3
$\begingroup$

Suppose to observe a random sample from a r.v. $Y_i=\min(T_i,C_i)$ where $T$ and $C$ are iid absolutely continuous distribution.

I would like to inference about a parameter of $T$ (for example, $\lambda$ in a exponential distribution). Therefore I calculate the pdf of $Y$:

$F_Y(y)=P(Y\le y) = P(\min(T,C) \le y) = 1-(1-F_T(y))(1-F_C(y)) $ that implies

\begin{align}f_Y(y) &= \frac{d}{dy}F_Y(y) = f_C(y)+f_T(y)-f_T(y)F_C(y)-F_T(y)f_C(y)\\ &= f_C(y)(1-F_T(y)) + f_T(y)(1-F_C(y)). \end{align}

Then $L(\lambda) = \prod_{i=1}^nf_Y(y_i)$.

Does my approach really answer the inferential question about a parameter of a distribution with random censoring? I can see that usually the likelihood is calculated as:

$$L(\lambda) = \prod_{i=1}^nf_C(y_i)^{\delta_i}(1-F_T(y_i))^{1-\delta_i}\prod_{i=1}^nf_T(y_i)^{1-\delta_i}(1-F_C(y_i))^{\delta_i}$$

where $\delta=1$ if the observation is censored and $\delta=0$ if not.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Yes, the likelihood is correct if the sample is i.i.d. $\endgroup$
    – Xi'an
    Commented Apr 8, 2018 at 9:57
  • $\begingroup$ I can't understand the difference between my solution and the approach used for example here : web.stanford.edu/~lutian/coursepdf/unit2.pdf ( section 2.1) . $\endgroup$
    – momomi
    Commented Apr 8, 2018 at 10:07
  • $\begingroup$ I've edited the question to make it clearer $\endgroup$
    – momomi
    Commented Apr 8, 2018 at 10:19
  • $\begingroup$ There are two likelihoods presented in the question: one observed and one completed with the censoring indicators (which are not observed). The second one is useful for resolutions like the EM algorithm. $\endgroup$
    – Xi'an
    Commented Apr 8, 2018 at 10:21
  • $\begingroup$ So the first one is the likelihood for a random sample when we do not know if the observed value was censored or not? But I would say that if we assume the censoring mechanism is random, we cannot say if the observed value is censored or not. $\endgroup$
    – momomi
    Commented Apr 8, 2018 at 10:42

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.