Prof Brian Ripley <[EMAIL PROTECTED]> writes:
> > x y > > 0 28 > > 0.03 21 > > 0.1 11 > > 0.3 15 > > 1 5 > > 3 4 > > 10 1 > > 30 0 > > 100 0 .... > > Where X is dose and Y is response. > > the relation is linear for log(response) = b log(dose) + intercept > > Is that log(*mean* response), that is a log link and exponential decay > with dose? > > > Response for dose 0 is a "control" = Ymax. So, What I want is the dose > > for 50% response. For instance, in example 1: > > > > Ymax = 28 (this is also an observation with Poisson error) > > Once you observe Ymax, Y is no longer Poisson. > > > So I want dose for response = 14 = approx. 0.3 > > What exactly is Ymax? Is it the response at dose 0? The mean response at > dose 0? The largest response? About the only thing I can actually > interpret is that you want to fit a curve of mean response vs dose, and > find the dose at which the mean response is half of that at dose 0. > That one is easy. > > I think you are confusing response with mean response, and we can't > disentangle them for you. I don't feel all that confused. Y is Poisson distributed with some mean depending on x. Ymax is a value at X=0, i.e. Poisson distr. with a mean as large as it can be. I think the main confusion here is trying to fit a functional relationship which doesn't extend to X=0. If you extrapolate a log-loglinear relation back to X=0, you get an infinite maximal response if b is negative, so this is going to be inconsistent with a finite Ymax. In some of the data sets I believe you actually do see a leveling off for very small doses. If you insist on this peculiar model, you'd end up with estimating the mean of Ymax by its observed value. Then you can get b and the intercept from the observations with X>0, and find your estimate of halving dose by solving log(Ymax/2) = b * log(dose50) + intercept i.e. dose50 = (log(Ymax/2)-intercept)/b. That's a nonlinear function of the estimates, so you'd need (at least) to employ the Delta method to find the approximate variance of the estimate. However, I'd suggest that you should look for a more realistic functional form of the relation, e.g. a logistic curve in log(x) or a Michalis-Menten style inhibition (mean(Y) = ymax/(1+dose/dose50) or variants thereof). These models are not (necessarily) GLMs, but I think you can fit them quite well with gnls() and a suitable variance function. [In fact the ymax/(1+dose/dose50) model is a GLM if you use an inverse link and reparametrize with 1/ymax, 1/(ymax*dose50), but inverse links are not built-in for the poisson() family, so you'd have to modify the code yourself.] -- O__ ---- Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
