# [R] inverse prediction and Poisson regression

Vincent Philion vincent.philion at irda.qc.ca
Fri Jul 25 15:25:13 CEST 2003

```Hi, ... and good morning!

;-)

On 2003-07-25 08:43:35 -0400 Spencer Graves <spencer.graves at PDF.COM> wrote:

> 	  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.

OK, I think my data can fit that description.

>
> 	  How is Y measured?

Y is the number of line intercepts which encounters mycelial growth. i/e if mycelia intercepts the line twice, 2 is reported. This follows poisson.

If it the number out N, with N approximately 500 (and you know N),
> then you have a logistic regression situation.

No, 500 spores can grow, but there is no "real" limit on the amount of growth possible, and so no limit on the number of intercepts. So this is why I adopted Poisson, not knowing how complicated my life would become!!!
;-)

In that case, section 7.2 in
> Venables and Ripley (2002) should do what you want.  If Y is a percentage
> increase

... But you may be right, that I'm making this just too complicated and that I should simply look at percentage... Any comments on that?

> 	  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.

OK, I see I will need stronger coffee to tackle this, but I will read this in depth today.

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
> solution.

In other words, "cheat" and model Y_0 with a "small" value = log(0.015) ? How would this affect the LD50 value calculated and the confidence intervals? I guess I could try several methods, but how would I go about choosing the right one? Criteria?

> 	  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)
>              Dose         SE
> p = 14: -2.451018 0.04858572
>> exp(LD50P100p[1,1]+c(-2,0,2)*LD50P100p[1,2])-0.015
> [1] 0.06322317 0.07120579 0.08000303

OK, I will need to try this (later today). I don't see "dose.p" in this?

again, many thanks,

--
Vincent Philion, M.Sc. agr.
Phytopathologiste
Institut de Recherche et de Développement en Agroenvironnement (IRDA)
3300 Sicotte, St-Hyacinthe
Québec
J2S 7B8

téléphone: 450-778-6522 poste 233
courriel: vincent.philion at irda.qc.ca
Site internet : www.irda.qc.ca

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