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For an exponential distribution of rate $\theta >0$, the probability density function and survivor function are $f(t)=\theta e^{-\theta t}$ and $F(t)=e^{-\theta t}$. Setwhere $D$ is the observed event count and $E$ is total person-time at risk. The log-likelihood, up to a constant, becomesWhen $D>0$ and $E>0$, its score function is $D/\theta -E$, which vanishes atIts second derivative is negative, so this is the unique maximum-likelihood estimator. This is exponential-rate estimation from censored exposure: the estimate is observed events divided by all observed exposure, including exposure from right-censored individuals.

If $D=0$ and $E>0$, $\ell =-\theta E$ decreases strictly. There is no maximizer in the open parameter space $\theta >0$; its supremum occurs as $\theta \downarrow 0$, or the extended boundary estimate is $\hat{\theta }=0$ if a zero rate is permitted. The formula should not be interpreted as an interior regular fit in this case.

Treating the two-year observation period as approximately complete, divide total person-time at risk by two to estimate the average number of children being observed. Dividing again by the number of families givesThus the estimate is about 1.1 eligible children per family. This is a person-time-based approximation for children contributing to the study, not a count of older children or an exact reconstruction of family composition if follow-up or eligibility changed.

Let $Y_I,Y_C$ be attendance totals and $T_I,T_C$ the observed person-time at risk. Under independent Poisson distributions for the totals, the estimated incidence rate ratio is $\hat{R}=(Y_I/T_I)/(Y_C/T_C)$. Treating the exposures as known, the supplied large-count approximation, or the delta method, givesIndependence adds the log-count variances, while fixed exposure terms add no variance. Form the normal confidence interval on the log scale and exponentiate:This specifies the calculation without evaluating it. Individual randomization makes a child the assignment unit, but an independent Poisson count model remains an additional assumption about outcomes, not a consequence of randomization alone.

For independent exponential survival times with total observed person-time at risk $T_{\bullet }$ and $d$ events, the maximum-likelihood rate is $\hat{h}=d/T_{\bullet }$. A two-group exponential model is a proportional-hazards model because both hazards are constant, so their ratio is constant.

HereThus the group-1 to group-0 hazard-ratio estimate isIt decreases continuously as $c$ increases. The parametric exponential likelihood uses exact exposure times through each arm's person-time, whereas the Cox partial likelihood uses only which subjects belong to each event's risk set. For $t_2<c<t_4$, changing $t_4$ alters group 0 person-time and hence $\hat{\lambda }$, but subject 4 is alone if it fails at $t_4$, so that event contributes one to the partial likelihood and $\hat{\theta }$ is unchanged.

In a constant hazard survival model, the maximum-likelihood estimator is the number of observed events divided by total person-time at risk. Treatment A contributes $3$ deaths in $1+2+3+3+5=14$ months, while Treatment B contributes $2$ deaths in $1+5+5+7+10=28$ months. Hence

Time to the first of $A$ and $B$ is exponential with rate $\theta +\varphi$ and mean $1/(\theta +\varphi )$. The expected total person-time at risk is

Under a country-specific constant hazard survival model, let $D_k$ be the number of deaths and $Y_k$ the total person-time at risk in country $k$. The likelihood is proportional toso the maximum-likelihood estimator is ${\hat{\lambda }}_k=D_k/Y_k$. Test $H_0:{\lambda }_0={\lambda }_1$ by a likelihood-ratio test, score test, or Wald test; equivalently, give $D_k$ a Poisson distribution with mean ${\lambda }_k{Y}_k$ and test the country coefficient in a Poisson regression with offset $\log Y_k$.

For constant hazard $h_i(t)=\lambda$, $S_i(t)=e^{-\lambda t}$. With $D={\sum }_i{d}_i$ and total person-time $T={\sum }_i{x}_i$,Thus the maximum-likelihood estimate is events divided by person-time at risk.

Without censoring, every $v_i=1$ and $\mathbb{E}{T}_i=1/\lambda$. ThereforeThis is events divided by total person-time at risk.

An unobserved survival time is right-censored when it is known to exceed the last event-free follow-up time. For a potential survival time $T$ and potential observation-ending time $C$, recordWhen $\delta =1$, the occurrence was observed at $Y$; when $\delta =0$, only $T>Y=C$ is known. An intact machine at the study's closing date and an event-free subject who stops attending follow-up can both yield right censoring, but the reasons for stopping matter for independent censoring. A subject right-censored at year two has contributed two years to person-time at risk and belongs to earlier risk sets. Neither recording a failure at year two nor discarding that subject preserves this information.

Under independent censoring and separate parameters for the observation mechanism, the part of the survival likelihood involving an absolutely continuous probability distribution for $T$ is $f(Y{)}^{\delta }S(Y{)}^{1-\delta }$. Thus a right-censored observation contributes the survival function $S(Y)$, rather than the event probability density function $f(Y)$. See Lu Tian's Stanford notes, Section 2.1, equations (2.1)–(2.2). The Kaplan–Meier six-subject cohort shows exactly how this distinction changes the calculation.