Consider two continuous functions f,g mapping the interval [0,S] of the real line into R. Let f also be strictly increasing. We are interested in the set of probability distributions on the interval [0,S] that maximize the expectation of f subject to the constraint that the expectation of g be no greater than a constant. We provide a sufficient condition on the pair (f,g) for the solution to this linear programming problem to be unique and show that this sufficient condition is satisfied "generically."Linear Programming

Nicola Persico

English

Research Papers in Economics

 
 
 
 
 
 
by 
 
 
http://ssrn.com/abstract=659143  
Nicola Persico 
 
 “Generic Uniqueness of the Solutions to a Continuous Linear  
Programming Problem” 
PIER Working Paper 05-010 
Penn Institute for Economic Research 
Department of Economics 
University of Pennsylvania 
3718 Locust Walk 
Philadelphia, PA 19104-6297 
pier@econ.upenn.edu 
http://www.econ.upenn.edu/pier Generic Uniqueness of the Solutions to a Continuous Linear
Programming Problem
Nicola Persico￿
February 14, 2005
Abstract
Let f : R ! R be strictly increasing. We are interested in the set of probability distributions
￿ on the interval [0;S] that solve the linear programming problem max￿
R S
0 f (p) d￿(p) subject
to
R S
0 g (p) d￿(p) ￿ C: We provide a su¢ cient condition on the pair (f;g) for the solution to
the linear programming problem to be unique and show that this su¢ cient condition is satis￿ed
￿generically.￿
Keywords: Linear Programming
JEL number: C60
1 The problem
Let [0;S] be an interval of the real line. Let the function f : [0;S] ! [0;1] be continuous and strictly
increasing. Let the function g : [0;S] ! R be continuous. Let M denote the set of probability
distributions on [0;S]. We wish to characterize the solutions to the following linear problem:
max
￿2M
Z S
0
f (p)d￿(p) (P)
s.t.
Z S
0
g (p)d￿(p) ￿ C:
To guarantee existence of a solution, in what follows we assume that minp2[0;S] g (p) ￿ C. We will
prove the following result showing that, ￿generically,￿the solution to problem P is unique.
Proposition 1 The solution to problem P is unique and puts positive probability on at most two
points whenever g ￿ f￿1 belongs to a set which is a dense G ￿ ￿ set in the space of all continuous
functions equipped with the supnorm.
￿Department of Economics, University of Pennsylvania, 3718 Locust Walk, Philadelphia PA 19104; email: per-
sico@ssc.upenn.edu. Web: http://www.ssc.upenn.edu/~persico.
12 Characterization of the solutions to problem P
Let T denote the set of continuous functions x(t) : [0;S] ! R with the property that at least one
of the hyperplanes that supports x(t) from below is tangent to x(t) at more than two points. In
what follows, the superscript C indicates set complements.
Lemma 2 If g ￿f￿1 belongs to T C, any solution to problem P places all the probability on one or
two points in [0;S]:
Proof: If ￿￿ is a solution to problem P, there exists a number ￿ ￿ 0 such that ￿￿ maximizes
the Lagrangean
L(￿;￿) =
Z S
0
[f (p) ￿ ￿g (p)]d￿(p) + ￿C:
The case ￿ = 0 is easily disposed of, since in that case the (unique) solution requires all the mass
to be places on S. Let us turn, then, to the case ￿ > 0. If ￿￿ is to maximize the Lagrangean, the
set of p￿ s on which ￿￿ may place positive probability is given by
A = argmax
p [f (p) ￿ ￿g (p)]:
By de￿nition of A, there is a number M such that
f (p) ￿ ￿g (p) = M for p 2 A
f (p) ￿ ￿g (p) < M for p = 2 A
If ￿￿ puts positive mass on more than two points, then the cardinality of A would have to exceed
2. Consider the transformation ’(p) = f￿1 (p=S). The function ’ is a one-to-one mapping of [0;S]
onto itself. We can therefore write
f (’(p)) ￿ ￿g (’(p)) = M for ’(p) 2 A
f (’(p)) ￿ ￿g (’(p)) < M for ’(p) = 2 A;
or, with the obvious meaning of symbols,
p
S ￿ ￿g (’(p)) = M for p 2 ’￿1 (A)
p
S ￿ ￿g (’(p)) < M for p = 2 ’￿1 (A):
Note that the set ’￿1 (A) has the same cardinality of A. Thus, if A has cardinality greater than
2, it means that the two numbers ￿ and M are such that the straight line identi￿ed by 1
￿
￿ p
S ￿ M
￿
is tangent to the function g
￿
f￿1 (p=S)
￿
at more than two points and never exceeds it. This means
that there is a tangent hyperplanes to the set
Y =
￿
(p;y) : y ￿ g
￿
f￿1 (p=S)
￿￿
2which makes contact with the set Y at more than two points. }
Lemma 3 If g ￿ f￿1 belongs to T C, the solution to problem P is unique.
Proof: If the constraint is not binding then clearly ￿￿ is unique. If the constraint is binding,
Lemma 2 guarantees that there are two points, call them pL and pH (here we allow for the
possibility that pL = pH); such that
￿￿(pL) + ￿￿(pH) = 1; (1)
where the constraint in problem (P) reads
g(pL) ￿ ￿￿(pL) + g(pH) ￿ ￿￿(pH) = C: (2)
The system of equations given by (1) and (2) identi￿es a unique pair ￿￿(pL);￿￿(pH) unless the
system has less than full rank, i.e., unless g(pL) = g(pH). But in that case, since f is strictly
increasing, the (unique) solution requires all the mass to be placed on maxfpL;pHg: }
3 Characterization of the set T
