[R] inverse prediction and Poisson regression
spencer.graves at pdf.com
Fri Jul 25 14:43:35 CEST 2003
The Poisson assumption means that Y is a number of independent events
from a theoretically infinite population occurring in a specific time or
place. The function "glm" with 'family="poisson"' with the default link
= "log" assumes that the logarithm of the mean of Y is a linear model in
the explanatory variable.
How is Y measured? Is it the number out of 500 who exceed a certain
threshold, or is it the average percentage increase in weight of the 500
or what? If it the number out N, with N approximately 500 (and you know
N), then you have a logistic regression situation. In that case,
section 7.2 in Venables and Ripley (2002) should do what you want. If Y
is a percentage increase
When dose = 0, log(dose) = (-Inf). Since 0 is a legitimate dose,
log(dose) is not acceptable in a model like this. You need a model like
Peter suggested. Depending on you purpose, log(dose+0.015) might be
sufficiently close to a model like what Peter suggested to answer your
question. If not, perhaps this solution will help you find a better
I previously was able to get dose.p to work in R, and I just now was
able to compute from its output. The following worked in both S-Plus
6.1 and R 1.7.1:
> LD50P100p <- print(LD50P100)
p = 14: -2.451018 0.04858572
 0.06322317 0.07120579 0.08000303
hope this helps. spencer graves
Vincent Philion wrote:
> Hello sir, answers follow...
> ... 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
> I'm not sure I understand what you mean by "mean", (no pun intended!)
but Y is a biologicial "growth". Only one "observation" for each X. But
this observation is from the growth contribution of about 500 individuals,
so I guess it is a "mean" response by design.
> the log link is for the Poisson regression, so the GLM is "response ~ log(dose), (family=poisson)"
> ...Response for dose 0 is a "control" = Ymax. So, What I want is the dose for 50% response.
> *** Once you observe Ymax, Y is no longer Poisson.
> I don't understand this? What do you mean? Please explain.
> ***What exactly is Ymax? Is it the response at dose 0?
> Correct. it is measured the same way as for any other Y. (It is also the largest response because the "dose" is always detrimental to growth)
> ***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's it. that sounds right! How? (Confidence interval on log scale and on real scale, etc) Given that the error on Y is Poisson and not "normal"
> ***That one is easy.
> *** I think you are confusing response with mean response, and we can't
> disentangle them for you.
> What else is needed?
> bye for now,
More information about the R-help