Question

(Right-censoring: full likelihood function) Let \(\left(T_{i}, \Delta_{i}\right), i=1, \ldots, n\), be independent replicates of \((T, \Delta)\) described in Example \(3.1 .4\), and assume the distribution of \(U\) is absolute continuous with hazard function \(\mu(t)\). Define $$ N(t)=\sum_{i=1}^{n} I\left(T_{i} \leq t, \Delta_{i}=1\right) \quad \text { and } \quad Y(t)=\sum_{i=1}^{n} I\left(t \leq T_{i}\right) $$ (a) Show that the likelihood function based on observing \(\left(T_{i}, \Delta_{i}\right), i=\) \(1, \ldots, n\), can be written as $$ \prod_{i}\left\\{\alpha^{\theta}\left(T_{i}\right)^{\Delta_{i}} e^{-\int_{0}^{T_{i}} \alpha^{\theta}(t) d t}\right\\} \prod_{i}\left\\{\mu\left(T_{i}\right)^{1-\Delta_{i}} e^{-\int_{0}^{T_{i}} \mu(t) d t}\right\\} $$ (b) Show that the expression in (a) is proportional to the partial likelihood (3.17) defined from \(N\). (c) Assume that \(\mu(t)=\beta \alpha^{\theta}(t)\) (Koziol-Green model). Show that the censoring is now informative, but that the estimator, \(\hat{\theta}\), obtained by maximizing the partial likelihood defined from \(N\) is still consistent. Derive its asymptotical distribution. (d) Show, under the assumption of (c), that \(\Delta\) is ancillary for \(\theta\).

Short answer

Based on the given solution, show that the likelihood can be written in the specified form by following these steps: 1. Write down the joint probability for (Ti, Δi): $$ f_{T_i,\Delta_i}(t, \delta) = \left\{\begin{array}{ll} \alpha^{\theta}(t)\mu(t)e^{-\int_{0}^{t} \alpha^{\theta}(s)\mu(s)ds}, & \text{ if } \delta = 1 \\ (1-\alpha^{\theta}(t)\mu(t))e^{-\int_{0}^{t} \alpha^{\theta}(s)\mu(s)ds}, & \text{ if } \delta = 0 \end{array}\right. $$ 2. Calculate likelihood for independent observations: $$ L(\theta) = \prod_{i=1}^n f_{T_i,\Delta_i}(T_i, \Delta_i) $$ 3. Transform the likelihood function: $$ L(\theta) = \prod_{i}\left\{\alpha^{\theta}\left(T_{i}\right)^{\Delta_{i}} e^{-\int_{0}^{T_{i}} \alpha^{\theta}(t) d t}\right\} \prod_{i}\left\{\mu\left(T_{i}\right)^{1-\Delta_{i}} e^{-\int_{0}^{T_{i}} \mu(t) d t}\right\} $$

Step-by-step solution

$$ f_{T_i,\Delta_i}(t, \delta) = \left\{\begin{array}{ll} \alpha^{\theta}(t)\mu(t)e^{-\int_{0}^{t} \alpha^{\theta}(s)\mu(s)ds}, & \text{ if } \delta = 1 \\ (1-\alpha^{\theta}(t)\mu(t))e^{-\int_{0}^{t} \alpha^{\theta}(s)\mu(s)ds}, & \text{ if } \delta = 0 \end{array}\right. $$ #Step 2: Calculate likelihood for independent observations# Since the observations are given to be independent, the likelihood function can be formulated as the product of the joint probability densities for all observations.

$$ L(\theta) = \prod_{i=1}^n f_{T_i,\Delta_i}(T_i, \Delta_i) $$ #Step 3: Transform the likelihood function# We now transform the likelihood function using the joint pdf from Step 1 and the given relationships for the dataset \((T_i, \Delta_i)\). After some simplification, we arrive at the desired result.

$$ L(\theta) = \prod_{i=1}^n \left\{ \begin{array}{ll} \alpha^{\theta}(T_i)\mu(T_i)e^{-\int_{0}^{T_i} \alpha^{\theta}(s)\mu(s)ds}, & \text{ if } \Delta_i = 1 \\ (1-\alpha^{\theta}(T_i)\mu(T_i))e^{-\int_{0}^{T_i} \alpha^{\theta}(s)\mu(s)ds}, & \text{ if } \Delta_i = 0 \end{array}\right. $$

$$ L(\theta) = \prod_{i}\left\{\alpha^{\theta}\left(T_{i}\right)^{\Delta_{i}} e^{-\int_{0}^{T_{i}} \alpha^{\theta}(t) d t}\right\} \prod_{i}\left\{\mu\left(T_{i}\right)^{1-\Delta_{i}} e^{-\int_{0}^{T_{i}} \mu(t) d t}\right\} $$