Solved – Re-parameterization of an asymmetric s-shaped function

Pace Procrastinator, this sort of thing can be done.

Consider the five-parameter logistic model. It has many parameterizations, but it's simple and natural to reduce them to something like

$$y = nu +tau left(1+e^{frac{x-mu }{sigma }}right)^{-rho }$$

with $rho gt 0$. We can interpret $mu$ and $nu$ as $x$ and $y$ locations and $sigma$ and $tau$ as $x$ and $y$ scales; $rho$ is the shape or asymmetry parameter.

Let $x_{+}$ be the location of one extremum of the second derivative and $x_{-}$ be the location of the other extremum. Then, solving for the zeros of the third derivative, I find them at

$$x_{pm} = mu + sigma log left(frac{(3 rho +1) pmsqrt{(rho +1) (5 rho +1) }}{2 rho ^2}right).$$

Whence, setting

$$kappa = exp frac{x_{+}-x_{-}}{sigma}$$

we find

$$rho = frac{3 kappa pmsqrt{kappa ^3+2 kappa ^2+kappa }}{kappa ^2-7 kappa +1},$$

taking the positive sign when $kappa$ exceeds the larger root of $1-7x+x^2=0$ (about $6.8541$) and the negative sign when $kappa$ is less than the smaller root (about $0.145898$)–other values of $kappa$ will not give a sigmoidal curve–and

$$mu = frac{x_{+} + x_{-}}{2} + sigma log(rho).$$

This allows a parameterization in terms of $(sigma, nu, tau, x_{-}, x_{+})$ (with some significant restrictions on $sigma$ needed to make the results valid).

Here is a plot of $y$ (in blue) and its third derivative (in red) based on these formulas with the parameters set to $(-1/4, 1/2, 1, 1, 0)$:

Indeed, the ascending zero of the third derivative occurs at $1$ and the descending zero at $0$, as specified.

This parameterization is not necessarily so messy. If your data are within a narrow range, for instance, the asymptotic values might not matter and you could just use a suitable polynomial. Without knowing more about the particular problem, it's hard to provide a specific recommendation.