Is anyone aware of a re-parameterization of any asymmetric s-shaped function (like, but not necessarily the 5 parameter logistic curve), where one of the parameters is the first inflection point of the first derivative (i.e. the maximum of the second derivative).

I mean point 1 in the upper figure:

enter image description here

(The picture shows the point mentioned above for the of case of a symmetric logistic function.)

So far, I had a look at various references (among others Ratkowsky, D. A. (1990). Handbook of Nonlinear Regression Models. New York, Dekker). However, I have only found parameterizations, where one of the parameters is the inflection point of the function (point 2 in the upper figure).

Unfortunately, calculating this point after the estimation is not a possible solution in my case as this parameter should be integrated in another equation that is estimated simultaneously.

  • $\begingroup$ Two comments: (i) In general it is not possible to reparameterise any asymmetric s-shaped curve in terms of this three points due to identifiability (e.g. for a one-parameter curve, these three points are a function of it); (ii) There may be many cases in which it is possible but these points cannot be found in closed-form, making the reparameterisation impractical. $\endgroup$ – user10525 Aug 14 '12 at 19:32

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

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

Whence, setting

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

we find

$$\rho = \frac{3 \kappa \pm\sqrt{\kappa ^3+2 \kappa ^2+\kappa }}{\kappa ^2-7 \kappa +1},$$

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 $(-1/4, 1/2, 1, 1, 0)$:

enter image description here

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.

  • $\begingroup$ Thanks for your help. As I am not sure about the parameter interpretation in your formula. Do you have any reference for me? I am curently using the following parameterization of the Richards curve, because of its interpretability of the parameters: en.wikipedia.org/wiki/Generalised_logistic_function If I would put the term you derived for c in the very first equation, this seems pretty messy to estimate, right? Do you see any potential for simplification? $\endgroup$ – majom Aug 15 '12 at 1:26
  • $\begingroup$ BTW: In the very end, I aim to be able to use this function to estimate a nonlinear mixed effects model. Thereby, alpha and the coefficient for the first inflection point of the first derivative should be allowed to vary on the group-level. $\endgroup$ – majom Aug 15 '12 at 1:28
  • $\begingroup$ @whuber Not in general (which is my point btw), it is easy to prove because there are no diffeomorphisms for larger dimensions to smaller dimensions (or viceversa). In the particular case you explain, I agree. (+1) $\endgroup$ – user10525 Aug 15 '12 at 9:16
  • $\begingroup$ @Proc That's a good point of view, but I think it does not apply here. The question asks for a change of coordinates for a five-dimensional manifold in which two of the five coordinate functions are prespecified, that's all. Locally and generically there will be a large family of answers. $\endgroup$ – whuber Aug 15 '12 at 12:49
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    $\begingroup$ Sorry that it took me a while to look into this. So far your solution worked great for a toy example, which I created for test purposes. Let's see if it also converges in a real world setting. Again, thanks for your help. $\endgroup$ – majom Sep 3 '12 at 7:18

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