At duration seven there are six at risk and two deaths, at duration thirteen there are four at risk and one death, at duration twenty-six there are three at risk and one death, and at duration twenty-seven there are two at risk and one death. The <Kaplan–Meier estimator> is therefore
$$
\widehat F_{48}(t)=
\begin{cases}
1,&0\leq t<7,\\
\dfrac23,&7\leq t<13,\\
\dfrac12,&13\leq t<26,\\
\dfrac13,&26\leq t<27,\\
\dfrac16,&t\geq27.
\end{cases}
$$
The censoring of D at duration thirty-seven causes no multiplicative drop.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 5 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-207/5/ii/solution