Embed Size (px)
Citation preview
Math Methods for Economicsand Microeconomic Theory
Cesar E. TamayoDepartment of Economics, Rutgers University
Class notes: fall 2010
Contents
I Math Methods 4
1 Topology and analysis 51.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Series, sequences and subsequences . . . . . . . . . . . . . . . . . . . . . . 61.3 Open and closed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Completeness, boundedness and compactness . . . . . . . . . . . . . . . . . 91.5 Continuity of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Continuity of correspondences (set-valued mappings) . . . . . . . . . . . . 141.7 Maximum theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.8 Fixed point theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2 Convex optimization 212.1 Convex sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Separating hyperplanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Convexity of functions and subgradients . . . . . . . . . . . . . . . . . . . 242.4 Support function theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.5 Kuhn-Tucker theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.5.1 Introduction to nonlinear programming . . . . . . . . . . . . . . . . 322.5.2 The Kuhn-Tucker conditions . . . . . . . . . . . . . . . . . . . . . . 322.5.3 Constraint quali�cations . . . . . . . . . . . . . . . . . . . . . . . . 352.5.4 Non-negativity and equality constraints . . . . . . . . . . . . . . . . 362.5.5 Su¢ ciency conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 38
II Microecomic Theory 40
3 Producer and consumer theory 413.1 Producer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.1.1 Production sets and technologies . . . . . . . . . . . . . . . . . . . 413.1.2 Cost minimization . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1.3 Pro�t maximization . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.2 Consumer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.2.1 Utility maximization . . . . . . . . . . . . . . . . . . . . . . . . . . 47
1
3.2.2 Expenditure minimization . . . . . . . . . . . . . . . . . . . . . . . 49
4 Game Theory and General Equilibrium 514.1 Game theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.1.1 Zero-sum games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.1.2 Non-zero-sum games and Nash Equilibrium . . . . . . . . . . . . . 514.1.3 The generalized game . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.2 General Equilibrium theory . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2.1 Prelminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2.2 Existence of equilibrium in pure exchange economies . . . . . . . . 564.2.3 Existence of equilibrium in production economies . . . . . . . . . . 584.2.4 Welfare theorems (pure exchange economies) . . . . . . . . . . . . . 614.2.5 Relaxing the Walrasian assumptions . . . . . . . . . . . . . . . . . 66
5 Decision making under uncertainty 705.1 Expected utility hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . 705.2 Risk aversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
5.2.1 Application: portfolio choice . . . . . . . . . . . . . . . . . . . . . . 725.3 Comparative risk aversion . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.3.1 Application: portfolio choice . . . . . . . . . . . . . . . . . . . . . . 765.3.2 Application: insurance . . . . . . . . . . . . . . . . . . . . . . . . . 76
5.4 First order stochastic dominance (FOSD) . . . . . . . . . . . . . . . . . . . 795.4.1 FOSD and precautionary savings . . . . . . . . . . . . . . . . . . . 815.4.2 FOSD and portfolio choice . . . . . . . . . . . . . . . . . . . . . . . 83
5.5 Likelihood ratio stochastic dominance . . . . . . . . . . . . . . . . . . . . . 845.6 Concave and second order stochastic dominance (SOSD) . . . . . . . . . . 85
5.6.1 Concave order SD and pro�t maximization . . . . . . . . . . . . . . 885.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
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 thePhD program at Rutgers University. The range of mathematical tools presented belowis wide, including topology, real analysis, convex optimization, �xed point theory andstochastic dominance, but attention is limited to their use in economic theory. The notesare almost entirely based on lectures by Professor Richard P. McLean, but all errorsand omissions are my sole responsability. Ocasionally, I also refer to examples andde�nitions found in Carter�s (2001) "Foundations of Mathematical Economics", Mas-Colellet. 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� X! R satisfyingthe following 8 x; y; z 2 X :
1:d (x; y) � 02:d (x; y) = d (y; x)3:d (x; y) = 0, x = y4:d (x; y) � d (x; z) + d (z; y)
For the case when X = Rn we have:
d1 (x;y) =nXi=1
jxi � yij
d2 (x;y) =
vuut nXi=1
(xi � yi)2
...
d1 (x;y) = maxi=1:::n
fjxi � yijg
De�nition 2 A metric space is a pair (X; d) where X is a non-? set and d is a metric onX:
De�nition 3 Let (X; d) and (Y; �) be metric spaces. The Box metric on X� Y is �((x; y); (�x; �y)) =maxfd(�x; x); �(�y; y)gDe�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¤ distanceis: Hd(A;B) = maxfmaxx2A d(x;B);maxx2B d(x;A)g: Note that each max function is wellde�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 fxng 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. Sincelimn!1 fxng = �x1 we know that 9 n1 such that n > n1 ) xn 2 B "
2(�x1). Likewise,
since limn!1 fxng = �x2 we know that 9 n2 such that n > n2 ) xn 2 B "2(�x2). So
choose m = max fn1; n2g 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 convergentwith limit zero, i.e., limn!1 xn = 0. To see why, choose " > 0 and n such that n > 1=".Now, n > n implies:
d (0; xn) = j0� xnj =1
n<1
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 con-vergent. To see this note that any element other than �1 or 1 cannot be a candidate fora limit since we can always �nd an open ball around it which contains no elements of thesequence. Furthermore, �1 or 1 cannot be limits of the sequence either since @ n such that8 " > 0; n > n) d (xn; 1) < " or d (xn;�1) < ":
De�nition 8 fxng1n=1 is a Cauchy Sequence if for any " > 0; 9 N s:t: 8 i; j � N; d(xi; xj) <".
De�nition 9 A subsequence of fxng1n=1 is a subset fxnmg1m=1 such that n1 < n2 < n3:::
Remark 1 Every sequence is a subsequence of itself.
6
Remark 2 fxng1n=1 is convergent, 9 �x such that every subsequence of fxng is convergentwith limit �x:
We note that if limfxng = x and limfyng = y, then limfxn+yng = x+y. Likewise formultiplication. It is also true that a sequence is convergent iif every one of its subsequenceis convergent with the same limitIn 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 thatthe sequence xn = (�1)n is not convergent.
Lemma 2 Let fxkg be a sequence in (X; d) and �x 2 X. If xk 2 B 1k(�x) 8 k then fxkg is
convergent with limit �x
Proof. Suppose that xk 2 B 1k(�x) 8 k:.Pick " > 0 and choose K such that K > 1
". Then
k > K ) xk 2 B 1k(�x). Also, k > K ) 1
K> 1
kso that:
B 1k(�x) � B 1
K(�x) � B"(�x)
) fxkg 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 inR1 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 isintS.
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 an-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 ballcentered 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) < "
=
�y 2 R2 j
q(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 whychoose any y 2 B"(�x) and note:
y 2 B"(�x)) j�x� 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 irrationalnumber.
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 anypoint 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 aclosure point of S , B"(�x) � Sc 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 if9 fxkg s:t: xk 2 S 8 k and fxkg ! �x.
De�nition 19 (sequential characterization of closed set) S in Rm is a closed set if when-ever fxng is a convergent sequence completely contained in S, it follows that limfxng 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= ? andB"(�x) \ Sc 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 withlimit 2 X.
9
Example 12 Let X = [0; 1) and d =Euclidean. Claim: "(X; d) is not complete". To seewhy consider the sequence xk = k
k+1and note that fxkg is a Cauchy sequence. This can
be shown by choosing " > 0 and noting that 9 N such that k > N ) d (xk; 1) < "=2 andm > 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:::xmg � X s:t: X �B"(x1) [ ::: [B"(xm)
De�nition 24 A set S is bounded if 9 B s:t: 8 x 2 S; kxk � B.
Lemma 4 If set S is bounded above, it has a smaller upper bound, or supremum: orK = 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 � X; then S is compact in (X; d)() (S; d) isa compact m.s.
De�nition 27 A m.s. (S; d) is compact () every sequence fxkg contains a subsequencefxkmg1m=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 thefolowing 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 opencover. Conversely, S is not compact if 9 an open cover (or collection), C from which wecannot 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 unionof sets.
Example 13 Let X = [0; 1), d =Euclidean metric and S = [0; 1). Claim: S is not com-pact. To see why we must exhibit an open cover from where no �nite subcovers containingS can be extracted. Consider:
C =��0;
n
n+ 1
�j n � 1
�10
and note that each�0; n
n+1
�is open and:
1[n=1
�0;
n
n+ 1
�= [0; 1)
i.e., every point in [0; 1) belongs to at least one of the sets�0; n
n+1
�. Thus, C is an open
cover for S. However, we cannot extract a �nite subcollection of C whose union containsS:
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 � Rn iscompact , S is closed and bounded.
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 convergentsequence is a Cauchy sequence, every convergent sequences has its limit in (S; d), thus, Sis 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. IfS 6= ? then by the remark above, we know that X = Rn ) (X; d) is complete and S � Xplus (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 andbounded in (X; d) :
Proposition 2 A metric space (S; d) is compact () every sequence fxkg contains asubsequence fxkmg1m=1 whose limit 2 (S; d) (i.e.fxkmg1m=1 is convergent in (S; d))
Proposition 3 Let qk = pkkpkk : Then kqkk = 1 and the set fqkj kqkk = 1g is compact, i.e.
9 qkm and �q s.t. k�qk = 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 � a metric on Y then a function isdenoted f : (X; d)! (Y; �) :
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 everyx 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 fxng such that xn ! �x:We must show that f (xn) is convergent in (Y; �) with limit f (�x) : Equivalently, we mustshow 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 continous9 some " > 0 such that 8 � > 0 we can �nd x 2 B� (�x) with the property that f (xn) =2B" (f (�x)). Therefore, 9 " > 0 such that 8 n we can �nd xn 2 B1=n (�x) but f (xn) =2B" (f (�x)). Thus from lemma 34, we conclude that xn ! �x but f (xn) =2 B" (f (�x)). Thusf (xn) is not convergent with limit f (�x) :
De�nition 31 f is uniformly continuous if for every "; 9 a � that "works" for any arbi-trary 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 compositionof continuous functions is continuous.
Proposition 5 Every uniformly continuous function is continuous.
Proof. Suppose that f : (X; d) ! (Y; �) is uniformly continuous. Choose " > 0: Then9 some � > 0 such that � (f (x) ; f (y)) < " whenever x 2 X, y 2 Y and d (x; y) < �: Inparticular, � (f (�x) ; f (y)) < " whenever d (�x; y) < � so that f is continuous at �x. Since f iscontinuous 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 continuousbut not uniformly continuous.
12
Proof. To see that f is continuous, choose " > 0 and �x > 0. Next, choose:
0 < � < min
��x2"
2;�x
2
�so that:
j�x� yj < � ) j�x� yj < �x
2) �x� y <
�x
2) �x
2< y
consequently:
j�x� yj < � )����1�x � 1y
���� = j�x� yj�xy
� �
�xy<2�
�x2< "
However, f is not uniformly continuous. To see why, choose " = 12: For each integer m > 0
choose xm = 1mand ym = 1
2m: Then:
jxm � ymj =1
2m
but: ���� 1m � 1
2m
���� = m
If � > 0, then choose m so that 12m
< �. Then jxm � ymj < � but�� 1m� 1
2m
�� = m > 12:
De�nition 32 f : (X; d) ! (Y; �) is uniformly continuous if, whenever xk is a Cauchysequence 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 sequencexk = 1� 1
kis a Cauchy sequence in X but f (xk) is not a Cauchy sequence in Y:
Example 19 If xk ! �x then jxkj ! j�xj 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) � h(x), thenf(x)! l as x! �x.
De�nition 33 The function f : Rn ! Rm de�ned as:
f (x) :
0BBB@f1 (x)f2 (x)...
fm (x)
1CCCAis continuous if and only if fi : Rn ! R1 is continuous for each i = 1; :::m
13
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 setfx 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 setfx 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 � 0gwhich 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. Onthe other hand, if e.g., 0 < � < 1 then the set fx 2 X j f (x) � �g = (�1; 0) which is nota 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 fykg such that yk 2 f (S) 8 k: This in turn implies that foreach k, 9 xk 2 S such that yk = f (xk). Since fxkg is a sequence in S and S is compact,9 a subsequence fxkmg and �x such that xkm ! �x. Moreover f continuous implies thatf (xkm) ! f (�x). But f (xkm) = ykm so that ykm ! f (�x) :Thus, fykmg is a subsequence offykg convergent with limit in f (S). Hence, f (S) is compact.
1.6 Continuity of correspondences (set-valued map-pings)
De�nition 36 Suppose that X 6= ?. A relation in X is a "rule" that associates with eachx 2 X a non-? subset of X, denoted � (x) :
De�nition 37 (preference relation) Let � : X� X be a relation.Then: i) � is re�exiveif x 2 � (x) ; ii) � is transitive if y 2 � (x) and z 2 � (y)) z 2 � (x) and iii)� is completeif x 6= y ) x 2 � (y) and/or y 2 � (x) : We say that � is a preference relation if it sats�esi)� 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)� (Y; �); is a rule thatassigns to every element x 2 X, a non-? subset �(x) of Y.
Proposition 7 Let (X; d) is a m.s. and if: i) � : X� X is re�exive, transitive andcomplete, 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 openset 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 � (Y; �)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 j y 2 � (x)g
De�nition 43 �: X� Y is closed graph (w.r.t. box metric) at �x if i) xk ! �x ii) yk ! �yand 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 � (x) = K 8 k is trivially continuous (i.e. UHCand LHC).
Lemma 7 If � is closed graph ) � is closed-valued
Example 21 (2.88, Carter) The correspondence � : R+� R de�ned by:
� (x) =
� �1x
if x > 0
f0g if x = 0
has closed graph but is not UHC. To see why, note that Gr(�) = f(x; 1=x) 2 R2jx > 0g [(0; 0) which is a closed set in R2; however, for every sequence fxkg such that �x ! 0; thesequence yk 2 � (xk) does not converge.
Example 22 The constant correspondence � (x) = (0; 1) is UHC but not closed-valuedand 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 �2:(X; d) � (Y; �) is UHC andcompact-valued, then �1 \ �2 is compact-valued.
15
Proposition 9 The product of UHC correspondences is UHC. Thus, the product of UHCand compact correspondences is UHC and compact.
Proposition 10 If � : (X; d) � (Y; �) is closed graph, then � is UHC whenever Y is acompact set.
Example 23 (2.89, Carter) Let P denote the domain of the budget correspondence, thatis, the set of all prices and incomes pairs for which some consumption is feasible:
P =
((p;m) 2 Rn�Rjmin
x2X
nXi=1
pixi � m
)
where X is the consumption set, the graph of the budget correspondence X (p;m) is givenby :
Gr (X (p;m)) =
((p;m;x) 2 P � X j
nXi=1
pixi � m
)which is a closed set in P �X. To see why, let fxkg be a sequence of consumption bundlesX (p;m). Since X (p;m) is bounded, xk ! �x for some �x 2 X: Thus,
p1x1k + :::+ pnxnk � m
so that:p1�x1 + :::+ pn�xn � m
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 correspondenceX (p;m) is UHC.
