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Concave Optimization for MicroeconomicTheory
Somdeb Lahiri
May 2003.
Preface
These lecture notes grew out of several years of teaching a course on AdvancedMicroeconomics, first at the Indian Institute of Management, Ahmedabad andsubsequently at the University of Witwatersrand, Johannesburg.When we were students at graduate schools in US (during the mid 1980’s), a typical firstyear graduate Microeconomics course, revolved around Hal Varian’s classic on thesubject: Microeconomic Analysis (first edition). The purpose of such a course was toprovide a bridge between a typical undergraduate Microeconomics course and GerardDebreu’s seminal contribution: Theory of Value. Most of advanced economic theorywhich culminated in research at the frontier, was until that point in time, inextricablybound to the existence and optimality results of the Arrow-Debreu general equilibriummodel and its immediate extensions. Thus, a course on graduate microeconomics such asthe one prescribed in Varian’s book, served an important need to make the transitionfrom undergraduate economic theory to advanced economic theory, less hostile for thestudents. As teachers of graduate microeconomics in the late 1980’s and early 1990’s,many of us (including myself) were ardent adherents of the cause espoused by Varian’sMicroeconomic Analysis.However, by the mid 1990’s it had become apparent, that research dealing with theexistence and optimality properties of competitive and related equilibria, was no longerthe sole pre-occupation of economic theory. Further, the volume of academic output oneach topic that was covered by a standard graduate level text book in Microeconomics,had become so enormous, that each topic started evolving as a separate course in its ownright (as for instance Information Economics), and sometimes (as in the case of gametheory) as a separate subject altogether. While teaching from standard graduate texts onMicroeconomics, I as an instructor, could not miss the feeling, that neither the studentsnor the instructor, were deriving very little by nibbling bits and pieces from sundrytopics. Such courses were contributing marginally more than what a good dictionary oneconomic theory would do, and by then there were several such dictionaries available inthe market. As an instructor, I was beginning to realize, that a first year graduatemicroeconomic theory course could survive and flourish, only if it centered on teachingstudents the tools of optimization theory along with a few robust applications.Optimization theory (: much more than fixed point theory) was the technique thatpervaded most topics discussed in microeconomics, and for most purposes, theoptimization theory that was required, involved maximizing “one or more” concaveobjective function (s) subject to linear constraints. This is the central message that I havetried to convey, in the lectures that follow.In preparing these lectures I have benefited immensely from the following sources:
(1) Concavity and Optimization in Microeconomics, by Paul Madden: BlackwellPublisher, 1986;
(2) Linear Programming and Economic Analysis, by Robert Dorfman, PaulSamuelson and Robert Solow: (The Rand Series), Mc Graw Hill, 1958;
(3) Resource Allocation Mechanisms, by Donald Campbell: Cambridge UniversityPress, 1987;
(4) Lectures on Mathematical Economics, by J.S. Lane: London School ofEconomics and Political Science, 2001;
(5) An Introduction to Mathematical Finance: Options and other Topics, by SheldonM. Ross : Cambridge University Press, 1999;
(6) Cooperative Microeconomics: A Game-Theoretic Introduction, by Herve Moulin:Prentice Hall (Harvester Wheatsheaf), 1995.
Chapter 9 of these lecture notes is concerned with the axiomatic analysis of the socialchoice rule which selects market equilibrium allocations corresponding to anexogenously given income distribution. This chapter which concerns the application ofconcave optimization within the framework of what is traditionally known as WalrasianSocial Choice Thory parallels results reported in:
(7) Walrasian social choice: Some simple axiomatic approaches, by Louis Gevers, publishedin: W. Heller et al., eds., Social Choice and Public Decision Making: Essays in Honor ofK.Arrow, Vol.1, pages 97-114: Cambridge University Press, London/New York, 1988;
and(8) A local independence condition for characterization of Walrasian allocations rule, by R.
Nagahisa, published in Journal of Economic Theory, Vol. 54, pages 106-123, 1991. However, our presentation of the axiomatic analysis differs slightly in its formalismsfrom these two papers.To be able to comprehend and do justice to the ensuing lecture notes, the reader shouldhave gone through a course on Advanced Calculus and some Linear Algebra.Ideally, the reader should be familiar with matter covered in an intermediatemicroeconomics course as well. While the mathematical requirements are a definiteprerequisite for these lectures, previous knowledge of intermediate microeconomics canin several instances be substituted by sufficient “mental maturity”, without doingconsiderable injustice to one’s understanding of what follows.While I would like to thank all my students, both past and present, whom I have had theprivilege of teaching Advanced Microeconomics, for making these “rough” lecture notespossible, at this point I would like to thank Gabor Szalontai, for pointing out numerouserrors in the text and Ralitza Dobreva and Eugene Peterson for fruitful participation in theclass room discussions. I would also like to thank Bruce Teubes, for raising someinteresting questions, which probably indicates that these lecture notes requireconsiderable polishing, before they can be deemed fit for unsupervised consumption.Diptesh Ghosh made very interesting critical observations about these notes andsuggested that the earlier version had too many “D’s”, representing several differentconcepts and thus leading to an inexcusable notational confusion. I have thus decided torepresent a convex domain by X in these lecture notes, in order to alleviate some miseryfor the reader. I would like to thank him and all my other students and colleagues, forextending their “social service” to me by way of time spent on these notes, and Isincerely hope that this list continues to grow longer with the passage of time.This Preface is my way of motivating the lectures that follow. Hence I would like tothank my wife, Rajyashree, for help in making this purpose linguistically clear andcorrect. Responsibility for errors that do remain, here as well as elsewhere, is obviouslymy own!
Contents
Preface
Part 1: The Theory of Concave Optimization
Chapter 1: Mathematical Preliminaries.Chapter 2: Concave Functions on Convex Domains.Chapter 3: Homogeneous Functions.Chapter 4: Concave Vector Optimization.
Part 2: Applications
Chapter 5: Elements of Consumer Choice Theory.Chapter 6: General Equilibrium and Welfare.Chapter 7: Arbitrage Pricing of Securities.
Part 3: The Way Ahead …
Chapter 8: Homogeneous Programming Games.Chapter 9: Market Equilibrium Allocations: Some Simple Axiomatics
Part 1
The Theory of Concave Optimization
Chapter 1
Mathematical PreliminariesLet ℜ denote the set of real numbers. ℜ+, ℜ++ denotes the set of non-negative and strictlypositive reals respectively. Given a non-empty subset Y of ℜ and any natural number n,let Yn denote the n-fold Cartesian product of Y. Let N denote the set of natural numbers.Any non-empty subset I of ℜ such that for all x,y ∈S and t a real number with 0 < t < 1,tx + (1-t)y ∈S, is called an interval. Thus, for a, b ∈ℜ with a ≤ b :(i) [ a, b] = {x∈ℜ / a ≤ x ≤ b}; and for a < b: (i) (a, b) = {x∈ℜ / a < x < b},(ii) (a, b] = {x∈ℜ / a < x ≤ b},(iii) [a, b) = {x∈ℜ / a ≤ x < b}, are all intervals. For a ∈ℜ:(i) [a, +∞) = {x∈ℜ / a ≤ x < + ∞}, (ii) (a, +∞) = {x∈ℜ / a < x < +∞},(iii) (-∞, a] = {x∈ℜ / -∞ < x ≤ a},(iv) (-∞, a)= {x∈ℜ / -∞ < x < a},(v) (-∞, ∞) = ℜ,are also intervals.In fact the above nine categories are the only types of intervals in ℜ.Given a, b ∈ℜ with a < b and any interval I of the form (a, b), (a, b], [a, b) or [a, b], theinterior of I, denoted int(I) is the interval (a, b). Given a ∈ℜ and any interval I of the form [a, +∞) or (a, +∞) its interior denoted int(I) isthe interval (a, +∞). If I is an interval of the form (-∞, a] or (-∞, a), then its interior is (-∞, a). The interior of ℜ is ℜ itself.The interior of [a, a] is the empty set.Given any interval I in ℜ and any x∈int(I), there exists δ > 0, such that (x-δ, x+δ) ⊂ I. Aninterval I is said to be open if I = int(I). By the definition of an interval, any open intervalis of necessity non-empty.
Given a natural number n a convex domain X is any subset of ℜn of the form ∏=
n
kkI
1
,
where each Ik, k= 1,…, n is an interval in ℜ. In particular a convex domain in ℜ is aninterval.
Given a convex domain X = ∏=
n
kkI
1
in ℜn, its interior denoted int(X) = ∏=
n
kkI
1
)int( .
A convex domain X in ℜn is said to be open if D = int(X).If x∈int(X), where X is a convex domain in ℜn, then there exists an open convex domainO in ℜn, such that x∈O ⊂ X. Given a convex domain X in ℜn, a continuous function f: X → ℜ is said to be concave iffor all x, y∈X and t∈(0,1): f(tx+(1-t)y) = f(y +t(x-y)) ≥ tf(x) + (1-t) f(y) = f(y) + t[ f(x) –f(y)].
Given an open convex domain X in ℜn, a continuous function f: X → ℜ is said to bestrictly concave if for all x, y∈X and t∈(0,1): f(tx+(1-t)y) > tf(x) + (1-t) f(y).Clearly, a strictly concave function is concave. However, the converse is not true ingeneral. For instance the function f: ℜ→ℜ defined by f(x) = x for all x∈ℜ is concave butnot strictly concave.Given a convex domain X in ℜn, a function f: X → ℜ is said to be convex if –f isconcave. Given an open convex domain X in ℜn, a function f: X → ℜ is said to bestrictly convex if -f is strictly concave.
Given a natural number n and i∈N= {1,…n}, let ei be the vector whose ..th
i coordinate is1 and all other coordinates are zero. Given a convex domain X in ℜn, a function f: X → ℜ, x∈int(X) and i∈N, let Dif(x) =
hxfhexf i
h
)()(lim0
−+→
, provided the limit on the right exists. If, Dif(x) exists for all i ∈N,
then the row vector Df(x) = (D1f(x),…, Dnf(x)) is called the gradient of f at x. f is said tobe continuously differentiable on int(X) if for i∈N, the function Dif : int(X) →ℜ iscontinuous.Suppose, f is continuously differentiable on int(X). Let x∈int(X) and i,j∈N. Let 2
, jiD f(x)
= h
xfDhexfD iji
h
)()(lim
0
−+→
, provided the limit on the right exists. If for all i,j∈N, the
function 2, jiD : int(X) → ℜ is well defined and continuous, then for all i,j∈N, 2
, jiD = 2,ijD .
Further, in such a situation f is said to be twice continuously differentiable on int(X) and
for x∈ int(X), the n×n matrix D2f(x) whose ..),(th
ji entry is 2, jiD f(x), is called the
Hessian of f at x.Given a convex domain X in ℜn, and a function f: X → ℜ:(i) if f is continuously differentiable in the interior of X, it is said to be C1 on int (X);(ii) if f is twice continuously differentiable in the interior of X, it is said to be C2 on
int (X).
Let f : X → ℜ be a function on a convex domain X and let g : I → ℜ be a function on aninterval I containing f(X). Then the function gοf : X → ℜ is defined as follows: for allx∈D, gοf(x) = g(f(x)).
Theorem 1.1: Let f : X → ℜ and g : X → ℜ be two concave functions on a convexdomain X and let a,b be non-negative real numbers. Then the function af + bg : X → ℜdefined by (af + bg)(x) = af(x) + bg(x) for all x∈X, is also concave. If in addition X isopen, g is strictly concave and b > 0, then af + bg is strictly concave.
Proof: Let f, g be concave and a,b be non-negative real numbers. Let x, y ∈X and t∈(0,1).Thus, (af + bg)(tx + (1-t)y) = af(tx + (1-t)y) + bg(tx + (1-t)y) ≥ a [ t f(x) + (1-t) f(y)] + b[ tg(x) + (1-t) g(y)] = t[a f(x) + bg(x)] + (1-t) [ af(y) + bg(y)] = t (af +bg)(x) + (1-t)(af+bg)(y). Thus, af + bg is concave.
Now suppose, g is strictly concave and b > 0. Let x, y ∈X with x ≠ y and t∈(0,1). Thus,(af + bg)(tx + (1-t)y) = af(tx + (1-t)y) + bg(tx + (1-t)y) > a [ t f(x) + (1-t) f(y)] + b[ t g(x)+ (1-t) g(y)] = t[a f(x) + bg(x)] + (1-t) [ af(y) + bg(y)] = t (af +bg)(x) + (1-t)(af +bg)(y).Thus, af + bg is strictly concave. Q.E.D.
We shall unless otherwise mentioned, treat vectors in ℜn as column vectors. Thetranspose of a vector x in ℜn will be denoted xT. Given two vectors x, y in ℜn we shall write:
(a) x ≥ y to denote xi ≥ yi for all i∈{1,…,n};(b) x > y to denote [x ≥ y & x ≠ y];(c) x > > y to denote xi > yi for all i∈{1,…,n}.
