Are there any reason to divide the MLE of a left truncated survival data with its surviving function at d, where d is the left truncation point?

Question

I am confused by this part when trying to revise the parts on survival analysis. As I understand if I want to estimate the MLE I can, in most cases, take the first derivative on that parameter and set the value to zero to find the maximum.

Even if I am interested in the fisher information, the survival function at d disappears under differentiation.

Is there any reason, assuming d is constant of x, to keep the 1/S(d) when constructing the MLE? Thanks

Answer 1

With left truncation of survival times, the idea is that the observation of an event at time $x$ is conditional upon having already survived until left-truncation time $Y_L$. You have no information about any events that might have happened prior to $Y_L$.

Expressing that idea in mathematical form, the contribution to likelihood from an event at time $x$ given left truncation at time $Y_L$ is:

$$L \propto f(x; \theta)/S(Y_L; \theta).$$

Here, $f(x; \theta)$ is the event density function at the event time $x$, $S(t; \theta)$ is the associated survival function, and $\theta$ is the vector of parameter values to be estimated.

So for a left-truncated observation it isn't just $S(Y_L; \theta)$ that you are optimizing, as your question suggests. That's combined with optimizing $f(x; \theta)$ at the observed event time, given that the individual had survived to time $Y_L$. You need to keep both in the likelihood.