Note that V (y)
Rl+ while Y
Rl+1
Proposition 23 Y convex ) V (y) convex.
41
An alternative notation (used by modern sources) is to label the elements of Y simply
as y and call yi 2 R+ outputs and yi 2 R+ inputs. Letting n = l + 1 we can use this
notation to write y = (y1 ; :::yn ) and the production possibility set as:
Y = fy 2 Rn jy is technologically feassibleg
Suppose that the nth commodity is the single output and that g : Rn 1 ! R is a production
function.Then "technologically feassible" would mean formally that:
yn
yn
g (y1 ; :::; yn 1 )
or
g (y1 ; :::; yn 1 ) 0
in which case the production possibility would be:
Y = fy 2 Rn jyn
or, if we relabel yn
g (y1 ; :::; yn 1 )
0g
g (y1 ; :::; yn 1 ) = F (y) :
Y = fy 2 Rn jF (y)
0g
the function F (y) is usually called the transformation function.
De…nition 69 A technology is said to satisfy free disposal if y 2Y and y0
De…nition 70 A technology is said to be monotonic if x 2V (y) and x0
y ) y0 2Y:
x ) x0 2V (y)
De…nition 71 A technology is said to be additive if y 2Y and y0 2Y ) y0 + y 2Y . In
other words, Y + Y
Y:
De…nition 72 A technology is said to satisfy nonincreasing returns to scale if y 2Y and
2 [0; 1] ) y 2Y:
De…nition 73 A technology is said to satisfy nondecreasing returns to scale if y 2Y and
1 ) y 2Y
De…nition 74 A technology is said to satisfy constant returns to scale if y 2Y and
0 ) y 2Y . That is, Y is a cone.
De…nition 75 A technology is said to satisfy "no free lunch" if Y \ Rn+ = f0g (all component of any y 2Y cannot be positive)
Exercise 55 (McLean, 78; Varian 1.10) Y satis…es additivity and nonincreasing returns to scale , Y is a convex cone.
42
Proof. ()) Since Y satis…es additivity, it follows that y 2 Y ) y+y 2 Y . Since Y satis…es
additivity and y 2 Y and y +y 2 Y it follows that y +y +y 2 Y . By induction, we conclude
that ky 2 Y 8 k > 1. Next, since Y satis…es NIRS then y 2 Y and k 2 [0; 1] ) ky 2 Y .
Summarizing, y 2 Y ) ky 2 Y 8 k 0 so that Y is a cone. To see that Y is convex choose
y; y 0 2 Y . Since Y satis…es NIRS it follows that k 2 [0; 1] ) ky 2 Y and (1 k) y 0 2 Y ;
and by additivity ky+ (1 k) y 0 2 Y so Y is convex.
(() Now suppose that Y is a convex cone:Since Y is a cone, ky 2 Y 8 k 0 so to see that
Y exhibits NIRS simply choose k 2 [0; 1] : Next, since Y is convex it follows that 8 y; y 0 2 Y
and k 2 [0; 1], ky+ (1 k) y 0 2 Y . But Y is a cone so ky 2 Y and (1 k) y 0 2 Y . Therefore
we see that whenever ky 2 Y and (1 k) y 0 2 Y it follows that ky+ (1 k) y 0 2 Y so Y
satis…es additivity.
Example 56 (Carter, 1.91) Suppose that an economy contains m producers and n commodities. The technology of each producer is summarized by its production possibility set
Y j Rn . Aggregate production (set) y is the sum of the net outputs of each of the producers
yj , that is:
y = y1 + ::: + ym
and the set of feasible aggregate production plans or the aggregate (economy-wide) production possibility set is:
Y = Y1 + ::: + Ym Rn
A su¢ cient condition for Y to be convex is that each Yj be convex.as is recorded in lemma
(8)
Remark 42 Note that although production posibility sets Yj ; Y are …xed and given by
technology, actual production yj and y depend on the price vector p. Thus, the sets yj and
y can in fact be thought of as correspondences yj (p) and y(p) where:
n
o
X
X
n
y(p) =
yj (p) = y 2R jy =
yj for some yj 2 yj (p)
3.1.2
Cost minimization
Support function approach
Using the input requirement set notation, and allowing for multi-product …rms (so that
y 2Rm ); the cannonical cost minimization problem can be cast as:
min w x
s:t:x 2V (y)
where w 2Rn is a vector of input prices. Note that if w 2 eb(V (y)) then arg minx2V (y) w x 6= ?:
In fact the (reverse) support function @ ~ V (y) (w) is the optimal (cost) value function.
43
Proposition 24 If V (y) is closed-convex and non-? and w 2 eb(V (y)) \ Rn+ = K convex
set, then:
c( ; y):K ! R is concave and homogeneous of degree 1
x 2 arg min w x , x 2 @c(w; y) = @ ~ V (y) (w)
x2V (y)
Moreover, if w 2 intK and c(w; y) is di¤erentiable:
arg min w x = frc(w; y)g = x
x2V (y)
|
{z
}
0
(Shephard s Lemma)
Example 57 (perfect substitutes) Let V (y) = x 2 R2+ jx1 + x2 y : This production technology results in either unique conrner solutions if the input price vector w 2 R2+
has coordenates wi > wj ; or in in…nitely many solutions if wi = wj : In fact, c(w; y) =
min fw1 ; w2 g y. Also, note that whenever c(w; y) is di¤erentiable, say when w1 > w2 Shephard’s lemma holds since c(w; y) = w2 y and therefore:
rc(w; y) =
0
y
so that there’s a unique supergradient equal to the gradient. Intuitively, since input 1 is
relatively more expensive and inputs are perfect substitutes, the …rm produces y using only
x2 :
Kuhn-Tucker approach
A similar method to solve cost minimization problems is to replace x 2 V (y) with g (x) y
where g ( ) is a production function. This way we face an optimization problem under
inequality and nonegativity constraints and a natural approach to deal with it is the KT
theory developed in section 2.5.
Example 58 Suppose that the …rm has production function V (y) = x 2 R2+ jx1
where i > 0 and 1 + 2 = 1. Then the problem can be cast as:
min w1 x1 + w2 x2
s:t: x1 y and x1 1 x2 2
xi 0 8 i
y
the KT conditions for this problem are:
w1
w1
2
1 x12
1
x2 2 0;
1 1
x2 2 x1 = 0;
1
2 1 x12
xi 0 , i 0 i = 1; 2;
1
2
y) = 0;
2 (x1 x2
1
1
44
1
2 1
w1
0
2 2 x12 x2
1
2 1
w1
x2 = 0
2 2 x12 x2
1
2
x1 x 2
y, x1 y
(x
y) = 0
1
1
y and x1 1 x2 2
y
…rst note that if there is to be any production in this problem, xi > 0 for both i = 1; 2. We
…rst look for a solution with i > 0 for i = 1; 2. Thus, we solve the system:
w1
1
1 x12
1
2
1
x2 2 = 0;
w1
x1 x2 = y
1
1
2
=0
x1 = y
2
1
2 x12 x2
2
which yields the solution:
1
= w1
1
2
w2 ;
2
=
w2
;
2
x1 = y = x2
w2
this will be a solution to the problem i¤ w11
. On the other hand, if w11 < w22 then
2
becomes infeasible so we look for a solution with 1 = 0 and rewrite the system as:
w1
1 x12
1
2
1
x2 2 = 0; w1
x1 1 x2 2 = y
2
2 x12 x2
1
2
1
1
=0
solving this new system yields:
2
=
w2
2
2
w1
1
,
x1 = y
1
w1
w2
1
2
and
w2
w1
x2 = y
2
2
1
1
now with 1 = 0, the condition 1 (x1 y) = 0 is trivially satis…ed. However, we still need
to check that condition x1 y is also satis…ed. Since we are looking a the special case when
w1
1 1
< w22 it follows that w
> 1 so that:
w2 2
1
x1
y=y
w1
w2
1
2
y=
2
and we have a solution to the problem even when
w1
w2
w1
1
<
2
1
1 y
0
2
w2
2
:
Example 59 (no interior solution) Suppose that V (y) = x 2 R2+ j
Then the problem can be cast as:
1 x1
+
2 x2
y .
min w1 x1 + w2 x2
s:t: 1 x1 + 2 x2 y
xi 0 8 i
Now, since this is a linear programming problem, we can use theorem (44) to conclude that
a KT pair for this LP will be a solution to the problem. Thus we look for x; 2 R2 R
such that:
w1
0
w2
0
1
2
w1
w2
1 x1 = 0;
2 x2 = 0
xi 0 8 i;
0
( 1 x1 + 2 x2 y) = 0
x
+
x
y 0
1 1
2 2
this system has no solution with both x1 ; x1 > 0. The intuition is simple; if the price
of input i relative to its productivity is smaller than the price of input j relative to its
productivity, the …rm’s optimal plan is to use only input i. That is, if w11 < w22 then solving
the system above would yield x2 < 0 so we replace condition w2
2 x2 = 0 with x2 = 0:
The solution to the problem would then be: x2 = 0, x1 = y1 and = w11 :
45
3.1.3
Pro…t maximization
Support function approach
n
Let y 2 Rm , x 2 Rn and let p 2 Rm
+ and w 2 R+ then the pro…t maximization problem
is:
maxfp y w xg
s:t: x 2 V (y) and y 2 Rn+
To make this a traditional support function problem, de…ne T = f(z; y)j z 2 V (y)g (that
is, z = x) and note that (x; y) solves the problem above , ( x; y) solves the problem:
maxfp y + w zg
s:t: (z; y) 2 T
Remark 43 The pro…les (w; p) for which the max problem has a solution is the barrier
cone b(T ):Furthermore, if (w; p) 2 b(T ) then (w; p) = T (w; p):
Proposition 25 Suppose T; b(T ) are convex. Let K = b(T )\[Rn+ Rm
+ ] so that (!; p) 2 K:
Then:
( ): K ! R is convex and homogeneous of degree 1
(x; y) 2 arg max fp y
w xg , ( x; y) 2 arg max fp y + w zg
x2V (y)
(z;y)2T
, x 2 @ (w; p) = @
moreover, if
T (w; p)
is di¤erentiable:
arg max fp y+w zg = f r w (w; p); r p (w; p)g = (x; y)
(z;y)2T
|
{z
}
0
(Hotelling s Lemma)
p
Example 60 Let V (y) = fx 2 R+ j x yg. Note that if w = 0, there’s no solution to
this problem. On the other hand, if p; w 2 R++ the problem is:
p
max p x wx
s:t: x 0
with solution:
x=
p
2w
2
and y =
p
2w
so that:
p
2w
p2
=
4w
(w; p) = p
46
w
p
2w
2
Note that (w; p) is convex as it should. To see why we can use corollary (7):
r (w; p) =
p 2
2w
p
2w
p2
2w3
p
2w2
)H =
p
2w2
1
2w
and since H .is PSD we conclude that (w; p) is convex. Finally note that (Hoteling’s
lemma):
"
#
@ (w;p)
p 2
x
@w
2w
=
=
@ (w;p)
p
y
@p
2w
Example 61 V (y) = fx 2 R+ jx yg. In this setup, constant returns to scale complicates
matters. Thus, if w > p the unique solution to the pro…t max problem is (0; 0). If p > w
the problem has no solution and if p = w there are in…nitely many solutions all of which
lead to (w; p) = 0:
Remark 44 Every function which is convex and homogeneous of degree one is a pro…t
function for some Y . Thus, we can recover the production technology set of a …rm from its
pro…t function.
3.2
3.2.1
Consumer theory
Utility maximization
The cannonical utility maximization problem:
(3.1)
max u (x)
s:t: p x y
xi 0
is not a linear problem and therefore we cannot apply the methods of support function
theory. However, we can still use the Kuhn-Tucker approach for solving NLP optimization
problems. Moreover, if the objective function u is concave, a straightforward extension of
theorem (52) implies that a KT pair would be a solution to the optimization problem. With
a solution at hand, we can compute the optimal value function which is usually referred as
the indirect utility function. In this case, the (reduced) KT conditions for optima become:
ru (x) + p 0;
h [y p x] =0;
i
@u
xj @x
(x)
p
i = 0;
j
y
p x 0
xi 0
0
Remark 45 Notice that max u (x) is equivalent to min u (x) so that the colinearity of
gradients condition is:
ru (x)
p
0
ru (x) + p
0
ru (x) + p 0
47
As in the cases for cost minimization and pro…t maximization, the optimal value function for the utility maximization problem:
v (p; y) = max u (x)
px y
which is called the indirect utility function, also has a nice interpretation. It gives the
maximum level of utility attainable for any given level of income at the prevailing prices.
The IUF also has some interesting properties described in what follows.