We now show that the set T is ￿small￿ in this topological sense: that its complement T C is a
dense G ￿ ￿ set, i.e., it is dense and the countable intersection of sets that are open, in the space
of continuous functions equipped with the supnorm.
To this end, let C (x(￿)) denote the convex hull of x(￿), i.e., the function that is given by the
intersection of all the epigraphs of the a¢ ne linear functions that lie below the function x(t) and
are tangent to it. Consider all open intervals ￿ of the real line such that for all t 2 ￿, the convex
hull Cx(t) coincides with a linear a¢ ne function; and such that this property is not enjoyed by any
open interval containing ￿: Among those, select all those intervals ￿0 such that x(t) ￿ Cx(t) = 0
for some t 2 ￿0. If there are no such ￿0; then x = 2 T . If there are such intervals, let the extremes of
the largest among these be denoted by ax and bx (if there are several largest intervals, pick one at
random.) Clearly, x(t) and C (x(t)) agree at t = ax;bx. Furthermore, denote
mx = arg min
t2(ax;bx)
￿
￿ ￿
￿t ￿
ax + bx
2
￿ ￿
￿ ￿
s.t. x(t) ￿ C (x(t)) = 0:
3Clearly, if x 2 T we have ax < mx < bx. The point mx represents the point of tangency of the
supporting hyperplane that is closest to the middle of the interval (ax;bx). Denote Ix = [ax;bx] For
x 2 T , Ix represents the largest interval at which any tangent hyperplane to x is tangent at more
than two points.
Denote
Tn;m =
￿
x 2 T such that (bx ￿ ax) ￿
1
n
and minfmx ￿ ax;bx ￿ mxg ￿
1
m
￿
:
Clearly,
T =
S
n￿1
m￿n
Tn;m;
and so to obtain the required characterization for T C we need to show that for each n;m, the set
(Tn;m)
C is open and that the set T C is dense in the set of all continuous functions endowed with
the supnorm. We start with showing dense.
Lemma 4 The set T C is dense in the set of continuous functions endowed with the supnorm.
Proof: We show that in the supnorm neighborhood of every continuous function there is a
function x0 2 T C. Fix, then, an arbitrary continuous function x. If x 2 T C there is nothing to
prove, so we can focus on the case x 2 T . Our task is to show that there is a function ~ x close to
x with the property that no supporting hyperplane to ~ x has more than two contact points with
~ x. Let H denote the set of hyperplanes that have more than two contact points with x. Elements
of H are identi￿ed by their slope h. For every hyperplane h 2 H, take the sup and the inf of all
its contact points t and call those ah and bh. Consider now a continuous function dg (t) which
is equal to 0 for every t unless t 2 (ah;bh) for some h 2 H, in which case dg (t) assumes values
strictly between zero and 1. Let ~ x" (t) ￿ [1 + " ￿ dg (t)]￿x(t). For any " > 0, the function ~ x" (t) has
exactly the same set of supporting hyperplanes as x(t). Moreover, by construction no hyperplane
is tangent to ~ x" (t) at more than two points. Since the function ~ x" (t) can be made arbitrarily close
to x(t) in the supnorm by choosing " to be small, the set T C is shown to be dense, as desired. }
Lemma 5 For each n ￿ 1;m ￿ n, the set (Tn;m)
C is open in the set of continuous functions
endowed with the supnorm.
Proof: The result will follow if we show that Tn;m is closed. To this end, let￿ s ￿x an arbitrary
sequence fxig that converges to some x in the supnorm. For each xi, consider the set Di =
4faxi;mxi;bxig: Since the interval [0;S] is compact in the Euclidean distance, the space of subsets
of [0;S] endowed with the Hausdor⁄distance is compact (see Theorem 3.71 in Aliprantis and Border
1999). The sequence of sets fDig therefore contains a convergent subsequence fDjg. Consequently,
the limits (in the euclidean distance) of axj;mxj; and bxj for j ! 1 exist. Denote these limit
points by a;m; and b respectively. It is clear that (b ￿ a) ￿ 1
n and minfm ￿ a;b ￿ mg ￿ 1
m.
Since the mapping C (￿) is continuous in the supnorm, we have
lim
j!1
C (xj (a)) = C (x(a)) and lim
j!1
C (xj (b)) = C (x(b))
For each j, the function C (xj (t)) is linear a¢ ne for t 2
￿
axj;bxj
￿
. Thus, for t 2 [a;b], the function
C (x(t)) is the limit in the supnorm of linear a¢ ne functions, hence a linear a¢ ne function itself.
To ￿nish the proof, we use the continuity of the mapping C (￿) to write
x(m) = lim
j!1
xj
￿
mxj
￿
= lim
j!1
C
￿
xj
￿
mxj
￿￿
= C (x(m)):
We have shown that C (x(t)) is linear a¢ ne for t 2 [a;b], and that x(m) = C (x(m)): This shows
that x 2 Tn;m, and thus that Tn;m is closed. }
4 Conclusion
We have shown that T C =
T
n￿1
m￿n
(Tn;m)
C is dense, and is the countable intersection of sets that are
open, in the space of continuous functions endowed with the supnorm. This fact, in conjunction
with with Lemma 3, proves Proposition 1.
References
[1] Aliprantis, Charalambos D., and Border, Kim C. In￿nite Dimensional Analysis. 2nd ed.
Springer Verlag, Heidelberg, 1999.
5

Generic Uniqueness of the Solutions to a Continuous Linear Programming Problem

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Generic Uniqueness of the Solutions to a Continuous Linear Programming Problem

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