Theorem 24 Suppose that �:(X; d)� (Y; �) is UHC and compact valued. If K is compactin (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 evenif xk is not convergent, it must contain a subsequence fxkmg with xkm ! �x and �x 2 K:Summarizing: 1) xkm ! �x, 2) yk 2 � (xk) so ykm 2 � (xkm). Now, � UHC and compactvalued implies that 9
�ykmt
1t=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 isconvergent and whose limit belongs to � (K) :
1.7 Maximum theorems
First we introduce the main theorem that relates topological assumptions of a maximizationproblem with the existence of optima.
16
Theorem 25 (Weierstrass) Suppose that (X; d) is a metric space and f : X! R is acontinuous function. If S � X is non-? and compact, then f attains a maximum and aminimum in S:
Proof. First, note that f (S) is non-? and compact in R. Hence, by the Heine-Boreltheorem f (S) is bounded and in particular, bounded from above. Next, we notice thatany set in R has a least upper bound or sup. Let sup f (S) = �. Now, let Ak = f (S) \�y 2 R j y � � � 1
k
. Note Ak 6= ? 8 k since there must be something between � � 1
kand
� which belongs to f (S) or else � would not be the least upper bound. So, choose fykgsuch that yk 2 f (S) \
�y 2 R j y � � � 1
k
which implies that � � yk � � � 1
k. Thus,
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 fattains a minimum in S:
Next, we study the properties of the maximizer correspondence and the optimal valuefunction of maximization prtoblems. Consider the general constrained maximization prob-lem:
max f(x; y)
s.t. y 2 �(x)
Theorem 26 (The continuous maximum) If f : X� Y! R is continuous (w.r.t. boxmetric) and �: X � Y is non-?, UHC, LHC, compact valued, then: 1) � : X� Ywhere �(x) = argmaxy2�(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 continuousfunction 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 quasicon-cave and � : Y� X is convex valued, then: 1) � : X� Y where �(y) = argmaxx2�(y) f(x; y)is convex-valued and 2) if f is strictly concave, then V : X! R where V (y) = maxff(x; y)jx 2 �(y)g is concave in x.
Proof. To see that � is convex-valued, choose x1; x2 2 �(y). Naturally, these two maxi-mizers yield the same maximum value. So let � = f(x1; y) = f(x2; y). Choose t 2 [0; 1] andnote 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 concludethat 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 allowsfor �exible domain sets, but requires heavy structure on the objectve function. The secondclass of FPT requires a highly structured domain sets but requires only mild assumptionsabout the objective function.
De�nition 45 f : (X; d) ! (X; d) is a contraction mapping if 9 � 2 [0; 1[ s.t.d(f(x); f(y)) � �d(x; y) 8 x; y 2 X:
Remark 16 A contraction is always a continuous function.
Theorem 28 (Banach Fixed Point) Let f : (X; d)! (X; d) be a contraction mappingand (X; d) be a complete metric space, then 9 a unique �xed point f (�x) = �x and if xk is asequence 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 aunique �xed point.
The Banach FPT guarantees the existence of a stationary point �x under the appropriatecontracion and completeness assumptions. However, stability around such stationary pointis another matter. A �xed point is globally stable if, starting from any point in the domainof f , we can generate a sequence that converges to �x. On the other hand, a �xed pointis called locally stable, if, starting from a neighborhood of �x (say B" (�x)) we can generatea sequence converging towards �x: The following proposition outlines the main conditionsunder which local stability can be established.
Proposition 11 Suppose that f : Rn ! Rn is a contraction de�ned in a complete metricspace (i.e., it has a �xed point �x). Then if Jf (�x) ; the Jacobian or matrix of partial deriva-tives 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. IfC� 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 violatethese assumptions and have a �xed point.
Exercise 30 (Excess demand theorem 2.6.1 Carter) De�ne the relative price set�n�1 =fp 2 Rnjpi � 0 and
Pni=1 pi = 1g and de�ne z : �n�1 ! Rn to be a continuous function
satisfying p � z (p) = 0 8 p 2�n�1. Then there exists p� 2 �n�1 such that z (p�) � 0.
18
Proof. First de�ne the following "adjustment function" gi : �n�1 ! �:
gi (p) =pi +max f0; zi (p)g
1 +Pn
j=1max f0; zj (p)g
and the corresponding g : �n�1 ! �n�1 as:
g (p) =�g1 (p) g2 (p) ::: gn (p)
�0then 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 theo-rem and justify the existence of p� such that g (p�) = p�. That is, for each i :
p�i =p�i +max f0; zi (p�)g
1 +Pn
j=1max f0; zj (p�)gor:
p�i
nXj=1
max f0; zj (p�)g = max f0; zi (p�)g
next, multiply both sides by zi (p�) :
zi (p�) p�i
nXj=1
max f0; zj (p�)g = zi (p�)max f0; zi (p�)g
and summ over i :nXi=1
zi (p�) p�i
nXj=1
max f0; zj (p�)g =nXi=1
zi (p�)max f0; zi (p�)g
now, since by assumptionPn
i=1 zi (p) pi = 0, the LHS is zero so that:
nXi=1
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 sumare 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 andcompact-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 InequalityProblem 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 problemminz2�(C) z � f (x) and note that z 7! z � f (x) is continuous and quasiconcave for eachx. Therefore, the maximum theorems imply that � (x) = minz2�(C) z � f (x) is non-?(Weierstrass), compact, UHC (Continuous Max) and convex (Concave Max): Next notethat � (x) � C for each x: That is, � : C � C satis�es all the assumptions required forKakutani�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 VIPfor�f;Rn+
�:
Theorem (28) establishes that, under the required topological conditions, the existenceof a unique �xed point is ensured. However, theorems (29) and (31) are merely existencetheorems. Uniqueness of �xed points, equilibria and solutions to system of equations require"monotonicity" conditions that generalize to multidimensional problems the notion of astrictly increasing function of one variable. For de�nition and examples of monotonicitysee 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 � 2 [0; 1].
Remark 19 Note that the convex combination �x+ (1� �)y is a particular linear combi-nation.
De�nition 47 The convex hull of S, denoted convS is the intersection of all convex su-persets of S :
convS =
(y 2 Rn j y =
mXi=1
�ixi
)forxi 2 S
and �i > 0;mXi=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 � B = fa+ b j a 2 A and b 2 Bgis a convex set
Lemma 9 Also, if A;B are convex, then the cartesian product A�B is a convex set.
21
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 isnormal to p � x = 0Remark 24 p "points" in the direction where p � x > 0Theorem 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; 8x 2 S
Proof. (step 1) De�ne f (x) = (x� �x)�(x� �x) and claim 9 x02 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, chooser > 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 x0on �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 x02S.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 xt2S since S is convex by assumption. Now:�xt � �x
���xt � �x
�= t2 (x� x0) � (x� x0) + 2t (x� x0) � (x0 � �x) + (x0 � �x) � (x0 � �x)
or:�xt � �x
���xt � �x
�� (x0 � �x) � (x0 � �x) = t2 (x� x0) � (x� x0) + 2t (x� x0) � (x0 � �x)
now, since (xt � �x) � (xt � �x)� (x0 � �x) � (x0 � �x) � 0 it follows that:t2 (x� x0) � (x� x0) + 2t (x� x0) � (x0 � �x) � 0
t (x� x0) � (x� x0) + 2 (x� x0) � (x0 � �x) � 0
and since limt!0 t (x� x0) � (x� x0)+2 (x� x0) � (x0 � �x) = 2 (x� x0) � (��p) we concludethat:
(x� x0) � (�p) � 0) �p � x � �p � x0
22
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:
De�nition 49 If A is a m� n matrix, posA =ny 2 Rm j y =
Pnj=1 ajxj and xj � 0
o:
Remark 25 Note that aj are the columns of A so posA is the cone formed by the column-vectors of A
Theorem 35 (Farka�s lemma) Let A be n �m matrix; then b 2 posA , y 2 Rm andATy � 0 imply bTy � 0
Proof. ()) Suppose b 2 posA then 9 �x 2 Rm+ such that b =Pn
j=1 aj�xj = A�x. Now, if
ATy � 0 then 0 � �xT�ATy
�= (A�x)T y = bTy:
(() (contrapositive and separating hyperplane theorem) Suppose that b =2 posA. SinceposA 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. Thusp � y0 � p � 0 = 0 and p � y0 � p�2y0 so that p � y0 � 0 and we conclude that p � y0 = 0which in turn implies that p � b < 0:Therefore, we conclude that ATy � 0 so ATp � 0 butpTb < 0. Summarizing, b =2 posA ) y 2 Rm and ATy � 0 but bTy < 0:
Proposition 12 (supporting hyperplane) Let S be non-? and convex, and supposethat 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 applythe separating hyperplane theorem are satis�ed. However, note that each xk requires adi¤erent �qk. So we conclude that for each xk, 9 �qk 6=0 such that �qk�xk� �qk�x 8 x 2 clSand since S � cls it follows that �qk�xk� �qk�x 8 x 2 S. Now, since �qk 6=0 then k�qkk 6= 0 sothat:
�qkk�qkk
� xk ��qkk�qkk
� x 8x 2 S
and since �qkk�qkk
= 1 it follows that:�qkk�qkk
2 A = fy 2Rnj kyk = 1g
note that A is the pre-image of f (y) = kyk = 1. Now this function is trivially continuousso 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, wecan extract a convergent subsequence:
�qkmk�qkmk
such that�qkmk�qkmk
! �p and �p 2A
Next, note that:
�qkmk�qkmk
� xk ��qkmk�qkmk
� x 8x 2 S ) �qkmk�qkmk
� �x � �qkmk�qkmk
� x 8x 2 S ) �p � �x � �p � x
23
Corollary 4 Let S be non-?; closed and convex, and suppose that �x 2 @S then 9 �p 6=0such that �p � x � �p � �x:
2.3 Convexity of functions and subgradients
De�nition 50 f is a convex function if: �f(x)+(1��)f(y) � f(�x+(1��)y) wheneverx; y 2 S and � 2 [0; 1].(� for concavity)
De�nition 51 Let f : S ! R and S � Rn. The epigraph of f is epi (f) = f(x; y) 2 Rn+1jx 2S and y � f (x)g
De�nition 52 Let f : S ! R and S � Rn. The hypograph of f is hyp (f) = f(x; y) 2 Rn+1jx 2S and y � f (x)g
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 andconvex on S then epif is closed and convex.
Proof. That epif is convex follows from the assumption that f is convex. Next, to see thatepif 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 thaty� � f (x�). Summarizing, (xk; yk) 2epif; (xk; yk) ! (x�; y�) and (x�; y�) 2epif so weconclude 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 eachx 2S then f is convex.
Proof. Choose x1;x2 2 S and � 2 [0; 1]. Let xt = tx1+(1� t)x2 and note that @f(xt) 6=?. Next choose p 2 @f(xt) and note:
f(x1)� f(xt) � p��x1 � xt
�f(x2)� f(xt) � p�
�x2 � xt
��nally multimply the �rst of these weak inequalities by t and the second one by 1� t; addthem together and obtain:
tf(x1) + (1� t) f(x2) � f (tx1 + (1� t)x2)
so we conclude that f is convex.
24
Proposition 15 Let g : Rn ! R and g(�x) � g(x) whenever x 2 B"(�x) for some " > 0; ifg is di¤erentiable. at �x, then rg(�x) = 0.
Corollary 5 Now let g(x) = �q � x if �q � �x � �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 continuousat �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 claimthat (�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) � q � x+ry
for each (x;y) 2 A; but since Gr(f) �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 thatr < 0 since r = 0 implies q � �x � q � x, which using corollary (5) implies that q =0 whichcontradicts the hypothesis that (q; r) 6= 0. Now, since r < 0 we can divide (2.1) by r toobtain:
f(x) � f(�x) +
��1rq
�| {z } �subgradient
(x� �x)
thus,��1rq�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 2intS. 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� �x) 8 x 2S
Now, since �x 2 intS then 9" > 0 such that B"(�x) � S. hence:
f(x) � f(�x) + q� (x� �x) 8 x 2B"(�x)q � �x�f(�x)| {z }
g(�x)
� q � x�f(x)| {z }g(x)
8 x 2B"(�x)
so that g (�x) � g (x) 8 x 2B"(�x). Finally, note that g is di¤erentiable at �x since f isdi¤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 onesolution.
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 increasingfunction. 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 C1.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 p12@f(x1) andp22@f(x2) then (p1 � p2) � (x1 � x2) � 0:
Proof. p12@f(x1)) f(x2) � f(x1) + p1 � (x2 � x1) while p22@f(x2)) f(x1) � f(x2) +p2 � (x1 � x2). Adding these two inequalities together we obtain (p1 � p2) � (x1 � x2) � 0:
Corollary 6 (monotonicity of gradients) Let f : S ! R and S � Rn open, convexand non-?: If f is di¤erentyiable and convex, then (rf (x1)�rf (x2)) � (x1 � x2) � 0 foreach x1;x2 2 S:
Remark 27 Note that if n = 1 then (rf (x1)�rf (x2))�(x1 � x2) � 0 becomes (f 0 (x1)� f 0 (x2))�(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� Y ! R is said to satisfy singlecrossing if s < t and x < y imply:
f (y; s)� f (x; s) < f (y; t)� f (x; t)
Remark 28 Supermodularity implies single crossing.
26
Proposition 18 (A.45, McLean) If f is di¤erentiable, the supermodularity property isequivalent to @2f
@x@t> 0:
Proof. Suppose that f : [a; b] � [0; 1] ! R is continuously di¤erentiable. Choose x < y
and s < t: Then for each z 2 [a; b], @2f@x@t
> 0 implies:
0 <
Z t
s
@2f
@x@t(z; w) dw =
@f
@x(z; t)� @f
@x(z; s)
and:
0 <
Z b
a
�@f
@x(z; t)� @f
@x(z; s)
�dz = f (y; t)� 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) � �g isa convex.set.
De�nition 58 A function f : S ! R is quasiconcave if fx 2 S j f(x) � �g is a convexset
De�nition 59 A function f : Rn ! Rm is strictly quasiconcave if f (x) � f (y) )f (�x+ (1� �)y) > f(y)
Proposition 19 If f is di¤erentiable on the open-convex set S � 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 � 2 [0; 1]. then:
f(�x2 + (1� �)x1) � �f(x2) + (1� �) f (x1)
f [� (x2 � x1)] + x1 � f (x1) � � [f(x2)� f (x1)]
f [� (x2 � x1)] + x1 � f (x1)
�� f(x2)� f (x1)
if � > 0 this is a Newton quotient de�ning the directional derivative of f in the directionof (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�) +rf(x�) � (x2�x�)= f(x�) +rf(x�) � ((1� �) (x2�x1))
f(x1) � f(x�) +rf(x�) � (x1�x�)= f(x�) +rf(x�) � (� (x1�x2))
27
multiply the �rst inequality by �, the second by (1� �) and add them together to get:
f(�x2 + (1� �)x1) � �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 fis convex if and only if its gradient is monotone i.e.:
f is strictly convex , [rf(x2)�rf(x1)] � (x2 � x1) > 0f is convex , [rf(x2)�rf(x1)] � (x2 � x1) � 0
Remark 29 In general, the product of two convex functions is not convex. However, iff; g are both convex, monotone and positive, f � g is convex.