Proposition 26 The indirect utility function v : Rn+ R+ ! R is quasiconvex. That is:
S = (p; y) 2 Rn+1
+ jv (p; y)
is convex 8
Proof. Choose (^
p; y^) and (p; y) both 2 S. Then choose t 2 [0; 1] let (pt ; y t ) = t (^
p; y^) +
(1 t) (p; y). Now (^
p; y^) 2 S ) v (^
p; y^)
and (p; y) 2 S ) v (p; y)
:Next, choose:
xt 2 arg max
u (x)
t
t
p x y
Now we can claim that xt is feasible for at least one of the (^
p; y^) or (p; y) problems. If
t
t
this were not the case p
^ x > y^ and p x > y, which, if multiplied by t and (1 t),
respectively, and added together imply:
t p
^ xt + (1
t) p xt
> t^
y + (1
t
t
p x > yt
t) y
a contradiction since by hypothesis xt solves the (pt ; y t ) problem. Thus, xt is feasible for
at least one of the (^
p; y^) or (p; y) problems:Therefore:
u xt
max fv (p; y) ; v (^
p; y^)g
v pt ; y t
and we conclude that pt ; y t 2 S so that S is a convex set and v ( ) is quasiconvex.
The following theorem helps us recover the solution to a utility maximization problem
from the indirect utility function:
Example 62 Suppose that a consumer lives in an n commodity world and has preferences
u (x) = x1 + ::: + xn with 0 < < 1; then the utility maximization problem for each level
of income y is:
max x1 + ::: + xn
n
X
s:t:
pi xi y
i=1
48
This problem is simpli…ed by the fact that u ( ) is strictly increasing and strictly concave
on each of its arguments. Strict concavity will guarantee that at the optimum, each xi > 0;
strict increasingness will guarantee that the constraint will bind so that > 0. Hence we
only need to look for an interior solution to the KT system:
xi
1
and
pi = 0
n
X
y
pi xi = 0
i=1
which yield solution:
=
Pn
y
i=1
pi
1
!
2
1
and
this yields an indirect utility function:
v (p; y) = y
n
X
i=1
2
4
1
xi = 4 P
1
pi
Pn
i=1
1
pi
1
pi
n
i=1
1
pi
1
3
5y
3
5
Theorem 63 (Roy’s identity) Suppose that x solves the problem in (3.1). If xi > 0 8
i, x; is a (reduced) KT pair, > 0 and x is regular, then:
xi =
3.2.2
@v (p; y) =@pi
@v (p; y) =@y
Expenditure minimization
Suppose that agents have a certain welfare target and want to …nd the minimum amount
of spending required to attain such level of utility. Then they face the problem:
min p x
s:t: u (x)
xi 0
Since the objective function is linear, this is a support function problem and we can use
the methods from section 2.4 to …nd a solution. The problem is identical to that of cost
minimization and, in this case yields an optimal value function e (p; ). As in any support
function problem, e (p; ) is homogeneous of degree one and concave. Moreover, from the
optimal value function we can recover the solution to the cost minimization by:
@e (p; )
= xi (p; )
@pi
49
Example 64 Suppose that Suppose that u : Rn+ ! R is continuous and monotonic in the
following sense: u(x) > u(y) whenever x > y where x > y means that xi > yi for each
i = 1; :::; n. If x solves the utility maximization problem:
max u (x)
s:t: p x
y, xi
0
show that x solves the expenditure maximization problem:
min p x
v (p; y) , xi
s:t: u (x)
0
where v (p; y) = u (x).
Proof. Suppose that x solves the utility maximization problem but does not solve the
expenditure maximization problem. Since x solves the utility maximization problem then
p x
y and u (x)
u (x) 8 x such that p x
y. Thus x is trivially feasible for the
expenditure min problem. But the hypothesis is that x does not solve the this problem
so there exists x0 such that u (x0 )
u (x) and p x0 < p x: This in turn implies that
p x0 < y or that x0 is feasible for the utility maximization problem. To complete the
argument, choose " > 0 small enough so that: p x0 < p (x0 + ") < y, then x0 + " > x0
which, by monotonicity of u ( ) ; implies that u (x0 + ") > u (x0 ) u (x), contradicting the
hypothesis that x solves the utility maximization problem.
Exercise 65 Consider a consumer with utility function u : Rn+ ! R. Furthermore, suppose u = f h where h : Rn+ ! R+ is homogeneous of degree one and f : Rn+ ! R+ is
a strictly increasing function satisfying f (0) = 0. Suppose that and ; 0 positive numbers
and that the expenditure minimization problems associated with utility levels and
and
0
are well defned for all nonnegative price vectors. Let e (p; ) denote the value of the
expenditure function for output price vector p and utility level : If u(x)
; x 0 and
p x = e (p; ); show that there exists a positive number such that p ( x) = e (p; 0 ).
Proof. First note that x solves the expenditure min problem for the utility level . Nex,
note also that by the assumptions of the problem f 1 is well de…ned. So, let:
=
f
f
1
50
( 0)
1( )
Chapter 4
Game Theory and General
Equilibrium
4.1
4.1.1
Game theory
Zero-sum games
De…nition 76 The standard n 1 simplex is the set
Remark 46 The standard n
= fx 2Rn jxi
0 and
Pn
i=1
xi = 1g
1 simplex is a closed, non-? and convex subset of Rn :
De…nition 77 A mixed strategy is a realization x of the random variable X that assigns
probabilities to each of the pure strategies (or actions) available to players. If M is the
action set for a player, these randomizations comprise the mixed extension of the action
set (M ).
Theorem 66 (Von-Neuman’s minimax) In a game with two players 1 and 2, each
with strategy set M and N, with the simplices (M ) Rm and (N ) Rn as the mixed
extensions of M and N, respectively, and A as the matrix of payo¤s:
max
x2 (M )
min xT Ay = min
y2N
y2 (N )
max xT Ay
x2M
Theorem 67 (Kakutani’s minimax) Suppose that C
Rn and D
Rn are non?; convex, compact. Suppose that f : C
D ! R is continuous and x !
7
f (x; y)
is quasiconcave while y 7! f (x; y) is quasiconvex. Then 9 (x; y) 2 C D such that
f (x; y) f (x; y) f (x; y). That is, (x; y) is a saddle point.
4.1.2
Non-zero-sum games and Nash Equilibrium
Preliminaries
The following game theoretic results are obtained in the context of a game with:
51
A set of players N = f1; 2; :::ng
A set of actions for each player: Ai
Payo¤ functions fi : A1
A2
An ! R
:::
De…nition 78 Allowing for mized strategies, a strategy pro…le x
^ is a vector of randomizations, one
containing x^o1 2 (A1 ) ; :::^
xn 2 (An ) where, as before,
n for each player, P
m
j
m j
(Ai ) = xi 2R jxi 0 and
j=1 xi = 1 :
De…nition 79 The strategy pro…le x
^ = (^
x1 ; x^2 :::^
xn ) 2 A1 A2 :::
equilibrium if, for each i 2 N one has that: x^i 2 arg max f (yi ; x
^ i)
An is a Nash
yi 2Ai
Remark 47 Recall that the continuity properties of correspondences are inherited by their
(Minkowski’s) sum and (cartesian) product, i.e., if 1 : X
Y and 2 : X
Z are both
upper hemicontinuous (UHC), then = 1 (x) + 2 (x) and = 1 (x)
2 (x) are UHC
too.
Remark 48 Recall that non-emptiness, compactness and convexity properties of sets are
also inherited by their (Minkowski’s) sum and (cartesian) product.
Theorem 68 (Nash, 1950) Suppose that:
1) :
2) :
3) :
Ai Rm is compact, convex, non-?
fi : A1 A2 ::: An ! R is continuous
xi 7 ! fi (xi ; x i ) is quasiconcave
then the game has a Nash Equilibrium.
Note that in order to …nd Nash equilibria, we need to be able to obtain the expected
payo¤ of each player for each strategy pro…le. Suppose that g ( ) describes the payo¤
function. Then we need to compute:
E [f (xi ; x i )] =
m1
X
i1 =1
:::
mn
X
gi (i1 ; :::in )
xi11 ... xinn :
in =1
Theorem (68) is the most widely used to justify the existence of equilibrium in strategic
form games. However, this result does not address the issue of uniqueness of equilibria.
There are mainly two avenues towards ensuring uniqueness, both of which rely on …xed
point arguments; the …rst one uses monotonicity of the objective function (as was hinted
in section 1.8) while the second uses a contraction mapping feature.
52
Exercise 69 (uniqueness via monotonicity) Let (u1 ; u2 ; A1 ; A2 ) comprise a 2-person
strategic form game. Suppose that Ai = R and that ui : R R ! R is twice di¤erentiable
and concave in xi for each x i . If the following two conditions hold:
@u2
@u1
(x1 ; x2 ) = 0 =
(x1 ; x2 )
@x1
@x2
!
2u
@ 2 u1
1
(z)
(z) @x@ 1 @x
@x21
2
(2) :
is negative de…nite 8z 2R2
@ 2 u1
@ 2 u1
(z)
(z)
@x1 @x2
@x2
(1) :
2
show that (x1 ; x2 ) is the unique Nash equilibrium of the game.
Proof. First note that xi 7! ui (x1 ; x2 ) concave along with the two conditions (1) (2)
imply that (x1 ; x2 ) is a Nash equilibrium for the game. Next, to see that it is unique, let
z = (x1 ; x2 ) and de…ne fi (z) = @ui (z) =@xi : Thus, condition (2) implies (by proposition
(21)) that f is strongly monotone. Therefore, by example (36) we know that the system
f (z) = 0 has a unique solution so that f (z) = 0 also has a unique solution and we
conclude that z = (x1 ; x2 ) is the unique Nash equilibrium of the game.
Exercise 70 (uniqueness via contraction) Consider the setup of theorem (68) and modify only the following: for each i 2 N , xi 7! fi (xi ; x i ) is strictly quasiconcave and
'i : A1 ::: An ! Ai is the best response mapping. Note that in this case strict quasiconcavity implies that 'i is single-valued, i.e. it is a best response function. Show that if
' is a contraction, the game has a unique Nash equilibrium.
Proof. First note that the conditions for the existence of an equilibrium are satis…ed.
Next, suppose that x and x0 are two Nash equilibria and x 6= x0 . Since ' is a contraction
we know that 9 < 1 such that d (' (x) ; ' (x0 ))
d (x; x0 ). However, if x and x0 are both
NE, then x =' (x) and x0 = ' (x0 ) implying that d (x; x0 )
d (x; x0 ) which is possible
0
only if x = x .
Next we provide Debreu’s (1952) version of the main theorem for the existence of Nash
equilibria in so-called "generalized" games. This is the key step towards the original proof
of the existence of equilibrium in pure exchange economies by Arrow and Debreu presented
in section 4.2.2.
4.1.3
The generalized game
Consider the case of a generalized game, also known as an "abstract economy". In this
setup, feasible actions are not independent as in the simple n person game, but there
exists a transition correspondence i : A i
Ai which maps action by players i into
feasible actions by player i; that is, it yields the set of actions that player i can take, given
what every one else is doing. Under the appropriate topological assumptions regarding i
the existence of equilibrium can be ensured as the following theorem states.
53
Theorem 71 (Debreu, 1952) Suppose that:
1)
2)
3)
4)
:
:
:
:
Ai Rmi is compact, convex, non-?
fi : A1 A2 ::: An ! R is continuous
xi 7 ! fi (xi ; x i ) is quasiconcave
Ai is UHC, LHC, compact,convex, non-?-valued
i : A i
then a Nash equilibrium for this generalized game (abstract economy) exists, i.e., 9 a pro…le
x
^ = (^
x1 ; x^2 :::^
xn ) 2 A1 A2 ::: An such that, for each i 2 N :
x^i 2 arg
max
yi 2
x i)
i (^
f (yi ; x
^ i)
Rmi compact, convex, non-? ) A1
Proof. (step 1) First note that Ai
m1
mn
R
::: R compact, convex, non-?.
(step 2) Next, for each (x1 ; :::xn ) 2 A1 ::: An , let:
i
(x i ) = arg
max
yi 2
i (x i )
:::
An
fi (yi ; x i )
now de…ne:
(x) =
1
(x2 ; :::xn )
:::
n
(x1 ; :::xn 1 )
and note:
(x)
A1
:::
An
(step3) Note that since fi (xi ; x i ) is continuous and i ( ) is non-?, UHC, LHC,
compact-valued, by the Continuous Maximum theorem (Theorem 26) we have that i (x i )
is non-?;compact-valued and UHC for each i. Furthermore, since xi 7 ! fi (xi ; x i ) is
quasiconcave and i ( ) is convex-valued, by the Concave Maximum theorem (Theorem
27) i (x i ) is convex-valued for each i. Therefore, (x) is also convex-valued.
(step 4) Since the set A1 ::: An is a non-?, compact, convex set and : A1 ::: An
A1 ::: An is a non-?, compact, convex-valued, UHC correspondence, by Kakutani’s
theorem, has a …xed point, i.e. 9 x
^ = (^
x1 ; :::; x^n ) A1 ::: An such that:
x
^ 2
(^
x)
x^i 2 i (^
x i) 8 i 2 N
x^i 2 arg max fi (yi ; x i ) 8 i 2 N
yi 2
i (x i )
so the "generalized" game has a Nash equilibrium.
