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Follow on to this question.

Using similar constraints:

The constraints regarding the system I'm modeling are (and looking for a general solution to):

  1. Marginal Utility for every unit of a good must be positive (or zero), finite, and declining (though never to below zero).
  2. The available quantities of all goods must be finite, though they can be arbitrarily large. 2.a There can be an arbitrarily large, though finite number of other goods in the system.
  3. Aggregate Utility must increase for all, while individual utility must decrease for all except one (the "monster") when a good of a certain class (let's say "cars") is transferred from any member of the group to the "monster".
  4. Condition 3 should be met for all transfers of "cars" from "innocents" (people not the monster) to the "monster", to the exhaustion of "cars" from the system.

The Difference

Let's replace the "monster" with a group of individuals, still in the minority.

Can a mathematical relationship be defined between their utility function over a good (x), and their representation in a population (p), that sets lower bounds for the existence of a class of "utility monsters".

For example, if members of the "monster" group a have function $u(x)=Ax+\frac{b}{x^2}$ and members of the innocent (i.e. non-monster group) have utility function $v(x)=Cx+\frac{d}{x^2}$, can we describe the ratio of $M:I$ (where $M$ is the size of the "monster" group and $I$ is the size of the "innocent" group) that allows $M$ to exist as a "utility monster" class solely in terms of $A$, $b$, $C$, and $d$?

This is basically a question in regards to constructing non-Pareto economies in simulation.

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Solution:

If we assume the Group Utility function equals:

$$G(x)= M\times u(qx) + I \times v(rx)$$

where:

$$u(x) = Ax+\frac bx$$

$$v(x) = Cx+\frac dx$$

and:

$$q+r=1$$ $$q,r,A,b,C,D \in\mathbb{R}(0,\infty)$$ $$x,M,I\in\mathbb{Z} [1,\infty)$$

We can state our goal as defining the boundaries of a situation where for some unfair allocation $M<I$ and $q>r$, $G(x)$ is greater than it would be if $r>q$ (a more "fair" allocation).

We can even simplify the question to finding a set of constraints on the variables $M,I,q,r,A,b,C,$ and $d$ such that would satisfy the inequality (which assumes $q$ and $r$ are equal and dropped):

$$M(Ax + \frac {b}{x}) > I(Cx+\frac{d}{x})$$

$$x\in \mathbb{Z}[1,\infty)$$

so taking the $\lim_{x\to\infty}$ of both sides, we can see that if $\frac{A}{C}>\frac{I}{M}$ at large $x$ the inequality is satisfied.

Conversely, at $x=1$, we get $M(A+b)$ and $I(C+d)$.

This would demonstrate that a non-Pareto optimal solution exists that still satisfies the idea of declining marginal utility and maximizes overall group utility for all distributions of goods for any class of goods where there are two preference functions $u(x)$ and $v(x)$ as long as they are distributed such that[1]:

$$\infty>AM>CI+d>0$$

Or in general a "monster class" can exist as long as the coefficients of its utility function are greater than that of an innocent class times it's fraction of the population times the total number of goods the class under consideration in the economy:

$$\{c_M|c_M>c_I\times\frac{I}{M}\times x\}$$

[1] Again, assuming that $MU \geq0$ is always true

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I believe that we would have to assume that there was no or unusual declining marginal utility for the monster group. Say we are talking about wealth, and a small group of people have increasing desire to become wealthier (ie always more), it could be the case that the pleasure that they take in having, say 327 million more dollars would be more than the cumulitive loss of 1 dollar per person in the US. But I think we need to assume an oddly shaped marginal utility graph for the monsters, otherwise this could only occur for a short time, ie there could be non-pareto transfers that would fit this scenario, but it could not continue for very long.

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    $\begingroup$ hmm... I have to think about that. what if we start adding units of x to the system, what are the constraints to insure that $u(n)>v(1)$... $\endgroup$ – Jason Nichols Nov 23 '14 at 23:09
  • $\begingroup$ fixed my answer $\endgroup$ – Jason Nichols Nov 23 '14 at 23:17

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