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I have (expect there exists) a function $f: (0,1) \to \mathbb R$ for which the following relationship holds: $$ f\left(\frac{2-p+p^2}{4-2p}\right) - f(p/2) =f\left(\frac{(1-p)(2+p)}{2-p}\right) - f(1-p), $$ for all $p\in(0,1)$. (Possibly there is a nice choice of $p$ that makes this look nicer.)

I know the function is smooth, monotonously increasing, and goes to infinity as its argument goes to 1 (respectively to $-\infty$ when the argument goes to 0.) That is, it's the inverse of a sigmoid function.

I think the above characterization should be enough to determine the function, at least up to some scaling and translation (if $f(x) = a g(x) + b$ clearly $a$ and $b$ cancel.)

Besides that I have no idea about how I might go about finding $f$. I don't even know how I could try to compute it numerically. Any help greatly appreciated.

Edit: Maybe it's nicer to state like this: $$ \begin{align} &f\left(p_1\frac{1-p_2p_3}{(1-p_2)(1-p_3)}\right)-f(p_1) \\ =&f\left(p_2\frac{1-p_1p_3}{(1-p_1)(1-p_3)}\right)-f(p_2) \\ =&f\left(p_3\frac{1-p_1p_2}{(1-p_1)(1-p_2)}\right)-f(p_3) \end{align} $$ for any choice of $p_1,p_2,p_3\ge 0, p_1+p_2+p_3=1$.

The formulation above comes from considering $p_1=p_2=p/2$ and $p_3=1-p$.

Thomas Ahle
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    It seems like you're trying to ask:

    Does there exists a monotonically increasing function, $f:(0,1)\longrightarrow \Bbb{R}$ so that $$f!\left(\frac{2-p+p^2}{2(2-p)}\right) - f(p/2)=

    f!\left(\frac{ 2+p-p^2}{2-p}\right) - f(1-p),$$ $\lim_{x\to 1^{-}} = \infty$ and $\lim_{x\to 0^{+}} = -\infty$

    Is that correct?

     – Chickenmancer Jul 28 '22 at 23:22
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    Thank you, yes. That's a better formulation. Let's cut the part about the limits, since I don't really care what happens there--it's only my intuition about how the function should behave. – Thomas Ahle Jul 29 '22 at 00:03
  • @ThomasAhle If you don't care about the limits, doesn't $f \equiv 0$ work? Do you mean "strictly increasing"? – mathworker21 Jul 31 '22 at 05:39
  • Ok, you are right. I need to either care about the limits or require it to be strictly monotonic. I think the later is the better choice. – Thomas Ahle Jul 31 '22 at 06:54

1 Answers1

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At least your “nicer” version has no solution. At $p_1 = 0.6, p_2 = 0.03, p_3 = 0.37$, we have

$$f(0.970938) - f(0.6) = f(0.092619) - f(0.03).$$

At $p_1 = 0.03, p_2 = 0.092619, p_3 = 0.877381$, we have

$$f(0.247722) - f(0.03) = f(0.758204) - f(0.092619).$$

At $p_1 = 0.247722, p_2 = 0.662382, p_3 = 0.089896$, we have

$$f(0.758204) - f(0.247722) = f(0.945929) - f(0.662382).$$

Adding these three equations and canceling yields

$$f(0.970938) - f(0.6) = f(0.945929) - f(0.662382).$$

But for $f$ to be strictly increasing, the left side must be greater than the right side.

Anders Kaseorg
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