4.2
4.2.1
General Equilibrium theory
Prelminaries
De…nition 80 (pure exchange economy) A pure exchange economy is de…ned by a set
of consumers N = f1; 2; :::ng, a set of endowments for each consumer ! i 2 RL+ and payo¤s
for each player/consumer ui : RL+ ! R:
54
De…nition 81 (improvable by coallition) Suppose that (x1 ; :::; xn ) is a feasible allocation of the pure exchange economy. Let S N . We say that S can improve upon x if, for
each i 2 S; 9 x0i 2 RL+ such that:
X
X
!i
x0i
i2S
i2S
ui (x0i )
> ui (xi )
De…nition 82 (core allocation) A feasible allocation is a core allocation if it cannot be
improved by any coallition.
Remark 49 When agents are given the opportunity to trade their endowments, the outcome of the exchange will lie inside the core of the economy.
De…nition 83 (Walrasian equilibrium) In a pure exchange economy a (n + 1) tuple:
(^
x1 ; :::; x
^n ; p) is a Walrasian equilibrium (W-Eq) if p 2 RL+ n f0g and:
1) : x
^i 2 arg max ui (xi ) (optimal choices)
2) :
n
X
i=1
x
^i
p xi p ! i
n
X
!i
(no excess demand)
i=1
Remark 50 Note that each ! i is a L 1 vector. Thus, the complete set of endowments
for the economy ! is a L N matrix whose typical element ! ki is the amount of good k
that player i is endowed with. The same note applies for the set of choices xi
Remark 51 Since ! i and xi are L 1 vectors, the second condition aboves implies:
1
1 0 Pn
0 Pn
1
1
!
x
^
i=1 i
i=1 i
C
C B
B
..
..
A
A @
@
Pn . L
Pn . L
!
x
^
i=1 i
i=1 i
that is, the sum of choices of a particular commodity accross players must be less than or
equal to the economy’s total endowment of such commodity.
Example 72 (W-Eq does not exist) Consider an economy with two agents. The economy is characterized by the following endowments and preferences:
! 1 = (a1 ; b1 ) = (1; 1) u1 (x1 ; y1 ) = x21 + y12
! 2 = (a2 ; b2 ) = (1; 1)
u2 (x2 ; y2 ) = x2 y2
A W-Eq for this economy does not exist. To see why, let p = (p; q) be the price vector and
note that the we can use the KT conditions to solve the problems:
max x21 + y12
s:t: px1 + qy1 p ! 1
max x2 y2
s:t: px2 + qy2 p ! 2
55
now an interior solution to the KT system would be:
2x1 p 1 = 0
2y1 q 1 = 0
px1 + q y1 = p + q
whose solution is:
x1 = p
p+q
p2 +q 2
x2 =
p+q
2pq
p+q
2p
y1 = q
p+q
p2 +q 2
y2 =
p+q
2q
1
note that if p > q then p
x2 q 2 = 0
y2 p 2 = 0
px2 + q y2 = p + q
p+q
p2 +q 2
= 2 pp+q
2 +q 2
+
p+q
2p
2
=
> 2 so that there is excess demand for good x.
Likewise, if p < q then q pp+q
+ p+q
> 2 so that there is excess demand for good y.
2 +q 2
2q
Finally, if p = q there are two solutions to agent 1’s problem in either of which there would
be excess demand for one good. Thus, a W-Eq for this economy does not exist.
So what is wrong with the last example? The main issue is that the utility function
of agent one (x1 ; y1 ) 7! u1 (x1 ; y1 ) is convex. In the following section we explore what is
required from preferences in order for a Walrasian equilibrium to exist.
4.2.2
Existence of equilibrium in pure exchange economies
In this section we present a version of the main existence theorem for the case of pure
exchange economies. This is nested in the more general "main theorem" found in Arrow
and Debreu (1954).
Theorem 73 (Arrow and Debreu, 1954) Suppose a pure exchange economy with ! i 2
RL+ n f0g and ui : RL+ ! R where ui ( ) is continuous and strictly quasiconcave on xi . Then
a Walrasian equilibrium exists for this economy.
Proof. (step 1: set up a generalized game) First, de…ne the L-simplex of relative prices:
=
p 2 RL+ j
and:
K = z 2 RL+ j z
L
P
pk = 1
k=1
n
P
! i + (1)
i=1
where (1) is a L 1 vector of ones. Next, for each p 2
of feasible actions:
= 0 (x1 ; :::; xn )
56
de…ne the zero-player with set
and payo¤ function:
u0 (p; x) =
n
P
p (xi
!i)
i=1
and note that now we have (n + 1) players: n consumers and the price-setter (auctioner).
Next, for a typical consumer i de…ne the budget set correspondence:
i
(p) = xi 2 RL+ j p xi
p !i
and the feasible action correspondence:
i
(p) =
i
(p) \ K
and note that the feasible action correspondence for player i is parametrized only by p
which is chosen by player zero.
(step 2: show the generalized game has a NE) Next, notice that 0 (x) = is obviously
non-? compact, convex, UHC, LHC. Note also that K compact and i (p) closed ) i (p)\
K = i (p) compact. Finally, one can easily show that i ( ) is UHC, LHC. On the other
hand u0 ( ) is continuous and quasiconcave in p since it’s linear and recall that ui ( ) is
continuous and quasiconcave on xi 8 i by assumption.
Summarizing, since all the objective functions are continuous and quasiconcave on the
decision variable, all the feasible action correspondences are UHC, LHC, convex, non?, compact-valued, the resulting optimal-value (best-response) correspondences is non-?;
compact, convex-valued, UHC and de…ned on non-?, compact, convex sets. Therefore, by
Kakutani’s theorem, the game has a NE, i.e., 9 an (n + 1)-tuple (^
x1 ; :::^
xn ; p
^ ) such that:
P
p
^ 2 arg max u0 (p; x
^) = ni=1 p (xi ! i )
p2
x)
0 (^
x
^i 2 arg max ui p
^; x
^ i ; xi
x2
p)
i (^
8i2N
(step 3: show the NE for the generalized game is a W-Eq) Now the claim is that
(^
x1 ; :::^
xn ; p
^ ) is a Walrasian equilibrium to the pure exchange economy. To see why, …rst
note that:
x
^i 2
i
and therefore:
max
max
p2
PL
p2
k=1
pk
(^
p) :) x
^i 2 i (^
p)
)p
^ (xi ! i )
P
) ni=1 p (xi
Pn
i=1
Pn
p (xi
i=1
Pn
xki
i=1
57
(xi
!i)
! ki
!i)
0
!i)
0
0
0
0
and therefore excess demand for all commodities in the economy is less than or equal to
zero. Next, to show that all consumers are optimizing, one must show that in this particular
case:
x
^i 2 arg max ui p
^; x
^ i ; xi ) x
^i 2 arg max ui p
^; x
^ i ; xi
xi 2
p)
i (^
to see this, suppose 9 x0i 2
that whenever 2 (0; ):
i
xi 2
p)
i (^
(^
p) such that ui p
^; x
^ i ; x0i > ui p
^; x
^ i; x
^i : Then 9
x0i + (1
)x
^i <
n
P
such
! i + (1)
i=1
2 K
and:
p [ x0i + (1
(p x0i ) + (1
(^
p ! i ) + (1
= p
^ !i
)x
^i ] =
) (p x
^i )
) (^
p !i)
therefore:
x0i + (1
)x
^i 2
i
(^
p)
but by assumption:
ui p
^; x
^ i ; x0i > ui p
^; x
^ i; x
^ i ) ui p
^; x
^ i ; x0i + (1
)x
^i > ui (^
xi )
which contradicts the result of step 2 that:
x
^i 2 arg max ui p
^; x
^ i ; xi
x2
p)
i (^
8i2N
Summarizing, x
^i 2 arg
^; x
^ i ; xi , therefore all the consumers are making
Pnmaxx2 i (^p) ui p
0 so there is no excess demand. Hence, the pure
optimal choices and i=1 (xi ! i )
exchange economy has a Walrasian equilibrium.
4.2.3
Existence of equilibrium in production economies
In this section we present a reduced-form version of Debreu’s (1959) celebrated theorem
for the existence of equilibrium in private ownership production economies. It is a reduced
version because we assume the existence of a "well behavied" net supply correspondence
(see below).
De…nition 84 A private ownership production economy is de…ned by a set of consumers
N = f1; 2; :::ng, a set of endowments for each consumer ! i 2 RL+ ; a set of …rms M =
f1;
j owned by consumer i and
Pn2; :::mg owned by consumers with ij as the share of …rm
L
i=1 ij = 1 and payo¤s for each player/consumer ui : R+ ! R:
58
De…nition 85 A Walrasian equilibrium in a private ownership production economy is a
(n + m + 1) tuple: (^
x1 ; :::; x
^n ; y
^1 ; :::; y
^m ; p) such that p 2 RL+ n f0g and for all i 2 N and
all j 2 M :
1) : x
^i 2 arg max ui (xi ) (consumers maximize)
x2
i (p)
2) : y
^j 2 arg max p yj
2) :
n
X
yj 2Yj
n
X
x
^i
i=1
(…rms maximize)
m
X
!i +
i=1
(no excess demand)
y
^j
i=1
n
P
where i (p) = xi 2 RL+ j p xi p ! i + m
j=1
aggregate production possibility set.
p y
^j
ij
o
and Y = Y1 + ::: + Ym is the
Remark 52 Note that the vector p contains the prices of all commodities in the economy,
both inputs and …nal goods.
Remark 53 Finally, a pure exchange economy is a special case of a production economy
where yj = f0g 8 j
Example 74 (decreasing returns) Suppose that there are two commodities and two
consumers with problems:
! 1 = (6; 0) ;
! 2 = (6; 0) ;
= 1=2 u1 (x1 ; y1 ) = x1 y1
1 = 1=2 u2 (x2 ; y2 ) = x2 y2
1
and a single …rm with production possibility set:
p
y = Y = (x; y) 2 R2 j x
y; x
so that the …rm”s problem is:
p
max q x
px s.t. x
0
the FOC for the …rm are:
xF =
q
2p
2
yF =
(p; q) =
q
2p
q2
4p
the FOC for the consumers are:
6p +
x1 =
q2
4p
2p
6p +
y1 =
1
2
1
2
2q
59
q2
4p
= x2
= y2
0
the market clearing conditions are:
2
x1 + x2 + xF = 2 4
2
y1 + y2 = 2 4
6p +
3
q2
4p
1
2
6p +
2p
q2
4p
1
2
2q
5+
q
2p
2
= 12
3
5 = q = yF
2p
(4.1)
(4.2)
A simple avenue to solve this system is as follows. Walras’ law says that (4.1) holds if
and only if (4.2) holds. Therefore, we can solve for the single variable q=p in equation
(4.2) and then verify that (4.1) holds under such price ratio. The solution to this system is
q=p = 4pwhich is the W-Eq price ratio. Finally, the production possibility frontier is given
by y = 12 x1 x2 :
Theorem 75 (Debreu, 1959) Suppose a private ownership production economy where,
for each consumer i; ui ( ) is continuous and concave and ! i 2 RL+ n f0g; for each …rm j, the
production possibility set Yj is closed, satis…es free disposal and no free lunch Yj \ RL+ = f0g .
Finally, suppose that for each j, 9 i :
RL+ non-?, UHC, compact, convex-valued and
for each p 2
one has that: i (p)
arg max p yj : Then, a Walrasian equilibrium
yj 2Yj
exists for this economy.
as in the proof of Theorem (73) and :
RL+ as (p) =
Proof. (step 1) De…ne
P
n
( ) is
j=1 i (p) (i.e., the Minkowski’s sum of the net-supply correspondences). Then
non-?; compact, convex-valued and UHC. Furthermore, compact ) ( ) is a compact
subset of RL+ . Therefore, 9 " > 0 such that z 2 ( ) ) z < ". Next, de…ne j (p) = p yj
where yj 2 i (p). Note that j :
R is continuous and since yj 2 i (p) ) yj 2
arg maxyj 2Yj p yj so that j (p) = p yj
p y0j for any yj0 2 Yj which given the no
free lunch assumption implies that j (p) 0 8 p 0.
(step 2) Next, let:
n
P
K = z 2 RL+ j z
!i + "
i=1
and for a typical consumer i de…ne the budget set correspondence:
(
)
m
X
xi 2 RL+ j p xi p ! i +
ij j (p)
i (p) =
j=1
and the feasible action correspondence:
i
(p) =
i
Note that K compact and i (p) closed )
is easily seen that K, i (p) are convex, so
60
(p) \ K
(p) \ K = i (p) compact. Moreover, it
i (p) \ K = i (p) is convex and one can
i
show that i ( ) is UHC, LHC. This, along with the fact that ui ( ) is continuous and
concave imply that for each p 2 ; the maximum value (best-response) correspondences
i (p) = arg maxxi 2 i (p) ui (x i ; p; xi ) is non-?, convex, compact, UHC (by the Maximum
Theorems).
de…ne the excess demand correspondence:
(step 3) Now, for each p 2
(p) =
n
X
n
X
i (p)
i=1
!i +
i=1
n
X
i
(p)
j=1
and notice that (p) is non-?, convex, compact, UHC (inherits its components properties).
