--- Joseph Coveney <[email protected]> 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
studies...
> 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
with:
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
-----------------------------------------
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
http://home.fsw.vu.nl/m.buis/
-----------------------------------------
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