The same rule applies to xT and yT as well.
Chapter 2Concave functions on convex domains
Lemma 2.1: Let f : X → ℜ be a continuous function on a convex domain X and let g : I→ ℜ be a non-decreasing concave function on an interval I containing f(X).
(i) If f is concave, then so is g ο f.(ii) If g is increasing, X is open and f is strictly concave, then so is g ο f.
Proof : (i) Let x,y ∈X and t∈(0, 1). Thus, f(tx+(1-t)y ≥ t f(x) + (1-t) f(y). Hence,g(f(tx+(1-t)y) ≥ g(t f(x) + (1-t) f(y)) ≥ t g(f(x)) + (1-t) g(f(y)). Thus, gοf is concave.
(ii)Let x,y ∈X with x ≠ y and t∈(0, 1). Thus, f(tx+(1-t)y > t f(x) + (1-t) f(y).Hence, g(f(tx+(1-t)y) > g(t f(x) + (1-t) f(y)) ≥ t g(f(x)) + (1-t) g(f(y)). Thus, gοf is strictlyconcave. Q.E.D.
Theorem 2.1: Let D be a convex domain in ℜn with non-empty interior. Let f: X →ℜ bea continuous function. Suppose that the restriction of f to int(X) is concave. Then, f isconcave.
Proof: Let f: X →ℜ be a continuous function and suppose that the restriction of f toint(X) is concave. Let x, y∈X and t∈(0, 1). Thus, there exists sequences <xk/ k∈N> and<yk/ k∈N> in int(X) , such that
∞→k
kxlim = x and ∞→k
kylim = y. By the concavity of the
restriction of f to int(X), f(txk + (1-t) yk) ≥ tf(xk) + (1-t) f(yk) for all k∈N. By thecontinuity of f on X, it follows by taking limits on either side of the inequality that f(tx +(1-t)y) ≥ tf(x) + (1-t)f(y). Thus, f is concave on D. Q.E.D.
Theorem 2.2: (i ) Let f: X →ℜ be a continuous function on a convex domain X of ℜn.For all x,y∈X, let ϕx,y : [0, 1] → ℜ be defined by ϕx,y(t) = f(x + t(y-x)). Then f is concave if and only if ϕx,y
is concave for all x,y ∈X.(ii) Let f: X →ℜ be a continuous function on an open convex domain X of ℜn. For allx,y∈X, with x ≠ y let ϕx,y : (0, 1) → ℜ be defined by ϕx,y(t) = f(x + t(y-x)). Then f isstrictly concave if and only if ϕx,y is strictly concave for all x,y ∈ X with x ≠ y.
Proof: (i) Suppose f is concave. Let x,y∈X, a, b ∈[0, 1] and t∈(0, 1).Thus, ϕx,y (ta + (1-t)b) = f ( x + (ta+(1-t)b)(y-x)) = f (t(x + a (y-x)) + (1-t)(x + b(y-x))) ≥tf (x + a(y –x)) + (1-t) f (x + b(y-x)) (: since f is concave) = t ϕx,y(a) + (1-t) ϕx,y (b). Thus,ϕx,y is concave.Now suppose, ϕx,y is concave for all x,y ∈X. Let x, y ∈X and t∈(0, 1). Thus, f(x + t(y-x))= ϕx,y (t) = ϕx,y(t.1 + (1-t).0) ≥ t ϕx,y(1) + (1-t) ϕx,y(0) = t f(y) + (1-t) f(x). Thus, f isconcave.Suppose f is strictly concave. Let x,y∈X with x ≠ y, a, b ∈(0, 1) with a ≠ b and t∈(0, 1).
Thus, ϕx,y (ta + (1-t)b) = f ( x + (ta+(1-t)b)(y-x)) = f (t(x + a (y-x)) + (1-t)(x + b(y-x))) >tf (x + a(y –x)) + (1-t) f (x + b(y-x)) (: since f is strictly concave and x +a(y-x) ≠ x+ b(y-x)) = t ϕx,y(a) + (1-t) ϕx,y (b). Thus, ϕx,y is strictly concave.Now suppose, ϕx,y is strictly concave for all x,y ∈X with x ≠ y. Let x, y ∈X with x ≠ yand t∈(0, 1). Since X is open, there exists σ > 0, such that x1 = x - σ(y-x) ∈ X and y1 = y
+ σ(y-x) ∈X. Thus, y1 – x1 = (2σ + 1) (y-x). Thus, x = x1 + 12 +σ
σ (y1 – x1) and y = y1 -
12 +σσ (y1 – x1) = x1+ y1 – x1 -
12 +σσ (y1 – x1) = x1 +
121
++
σσ (y1 – x1). Further, x + t(y-x)
= x1 + 12 +σ
σ (y1 – x1) + 12 +σ
t (y1 – x1) = x1 + 12 +
+σ
σ t (y1 – x1).
For the sake of notational simplicity, let us agree that ϕ : (0, 1) →ℜ is defined by ϕ(a) = f
(x1 + a (y1 – x1)) for all a ∈(0, 1). It is easily checked that 0 < 12 +σ
σ < 12 +
+σ
σ t < 12
1+
+σ
σ
< 1. Further, 12 +
+σ
σ t = t (12
1+
+σ
σ ) + (1-t) (12 +σ
σ ). Thus, by the strict concavity of ϕ , ϕ(
12 ++
σσ t ) > t ϕ (
121
++
σσ ) + (1-t) ϕ (
12 +σσ ). Thus, f (tx + (1-t)y) > t f(x) + (1-t) f(y). Thus,
f is strictly concave. Q.E.D.
As a consequence of Theorem 2.1 we have the following theorem.
Theorem 2.3: Let f : X → ℜ be a function which is continuous on X and C1 on int(X).Then, f is concave if and only if for all x∈int(X) and y∈X, f(y) ≤ f(x) + Df(x).(y-x).
Proof: Let f(y) ≤ f(x) + Df(x). (y-x) for all x∈int(X) and y∈X. Let x, y ∈int(X) and t∈(0,1). Thus, x+ t (y – x) ∈int(X). Hence, f(x) ≤ f (x+ t (y – x)) – t Df (x+ t (y – x)).(y- x)and f(y) ≤ f (x+ t (y – x)) +(1- t) Df (x+ t (y – x)).(y- x). Hence multiplying both sides ofthe first inequality by (1-t) and those of the second by t, and adding them up yields (1-t)f(x) + t f(y) ≤ f (x+ t (y – x)). Thus, the restriction of f to int(X) is concave. It follows as aconsequence of Theorem 2.1 that f is concave.Now suppose that f is C1 and concave. Let x ∈int(X) and y∈X. Hence for all t ∈ (0, 1): x+ t(y-x) ∈int(X). Thus, f (x) + t (f(y) – f(x)) ≤ f(x+t(y-x)) for all t ∈(0,1). Thus, f(y) – f(x)
≤ t
xfxytxf )())(( −−+ for all t∈(0, 1). Since, f is C1 on int(X), f(y) – f(x) ≤ Df(x). (y-
x). Q.E.D.
Corollary of Theorem 2.3: Let f: X→ℜ be a continuous function on a convex domain Xin ℜn, which is C2 on int(X). Then, f is concave if and only if for all x∈int(I) and h ∈ ℜn:hTD2f(x)h ≤ 0 (i.e. D2f(x) is negative semi-definite).
Proof: Let f be continuous on a convex domain X of ℜn and C2 on int(X). Suppose f isconcave. Let x∈int(X) and h∈ ℜn \ {0}. Hence there exists a > 0, such that x + bh
∈int(X) for all –a < b < a. Now, for –a < b < a, f(x + bh) = f(x) + bDf(x).h +b2hTD2f(x+δ(b)bh)h, where 0 < δ(b) < 1. Since, f is concave, by Theorem 2.3, for all –a < b < a, f(x+ bh) ≤ f(x) + bDf(x).h. Thus, b2hTD2f(x +δ(b)bh)h ≤ 0, for all –a < b < a. Let <bk/ k∈N>be a sequence of positive real numbers, such that
∞→klim bk = 0. Thus, hTD2f(x +δ(bk)bkh)h
≤ 0 for all k∈N. Proceeding to the limit as k tends to infinity, we get by virtue of thetwice continuous differentiability of f on C2, that hT D2f(x)h ≤ 0.Now, suppose that for all x∈ int(X) and h ∈ ℜn, hT D2f(x)h ≤ 0. Let x, y∈int(X). Thus,f(y) = f(x) + Df(x).(y –x) + (y-x)TD2f (x + δ(y-x)).(y-x) for some 0 < δ < 1. Since, (y-x)TD2f (x + δ(y-x)).(y-x) ≤ 0, we get f(y) ≤ f(x) + Df(x).(y –x). By Theorem 2.3, therestriction of f to int(X) is concave. It thus follows as a consequence of Theorem 2.1 thatf is concave. Q.E.D.
As a consequence of Theorem 2.2 we have the following theorem.
Theorem 2.4: Let f : X → ℜ be a function which is continuous on D and C1 on the openconvex domain X. Then, f is strictly concave if and only if for all x, y∈D with x ≠ y, f(y)< f(x) + Df(x).(y-x).
Proof: Let f(y) < f(x) + Df(x). (y-x) for all x, y∈D with x ≠ y. Let x, y ∈D and t ∈(0,1). Hence, f(x) < f (x+ t (y – x)) – t Df (x+ t (y – x)).(y- x) and f(y) < f (x+ t (y – x)) +(1- t)Df (x+ t (y – x)).(y- x). Hence multiplying both sides of the first inequality by (1-t) andthose of the second by t, and adding them up yields (1-t)f (x) + t f(y) < f (x+ t (y – x)). Now suppose that f is C1 and strictly concave. Let x, y ∈D with x ≠ y. By Theorem 2.3,f(y) – f(x) ≤ Df(x). (y-x). Towards a contradiction suppose f(y) – f(x) = Df(x). (y-x). By
Theorem 2.2, f (2
yx + ) ≤ f(x) +21 Df(x). (y-x) = f(x) +
21 [ f(y) – f(x)], contradicting the
strict concavity of f. Thus, f(y) – f(x) < Df(x). (y-x). Q.E.D.
Corollary of Theorem 2.4: Let f: X→ℜ be a continuous function on an open convexdomain X in ℜn, which is C2 on D. If for all x∈D and h ∈ ℜn\ {0}: hTD2f(x)h < 0 (i.e.D2f(x) is negative definite), then f is strictly concave. However, the converse is not true.
Proof: Suppose that for all x∈ D and h ∈ ℜn \{0}, hT D2f(x)h < 0. Let x, y∈D with x ≠ y.Thus, f(y) = f(x) + Df(x).(y –x) + (y-x)TD2f (x + δ(y-x)).(y-x) for some 0 < δ < 1. Since,(y-x)TD2f (x + δ(y-x)).(y-x) < 0, we get f(y) < f(x) + Df(x).(y –x). It thus follows as aconsequence of Theorem 2.4 that f is strictly concave. To show that the converse is not true, consider the function f : ℜ → ℜ defined by f(x) = -x4 for all x ∈ ℜ. It is easy to verify, by drawing a graph that f is strictly concave. A morerigorous way of establishing the same observation would be the following: Let h :ℜ → ℜ defined by h(x) = -x2 for all x ∈ ℜ. Since D2h(x) = -2 < 0 for all x∈ ℜ, it followsfrom the first part of the theorem that h is strictly concave. Hence the function – h:ℜ → ℜ defined by (-h)(x) = x2 for all x ∈ ℜ is strictly convex. Now the function g :ℜ → ℜ defined by g(x) = x4 for all x ∈ ℜ is the composition function h*ο (-h) : ℜ → ℜ,where h*:ℜ+→ℜ is defined by h*(x) = x2 for all x∈ℜ. Now, h* is strictly increasing and
being the restriction of –h to ℜ+, is strictly convex as well. Let x,y∈ℜ with x ≠y and lett∈(0, 1). Thus, g(tx +(1-t)y) = h*(-h(tx+(1-t)y)). Now, by the strict convexity of –h, -h(tx+(1-t)y) < t(-h)(x) + (1-t)(-h)(y). Since, h* is strictly increasing and strictly convex, h* (-h(tx+(1-t)y)) < h* (t(-h)(x) + (1-t)(-h)(y)) < th*(-h(x)) + (1-t)h*(-h(y)) = tg(x) + (1-t)g(y). Thus, g is strictly convex. Hence, f = -g, is strictly concave. However, D2f(0) =0. Q.E.D.
Let f : X → ℜ be a function on a convex domain X in ℜn. A point x*∈X is said to be:(a) a global maximum of f if f(x*) ≥ f(x) for all x ∈X;(b) a strict global maximum of f if f(x*) > f(x) for all x ∈X\ {x*};(c) a local maximum of f if there exists an open convex domain O: (i) x*∈O ⊂ X; (ii)
f(x*) ≥ f(x) for all x∈O;(d) a strict local maximum of f if there exists an open convex domain O: (i) x*∈O ⊂
X; (ii) f(x*) > f(x) for all x∈O\ {x*}.