Next suppose p 2
and ^
z 2 (p), then 9 x
^i 2 i (p) and y
^j 2 i (p) such that:
^
z =
n
X
n
X
x
^i
i=1
p ^
z = p
=
"
n
X
i=1
n
X
x
^i
i=1
p x
^i
=
i=1
=
n
X
i=1
!i +
i
m
X
p !i +
i=1
n
X
p x
^i
"
#
m
X
p y
^j
j=1
i=1
j=1
p !i +
i=1
p x
^i
y
^j
j=1
p !i +
m
n
X
X
n
X
i=1
0
since we know that x
^i 2
implies that p ^
z 0)^
z
y
^j
j=1
n
X
i=1
n
X
i=1
n
X
!i +
n
X
(p)
i (p)
0: Summarizing:
i
ij
j
ij
#
!
j
(p)
(p)
(p). But p 2
)p
0 which in turn
1) : x
^i 2 arg max ui (xi ) (consumers maximize)
x2
i (p)
2) : y
^j 2 arg max p yj
2) :
n
X
i=1
yj 2Yj
x
^i
n
X
i=1
!i +
m
X
(…rms maximize)
y
^j
(no excess demand)
i=1
Therefore the only thing we’re missing for a Walrasian equilibrium is that is x
^i 2 arg maxx2 i (p) ui (xi ).
But form step 3 in Theorem 18 above this is easily shown by contradiction since ui ( ) is
concave and thus, strictly quasiconcave.
4.2.4
Welfare theorems (pure exchange economies)
Prelminaries
61
De…nition 86 A function f : Rn ! Rm is weakly monotonic if x > y ) f (x) > f (y)
i.e. if the vector x is greater than the vector y component by component
De…nition 87 A function f : Rn ! Rm is strongly monotonic if x
y and xi > yi
for some i ) f (x) > f (y) i.e. if at least one component of the vector x is greater than
its corresponding component in the vector y.
Remark 54 Naturally, strongly monotonic)weakly monotonic
De…nition 88 A pure exchange economy is de…ned by a set of consumers N = f1; 2; :::ng,
a set of endowments for each consumer ! i 2 RL+ and payo¤s for each player/consumer
ui : RL+ ! R:
De…nition 89 In a pure exchange economy, a feasible allocation x
^ = (^
x1 ; :::; x
^n ) is weakly
0
0
0
0
xi ) for all i (i.e., not all
Pareto optimal (WPO) if @ x = (x1 ; :::; xn ) such that ui (xi ) > ui (^
players/consumers can be made better o¤ ).
De…nition 90 In a pure exchange economy, a feasible allocation x
^ = (^
x1 ; :::; x
^n ) is strongly
0
0
0
0
ui (^
xi ) for all i 2 N and
Pareto optimal (SPO) if @ x = (x1 ; :::; xn ) such that ui (xi )
0
uj xj > uj (^
xj ) for at least one j 2 N (i.e. nobody can be made better o¤ without hurting
some player).
In other words, the set of WPO allocations is:
W P O = x 2RL+ j @ x0 = (x01 ; :::; x0n ) such that ui (x0i ) > ui (^
xi ) 8 i
while the set of SPO allocations is:
SP O = x 2RL+ j @ x0 = (x01 ; :::; x0n ) such that ui (x0i )
ui (^
xi ) 8 i and uj x0j > uj (^
xj ) for some j
Remark 55 Naturally, x 2SP O ) x 2W P O, that is the set of SP O
Proposition 27 If ui ( ) is strongly monotonic for all i, then W P O
W P O = SP O or x
^ is W P O , x
^ is SP O:
W P O.
SP O so that
Proposition 28 Every core allocation is WPO.
Proof. If x 2 core then x cannot be improved upon by any coallition. Apply de…nition
(81).Done!
Proposition 29 Every competitive equilibrium allocation belongs to the core of the economy.
Proof. The mechanics of this proof is the same as that used to prove the …rst welfare
theorem so see next section.
62
First fundamental theorem of welfare economics
Theorem 76 Every Walrasian equilibrium allocation is weakly Pareto optimal.
Proof. (always by contradiction) Suppose that x
^ = (^
x1 ; :::; x
^n ; p) is a Walrasian equilibrium but it’s not WPO. Then 9 some x0 feasible such that 8 i:
xi )
ui (x0i ) > ui (^
and:
n
X
n
X
x0i
i=1
and therefore:
p
"
n
X
x0i
i=1
!i
i=1
#
p
"
but x
^ is a Walrasian equilibrium so:
n
X
!i
i=1
#
(4.3)
x
^i 2 arg max ui (xi )
p xi p ! i
which means that:
ui (x0i ) > ui (^
xi ) ) p x0i > p ! i
" n
#
" n
#
X
X
)p
x0i > p
!i
i=1
i=1
contradicting inequality (4.3).
Pn
Pn
0
0
0
x
]
>
p
[
Remark
56
Note
that
p
x
>
p
!
;
x
>
!
just
as
p
[
i
i
i
i
i
i=1 ! i ] ;
i=1
Pn
Pn
0
the contradiction conveys is the fact that for at least some
i=1 ! i . WhatP
i=1 xi >
Pn
Pn
0
k
;
that
is,
even
though
!
commodity k P
it is true that ni=1 xk0
i
i >
i=1 xi may not be
i=1
n
greater than i=1 ! i component by component, it is larger in at least one element. But
this is enough to violate feasibility as is detailed in remark (51).
Theorem 77 If the consumers’objective functions (player payo¤s) are strongly monotonic,
then any Walrasian equilibrium allocation is strongly Pareto optimal.
Proof. Suppose that x
^ = (^
x1 ; :::; x
^n ; p) is a Walrasian equilibrium but x
^ it’s not SPO.
0
Since x
^ is not SPO, 9 x feasible, such that:
ui (x0i )
ui (^
xi ) 8 i
and:
uj x0j > uj (^
xj ) for some j
63
So, chose and …x some (vector) " arbitrarily small and let:
yi = x0i + "
now, by strong monotonicity:
yi > x0i ) ui (yi ) > ui (x0i )
ui (^
xi )
but since x
^ is a Walrasian equilibrium, then it follows that:
ui (yi ) > ui (^
xi ) ) p y i > p ! i
so that:
which, implies:
p yi > p ! i
p [x0i + "] > p ! i
PL
> p !i
p x0i + "
k=1 pk
p x0i
p ! i for all i 6= j
and:
p x0j > p ! j for some j
summing over i (including j) we conclude:
Pn
Pn
0
i=1 [p ! i ]
i=1 [p xi ] >
P
Pn
p i=1 x0i > p ni=1 ! i
which, again, implies that for at least some commodity k it is true that
contradicting the hypothesis that x0i is feasible.
Pn
i=1
xk0
i >
Pn
i=1
! ki ,
Second fundamental theorem of welfare economics
Theorem 78 If the consumers’ (players’) objective functions (payo¤s) are continuous,
quasiconcave and weakly monotonic, then any given WPO allocation x
^ can be descentrilized,
i.e., 9 p such that (^
x; p) is a Walrasian equilibrium.
Proof. Suppose that x
^ = (^
x1 ; :::; x
^n ) is WPO and x
^i 2 RL+ , then we must show that 9
p 2 RL+ n f0g such that:
ui (xi ) > ui (^
xi ) ) p xi > p x
^i
(step 1: …nd a hyperplane bounding the consumers’preferred set) For each i de…ne:
Si = xi 2 RL+ j ui (xi ) > ui (^
xi )
64
and note that Si is non-? since 9 " s.t. x
^i +" = x0i 2 RL+ and by monotonicity of u ( ) ;
xi ). The set Si is the halfspace containing all comodity bundles strictly
ui (x0i ) > ui (^
preferred by agent i to x
^i : Also note that by quasiconcavity of u ( ) ; Si is convex for each
i (the set xi 2 RL+ j ui (xi ) >
is convex 8 ). Next, let S = S1 + ::: + Sn and note that
S is convex, non-?: Now, de…ne:
0
1
1
1
x
+
:::
+
x
n
1
n
X
B
C
..
^
z=
x
^i = @
A
.
L
L
i=1
x1 + ::: + xn
and claim that for " > 0, ^
z " 2
= clS. To see this, note that if ^
z " 2 clS then 9 a
sequence fz (k)g1
such
that
z
(k)
2
S
for
every
k
and
z
(k)
!
^
z ", which in turn
k=1
means that 9 k such that:
P
Pn
z (k ) < ^
z = ni=1 x
^i
i=1 ! i
so z (k ) is feasible and z (k ) 2 S which means that ui (z (k )) > ui (^
xi ) for every i and this
contradicts the assumption that x
^i is WPO. Summarizing, S is non-?, closed, convex, and
^
z "2
= clS therefore, by Minkowski’s theorem, 9 p 6= 0 such that p z p ^
z 8 z 2 S.
(step 2: show that p is nonegative) To show that p is nonegative, we must show that
pk
0 8 k = 1; :::L: To see this, …rst choose and …x k and " > 0 and de…ne (j) as the
zero vector with 1 as its j-th coordenate:
yi = x
^i + (k) + "
L
P
(l)
l=1
l6=k
therefore by monotonicity of u ( ):
ui (yi ) > ui (^
xi )
) yi 2 S
so that using the …nal result from step 1:
P
p ( ni=1 yi )
P
p ( ni=1 x
^i )
hence:
p
"
x
^i + (k) + "
"
n pk + "
#
L
P
(l)
l6=k
L
P
pl
l6=k
P
p ( ni=1 x
^i )
!#
0
pk
0
65
(step 3a: show that p is an equilibrium price vector) Next we show that ui (xi ) >
ui (^
xi ) ) p xi p x
^i . To see this suppose for some i that ui (xi ) > ui (^
xi ) which implies
xi 2 Si and for all j 6= i let yj = x
^j + " so that uiP
(yj ) > ui (^
xj ) by monotonicity, implying
that yj 2 Sj for each and all j. Notice that xi + ni=1 yj 2 S, so:
P
P
p [xi + ni=1 yi ]
p [ ni=1 x
^i ]
hP
i
P
p xi + p
xj + ")
p x
^i + p
^j
j6=i (^
j6=i x
p xi
p x
^i
^i ;
(step 3b: show that p is an equilibrium price vector) Finally, to se that p xi 6= p x
suppose p xi = p x
^i (and …nd a contradiction). From the previous step we know
ui (xi ) > ui (^
xi ) : By continuity of u ( ) 9
2 (0; 1) such that ui ( xi ) > ui (^
xi ); but
0 < < 1 ) p xi < p x
^i so let yi = xi and note that ui (yi ) > ui (^
xi ) but p yi < p x
^i
which contradicts the result of step 3. Summarizing, we have shown that if x
^ is WPO, 9
xi ) ) p xi p x
^i
p 2 RL+ n f0g such that ui (xi ) > ui (^
4.2.5
Relaxing the Walrasian assumptions
The welfare theorems rely heavily on the Walrasian equilibrium assumptions; agents are
price takers (perfect competition), perfect information and complete markets. To see the
crucial role that these assumptions play, consider the following cases in which equilibria
may not be Pareto optima.
Example 79 (not everyone is price taker) Consider the following two consumer, two
good pure exchange economy.
u1 (x1 ; y1 ) = x1 y1
u2 (x2 ; y2 ) = x2 y2
! 1 = (3; 1)
! 2 = (1; 3)
The Walrasian equilibrium for this economy is found by solving:
x1 =
3p + q
= x2 ,
2p
y1 =
3p + q
= y2
2q
whose solution is easily seen to be p=q = 1 and (x1 ; x2 ; y1 ; y2 ) = (2; 2; 2; 2). Next, consider
the case in which agent 1 is a price taker but agent 2 is a price setter. In particular, agent
1 has demands x1 (p; q) = 3p+q
and y1 (p; q) = 3p+q
but 2 chooses the nonnegative prices
2p
2q
p; q at which they trade. If these prices yield feasible demands for agent 1, he consumes the
bundle (x1 (p; q) ; y1 (p; q)) and agent 2 consumes the bundle (4 x1 (p; q) ; 4 y1 (p; q)). If
1’s demands are infeasible at the prices that 2 o¤ered each agent consumes its endowment.