De�nition 60 Let A be n� n. We say that A is PSD if xTAx � 0 for all x 6= 0
Remark 30 In R1 we say that f(�) is non-decreasing , f 0(�) � 0: In Rn we replacenon-decreasing by monotonicity, f 0(�) by the Jacobian, Jf (�); and � 0 with positive-semi-de�niteness (PSD).
Proposition 21 g : Rn � 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 thefunction � (t) = g ((1� t)x1 + tx2) � (x2 � x1). Since � (t) is a function of t and a scalarfor each t, we can invoke the Mean Value Theorem (MVT) to deduce that 9 t� 2 (0; 1)such that:
� (1)� � (0)
1� 0 = �0 (t�)
writing out this expression for �0 (t�) we get:
[g (x2)� g (x1)] � (x2 � x1) = (x2 � x1)T Jg ((1� t�)x1 + t�x2) (x2 � x1)| {z }�0(t�)
and since Jg is PSD we conlcude that:
[g (x2)� g (x1)] � (x2 � 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) + t�x)�(�x+�y � �x). Again using the MVT we obtain:
[g (�x+�y)� g (�x)] � (�x+�y � x1)| {z }�0 since g is assumed to be monotone
= (�y)T Jg (z�) (�y)
where z� = (1� t�) (�x+�y) + t��x. We want to show that (�y)T Jg (�x) (�y) � 0 or equiv-alently yTJg (�x)y � 0. So let �! 0 so that z� ! �x.and we arrive at the conclusion thatyTJg (�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) = xTAx+ bxthen 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:
maxq � xsubject to x 2 S
which is the cannonical form of a type of convex optimization problem; the support functionproblem.
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 InequalityProblem (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+.
Since 0 2 Rn+ it follows that f (�x) � (0� �x) � 0 ) f (�x) � �x � 0. But Rn+ is a cone so�x 2Rn+ ) 2�x 2Rn+. Therefore, f (�x) � (2�x� �x) � 0) f (�x) ��x � 0 from which we concludethat 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 thatf (�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 concludethat �x solves the VIP for
�f;Rn+
�:
De�nition 63 (barrier cone) Let S � Rn be non-?. The barrier cone of S, b (S) isde�ned as follows: �p 2 b(S) if and only if the problem maxx �p � x subject to x 2 S has asolution.
Remark 33 b (S) � Rn and 0 2 b (S) since maxq � x = 0 � x makes any x a maximizer.
Example 38 Suppose that S = fx 2R2j kxk < 1g, then b (S) = f0g.
29
Example 39 On the other hand, if S = fx 2R2j kxk � 1g, then b (S) = R2.
Example 40 Suppose that S = fx 2R2j x1 � 0 ;x2 � 0g (i.e., the third quadrant). To �ndb (S) in this case, �rst note that if p has a negative coordinate, @ a maximizer. Thus, pmust have both non-negative coordinates, i.e., b (S) = fp 2R2j p1 � 0 ;p2 � 0g. Naturally,if pi > 0 and pj = 0 the problem would have in�nitely many maximizers. On the otherhand, if p1 > 0 and p2 > 0 the problem has a unique maximizer. The support function isfound 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 suchthat p � �x � p � x 8 x 2S. Therefore, � � 0) � (p � �x)�� (p � x) or (�p) ��x � (�p) � x sothat 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 �s (tp) = t�s (p) :
Remark 36 �s is a convex function, whenever b (S) is a convex set.
Theorem 41 (support function) Let S � Rn convex and �p 2 b(S) then �x 2 argmaxx2S �p � x , �x 2 @�s(�p):
Proof. ()) Suppose that �x 2 arg maxx2S �p �x. Then �s(�p) = �p ��x. Next choose q 2 b(S)and notice that
�s(q) � q � �x
= q � �x+=0z }| {
(�s(�p)� �p � �x)= �s(�p) + �x� (q� �p)) �x 2@�s(�p)
(() Suppose that �x 2@�s(�p). This implies that �p 2 b(S): Now, by the de�nition ofsubgradient:
�s(p) � �s(�p) + �x� (p� �p) 8 p 2 b(S)choosing p = 0 and p =2�p we note that 2�p 2 b(S) since �p 2 b(S) and b(S) is a cone.Thus:
�s(0) = 0
� �s(�p) + �x� (0� �p)) �x � �p � �s(�p)
30
while:
�s(2�p) = 2�s(�p) (by homogeneity)
� �s(�p) + �x� (2�p� �p)= �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 wemust show that �x 2S. We do this by contradiction and using the separating hyperplanetheorem. 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 havethat:
�s(w) < w � �x
= w � �x+=0z }| {
(�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 �sis di¤erentiable at �p then argmaxx2S x � �p = fr�s(�p)g = �x is the unique solution to themaximization problem.
Example 42 Let S =�x 2R2+j x1x2 � 1
then b (S) = fp 2R2jp1 < 0,p2 < 0g [ f0g and
intb (S) = fp 2R2jp1 < 0, p2 < 0g. Also, �s(p1; p2) = �2pp1; p2. Notice that �s(p1; p2) is
di¤erentiable at intb (S) so that:
�x1 =@�s(p1; p2)
@p1=
rp2p1
�x2 =@�s(p1; p2)
@p2=
rp1p2
and (�x1; �x2) =�q
p2p1;q
p1p2
�is the solution to the max problem.
So far we have de�ned all these objects in terms of a maximization problem. There isa natural equivalence with the minimization problem:
De�nition 65 (supergradient) Let f : S ! R , then p is a supergradient for f at �x 2 Sif 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 hasa 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 Kuhn-Tucker theory
2.5.1 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 gradientof the restriction. After eliminating � this method yields the familiar result:
@f (�x1; �x2) =@�x1@f (�x1; �x2) =@�x2
=@g (�x1; �x2) =@�x1@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; gdi¤erentiable. The cannonical NLP problem is:
min f(x)
s:t: g1(x) � 0... (2.2)
gm(x) � 0
32
A Kuhn-Tucker pair for this problem is a pair��x; ��
�2 Rn � Rn satisfying:
1) gi(�x) � 0 8 i = 1; :::;m (2.3)
2) ��i � 0 8 i = 1; :::;m (2.4)
3) rf(�x) =X
i
��irgi(�x) (2.5)
4) ��i � gi(�x) = 0 8 i = 1; :::;m (2.6)
(complementary slackness)
Example 43 (linear programming) Let the objective function be f (x1; x2) = c1x1 +c2x2 and the constraints be g1 (x1; x2) = a11x1+a12x2�b1 and g2 (x1; x2) = a21x1+a22x2�b2.Then the LP problem is:
min c � xs:t: aT1 x �b1aT2 x �b2
a KT pair for this problem would be��x; ��
�2 R2 � R2 satisfying:
c =��1a1 + ��2a2; �� 2R2+aT2 �x�b2 � 0; ��1
�aT2 �x�b2
�= 0
aT2 �x�b2 � 0; ��2�aT2 �x�b2
�= 0
Theorem 44 (Kuhn-Tucker for LP) Let c 2Rn; A be m�n and b 2Rm: Then �x solvesthe LP problem:
min c � xs:t: Ax� b � 0
if and only if, 9 �� 2Rm+ such that:c =AT �� (2.7)
A�x� b � 0 (2.8)��� [A�x� b] = 0 (2.9)
Proof. (() Suppose that��x; ��
�2 Rn � Rm satis�es the KT conditions. We must show
that any other x such that Ax� b � 0) c � x � c � �x: Now, from (2.7):
c =AT ��) cTx =
�AT ��
�Tx
so that:
cTx =�AT ��
�Tx
=�AT ��
�Tx+ bT ��� bT ��
= (Ax� b)T ��+ bT ��� bT ��
= AT ��
= cT�x
33
()) 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 sat-
isfy rcT�x = 0) c =AT �� we simply need �� = 0:Now, if I (�x) 6= ?, suppose that y 2Rnsatis�es aTi y � 0. Then we can claim that cTy � 0; to see this, choose t > 0 such thataTi x+ta
Ti 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) � cTx) cTy � 0. Now, applying Farka�slemma (proposition (35)) we conclude that for each i 2 I (�x) ;9 ��i � 0 such that c 2posAthat is:
c =Xi2I(�x)
ai��i
or c =AT ��: To complete the argument let ��i = 0 8 i =2 I (�x) :
Theorem 45 (duality of linear programming) Let A be m � n; b 2Rm and c 2Rn:The problem:
min c � x (2.10)
s:t: Ax� b � 0
has a solution if and only if the problem:
maxb � y (2.11)
s:t: ATx = 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 forthe problem in (2.10). Then:
c =AT�y, A�x� b � 0�y � 0 and �y� [A�x� b] = 0
therefore, since max f (x) is equivalent to min�f (x) ; it follows that (�y; �x) is a reducedKT pair for the problem:
min�b � y (2.12)
s:t: ATx = c
yi � 0
34
where the complementary slackness condition for the problem in (2.10) becomes the colin-earity of gradients condition for the problem in (2.12), the colinearity of gradients conditionfor the problem in (2.10) guarantees that the equality restriction in (2.12) is satis�ed andthe condition that the multiplier is nonnegative in (2.10) guarantees that the nonnegativityrestriction 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 theproblem:
minrf(�x) � xs: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) � 0
If �x satis�es the GCQ then 9 �� 2Rm such that��x; ��
�is a KT pair; that is,
��x; ��
�satisfy
the KT conditions (2.3)-(2.6).
Example 47 (GCQ is not satis�ed) Let the problem be:
minx1
s:t: x2 � 0; x31 � x2 � 0
the unique solution to this problem is �x =(0; 0) and:
rf (�x) =�10
�, rg1 (�x) =
�01
�, rg2 (�x) =
�3�x21�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 9x0 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) � 0 (2.13)
xi � 0 8 i
In order to ignore the multipliers for these restrictions, use the (reduced) K-T system:
rf(�x)�Pm
i=1��irgi(�x) � 0 (1)
��i � 0 (2) gi(�x) � 0 (3)
xj � 0 (4) ��igi(�x) = 0 (5)h@fxJ(�x)�
Pmi=1��i@gxJ(�x)ixj = 0 (6)
Example 51 Let the problem be:
minx1 + x2
s:t: x21 + 4x22 � 4
xi � 0 8 i
the KT conditions are:
�� � 0, �xi � 0 8 i�x21 + 4�x
22 � 4 � 0; �� (�x21 + 4�x
22 � 4) = 0�
1� 2���x1��x1 = 0;
�1� 8���x2
��x2 = 0�
11
�� ��
�2�x18�x2
�� 0
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:
1� 2���x1 = 0) �x1 =1
2���1
2��
�2� 4 = 0
so that:�� =
1
4and �x1 = 2
Case 2: �x1 = 0, �x2 > 0. Now we replace the second-to-last condition with �x1 = 0 and solve:
1� 8���x2 = 0) �x2 =1
8��
4
�1
8��
�2� 4 = 0
so that:�� =
1
8and �x2 = 1
Case 3: Finally we look for an interior solution (i.e. �x1 > 0, �x2 > 0) so we solve:
1� 2���x1 = 0) �x1 =1
2��
1� 8���x2 = 0) �x2 =1
8���1
2��
�2+ 4
�1
8��
�2= 4
so that:�� =
p5
8, �x2 =
4p5, �x2 =
1p5
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:
1 < 2 <4p5+
1p5
Remark 39 In the last example we found 3 KT pairs but only one of them is a solutionto 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:
min f(x) (2.14)
s:t: g1(x) � 0:::gm(x) � 0 (2.15)
h1(x) = 0:::hr(x) = 0 (2.16)
then the KT pair is��x; ��; �
�conditions for this problem are:
1) rf(�x) =Xm
i=1
��irgi(�x) +Xr
i=1� irhi(�x) (2.17)
2) ��i � 0 8 i = 1; :::;m (2.18)
3) gi(�x) � 0 8 i = 1; :::;m (2.19)
4) ��i � gi(�x) = 0 8 i = 1; :::;m (2.20)
4) hi(�x) = 0 8 i = 1; :::; r (2.21)
Remark 40 Note that the problem min f(x) s.t. h(x) = 0 is equivalent to the problemmin 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 withinequality 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 solutionto the problem. Additionally, we can �nd conditions under which a KT pair is a solutionto 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, when-ever (�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 wehave:
rf(�x) =X
i
��irgi(�x)
Now choose x satisfying gi(x) � 0 for each i. Then:
rf(�x) � (x� �x) =X
i
��irgi(�x) � (x� �x)
but since gi(�) is concave, it follows that rgi(�x) � (x� �x) � gi(x)� gi(�x) thus:
rf(�x) � (x� �x) �X
i
��i [gi(x)� gi(�x)]
=X
i
��igi(x)| {z }�0
�X
i
��igi(�x)| {z }=0 by KT cond.
� 0
38
thus, we have rf(�x) � (x� �x) � 0. But f is convex so that:
f(x)� 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 ), thenwhenever (�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 practicalissues of proudction, cost and expenditure minimization and pro�t and utility maximiza-tion.
3.1 Producer theory
3.1.1 Production sets and technologies
In classical producer theory, a �rm produces a single output with n di¤erent inputs. If welet 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 productionpossibility set is the set:
Y =�(y;�x) 2 R� Rlj (y;�x) is technologically feassible
For some production function g (x) this can be written as:
Y =�(y;�x) 2 Rl+1jg (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 isstandard practice to assume that V (y) is a convex set 8 y so that technology can besaid to be convex. A less restrictive assumption would be to assume that the productionpossibility 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 simplyas y and call yi 2 R+ outputs and yi 2 �R+ inputs. Letting n = l + 1 we can use thisnotation to write y =(y1; :::yn) and the production possibility set as:
Y = fy 2 Rnjy is technologically feassibleg
Suppose that the nth commodity is the single output and that g : Rn�1 ! R is a productionfunction.Then "technologically feassible" would mean formally that:
yn � g (y1; :::; yn�1)
or
yn � g (y1; :::; yn�1) � 0
in which case the production possibility would be:
Y = fy 2 Rnjyn � g (y1; :::; yn�1) � 0g
or, if we relabel yn � g (y1; :::; yn�1) = F (y) :
Y = fy 2 RnjF (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 � y) y02Y:
De�nition 70 A technology is said to be monotonic if x 2V (y) and x0 � x) x02V (y)
De�nition 71 A technology is said to be additive if y 2Y and y02Y ) y0 + y 2Y . Inother 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 com-ponent of any y 2Y cannot be positive)
Exercise 55 (McLean, 78; Varian 1.10) Y satis�es additivity and nonincreasing re-turns 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�esadditivity and y 2 Y and y+y 2 Y it follows that y+y+y 2 Y . By induction, we concludethat 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 choosey; y0 2 Y . Since Y satis�es NIRS it follows that k 2 [0; 1] ) ky 2 Y and (1� k) y0 2 Y ;and by additivity ky+ (1� k) y0 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 thatY exhibits NIRS simply choose k 2 [0; 1] : Next, since Y is convex it follows that 8 y; y0 2 Yand k 2 [0; 1], ky+ (1� k) y0 2 Y . But Y is a cone so ky 2 Y and (1� k) y0 2 Y . Thereforewe see that whenever ky 2 Y and (1� k) y0 2 Y it follows that ky+ (1� k) y0 2 Y so Ysatis�es additivity.
