# Two types of left censoring in a time-to-event model

I have (population) survey data where I know one of three things about each respondent's experience of an event (A):

1. The event A occurred at an age specified by the respondent.
2. The event A is a precursor to another event B; the respondent was not asked for an age for event A, but was asked (and recorded) an age at event B. This is a left censoring time, whose distribution cannot be assumed independent of event A.
3. The event A is known to have occurred, but the age at the event was not asked. This means the age at survey (which is known) is the only information on the event time, and is a left censoring time, whose distribution can be assumed independent of the time of event A.

We can assume the time of event A is independent of the type of answer given.

I think I understand that when there is one type of (independent) censoring, the distribution of the censoring time can be factored out in the likelihood expression for the parameters (e.g. https://stats.stackexchange.com/a/22605/175515 or see these slides on Censoring mechanisms). But this isn't true when there are two different censoring-time distributions, nor is it true when the censoring time is not random.

Is there any hope?

For those respondents in 1. I actually have the time for both event A and event B, meaning I could also propose and fit some model for the distribution of the 'censoring' time (conditional on the time of event A).

## Possible solution

Along the lines of proposing a model for event B given the data where both are available.

Assume something that relatively trivial, that the time of event B, $t_b$ has a degenerate distribution given $t_a$, the time of event A; that is

$$f_{B|A}(t_b|t_a) = \delta \left(t_b - \mu(t_a) \right)$$

for some function $\mu$ which is monotonically increasing, and $\mu(t_a) > t_a$. Such a parameter can be estimated via MLE relatively easily.

Assume some form of the failure/event density $f_A(t_a; \theta)$ for some parameters $\theta$. For the observations $t_{b,i}$ from the second case, the likelihood expression would be;

\begin{align} L( \theta ) &= \prod_{i} \int_{ \hat{t}_a < t_{b,i} } f_{B|A}( t_{b,i}, \hat{t}_a) f_A( \hat{t}_a \vert \theta ) \, \mathrm{d} \hat{t}_a, \\ &= \prod_{i} \int_{ \hat{t}_a < t_{b,i} } \delta \left( t_{b,i} - \mu(\hat{t}_a) \right) f_A( \hat{t}_a \vert \theta ) \, \mathrm{d} \hat{t}_a, \\ &= \prod_{i} f_A \left( \mu^{-1}(t_{b,i}) \vert \theta \right). \end{align}

The last two steps are obviously particular to this example, and follow the assumptions on $\mu$.

The terms in the product are equal to the probability of observing the corresponding event A (precursor) for each observed event B.