Since agent 2 can be better o¤ by trading, he will not o¤er p = 0 or q = 0 since these yields
infeasible demands for agent 1 and forces both to consume their endowments. Thus, since
66
both o¤ered prices are positive, only their ration matters so WLOG suppose q = 1. Then
agent 1’;s demands would be:
x1 =
3
1
+
2 2p
and
y1 =
3p 1
+
2
2
if we solve for p in one of these equations and replace in the other, we obtain the agent 2’s
o¤er curve:
1
3
3
y1
=
x1
2
2
4
Now, agent 2’s consumption bundle would be:
3
1
5
+
=
2 2p
2
3p 1
7
+ =
2
2
2
x2 = 4
y2 = 4
1
2p
3p
2
now, to compute agent 2’s best price o¤er to agent one we solve the problem:
max x2 y2
x2 ;y2
s:t: x2 = 4
y2 = 4
3
y1
x1
2
x1
y1
1
3
=
2
4
or equivalently, replacing x1 ; y1 and expanding the o¤er equation:
max (4
x1 ) (4
x1 ;y1
s:t
3
y1
2
x1 y 1
y1 )
1
x1 = 0
2
the Lagrangian for this problem is:
L = (4
with FOC:
x1 ) (4
y1 )
x1 y1
3
y1
2
1
x1
2
2y1 8
2x1 8
=
) 14x1 = 10y1 + 16
1 2y1
3 2x1
since we are looking for the price p that maximizes 2’s welfare, we can replace y1 ; x1 with
the expressions found above so 14x1 = 10y1 + 16 becomes:
14
3
1
+
2 2p
= 10
67
3p 1
+
2
2
+ 16
and solving for p yields p =
are:
x2
x1
p
7=15. Thus, the allocations when agent 2 is a price setter
5
=
2
3
=
2
r
1 15
,
2 7
r
1 15
,
2 7
7
y2 =
2
1
y1 =
2
r
3 7
2 15
r
3 7
2 15
and notice that agent 2 is better o¤ as price setter (PS) since:
u2 (W E) = u2 (x2 ; y2 ) = 2
u2 (P S) = u2 (x2 ; y2 ) =
2=4
r !
5 1 15
2 2 7
7
2
3
2
r
7
15
!
>4
however, this allocation is not WPO; to see why recall that at WPO allocations both agents
must have equal MRS but in this case:
r
r
@u1
@u1
7
7
@x1
@x1
=
and
>
@u1
@u
1
15
15
@y1 (x ;y )
@y1 (x ;y )
1 1
2 2
Example 80 (imperfect competition: Walras is not Nash) Consider an industry with
demand curve P = 2 Q
i : The
P in which every …rm has the same cost structure Ci (Q) = QP
market output is Q = i Qi and the associated market clearing price is P = 2
i Qi .
Consequently, …rm i’s pro…t is given by:
#
"
X
Qi Qi
i (Q) = 1
i
in a competitive world, the industry price would equal the marginal cost:
P = M C ) P = @Ci (Q) =@Qi = 1
and therefore pro…ts would be zero. The competitive equilibrium allocation is symmetric
and is found by solving:
P
Q )1 = 2
1
) Qi =
n
= 2
nQi
However, the pro…le Q = (Q1 ; Q2 ; :::Qn ) is not a Nash equilibrium. To see why note that
the best response of …rm i to what every other …rm does when they choose the competitive
equilibrium production is given by:
^ i = arg max [1
Q
Qi 0
(n
68
1) Qi
Qi ] Qi
^ i = 1 6= Qi from where we conclude that
which given that Qi = n1 can be found to be Q
2n
Q = (Q1 ; Q2 ; :::Qn ) is not a Nash equilibrium. In fact, we know that the NEq is the solution
to the Cournot model which is given by:
"
#
X
Qi = arg max 1
Qi Qi
Qi 0
with equilibrium pro…le Qi =
1
n+1
i
for each i and market price given by P =
69
2+n
:
1+n
Chapter 5
Decision making under uncertainty
5.1
Expected utility hypothesis
Suppose that D is the space of distribution functions and that % is a preference relation
on D. Then if F 2 D then F is a CDF. In a world of n comodities, we say that:
x % y , u (x)
u (y)
to have an analogous de…nition for distribution functions we need an objective function
that satis…es the expected utility hypothesis (EUH):
R
De…nition 91 u : R ! R satis…es the EUH if 9 V s.t. V (F ) = u(x)f (x)dx where x
is a realization of (continuous) RV X, X F ( ); f (x) = dF (x) is a Riemann integrable
density and F 2 D the distribution space.
R
P
Remark 57 If X is a discrete RV we use
instead of in the above de…nition.
Theorem 81 u is a Von-Neumann-Morgenstain (vNM) utility function , u satis…es the
EUH.
Remark 58 This notation is by no means standard; for instance, u is what Mas-Colell,
Whinston and Green call Bernoulli utility function; they assign the label vNM utility to
where (x) = E [u (x)] :
w
De…nition 92 Continuity axiom: If Fn ! F and Fn % G ) F % G
De…nition 93 % on D satis…es the independence axiom if the preference order of two
distributions remains unaltered by convex combinations of each of the two with a third
distribution, i.e., F % G , F + (1
)H % G + (1
)H:
Now we can de…ne a preference relation for D analogous to that de…ned on the commodity space:
70
Proposition 30 If % on D satis…es order, continuity and independence axioms, then 9
V : D ! R such that:
F % G , V (F ) V (G)
If t 2 (0; 1) and F; G 2 D, then:
V (tF + (1 t)G) = tV (F ) + (1 t)G
9 u : R ! R that satis…es the EUH
(5.1)
(5.2)
(5.3)
We can also de…ne this preference relation on the space of random variables. In particular, suppose that ( ; A; P ) is a probability space and Lo is the space of random variables,
then % can be de…ned on Lo by:
X % Y , E(u X)
, E(u(X))
5.2
E(u Y )
E(u(Y ))
Risk aversion
De…nition
94 (risk aversion I) An agent is risk averse if E [u (x)]
R
u (x)f (x)dx
De…nition 95 (risk aversion II) An agent is risk averse if FftX+(1
t)FfY g
R
u (E [x]) or u(x)f (x)dx
t)Y g
% tFfXg + (1
De…nition 96 (risk aversion III) An agent is risk-averse if u ( x + (1
(1
) u (y) i.e., if his utility function is concave (u0 > 0 and u00 < 0).
)y) > u (x)+
Remark 59 Note that if F is a degenerate distribution (i.e, f (x = t) = 1 for some real
#), then V (Ft ) = u(t)
In the last de…nition, we can take things a step further. If u is a vNM utility representation for % then:
V FftX+(1 t)Y g
V tFfXg + (1 t)FfY g
= u (tx + (1 t)y) (no uncertainty)
= tV FfXg + (1 t) V FfY g
De…nition 97 Suppose that u : R ! R is a Von Neumann–Morgenstern utility function
satisfying u0 ( ) > 0 then, the Arrow-Pratt measure of absolute risk aversion (ARA) at x is
de…ned as:
u00 (x)
(x) = 0
u (x)
and the Arrow-Pratt measure of relative risk aversion (RRA) is de…ned as:
(x) x =
71
u00 (x)
x
u0 (x)
Example 82 The utility function u(x) =
(x) = k:
e
kx
exhibits constant ARA with RA coe¢ cient
Remark 60 Notice that if ' ( ) is increasing and
is non-decreasing.
5.2.1
( ) non-decreasing then z 7!
(' (z))
Application: portfolio choice
Let the cannonical portfolio choice problem with one risky (Y ), one riskless asset (r) and
initial wealth w be de…ned as:
max f (x) = max E [u (xY + (w
x
x
s:t: 0
x
(5.4)
x)r)]
w
Proposition 31 Suppose there exists an interval (0; w ) on which the following is true:
for every w 2 (0; w ) ; E [Y ]
r and E [u0 (xY + (w x)r)] < 0 which implies that 0 <
x (w) < w for w 2 (0; w ). Furthermore, suppose that u00 ( ) < 0: If x = arg max E [u (xY + (w
x
then:
1) :
2) :
3) :
dx(w)
dw
dx(w)
( ) non-increasing )
dw
dx(w)
=0
( ) constant )
dw
( ) non-decreasing )
0
0
Proof. We prove 1) since 2)-3) follow inmediately. First note that 0 < x (w) < w ) the
constraint does not bind and we have an interior solution (i.e. in any KT system x > 0
and = 0). Thus f 0 (x) = 0 8 w 2 (0; w ) and x is a function of w, i.e., x = x (w). Hence
by the F.O.C.:
E [u0 (rw + x(Y r)) (Y r)] = 0
di¤erentiating w.r.t. w :
E u00 (rw + x(Y
dx
(Y r)
= 0
dw
dx
rE fu00 (rw + x(Y r)) (Y r)g
=
00
2
E [u (rw + x(Y r)) (Y r) ]
dw
r)) (Y
r) r +
now, to sign this expression notice that the denominator is positive since u00 ( ) < 0 and
necessarily (Y r)2 > 0: Hence we want to sign the numerator so let:
' (y) = rw + x(y
72
r)
x)r)] ;
where y is a realization of RV Y: Note that y 7! ' (y) is strictly increasing and since ( )
is non decreasing we have that y 7! (' (y)) is non-decreasing. Thus by the de…nition of
non decreasing functions:
[ (' (y))
(' (r))] [y
(' (y)) (y
r]
r)
0
(' (r)) (y
r)
and note that ' (r) = rw so that:
u00 (' (y))
(y
u0 (' (y))
u00 (rw + x(y
E [u00 (rw + x(Y
r)) (y
r)) (Y
u00 (rw)
(y r)
u0 (rw)
u00 (rw) 0
u (' (y)) (y r)
u0 (rw)
u00 (rw)
E [u0 (' (Y )) (Y r)]
0
{z
}
u (rw) |
r)
r)
r)]
=0 by FOC above
00
E [u (rw + x(Y
and therefore
dx
dw
r)) (Y
r)]
0
0.
Example 83 Suppose that u (x) = e kx . Then E [u (xY + (w x)r)] = E [ exp ( k (xY + (w
which can be written exp ( kwr) E [ exp ( kx (Y r))] so that w is out of the expectation and therefore dx(w)
= 0:
dw
Proposition 32 Suppose there exists an interval (0; w ) on which the following is true:
for every w 2 (0; w ) ; E [Y ]
r and E [u0 (xY + (w x)r)] < 0 . Furthermore, suppose
that u00 ( ) < 0: If x = arg max E [u (xY + (w x)r)] ; then:
x
d x(w)
dw
w
d x(w)
2) : z 7! (z) z non-increasing )
dw
w
d x(w)
3) : z 7! (z) z constant )
=0
dw
w
1) : z 7! (z) z non-decreasing )
Proof. Let ^ (w) =
x(w)
w
and proceed as before:
E [u0 (rw + x(Y r)) (Y
x
E u0 w r + (Y r) (Y
w
0
E fu [wr + w^ (w) (Y r)] (Y
h
73
r)] = 0
i
r) = 0
r)g = 0
0
0
x)r))
di¤erentiate w.r.t. w :
E u00 (w (r + ^ (w) (Y
d^ (w)
(Y r)w
= 0
dw
E fu00 (wr + w^ (w) (Y r)) [r + ^ (w) (Y r)] (Y r)g
d^
=
0
2
wE [u (wr + w^ (w) (Y r)) (Y r) ]
dw
r))) (Y
r) r + ^ (w) (Y
r) +
as before, the denominator is positive so the numerator signs this expression. Let:
' (y) = r + ^ (w) (y
r)
so that:
E fu00 (wr + w^ (w) (Y
r)) [r + ^ (w) (Y
r)] (Y
r)g = E [u00 (w' (Y )) (Y
r)w' (Y )]
to sign this expression we again use the facts: z !
7
(z) z = (z) non-decreasing and
y 7! ' (y) strictly increasing to conclude that z 7! (' (z)) and thus:
[ (w' (y))
(w' (r))] [y
r]
0
(w' (r)) (y
r)
so that:
u00 (w' (y)) w' (y)
(y r)
u0 (w' (y))
E [u00 (w' (y)) w' (y) (y r)]
(w' (r)) E [u0 (w' (y)) (y
|
{z
=0 by FOC above
00
E [u (w' (y)) w' (y) (y
and therefore
d^
dw
0)
d
dw
x(w)
w
r)]
0
r)]
}
0:
Example 84 (CRRA) Suppose
that u (x) = x1 then
i E [u (xYh+ (w x)r)] = E (xY
i + (w
h
1
1
x
x
1
= w E r + w (Y r)
and
which can be written as E w r + w (Y r)
again,
5.3
d
dw
x(w)
w
= 0:
Comparative risk aversion
De…nition 98 Agent 2 is at least as risk averse as agent 1 if:
2 (x) =
u002 (x)
u02 (x)
u001 (x)
=
u01 (x)
1
(x)
Theorem 85 (Pratt; Econometrica, 1964) Suppose that u1 ( ) and u1 ( ) are C2 with
u0i ( ) > 0. Then the following are equivalent:
u002 (x)
u001 (x)
u02 (x)
u01 (x)
9 a C2 concave, increasing ( ) s.t. u2 =
74
(A)
u1
(B)
x)r)1
Proof. (B) ) (A) : Suppose u2 (x) =
0
u02 (x) =
u002 (x) =
0
(u1 (x)) 8x with
(u1 (x)) u01 (x)
2
00
(u1 (x)) [u01 (x)] +
0
( ) > 0 and
00
( ) < 0. Then:
(u1 (x)) u001 (x)
dividing these two equations:
00
u001 (x)
=
u01 (x)
u002 (x)
u02 (x)
(u1 (x)) u01 (x)
0
(u1 (x))
0
snce 0 ( ) > 0, 00 ( ) < 0 and u01 ( ) > 0.