Example 56 (Carter, 1.91) Suppose that an economy contains m producers and n com-modities. The technology of each producer is summarized by its production possibility setY j � Rn. Aggregate production (set) y is the sum of the net outputs of each of the producersyj, that is:
y = y1 + :::+ ym
and the set of feasible aggregate production plans or the aggregate (economy-wide) produc-tion 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 bytechnology, actual production yj and y depend on the price vector p. Thus, the sets yj andy can in fact be thought of as correspondences yj(p) and y(p) where:
y(p) =X
yj(p) =ny 2Rnjy =
Xyj for some yj 2 yj (p)
o3.1.2 Cost minimization
Support function approach
Using the input requirement set notation, and allowing for multi-product �rms (so thaty 2Rm); the cannonical cost minimization problem can be cast as:
minw � xs:t:x 2V (y)
wherew 2Rn is a vector of input prices. Note that ifw 2eb(V (y)) then argminx2V (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 convexset, then:
c(�;y):K ! R is concave and homogeneous of degree 1
�x 2 arg minx2V (y)
w � x , �x 2 @c(w;y) = @~�V (y)(w)
Moreover, if w 2 intK and c(w;y) is di¤erentiable:
arg minx2V (y)
w � x = frc(w; y)g = �x| {z }(Shephard0s Lemma)
Example 57 (perfect substitutes) Let V (y) =�x 2 R2+jx1 + x2 � y
: This produc-
tion 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; w2g y. Also, note that whenever c(w; y) is di¤erentiable, say when w1 > w2 Shep-hard�s lemma holds since c(w; y) = w2y and therefore:
rc(w; y) =�0y
�so that there�s a unique supergradient equal to the gradient. Intuitively, since input 1 isrelatively more expensive and inputs are perfect substitutes, the �rm produces y using onlyx2:
Kuhn-Tucker approach
A similar method to solve cost minimization problems is to replace x 2 V (y) with g (x) � ywhere g (�) is a production function. This way we face an optimization problem underinequality and nonegativity constraints and a natural approach to deal with it is the KTtheory developed in section 2.5.
Example 58 Suppose that the �rm has production function V (y) =�x 2 R2+jx1 � y and x�11 x
�22 � y
where �i > 0 and �1 + �2 = 1. Then the problem can be cast as:
minw1x1 + w2x2
s:t: x1 � y and x�11 x�22 � y
xi � 0 8 i
the KT conditions for this problem are:
w1 � ��1 � ��2�1�x�1�112 �x�22 � 0; w1 � ��2�2�x�112 �x�2�12 � 0�w1 � ��1 � ��2�1�x�1�112 �x�22
�x1 = 0;
�w1 � ��2�2�x�112 �x�2�12
�x2 = 0
xi � 0 , ��i � 0 i = 1; 2; x�11 x�22 � y, x1 � y
��2 (x�11 x
�22 � y) = 0; ��1 (�x1 � y) = 0
44
�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 � ��2�1�x�1�112 �x�22 = 0; w1 � ��2�2�x�112 �x�2�12 = 0x�11 x
�22 = y �x1 = y
which yields the solution:
��1 = w1 � �1�2w2; ��2 =
w2�2; �x1 = y = �x2
this will be a solution to the problem i¤ w1�1� w2
�2. On the other hand, if w1
�1< w2
�2then ��1
becomes infeasible so we look for a solution with ��1 = 0 and rewrite the system as:
w1 � ��2�1�x�1�112 �x�22 = 0; w1 � ��2�2�x�112 �x�2�12 = 0x�11 x
�22 = y
solving this new system yields:
��2 =
�w2�2
��2 �w1�1
��1, �x1 = y
�w1�1w2�2
��2and �x2 = y
�w2�2w1�1
��1now with ��1 = 0, the condition ��1 (�x1 � y) = 0 is trivially satis�ed. However, we still needto check that condition x1 � y is also satis�ed. Since we are looking a the special case whenw1�1< w2
�2it follows that w1�1
w2�2> 1 so that:
�x1 � y = y
�w1�1w2�2
��2� y =
��w1�1w2�2
��2� 1�y � 0
and we have a solution to the problem even when w1�1< w2
�2:
Example 59 (no interior solution) Suppose that V (y) =�x 2 R2+j�1x1 + �2x2 � y
.
Then the problem can be cast as:
minw1x1 + w2x2
s:t:�1x1 + �2x2 � y
xi � 0 8 iNow, since this is a linear programming problem, we can use theorem (44) to conclude thata KT pair for this LP will be a solution to the problem. Thus we look for
��x;���2 R2 � R
such that:w1 � ���1 � 0 w2 � ���2 � 0�
w1 � ���1�x1 = 0;
�w2 � ���2
�x2 = 0
xi � 0 8 i; �� � 0�� (�1x1 + �2x2 � y) = 0 �1x1 + �2x2 � y � 0
this system has no solution with both x1; x1 > 0. The intuition is simple; if the priceof input i relative to its productivity is smaller than the price of input j relative to itsproductivity, the �rm�s optimal plan is to use only input i. That is, if w1
�1< w2
�2then 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 =y�1and �� = w1
�1:
45
3.1.3 Pro�t maximization
Support function approach
Let y 2 Rm , x 2 Rn and let p 2 Rm+ and w 2 Rn+ then the pro�t maximization problemis:
maxfp � y �w � xgs: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 (thatis, z = �x) and note that (�x; �y) solves the problem above , (��x; �y) solves the problem:
maxfp � y +w � zgs:t: (z;y) 2 T
Remark 43 The pro�les (w;p) for which the max problem has a solution is the barriercone b(T ):Furthermore, if (w; �p) 2 b(T ) then �(w;p) = �T (w;p):
Proposition 25 Suppose T; b(T ) are convex. LetK = b(T )\[Rn+�Rm+ ] so that (!;p) 2 K:Then:
�(�): K ! R is convex and homogeneous of degree 1
(�x; �y) 2 arg maxx2V (y)
fp � y �w � xg , (��x; �y) 2 arg max(z;y)2T
fp � y +w � zg
, �x 2 @�(w;p) = @�T (w;p)
moreover, if � is di¤erentiable:
arg max(z;y)2T
fp�y+w�zg = f�r�w(w;p);r�p(w;p)g = (�x; �y)| {z }(Hotelling0s Lemma)
Example 60 Let V (y) = fx 2 R+jpx � 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:
max ppx� wx
s:t: x � 0
with solution:�x =
� p
2w
�2and �y =
p
2wso that:
�(w;p) = pp
2w� w
� p
2w
�2=
p2
4w
46
Note that �(w;p) is convex as it should. To see why we can use corollary (7):
r�(w;p) =���p2w
�2p2w
�) H� =
�p2
2w3� p2w2
� p2w2
12w
�and since H�.is PSD we conclude that �(w;p) is convex. Finally note that (Hoteling�slemma): "
�@�(w;p)@w
@�(w;p)@p
#=
� �p2w
�2p2w
�=
��x�y
�Example 61 V (y) = fx 2 R+jx � yg. In this setup, constant returns to scale complicatesmatters. Thus, if w > p the unique solution to the pro�t max problem is (0; 0). If p > wthe problem has no solution and if p = w there are in�nitely many solutions all of whichlead to �(w;p) = 0:
Remark 44 Every function which is convex and homogeneous of degree one is a pro�tfunction for some Y . Thus, we can recover the production technology set of a �rm from itspro�t function.
3.2 Consumer theory
3.2.1 Utility maximization
The cannonical utility maximization problem:
maxu (x) (3.1)
s:t: p � x � y
xi � 0is not a linear problem and therefore we cannot apply the methods of support functiontheory. However, we can still use the Kuhn-Tucker approach for solving NLP optimizationproblems. Moreover, if the objective function u is concave, a straightforward extension oftheorem (52) implies that a KT pair would be a solution to the optimization problem. Witha solution at hand, we can compute the optimal value function which is usually referred asthe indirect utility function. In this case, the (reduced) KT conditions for optima become:
ru (�x) + ��p � 0; y � p � �x �0�� [y � p � �x] =0; �xi � 0
�xj
h@u@xj(�x)� ��pi
i= 0; � � 0
Remark 45 Notice that maxu (x) is equivalent to min�u (x) so that the colinearity ofgradients 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 func-tion for the utility maximization problem:
v (p; y) = maxp�x�y
u (x)
which is called the indirect utility function, also has a nice interpretation. It gives themaximum 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; yt) = 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 maxpt�x�yt
u (x)
Now we can claim that xt is feasible for at least one of the (p; y) or (�p; �y) problems. Ifthis were not the case p � xt > y and �p � xt > �y, which, if multiplied by t and (1� t),respectively, and added together imply:�
t�p � xt
�+ (1� t)
��p � xt
��> ty + (1� t) �y
pt � xt > yt
a contradiction since by hypothesis xt solves the (pt; yt) problem. Thus, xt is feasible forat least one of the (p; y) or (�p; �y) problems:Therefore:
u�xt�� max fv (�p; �y) ; v (p; y)g � �
v�pt; yt
�� �
and we conclude that pt; yt 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 preferencesu (x) = x�1 + ::: + x�n with 0 < � < 1; then the utility maximization problem for each levelof income y is:
maxx�1 + :::+ x�n
s:t:
nXi=1
pixi � y
48
This problem is simpli�ed by the fact that u (�) is strictly increasing and strictly concaveon 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 weonly need to look for an interior solution to the KT system:
��x��1i � ��pi = 0 and y �nXi=1
pi�xi = 0
which yield solution:
�� = �
yPn
i=1 p�
��1i
!��1and �xi =
24 p1
��1iPn
i=1 p�
��1i
35 ythis yields an indirect utility function:
v (p; y) = ynXi=1
24 p1
��1iPn
i=1 p�
��1i
35Theorem 63 (Roy�s identity) Suppose that �x solves the problem in (3.1). If �xi > 0 8i,��x;���is a (reduced) KT pair, �� > 0 and �x is regular, then:
�xi =�@v (p; y) =@pi@v (p; y) =@y
3.2.2 Expenditure minimization
Suppose that agents have a certain welfare target � and want to �nd the minimum amountof spending required to attain such level of utility. Then they face the problem:
minp � xs:t: u (x) � �
xi � 0
Since the objective function is linear, this is a support function problem and we can usethe methods from section 2.4 to �nd a solution. The problem is identical to that of costminimization and, in this case yields an optimal value function e (p; �). As in any supportfunction problem, e (p; �) is homogeneous of degree one and concave. Moreover, from theoptimal value function we can recover the solution to the cost minimization by:
@e (p; �)
@pi= �xi (p; �)
49
Example 64 Suppose that Suppose that u : Rn+ ! R is continuous and monotonic in thefollowing sense: u(x) > u(y) whenever x > y where x > y means that xi > yi for eachi = 1; :::; n. If �x solves the utility maximization problem:
maxu (x) s:t: p � x � y, xi � 0
show that �x solves the expenditure maximization problem:
minp � x s:t: u (x) � v (p; y) , xi � 0
where v (p; y) = u (�x).
Proof. Suppose that �x solves the utility maximization problem but does not solve theexpenditure maximization problem. Since �x solves the utility maximization problem thenp � �x � y and u (�x) � u (x) 8 x such that p � x � y. Thus �x is trivially feasible for theexpenditure min problem. But the hypothesis is that �x does not solve the this problemso there exists x0 such that u (x0) � u (�x) and p � x0 < p � �x: This in turn implies thatp � x0 < y or that x0 is feasible for the utility maximization problem. To complete theargument, 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 thehypothesis that �x solves the utility maximization problem.
Exercise 65 Consider a consumer with utility function u : Rn+ ! R. Furthermore, sup-pose u = f � h where h : Rn+ ! R+ is homogeneous of degree one and f : Rn+ ! R+ isa strictly increasing function satisfying f(0) = 0. Suppose that and �; �0 positive numbersand 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 theexpenditure function for output price vector p and utility level : If u(�x) � �; �x � 0 andp � �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�1 (�0)
f�1 (�)
50
Chapter 4
Game Theory and GeneralEquilibrium
4.1 Game theory
4.1.1 Zero-sum games
De�nition 76 The standard n�1 simplex is the set � = fx 2Rnjxi � 0 andPn
i=1 xi = 1g
Remark 46 The standard n� 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 assignsprobabilities to each of the pure strategies (or actions) available to players. If M is theaction set for a player, these randomizations comprise the mixed extension of the actionset �(M).
Theorem 66 (Von-Neuman�s minimax) In a game with two players 1 and 2, eachwith strategy set M and N, with the simplices �(M) � Rm and �(N) � Rn as the mixedextensions of M and N, respectively, and A as the matrix of payo¤s:
maxx2�(M)
�miny2N
xTAy
�= min
y2�(N)
�maxx2M
xTAy
�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 thatf (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 random-izations, one for each player, containing x1 2 �(A1) ; :::xn 2 �(An) where, as before,�(Ai) =
nxi2Rmjxji � 0 and
Pmj=1 x
ji = 1
o:
De�nition 79 The strategy pro�le x = (x1; x2:::xn) 2 A1 � A2 � ::: � An is a Nashequilibrium if, for each i 2 N one has that: xi 2 arg max
yi2Aif (yi; x�i)
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 bothupper hemicontinuous (UHC), then = 1 (x) + 2 (x) and = 1 (x)� 2 (x) are UHCtoo.
Remark 48 Recall that non-emptiness, compactness and convexity properties of sets arealso inherited by their (Minkowski�s) sum and (cartesian) product.
Theorem 68 (Nash, 1950) Suppose that:
1) : Ai � Rm is compact, convex, non-?2) : fi : A1 � A2 � :::� An ! R is continuous3) : 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 expectedpayo¤ 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)] =
m1Xi1=1
:::
mnXin=1
gi (i1; :::in) ��xi11 ... x
inn
�:
Theorem (68) is the most widely used to justify the existence of equilibrium in strategicform 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 �xedpoint arguments; the �rst one uses monotonicity of the objective function (as was hintedin 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-personstrategic form game. Suppose that Ai = R and that ui : R� R! R is twice di¤erentiableand concave in xi for each x�i. If the following two conditions hold:
(1) :@u1@x1
(�x1; �x2) = 0 =@u2@x2
(�x1; �x2)
(2) :
@2u1@x21
(z) @2u1@x1@x2
(z)@2u1@x1@x2
(z) @2u1@x22
(z)
!is negative de�nite 8z 2R2
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 weconclude that �z =(�x1; �x2) is the unique Nash equilibrium of the game.