Theorem 2.5: Let f: X → ℜ be a concave function on a convex domain X in ℜn. If x* is alocal maximum of f, then it is a global maximum. If it is a strict local maximum of f, thenit is a strict global maximum of f. If in addition, f is strictly concave, then a globalmaximum of f is a strict global maximum of f.
Proof: Suppose, there exists an open convex domain O ⊂ X, such that x*∈O and f(x*) ≥f(x) for all x∈O. Let y ∈X\O and towards a contradiction suppose f(y) > f(x*). Since, f isconcave, for all t∈(0,1): f(x*+t(y-x*)) ≥ f(x*) + t[f(y) – f(x*)] > f(x*). For, t > 0sufficiently small, x*+t(y-x*) ∈O, contradicting the assumption we made about x*.Now suppose, there exists an open convex domain O ⊂ X, such that x*∈O and f(x*) >f(x) for all x∈O\{x*}. By the previous argument x* is a global maximum. Let y ∈X\O andtowards a contradiction suppose f(y) = f(x*). Since, f is concave, for all t∈(0,1): f(x*+t(y-x*)) ≥ f(x*) + t[f(y) – f(x*)] = f(x*). For, t > 0 sufficiently small, x*+t(y-x*) ∈O,contradicting the assumption we made about x*.Finally suppose f is strictly concave, so that X is open. Let x* be a global maximum of f.Towards a contradiction suppose there exists y ∈D\{x*}, such that f(y) = f(x*). Since X is
an open convex domain, 21 (x* + y) ∈X, and since f is strictly concave,
f (21 (x* + y)) >
21 f(x*) +
21 f( y) = f(x*), contradicting our assumption about x*. Q.E.D.
Theorem 2.6: Let f: X → ℜ be a concave and continuously differentiable function on anopen convex domain X in ℜn. x* ∈X is a global maximum of f if and only if Df(x*) = 0.
Proof: Let x*, x ∈X. Since f is concave and X is open, by Theorem 2.3, f(x) ≤ f(x*) +Df(x*).(x-x*). Thus, Df(x*) = 0 implies f(x*) ≥ f(x). Thus, Df(x*) = 0 implies x* is a globalmaximum of f.If x* is a global maximum of f, then since f is C1 and X is open, Df(x*) = 0. Q.E.D.
Theorem 2.7: Let f : X →ℜ be concave on a convex domain X and C1 on int(X). Letx∈int(X), y∈X and Df(x).(y-x) > 0. Then, there exists a ∈(0, 1]: f(x+ b (y-x)) > f(x) forall 0 < b ≤ a.
Proof: Suppose that for all a ∈(0, 1] it is the case that f(x + a(y-x)) ≤ f(x). Now, for alla∈(0, 1], there exists 0 < θ(a) < 1, such that f(x + a(y-x)) = f(x) + a Df(x + θ(a)a(y-x)).(y-x). Since, f(x +a(y-x)) ≤ f(x) for all a ∈(0, 1], it must be the case that Df(x + θ(a)a(y-x)).(y-x) ≤ 0, for all a ∈(0, 1]. Proceeding to the limit as a tends to zero, it follows fromthe continuous differentiability of f and the fact that 0 < θ(a) < 1 for all a∈(0, 1] thatDf(x).(y-x) ≤ 0, contrary to our assumption. Thus, there exists a∈(0, 1] such that f(x+ a
(y-x)) > f(x). Now if, 0 < b < a, then f(x + b(y-x)) = f (x + ab ( x+ a(y-x) –x) ≥ f(x) +
ab [f(x+ a(y-x)) –f(x)] > f(x). This proves the theorem. Q.E.D.
Chapter 3Homogeneous Functions
In this chapter we will consider a convex domain X which is either ℜn or n+ℜ or n
++ℜ . Acontinuous function f : X→ℜ on a convex domain X, which is either ℜn or n
+ℜ or n++ℜ , is
said to be homogeneous of degree k, where k is any non-negative real number, if for allx∈X and t > 0, f(tx) = tkf(x). Clearly, any such function must satisfy the property f(0) = 0if 0 ∈X.If n = 1, then any function f which is homogeneous of degree k must be of the form f(x) =xkf(1), for all x∈X. This follows from the fact that f(x) = f(x.1) = xkf(1) for all x∈X.
Theorem 3.1(Euler’s Theorem) Let f : X →ℜ be a function on a convex domain X whichis either ℜn or n
+ℜ or n++ℜ and which is homogeneous of degree k. Suppose f is C1 on
int(X). Then, for all x ∈ int(X), k f(x) =Df(x)x.
Proof : Let x ∈int(X). For t > 0, f(tx) = tkf(x). Differentiating the above equation with respect to t, we get ktk-1f(x) =Df(tx).Putting t = 1, we get k f(x) = Df(x)x, as was required to be proved. Q.E.D.
The following result is worth noting:
Theorem 3.2: Let f: ℜn →ℜ be a C1 function. If f is homogeneous of degree one, then forall x∈ℜn: f(x) = Df(0)x.
Proof: Let f be as stated in the statement of the theorem and let x∈ℜn. Let t > 0 be a realnumber and ei the ith unit coordinate vector in ∈ℜn(i.e. the vector in ∈ℜn which has 1 inits ith coordinate and zero elsewhere.
Now, 0
lim→h h
txfhetxf i )()( −+ = 0
lim→h
tht
xfethxft i )]()([ −+
(: since f is homogeneous of
degree one) = 0
lim→h
th
xfethxf i )()( −+
= 0
lim→h h
xfhexf i )()( −+ . Thus, Df(tx) = Df(x) for
all x∈ℜn. Since f is C1, Df(0) = 0
lim→t
Df(tx) = Df(x) for all x∈ℜn.
By Theorem 3.1, f(x) = Df(x).x = Df(0).x Q.E.D.Note, that while proving theorem 3.2, we implicitly established in the first part thefollowing result: Let f: X →ℜ be a C1 function where X is either ℜn or n
+ℜ or n++ℜ . If f is homogeneous of
degree one, then for all x∈int(X) and t > 0: Df(tx) = Df(x).
For the remaining discussion in this chapter, we will consider the case n =2.
Example 1: Let a ∈ℜ2 and for z = (x,y)∈X, let f(z) = aTz. Such an f is homogeneous ofdegree one and is called a linear function on X. It is easily verified that f is concave aswell.
Example 2: Let a, b ≥ 0. The function f : X → ℜ defined by f(x,y) = xayb is homogeneousof degree a + b. If 1 > a > 0 and b = 1 – a, then f is said to be a Cobb-Douglas function onX.
Example 3: Let X = 2++ℜ . The function f : X → ℜ defined by f(x,y) =
[ ] ρρρ δδ1
)1( −−− −+ yx , where:(a) ρ > -1 and ρ ≠ 0,(b) 0 < δ < 1,
is called a Constant Elasticity of Substitution (CES) function. A CES function ishomogeneous of degree one.
As a consequence of Theorem 3.1, we have the following theorem.Theorem 3.3 : Let f : X → ℜ be homogeneous of degree one and C1 on int(X). Let (x, y) ∈int(X) and t > 0. Then, for all t > 0:
f(tx, ty) = f(x, y) + (t – 1)[ x x
yxf∂
∂ ),( + y y
yxf∂
∂ ),( ].
Proof: For all t > 0, f (tx, ty) = tf (x, y) = f(x,y) + (t-1) f(x, y) = f(x,y) + (t-1) [ x
xyxf
∂∂ ),( + y
yyxf
∂∂ ),( ], by Theorem 3.2 and the fact that f is a function which is C1 on
int(X) and homogeneous of degree one on D. Q.E.D.
Consider the Cobb-Douglas function f : 2++ℜ → ℜ, defined by f(x, y) = xay1-a, for all
x∈ 2++ℜ , where 0 <a < 1. Let (x, y) ∈ 2
++ℜ and suppose f (x, y) = c1-a > 0. Thus, y = a
a
x
c
−1
.
Let y(ξ ) = a
a
c
−1ξ for all ξ > 0. Thus, y(x ) = y and f (ξ, y(ξ)) = c1-a, for all ξ > 0. Further,
y : ℜ++ → ℜ++ as defined above is C1, and Dy(x) = - xa
ay)1( −
.
For, r > 0, y(rx) = a
a
r −1
1 y(x).
Now, [ y(rx) – y(x)] – (r -1)xDy(x) = y(x)[ a
a
r −1
1 - 1] – (r -1)xDy(x) = y(x)[ a
a
r −1
1 - 1] + (r
-1) )1()(
axay
− = y(x) [
aa
r −1
1 - 1 + )1( a
a−
(r -1)].
Let h : ℜ++ → ℜ++ be defined by h(r) = a
a
r −1
1 - 1 + )1( a
a−
(r -1) for all r > 0. Thus,
[ y(rx) – y(x)] – (r -1)xDy(x) = y(x) h(r) for all r > 0. Further, h(1) = 0 and Dh (r) =
- )1( a
a−
aa
r −1
1 + )1( a
a−
= )1( a
a−
( 1 - a
a
r −1
1 ). Dh (1) = 0, Dh (r) < 0, for 0 < r < 1 and
Dh(r) > 0, for 1 < r. Thus, h attains a strict global minimum at r = 1. Thus, [ y(rx) – y(x)]
– (r -1)xDy(x) > 0, for all r > 0 with r ≠1. Thus, [ y(rx) – y(x)] + (r -1)x
yyxf
xyxf
∂∂
∂∂
),(
),(
> 0, for
all r > 0 with r ≠ 1. Since y
yxf∂
∂ ),( = y
c a−1
> 0, we get [ y(rx) – y(x)] y
yxf∂
∂ ),( + (r -
1)xx
yxf∂
∂ ),( > 0, for all r > 0 with r ≠ 1.
Theorem 3.4: The Cobb- Douglas function f : 2+ℜ →ℜ is concave.
Proof : Let (x*, y*), (x,y) ∈ 2++ℜ . Suppose, (x*, y*) ≠ (x,y). Let t
yxfyxf
=),(
),(** > 0. By
Theorem 3.3, f(tx*, ty*) = f(x*,y*) + (t-1)[ x* x
yxf∂
∂ ),( **
+ y* y
yxf∂
∂ ),( **
]. From, what we
have obtained above, [ y – ty*] y
tytxf∂
∂ ),( **
+ ( *txx -1)tx*
xtytxf
∂∂ ),( **
≥ 0, since r = *txx in
this case. Thus, [ y – ty*] y
tytxf∂
∂ ),( **
+ (x -tx* )x
tytxf∂
∂ ),( **
≥ 0. Now, y
tytxf∂
∂ ),( **
=
*
** ),(ty
tytxf = *
** ),(ty
yxtf (: since f is homogeneous of degree one) = *
** ),(y
yxf =
yyxf
∂∂ ),( **
. Similarly x
tytxf∂
∂ ),( **
= x
yxf∂
∂ ),( **
. Thus, (y – ty*)y
yxf∂
∂ ),( **
+ (x -tx* )
xyxf
∂∂ ),( **
≥ 0. Thus, yy
yxf∂
∂ ),( **
+ xx
yxf∂
∂ ),( **
≥ ty*
yyxf
∂∂ ),( **
+ tx* x
yxf∂
∂ ),( **
Hence, f (x,y) = f(tx*, ty*) = f(x*,y*) + (t-1)[ x* x
yxf∂
∂ ),( **
+ y* y
yxf∂
∂ ),( **
] ≤ f(x*,y*) +
(x -x*) x
yxf∂
∂ ),( **
+ (y-y*) y
yxf∂
∂ ),( **
. By Theorem 2.3, the restriction of f to 2++ℜ is
concave. Thus, by Theorem 2.1, f is concave. Q.E.D.Now consider the CES function f : 2
++ℜ → ℜ defined by
f(x,y) = [ ] ρρρ δδ1
)1( −−− −+ yx , where:
(a) ρ > -1 and ρ ≠ 0,(b) 0 < δ < 1.
Let (x,y) ∈ 2++ℜ and suppose that [ ] ν−),( yxf = c > 0. Thus, [ ]ρρ δδ −− −+ yx )1( = c.
Thus, y-ν = )(1
1 νδδ
−−−
xc . Let y (ξ) = ννξδδ
−−−−
1
)](1
1[ c for 0 < ξ < νδ
c.
This defines a C1 function y : (0, νδ
c) → ℜ++.
Now, -νy-ν-1 Dy(x) = δ
νδ−1
x-ν-1. Thus, Dy(x) = 1)(1
+
−− ν
δδ
xy .
Forνδ
cx1 > r > 0, y (rx) = ν
ν
νδδ
−−
−−
1
)](1
1[rxc .
Thus, y(rx) – y(x) – x(r-1)Dy(x) = y(rx) – y(x) + x(r-1) 1)(1
+
−ν
δδ
xy .
