Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 1

For $t\ge 0$ and $\theta >0$, the exponential distribution has probability density function $f(t)=\theta e^{-\theta t}$. Integrating the probability density function gives the survivor function $S(t)=P(T>t)=e^{-\theta t}$. The hazard function divides instantaneous failure density by the probability of having survived, and the integrated hazard accumulates this rate. ThusThe probability density function is zero for $t<0$, and the survivor function is one there.

Write $X=\min (T,c)$ and $\Delta =1_{\{T\le c\}}$ for the right-censored survival time and observed-failure indicator. Since the exponential distribution is continuous, including or excluding equality at $c$ makes no difference. Using the tail-integral expression for an expectation,Consequently the expected integrated hazard at the observed endpoint satisfiesAdministrative censoring therefore stops both the accumulated exposure and the opportunity to record a failure.

For individual $i$, let $X_i=\min (T_i,c_i)$ and ${\Delta }_i=1_{\{T_i\le c_i\}}$. The preceding expectation calculation applies separately at each fixed censoring time:Summing and using linearity of expectation, with $D={\sum }_i{\Delta }_i$ the failure count, givesThis identity needs the stated marginal exponential distributions, but not independence between individuals; independence will be needed for the product likelihood.

Replace the expected total person-time and failure count in the preceding identity by their observed values, $E={\sum }_i{x}_i$ and $d$. The resulting estimating equation is $\theta E=d$, soThis events divided by exposure estimator is failures per unit observed person-time. We assume positive total exposure; if $d=0$, it gives the boundary estimate zero.

For independent individuals under fixed administrative censoring, an observed failure contributes the probability density function $\theta e^{-\theta x_i}$, while a censored observation contributes the survivor function $e^{-\theta x_i}$. Hence the survival likelihood and log-likelihood areFor $d>0$, the score function ${\ell }^{\prime }(\theta )=d/\theta -E$ vanishes at $d/E$, and ${\ell }^{\prime \prime }(\theta )=-d/{\theta }^2<0$. This is the unique maximum over $\theta >0$:Thus maximum likelihood estimation agrees with the observed-exposure estimating equation. If there are no failures, the likelihood decreases with $\theta$: its supremum is approached as $\theta \downarrow 0$, or attained at zero if that boundary is admitted.

Differentiate the log-likelihood again. For $d>0$, the positive interior maximum-likelihood estimate givesThe observed information is $-{\ell }^{\prime \prime }(\hat{\theta })=d/{\hat{\theta }}^2$, giving the usual local standard error approximation $\hat{\theta }/\sqrt{d}$. For $d=0$ there is no positive interior maximum, so the displayed substitution and this standard error approximation do not apply; at every positive $\theta$ the second derivative is zero.