Exercise 70 (uniqueness via contraction) Consider the setup of theorem (68) and mod-ify 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 quasi-concavity 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 contractionwe know that 9 � < 1 such that d (' (�x) ; ' (x0)) � �d (�x;x0). However, if �x and x0 are bothNE, then �x =' (�x) and x0 = ' (x0) implying that d (�x;x0) � �d (�x;x0) which is possibleonly if �x = x0.
Next we provide Debreu�s (1952) version of the main theorem for the existence of Nashequilibria in so-called "generalized" games. This is the key step towards the original proofof the existence of equilibrium in pure exchange economies by Arrow and Debreu presentedin section 4.2.2.
4.1.3 The generalized game
Consider the case of a generalized game, also known as an "abstract economy". In thissetup, feasible actions are not independent as in the simple n�person game, but thereexists a transition correspondence �i : A�i � Ai which maps action by players �i intofeasible actions by player i; that is, it yields the set of actions that player i can take, givenwhat every one else is doing. Under the appropriate topological assumptions regarding �ithe existence of equilibrium can be ensured as the following theorem states.
53
Theorem 71 (Debreu, 1952) Suppose that:
1) : Ai � Rmi is compact, convex, non-?2) : fi : A1 � A2 � :::� An ! R is continuous3) : xi 7�! fi (xi;x�i) is quasiconcave
4) : �i : A�i � Ai is UHC, LHC, compact,convex, non-?-valued
then a Nash equilibrium for this generalized game (abstract economy) exists, i.e., 9 a pro�lex = (x1; x2:::xn) 2 A1 � A2 � :::� An such that, for each i 2 N :
xi 2 arg maxyi2�i(x�i)
f (yi; x�i)
Proof. (step 1) First note that Ai � Rmi compact, convex, non-? ) A1 � ::: � An �Rm1 � :::� Rmn compact, convex, non-?.(step 2) Next, for each (x1; :::xn) 2 A1 � :::� An, let:
�i (x�i) = arg maxyi2�i(x�i)
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) isquasiconcave and �i (�) is convex-valued, by the Concave Maximum theorem (Theorem27) �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�stheorem, � has a �xed point, i.e. 9 x =(x1; :::; xn) � A1 � :::� An such that:
x 2 � (x)
xi 2 �i (x�i) 8 i 2 Nxi 2 arg max
yi2�i(x�i)fi (yi;x�i) 8 i 2 N
so the "generalized" game has a Nash equilibrium.
4.2 General Equilibrium theory
4.2.1 Prelminaries
De�nition 80 (pure exchange economy) A pure exchange economy is de�ned by a setof consumers N = f1; 2; :::ng, a set of endowments for each consumer !i 2 RL+ and payo¤sfor each player/consumer ui : RL+ ! R:
54
De�nition 81 (improvable by coallition) Suppose that (x1; :::;xn) is a feasible alloca-tion of the pure exchange economy. Let S � N . We say that S can improve upon x if, foreach i 2 S; 9 x0i 2 RL+ such that: X
i2Sx0i �
Xi2S!i
ui (x0i) > ui (xi)
De�nition 82 (core allocation) A feasible allocation is a core allocation if it cannot beimproved by any coallition.
Remark 49 When agents are given the opportunity to trade their endowments, the out-come 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; :::; xn;p) is a Walrasian equilibrium (W-Eq) if p 2 RL+n f0g and:
1) : xi 2 arg maxp�xi�p�!i
ui (xi) (optimal choices)
2) :nXi=1
xi �nXi=1
!i (no excess demand)
Remark 50 Note that each !i is a L � 1 vector. Thus, the complete set of endowmentsfor the economy ! is a L � N matrix whose typical element !ki is the amount of good kthat 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:0B@Pn
i=1 x1i
...Pni=1 x
Li
1CA �
0B@Pn
i=1 !1i
...Pni=1 !
Li
1CAthat is, the sum of choices of a particular commodity accross players must be less than orequal to the economy�s total endowment of such commodity.
Example 72 (W-Eq does not exist) Consider an economy with two agents. The econ-omy is characterized by the following endowments and preferences:
!1 = (a1; b1) = (1; 1) u1 (x1; y1) = x21 + y21!2 = (a2; b2) = (1; 1) u2 (x2; y2) = x2y2
A W-Eq for this economy does not exist. To see why, let p =(p; q) be the price vector andnote that the we can use the KT conditions to solve the problems:
maxx21 + y21 maxx2y2s:t: px1 + qy1 � p � !1 s:t: px2 + qy2 � p � !2
55
now an interior solution to the KT system would be:
2�x1 � p��1 = 0 �x2 � q��2 = 02�y1 � q��1 = 0 �y2 � p��2 = 0
p�x1 + q�y1 = p+ q p�x2 + q�y2 = p+ q
whose solution is:��1 = 2
p+qp2+q2
��2 =p+q2pq
�x1 = p�
p+qp2+q2
��x2 =
p+q2p
�y1 = q�
p+qp2+q2
��y2 =
p+q2q
note that if p > q then p�
p+qp2+q2
�+ p+q
2p> 2 so that there is excess demand for good x.
Likewise, if p < q then q�
p+qp2+q2
�+ p+q
2q> 2 so that there is excess demand for good y.
Finally, if p = q there are two solutions to agent 1�s problem in either of which there wouldbe 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 functionof agent one (x1; y1) 7! u1 (x1; y1) is convex. In the following section we explore what isrequired 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 pureexchange economies. This is nested in the more general "main theorem" found in Arrowand Debreu (1954).
Theorem 73 (Arrow and Debreu, 1954) Suppose a pure exchange economy with !i 2RL+n f0g and ui : RL+ ! R where ui (�) is continuous and strictly quasiconcave on xi. Thena 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
LPk=1
pk = 1
�and:
K =�z 2 RL+ j z �
nPi=1
!i + (1)
�where (1) is a L � 1 vector of ones. Next, for each p 2� de�ne the zero-player with setof feasible actions:
� =�0 (x1; :::;xn)
56
and payo¤ function:
u0 (p;x) =nPi=1
p � (xi � !i)
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 pwhich 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 otherhand u0 (�) is continuous and quasiconcave in p since it�s linear and recall that ui (�) iscontinuous 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, byKakutani�s theorem, the game has a NE, i.e., 9 an (n+ 1)-tuple (x1; :::xn; p) such that:
p 2 arg maxp2�0(x)
u0 (p; x) =Pn
i=1 p � (xi � !i)
xi 2 arg maxx2�i(p)
ui�p; x�i;xi
�8 i 2 N
(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, �rstnote that:
xi 2 �i (p) :) xi 2 �i (p)) p � (xi � !i) � 0)Pn
i=1 p � (xi � !i) � 0
and therefore:
maxp2�
Pni=1 p � (xi � !i) � 0
maxp2�
PLk=1 pk
�Pni=1
�xki � !ki
��� 0Pn
i=1 (xi � !i) � 0
57
and therefore excess demand for all commodities in the economy is less than or equal tozero. Next, to show that all consumers are optimizing, one must show that in this particularcase:
xi 2 arg maxxi2�i(p)
ui�p; x�i;xi
�) xi 2 arg max
xi2�i(p)ui�p; x�i;xi
�to see this, suppose 9 x0i 2 �i (p) such that ui
�p; x�i;x
0i
�> ui
�p; x�i; xi
�: Then 9 �� such
that whenever � 2 (0; ��):
�x0i + (1� �) xi <nPi=1
!i + (1)
2 K
and:
p � [�x0i + (1� �) xi] = � (p � x0i) + (1� �) (p � xi)� � (p � !i) + (1� �) (p � !i)= p � !i
therefore:�x0i + (1� �) xi 2 �i (p)
but by assumption:
ui�p; x�i;x
0i
�> ui
�p; x�i; xi
�) ui
�p; x�i; �x
0i + (1� �) xi
�> ui (xi)
which contradicts the result of step 2 that:
xi 2 arg maxx2�i(p)
ui�p; x�i;xi
�8 i 2 N
Summarizing, xi 2 argmaxx2�i(p) ui�p; x�i;xi
�, therefore all the consumers are making
optimal choices andPn
i=1 (xi � !i) � 0 so there is no excess demand. Hence, the pureexchange 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 theoremfor the existence of equilibrium in private ownership production economies. It is a reducedversion 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 consumersN = f1; 2; :::ng, a set of endowments for each consumer !i 2 RL+; a set of �rms M =f1; 2; :::mg owned by consumers with �ij as the share of �rm j owned by consumer i andPn
i=1 �ij = 1 and payo¤s for each player/consumer ui : RL+ ! R:
58
De�nition 85 A Walrasian equilibrium in a private ownership production economy is a(n+m+ 1)�tuple: (x1; :::; xn; y1; :::; ym;p) such that p 2 RL+n f0g and for all i 2 N andall j 2M :
1) : xi 2 arg maxx2�i(p)
ui (xi) (consumers maximize)
2) : yj 2 arg maxyj2Yj
p � yj (�rms maximize)
2) :nXi=1
xi �nXi=1
!i +
mXi=1
yj (no excess demand)
where �i (p) =nxi 2 RL+ j p � xi � p � !i +
Pmj=1 �ij
�p � yj
�oand Y = Y1+ :::+Ym is the
aggregate production possibility set.
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 economywhere yj = f0g 8 j
Example 74 (decreasing returns) Suppose that there are two commodities and twoconsumers with problems:
!1 = (6; 0) ; �1 = 1=2 u1 (x1; y1) = x1y1!2 = (6; 0) ; �1 = 1=2 u2 (x2; y2) = x2y2
and a single �rm with production possibility set:
y = Y =�(x; y) 2 R2j
px � y; x � 0
so that the �rm�s problem is:
max qpx� px s.t. x � 0
the FOC for the �rm are:
�xF =�q2p
�2�yF =
q2p
� (p; q) = q2
4p
the FOC for the consumers are:
�x1 =6p+ 1
2
�q2
4p
�2p
= �x2
�y1 =6p+ 1
2
�q2
4p
�2q
= �y2
59
the market clearing conditions are:
�x1 + �x2 + �xF = 2
246p+ 12
�q2
4p
�2p
35+ � q
2p
�2= 12 (4.1)
�y1 + �y2 = 2
246p+ 12
�q2
4p
�2q
35 = q
2p= �yF (4.2)
A simple avenue to solve this system is as follows. Walras� law says that (4.1) holds ifand 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 isq=p = 4 which is the W-Eq price ratio. Finally, the production possibility frontier is givenby y =
p12� 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, theproduction 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 andfor each p 2� one has that: �i (p) � arg max
yj2Yj
�p � yj
�: Then, a Walrasian equilibrium
exists for this economy.
Proof. (step 1) De�ne � as in the proof of Theorem (73) and � : � � RL+ as � (p) =Pnj=1 �i (p) (i.e., the Minkowski�s sum of the net-supply correspondences). Then � (�) is
non-?; compact, convex-valued and UHC. Furthermore, � compact) � (�) is a compactsubset of RL+. Therefore, 9 " > 0 such that z 2 � (�)) z < ". Next, de�ne �j (p) = p � yjwhere yj 2 �i (p). Note that �j : � � R is continuous and since yj 2 �i (p) ) yj 2argmaxyj2Yj
�p � yj
�so that �j (p) = p � yj � p � y0j for any y0j 2 Yj which given the no
free lunch assumption implies that �j (p) � 0 8 p � 0.(step 2) Next, let:
K =�z 2 RL+ j z �
nPi=1
!i + "
�and for a typical consumer i de�ne the budget set correspondence:
�i (p) =
(xi 2 RL+ j p � xi � p � !i +
mXj=1
�ij�j (p)
)
and the feasible action correspondence:
�i (p) = �i (p) \ K
Note that K compact and �i (p) closed ) �i (p) \ K =�i (p) compact. Moreover, itis easily seen that K, �i (p) are convex, so �i (p) \ K =�i (p) is convex and one can
60
show that �i (�) is UHC, LHC. This, along with the fact that ui (�) is continuous andconcave imply that for each p 2�; the maximum value (best-response) correspondences�i (p) = argmaxxi2�i(p) ui (x�i;p;xi) is non-?, convex, compact, UHC (by the MaximumTheorems).(step 3) Now, for each p 2� de�ne the excess demand correspondence:
� (p) =nXi=1
�i (p)�nXi=1
!i +
nXj=1
�i (p)
and notice that � (p) is non-?, convex, compact, UHC (inherits its components properties).Next suppose p 2� and z 2 � (p), then 9 xi 2 �i (p) and yj 2 �i (p) such that:
z =nXi=1
xi �nXi=1
!i +
nXj=1
yj
p � z = p�"
nXi=1
xi �nXi=1
!i +mXj=1
yj
#
=nXi=1
p � xi �nXi=1
p � !i +mXj=1
p � yj
=nXi=1
p � xi �nXi=1
p � !i +mXj=1
nXi=1
�ij
!�j (p)
=nXi=1
"p � xi � p � !i +
nXi=1
�ij�j (p)
#� 0
since we know that xi 2 �i (p) � �i (p) � �i (p). But p 2�) p � 0 which in turnimplies that p � z � 0) z � 0: Summarizing:
1) : xi 2 arg maxx2�i(p)
ui (xi) (consumers maximize)
2) : yj 2 arg maxyj2Yj
p � yj (�rms maximize)
2) :nXi=1
xi �nXi=1
!i +mXi=1
yj (no excess demand)
Therefore the only thing we�re missing for aWalrasian equilibrium is that is xi 2 argmaxx2�i(p) ui (xi).But form step 3 in Theorem 18 above this is easily shown by contradiction since ui (�) isconcave 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 > yifor some i ) f (x) > f (y) i.e. if at least one component of the vector x is greater thanits 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/consumerui : RL+ ! R:
De�nition 89 In a pure exchange economy, a feasible allocation x =(x1; :::; xn) is weaklyPareto optimal (WPO) if @ x0=(x01; :::;x0n) such that ui (x0i) > ui (xi) for all i (i.e., not allplayers/consumers can be made better o¤ ).
De�nition 90 In a pure exchange economy, a feasible allocation x =(x1; :::; xn) is stronglyPareto optimal (SPO) if @ x0=(x01; :::;x0n) such that ui (x0i) � ui (xi) for all i 2 N anduj�x0j�> 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:
WPO =�x 2RL+j @ x0=(x01; :::;x0n) such that ui (x0i) > ui (xi) 8 i
while the set of SPO allocations is:
SPO =�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 2SPO ) x 2WPO, that is the set of SPO � WPO.
Proposition 27 If ui (�) is strongly monotonic for all i, then WPO � SPO so thatWPO = SPO or x is WPO , x is SPO:
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 econ-omy.
Proof. The mechanics of this proof is the same as that used to prove the �rst welfaretheorem 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; :::; xn; �p) is a Walrasian equilib-rium but it�s not WPO. Then 9 some x0 feasible such that 8 i:
ui (x0i) > ui (xi)
and:nXi=1
x0i �nXi=1
!i
and therefore:
�p�"
nXi=1
x0i
#� �p�
"nXi=1
!i
#(4.3)
but x is a Walrasian equilibrium so:
xi 2 arg max�p�xi��p�!i
ui (xi)
which means that:ui (x
0i) > ui (xi)) �p � x0i > �p � !i
) �p�"
nXi=1
x0i
#> �p�
"nXi=1
!i
#contradicting inequality (4.3).
