Property of the solution for a specific system of non-linear equations.

Question

We are stuck on a proof, and would appreciate any help:

Let $\gamma >1$ be a known scalar and let $g,h=1,...,G$ and $s=1,...,S$.

Let $\pi _{gs}$, $\beta _{s}$ and $y_{g}$ be known variables with $\pi _{gs},\beta _{s},y_{g}\in (0,1)$ and $\sum_{s=1}^{S}\pi _{gs}=\sum_{s=1}^{S}\beta _{s}=\sum_{g=1}^{G}y_{g}=1$.

$x_{g}$ for $g=1,...,G$ and $w_{s}$ for $s=1,...,S$ are real positive numbers that satisfy the following equations

$ 1=\sum_{s=1}^{S}\pi _{gs}\beta _{s}x_{g}^{\gamma }w_{s}\text{ for all }% g=1,...,G $

and

$ 1=\sum_{g=1}^{G}\pi _{gs}y_{g}w_{s}x_{g}\text{ for all }s=1,...S $

We want to show that $\left( I_{g}-I_{h}\right) \left( x_{g}-x_{h}\right) \leq 0$ for all $g,h=1,...,G$, where $I_{g}\equiv \sum_{s=1}^{S}\frac{\pi _{gs}\beta _{s}}{\sum_{h}\pi _{hs}y_{h}}$.

Note that a necessary condition for this property is that $I_{g}=I_{h}$ iff $% x_{g}=x_{h}$, also something we would like to show.

Matlab simulations indicate that both of these properties hold.

Answer 1

This is perhaps obvious, but it may help to get started: If $\pi _{gs} = \pi _{s}$ for all g then $I_{g} = 1$ for all g, and a solution is $x_{g} = 1$ for all g and $w_{s} = 1/\pi _{s}$ for all s.

Answer 2

This may help. Think of the problem as showing a property of the solution to a fixed point problem. The fixed point problem is $f(x) = x$, where $$f _{g}(x) = (\sum_{s=1}^{S}\frac{\pi _{gs}\beta _{s}}{\sum_{k}\pi _{ks}y_{k}x_{k}})^{-1/\gamma }$$

It can be easily shown that this is a contraction mapping that goes from the space $X\equiv\{x\in\ R^G |x_g \geq I_h ^{(1/(1-\gamma))}$for all $g,h$} to that same space. You can then use Banach's Fixed Point Theorem to show that there is a unique solution that can be obtained as the limit of a sequence that starts (for example) at the vector $x$ with $x_g = 1$ for all $g$ (which is in X) with element n in this sequence being $x_n = f(x_{n-1})$. If you can show that all along this sequence your property holds, then you know it must hold in the limit.