revised link function of logistic regression model?

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My question is: If I changed the function of the survival probability to 
the following: P(y=1|x) = (exp(b0+b1*x) / (1+exp(b0+b1*x)) ) ^t, how 
should I do the regression using the glm? How can I add the "t" into the 
function as a predictor variable? Or should I revise the link function?
I shall hazard a guess and say that you can't, exactly as you have 
specified it.  You could fit such a model by ML, but you would have to 
hand code it using nlme (?) or directly maximizing the likelihood using 
optim() etc.  But I suspect there are many similar models that might be 
better behaved.  Are you trying to fit a parametric survival model in a 
roundabout fashion?

Cheers, David Duffy.
| David Duffy (MBBS PhD)                                         ,-_|\
| email: davidD at qimr.edu.au  ph: INT+61+7+3362-0217 fax: -0101  /     *
| Epidemiology Unit, Queensland Institute of Medical Research   \_,-._/
| 300 Herston Rd, Brisbane, Queensland 4029, Australia  GPG 4D0B994A v
Hey, I have a problem using the glm function for the logistic regression model.

For the logistic model, the survival probability is: P(y=1|x) = exp(b0+b1*x) / (1+exp(b0+b1*x))
The link function is: b0+b1*x
Your terminology is a bit off:  b0+b1*x is the linear predictor, g(p)
= log(p/(1-p)) is the link function and g^1(eta) =
exp(eta)/(1+exp(eta)) is the inverse link.
My question is: If I changed the function of the survival probability to the following: P(y=1|x) = (exp(b0+b1*x) / (1+exp(b0+b1*x)) ) ^t, how should I do the regression using the glm? How can I add the "t" into the function as a predictor variable? Or should I revise the link function?
The glm function allows for a user-written family but, unfortunately,
doesn't allow for additional covariates or even constants.  It is
possible to "smuggle" them  in at the time that the family is
instantiated.  See the glm.nb function in the MASS package for an
example.
Give an example here:

y=c(0,1,1,0,1,1,1,1,0,0,1,0,1,0,0)
x1=rnorm(length(y))
t=sample(1:30,length(y),replace=TRUE)
(dat1=data.frame(y,x1,t))
? y ? ? ? ? x1 ?t
1 ?0 ?0.6486104 23
2 1 0.6989650 ?9
3 ?1 -1.6542582 ?9
4 ?0 -0.2465761 30
5 ?1 ?1.7133463 17
6 1 -0.3536826?20
7 1 0.4766255 ?3
8 1 -0.3549167 ?2
9 0 0.3203217?22
10 0 ?0.6136343 ?3
11 1 -0.3993963 29
12 0 ?0.8339387 ?3
13 1 0.5359685?17
14 0 ?0.2322983 ?4
15 0 0.7317251?21
rm(y,x1,t)
res1=glm(y~x1,family=binomial(link ="logit"), data=dat1) ?### How can I add the "t" into the function as a predictor variable
summary(res1)
Call:
glm(formula = y ~ x1, family = binomial(link = "logit"), data = dat1)
Deviance Residuals:
? ?Min ? ? ? 1Q ? Median ? ? ? 3Q ? ? ?Max
-1.3999 ?-1.1258 ? 0.6282 ? 1.0614 ? 1.5764
Coefficients:
? ? ? ? ? ?Estimate Std. Error z value Pr(>|z|)
(Intercept) ? 0.3317 ? ? 0.5853 ? 0.567 ? ?0.571
x1 ? ? ? ? ? -0.7200 ? ? 0.7948 ?-0.906 ? ?0.365
(Dispersion parameter for binomial family taken to be 1)
? ?Null deviance: 20.728 ?on 14 ?degrees of freedom
Residual deviance: 19.792 ?on 13 ?degrees of freedom
AIC: 23.792
Number of Fisher Scoring iterations: 4

Jian

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