Remark 56 Note that �p � x0i > �p � !i ; x0i > !i just as �p� [Pn
i=1 x0i] > �p� [
Pni=1!i] ;Pn
i=1 x0i >
Pni=1!i. What the contradiction conveys is the fact that for at least some
commodity k it is true thatPn
i=1 xk0i >
Pni=1 !
ki ; that is, even though
Pni=1 x
0i may not be
greater thanPn
i=1!i component by component, it is larger in at least one element. Butthis 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; :::; xn; �p) is a Walrasian equilibrium but x it�s not SPO.Since x is not SPO, 9 x0 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 (x
0i) � ui (xi)
but since x is a Walrasian equilibrium, then it follows that:
ui (yi) > ui (xi)) �p � yi > �p � !i
so that:
�p � yi > �p � !i�p� [x0i + "] > �p � !i
�p � x0i + "�PL
k=1 pk
�> �p � !i
which, implies:�p � x0i � �p � !i for all i 6= j
and:�p � x0j > �p � !j for some j
summing over i (including j) we conclude:Pni=1 [�p � x0i] >
Pni=1 [�p � !i]
�p�Pn
i=1 x0i > �p�
Pni=1!i
which, again, implies that for at least some commodity k it is true thatPn
i=1 xk0i >
Pni=1 !
ki ,
contradicting the hypothesis that x0i is feasible.
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; :::; xn) is WPO and xi 2 RL+, then we must show that 9�p 2 RL+n f0g such that:
ui (xi) > ui (xi)) �p � xi > �p � xi(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. xi+" = x0i 2 RL+ and by monotonicity of u (�) ;ui (x
0i) > ui (xi). The set Si is the halfspace containing all comodity bundles strictly
preferred by agent i to xi: Also note that by quasiconcavity of u (�) ; Si is convex for eachi (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:
z =nXi=1
xi =
0B@x11 + :::+ x1n
...xL1 + :::+ xLn
1CAand claim that for " > 0, z� " =2 clS. To see this, note that if z� " 2 clS then 9 asequence fz (k)g1k=1 such that z (k) 2 S for every k and z (k) ! z� ", which in turnmeans that 9 k� such that:
z (k�) < z =Pn
i=1 xi �Pn
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 thiscontradicts the assumption that xi is WPO. Summarizing, S is non-?, closed, convex, andz� " =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 thezero vector with 1 as its j-th coordenate:
yi = xi + � (k) + "LPl=1l 6=k
� (l)
therefore by monotonicity of u (�):
ui (yi) > ui (xi)
) yi 2 S
so that using the �nal result from step 1:
�p � (Pn
i=1 yi) � �p� (Pn
i=1 xi)
hence:
�p �"xi + � (k) + "
LPl 6=k
� (l)
#� �p� (
Pni=1 xi)
n
"�pk + "
LPl 6=k
pl
!#� 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 � xi. To see this suppose for some i that ui (xi) > ui (xi) which impliesxi 2 Si and for all j 6= i let yj = xj + " so that ui (yj) > ui (xj) by monotonicity, implyingthat yj 2 Sj for each and all j. Notice that xi +
Pni=1 yj 2 S, so:
�p � [xi +Pn
i=1 yi] � �p � [Pn
i=1 xi]
�p � xi + �p �hP
j 6=i (xj + ")i� �p � xi + �p �
Pj 6=i xj
�p � xi � �p � xi
(step 3b: show that �p is an equilibrium price vector) Finally, to se that �p � xi 6= �p � xi;suppose �p � xi = �p � xi (and �nd a contradiction). From the previous step we knowui (xi) > ui (xi) : By continuity of u (�) 9 � 2 (0; 1) such that ui (�xi) > ui (xi); but0 < � < 1) �p ��xi < �p � xi so let yi = �xi and note that ui (yi) > ui (xi) but �p �yi < �p � xiwhich contradicts the result of step 3. Summarizing, we have shown that if x is WPO, 9�p 2 RL+n f0g such that ui (xi) > ui (xi)) �p � xi � �p � xi
4.2.5 Relaxing the Walrasian assumptions
The welfare theorems rely heavily on the Walrasian equilibrium assumptions; agents areprice takers (perfect competition), perfect information and complete markets. To see thecrucial role that these assumptions play, consider the following cases in which equilibriamay not be Pareto optima.
Example 79 (not everyone is price taker) Consider the following two consumer, twogood pure exchange economy.
u1 (x1; y1) = x1y1 !1 = (3; 1)u2 (x2; y2) = x2y2 !2 = (1; 3)
The Walrasian equilibrium for this economy is found by solving:
x1 =3p+ q
2p= x2, y1 =
3p+ q
2q= y2
whose solution is easily seen to be p=q = 1 and (�x1; �x2; �y1; �y2) = (2; 2; 2; 2). Next, considerthe case in which agent 1 is a price taker but agent 2 is a price setter. In particular, agent1 has demands x1 (p; q) =
3p+q2p
and y1 (p; q) =3p+q2q
but 2 chooses the nonnegative pricesp; q at which they trade. If these prices yield feasible demands for agent 1, he consumes thebundle (x1 (p; q) ; y1 (p; q)) and agent 2 consumes the bundle (4� x1 (p; q) ; 4� y1 (p; q)). If1�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 yieldsinfeasible 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. Thenagent 1�;s demands would be:
x1 =3
2+1
2pand y1 =
3p
2+1
2
if we solve for p in one of these equations and replace in the other, we obtain the agent 2�so¤er curve: �
x1 �3
2
��y1 �
1
2
�=3
4
Now, agent 2�s consumption bundle would be:
x2 = 4� 32+1
2p=5
2� 1
2p
y2 = 4� 3p2+1
2=7
2� 3p2
now, to compute agent 2�s best price o¤er to agent one we solve the problem:
maxx2;y2
x2y2
s:t: x2 = 4� x1
y2 = 4� y1�x1 �
3
2
��y1 �
1
2
�=3
4
or equivalently, replacing x1; y1 and expanding the o¤er equation:
maxx1;y1
(4� x1) (4� y1)
s:t x1y1 �3
2y1 �
1
2x1 = 0
the Lagrangian for this problem is:
L = (4� x1) (4� y1)� �
�x1y1 �
3
2y1 �
1
2x1
�with FOC:
2y1 � 81� 2y1
=2x1 � 83� 2x1
) 14x1 = 10y1 + 16
since we are looking for the price p that maximizes 2�s welfare, we can replace y1; x1 withthe expressions found above so 14x1 = 10y1 + 16 becomes:
14
�3
2+1
2p
�= 10
�3p
2+1
2
�+ 16
67
and solving for p yields p� =p7=15. Thus, the allocations when agent 2 is a price setter
are:
x�2 =5
2� 12
r15
7, y�2 =
7
2� 32
r7
15
x�1 =3
2� 12
r15
7, y�1 =
1
2� 32
r7
15
and notice that agent 2 is better o¤ as price setter (PS) since:
u2 (WE) = u2 (�x2; �y2) = 2� 2 = 4
u2 (PS) = u2 (x�2; y
�2) =
5
2� 12
r15
7
!� 7
2� 32
r7
15
!> 4
however, this allocation is not WPO; to see why recall that at WPO allocations both agentsmust have equal MRS but in this case:
@u1@x1@u1@y1
�����(x�1;y�1)
=
r7
15and
@u1@x1@u1@y1
�����(x�2;y�2)
>
r7
15
Example 80 (imperfect competition: Walras is not Nash) Consider an industry withdemand curve P = 2�Q in which every �rm has the same cost structure Ci (Q) = Qi: Themarket output is Q =
PiQi and the associated market clearing price is P = 2 �
PiQi.
Consequently, �rm i�s pro�t is given by:
�i (Q) =
"1�
Xi
Qi
#Qi
in a competitive world, the industry price would equal the marginal cost:
P =MC ) P = @Ci (Q) =@Qi = 1
and therefore pro�ts would be zero. The competitive equilibrium allocation is symmetricand is found by solving:
P = 2�Q)1 = 2� nQ�i
) Q�i =1
n
However, the pro�le Q�=(Q�1; Q�2; :::Q
�n) 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 competitiveequilibrium production is given by:
Qi = argmaxQi�0
[1� (n� 1)Q�i �Qi]Qi
68
which given that Q�i =1ncan be found to be Qi = 1
2n6= Q�i from where we conclude that
Q�=(Q�1; Q�2; :::Q
�n) is not a Nash equilibrium. In fact, we know that the NEq is the solution
to the Cournot model which is given by:
�Qi = argmaxQi�0
"1�
Xi
Qi
#Qi
with equilibrium pro�le �Qi = 1n+1
for each i and market price given by P = 2+n1+n
:
69
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 relationon 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 functionthat satis�es the expected utility hypothesis (EUH):
De�nition 91 u : R ! R satis�es the EUH if 9 V s.t. V (F ) =Ru(x)f(x)dx where x
is a realization of (continuous) RV X, X � F (�); f (x) = dF (x) is a Riemann integrabledensity and F 2 D the distribution space.
Remark 57 If X is a discrete RV we usePinstead of
Rin the above de�nition.
Theorem 81 u is a Von-Neumann-Morgenstain (vNM) utility function , u satis�es theEUH.
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)] :
De�nition 92 Continuity axiom: If Fnw! F and Fn % G ) F % G
De�nition 93 % on D satis�es the independence axiom if the preference order of twodistributions remains unaltered by convex combinations of each of the two with a thirddistribution, 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 com-modity space:
70
Proposition 30 If % on D satis�es order, continuity and independence axioms, then 9V : D ! R such that:
F % G, V (F ) � V (G) (5.1)
If t 2 (0; 1) and F;G 2 D, then:V (tF + (1� t)G) = tV (F ) + (1� t)G (5.2)
9 u : R! R that satis�es the EUH (5.3)
We can also de�ne this preference relation on the space of random variables. In partic-ular, 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 � Y ), E(u(X)) � E(u(Y ))
5.2 Risk aversion
De�nition 94 (risk aversion I) An agent is risk averse if E [u (x)] � u (E [x]) orRu(x)f(x)dx �
u�R(x)f(x)dx
�De�nition 95 (risk aversion II) An agent is risk averse if FftX+(1�t)Y g % tFfXg + (1�t)FfY g
De�nition 96 (risk aversion III) An agent is risk-averse if u (�x+ (1� �)y) > �u (x)+(1� �)u (y) i.e., if his utility function is concave (u0 > 0 and u00 < 0).
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 represen-tation for % then:
V�FftX+(1�t)Y g
�= u (tx+ (1� t)y) (no uncertainty)
V�tFfXg + (1� t)FfY g
�= tV
�FfXg
�+ (1� t)V
�FfY g
�De�nition 97 Suppose that u : R ! R is a Von Neumann�Morgenstern utility functionsatisfying u0 (�) > 0 then, the Arrow-Pratt measure of absolute risk aversion (ARA) at �x isde�ned as:
� (�x) =�u00 (�x)u0 (�x)
and the Arrow-Pratt measure of relative risk aversion (RRA) is de�ned as:
� (�x) �x =�u00 (�x)u0 (�x)
�x
71
Example 82 The utility function u(x) = �e�kx exhibits constant ARA with RA coe¢ cient�(x) = k:
Remark 60 Notice that if ' (�) is increasing and (�) non-decreasing then z 7! (' (z))is non-decreasing.
5.2.1 Application: portfolio choice
Let the cannonical portfolio choice problem with one risky (Y ), one riskless asset (r) andinitial wealth w be de�ned as:
maxx
f (x) = maxx
E [u (xY + (w � x)r)] (5.4)
s:t: 0 � x � 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 = argmax
xE [u (xY + (w � x)r)] ;
then:
1) : � (�) non-decreasing ) d�x(w)
dw� 0
2) : � (�) non-increasing ) d�x(w)
dw� 0
3) : � (�) constant) d�x(w)
dw= 0
Proof. We prove 1) since 2)-3) follow inmediately. First note that 0 < �x (w) < w ) theconstraint does not bind and we have an interior solution (i.e. in any KT system �x > 0and �� = 0). Thus f 0 (�x) = 0 8 w 2 (0; w�) and �x is a function of w, i.e., �x = �x (w). Henceby the F.O.C.:
E [u0 (rw + �x(Y � r)) (Y � r)] = 0
di¤erentiating w.r.t. w :
E
�u00 (rw + �x(Y � r)) (Y � r)
�r +
d�x
dw(Y � r)
��= 0
rE fu00 (rw + �x(Y � r)) (Y � r)g�E [u00 (rw + �x(Y � r)) (Y � r)2]
=d�x
dw
now, to sign this expression notice that the denominator is positive since u00 (�) < 0 andnecessarily (Y � r)2 > 0: Hence we want to sign the numerator so let:
' (y) = rw + �x(y � r)
72
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 ofnon decreasing functions:
[� (' (y))� � (' (r))] [y � r] � 0
� (' (y)) (y � r) � � (' (r)) (y � r)
and note that ' (r) = rw so that:
�u00 (' (y))u0 (' (y))
(y � r) � �u00 (rw)u0 (rw)
(y � r)
u00 (rw + �x(y � r)) (y � r) � u00 (rw)
u0 (rw)u0 (' (y)) (y � r)
E [u00 (rw + �x(Y � r)) (Y � r)] � u00 (rw)
u0 (rw)E [u0 (' (Y )) (Y � r)]| {z }
=0 by FOC above
E [u00 (rw + �x(Y � r)) (Y � r)] � 0
and therefore d�xdw� 0.
