Positive Solutions of Real Homogeneous Algebraic Inequalities

Abstract

We show a simple and explicit reduction of the problem of the existence of positive solution of a system of real algebraic homogeneous inequalities in $R_{ + }^{n}$ to solvability of a standard concave programming problem, which in turn is equivalent to checking whether a particular constraint of a finite system of convex inequalities is redundant. We illustrate this result on two well-known problems in mathematical economics: the weak separability problem and the collective consumption behavior for the homogeneous utilities.

Appendix

Proposition 1

The complete PH-separability holds if and only if the system of (4)–(6) is consistent.

Proof

At first, recall that Euler’s homogeneous function theorem can be reformulated as follows for the case of PH well-behaved functions $f\left( \cdot \right):R_{ + }^{n} \to R_{ + }$. If $x_{0} \in R_{ + }^{n}$ and $p \in \partial f\left( {x_{0} } \right)$, i.e. p belongs to a supergradient set of $f\left( \cdot \right)$ at x₀, then $px_{0} = f\left( {x_{0} } \right)$. Indeed, it follows from concavity of $f\left( \cdot \right)$ that the inequality $f\left( x \right) - f\left( {x_{0} } \right) \ge p\left( {x - x_{0} } \right)$ holds for all $x \in R_{ + }^{n}$. Let $x =\uplambda\,x_{0} ,\;\uplambda > 0$. As $f\left( \cdot \right)$ is positive homogeneous, then for all positive λ it holds that $\left( {\uplambda - 1} \right)f\left( {x_{0} } \right) \geq \left( {\uplambda - 1} \right)px_{0}$, hence, $px_{0} = f\left( {x_{0} } \right)$.

Necessity. Assume that statistics is complete PH-separable. By definition it means that there exists a pair of a well-behaved PH functions $u_{0} \left( {q,z} \right)$ and $u_{1} \left( y \right)$ such that

$$ \left( {q^{t} ,y^{t} } \right) \in {\text{Arg}}\;\hbox{max} \left\{ { u_{1} \left( {q,u_{1} \left( y \right)} \right) \left| { p^{t} q} \right. + x^{t} y \le p^{t} q^{t} + x^{t} y^{t} ,\;\;q \ge 0,\;\;y \ge 0} \right\} ,\quad t = 1, \ldots ,T. $$

(11)

As PH well-behaved functions are monotonic, then it follows from (11) that

$$ \left( {q^{t} } \right) \in {\text{Arg}}\;\hbox{max} \left\{ { u_{1} \left( y \right) \left| { x^{t} y \le x^{t} y^{t} } \right.,\;y \ge 0} \right\} ,\quad t = 1, \ldots ,T. $$

Define the Young transforms:

$$ v_{0} \left( {p,s} \right)\mathop = \limits^{\text{def}} \mathop {\inf }\limits_{{\left\{ {q \ge o,z \ge 0\left| {u_{0} \left( {q,z} \right)} \right. > 0} \right\}}} \frac{pq + sz}{{u_{0} \left( {q,z} \right)}},\;\;\;\;\;\;v_{1} \left( x \right)\mathop = \limits^{\text{def}} \mathop {\inf }\limits_{{\left\{ {y \ge o\left| {u_{1} \left( y \right)} \right. > 0} \right\}}} \frac{xy}{{u_{1} \left( y \right)}}. $$

By definition it holds

$$ \begin{aligned} & v_{1} \left( {x^{t} } \right) u_{1} \left( {y^{t} } \right) = x^{t} y^{t} , \\ & v_{1} \left( {x^{\uptau} } \right) u_{1} \left( {y^{t} } \right) \le x^{\uptau} y^{t} , \\ & v_{0} \left( {p^{t} ,v_{1} \left( {x^{t} } \right)} \right) u_{0} \left( {q^{t} ,u_{1} \left( {y^{t} } \right)} \right) = p^{t} q^{t} + v_{1} \left( {x^{t} } \right) u_{1} \left( {y^{t} } \right), \\ & v_{0} \left( {p^{\uptau} ,v_{1} \left( {x^{\uptau} } \right)} \right) u_{0} \left( {q^{t} ,u_{1} \left( {y^{t} } \right)} \right) \le p^{\uptau} q^{t} + v_{1} \left( {x^{\uptau} } \right) u_{1} \left( {y^{t} } \right), \\ & t,\uptau = 1, \ldots ,T. \\ \end{aligned} $$

Now necessity follows as $\uplambda_{t} = \frac{1}{{v_{1} \left( {x^{t} } \right)}},\;\;\upmu_{t} = \frac{1}{{v_{0} \left( {p^{t} ,v_{1} \left( {x^{t} } \right)} \right)}},\;\;t = 1, \ldots ,T$ are solutions of the system of (4)–(6).

Sufficiency. Set $u_{1} \left( y \right)\mathop = \limits^{\text{def}} \hbox{min} \left( {\uplambda_{t} x^{t} y\left| {t = 1, \ldots ,T} \right.} \right)$. By homogeneous Afriat’s theorem it holds $\left( {y^{t} } \right) \in {\text{Arg}}\;\hbox{max} \left\{ { u_{1} \left( y \right) \left| { x^{t} y \le x^{t} y^{t} } \right.,\;\;y \ge 0} \right\} ,\;\;t = 1, \ldots ,T$.

Set $u_{0} \left( {q,z} \right)\mathop = \limits^{\text{def}} \hbox{min} \left( {\upmu_{t} \left( {p^{t} q + \frac{1}{{\uplambda_{t} }}z} \right)\left| {t = 1, \ldots ,T} \right.} \right)$.

Let $q \ge 0,\;\;y \ge 0,\;\;p^{t} q + x^{t} y \le p^{t} q^{t} + x^{t} y^{t}$. Then it holds

$$ \begin{aligned} & p^{t} q^{t} + x^{t} y^{t} = p^{t} q^{t} + \frac{1}{{\uplambda_{t} }}u_{1} \left( {y^{t} } \right) = \frac{1}{{\upmu_{t} }}u_{0} \left( {q^{t} ,u_{1} \left( {y^{t} } \right)} \right), \\ & x^{t} y \geq \frac{1}{{\uplambda_{t} }}u_{1} \left( y \right),\;\;\;p^{t} q + x^{t} y \ge p^{t} q + \frac{1}{{\uplambda_{t} }}u_{1} \left( y \right) \ge \frac{1}{{\upmu_{t} }}u_{0} \left( {q,u_{1} \left( y \right)} \right). \\ \end{aligned} $$

And it follows that $u_{0} \left( {q^{t} ,u_{1} \left( {y^{t} } \right)} \right) \ge u_{0} \left( {q,u_{1} \left( y \right)} \right)$ and, thus, (11) holds.