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Is there a class of functions that satisfies the following properties?

  • $\lim_{x \to -\infty}f(x)=-k, \lim_{x \to \infty}f(x)=k$
  • $f(x)<f(y) \Longleftrightarrow x<y $
  • $f(x_1)+f(x_2)+\ldots+f(x_n)=0 \Longleftrightarrow x_1+x_2+\ldots +x_n=0$

I wanted to find such a function after playing a game of monopoly, in which we wanted to redistribute income fairly when the game began getting out of hand. Specifically, each of the above properties correspond to

  • Bounding the final cash values in $(-k, k)$
  • Preserving the ordering of cash values (i.e. If I had more than you before the redistribution, I would have more than you after)
  • Eliminating the need for money to be taken or returned to the bank.

Help my monopoly game redistribute money fairly!

Vincent Tjeng
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    I think the first property is not necessary unless you have a game with infinitely many players (and infinite total cash). You could just use a scaling function $f(x) = \dfrac{kx}{\max{X}}$ (where ${X}$ is the set of amounts of money currently with the players). – M. Vinay Jun 14 '14 at 04:31

2 Answers2

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There is no function satisfying the first and third conditions simultaneously. If a function satisfies the first condition, then there exists $A>0$ such that $|f(x)+k|<k/3$ when $x<-A$ and $|f(x)-k|<k/3$ when $x>A$. This means, for all $x_1,x_2>A$, $$ f(x_1) + f(x_2) + f(-x_1-x_2) > \frac{2k}3 + \frac{2k}3 - \frac{4k}3 = 0, $$ contradicting the third condition.

Note that this proof doesn't require $f$ to be increasing or even continuous.

Greg Martin
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$$f(x) = \dfrac{2k}{\pi}\arctan(x)$$

Plot for $k = 1$:
Graph

Edit: Greg Martin has proved that the first and third properties cannot be satisfied simultaneously. The above function satisfies the first two properties, but not the third. But it is symmetric about the origin, so that whenever $x_i = -x_j$, $f(x_i) = -f(x_j)$.

M. Vinay
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  • You can also multiply $x$ (in the $\arctan(x)$) by a scaling constant so the inflection points (that you can see in graph above) get shifted further apart (so you have a slower-growing function). – M. Vinay Jun 14 '14 at 04:14
  • Can you prove that this satisfies the 3rd property? – Gerry Myerson Jun 14 '14 at 05:02
  • @GerryMyerson Hm, actually I don't think it satisfies the third property. It only comes close to it because of the symmetry (about the origin). Meaning it's an odd function. Whenever the $x_i$s are symmetrically distributed, so their sum is $0$, the $f(x_i)$s will also by symmetrically distributed, and their sum will be $0$. – M. Vinay Jun 14 '14 at 05:12
  • But when it's a case like $x_1 < 0, x_2 = x_3 = -x_1/2$, then $x_1 + x_2 + x_3 = 0$, but $f(x_1) + f(x_2) + f(x_3) \ne 0$ (as $2\arctan(x) \ne \arctan(2x)$). – M. Vinay Jun 14 '14 at 05:14
  • So then it's not an answer to the question. – Gerry Myerson Jun 14 '14 at 05:14
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    @GerryMyerson Maybe the answer is "There is no such function", and this function is one that comes close enough. I suspect the third condition implies linearity of $f(x)$, which conflicts with the first condition. I'll try to prove this and add it to the answer. Thanks for pointing it out. – M. Vinay Jun 14 '14 at 05:17
  • @GerryMyerson is it true that only linear functions can satisfy the third condition? – Vincent Tjeng Jun 15 '14 at 15:37
  • I don't know. Why not post that as a new question? – Gerry Myerson Jun 15 '14 at 23:59