Let h : (0, νδ
cx1 ) → ℜ be the C1 function defined by h(r) = y(rx) – y(x) +
x(r-1) 1)(1
+
−ν
δδ
xy for all 0 < r <
νδ
cx1 . h(1) = 0. Further, Dh(r) = 1))((
1+
−− ν
δδ
rxrxyx +
x 1)(1
+
−ν
δδ
xy . Thus, Dh(r) = x
δδ−1
[ 1)( +ν
xy - 1))(( +ν
rxrxy ] for all 0 < r <
νδ
cx1 . Clearly,
Dh(1) = 0. Since Dy(ξ) < 0, for all 0 < ξ < νδ
c, 1)( +ν
xy < 1))(( +ν
rxrxy for 0 < r < 1 and
1)( +ν
xy > 1))(( +ν
rxrxy for 1 < r <
νδ
cx1 . Thus, Dh(r) < 0 for 0 < r < 1 and Dh(r) > 0 for 1
< r <νδ
cx1 . Thus, h attains a strict global minimum at r = 1. Using methods similar to
the ones in the proof of Theorem 3.4, we may conclude the following:
Theorem 3.5: The CES function f : 2++ℜ →ℜ is concave.
When ν = 1, the CES function becomes f(x,y) = δx + (1-δ)y, for all (x, y) ∈ 2++ℜ .
Now suppose, f(x,y) = [ ] ρρρ δδ1
)1( −−− −+ yx , where:(a) ν > -1 and ν ≠ 0,(b) 0 < δ < 1.
Thus, log f(x,y) = ν1
− log [ δ x-ν + (1-δ) y-ν].
As, ν → 0, the denominator as well as the numerator tend to zero. Thus, by applyingL’Hospital’s rule we get
0lim
→ν ν1
− log [ δ x-ν + (1-δ) y-ν] = 0
lim→ν ν
1− log [ δ e-νlog x + (1-δ) e-νlog y] =
0lim
→ν νν
νν
δδδδ
−−
−−
−+−+
yxyyxx
)1(log)1(log = δ log x + (1-δ) log y.
Hence as ν → 0, log f(x,y) →δ log x + (1-δ) log y. Thus, as ν → 0, f(x,y) → xδy(1-δ) forall (x,y) ∈ 2
++ℜ .Given a concave function f : X → ℜ which is C1 on int(X), the function σf : int(X) → ℜdefined by:
σf(x, y) = )log(
]),(),(log[
xyd
yxfyxfd
y
x
, where fx(x,y) = x
yxf∂
∂ ),( and fy(x, y) = y
yxf∂
∂ ),( for all (x,y)
∈ int(X) is called the elasticity of substitution.Suppose f(x,y) = xδy(1-δ) for all (x, y)∈ 2
++ℜ where 0 < δ < 1.
Then, fx(x,y) = ),( yxfxδ and fy(x, y) = ),( yxf
yδ . Thus,
xy
yxfyxf
y
x
)1(),(),(
δδ−
= and hence,
σf(x, y) = )log(
]),(),(log[
xyd
yxfyxfd
y
x
= 1, for all (x, y) ∈ 2++ℜ .
Suppose, f(x,y) = [ ] ρρρ δδ1
)1( −−− −+ yx for all (x, y) ∈ 2++ℜ , where:
(a) ν > -1 and ν ≠ 0,(b) 0 < δ < 1.
Thus, ),(),(
yxfyxf
y
x = 1)(1
+
−ν
δδ
xy . Thus log [
),(),(
yxfyxf
y
x ] = log [δ
δ−1
] + (ν+1) log[ xy ], so
that σf(x, y) = )log(
]),(),(log[
xyd
yxfyxfd
y
x
= ν+1, for all (x, y) ∈ 2++ℜ .
If σf(x, y) is high, then as we move upwards along a contour of f (through out which thevalue of f is constant), the curve becomes steeper.
Chapter 4Concave Vector Optimization
Let X be a convex domain in ℜn. An array <(f1,…, fk), A, c>, where k is a naturalnumber, fi : X → ℜ is a concave function for each i ∈{1,…, k}, A is a matrix of realnumbers comprising m rows and n columns and c is a vector in ℜm is called a concaveoptimization problem. If in addition each fi is C1 on int(X), for i∈{1,…, k}, then the array<(f1,…, fk), A, c> is called a differentiable concave optimization problem.Given a concave optimization problem <(f1,…, fk), A, c>, any element x ∈X satisfyingAx ≤ c, is said to be feasible for the problem.A feasible element x* for a concave optimization problem <(f1,…, fk), A, c> is said to bePareto Optimal, if there does not exist any other element x which is feasible for theproblem and satisfies fi(x) > fi(x*) for all i∈{1,…, k}. An interesting application of Theorems 2.3 and 2.7 is the following:
Theorem 4.1: Let <(f1,…, fk), A, c> be a differentiable concave optimization problem ona convex domain X of ℜn. Let x* ∈int(X). Then x* is Pareto Optimal for <(f1,…, fk), A,c> if and only if x* is Pareto optimal for the concave optimization problem <(g1,…, gk),A, c>, where for each i∈{1,…, k}, gi : X → ℜ is defined as gi(x) = Dfi (x*)x, for all x∈X.
Proof : Suppose x* ∈int(X) is Pareto optimal for <(f1,…, fk), A, c>. Towards acontradiction suppose that there exists y∈X such that Ay ≤ c and Dfi (x*)y > Dfi (x*)x* for all i ∈{1,…, k}. By Theorem 2.7, there exists for each i∈{1,…, k}, thereexists 1 ≥ ai > 0, such that f (x*+ b(y-x*)) > f(x*) for all 0 < b ≤ ai. Let a = min{ai/i∈{1,…, k}}. Further, x*+ b(y-x*) ∈D and satisfies A(x*+ b(y-x*)) ≤ c for all 0 < b ≤ a.This contradicts our assumption about x*.Now suppose x*∈ int(X) is Pareto Optimal for <(g1,…, gk), A, c>. Towards acontradiction suppose that there exits y∈X such that Ay ≤ c and fi (y) > fi (x*) for alli∈{1,…,k}. By Theorem 2.3, fi (y) ≤ fi (x*) + Dfi (x*)(y-x*), for all i∈{1,…, k}. Hence, fi (y) > fi (x*) for all i∈{1,…, k}implies fi (x*) < fi (x*) + Dfi (x*)(y-x), i.e. Dfi (x*)(y-x*) >0, contrary to hypothesis. Q.E.D.
Given a concave optimization problem <(f1,…, fk), A, c>, let L : X×( k+ℜ \ {0})× m
+ℜ → ℜ
be defined as follows : for all (x, µ, λ) ∈ X×( k+ℜ \ {0})× m
+ℜ , let L(x, µ, λ) = ∑=
k
iii xf )(µ +
λT[ c – Ax]. L is called the generalized Lagrangean for <(f1,…, fk), A, c>.
Theorem 4.2: Let <(f1,…, fk), A, c> be a concave optimization problem on a convexdomain X of ℜn. Let (x*, µ, λ*) ∈ X × ( k
+ℜ \{0})× m+ℜ be such that L(x, µ, λ*) ≤ L(x*, µ,
λ*) ≤ L(x*, µ, λ) for all (x, λ)∈ X × m+ℜ . Then
(i) x* is Pareto Optimal for <(f1,…, fk), A, c>;(ii) (λ*)T[ c- Ax*] = 0.
Proof: Let (x*, µ, λ*) ∈ X ×( k+ℜ \{0})× m
+ℜ be such that L(x, µ, λ*) ≤ L(x*, µ, λ*) ≤ L(x*,
µ, λ) for all (x, λ)∈ X× m+ℜ . Hence ∑
=
k
iii xf )( *µ + (λ*)T[ c – Ax*] ≤ ∑
=
k
iii xf )( *µ + λT[ c –
Ax*] for all λ∈ m+ℜ . Thus, (λ*)T[ c – Ax*] ≤ λT[ c – Ax*] for all λ∈ m
+ℜ . Let Ai denotethe ith row of A and towards a contradiction suppose ci < Aix*. Then by letting ei denotethe unit ith coordinate vector in ℜm (i.e. the vector which has 1 in the ith coordinate and0’s elsewhere) and by letting λ = rei, where r is any non-negative real number, λT[ c –Ax*] can be made to violate (λ*)T[ c – Ax*] ≤ λT[ c – Ax*] for all λ∈ m
+ℜ , simply bychoosing r sufficiently large. Thus, Ax* ≤ c. By taking λ = 0, we get, (λ*)T[ c- Ax*] ≤ 0.Since λ* ∈ m
+ℜ and c- Ax* ≥ 0, we get (λ*)T[ c- Ax*] ≥ 0. Hence, (λ*)T[ c- Ax*] = 0. Thisproves (ii).
Now, L(x, µ, λ*) ≤ L(x*, µ, λ*) for all x∈X implies ∑=
k
iii xf )(µ + (λ*)T[ c – Ax] ≤
∑=
k
iii xf )( *µ + (λ*)T[ c – Ax*] for all x∈X. By (ii), (λ*)T[ c- Ax*] = 0. Thus, ∑
=
k
iii xf )(µ +
(λ*)T[ c – Ax] ≤ ∑=
k
iii xf )( *µ for all x∈X. Thus, x∈X, Ax ≤ c implies ∑
=
k
iii xf )(µ ≤
∑=
k
iii xf )(µ + (λ*)T[ c – Ax] ≤ ∑
=
k
iii xf )( *µ . Thus, x∈X, Ax ≤ c implies ∑
=
k
iii xf )(µ ≤
∑=
k
iii xf )( *µ . Towards a contradiction suppose there exists x∈X, Ax ≤ c such that fi (x) >
fi(x*) for all i∈{1,…, k}. Since µ ∈ k+ℜ \{0}, ∑
=
k
iii xf )(µ > ∑
=
k
iii xf )( *µ contrary to what
we obtained above. This proves (i). Q.E.D.
It follows directly from Theorems 1.1,2.6 and 4.2, that the following is true:
Theorem 4.3: Let <(f1,…, fk), A, c> be a differentiable concave optimization problem ona convex domain X of ℜn. Let (x*, µ, λ*) ∈ int(X)× ( k
+ℜ \{0})× m+ℜ be such that
(i) ∑=
k
iii xDf
1
*)(µ - (λ*)TA = 0; (ii) (λ*)T[ c- Ax*] ≤ λT[ c- Ax*] for all λ∈ m+ℜ .
Then x* is Pareto Optimal for <(f1,…, fk), A, c>;
Proof: By Theorem 1.1, it follows that the function ∑=
k
iii f
1µ : X → ℜ is concave. Further,
it is C1 on int(X). By (i) and Theorem 2.6, it follows that L(x, µ, λ*) ≤ L(x*, µ, λ*) for allx ∈ X. By (ii) (λ*)T[ c- Ax*] ≥ λT[ c- Ax*] for all λ∈ m
+ℜ . Thus, L(x*, µ, λ*) ≤ L(x*, µ, λ)for all λ∈ m
+ℜ . Hence, it follows from the Theorem 4.2, that x* is Pareto Optimal for<(f1,…, fk), A, c>. Q.E.D.
Farkas’s Theorem : Let A be a matrix of real numbers, comprising m rows and ncolumns, and let ‘c’ be a real m-vector. Then exactly one of the following is true:EITHER There exists x ∈ n
+ℜ , such that Ax = c;OR There exists y∈ℜm such that yTA ≥ 0 and yTc < 0.
Theorem 4.4: Let <(f1,…, fk), A, c> be a differentiable concave optimization problem ona convex domain X of ℜn. Suppose x* ∈ int(X) is Pareto Optimal for <(f1,…, fk), A, c>.
Then there exists Let (µ, λ*) ∈ ( k+ℜ \{0})× m
+ℜ be such that (i) ∑=
k
iii xDf )( *µ - (λ*)TA = 0;
(ii) (λ*)T[ c- Ax*] = 0 ≤ λ[ c- Ax*] for all λ∈ m+ℜ .
Proof : By Theorem 4.1, x* ∈ int(X) is Pareto Optimal for <(g1,…, gk), A, c>, where foreach i∈{1,…, k}, gi : X → ℜ is defined as gi(x) = Dfi (x*)x, for all x∈X.Case 1: Ax* = c. Suppose, there exists h∈ℜn, such that Ah ≤ 0 and Dfi(x*)h > 0, for all i ∈{1,…,k}. Sincex* ∈int(X), there exists 0 < a < 1 such that x* + bh ∈X for all 0 < b ≤ a. Thus, A(x* + bh)≤ c and Dfi(x*)(x*+ bh) > 0, for all i ∈{1,…,k} contradicting the Pareto optimality of x*
for <(g1,…, gk), A, c>. Thus, Ah ≤ 0 and Dfi(x*)h > 0, for all i ∈{1,…,k} has no solutionh∈ℜn. Hence, there does not exist h∈ℜn and θ > 0, such that Ah ≤ 0 and Dfi(x*)h - θ ≥ 0,for all i ∈{1,…,k}. Let BT be the (m+k)× (n+1) matrix of real numbers, BT =
−
−exDf
A|)(0|
* , where Df(x*) is the k ×n matrix, whose ith row is Dfi(x*) and e is the
vector in ℜn all whose components are equal to one. Hence, there does not exist a vector
θh
∈ℜn×ℜ such that BT
θh
≥ 0, and (0,…,0,-1)
θh < 0. By Farkas’s Theorem, there
exists (λ*, µ) ∈ m+ℜ × k
+ℜ such that ((λ*)T, µT) BT = (0,…,0,-1). Thus, (λ*)TA + µTDf(x*) =
0 and -µTe = -1. Thus, ∑=
k
iii xDf )( *µ - (λ*)TA = 0 and ∑
=
k
iiµ = 1. Further, Ax* = c implies
(λ*)T[ c- Ax*] = 0 = λT[ c- Ax*] for all λ∈ m+ℜ .
