[R] Is there an equivalent to predict(..., type="linear") of a Proportional hazard model for a Cox model instead?

[Ben Rhelp]

I manage to achieve similar results with a Cox model as follows but I don't 
really understand why we have to take the inverse of the linear prediction with 
the Cox model and why we do not need to divide by the number of days in the year 
anymore?

Am I getting a similar result out of pure luck?

thanks in advance,

Ben

# MASS example with the proportional hazard model
par(mfrow = c(1, 2));
(aids.ps <- survreg(Surv(survtime + 0.9, status) ~ state + T.categ + 
pspline(age, df=6), data = Aidsp))

zz <- predict(aids.ps, data.frame(state = factor(rep("NSW", 83), levels = 
levels(Aidsp$state)),
    T.categ = factor(rep("hs", 83), levels = levels(Aidsp$T.categ)), age = 
0:82), se = T, type = "linear")
plot(0:82, exp(zz$fit)/365.25, type = "l", ylim = c(0, 2), xlab = "age", ylab = 
"expected lifetime (years)")
lines(0:82, exp(zz$fit+1.96*zz$se.fit)/365.25, lty = 3, col = 2)
lines(0:82, exp(zz$fit-1.96*zz$se.fit)/365.25, lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015)


# same example but with a Cox model instead
(aids.pscp <- coxph(Surv(survtime + 0.9, status) ~ state + T.categ + 
pspline(age, df=6), data = Aidsp))
zzcp <- predict(aids.pscp, data.frame(state = factor(rep("NSW", 83), levels = 
levels(Aidsp$state)),
    T.categ = factor(rep("hs", 83), levels = levels(Aidsp$T.categ)), age = 
0:82), se = T, type = "lp")
plot(0:82, 1/exp(zzcp$fit), type = "l", ylim = c(0, 2), xlab = "age", ylab = 
"expected lifetime (years)")
lines(0:82, 1/exp(zzcp$fit+1.96*zzcp$se.fit), lty = 3, col = 2)
lines(0:82, 1/exp(zzcp$fit-1.96*zzcp$se.fit), lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015)