Example 83 Suppose that u (x) = �e�kx. Then E [u (xY + (w � x)r)] = E [� exp (�k (xY + (w � x)r))]which can be written � exp (�kwr)E [� exp (�kx (Y � r))] so that w is out of the expec-tation and therefore d�x(w)
dw= 0:
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, supposethat u00 (�) < 0: If �x = argmax
xE [u (xY + (w � x)r)] ; then:
1) : z 7! � (z) z non-decreasing ) d
dw
��x(w)
w
�� 0
2) : z 7! � (z) z non-increasing ) d
dw
��x(w)
w
�� 0
3) : z 7! � (z) z constant) d
dw
��x(w)
w
�= 0
Proof. Let � (w) = �x(w)wand proceed as before:
E [u0 (rw + �x(Y � r)) (Y � r)] = 0
Ehu0�w�r +
�x
w(Y � r)
��(Y � r)
i= 0
E fu0 [wr + w� (w) (Y � r)] (Y � r)g = 0
73
di¤erentiate w.r.t. w :
E�u00 (w (r + � (w) (Y � r))) (Y � r)
�r + � (w) (Y � r) +
d� (w)
dw(Y � r)w
��= 0
E fu00 (wr + w� (w) (Y � r)) [r + � (w) (Y � r)] (Y � r)g�wE [u0 (wr + w� (w) (Y � r)) (Y � r)2]
=d�
dw
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 andy 7! ' (y) strictly increasing to conclude that z 7! (' (z)) and thus:
[ (w' (y))� (w' (r))] [y � r] � 0
so that:
�u00 (w' (y))w' (y)
u0 (w' (y))(y � r) � (w' (r)) (y � r)
E [u00 (w' (y))w' (y) (y � r)] � � (w' (r))E [u0 (w' (y)) (y � r)]| {z }=0 by FOC above
E [u00 (w' (y))w' (y) (y � r)] � 0
and therefore d�dw� 0) d
dw
��x(w)w
�� 0:
Example 84 (CRRA) Suppose that u (x) = x1�� then E [u (xY + (w � x)r)] = E�(xY + (w � x)r)1��
�which can be written as E
h�w�r + x
w(Y � r)
��1��i= w1��E
h�r + x
w(Y � r)
�1��iand
again, ddw
��x(w)w
�= 0:
5.3 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 withu0i (�) > 0. Then the following are equivalent:
�u002 (�x)u02 (�x)
� �u001 (�x)u01 (�x)
(A)
9 a C2 concave, increasing (�) s.t. u2 = � u1 (B)
74
Proof. (B)) (A) : Suppose u2 (x) = (u1 (x)) 8x with 0 (�) > 0 and 00 (�) < 0. Then:
u02 (x) = 0 (u1 (x))u01 (x)
u002 (x) = 00 (u1 (x)) [u01 (x)]
2+ 0 (u1 (x))u
001 (x)
dividing these two equations:
u002 (x)
u02 (x)� u001 (x)
u01 (x)=
00 (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 = u�11 (y). Hence we wantto show:
u2�u�11 (y)
�=
�u1�u�11 (y)
�| {z }�
y
= (y)
we can claim that 0 > 0 and 00 < 0. To see this note:
0 (y) = u02�u�11 (y)
� ddy
�u�11 (y)
�and since:
y = u1�u�11 (y)
�) dy = u01
�u�11 (y)
�d�u�11 (y)
�)
d�u�11 (y)
�dy
=1
u01�u�11 (y)
�(and bear in mind that u01 > 0)
d(u�11 (y))dy
> 0) so replacing:
0 (y) =u02�u�11 (y)
�u01�u�11 (y)
� > 0since u0i (�) > 0. Next, let ' (z) =
u02(z)u01(z)
so that:
'0 (z) =u002 (z)u
01 (z)� u001 (z)u
02 (z)
[u01 (z)]2 =
u02 (z)
u01 (z)
�u002 (z)
u02 (z)� u001 (z)
u01 (z)
�� 0
�nally note that:
0 (y) = '�u�11 (y)
� 00 (y) = '0
�u�11 (y)
�| {z }�0
d
dy
�u�11 (y)
�| {z }
>0
so that 00 � 0:
75
5.3.1 Application: portfolio choice
Proposition 33 Suppose that two agents with identical wealth face the same problem as(5.4). That is:
maxx
fi (x) = maxxE [ui (xY + (w � x)r)] (5.5)
s:t: 0 � x � w
for i = 1; 2 with u0 > 0 and u00 < 0. If 0 < �xi < w for each i, then �1 (x) � �2 (x) )�x1 � �x2:
Proof. Since fi (x) is concave for each i it su¢ ces to show that f 02 (�x1) � 0. To see this,note:
f 02 (�x1) = E f 0 [u1 (rw + �x1 (Y � r))]u01 (rw + �x1 (Y � r)) (Y � r)gnow, since 00 < 0 and u0 > 0 it follows that y 7! 0 [u1 (rw + �x1 (Y � r))] is nonincreasing.Thus:
f 0 [u1 (rw + �x1 (y � r))]� 0 [u1 (rw)]g (y � r) � 0so:
0 [u1 (rw + �x1 (y � r))] (y � r) � 0 [u1 (rw)] (y � r)
now multiply both sides by u01 (rw + �x1 (Y � r)):
0 [u1 (rw + �x1 (y � r))] (y � r)u01 (rw + �x1 (Y � r)) � 0 [u1 (rw)] (y � r)u01 (rw + �x1 (Y � r))
then take expectations:
E f 0 [u1 (rw + �x1 (y � r))] (y � r)u01 (rw + �x1 (Y � r))g � E f 0 [u1 (rw)] (y � r)u01 (rw + �x1 (Y � r))g
and notice that E f(y � r)u01 (rw + �x1 (Y � r))g = f 01 (�x1) = 0 by the FOC from agent 1�sproblem. Therefore:
E f 0 [u1 (rw + �x1 (y � r))] (y � r)u01 (rw + �x1 (Y � r))g � 0
f 02 (�x1) � 0
which given that f is concave, implies that �x1 � �x2:
5.3.2 Application: insurance
Theorem 86 (Jensen�s inequality) If f is convex and Y is a random variable, thenE (f (Y )) � f (E (Y )) that is:Z
f (y) g (y) dy � f
�Zg (y) dy
�where g is the (Riemann integrable) density of RV Y:
76
Suppose that agents are endowed with wealth w in period t and in period t + 1 facethe occurence of a damage of amount d with probability p: The expected wealth withoutinsurance is:
p (w � d) + (1� p)w = w � pd
On the other hand, if the agent buys (full coverage) insurance at a premium �; his expectedwealth 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 callthe vNM representation of u) the cannonical insurance problem of agent i is thereforeformulated as:
max0��i�1
�i (w � �i) = pui (w � d)� (1� p)ui (w)
Theorem 87 (Pratt; Econometrica, 1964) Suppose that u1 (�) and u2 (�) are C2 withu0i (�) > 0 and u00i (�) < 0. If �i is the maximum insurance premium that agent i is willingto pay and the loss-inducing RV X has E [X] = 0, then:
�u002 (�x)u02 (�x)
� �u001 (�x)u01 (�x)
, �2 (x) � �1 (x)
Proof. ()) Since �u002 (�x)
u02(�x)� �u001 (�x)
u01(�x), by theorem (85) we know 9 (�) increasing and concave
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 � t � 1 and de�ne X = ty; note thatE [Y ] = 0) E [X] = 0. Next, let:
Gi (t; �) = ui (w � �)� E (ui (tY + w))
and note Gi (0; 0) = 0. Furthermore:
@Gi@�
����(0;0)
= �u0i (w � �)j�=0 = �u0i (w) 6= 0
77
From the implicit function theorem 9 �i : B" (0)! R s.t. �i is C2 and Gi (t; �i (t)) = 0 8t 2 B" (0). A sedcond-order Taylor approx. around zero yields:
�i (t) = �i (0) + �0i (0) t+t2
2�00i (Ci (t))
where Ci (t) 2 (0; t) 8 t 2 B" (0). Next, note that ui (w � �i (0)) = E [ui (w � 0)] =E [ui (w)] = ui (w). Moreover, ui (w � �i (0)) = E [ui (w �X)] = E [ui (w � tY )] so that:
u0i (w � �i (t)) [��0i (t)] = E [u0i (w � tY )Y ]
u00i (w � �i (t)) [��0i (t)]2+ u0i (w � �i (t)) [��00i (t)] = E
�u00i (w � tY )Y 2
�evaluating each of these expressions at zero:
u0i (w) [��0i (0)] = u0i (w)E [Y ]| {z }=0
u0i (w) [��0i (0)] = 0) �0i (0) = 0
and:
u0i (w) [��00i (0)] = u00i (w)E�Y 2�
�00i (0) =�u00i (w)u0i (w)
E�Y 2�
now, replacing �0i (0) = 0 and �i (0) = 0 in the Taylor expansion:
�i (t) =t2
2�00i (Ci (t))) lim
t!0�i (t) = lim
t!0
t2
2�00i (Ci (t))
but from �2 (t) � �1 (t) :
�2 (t) � �1 (t)
t2
2�002 (C2 (t)) � t2
2�001 (C1 (t))
limt!0
t2
2�002 (C2 (t)) � lim
t!0
t2
2�001 (C1 (t))
�002 (0) � �001 (0)
�u002 (w)u02 (w)
E�Y 2�� �u001 (w)
u01 (w)E�Y 2�
and therefore we conclude that �u002 (x)
u02(x)� �u001 (x)
u01(x), i.e., 2 is more risk averse than 1.
78
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, RVX 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 �rstorder stochastic dominance.
Remark 61 X (!) � Y (!)) FX (t) � FY (t) but the converse is false.
Example 88 To ilustrate the last remark, let X � [0; 1] and Y = 1 � X � [0; 1] thenFX (t) = FY (t) but X (!) 6= Y (!) :
De�nition 99 The RV X dominates RV Y in the First-order stochastic sense denotedFX %1 FY if FX (t) � FY (t) 8 t 2 R:
Remark 62 Note that %1is a preference relation on D which is transitive and re�exivebut not complete.
Proposition 34 (FOSD equivalence I) The following are equivalent:
FX %1 FY (5.6)
9 (;A; P ) and X; Y such that: (5.7)
Xd= F , Y d
= G and
X (!) � Y (!) 8 ! 2
Theorem 89 (FOSD equivalence II) The following are also equivalent:
X % 1Y (5.8)
E [u (X)] � E [u (Y )] (5.9)
whenever u (�) is non-decreasing
Proof. Assume that Y d= G and X d
= F and that Y;X have Riemann integrable densitiesdenoted f and g: Moreover, assume that 8 x =2 [a; b] ) f (x) = 0 = g (x) which impliesthat 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 nondecreasingand C1. Then:
E [u (X)] =Z b
a
u (t) f (t) dt
79
using integration by parts (recall f = dF ):
E [u (X)] = [u (t)F (t)]ba �Z b
a
u0 (t)F (t) dt
= u (b)�Z b
a
u0 (t)F (t) dt
similarly:
E [u (Y )] = [u (t)G (t)]ba �Z b
a
u0 (t)G (t) dt
= u (b)�Z b
a
u0 (t)G (t) dt
so that:
E [u (X)]� E [u (Y )] =Z b
a
u0 (t)G (t) dt�Z b
a
u0 (t)F (t) dt
=
Z b
a
u0 (t) [G (t)� F (t)]| {z }�0 since X%1Y
dt
) E [u (X)] � E [u (Y )]
(() Now suppose that E [u (X)] � E [u (Y )] whenever u is nondecreasing and expectationsexist. 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 � t0; if x < t
and note:
E [u (X)] =Z b
a
u (s) f (s) ds
=
Z t
a
u (s) f (s) ds| {z }=0
+
Z b
t
u (s) f (s) ds
=
Z b
t
f (s) ds = F (b)� F (t)
= 1� F (t)
an identical argument establishes that:
E [u (Y )] = 1�G (t)
80
so that:
E [u (X)] � E [u (Y )]) F (t) � G (t)
) X %1 Y
Example 90 Suppose that:
F (x) =
8<:1; if x � 1x; if 1
2� x < 1
0; if x < 12
G (x) =
8<:1; if x � 1x; if 0 � x < 10; 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,de�ne Y � U [0; 1] and let X = max f1=2; Y g. Then Y d
= G and X d= F . Now, notice that
dG(t) = 1 so that E (u (Y )) =R 10u (t) dt so:
E [u (X)] = E�u
�max
�1
2; Y
���=
Z 1
0
u
�max
�1
2; t
��dt
=
Z 12
0
u
�max
�1
2; t
��dt+
Z 1
12
u
�max
�1
2; t
��dt
=
Z 12
0
u
�1
2
�dt+
Z 1
12
u (t) dt = u
�1
2
�Z 12
0
dt+
Z 1
12
u (t) dt
= u
�1
2
�[t ]
120 +
Z 1
12
u (t) dt = u
�1
2
�1
2+
Z 1
12
u (t) dt
�Z 1
2
0
u (t) dt+
Z 1
12
u (t) dt =
Z 1
0
u (t) dt
= 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 inperiod 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 problemfaced becomes:
max0�x�w
u (w � x) + E [u (xY )]
Suppose that u0 > 0 and u00 < 0: Now, if we want to compare the amount invested in tworisky assets with gross returns Y1 and Y2 by the same individual we want to compare the
81
solutions to the problems:
max0�x�w
f1 (x) = u (w � x) + E [u (xY1)]
max0�x�w
f2 (x) = u (w � x) + E [u (xY2)]
Remark 63 Notice that the objective function fi (x) is concave. To se why, note:
f 00i (x) = u00 (w � x) + E�u00 (xY1)Y
21
�� 0 since u00 < 0
Now, suppose that we have an interior solution to each of these problems, i.e.:
�x1 2 arg max0�x�w
f1 (x)
�x2 2 arg max0�x�w
f2 (x)
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 isthe 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)
@x= u0 (xy) y
@2� (x; y)
@x@y= u0 (xy) + u00 (xy) yx
and notice that RRA < 1 implies that for each z:��u00(z)u0(z)
z
�< 1
) �u00(z)z < u0(z)
) 0 < u0(z) + u00(z)z
so that:@2� (x; y)
@x@y= u0 (xy) + u00 (xy) yx > 0
82
which in turn implies that y 7! u0 (xy) y is an increasing function.step 2: Note that since �xi 2 argmax0�x�w fi (x) (and is interior) it follows that:
f 01 (�x1) = 0
= �u0 (w � x) + E [u0 (�x1Y1)Y1]� �u0 (w � x) + E [u0 (�x1Y2)Y2]= f 02 (�x1)
thus, since fi is concave f 02 (�x1) � f 02 (�x2)) �x1 � �x2:
Remark 64 This problem ilustrates the use of supermodularity of single crossing propertyintroduced 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 riskyassets. We want to study if agents invest more in risky assets that dominate (in returns)other risky assets. The problem is therefore:
maxx
fi (x) = maxxE [u (wr + x (Yi � r))] (5.10)
s:t: 0 � x � 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)
@x= u0 (wr + x (y � r)) (y � r)
@2� (x; y)
@x@y= u0 (wr + x (y � r)) + u00 (wr + x (y � r)) (y � r)x
now, from the assumption of RRA < 1:��u00(z)u0(z)
z
�< 1
�u00 (wr + x (y � r)) [wr + x (y � r)]
u0 (wr + x (y � r))< 1
�u00 (wr + x (y � r))wr < u0 (wr + x (y � r))u00 (wr + x (y � r)) (y � r)x
83
now, u00 < 0) �u00 (wr + x (y � r)) > 0) �u00 (wr + x (y � r))wr > 0. Thus:
0 < �u00 (wr + x (y � r))wr
< u0 (wr + x (y � r))u00 (wr + x (y � r)) (y � r)x
so that @2�(x;y)@x@y
� 0 and we conclude that y 7! u0 (wr + x (y � r)) (y � r) is increasing.Now, since �xi 2 argmax0�x�w fi (x) (and is interior) it follows that:
f 01 (�x1) = 0
= E [u0 (wr + x (Y1 � r)) (Y1 � r)]
� E [u0 (wr + x (Y2 � r)) (Y2 � r)]
= f 02 (�x1)
as in the last application, since fi is concave f 02 (�x1) � f 02 (�x2)) �x1 � �x2:
5.5 Likelihood ratio stochastic dominance
De�nition 100 Suppose that X and Y are RV with Riemann integrable (positive) densities
f and g (i.e., X d= F and Y d
= G), we say that X dominates Y in the likelihood ratiostochastic 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 saythat X %LR Y if:
[f (s) g (t)� f (t) g (s)] (t� s) � 0or:
f (s)
g (s)(t� s) � f (t)
g (t)(t� 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
s
f (s) g (t) dt �Z 1
s
f (t) g (s) dt
f (s) [1�G (s)] � g (s) [1� F (s)] 8 s 2 R
84
now integrate the same expression over s on both sides:Z 1
t
f (s) g (t) ds �Z 1
t
f (t) g (s) ds
g (t)F (t) � f (t)G (t) 8 t 2 R
so that:F (x)
G (x)� f (x)
g (x)� 1� F (x)
1�G (x)
and we conclude that F (x) � G (x) so that X %1 Y:
Example 93 (portfolio choice) Come back to the problem of portfolio choice with two
risky assets introduced in section 5.4.2: Suppose that Y1d= F and Y2
d= G:Then if:
�xi 2 arg max0�x�w
fi (x)
0 < �xi < w
it follows that:
f 01 (�x2) = E [u0 (wr + �x2 (Y1 � r)) (Y1 � r)]
=
Z[u0 (wr + �x2 (t� r)) (t� r)] f (t) dt
=
Z �u0 (wr + �x2 (t� r))
�(t� r)
�f (t)
g (t)
��g (t)
�dt
�Z �
u0 (wr + �x2 (t� r))
�(t� r)
�f (r)
g (r)
��g (t)
�dt
=f (r)
g (r)
Zfu0 (wr + �x2 (t� r)) (t� r) g (t)g dt
= E [u0 (wr + �x2 (Y2 � r)) (Y2 � r)]
= f 02 (�x2) = 0
therefore, f 01 (�x2) � 0 and since fi is concave we conclude that �x2 � �x1:
5.6 Concave and second order stochastic dominance(SOSD)
The �rst and likelihood ratio orders are preference relations over distributions according toexpected returns. On the other hand, we can de�ne preference relations over distributionsaccording 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 (5.11)
E (X) = E (Y ) andZ t
�1F (s) ds �
Z t
�1G (s) ds (5.12)
and:
The following are equivalent:
X %2 Y (5.13)Z t
�1F (s) ds �
Z t
�1G (s) ds (5.14)
Proof. Assume that Y d= G and X d
= F; and that Y;X have Riemann integrable densitiesdenoted f and g: Moreover, assume that 8 x =2 [a; b] ) f (x) = 0 = g (x) which impliesthat F (a) = 0 = G (a) and F (b) = 1 = G (b) :(5:11) 5:12 and 5:13) 5:14) Case 1: suppose that a � t. Then f (t) = 0 = g (t) andF (t) = 0 = G (t). Thus, trivially
R t�1 F (s) ds �
R t�1G (s) ds.