(A) ) (B) : We must show that if (A) is satis…ed, 9 ( ) s.t. u2 (x) = (u1 (x)) : Now,
notice that u0i ( ) > 0 ) u1 ( ) is invertible, i.e, u1 (x) = y ) x = u1 1 (y). Hence we want
to show:
u2 u1 1 (y) =
u1 u1 1 (y)
|
{z
}
=
(y)
y
0
we can claim that
00
> 0 and
0
< 0. To see this note:
(y) = u02 u1 1 (y)
d
u 1 (y)
dy 1
and since:
y = u1 u1 1 (y) ) dy = u01 u1 1 (y) d u1 1 (y)
)
d u1 1 (y)
1
= 0
dy
u1 u1 1 (y)
(and bear in mind that u01 > 0 )
d(u1 1 (y))
dy
0
since u0i ( ) > 0. Next, let ' (z) =
'0 (z) =
> 0) so replacing:
u02 u1 1 (y)
>0
(y) = 0
u1 u1 1 (y)
u02 (z)
u01 (z)
so that:
u002 (z) u01 (z) u001 (z) u02 (z)
u02 (z) u002 (z)
=
u01 (z) u02 (z)
[u01 (z)]2
…nally note that:
0
00
(y) = ' u1 1 (y)
d
(y) = '0 u1 1 (y)
u 1 (y)
|
{z
} dy 1
|
{z
}
0
>0
so that
00
0:
75
u001 (z)
u01 (z)
0
5.3.1
Application: portfolio choice
Proposition 33 Suppose that two agents with identical wealth face the same problem as
(5.4). That is:
max fi (x) = max E [ui (xY + (w
x
x
s:t: 0
x
(5.5)
x)r)]
w
for i = 1; 2 with u0 > 0 and u00 < 0. If 0 < xi < w for each i, then
x1 x2 :
1
Proof. Since fi (x) is concave for each i it su¢ ces to show that f20 (x1 )
note:
f20 (x1 ) = E f 0 [u1 (rw + x1 (Y r))] u01 (rw + x1 (Y r)) (Y
now, since
Thus:
00
0
< 0 and u0 > 0 it follows that y 7!
f
0
[u1 (rw + x1 (y
r))]
0
r))] (y
r)
[u1 (rw + x1 (Y
[u1 (rw)]g (y
(x)
2
(x) )
0. To see this,
r)g
r))] is nonincreasing.
r)
0
[u1 (rw)] (y
r)
so:
0
[u1 (rw + x1 (y
now multiply both sides by u01 (rw + x1 (Y
0
[u1 (rw + x1 (y
r))] (y
0
r)):
r) u01 (rw + x1 (Y
0
r))
[u1 (rw)] (y
r) u01 (rw + x1 (Y
r))
then take expectations:
Ef
0
[u1 (rw + x1 (y
and notice that E f(y
problem. Therefore:
Ef
0
r))] (y
r) u01 (rw + x1 (Y
r) u01 (rw + x1 (Y
[u1 (rw + x1 (y
r))] (y
Ef
0
[u1 (rw)] (y
r) u01 (rw + x1 (Y
r))g = f10 (x1 ) = 0 by the FOC from agent 1’s
r) u01 (rw + x1 (Y
f20
which given that f is concave, implies that x1
5.3.2
r))g
r))g
(x1 )
0
0
x2 :
Application: insurance
Theorem 86 (Jensen’s inequality) If f is convex and Y is a random variable, then
E (f (Y )) f (E (Y )) that is:
Z
Z
f (y) g (y) dy f
g (y) dy
where g is the (Riemann integrable) density of RV Y:
76
r))g
Suppose that agents are endowed with wealth w in period t and in period t + 1 face
the occurence of a damage of amount d with probability p: The expected wealth without
insurance is:
p (w d) + (1 p) w = w pd
On the other hand, if the agent buys (full coverage) insurance at a premium ; his expected
wealth is the same whether the damage occurs or not:
p (w
d
+ d) + (1
p) (w
) = (w
)
Now, if insurance companies o¤er coverage at a "fair premium" then pd = and w pd =
w
so that if i (x) = E (ui (x)) (i.e., is what Mas-Colell, Whiston and Green call
the vNM representation of u) the cannonical insurance problem of agent i is therefore
formulated as:
max i (w
d) (1 p) ui (w)
i ) = pui (w
0
i
1
Theorem 87 (Pratt; Econometrica, 1964) Suppose that u1 ( ) and u2 ( ) are C2 with
u0i ( ) > 0 and u00i ( ) < 0. If i is the maximum insurance premium that agent i is willing
to pay and the loss-inducing RV X has E [X] = 0, then:
u001 (x)
,
u01 (x)
u002 (x)
u02 (x)
u00 (x)
2
(x)
1
(x)
u00 (x)
1
Proof. ()) Since u0 2(x)
, by theorem (85) we know 9 ( ) increasing and concave
u01 (x)
2
s.t. u2 =
u1 . Next, by Jensen’s inequality, ( ) concave ) E [ (Z)]
(E [Z])
therefore:
u2 (w
2
(x)) = E [u2 (w + x)]
= E [ (u1 (w + x))]
[E (u1 (w + x))]
=
[u1 (w
1 (x))]
= u2 (w
1 (x))
(()First, choose a RV Y s.t. E [Y ] = 0, let 0
E [Y ] = 0 ) E [X] = 0. Next, let:
Gi (t; ) = ui (w
)
t
1 and de…ne X = ty; note that
E (ui (tY + w))
and note Gi (0; 0) = 0. Furthermore:
@Gi
@
=
u0i (w
)j
(0;0)
77
=0
=
u0i (w) 6= 0
From the implicit function theorem 9 i : B" (0) ! R s.t. i is C2 and Gi (t;
t 2 B" (0). A sedcond-order Taylor approx. around zero yields:
i (t) =
i (0) +
0
i
(0) t +
t2
2
00
i
i
(t)) = 0 8
(Ci (t))
where Ci (t) 2 (0; t) 8 t 2 B" (0). Next, note that ui (w
0)] =
i (0)) = E [ui (w
E [ui (w)] = ui (w). Moreover, ui (w
(0))
=
E
[u
(w
X)]
=
E
[u
(w
tY
)]
so
that:
i
i
i
u00i (w
i
(t)) [
u0i (w
2
0
0
i (t)] + ui (w
0
i
00
i
(t)) [
i (t)) [
i
(t)] = E [u0i (w tY ) Y ]
(t)] = E u00i (w tY ) Y 2
evaluating each of these expressions at zero:
u0i (w) [
0
i
u0i
0
i
(0)] = u0i (w) E [Y ]
| {z }
=0
(w) [
0
i
(0)] = 0 )
(0) = 0
and:
u0i (w) [
now, replacing
0
i
(0) = 0 and
i
t2
2
00
i
i
but from
2
(t)
1
(t) =
00
i
(0)] = u00i (w) E Y 2
u00i (w)
00
E Y2
(0)
=
i
u0i (w)
(0) = 0 in the Taylor expansion:
(Ci (t)) ) lim
t!0
i
t2
t!0 2
(t) = lim
00
i
(Ci (t))
(t) :
2
(t)
1
2
2
t
2
t2
lim
t!0 2
00
2
00
2
u002 (w)
E
u02 (w)
u00
2 (x)
u02 (x)
00
1
(C1 (t))
t2 00
lim
1 (C1 (t))
t!0 2
00
1 (0)
u001 (w)
E Y2
u01 (w)
(C2 (t))
00
2
and therefore we conclude that
t
2
(C2 (t))
(t)
(0)
Y2
u00
1 (x)
,
u01 (x)
78
i.e., 2 is more risk averse than 1.
5.4
First order stochastic dominance (FOSD)
Suppose we have a probability space ( ; A; P ) and suppose that X (!) Y (!) 8 ! 2 .
Then we’re saying that the RV X is larger than RV Y in every conceibable state of nature.
If X; Y represented some measure of payo¤, then we would say that X % Y that is, RV
X is prefered in some sense to RV Y:
Alternatively, suppose that X FX and Y
FY , then FX (t) FY (t) or 1 FX (t)
1 FY (t) would also imply that F % G in some sense. It turns out that the appropriate
"sense" is the return (or payo¤) sense and we call this type of preference a relation of …rst
order stochastic dominance.
Remark 61 X (!)
Y (!) ) FX (t)
FY (t) but the converse is false.
Example 88 To ilustrate the last remark, let X
FX (t) = FY (t) but X (!) 6= Y (!) :
[0; 1] and Y = 1
X
[0; 1] then
De…nition 99 The RV X dominates RV Y in the First-order stochastic sense denoted
FX %1 FY if FX (t) FY (t) 8 t 2 R:
Remark 62 Note that %1 is a preference relation on D which is transitive and re‡exive
but not complete.
Proposition 34 (FOSD equivalence I) The following are equivalent:
FX %1 FY
9 ( ; A; P ) and X; Y such that:
d
(5.6)
(5.7)
d
X = F , Y = G and
X (!) Y (!) 8 ! 2
Theorem 89 (FOSD equivalence II) The following are also equivalent:
X % 1Y
E [u (X)]
E [u (Y )]
(5.8)
(5.9)
whenever u ( ) is non-decreasing
d
d
Proof. Assume that Y = G and X = F and that Y; X have Riemann integrable densities
denoted f and g: Moreover, assume that 8 x 2
= [a; b] ) f (x) = 0 = g (x) which implies
that F (a) = 0 = G (a) and F (b) = 1 = G (b) :
()) Suppose that X %1 Y so that F (t) G (t) 8 t 2 R. Next, suppose u is nondecreasing
and C1 . Then:
Z b
E [u (X)] =
u (t) f (t) dt
a
79
using integration by parts (recall f = dF ):
Z
(t)]ba
E [u (X)] = [u (t) F
Z
= u (b)
b
u0 (t) F (t) dt
a
b
u0 (t) F (t) dt
a
similarly:
Z
[u (t) G (t)]ba
E [u (Y )] =
Z
= u (b)
b
u0 (t) G (t) dt
a
b
u0 (t) G (t) dt
a
so that:
E [u (X)]
Z
E [u (Y )] =
b
0
u (t) G (t) dt
a
Z
=
Z
b
u0 (t) F (t) dt
a
b
u0 (t) [G (t) F (t)]dt
|
{z
}
a
0 since X%1 Y
) E [u (X)]
E [u (Y )]
(() Now suppose that E [u (X)] E [u (Y )] whenever u is nondecreasing and expectations
exist. There are three cases to consider. Case 1: If t a then F (t) = 0 = G (t). Case 2:
if t b then F (t) = 1 = G (t). Finally, if t 2 (0; 1) then let:
u (t) =
1; if x t
0; if x < t
and note:
E [u (X)] =
Z
b
u (s) f (s) ds
a
=
=
Z
t
|a
Z b
u (s) f (s) ds +
{z
}
Z
b
u (s) f (s) ds
t
=0
f (s) ds = F (b)
t
= 1
F (t)
an identical argument establishes that:
E [u (Y )] = 1
80
G (t)
F (t)
so that:
E [u (X)]
E [u (Y )] ) F (t)
) X %1 Y
Example 90 Suppose that:
8
< 1; if x 1
x; if 12 x < 1
F (x) =
:
0; if x < 21
G (t)
8
< 1; if x 1
x; if 0 x < 1
G (x) =
:
0; if x < 0
and note that F %1 G. To see why, note that when x 1=2 and when x < 0 then F = G.
Now, when x 2 (0; 1=2) then F < G. Thus, F (t) G (t) 8 t 2 R so that F %1 G. Next,
d
d
de…ne Y
U [0; 1] and let X =R max f1=2; Y g. Then Y = G and X = F . Now, notice that
1
dG(t) = 1 so that E (u (Y )) = 0 u (t) dt so:
E [u (X)] = E u max
1
;Y
2
=
Z
1
u max
0
1
;t
2
dt
Z 1
1
1
u max
u max
;t
dt +
;t
dt
=
1
2
2
0
2
Z 1
Z 1
Z 1
Z 1
2
2
1
1
u (t) dt = u
u (t) dt
=
u
dt +
dt +
1
1
2
2
0
0
2
2
Z 1
Z 1
1
1
1 1
2
= u
+
[t]0 +
u (t) dt = u
u (t) dt
1
1
2
2
2
2
2
Z 1
Z 1
Z 1
2
u (t) dt +
u (t) dt =
u (t) dt
Z
1
2
1
2
0
0
= E [u (Y )]
5.4.1
FOSD and precautionary savings
Suppose that a single agent that lives for two periods and is endowed with wealth w in
period 1 must decide how much to invest in a risky asset that yields returns in period 2.