Case 2: Ax* < < c. Suppose, there exists h∈ℜn, such that Dfi(x*)h > 0, for all i ∈{1,…,k}. Since x* ∈int(X)and since Ax* < < c, there exists 0 < a < 1 such that x* + bh ∈X and A(x* + bh) ≤ c for all0 < b ≤ a. Thus, Dfi(x*)(x*+ bh) > 0, for all i ∈{1,…,k}and A(x* + bh) ≤ c contradictingthe Pareto optimality of x* for <(g1,…, gk), A, c>. Thus, Dfi(x*)h > 0, for all i ∈{1,…,k}has no solution h∈ℜn. Now suppose there exists h∈ℜn, such that Dfi(x*)h < 0, for all i∈{1,…,k}. Since x* ∈int(X) and since Ax* < < c, there exists 0 < a < 1 such that x* - bh∈X and A(x* - bh) ≤ c for all 0 < b ≤ a. Thus, Dfi(x*)(x*- bh) > 0, for all i ∈{1,…,k}andA(x* - bh) ≤ c contradicting the Pareto optimality of x* for <(g1,…, gk), A, c>. Thus,Dfi(x*)h < 0, for all i ∈{1,…,k} has no solution h∈ℜn. Thus, Dfi(x*)h = 0, for all i∈{1,…,k}. Thus, taking λ* = 0 and µi = 1 for all i ∈{1,…,k}, we arrive at the desiredresult. Further, Ax* << c implies (λ*)T[ c- Ax*] = 0 ≤ λT[ c- Ax*] for all λ∈ m
+ℜ .
Case 3: Ax* < c. In this case let us partition the matrix A and the vector c into two matrices A1, A2 and
vectors c1, c2 respectively as follows: A =
2
1
AA and c =
2
1
cc such that A1x* = c1 and A2
x* < < c2. Suppose, there exists h∈ℜn, such that A1h ≤ 0 and Dfi(x*)h > 0, for all i∈{1,…,k}. Since, x*∈int(X) and A2x* < < c2, there exists 0 < a ≤ 1, such that A2 (x* + bh)≤ c2 and x*+bh ∈int(X) for all 0 < b ≤ a. Further, A1 (x* + bh) ≤ c1 for all 0 < b ≤ a. Butthen, this contradicts x* is Pareto Optimal for <(g1,…, gk), A, c>. Thus, A1 h ≤ 0 andDfi(x*)h > 0, for all i ∈{1,…,k} has no solution h∈ℜn. Thus, by Case 1, there exists avector *
1λ ≥ 0, having as many components as the number of rows of A1, and µ ∈ k+ℜ such
that ∑=
k
iii xDf )( *µ - ( *
1λ )TA1 = 0 and ∑=
k
iiµ = 1. Further, A1x* = c1 implies ( *
1λ )T[ c1 – A1x*]
= 0. Let *2λ = 0 be a vector, having as many components as the number of rows of A2 and
let λ* =
*2
*1
λλ . Thus, (µ, λ*) ∈ ( k
+ℜ \{0})× m+ℜ and [(i) ∑
=
k
iii xDf )( *µ - (λ*)TA = 0; (ii)
(λ*)T[ c- Ax*] = 0≤ λT[ c- Ax*] for all λ∈ m+ℜ , as was required. Q.E.D.
Part 2
Applications
Chapter 5Elements of Consumer Choice Theory
A consumer is an economic entity, which/who gains satisfaction from consumption ofcommodities and which/who uses the resources available (income) to buy commodities inorder to maximize the satisfaction gained from consumption.To make things precise we assume that there are two infinitely divisible commodities orgoods available. Let (x, y) ∈ 2
+ℜ denote a typical consumption vector, where x is theamount of the first good consumed and y is the amount of the second good consumed. 2
+ℜis called the consumption set of the consumer. The consumer is assumed to havepreferences between alternative consumption vectors in the consumption set. We assumethat the consumer has a concave utility function U: 2
+ℜ → ℜ which is C1 on 2++ℜ such that
for all (x, y) and (x1, y1)∈ 2+ℜ : U(x, y) ≥ U(x1, y1) if and only if the consumer does not
prefer the consumption vector (x1, y1) to (x, y).We assume that the consumer has an exogenously given income w > 0, and the consumercan purchase a unit of the two goods at prices 1p > 0 and 2p > 0 respectively. The vector
p =
2
1
pp
∈ 2++ℜ is called the price vector. The purchase of the bundle (x,y) at the price
vector p, entails an expenditure of pT
yx
, and this expenditure cannot exceed the
consumer’s income. Given the price vector p, the budget set B(p) = {(x,y)∈ 2+ℜ / pT
yx
≤
w} and the consumer can acquire only consumption bundles which belong to the budgetset. We also assume that the price vector p is a parameter for the choice problem faced bythe consumer.The traditional model of consumer behavior is that the consumer choose a consumptionbundle which solves the following optimization problem:Maximize U(x, y)Subject to (x,y) ∈B(p).
We call this problem the utility maximization (UM) problem faced by the consumer.
We make the following assumption on U:Assumption: for all (x,y) ∈ 2
++ℜ : DU(x,y) > > 0.Under the above Assumption the following observations are immediate:(i) If (x*, y*) ∈ 2
++ℜ , (x, y) ∈ 2+ℜ and (x*, y*) > (x, y) then U(x*, y*) > U(x, y);
(ii) If (x*, y*) , (x, y) ∈ 2+ℜ \ 2
++ℜ and (x*, y*) > (x, y) then U(x*, y*) ≥ U(x, y).Let us first prove (i). Since (x*, y*) > (x, y) we can without loss of generality assume x* >x. Thus, for all x*- x > δ > 0, x* > x + δ > 0.
Case 1: y = y* > 0. Fix 0 < δ < x* - x. By Theorem 1.2, U(x + ε, y*) ≤ U(x + δ , y*) +(ε − δ) Ux(x+δ, y*), whenever x*- x > δ, ε > 0. Suppose δ > ε. Since Ux(x+ δ , y*) > 0, U(x+ δ, y*) > U(x + ε , y*). Further by Theorem 2.3, U(x + δ, y*) ≤ U(x*, y*) + (x+ δ − x*)Ux(x*, y*) . Since x+ δ − x* < 0, and Ux(x*, y*) > 0, we get that U(x + δ, y*) < U(x*, y*).Now, U(x + δ, y*) > U(x + ε , y*) for 0 < ε < δ, and the continuity of U implies that U(x, y) =
0lim
→ε U(x + ε, y) ≤ U(x + δ, y) < U(x*, y*).
Case 2: y* > y. In this case, Thus, for all min{x*- x, y* -y} > δ > 0, x* > x + δ > 0 and y* >y + δ > 0. Fix 0 < δ < min{x*- x, y* -y}. In that case, by Theorem 2.3, U(x + ε, y + ε) ≤
U(x + δ , y + δ ) - DU(x+ δ, y+δ ))
−−
εδεδ
, whenever min{x*- x, y* -y} > δ, ε > 0.
Suppose δ > ε. Since DU(x+ δ, y + δ)) > > 0, U(x + δ, y + δ) > U(x + ε , y + ε). Further,
U(x + δ, y + δ) ≤ U(x*, y*) + DU(x*, y* )
−+−+
*
*
yyxx
δδ
. Since x+ δ − x*, y+ δ − y*< 0,
and DU(x*, y* ) > > 0, we get that U(x + δ, y + δ) < U(x*, y*). Now, U(x + δ, y + δ) >U(x + ε , y + ε) for 0 < ε < δ, and the continuity of U implies that U(x, y) =
0lim
→εU(x + ε,
y + ε) ≤ U(x + δ, y + δ) < U(x*, y*). This proves (i).Now let us prove (ii). (x*, y*) , (x, y) ∈ 2
+ℜ \ 2++ℜ and (x*, y*) > (x, y) implies either x* >
x and y* = y = 0 or x* = x = 0 and y* > y. Without loss of generality assume that x* > xand y* = y = 0. Thus, for all x*- x > ε > 0, x* > x + ε > 0.By Case 1 of (i) above, we getthat U(x + ε, ε) < U(x*, ε). The continuity of U implies that U(x, 0) =
0lim
→εU(x + ε, ε) ≤
0lim
→ε U(x* + ε, ε)= U(x*, 0). This proves (ii).
Note: A consequence of the concavity assumption on preferences, is that the consumer“prefers mixtures to extremes”.
Theorem 5.1: Let (x*, y*) ∈ 2++ℜ and suppose there exists p∈ 2
++ℜ such that
X
Y
x
y
pp
yxUyxU
=),(),(
**
**
. Let (x, y) ∈ 2+ℜ be such that U(x, y) > U(x*, y*). Then, pT
−−
*
*
yyxx
> 0.
Proof: X
Y
x
y
pp
yxUyxU
=),(),(
**
**
implies that there exists λ > 0, such that y
y
pyxU ),( **
=
x
x
pyxU ),( **
= λ. Clearly, (2
*xx + ,2
*yy + ) ∈ 2++ℜ . By Theorem 2.3, U(
2
*xx + ,2
*yy + ) ≤
U(x*, y*) + 21 DU(x*, y*)
−−
*
*
yyxx
. Thus, U(2
*xx + ,2
*yy + ) ≤ U(x*, y*) + 21
λpT
−−
*
*
yyxx
.
Since U is concave and U(x, y) > U(x*, y*), it follows U(2
*xx + ,2
*yy + ) ≥ 21 U(x, y) +
21 U(x*, y*) > U(x*,y*). Hence, it must be the case that
21
λpT
−−
*
*
yyxx
> 0, i.e.
pT
−−
*
*
yyxx
> 0 . Q.E.D.
Now suppose, (x*, y*) ∈ 2+ℜ is such that pT
*
*
yx
< w. Thus, for ε > 0, sufficiently small
pT
++
εε
*
*
yx
≤ w. By Observation 1 above, U(x*+ε, y*+ε) > U(x*, y*). Thus,
(x*, y*) does not solve UM. Hence, if (x*, y*) solves UM, it is necessary that pT
*
*
yx
= w.
Now, suppose (x*, y*) ∈ 2++ℜ is such that
X
Y
x
y
pp
yxU
yxU>
),(
),(**
**
. Thus, X
Y
pp Ux (x*, y*) – Uy
(x*, y* ) < 0. Now, for ε > 0, sufficiently small x* - εX
Y
pp > 0. Let y(ε) = y* + ε and x(ε) =
x* - εX
Y
pp . By Theorem 2.3, U(x*, y*) ≤ U(x(ε), y(ε)) + ε D U(x(ε), y(ε))
−1/ XY pp
,
whenever (x (ε), y (ε)) ∈ 2++ℜ . Now, D U(x(ε), y(ε))
−1/ XY pp
= X
Y
pp Ux (x(ε), y(ε)) – Uy
(x(ε), y(ε)) < 0, for ε > 0 sufficiently small, since U is C1 on 2++ℜ . Thus, U(x*, y*) <
U(x(ε), y(ε)) for ε > 0 sufficiently small. Further, pT
)()(
εε
yx
= pT
*
*
yx
for all ε. Thus, (x*,
y*) ∈ 2++ℜ cannot solve (UM).
On the other hand, if X
Y
x
y
pp
yxUyxU
<),(),(
**
**
, then X
Y
pp Ux (x*, y*) – Uy (x*, y* ) > 0. For ε > 0
sufficiently small, y* - ε > 0. Let y(ε) = y* - ε and x(ε) = x* + εX
Y
pp . By Theorem 2.3,
U(x*, y*) ≤ U(x(ε), y(ε)) + ε D U(x(ε), y(ε))
−1/ XY pp
, whenever (x (ε), y (ε)) ∈ 2++ℜ .
Now, D U(x(ε), y(ε))
−1/ XY pp
= -X
Y
pp Ux (x(ε), y(ε)) + Uy (x(ε), y(ε)) < 0, for ε > 0
sufficiently small, since U is C1 on 2++ℜ . Thus, U(x*, y*) < U(x(ε), y(ε)) for ε > 0
sufficiently small. Further, pT
)()(
εε
yx
= pT
*
*
yx
for all ε. Thus, (x*, y*) ∈ 2++ℜ cannot solve
(UM).
