Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 207 / 4

Throughout the survival questions, use the paper's notation $F(t)=\mathbb{P}(T>t)$ for the survivor function, not for a cumulative distribution function. To construct the Kaplan–Meier estimator, sort the distinct observed event times $a_j$. Let $r_j$ be the size of the risk set immediately before $a_j$, and $d_j$ the number of events at that time. The estimated conditional chance of surviving that event time is $1-d_j/r_j$. Multiplying these conditional probabilities givesA right censoring time reduces later risk sets but causes no jump itself. The usual interpretation as a population survivor function uses independent censoring.

The risk sets at the six event times are $(7,6,5,3,2,1)$, each with one event: the censoring between $t_3$ and $t_4$ removes the seventh individual before the fourth event. Thus the survivor estimate after the fifth event isConditional on reaching $t_4$, the event mass at $t_4$ is $1/3$, that at $t_5$ is $(2/3)(1/2)$, and that at $t_6$ is $(2/3)(1/2)(1)$. Hence the conditional event probabilities are

Now treat the seventh observation as an actual event at $t_4$, not as a censoring at that time. The risk sets are $(7,6,5,4,2,1)$ and the event counts are $(1,1,1,2,1,1)$. The Kaplan–Meier estimator therefore givesConditional on reaching $t_4$, the two events at that time have total probability $2/4=1/2$. Surviving them has probability $1/2$, which is then divided equally between the remaining two event times. Thus

For fractional event imputation, let $q_j$ be the fractional event weight assigned to the seventh individual at $t_j$. Its event contribution at $t_j$ is $q_j$. It belongs to the risk set at that time precisely for allocations at $t_j$ or later, so its fractional risk set contribution is ${\sum }_{j^{\prime }\ge j}{q}_{j^{\prime }}$. This justifies both parts of the hint.

For $(q_4,q_5,q_6)=(1/2,1/4,1/4)$, the fractional counts at the last three times areEarlier Kaplan–Meier estimator factors still give survival $4/7$ immediately before $t_4$. ConsequentlyThe three values are ${\hat{F}}_5^0=1/7\simeq 0.1429$, ${\hat{F}}_5^1=5/28\simeq 0.1786$, and ${\hat{F}}_5^{\ast }=4/21\simeq 0.1905$, in strictly increasing order. Spreading the censored individual's mass beyond $t_4$ raises the survivor estimate from the immediate-event imputation and moves it toward the original Kaplan–Meier estimator.

Assign the seventh individual's event weights $(1/3,1/3,1/3)$, the conditional probabilities obtained from the original Kaplan–Meier estimator. The last three fractional counts areTheir conditional event fractions are $1/3$, $1/2$, and $1$, exactly as in the original censored-data calculation. Hence the fractional allocation reproduces the original estimate:This illustrates self-consistency of Kaplan–Meier estimation: distributing the unobserved event according to the current conditional fitted event distribution leaves the fitted distribution unchanged. The first imputation in part (b) was not such a self-consistent allocation.