Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 207 / 4

Place discrete hazards $q_j$ at the ordered distinct event times $t_j$. If $d_j$ events occur among $r_j$ individuals at risk immediately before $t_j$, the nonparametric empirical likelihood isEquivalently, an observed event at $x$ contributes the probability mass at $x$, while a right-censored observation contributes the survivor probability beyond its censoring time. Maximization gives ${\hat{q}}_j=d_j/r_j$ and the Kaplan–Meier estimator

Left truncation means an individual is observed only conditional on surviving beyond an entry time $L$. Ignoring it overrepresents long survivors and creates survivorship bias. A subject with event or censoring time $X>L$ contributes its usual likelihood divided by $S(L)$; in risk-set form, that individual enters each $r_j$ only for event times satisfying $L<t_j\le X$.

The professors enter observation only when elected. Anyone dying before election cannot appear, so the birth-to-death sample is left-truncated at varying, outcome-related ages and is not an inception cohort from birth. Every listed professor was elected before age $60$ and survived to $60$, so all six are under observation from the common origin age $60$. The data can therefore estimate survival from age $60$ without delayed entry.

For $T=$ age at death minus $60$, the observations areThere are three distinct event times and hence three independent discrete hazards $q_2,q_7,q_{15}$. Their risk-set sizes are $6,5,3$, soThe maximum-likelihood estimates are ${\hat{q}}_2=1/6$, ${\hat{q}}_7=1/5$, and ${\hat{q}}_{15}=1/3$.

The estimated survivor function isuntil the last observed time. It first falls to at most one half at $t=15$, so

After correction, the records from age $60$ arewhere each pair is delayed-entry time and event or censoring time. The risk sets at event times $2,7,35$ all have size three. Thus the left-truncation-adjusted empirical likelihood isEach estimated event hazard is $1/3$.

The corrected Kaplan–Meier estimator is $2/3$ immediately after $t=2$ andimmediately after $t=7$. It first crosses one half there, so

Proton and Prune are censored after the estimated survivor curve has already crossed one half. Learning their later death times may add jumps in the far tail but cannot change the first crossing, so the nonparametric median remains seven years. An exponential distribution instead estimates one constant hazard from the total event count divided by total time at risk. Replacing censoring by observed later deaths changes both quantities and therefore changes the fitted rate and its median $\log 2/\hat{\lambda }$.