[R] inverse prediction and Poisson regression
Prof Brian Ripley
ripley at stats.ox.ac.uk
Fri Jul 25 14:38:00 CEST 2003
Ymax is the maximum observation in your example, and also the observation
at zero. I was asking which you meant: if you meant Y at 0 (and I think
you do) then it is somewhat misleading notation.
You have a set of Poisson random variables Y_x at different values of x.
Poisson random variables have a mean (I am using standard statistical
terminilogy), so let's call that mu(x). Then you seem to want the value
of x such that mu(x) = mu(0)/2 *or* mu(x) = Y_0/2, and I don't know
which, except that in your model mu(0) would be infinity, and so the
model cannot fit your data (finite values of Y_0 have zero probability).
On Fri, 25 Jul 2003, 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)"
So you have -Inf as the explanatory variable at zero dose?
> ...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.
That was understanding Ymax to be the maximum Y, which is what it looks
> ***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)
The last is not true, given your assumptions, It could have the largest
mean response, but 0 is a possible value for Y_0.
> ***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.
Fit a model for the mean response (one that actually can fit your data),
and solve the estimated mu(x) = mu()/2 or Y_0/2. That gives you an
estimate, and the delta method will give your standard errors.
> *** I think you are confusing response with mean response, and we can't
> disentangle them for you.
> What else is needed?
> bye for now,
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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