Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 207 / 6

With constant hazard $\theta$, observation $i$ contributesWriting $D={\sum }_i{v}_i$ and $T={\sum }_i{x}_i$, the log-likelihood is $D\log \theta -\theta T$, whose derivative is $D/\theta -T$. Hencethe number of events divided by total person-time.

Left truncation means that an individual enters observation only after a delayed entry time $L_i$ and is included only if the event time exceeds $L_i$. Earlier failures are absent from the dataset, so analysis must condition on survival to entry.

In the Kaplan–Meier estimator, replace the ordinary risk set at event time $t$ byThe product-limit factor remains $1-d_j/|R(t_j)|$, but only individuals who have entered and not yet exited contribute to its denominator.

Conditioning on survival to $L_i$ changes the constant-hazard likelihood contribution toThereforeusing only person-time accumulated after entry.

For period survival analysis over a calendar window, patients diagnosed earlier enter at the window's start only if still alive. Their survival times are left-truncated at the elapsed duration from diagnosis to that date, while follow-up is censored at the window's end. This estimates survival using mortality experience in the chosen recent period without pretending that earlier survivors were under observation from diagnosis.

There are seven deaths by day 28. Their total time plus 28 days from each of the ten later observations gives first-interval exposureAmong the ten observations beyond day 28, eight are deaths, and their exposure beyond 28 is

Under the two-rate alternative, maximize the piecewise-exponential survival model likelihood separately to obtain the estimates in part (i). Under $H_0:{\theta }_1={\theta }_2=\theta$, there are 15 deaths and 2200 days of exposure, so $\hat{\theta }=15/2200$. Twice the maximized log-likelihood difference isUnder the null this likelihood-ratio test statistic is asymptotically ${\chi }_1^2$.