I am currently trying to read through examples from http://www.ats.ucla.edu/stat/r/examples/asa/asa_ch1_r.htm. One of the models I saw was:
survreg( Surv(time, censor) ~ age, dist="exponential").
I know that for the case where there ISN'T a covariate (age in this case), we have:
survreg( Surv(time, censor) ~ 1, dist="exponential"), and the log-likelihood here is:
$l(\theta) = \log(\theta)-\theta t$.
Would anyone know how this log likelihood would change with a covariate? Thanks!
I think it would be more accurate to say that:
$l(\theta) = \sum( \log(\theta) )- \sum( \theta Y_i)$
$l(\theta) = n_u \log(\theta) )- \sum( \theta Y_i)$ where u's are uncensored and Y's are observation times
When there is no covariate;
$\delta \log L/ \delta \lambda = n_u/\lambda - w$ where $w = \sum Y_i$ (i.e total person-years, censored or not)
So the MLE for $\lambda$ is just what you would expect: events per total person years of observation. For models with covariate, you substitute $\lambda(t|X_i)$ with $\lambda(t)*exp(beta_1*X)$ and then redo the summations. On the log hazard scale it turns into a linear regression.