In view of the above we have the following theorem:
Theorem 5.2: Let (x*, y*) ∈ 2++ℜ . A necessary and sufficient condition for (x*, y*) to solve
(UM) is that [ pT
*
*
yx
= w & X
Y
x
y
pp
yxUyxU
=),(),(
**
**
].
Proof: Sufficiency follows from Theorem 5.1. Necessity follows from what we haveobtained above. Q.E.D.
Chapter 6General Equilibrium and Welfare
Suppose (as in Chapter 5), that there are two infinitely divisible goods which can beconsumed in non-negative amounts x and y respectively. Let N = {1,…, n} be the set ofconsumers, where n is some natural number. For i∈N, let Ui: 2
+ℜ → ℜ be a concaveutility function which is C1 on 2
++ℜ and such that for all (x,y) ∈ 2++ℜ : DUi(x,y) > > 0. Let
ω1 > 0 be the total amount of the first good and ω2 > 0 be the total amount of the secondgood, that is available for consumption amongst the agents. If the income of consumer i ∈N is wi > 0, then at a price vector p∈ 2
++ℜ , consumer isassumed to solve the following problem:Maximize Ui(xi, yi)Subject to (xi, yi)∈B(p, wi),
where Bi(p, wi) = {(xi, yi) ∈ 2+ℜ / pT
i
i
yx
≤ wi}, is the budget set of consumer i at price
vector p and income wi > 0. The above problem denoted (UMi) is the utility maximization problem faced byconsumer ‘i’ at (p, wi).An element (w1,…, wn) ∈ n
++ℜ is said to be an income distribution, where wi > 0, denotesthe income of consumer i. An allocation is an array <(xi, yi) / i∈N> such that (xi, yi) ∈ 2
+ℜ for all i∈N.An allocation <(xi, yi) / i∈N> is said to be feasible if ∑
∈Ni
ix ≤ Xw and ∑∈Ni
iy ≤ Yw .
A price allocation pair (p, <(xi, yi) / i∈N>) is said to be a market equilibrium for theincome distribution (w1,…, wn) ∈ n
++ℜ if:(1) <(xi, yi) / i∈N> is a feasible allocation;(2) for all i ∈ N: (xi, yi) solves (UMi) at (p, wi);
(3) pT
2
1
ωω
= ∑∈Ni
iw (: the worth of the initial endowment is equal to the to the total
amount of money that is available ).A feasible allocation <(x*i, y*i) / i∈N> is said to be Pareto Optimal if there is no otherfeasible allocation <(xi, yi) / i∈N>, such that for all i∈N: Ui(xi, yi) > Ui(x*i, y*i).A feasible allocation <(x*i, y*i) / i∈N> is said to be Strongly Pareto Optimal if there is noother feasible allocation <(xi, yi) / i∈N>, such that Ui(xi, yi) ≥ Ui(x*i, y*i) for all i∈N withstrict inequality holding for at least one i∈N.
We now establish what is known as the First Fundamental Theorem of WelfareEconomics.
Theorem 6.1: Let (p, <(x*i, y*i) / i∈N>) be a market equilibrium at an income distribution(w1,…, wn)∈ n
++ℜ . Then <(x*i, y*i) / i∈N> is Strongly Pareto Optimal.
Proof: Suppose not. Then there exists another feasible allocation <(xi, yi) / i∈N>,such that Ui(xi, yi) ≥ Ui(x*i, y*i) for all i∈N with strict inequality holding for at least onei∈N. Suppose Uj(xj, yj) > Uj(x*j, y*j). Since for all i ∈ N: (x*i, y*i) solves
(UMi), it must be the case that pT
j
j
yx
> wj. Suppose for some i∈N, it is the case that
pT
i
i
yx
< wi. Clearly, Ui(xi, yi) = Ui(x*i, y*i). For ε > 0, sufficiently small, pT
++
εε
i
i
yx
≤
wi. Further, by Observation (i) of Chapter 5, Ui(xi+ε, yi +ε) > Ui(xi, yi) = Ui(x*i, y*i),
contradicting (x*i, y*i) solves (UMi). Thus, pT
i
i
yx
≥ wi for all i∈N. Thus∑∈
Nii
iT
yx
p
>∑∈Ni
iw . However, feasibility of <(xi, yi) / i∈N> implies ∑∈
Nii
i
yx
≤
Y
X
ww
so that pT
∑∈
Nii
i
yx
≤ pT
Y
X
ww
= ∑∈Ni
iw (: by (3) in the definition of a market equilibrium),
contradicting what we obtained earlier. Thus, <(x*i, y*i) / i∈N> is Strongly ParetoOptimal. Q.E.D.
A partial converse of Theorem 6.1 is the following result known as the SecondFundamental Theorem of Welfare Economics.
Theorem 6.2: Let <(x*i, y*i) / i∈N> be a Pareto Optimal allocation. Suppose (x*i, y*i) ∈2
++ℜ for all i∈N. Then there exists an income distribution (w1,…, wn)∈ n++ℜ and a price
vector p ∈ 2++ℜ such that (p, <(x*i, y*i) / i∈N>) is a market equilibrium for the income
distribution (w1,…, wn)∈ n++ℜ .
Proof: Let D =( 2++ℜ )n. For each i ∈N, let Ai be the 2×2 matrix
1001
. Let A be the
2×2n matrix [ ]nAAA ,...,, 21 and let c be the column vector
2
1
ωω
. Thus, <(x*i, y*i) / i∈N>
is a Pareto Optimal solution for the differentiable concave optimization problem <(f1,…,fn), A, c>, where for each i ∈N, fi : X→ ℜ is defined by fi(<(xj, yj) / j∈N>) = Ui(xi, yi).Since z* = <(x*i, y*i) / i∈N>∈ int(X) is Pareto Optimal for <(f1,…, fk), A, c>, by Theorem4.4, there exists (µ, λ*) ∈ ( n
+ℜ \{0})× 2+ℜ be such that
(i) ∑=
n
iii xDf )( *µ - (λ*)TA = 0; (ii) (λ*)T[ c- Ax*] = 0. Hence, ),( ** iii
i yxDUµ - (λ*)T
= ),( ** iiii yxDUµ - (λ*)TAi = 0 for all i∈N and (λ*)T[ c- Ax*] = 0. Without loss of
generality, suppose µ1 > 0. Thus, ),( 1*1*11 yxDUµ - (λ*)T = 0. Since ),( 1*1*1 yxDU > > 0, it
follows that λ* > > 0. Hence, µi > 0 for all i∈N Thus, Let p = λ*. Thus, pT
2
1
ωω
= pTAx*
= pT
∑∑
∈
∈
Ni
iNi
i
y
x*
*
. Let wi = pT
i
i
yx
*
*
> 0, for all i ∈N. Thus, ),( ** iiii yxDUµ = (p)T for all
i∈N, implies ),(),(
**
**
iiix
iiiy
yxUyxU
= 1
2
pp for all i∈N. This, along with wi = pT
i
i
yx
*
*
> 0, for all i
∈N implies (x*i, y*i) solves (UMi) at (p, wi ) for all i∈N. Since, pT
2
1
ωω
= pT
∑∑
∈
∈
Ni
iNi
i
y
x*
*
=
∑∈Ni
im , it follows that (p, <(x*i, y*i) / i∈N>) is a market equilibrium for the income
distribution (w1,…, wn)∈ n++ℜ . Q.E.D.
Theorem 6.3: Given ,py , x ++nn
++ ++++ ℜℜℜ∈ℜ∈ εωε 2 , w =(w1,…, wn)∈ n++ℜ .
with , = yx and w = p. ii
Mii
Miωω
εε),(ΣΣ there exists for i = 1,…, n: ui : ℜ→ℜ+
2 and
0 < αi < 1 such that:(i) for all i = 1,…,n: ui(xi,yi) = ii ii yx αα −1)()( ;(ii) (p, <(xi, yi) / i∈N>) is the unique market equilibrium for the income distribution
(w1,…, wn)∈ n++ℜ .
Proof:- Let w / x p = iii1α for i = 1,…,n. For i = 1,…,n , let ui : ℜ→ℜ+
2 be defined byui(a,b) = ii ba αα −1 , whenever a ≥ 0 and b ≥ 0.
It easy to see that (p, <(xi, yi) / i∈N>) is a market equilibrium for the income distribution(w1,…, wn)∈ n
++ℜ . Let (q ,<(zi, vi) / i∈N>) be any other market equilibrium for the income distribution(w1,…, wn)∈ n
++ℜ .Thus, w / z q = iii
1α for all i = 1,…,n.If },{1,jandniexists there x,z 2},...,1{ ∈∈≠ such that j
ijiz x .≠ Without loss of
generality assume . x > z jj11
Then j jq < p . Thus x > z kj
kj for all k∈{1,…,n}. . = x > z j
kj
Mk
kj
Mkω∴ ΣΣ
εε
Hence z is not
feasible for E contradicting that z is a w market equilibrium for E. Q.E.D.
Chapter 7Arbitrage Pricing of Securities
Consider a two period economy with ‘0’ denoting the current period and ‘1’ denoting thefuture period. Suppose there are m+1 assets and n possible future states of nature wherem and n are both natural numbers. Let aij ∈ℜ (: the set of real numbers) be the payofffrom one unit of investment in period ‘0’ in asset j if state of nature i prevails in period‘1’. We assume that the (m+1)st is riskless, i.e. ai,m+1 =1 for all i =1,…,n.A vector p ∈ℜm is called a price vector if pj denotes the price of the jth asset, for j =1,..m.The price of the (m+1)st asset is assumed to be 1.A portfolio is avector (x,v) ∈ℜm ×ℜ, where the jth coordinate of x, denotes the amount ofinvestment in the jth asset for j =1,…,m and v denotes the investment in the riskless asset.The yield vector from a portfolio (x,v) is A [ ]v
x , where the portfolio is now represented asa column vector [ ]v
x .
A price vector p ∈ℜm of the ‘m’ risky assets is said to satisfy the no-arbitrage condition ,
if there does not exist a portfolio (x,v) for which j
m
1jijxa∑
=
- ∑=
m
1jjjxp > 0, for all i = 1,…,m.
Theorem 7. 1: A price vector p ∈ℜm satisfies the no-arbitrage condition if and only if
there exists q ∈ n+ℜ , with ∑
=
n
1iiq = 1 such that i
n
1iijqa∑
=
= pj for all j =1,…,m.
Proof : Let B be the m ×n matrix where for j = 1,…,m and i =1,…,n, the (j,i)th entry of B,bij = aij. Let ξ be the vector in ℜn, all whose coordinates are equal to 1.
By Farkas’s Theorem,
ξTB q =
1p has no non-negative solution q, if and only if there
exists x∈ℜm and v∈ℜ such that ( xT,v)
ξTB ≥ 0 and pTx +v < 0.
Now, ( xT,v)
ξTB ≥ 0 and pTx +v < 0 holds if and only if xTB +vξ ≥ 0 and pTx +v < 0
holds. However, the statement [there exists x∈ℜm and v∈ℜ such that xTB +vξ ≥ 0 andpTx +v < 0] is equivalent to the statement [there exists x∈ℜm such that xTB >> (pTx) ξT,
which in turn is equivalent to the statement [there exists x∈ℜm such that j
m
1jijxa∑
=
-
∑=
m
1jjjxp > 0, for all i = 1,…,m.]. Q.E.D.
If corresponding to the n future states of nature, there are n assets , such that the payofffrom one unit of investment at time 0 in asset i is 1, if the state of nature at time 1 is i, and0 otherwise, then each of these n assets is called an Arrow- Debreu security.
A vector z ∈ℜn is called an Arrow- Debreu price vector for the price vector p∈ℜm if
∑=
n
1iijiaz = pj for j =1,…,m.
Consider a securities market where there is just one risky-asset, such that one unit ofinvestment in that risky asset in period zero yields ai units of return in period one if stateof nature ‘i’ is realized. Let p be the price of the security and let z be the price vector in
the Arrow- Debreu market. Suppose ∑=
n
1iiiaz > p. Then an investor could buy one unit of
the security in the primary market, sell ai units of the ith Arrow-Debreu security in period
zero and make a profit of ∑=
n
1iiiaz - p in period zero. In period one, if state of nature ‘i’ is
realized he receives ai from the primary market, which he pays off as dividend to theperson he sold the Arrow-Debrue security. Since by replicating this activity his profit getsmultiplied, there would eventually be an excess demand for the security in the primary
market and an excess supply in the Arrow-Debreu market. Now, suppose ∑=
n
1iiiaz < p.
Then an investor could sell one unit of the security in the primary market, buy ai units of
the ith Arrow-Debreu security in period zero and make a profit of p -∑=
n
1iiiaz in period
zero. In period one, if state of nature ‘i’ is realized he receives ai from the Arrow- debruesecurity that he has invested in, which he pays off as dividend to the person he sold thesecurity to in the primary market. Since by replicating this activity his profit getsmultiplied, there would eventually be an excess supply of the security in the primarymarket and an excess demand in the Arrow-Debreu market. Thus, Arrow-Debrue prices ensure that inter market arbitrage possibilities are ruled out.
