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Re: st: reverse prediction - confidence interval for x at given y in nonlinear model

From   Maarten buis <>
Subject   Re: st: reverse prediction - confidence interval for x at given y in nonlinear model
Date   Fri, 26 Oct 2007 17:59:22 +0100 (BST)

--- Joseph Coveney <> wrote:
> If I'm not mistaken, the four-parameter logistic model Rosy used is
> for the *logarithm* of dose and *logarithm* of ED50, and not the dose
> and ED50, per se (cf. Maarten's y-axis values). 

I am studying long term trends in inequality of educational opportunity
between children of different socioeconomic background, so I am not
very up to date on common parameterizations in biological and medical

> Perhaps the parameterization is numerically stabler, too, in some
> sense.

When I estimated this model on my generated data I found that it wasn't
very stable. However it is not very surprising as it uses 4 parameters
to estimate a single curve and 2 of these parameters refer to
asymptotes, i.e. parts of the curve at the extreme left and the extreme
right where there are few observations.

Next Joseph compares the model estimated by Rosy:
The model is  y= b0 + b1/(1 + exp(-b2*(x-b3))) + error

b0 = baseline outcome
b1 = Emax i.e. largest change from baseline
b2 = Hill or slope coefficient
b3 = ED50 i.e. value of x (dose) required to produce half-maximal
effect, that is x required for y=b0 + b1 / 2

E = Emin + Emax * Dose^Hill / (Dose^Hill + ED50^Hill)

E = response
Emin = response at zero dose
Emax = asymptotic response at infinite dose, 
Dose = untransformed dose
Hill = the coefficient of receptor cooperativity
ED50 = dose yielding a response that is Emin + half of Emax

Joseph asks:
> does the -nl log4:- four-parameter logistic model give rise to biased
> estimates of ED50 (ED10, ED90, etc.) and confidence intervals in the
> original measurement scale with nonasymptotic sample sizes?  That is,
> should a pharmacologist ever use -nl log4:- in lieu of the model 
> shown just above?

The two models are equivalent. To see this notice first that they have
a very similar structure: 

y = minimum + maximum * S-shaped curve

whereby the S-have curve has a minimum of 0 and a maximum of 1. b0 in
Rosy's model corresponds to Emin in Joseph's model, and b1 in Rosy's
model corresponds to Emax in Josephs model. So all I need to show is
that the two S-shaped curves are the same: 

1/(1 + exp(-b2*(x-b3))) = Dose^Hill / (Dose^Hill + ED50^Hill)

step 1:
Dose^Hill / (Dose^Hill + ED50^Hill) = 
e^{Hill*ln(Dose)} / (e^{Hill*ln(Dose)} + e^{Hill * ln(ED50)})

( using the rules: x = e^{ln(x)} and (e^x)^b = e^{b*x} )

Step 2: Lets call ln(Dose) x, Hill b2, and ln(ED50) b3
e^{b2*x} /(e^{b2*x} + e^{b2*b3)

Step 3: divide the numerator and the denomenator by e^{b2*x}
e^{b2*x - b2*x}/e^{b2*x - b2*x} + e^{b2*b3 - b2*x}

( using the rule e^a / e^b = e^{a-b} )

This simplifies to:
1/(1 + e^{-b2*(x-b3)})

( using the rule e^0 = 1 )

Hope this helps,

Maarten L. Buis
Department of Social Research Methodology
Vrije Universiteit Amsterdam
Boelelaan 1081
1081 HV Amsterdam
The Netherlands

visiting address:
Buitenveldertselaan 3 (Metropolitan), room Z434

+31 20 5986715

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