Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 34 / 3 / b / Solution

Let $D_k={\sum }_{j:g_j=k}{v}_j$ be the observed event count and $T_k={\sum }_{j:g_j=k}{x}_j$ the total person-time in group $k$. Both groups are present; suppose each has $T_k>0$. The grouped log-likelihood isWhen $D_k>0$, its derivative is $D_k/{\beta }^{(k)}-T_k$ and its second derivative is $-D_k/({\beta }^{(k)}{)}^2<0$. Therefore the maximum-likelihood estimates areBoth events and censored observations contribute to $T_k$. If $D_k=0$, the likelihood decreases with the positive rate: its supremum is at the boundary ${\beta }^{(k)}\downarrow 0$, rather than at a finite strictly positive estimate. Writing $D_k/T_k=0$ then describes the closure estimate.

Under the null hypothesis of a common rate, the same maximization givesLet $D=D_0+D_1$ and $T=T_0+T_1$. The maximized linear terms cancel, since $T_k{\hat{\beta }}^{(k)}=D_k$ and $T\tilde{\beta }=D$. Thus the likelihood-ratio test usesInterpret a zero-event summand by continuity as zero. If $D=0$, both likelihood suprema are equal and $W=0$: no rate comparison is informed by observed events.

Under positive common hazard, independent censoring and the usual increasing-information conditions in both groups, Wilks theorem gives $W\Rightarrow {\chi }_1^2$, because the unrestricted model has two rate parameters and the null has one. Reject for large $W$; at 5%, the usual cutoff is approximately $3.84$. The chi-squared calibration is asymptotic and can be poor with sparse event counts or boundary estimates.