Thanks David and Rubén!
@David: indeed 15 betas I forgot the interaction terms, thanks for correcting!
@Rubén: 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
René
Zitat von "David Winsemius" <dwinsem...@comcast.net>:
On Dec 1, 2011, at 8:24 AM, René 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,
René
--
David Winsemius, MD
West Hartford, CT
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