Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 4

Assume independent censoring: the censoring mechanism contributes no factor involving the lifetime rate $\theta$. The exponential density and survival function are $f_{\theta }(x)=\theta e^{-\theta x}$ and $S_{\theta }(x)=e^{-\theta x}$. An observed event contributes the density; a right-censored lifetime contributes the probability of surviving its censoring time. Thus the likelihood for $\theta$, up to censoring factors independent of it, isFor $D>0$ and $T>0$, ${\ell }^{\prime }(\theta )=D/\theta -T$ vanishes at $D/T$, and ${\ell }^{\prime \prime }(\theta )=-D/{\theta }^2<0$ proves the maximum:Censored individuals add follow-up time to the denominator but no event to the numerator. If $D=0$ and $T>0$, the likelihood decreases for $\theta >0$ and has only a supremum as $\theta \downarrow 0$; zero is an extended boundary estimate, not a positive-rate exponential MLE. The derivation is for independent individuals entering at time zero; delayed entry would require conditional survival contributions and exposure measured from entry.

Within month two, each individual contributes only the time spent at risk between times one and two. There are $104-6-1=97$ complete one-month contributions. The six event contributions are $0.08,0.16,0.22,0.50,0.63,0.72$, and the censored contribution is $0.69$. ConsequentlyThe month-specific likelihood factor in a piecewise-exponential survival model is ${\theta }_2^6{e}^{-{\theta }_2{T}_2}$, givingThe eight individuals no longer at risk at time one contribute no month-two exposure. Using either all 112 individuals or all 104 as full-month observations would give the wrong denominator.