If x is the amount of wealth invested on the risky asset (with return) Y then the problem
faced becomes:
max u (w x) + E [u (xY )]
0 x w
0
00
Suppose that u > 0 and u < 0: Now, if we want to compare the amount invested in two
risky assets with gross returns Y1 and Y2 by the same individual we want to compare the
81
solutions to the problems:
max f1 (x) = u (w
x) + E [u (xY1 )]
max f2 (x) = u (w
x) + E [u (xY2 )]
0 x w
0 x w
Remark 63 Notice that the objective function fi (x) is concave. To se why, note:
fi00 (x) = u00 (w x) + E u00 (xY1 ) Y12
0 since u00 < 0
Now, suppose that we have an interior solution to each of these problems, i.e.:
x1 2 arg max f1 (x)
0 x w
x2 2 arg max f2 (x)
0 x w
0 < xi < w
since the objective function is concave, it su¢ ces to show that fi (xj ) 0 to conclude that
xj
xi . To do so, we need to know how risk averse is the agent in question and what is
the stochastic order relation between the two risky assets Y1 and Y2 .
Claim 91 If the decision-maker’s preferences satisfy RRA < 1 and if Y1 %1 Y2 then
x1 x2 .
Proof. step 1: First, de…ne:
(x; y) = u (xy)
and di¤erentiate with respect to x and then with respect to y :
@ (x; y)
= u0 (xy) y
@x
@ 2 (x; y)
= u0 (xy) + u00 (xy) yx
@x@y
and notice that RRA < 1 implies that for each z:
u00 (z)
z <1
u0 (z)
) u00 (z)z < u0 (z)
) 0 < u0 (z) + u00 (z)z
so that:
@ 2 (x; y)
= u0 (xy) + u00 (xy) yx > 0
@x@y
82
which in turn implies that y 7! u0 (xy) y is an increasing function.
step 2: Note that since xi 2 arg max0 x w fi (x) (and is interior) it follows that:
f10 (x1 ) = 0
=
u0 (w
u0 (w
= f20 (x1 )
thus, since fi is concave f20 (x1 )
x) + E [u0 (x1 Y1 ) Y1 ]
x) + E [u0 (x1 Y2 ) Y2 ]
f20 (x2 ) ) x1
x2 :
Remark 64 This problem ilustrates the use of supermodularity of single crossing property
introduced by de…nition (56).
5.4.2
FOSD and portfolio choice
Suppose that the decision maker faces a problem similar to that in (5.4) but with two risky
assets. We want to study if agents invest more in risky assets that dominate (in returns)
other risky assets. The problem is therefore:
max fi (x) = max E [u (wr + x (Yi
x
x
s:t: 0
x
(5.10)
r))]
w and i = 1; 2
with u0 > 0, u00 < 0 and w; r > 0:
Claim 92 If RRA < 1 and if Y1 %1 Y2 then x1
x2 :
Proof. Proceed as in the last proof and de…ne:
(x; y) = u (wr + x (y
r))
so that:
@ (x; y)
= u0 (wr + x (y
@x
@ 2 (x; y)
= u0 (wr + x (y
@x@y
r)) (y
r)
r)) + u00 (wr + x (y
r)) (y
r) x
now, from the assumption of RRA < 1:
u00 (wr + x (y
u00 (z)
z <1
u0 (z)
u00 (wr + x (y r)) [wr + x (y r)]
<1
u0 (wr + x (y r))
r)) wr < u0 (wr + x (y r)) u00 (wr + x (y
83
r)) (y
r) x
now, u00 < 0 )
u00 (wr + x (y
u00 (wr + x (y
r)) > 0 )
0 <
u00 (wr + x (y r)) wr
< u0 (wr + x (y r)) u00 (wr + x (y
r)) wr > 0. Thus:
r)) (y
r) x
2
(x;y)
so that @ @x@y
0 and we conclude that y 7! u0 (wr + x (y r)) (y
Now, since xi 2 arg max0 x w fi (x) (and is interior) it follows that:
f10 (x1 ) = 0
= E [u0 (wr + x (Y1
E [u0 (wr + x (Y2
= f20 (x1 )
r)) (Y1
r)) (Y2
5.5
r)]
r)]
f20 (x2 ) ) x1
as in the last application, since fi is concave f20 (x1 )
r) is increasing.
x2 :
Likelihood ratio stochastic dominance
De…nition 100 Suppose that X and Y are RV with Riemann integrable (positive) densities
d
d
f and g (i.e., X = F and Y = G), we say that X dominates Y in the likelihood ratio
stochastic order, X %LR Y if:
s<t)
f (s)
g (s)
f (t)
g (t)
Remark 65 Note that the LR order is a monotone property, so we can equivalently say
that X %LR Y if:
[f (s) g (t) f (t) g (s)] (t s) 0
or:
f (s)
(t
g (s)
f (t)
(t
g (t)
s)
s)
Remark 66 LRSD asserts that higher values of "money" are more likely under f
Proposition 35 X %LR Y ) X %1 Y
Proof. First, note that since X %LR Y then:
s < t ) f (s) g (t)
f (t) g (s)
now integrate over t on both sides (recall limit operations preserve inequalities):
Z 1
Z 1
f (s) g (t) dt
f (t) g (s) dt
s
f (s) [1
s
G (s)]
g (s) [1
84
F (s)] 8 s 2 R
now integrate the same expression over s on both sides:
Z 1
Z 1
f (t) g (s) ds
f (s) g (t) ds
t
t
g (t) F (t)
so that:
F (x)
G (x)
and we conclude that F (x)
f (t) G (t)
f (x)
g (x)
1
1
8t2R
F (x)
G (x)
G (x) so that X %1 Y:
Example 93 (portfolio choice) Come back to the problem of portfolio choice with two
d
d
risky assets introduced in section 5.4.2: Suppose that Y1 = F and Y2 = G:Then if:
xi 2 arg max fi (x)
0 x w
0 < xi < w
it follows that:
f10 (x2 ) = E [u0 (wr + x2 (Y1 r)) (Y1 r)]
Z
=
[u0 (wr + x2 (t r)) (t r)] f (t) dt
Z
f (t)
g (t) dt
=
u0 (wr + x2 (t r)) (t r)
g (t)
Z
f (r)
u0 (wr + x2 (t r)) (t r)
g (t) dt
g (r)
Z
f (r)
=
fu0 (wr + x2 (t r)) (t r) g (t)g dt
g (r)
= E [u0 (wr + x2 (Y2 r)) (Y2 r)]
= f20 (x2 ) = 0
therefore, f10 (x2 )
5.6
0 and since fi is concave we conclude that x2
x1 :
Concave and second order stochastic dominance
(SOSD)
The …rst and likelihood ratio orders are preference relations over distributions according to
expected returns. On the other hand, we can de…ne preference relations over distributions
according to risk and to risk and return. The concave and second orders do precisely that.
85
Theorem 94 Suppose that E (X) and E (Y ) are …nite.
The following are equivalent:
X %co Y
Z t
Z t
F (s) ds
G (s) ds
E (X) = E (Y ) and
1
(5.11)
(5.12)
1
and:
The following are equivalent:
X %2 Y
Z t
Z t
F (s) ds
G (s) ds
1
d
(5.13)
(5.14)
1
d
Proof. Assume that Y = G and X = F; and that Y; X have Riemann integrable densities
denoted f and g: Moreover, assume that 8 x 2
= [a; b] ) f (x) = 0 = g (x) which implies
that F (a) = 0 = G (a) and F (b) = 1 = G (b) :
(5:11 ) 5:12 and 5:13 ) 5:14) Case
t. Then f (t) = 0 = g (t) and
R t 1: suppose that
Rt a
F (t) = 0 = G (t). Thus, trivially 1 F (s) ds
G (s) ds.
1
Case 2: Suppose that b t. Then:
Z t
Z t
F (s) ds
F (s) ds =
1
=
Z
a
b
F (s) ds +
Z
t
F (s) ds
b
a
Rt
now, since t b ) F (t) = 1 then b F (s) ds = t b. Next integrating by parts the …rst
expression on the RHS:
Z b
Z b
b
sF (s)
F (s) ds = [sF (s)]a
a
a
= b
E (X)
so that:
Z
t
F (s) ds = b
1
= t
a similar argument establishes that:
Z t
G (s) ds = t
1
86
E (X) + t
E (X)
E (Y )
b
Therefore:
Z
Z
t
G (s) ds
1
t
F (s) ds = E (X)
E (Y )
1
Rt
Rt
thus, if X %co Y then E (X) = E (Y ) and 1 G (s) ds = 1 F (s) ds: If, on the other
Rt
Rt
hand, X %2 Y then E (X) E (Y ) and thus 1 G (s) ds
F (s) ds:
1
Case 3: Suppose that a < t < b. Then consider:
t; if x t
x; if x < t
u (t) =
then:
E [u (X)] =
=
Z
Z
1
u (x) f (x) dx =
1
t
Z
=
=
=
b
u (x) f (x) dx
a
b
u (x) f (x) dx +
u (x) f (x) dx
t
Z b
Z t
tf (x) dx
xf (x) dx +
t
a
Z t
t
F (x) dx + t (F (b)
[x F (x)]0
a
Z t
tF (x)
F (x) dx + t tF (t)
a
Z t
t
F (x) dx
a
=
Z
F (t))
a
and a similar argument establishes that:
E [u (Y )] = t
Z
t
G (x) dx
a
now if X %co Y then u concave implies that E [u (X)] E [u (Y )] which in turn implies:
Z t
Z t
F (x) dx
G (x) dx
a
a
and if X %2 Y then u nondecreasing, concave implies again that E [u (X)]
(5:12 ) 5:11 and 5:14 ) 5:13) Assume that u is C2 and note that:
Z 1
Z b
E [u (X)] =
u (t) f (t) dt =
u (t) f (t) dt
1
= [u (t) F (t)]ba
(
= u (b)
= u (b)
Z
a
b
a
0
u (t)
u0 (b) []
E [u (Y )].
u0 (t) F (t) dt
Z
b
t
F (s) ds
a
a
87
Z
a
b
00
u (t)
Z
a
t
)
F (s) ds dt
De…nition 101 G is a mean-preserving spread of F , y = x+", with E ["jX] = 0; X
and Y
G:
Lemma 16 If X
spread of F:
5.6.1
F and Y
G and
=
G
F
then X %2 Y , G is a mean-preserving
F
Concave order SD and pro…t maximization
Suppose that a multiproduct …rm uses a single input (labor) to produce n di¤erent commodities (y1 ; :::yn ). Suppose that xi is the amlount of labor used in the production of
commodity i and for simplicity suppose that the wage (price of the only input) is w = 1:
Furthermore suppose that production technology for every good is identical, that is:
(xi )1
yi
with 0 <
i
i
8 i = 1; :::; n
< 1; then the …rm’s pro…t maximization problem is:
max
xi
n
X
1
n
X
i
pi (xi )
i=1
with associated optimal production plan:
xi = ((1
and pro…t function
xi
i=1
i ) pi )
1
i
(p):
(p1 ; :::; pn ) =
n
X
1
pi [pi (1
i )]
i=1
i
i
n
X
((1
i ) pi )
1
i
i=1
Now suppose that the p vector varies with the plant locations. In particular suppose that in
location A the price vector would be the realization of the random vector pA = (X1 ; :::; Xn )
and in location B the price vector would be the realization of the random vector pB =
(Y1 ; :::; Yn ). Suppose that each Xi and each Yi has …nite expectation and suppose that Xi
%co Yi .
Claim 95 The …rm manager would want to build his plant in location B
Proof. First note that 1= i ) pi 7! (p1 ; :::; pn ) is convex for each i (as any pro…t
function). Hence, pi 7!
(p1 ; :::; pn ) is concave for each i. Therefore, Xi %co Yi and
( ) concave imply:
E(
(X1 ; :::; Xn ))
E ( (X1 ; :::; Xn ))
E
pA
E(
(Y1 ; :::; Yn ))
E ( (Y1 ; :::; Yn ))
E
pB
and therefore, the …rm’s manager would want to build his plant in location B:
88
5.7
Summary
Remark 67 The implications of SD orders are as follows:
X %LR Y ) X %1 Y ) X %2 Y
and
X %C Y ) X %2 Y
Summary 96 Let X; Y have respectively distribution functions F; G then:
"better than"
in terms of returns
in terms of risk
in terms of risk and returns
Stoch. Order
X %1 Y
X %C Y
X %2 Y
function
u ( ) non-decreasing
u ( ) concave
u ( ) non-decreasing, concave
in terms of RV
E [u (X)]
in terms of DF
F (t)
E [u (Y )]
G (t)
E [u (X)]
Rt
F (s) ds
1
89
E [u (Y )]
Rt
G (s) ds
1
E [u (X)]
Rt
F (s) ds
1
E [u (Y )]
Rt
1
G (s) ds
Appendix A
Review of functions, di¤erentiation
and integration
1. After t periods of continuously compounding growth rate r per period, quantity A is
Aert :
k
1
e
lim (1 + )n and in general; ek = lim (1 + )n
n!1
n!1
n
n
p
2. eix = cos(x) + i sin(x), where i =
1
3. Every polynomial is a continuous function. Since its derivative is also a polynomial
of one-less degree, it is also continuous. Hence, every polynomial is C 1 .