Case 2: Suppose that b � t. Then:Z t
�1F (s) ds =
Z t
a
F (s) ds
=
Z b
a
F (s) ds+
Z t
b
F (s) ds
now, since t � b ) F (t) = 1 thenR tbF (s) ds = t � b. Next integrating by parts the �rst
expression on the RHS: Z b
a
F (s) ds = [sF (s)]ba �Z b
a
sF (s)
= b� E (X)
so that: Z t
�1F (s) ds = b� E (X) + t� b
= t� E (X)
a similar argument establishes that:Z t
�1G (s) ds = t� E (Y )
86
Therefore: Z t
�1G (s) ds�
Z t
�1F (s) ds = E (X)� E (Y )
thus, if X %co Y then E (X) = E (Y ) andR t�1G (s) ds =
R t�1 F (s) ds: If, on the other
hand, X %2 Y then E (X) � E (Y ) and thusR t�1G (s) ds �
R t�1 F (s) ds:
Case 3: Suppose that a < t < b. Then consider:
u (t) =
�t; if x � tx; if x < t
then:
E [u (X)] =
Z 1
�1u (x) f (x) dx =
Z b
a
u (x) f (x) dx
=
Z t
a
u (x) f (x) dx+
Z b
t
u (x) f (x) dx
=
Z t
a
xf (x) dx+
Z b
t
tf (x) dx
= [x� F (x)]t0 �Z t
a
F (x) dx+ t (F (b)� F (t))
= tF (x)�Z t
a
F (x) dx+ t� tF (t)
= t�Z t
a
F (x) dx
and a similar argument establishes that:
E [u (Y )] = t�Z t
a
G (x) dx
now if X %co Y then u concave implies that E [u (X)] � E [u (Y )] which in turn implies:Z t
a
F (x) dx �Z t
a
G (x) dx
and if X %2 Y then u nondecreasing, concave implies again that E [u (X)] � E [u (Y )].(5:12) 5:11 and 5:14) 5:13) Assume that u is C2 and note that:
E [u (X)] =
Z 1
�1u (t) f (t) dt =
Z b
a
u (t) f (t) dt
= [u (t)F (t)]ba �Z b
a
u0 (t)F (t) dt
= u (b)�(��
u0 (t)
Z t
a
F (s) ds
��ba
�Z b
a
u00 (t)
�Z t
a
F (s) ds
�dt
)= u (b)� u0 (b) []
87
De�nition 101 G is a mean-preserving spread of F , y = x+", with E ["jX] = 0; X � Fand Y � G:
Lemma 16 If X � F and Y � G and �G = �F then X %2 Y , G is a mean-preservingspread of F:
5.6.1 Concave order SD and pro�t maximization
Suppose that a multiproduct �rm uses a single input (labor) to produce n di¤erent com-modities (y1; :::yn). Suppose that xi is the amlount of labor used in the production ofcommodity 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:
yi � (xi)1��i 8 i = 1; :::; n
with 0 < �i < 1; then the �rm�s pro�t maximization problem is:
maxxi
nXi=1
pi (xi)1��i �
nXi=1
xi
with associated optimal production plan:
�xi = ((1� �i) pi)1�i
and pro�t function � (p):
� (p1; :::; pn) =nXi=1
pi [pi (1� �i)]1��i�i �
nXi=1
((1� �i) pi)1�i
Now suppose that the p vector varies with the plant locations. In particular suppose that inlocation 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�tfunction). Hence, pi 7! �� (p1; :::; pn) is concave for each i. Therefore, Xi %co Yi and�� (�) concave imply:
E (�� (X1; :::; Xn)) � E (�� (Y1; :::; Yn))E (� (X1; :::; Xn)) � E (� (Y1; :::; Yn))
E���pA��
� 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)] � E [u (Y )] E [u (X)] � E [u (Y )] E [u (X)] � E [u (Y )]
in terms of DF F (t) � G (t)R t�1 F (s) ds �
R t�1G (s) ds
R t�1 F (s) ds �
R t�1G (s) ds
89
Appendix A
Review of functions, di¤erentiationand integration
1. After t periods of continuously compounding growth rate r per period, quantity A isAert:
e � limn!1
(1 +1
n)n and in general; ek = lim
n!1(1 +
k
n)n
2. eix = cos(x) + i sin(x), where i =p�1
3. Every polynomial is a continuous function. Since its derivative is also a polynomialof one-less degree, it is also continuous. Hence, every polynomial is C1.
4. Recall that lnx = y , ey = x; elnx = x and ln ex = x:
5. Recall that�ddxeg(x)
�= g0 (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 errorof such approximation tends to zero as one approaches �x from any direction, and insuch case, g(�) is the derivative. That is:
limx!0
f(�x+ x)� f(�x)� g(x)
kxk = 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) functionand dx a small vector.
8. Two functions f; g are tangent if limx!0f(�x+x)�h(�x+x)
kxk = 0
9. The partial derivative of f is de�ned: limt!0h(�x+t)�h(�x)
twhile the directional derivative
is ~Dxh(�x) = limt!0h(�x+tx)�h(�x)
t: In fact, the partial derivative w.r.t. xi is simply
90
the directional derivative in the direction paralell to the xi axis, that is:@F (�x)�xi
=
rF (�x) � e1 = ~De1h(�x) where e1 = (0; 0; ::i; :::; 0)
10. Rolle�s Theorem : If f : [a; b]! R1 is C1 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 ifa; 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)� f(a)
(b� a)(A.2)
12. Inverse Function Theorem in R1. Let f be C1 on I 2 R1. If f 0(x) 6= 0 8 x in I,then: (i) f is invertible, (ii) Its inverse, g is C1 on f(I) and (iii) 8z in the domainof g, g0(z) = 1=f 0(g(z)).
13. Implicit Function Theorem in R2: Given the implicit function G(x; y) = c then:
y0(x0) = �@G(x0; y0)=@x
@G(x0; y0)=@y(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) =tkf(x1; :::; xn). If k = 1; k > 1 and k < 1, f has constant, increasing and decreasingreturns 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 Theo-rem).
18. If f is Cn on an open interval I containing �, given any x 2 I (Taylor�s theorem) :
f(x) = f(�)+1
1!(x��)f 0(�)+ 1
2!(x��)2f 00(�)+:::+ 1
(n� 1)!(x��)n�1f (n�1)(�)+En; with
En =1
n!(x� �)nfn(�)
19. Let f be continuous on the compact (i.e. bounded and closed) I = [a; b], then if P issome partition of [a; b] and S(P ) the sum of areas for partition P , we de�ne:Z b
a
f(x)dx = supPS(P )
91
20. Let F be the anti-derivative of f , then for some interval [a; b]:
F (b)� F (a) =
Z b
a
f(x)dx = lim�!0
NXi=1
f(xi)�
for � = (b� a)=N . The last term (Riemann sum) closes the Fundamental Theoremof Calculus.
21. A (di¤erentiable) function f : Rn ! R1 has gradient ; i:e: : a vector of n �rst-orderpartial derivatives.
22. A (di¤erentiable) function f : Rn ! Rm has Jacobian; i:e: : a matrix of m vectorseach with n entries of �rst-order partial derivatives.
23. Some useful integrals:Zxndx =
xn�1
n+ 1+ C
Z1
xdx = ln x+ C
Zexdx = ex + C
Z �ax2 + bx+ c
�=ax3
3+bx2
2+ cx
Z(f(x))nf 0(x)dx =
1
n+ 1(f(x))n+1 + CZ b
a
u(x)v0(x)dx = [u(x)v(x)]ba �Zu0(x)v(x)dx (integration by parts)
92
Appendix B
Review of vectors and matrix algebra
1. Rank : If A is m�n, then rkA = rkA0 � minfm;ng, where A0 is RREF of A. OJO:If rkA = m = n, A0 = I
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 � Rm is a subspace of Rm if �x + �y 2 S whenever x; y 2 S and�; �;2 R
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) � R(B); N(A) � N(BA).
8. For any Am�n; R(A) = Rm , rkA = m.
9. For any Am�n; N(A) = 0, rkA = n.
10. Span: Let S (subspace of Rm); a collection of vectors fa1; :::; ang in S spans S ifeach x 2 S is a linear combination of the members of fa1; :::; ang.
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; :::; ang inRm is linearly independent if 8 �1; :::�n 2R:nPi=1
�iai = 0 only for �1 = � � � = �n = 0.
13. Basis: If S is subspace of Rm, a set fa1; :::; ang � S is a basis for S if (i) it spans Sand (ii) it is linearly independent.
14. Furthermore, if A = [a1... : : :
...an]; R(A) = S and rkA = n (i.e., A is Full ColumnRank).
93
15. Dimension: (dimS) is number of elements in a basis for S.
16. Let S (subspace of Rm) have basis fa1; :::; ang. If fb1; :::; brg is linearly indep., thenr � n.
Corollary: Any m+ 1 vectors in Rm are linearly dependent.Corollary: If fa1; :::; ang and fb1; :::; brg are basis for S, then n = r.
17. Fundamental Theorem of Linear Algebra (I): For any Am�n:
i) dim R(AT ) = rkAii) 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�1g? = 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 =MT . A matrix is called Idempotent if MM =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) = cTx for c 2 Rn; rf = c. If f(x) = xTAx;rf =(A+ AT )x(ii) If A is m�n and rkA = n, then ATA is nonsingular (i.e., is n�n) and has rankn.
Given x 2 S = R(A) and y 2 Rm, the projection problem is:
minu2SpPn
i=1 (yi � ui)2 = kyuk = d(y; u) =p(y � u)T (y � u) =, which is equiva-
lent to minx2Rn(y � Ax) � (y � Ax) (recall minpx2 , min x2).
The f:o:c: being�2ATy+2ATAx = 0; x = (ATA)�1ATy; or u = Ax = A(ATA)�1ATy =My, where M = A(ATA)�1AT is called the projection matrix for R(M) (M beingsymetric and idempotent).
24. For An�n; n � 2; detA =Pn
j=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 Awith 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�1b.
27. A diagonal matrix D with fd1; d2; :::; dng as diagonal has detD =Qni=1 di.
28. For An�n, r is eigenvalue of A if det(A� rI) = 0, i.e., A� rI is singular. Note thatAn�n has n eigenvalues which are solutions to the nth degree polynomial equation inr, 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¤erentcolumn vectors is 0 or orthogonal and the norm of each column vectos is 1). Notethat detP = �1.
32. Let S (subspace of Rm) have basis fa1; :::; ang. This basis is orthonormal if aiaj = 0whenever i 6= j and aiaj = 1 if i = j
Review of Vector and Matrix Operations
� The general quadratic form Q(x1; :::; xn) =P
i�j aijxixj (aijxixj is a 2nd degreemonomial) can be expresed as xTAx where A is unique symetric matrix whose ij thelement is 1
2aij
� Matrix A is positive semide�nite if xTAx � 0 whenever x 6= 0 and negative semidef-inite if xTAx � 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 detA > 0 and negative de�niteif a11 < 0 and detA > 0:
� A diagonal matrix A will be positive semide�nite if ai � 0 8 i (i.e. all diagonalelements � 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 whichsolve 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 = [x1x2:::xn]
T : Note eigenvectors are not unique.
� If P is a matrix whose columns are eigenvectors of A, then P�1AP = rIn
Let A =
0@ a11 a12
a21 a22
1A ; then
AA =
0BB@a211 + a12a21 a12(a11 + a22)
a21(a11 + a22) a222 + a21a12
1CCA ; ATA =
0BB@a211 + a212 a12a11 + a22a12
a21a11 + a22a12 a222 + a221
1CCAWhile for the (n� 1) vectors x and y:
� x � y = yTx = xTy =Pn
i=1 xiyi
� x � x = xTx = xTx =Pn
i=1 x2i
� Recall that ifA is a symmetric (comformable) matrix (x�y)TA(x�y) = (xTAx�2xTAy + yTAy)
� kxk = (x � x)1=2 =pPn
i=1 x2i and therefore x � x = kxk
2
� xkxk
= 1 (read x is normalized)� Cauchy-Schwarz Inequality for vectors jx � yj � kxk kyk
� x � x � 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)= kxk2
� 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