If z is an Arrow- Debreu price vector for the price vector p, which satisfies 1zn
1ii =∑
=
,
then it is easy to verify that for all (x,v) ∈ℜm × ℜ: A [ ]vx : zT A [ ]v
x = pTx + v. This occursbecause zT A [ ]v
x = zT ]|B[ T ξ [ ]vx = zTBTx + zTξv = pTx + v, since p = Bz and zTξ = 1.
It follows directly from Theorem 7.1, that a vector z ∈ n+ℜ for the Arrow- Debreu
securities, which satisfies the no-arbitrage condition, if and only if 1zn
1ii =∑
=
. This is
because, of the specific nature of Arrow- Debreu securities.An Arrow- Debreu price vector z for the price vector p∈ℜm is said to satisfy the no-
arbitrage condition if 1zn
1ii =∑
=
.
Let C = BT (i.e. the matrix comprising the first m columns of A). The asset market is saidto be complete if the rank of C is n.Thus, if an asset market is complete, then any yield vector can be obtained by appropriatechoices of investments in the risky assets alone. In fact, it can be easily shown byapplying a simple induction argument on the cardinality of P, where P is any non-emptysubset of {1,…,n} and by letting ℜP = {x∈ℜn/ xi = 0 if i∉P}, that the condition [any
yield vector can be obtained by appropriate choices of investments in the risky assetsalone] is equivalent to the condition [any non-negative yield vector can be obtained byappropriate choices of investments in the risky assets alone].
The following theorem is a direct consequence of Theorem 7.1:
Theorem 7.2: There exists an Arrow-Debreu price vector z satisfying the no-arbitragecondition if and only if there exists a price vector p ∈ℜm in the primary market whichalso satisfies the then so does the price vector p ∈ℜm in the primary market. If a pricevector p ∈ℜm in a complete asset market satisfies the no-arbitrage condition, then thecorresponding Arrow-Debrue price vector is unique and satisfies the no-arbitragecondition.
Proof: Let z be an Arrow Debrue price vector which satisfies the no-arbitrage condition.
Then, by Theorem 7.1, it must be the case that z∈ n+ℜ and 1z
n
1ii =∑
=
. Let, p = Bz. Thus z is
an Arrow-Debrue price vector for p. Since, p = Bz, it follows by Theorem 7.1, that psatisfies the no-arbitrage condition.Now p be a vector of asset prices in the primary market which satisfies the no arbitrage
condition. By Theorem 7.1, there exists q ∈ n+ℜ , with ∑
=
n
1iiq = 1 such that Bq = p. Thus, q
is an Arrow-Debreu price vector. If the asset market is complete then, the rank of B is ‘n’.Thus C = BT has rank n. Let C1 be a set of n linearly independent columns of C. Let p1 bethe vector obtained from p by retaining only those coordinates which correspond to thecolumns included in C1. Thus, zTC1 = (p1)T and the non singularity of C1 implies, zT =(C1)-1(p1)T. This proves the theorem. Q.E.D.
In view of Theorem 7.2, it is desirable to obtain conditions under which there exists anArrow- Debreu price vector which satisfies the no-arbitrage condition, thus implying thatthe price vector p of the assets satisfy the no-arbitrage condition as well.
Theorem 7.3 : Suppose the following problem:(M) minimize pTx + v
subject to A ≥ y (x,v) ∈ℜm ×ℜ
has an optimal solution for some y ∈ℜn.Then p satisfies the no-arbitrage condition. Further, there exists an Arrow- Debreu pricevector for p which satisfies the no-arbitrage condition as well. Further, if the asset marketis complete, then such a price vector is unique.
Proof: Suppose, problem (M) has an optimal solution u*= (x*,v*)∈ℜm ×ℜ and towardsa contradiction suppose that there does not exist an Arrow- Debreu price vector for pwhich satisfies the no-arbitrage condition.Thus zTA = (pT,1) has no non-negative solution z.Thus, by Farkas’s Theorem, there exists u = (x1,v1) ∈ℜm ×ℜ such that Au ≥ 0 and pTx1 +v1 < 0.Thus, A(u+u*) ≥ y and pT(x*+x1) + v*+v1 < pTx* + v*, contradicting that (x*,v*) solves(M). Thus there exists an Arrow- Debreu price vector z for p which satisfies the no-arbitrage condition.The rest of the theorem follows from Theorem 7.1. Q.E.D.
PART 3
THE WAY AHEAD …
Chapter 8Homogeneous Programming Games
Let N = {1,…,n} be a set of agents (players), where n is a natural number. Let P(N)denote the power set of N. A cooperative game in characteristic function form hereafterreferred to simply as a game, is a function v: P(N)→ ℜ such that v(φ) = 0. A non-empty subset of N is called a coalition. Given a coalition S, the number v(S) canbe regarded as the worth of coalition S in the game v.Given a game v, an element x∈ nℜ is called a pay-off distribution. A pay-off distributionis said to be efficient if ∑
∈Naax = v(N).
An efficient pay-off distribution of a game v is said to belong to the core of v, if for all S⊂ N: [ S ≠ φ] implies [∑
∈Saax ≥ v(S)].
Let C(v) denote the set of pay-off distributions that belong to the core of v.Consider an economy consisting of n agents and a single infinitely divisible primaryresource. Each agent a ∈N = {1,…, n} is endowed with ca ∈ℜ units of the resource. Thisresource can be used to produce m goods. To produce one unit of good j, the availabletechnology requires aj units of the resource. Let A = (a1,…,am) be the row vector denotingthe input requirements to produce one unit of each good. Thus to produce a bundley∈ m
+ℜ of the goods the economy requires Ay units of the resource. Each agent caneither choose to contribute to the production process or abstain from it. If a coalition S ofagents decided to participate in the production process then the available amount of theresource is given by ∑
∈Sa
ac . Let X be either ℜm or m+ℜ or m
++ℜ . The revenue function for
the economy is a real valued function f on X, which is homogeneous of degree ‘k’ on X,where 'k' is a real number greater than or equal to one. If a coalition S of the agentsparticipate in production and produce a bundle y∈ m
+ℜ of the goods, the revenue thataccrues solely to coalition S is f(y). Hence if a coalition S participates in production, it isnatural for the coalition to solve the following problem:Maximize f(y)Subject to Ay ≤ ∑
∈Sa
ac ,
y ≥ 0.Let V(S) be the set of solutions for the above problem.An array <f, A, (ca)a∈N> is called a private ownership production economy,where:
(i) f: X → ℜ is a function which is homogeneous of degree k ≥ 1 on X; further, ifk > 1, then f(y) ∈ℜ+ for all y∈X;
(ii) X is either m++ℜ , m
+ℜ or ℜm;(iii) A is m-dimensional vector of real numbers;(iv) for every a ∈N = {1,…, n},ca ≥ 0 with c = ∑
∈Na
ac > 0.
.
Given a private ownership production economy <f, A, (ca)a∈N>, and any coalition S, letV(S) be the set of solutions to the following problem:
Maximize f(y)Subject to Ay ≤ ∑
∈Sa
ac ,
y ≥ 0.A game v is said to be a homogeneous programming game if there exists a privateownership production economy <f, A, (ca)a∈N> such that for every coalition S, thereexists y∈ V(S) such that v(S) = f(y). In this case we say that the homogeneousprogramming game v is generated by the private ownership production economy <f, A,(ca)a∈N>. Theorem 8.1: Let v be a homogeneous programming game generated by the privateownership production economy <f, A, (ca)a∈N>. Then C(v) ≠ φ.
Proof: Let y* ∈ V(N) such that v(N) = f(y*). Thus, y* solves:Maximize f(y)Subject to Ay ≤ c =∑
∈Na
ac ,
y ≥ 0.
Let xa = cca
f(y*) for all a∈N. Thus, ∑∈Na
ax = f(y*) = v(N).
Now, for any coalition S, ∑∈Sa
ax = c1 (∑
∈Sa
ac ) f(y*). Suppose r = ∑∈Sa
ax / ∑∈Na
ax . Thus, r
=c1 ∑
∈Sa
ac ≥ 0.
Further, ∑∈Sa
ax = rf(y*).
Case 1: f is homogeneous of degree one.
Since y* solves Maximize f(y)Subject to Ay ≤ c,y ≥ 0.and since f is homogeneous of degree one on X, ry* solves Maximize f(y)Subject to Ay ≤ rc = ∑
∈Sa
ac ,
y ≥ 0.whenever r ≥ 0. Thus, ry*∈V(S). Thus, ∑
∈Saax = rf(y*) = f(ry*) = v(S).
Thus, x∈ C(v). Case 2: f is homogeneous of degree k > 1. Thus, f(y) ≥ 0 for all y∈X.
Let <g, A, (ca)a∈N> be the private ownership production economy, where g(y) = k1
)]y(f[for all y∈X.Since g is a monotone increasing transformation of f, y* solves:Maximize g(y)Subject to Ay ≤ c =∑
∈Na
ac ,
y ≥ 0.Since, g is homogeneous of degree one on X, ry* solves Maximize g(y)Subject to Ay ≤ rc = ∑
∈Sa
ac ,
y ≥ 0.
However, the above implies that ry* solves Maximize f(y)Subject to Ay ≤ rc = ∑
∈Sa
ac ,
y ≥ 0.Thus, v(S) = f(ry*) = rkf(y*) ≤ rf(y*) (: since 0 ≤ r ≤ 1 and k > 1) = ∑
∈Saax . Thus, C(v) ≠ φ.
Q.E.D.
Example: Let m = 2 and let f: ℜ→ℜ+2 be defined as f(y) = α−α 1
21 yy for all y∈X = 2+ℜ . Let
<f, A, (ca)a∈N> be a private ownership production economy.
Let y* = (α1a
c , (1-α)2a
c ). Clearly, y* solves
Maximize f(y)Subject to Ay ≤ c =∑
∈Na
ac ,
y ≥ 0.
Further, f(y*) = c[ α−α α−α 1
21
)a
1()a
( . Let xa = ca[ α−α α−α 1
21
)a
1()a
( ]for all a∈N. Let v be the
homogeneous programming game generated by <f, A, (ca)a∈N>. Then, x∈C(v).
An array <f, A, (ca)a∈N>, where:1. f: ℜm → ℜ is a C1 function which is homogeneous of degree one on ℜm;2. A is a m- dimensional row vector matrix real numbers;3. for every a ∈N = {1,…, n},ca ≥ 0, with c = ∑
∈Na
ac > 0
is called a private ownership linear production economy.If <f, A, (ca)a∈N> is a private ownership linear production economy, then by Theorem 3.2,f(y) = Df(0).y for all y ∈ℜm.
Given a private ownership linear production economy <f, A, (ca)a∈N>, and any coalitionS, let V(S) be the set of solutions to the following problem:Maximize f(y)Subject to Ay ≤ ∑
∈Sa
ac .
y ≥ 0.A game v is said to be a linear programming game if there exists a private ownershiplinear production economy <f, A, (ca)a∈N> such that for every coalition S, there exists y∈V(S) such that v(S) = f(y). In this case we say that the linear programming game v isgenerated by the private ownership linear production economy <f, A, (ca)a∈N>.
Corollary of Theorem 8.1: Let v be a linear programming game generated by the privateownership linear production economy <f, A, (ca)a∈N>. Then C(v) ≠ φ.
Chapter 9Market Equilibrium Solutions : Some Simple Axiomatics
An important problem in social choice theory is one of allocating a given bundle ofresources among a finite number of agents. The study of such problems is alsoknown as problems of fair division. There are basically two approaches to solving such problems. In one approach, we areinterested in first choosing a vector of utilities for the agents, and then choosing anallocation, conformable with the utility vector. There is of course the underlyingassumption of agents being equipped with utility functions, which expresses theirpreferences for consumption bundles. The other approach to the problem proceeds byoperating some variant of the market mechanism from an initial distribution of income.The distribution of income is part of the process of resolving the problem. One suchmethod is equal division of income. The equal-income method is treated as a distinct category all by itself, which is notamenable to non-symmetric generalizations. This is true, at least in so far as the axiomatictheory of solutions to such problems go. Our purpose here is to obtain an axiomaticcharacterization for the entire class of solutions consisting of the equal income marketequilibrium solution as well as its non-symmetric variants.
We consider economies with a finite number of agents. Let N = {1,…,n} be the set ofagents for some natural number n ≥ 2.An economy is a list E = < (ui)i∈N, ω> where ω ∈ 2
++ℜ is the total initial endowment ofgoods in the economy and ui : 2
+ℜ →ℜ is the utility function of agent i∈N. Let ℑ1 be theset of economies E = < (ui)i∈N, ω>, such that for all i∈N:(i) ui is concave;(ii) ui weakly increasing; i.e. for all a,a*,b,b* ≥ 0: [a > a*, b > b*] implies [ui(a,b) >
ui(a*, b*)].(iii) ui continuously differentiable in 2
++ℜ , with all partial derivatives strictly positive;(iv) a,a*,b,b* ≥ 0: [ (a,b) ∈ 2
++ℜ , (a*,b*) ∈ 2+ℜ \ 2
++ℜ ] implies [ui(a,b) > ui(a*, b*)].