4. Recall that ln x = y , ey = x; eln x = x and ln ex = x:
5. Recall that
d g(x)
e
dx
= g 0 (x) eg(x)
6. Recall that f : Rn ! Rm is di¤erentiable at x if 9 a best linear approximation at
x: That is, if 9 g : X ! Y such that is: f (x + x) f (x) + g(x); the relative error
of such approximation tends to zero as one approaches x from any direction, and in
such case, g( ) is the derivative. That is:
lim
f (x + x)
x!0
f (x)
kxk
g(x)
=0
(A.1)
7. If dx is a (in…nitesimally) small vector in Rn , we call df
f (x + dx) f (x)
Df (x)(dx) the total di¤erential of f , with Df (x) being the linear (or a¢ ne) function
and dx a small vector.
8. Two functions f; g are tangent if limx!0
f (x+x) h(x+x)
kxk
=0
9. The partial derivative of f is de…ned: limt!0 h(x+t)t h(x) while the directional derivative
~ x h(x) = limt!0 h(x+tx) h(x) : In fact, the partial derivative w.r.t. xi is simply
is D
t
90
the directional derivative in the direction paralell to the xi axis, that is:
~ e1 h(x) where e1 = (0; 0; ::i; :::; 0)
rF (x) e1 = D
@F (x)
xi
=
10. Rolle’s Theorem : If f : [a; b] ! R1 is C 1 and if f (a) = f (b) = 0 then 9 c 2 (a; b)
s.t. f 0 (c) = 0
11. Mean Value Theorem: let f : U ! R1 with U connected interval (set), then if
a; b 2 U , then 9 c 2 (a; b) s.t. f (b) f (a) = f 0 (c)(b a) or:
f 0 (c) =
f (b)
(b
f (a)
a)
(A.2)
12. Inverse Function Theorem in R1 . Let f be C 1 on I 2 R1 . If f 0 (x) 6= 0 8 x in I,
then: (i) f is invertible, (ii) Its inverse, g is C 1 on f (I) and (iii) 8z in the domain
of g, g 0 (z) = 1=f 0 (g(z)).
13. Implicit Function Theorem in R2 : Given the implicit function G(x; y) = c then:
@G(x0 ; y0 )=@x
@G(x0 ; y0 )=@y
y 0 (x0 ) =
(A.3)
that is, the slope of the function y = f (x) at point (x0; y0 ) is the ratio of the derivatives
(marginal products) evaluated at the poitn (x0; y0 ):
14. A function f is homogeneous of degree k if for any k 2 R; f (tx1 ; :::; txn ) =
tk f (x1 ; :::; xn ). If k = 1; k > 1 and k < 1, f has constant, increasing and decreasing
returns to scale, respectively.
15. The domain of a homogeneous function must be a cone (i.e. if x 2 B; tx 2 B; 8t)
16. If f (x) is homogeneous of degree k, g(x) = rf (x) is homogeneous of degree (k
1).
17. If f (x) is homogeneous of degree k, then 8x : x rf (x) = kf (x) (Euler’s Theorem).
18. If f is C n on an open interval I containing , given any x 2 I (Taylor’s theorem) :
1
f (x) = f ( )+ (x
1!
1
)f 0 ( )+ (x
2!
)2 f 00 ( )+:::+
En =
1
(x
n!
1
(n
1)!
(x
)n 1 f (n
1)
( )+En ; with
)n f n ( )
19. Let f be continuous on the compact (i.e. bounded and closed) I = [a; b], then if P is
some partition of [a; b] and S(P ) the sum of areas for partition P , we de…ne:
Z b
f (x)dx = sup S(P )
P
a
91
20. Let F be the anti-derivative of f , then for some interval [a; b]:
Z b
N
X
f (xi )
f (x)dx = lim
F (b) F (a) =
!0
a
i=1
for = (b a)=N . The last term (Riemann sum) closes the Fundamental Theorem
of Calculus.
21. A (di¤erentiable) function f : Rn ! R1 has gradient; i:e: : a vector of n …rst-order
partial derivatives.
22. A (di¤erentiable) function f : Rn ! Rm has Jacobian; i:e: : a matrix of m vectors
each with n entries of …rst-order partial derivatives.
23. Some useful integrals:
Z
Z
Z
1
xn 1
n
+C
dx = ln x + C
ex dx = ex + C
x dx =
n+1
x
Z
Z
ax3 bx2
1
2
ax + bx + c =
+
+ cx
(f (x))n f 0 (x)dx =
(f (x))n+1 + C
3
2
n+1
Z
Z b
b
0
u0 (x)v(x)dx (integration by parts)
u(x)v (x)dx = [u(x)v(x)]a
a
92
Appendix B
Review of vectors and matrix algebra
1. Rank: If A is m n, then rkA = rkA0
If rkA = m = n, A0 = I
minfm; ng, where A0 is RREF of A. OJO:
2. If rkA = m , 8 b 2 Rm ; 9 x 2 Rn s:t: Ax = b (at least one solution).
3. rkA = n , 8 b 2 Rm , if Ax = Ay = b, then x = y (at most one solution).
4. Subspace: S
; ;2 R
Rm is a subspace of Rm if
x + y 2 S whenever x; y 2 S and
5. The (subspace) range of A is the set R(A) = fb 2 Rm : 9 x 2 Rn s:t: Ax = bg
6. The (subspace) null of A is the set N (A) = fx 2 Rn : Ax = 0g
7. N (A) = N (A0 ); R(A) 6= R(A0 ); R(BA)
8. For any Am
n;
R(A) = Rm , rkA = m.
9. For any Am
n;
N (A) = 0 , rkA = n.
R(B); N (A)
N (BA).
10. Span: Let S (subspace of Rm ); a collection of vectors fa1 ; :::; an g in S spans S if
each x 2 S is a linear combination of the members of fa1 ; :::; an g.
11. For any matrix A, the columns of A span R(A) and the rows of A span R(AT ).
12. Linear Independence: A set fa1 ; :::; an g in Rm is linearly independent if 8
R:
n
P
= n = 0.
i ai = 0 only for 1 =
1 ; ::: n
i=1
13. Basis: If S is subspace of Rm , a set fa1 ; :::; an g
and (ii) it is linearly independent.
S is a basis for S if (i) it spans S
. .
14. Furthermore, if A = [a1 .. : : : ..an ]; R(A) = S and rkA = n (i.e., A is Full Column
Rank).
93
2
15. Dimension: (dimS) is number of elements in a basis for S.
16. Let S (subspace of Rm ) have basis fa1 ; :::; an g. If fb1 ; :::; br g is linearly indep., then
r n.
Corollary: Any m + 1 vectors in Rm are linearly dependent.
Corollary: If fa1 ; :::; an g and fb1 ; :::; br g are basis for S, then n = r.
17. Fundamental Theorem of Linear Algebra (I): For any Am
n:
i) dim R(AT ) = rkA
ii) rkA = rkAT
iii) dim N (A) = n rkA.
18. The orthogonal complement of S, S ? = fy 2 Rm : y x = 0 8x 2 Sg (Note that:
f0n 1 g? = Rm and vice-versa. Also, S \ S ? = f0g ).
19. Fundamental Theorem of Linear Algebra (II): For any Am
n:
i) R(A)? = N (A)
ii) R(A) = N (AT )? .
Corollary: (S ? )? = S.
20. A matrix is called Symetric if M = M T . A matrix is called Idempotent if M M =
M.
21. An
n
is invertible if 9 Cn
n
s:t: AC = CA = I. If A and B are invertible, so is AB.
22. If A is invertible, then so is AT and (AT )
1
= (A 1 )T .
23. For the Projection problem we need:
(i) f (x) : Rn ! R. If f (x) = cT x for c 2 Rn ; rf = c. If f (x) = xT Ax; rf =
(A + AT )x
(ii) If A is m n and rkA = n, then AT A is nonsingular (i.e., is n n) and has rank
n.
Given x 2 S = R(A) and y 2 Rm , the projection problem is:
p
pPn
ui )2 = kyuk = d(y; u) = p(y u)T (y u) =, which is equivaminu2S
i=1 (yi
lent to minx2Rn (y Ax) (y Ax) (recall min x2 , min x2 ).
The f:o:c: being 2AT y+2AT Ax = 0; x = (AT A) 1 AT y; or u = Ax = A(AT A) 1 AT y =
M y, where M = A(AT A) 1 AT is called the projection matrix for R(M ) (M being
symetric and idempotent).
P
24. For An n ; n 2; detA = nj=1 ( 1)1+j (a1j )(detA1j )
94
25. For the system Ax = x, if A is nonsingular:
(a) A
1
= (1=detA)adjA and,
(b) The unique solution to the system is xi = detBi =detA where Bi is the matrix A
with b replacing the ith column of A. (Cramer’s rule)
26. A is invertible , A is nonsingular , detA 6= 0 , rkA = n , 8b 9 a unique x s.t.
Ax = b which is x = A 1 b.
Q
27. A diagonal matrix D with fd1 ; d2 ; :::; dn g as diagonal has detD = ni=1 di .
28. For An n , r is eigenvalue of A if det(A rI) = 0, i.e., A rI is singular. Note that
An n has n eigenvalues which are solutions to the nth degree polynomial equation in
r, det(A rI) = 0.
29. For An
n
with eigenvalue r, 0n
1
6= v 2 Rm is an eigenvector if (A
rI)v = 0.
30. If the eigenvalues of A are distinct, A is diagonalizable.
31. A n n matrix P is said to be orthogonal if P T = P (i.e., the of any two di¤erent
column vectors is 0 or orthogonal and the norm of each column vectos is 1). Note
that detP = 1.
32. Let S (subspace of Rm ) have basis fa1 ; :::; an g. This basis is orthonormal if ai aj = 0
whenever i 6= j and ai aj = 1 if i = j
Review of Vector and Matrix Operations
P
The general quadratic form Q(x1 ; :::; xn ) =
i j aij xi xj (aij xi xj is a 2nd degree
T
monomial) can be expresed as x Ax where A is unique symetric matrix whose ij th
element is 12 aij
Matrix A is positive semide…nite if xT Ax 0 whenever x 6= 0 and negative semidefinite if xT Ax 0 whenever x 6= 0 (if inequalities are strict, remove "semi").
The 2x2 matrix of ones (i.e.aij = 2) is positive semide…nite.
The 2x2 matrix A is positive de…nite if a11 > 0 and det A > 0 and negative de…nite
if a11 < 0 and det A > 0:
A diagonal matrix A will be positive semide…nite if ai
0 8 i (i.e. all diagonal
elements 0) and negative semide…nite if ai
0 8 i (again, for strict inequalities,
remove "semi")
The eigenvalues (only for square matrices) of A (n n matrix) are those r which
solve det(A rIn ) = 0. Note A has at most n distinct eigenvalues.
95
To …nd eigenvectors for A: for each r solve the system (A rI)x = 0, where x = [x1
x2 :::xn ]T : Note eigenvectors are not unique.
If P is a matrix whose columns are eigenvectors of A, then P
0
Let A = @
0
B
AA = B
@
a11 a12
a21 a22
a211 + a12 a21
While for the (n
AP = rIn
1
A ; then
a12 (a11 + a22 )
a21 (a11 + a22 )
1
a222
+ a21 a12
1
C
C;
A
0
B
AT A = B
@
a211 + a212
a21 a11 + a22 a12
a12 a11 + a22 a12
a222
+
a221
1
C
C
A
1) vectors x and y:
x y = yT x = xT y =
x x = xT x = xT x =
Pn
i=1
Pn
i=1
xi yi
x2i
Recall that if A is a symmetric (comformable) matrix (x y)T A(x y) = (xT Ax 2xT Ay + yT Ay)
pPn
2
2
kxk = (x x)1=2 =
i=1 xi and therefore x x = kxk
x
kxk
= 1 (read x is normalized)
Cauchy-Schwarz Inequality for vectors jx yj
x x
kxk kyk
0 whenever x 6= 0 (i.e., x x = 0 , x = 0)
x y=y x
Note that (x y) (x y) =(x y)2
Note also that (x
y) (x
y) =
P
(xi yi )2
x is the projection of y under x if x x = x y , where
x y = 0 , x and y are orthogonal (perpendicular)
x y = (x y) = x y
x (y + z) = x y + x z and (x + y) z = x z + y z
96
= (x y)= kxk2