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a journal article has a method for designing experiments to be fit to a 4-parameter logistic model. The model used is $y= D + \frac{A - D}{1 + (\frac{x}{C}) ^ B}$
A = upper asymptote
B = maximum slope
C = x value when y = 50% of maximum (i.e. 1/2 of upper asymptote)
D = lower asymptote
Using pilot data, further experiments are optimally designed by plugging the preliminary parameter estimates into equations presented in the article.

However, the nonlinear modeling software that I have access to parameterizes the 4-parameter logistic model differently. The model used is $y = D + \frac{A - D}{1 + e^{B(x-C)}}$

Once I estimate the parameters from the software, how do I translate these to the exponential parameterization used by the software? Thank you.

Paul
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1 Answers1

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These models are not the same, because the first is a rational function of $x$ and the second is an exponential function. The second one is truly a "logistic" model but the first is not. Moreover, the claims about $B$ and $C$ in the first model are not true. The maximum slope depends on $C$, which is a scale parameter, not a location parameter. Moreover, although the maximum slope does depend on $B$, it does so in a complicated fashion. For instance, when $B = 2$, the maximum slope equals $\frac{9}{8\sqrt{3}}$.

If you write $x = \log(z)$ and $C = \log(\gamma)$ then the second model is

$$y = D + \frac{A-D}{1 + \exp(B(\log(z)-\log(\gamma))} = D + \frac{A-D}{1 + (z/\gamma)^B},$$

which does have the form of the first model. In other words, in the first model the logarithm of $z$ has a logistic form.

whuber
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  • Thank you for that explanation. So what I gather is that when the software returns the pilot parameters, they can be plugged directly into the journal formulas, except for C. The software returns gamma = exp(C). To translate into the journal article's equation, take log(gamma) to get C. ** You've helped a lot, thanks again ** – Paul Apr 07 '11 at 15:06