how to get inflection point in binomial glm

Ren?,

Yes, to fit a re-parameterized logistic model I think you'd have to code the whole enchilada yourself, not relying on glm (but not nls() as nls() deals with least squares minimization whereas here we want to minimize a minus log binomial likelihood).

I did that and have the re-parameterized logistic model in a package I wrote for a colleague (this package has the logistic fit fully functional and documented).
My code though only considers one continuous predictor.

If you want I may email you this package and you figure out how to deal with the categorical predictor.
One thing I believe at this point is that you'd have to do the inference on the continuous predictor _conditional_ on certain level(s) of the categorical predictor.

Rub?n

-----Mensaje original-----
De: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] En nombre de Ren? Mayer
Enviado el: jueves, 01 de diciembre de 2011 20:34
Para: David Winsemius
CC: r-help Help
Asunto: Re: [R] how to get inflection point in binomial glm

Thanks David and Rub?n!

@David: indeed 15 betas I forgot the interaction terms, thanks for correcting!

@Rub?n:  the re-parameterize would be done within nls()? how to do this practically with including the factor predictor?

and yes, we can solve within each group for Y=0 getting

0=b0+b1*X |-b0
-b0=b1*X |/b1
-b0/b1=X

but I was hoping there might a more general solution for the case of multiple logistic regression.


HTH

Ren?

Zitat von "David Winsemius" <dwinsemius at comcast.net>:

On Dec 1, 2011, at 8:24 AM, Ren? Mayer wrote:

Dear All,

I have a binomial response with one continuous predictor (d) and one 
factor (g) (8 levels dummy-coded).

glm(resp~d*g, data, family=binomial)

Y=b0+b1*X1+b2*X2 ... b7*X7

Dear Dr Mayer;

I think it might be a bit more complex than that. I think you should 
get 15 betas rather than 8. Have you done it?

how can I get the inflection point per group, e.g., P(d)=.5

Wouldn't that just be at d=1/beta in each group? (Thinking, perhaps 
naively, in the case of X=X1 that

(Pr[y==1])/(1-Pr[y==1])) = 1 = exp( beta *d*(X==X1) )  # all other 
terms = 0

And taking the log of both sides, and then use "middle school" math to solve.

Oh, wait. Muffed my first try on that for sure.  Need to add back both 
the constant intercept and the baseline "d" coefficient for the
non-b0 levels.

(Pr[y==1])/(1-Pr[y==1])) = 1 = exp( beta_0 + beta_d_0*d +
                                    beta_n + beta_d_n *d*(X==Xn) )

And just

(Pr[y==1])/(1-Pr[y==1])) = 1 = exp( beta_0 + beta_d_0*d ) # for the 
reference level.

This felt like an exam question in my categorical analysis course 25 
years ago. (Might have gotten partial credit for my first stab, 
depending on how forgiving the TA was that night.)

I would be grateful for any help.

Thanks in advance,
Ren?

--

David Winsemius, MD
West Hartford, CT

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.