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From |
Maarten buis <maartenbuis@yahoo.co.uk> |

To |
statalist@hsphsun2.harvard.edu |

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 <jcoveney@bigplanet.com> 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/ ----------------------------------------- ___________________________________________________________ Want ideas for reducing your carbon footprint? Visit Yahoo! For Good http://uk.promotions.yahoo.com/forgood/environment.html * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**Re: st: reverse prediction - confidence interval for x at given y in nonlinear model***From:*"Joseph Coveney" <jcoveney@bigplanet.com>

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