Let ℑ2 be the set of economies E = < (ui)i∈N, ω>, such that the following is true for somep∈ 2
++ℜ :
For all i∈N: ui(a,b) = pT
ba
for all (a,b)∈ 2+ℜ .
Let ℑ = ℑ1∪ℑ2.Given E = < (ui)i∈N, ω>∈ℑ we defineA(E) = { <(xi,yi)/i∈N>/ (xi,yi)∈ 2
+ℜ for all i∈N and∑∈Ni
ii )y,x( = ω}
A(E) is the set of feasible allocations for E∈ℑ. Let A0(E) = { <(xi,yi)/i∈N>∈ A(E)/(xi,yi)∈ 2
++ℜ for all i∈N}.Let w∈ n
++ℜ be a given income distribution.
Given E = < (ui)i∈N, ω>∈ℑ a feasible allocation <(xi,yi)/i∈N> is said to be a marketequilibrium allocation for E is there exists p∈ 2
++ℜ such that (p, <(xi,yi)/i∈N> is a marketequilibrium for the income distribution w. An interior market equilibrium allocation forE = < (ui)i∈N, ω>∈ℑ is a market equilibrium allocation <(xi,yi)/i∈N> such that for alli∈N: (xi,yi)∈ 2
++ℜ . Given E∈ℑ, let G(E) be the set of all market equilibrium allocations for E and G0(E) bethe set of all interior market equilibrium allocations for E. Thus, G0(E) = G(E)∩A(E).
Given E = < (ui)i∈N, ω> ∈ℑ an <(xi,yi)/i∈N>∈A(E) is said to be efficient for E, if<(xi,yi)/i∈N> is Strongly Pareto Optimal for the economy described by E.Given E∈ℑ the set of efficient allocations is non-empty since A(E) is compact and theutility functions are continuous.
It is easy to verify that, if E = < (ui)i∈N, ω> and (p, <(xi,yi)/i∈N> is a market equilibrium
for the income distribution w, then pT
i
i
yx
= wi for all i∈N.
Proposition 9.1:- Let E = < (ui)i∈N, ω>∈ℑ1 and <(xi,yi)/i∈N> be a marketequilibrium allocation for E. Then (xi,yi) ∈ 2
++ℜ for all i∈N.Proof:- Let E be as above and let <(xi,yi)/i∈N> be a market equilibrium allocationfor E. Since wi > 0, if p∈ 2
++ℜ is the price vector associated with <(xi,yi)/i∈N>,then there exists z∈ 2
++ℜ such that pTz ≤ wi. If (xi,yi)∈ 2+ℜ \ 2
++ℜ ,then, (z),u<)x(u iii contradicting that (p, <(xi,yi)/i∈N>) is a market equilibriumwith respect to the income distribution w. Thus, (xi,yi) ∈ 2
++ℜ for all i∈N. Q.E.D.
Remark: If E = < (ui)i∈N, ω>∈ℑ2 be such that or all i∈N: ui(a,b) = pT
ba
for all
(a,b)∈ 2+ℜ , then G(E) = {<(xi,yi)/i∈N>∈A(E)/ for all i∈N, pT
i
i
yx
= pTεi} and
G0(E) = {<(xi,yi)/i∈N>∈A0(E)/ for all i∈N, pT
i
i
yx
= pTεi}.
Given E ∈ ℑ, let [A0(E)] denote the set of all non-empty subsets of A0(E).
A solution on ℑ is a function F: ℑ → Uℑ∈E
0 )]E(A[ , such that for all E∈ℑ: F(E) ⊂
A0(E). A solution F on ℑ is said to satisfy Local Independence (LI) if for all<(xi,yi)/i∈N>∈F(E): [j∈N,(xj,yj)∈ 2
++ℜ and pj ∈ 2++ℜ supports uj at (xj,yj)] implies
<(xi,yi)/i∈N>∈F( E ), where E = < (vk)k∈N, ω>∈ℑ is such that vk = uk for all k≠ jand pj ∈ 2
++ℜ supports uj at (xj,yj).
It is easily seen that G0 satisfies LI on ℑ.
The next condition we invoke is a type of individual rationality condition.A solution F on ℑ is said to satisfy w-Individual Rationality (w-IR) if for all E =
< (ui)i∈N, ω>∈ℑ and <(xi,yi)/i∈N>∈F(E) : ui(xi,yi) ≥ ui( ω∑∈Nj
j
i
ww ) for all i∈N.
εi ≡ ω∑∈Nj
j
i
ww is often referred to as agent i's entitlement in E under the income
distribution w.
We postulate now the following property for a solution F on ℑ called non-discrimination of Pareto indifferent solutions.
A solution F on ℑ is said to satisfy Non-Discrimination of Pareto IndifferentSolutions (ND) if for all E ∈ℑ, <(xi,yi)/i∈N>∈F(E) and <(x*i,y*i)/i∈N>∈A0(E):[ui(xi,yi) = ui(x*i,y*i) for all i∈N] implies <(x*i,y*i)/i∈N>∈F(E).It is easy to see that G0 satisfies ND for all w ∈ n
++ℜ .
The following property is called Monotonicity.A solution F on ℑ is said to satisfy Monotonicity (MON) if for all E1= < (ui)i∈N,ω>∈ℑ, E2 = < (vi)i∈N, ω> ∈ℑ and <(xi,yi)/i∈N>∈F(E1): [{(a,b)∈ 2
+ℜ / ui(xi,yi) ≥ ui(a,b)}⊂ {(a,b)∈ 2+ℜ / vi(xi,yi) ≥ vi(a,b)}for all i∈N]
implies <(xi,yi)/i∈N>∈F(E2).
Lemma 9.1: Let E = < (ui)i∈N, ω>∈ℑ, <(xi,yi)/i∈N> ∈ A(E) and p∈ 2++ℜ .
If for all i∈N, pT
i
i
yx
≥ pTεi then for all i∈N, pT
i
i
yx
= pTεi.
Proof: Obvious.
Lemma 9.2: Suppose F on ℑ satisfies ND and w-IR.
Let E = < (ui)i∈N, ω>∈ℑ be such that p∈ 2++ℜ and for all i∈N, p supports ui at
ω∑∈Nj
j
i
ww (i.e. for all i∈N: [(a,b)∈ 2
+ℜ , ui(a,b) ≥ ui(εi)] implies [pT
ba
≥ pTεi].
Then F(E) = {<(xi,yi)/i∈N> ∈ A0(E)/ ui(xi,yi) = ui(εi) for all i∈N}.
Proof: By w-IR, <(xi,yi)/i∈N> ∈ F(E) implies ui(xi,yi) ≥ ui(εi) for all i∈N, where ei
is agent is entitlement.
If p supports εi for all i∈N, and if for some j∈N, uj(xj,yj) > uj(εj) then pT
j
j
yx
>
pTεj.
Clearly pT
i
i
yx
≥ pTεi , for all i∈N.
pT ∑∈
Nii
i
yx
> pTi
Ni∑∈
ε contradicting <(xi,yi)/i∈N> ∈ A(E).
Thus, <(xi,yi)/i∈N> ∈ F(E) implies ui(xi,yi) = ui(εi) for all i∈N.By ND, [<(xi,yi)/i∈N> ∈ F(E)] if and only if [<(xi,yi)/i∈N> ∈ A0(E) and ui(xi,yi) =ui(εi) for all i∈N].
Q.E.D.Proposition 9.2: Let F be any solution on ℑ which satisfies MON, w-IR and ND.Then, G0(E) ⊂ F(E) for all E∈ℑ.
Proof: Consider any E = < (ui)i∈N, ω>∈ℑ and any <(xi,yi)/i∈N> ∈ G0(E).
Thus, pT
j
j
yx
= pTεj for all j∈N, where (p, <(xi,yi)/i∈N>) is a market equilibrium
with respect to the income distribution w.Since (p, <(xi,yi)/i∈N>) is a market equilibrium with respect to the income
distribution w, for all i∈N and (a,b)∈ 2+ℜ : [pT
i
i
yx
≥ pT
ba
] implies [ui (xi,yi) ≥
ui(a,b)].
Let E1 = < (vi)i∈N, ω>∈ℑ2 be such that for all i∈N: vi(a,b) = pT
ba
for all (a,b)∈
2+ℜ .
By Lemma 9.2, <(xi, yi)/ i∈N> ∈F(E1).By MON, <(xi, yi)/ i∈N> ∈F(E). Q.E.D.
Proposition 9.3: Let F be any solution on ℑ which satisfies LI, w-IR and ND.Then, G0(E) ⊂ F(E) for all E∈ℑ.Proof: Consider any E = < (ui)i∈N, ω>∈ℑ and any <(xi,yi)/i∈N> ∈ G0(E). Case 1: E∈ℑ2. Then by Lemma 9.2, the fact that F satisfies w-IR and ND impliesthat F(E) = G0(E).
Case 2: E∈ℑ1. Thus, pT
j
j
yx
= pTεj for all j∈N, where (p, <(xi,yi)/i∈N>) is a
market equilibrium with respect to the income distribution w.
Since (p, <(xi,yi)/i∈N>) is a market equilibrium with respect to the income
distribution w, for all i∈N and (a,b)∈ 2+ℜ : [pT
i
i
yx
≥ pT
ba
] implies [ui (xi,yi) ≥
ui(a,b)].
Let E1 = < (vi)i∈N, ω>∈ℑ2 be such that for all i∈N: vi(a,b) = pT
ba
for all (a,b)∈
2+ℜ .
By Lemma 9.2, <(xi, yi)/ i∈N> ∈F(E1).By LI, <(xi, yi)/ i∈N> ∈F(E). Q.E.D.
In order to obtain the reverse inclusion, we need a weak version of ParetoOptimalityA solution F on ℑ is said to satisfy Binary Efficiency (BE) if for all E = < (ui)i∈N,ω>∈ℑ ,<(xi,yi)/i∈N> ∈ F(E) and j,k∈N the following is true: there does not exist(x*j,y*j), (x*k,y*k) ∈ 2
+ℜ , with x*j + x*k = xj + xk, y*j + y*k = yj + yk, uj(x*j,y*j) ≥uj(xj,yj), uk(x*k,y*k) ≥ uk(xk,yk) with at least one of the two inequalities being strict.Once again Binary Efficiency being a weaker version of Pareto Optimality is seento be easily satisfied by G.
Proposition 9.4: If a solution F on satisfies (BE), (w-IR) and (LI) then F(E) ⊂G0(E) for all E∈ℑ.Proof:- Let E = < (ui)i∈N, ω>∈ℑ .
Case 1: E∈ℑ2. Suppose, For all i∈N: ui(a,b) = pT
ba
for all (a,b)∈ 2+ℜ . By BE and
w-IR, F(E) = {<(xi,yi)/i∈N>∈A(E)/ for all i∈N, pT
i
i
yx
= pTεi} = G0(E).
Case 2: E∈ℑ2. Let <(xi,yi)/i∈N> ∈ F(E). By BE and Theorem 6.2, for all i,j ∈Nwith i ≠j, there exists p{i,j}∈ 2
++ℜ , such that for k∈{i,j}, p{i,j} supports uk at (xk,yk).If n =2, then this implies that <(xi,yi)/i∈N> is efficient for E. If n > 2, let i,j,k ∈N,with i≠j≠k. Thus, p{i,j} supports uj at (xj,yj), p{j,k} supports uj at (xj,yj),and thecontinuous differentiability of uj implies that there exists λ{i,j,k} > 0, such that p{i,j}
= λ{i,j,k}p{j,k}. Thus, there exists p∈ 2++ℜ such that (a) pTω = ∑
∈Niiw ; (b) for all i∈N,
p supports ui at (xi,yi). Let E1 = < (vi)i∈N, ω>∈ℑ2 be such that for all i∈N: vi(a,b) =
pT
ba
for all (a,b)∈ 2+ℜ . By LI, <(xi,yi)/i∈N>∈F(E1). Since F satisfies (w-IR),
pT
i
i
yx
≥ pTεi = wi for all i∈N. By Lemma 9.1, pT
i
i
yx
= pTεi = wi for all i∈N.
Thus, (p, <(xi,yi)/i∈N>) is a market equilibrium with respect to the incomedistribution w. Hence, <(xi,yi)/i∈N> ∈ G0(E). Q.E.D.
As a consequence of Propositions 9.3 and 9.4 we have the followingcharacterisation theorem.
Theorem 1: Let F be any solution on ℑ which satisfies LI, w-IR and ND. Then,F(E) = G0(E) for all E∈ℑ.