Why Censoring Changes a Likelihood Comparison
In insurance and survival studies, an observation may end before the event of interest is observed. A hypothetical policy could remain claim-free when the study closes, or a component could still be operating when monitoring ends. Such a record is not an event at the end of follow-up: it tells us that the event time exceeds the observed follow-up time. A likelihood comparison must preserve that information.
For a right-censored observation, the likelihood contribution is a survival probability, not an event density. Thus a dataset containing both observed events and censored records contributes a mixture of density and survival terms. Candidate models must be evaluated against the same observed records, with the same event and censoring indicators.
This convention assumes continuous event times, so the distinction between \(X\le C\) and \(X<C\) has probability zero. The censoring time must be independent of the event time, or the censoring mechanism must otherwise be modeled appropriately. If censoring is informative, treating the censoring mechanism as irrelevant can give an invalid event-time likelihood.
Constructing the Observed-Data Likelihood
For independent observations \((T_i,\Delta_i)\), \(i=1,\ldots,n\), the event-time likelihood under a candidate model with parameter \(\theta\) is
An observed event at time \(t\) contributes \(f_X(t;\theta)\): the event occurred near \(t\). A right-censored record at time \(c\) contributes \(S_X(c;\theta)\): the event had not occurred by \(c\). In log-likelihood form, multiplication becomes addition:
The density term has units inverse to the time unit, while a survival probability is dimensionless. Likelihood comparisons remain meaningful when every candidate uses the same observations and time units. Changing time units requires transforming event densities consistently; changing units for only one model does not produce a valid comparison.
For every observation, identify the observed time and whether an event was observed or the record was censored.
Use the candidate density for an event and the candidate survival function for a right-censored record.
Multiply them, or add their logarithms, separately for each candidate model.
Use the likelihoods at specified parameters or maximize each likelihood over its parameters, stating which comparison is being made.
Why the Censored Record Contributes Survival
Suppose \(X\) is the event time and \(C=c\) is a known administrative censoring time. Observing a censoring record at \(c\) means the event time is greater than \(c\), so the probability of the observed event-time information is \(P(X>c)=S_X(c)\). It is not the probability of an event at \(c\). By contrast, observing an event at \(t\) gives a density contribution \(f_X(t)\), because an exact value from a continuous distribution has probability zero.
If the censoring time itself is random and independent of \(X\), the joint likelihood for an event record at \(t\) contains \(f_X(t)S_C(t)\); a censored record at \(c\) contains \(f_C(c)S_X(c)\). The factors involving the censoring distribution do not involve the event-model parameters when that censoring distribution is common and independent. They can therefore be omitted when comparing event-time models. The event-model contributions that remain are precisely \(f_X(t)\) for events and \(S_X(c)\) for censored records.
Worked Example: Comparing Two Fully Specified Models
A hypothetical equipment study records event times in years: an event at \(1\), an event at \(2\), and right-censoring at \(3\). Compare an exponential model with rate \(\lambda=0.5\) per year to a Weibull model with shape \(k=2\) and scale \(\eta=2\) years. Treat both parameter choices as specified, not estimated.
Method. Construct the three contributions under each model, using densities for the two events and a survival probability for the censored record.
For an exponential distribution, \(f_X(t)=\lambda e^{-\lambda t}\) and \(S_X(t)=e^{-\lambda t}\). The likelihood is
The log-likelihood is \(\log L_{\mathrm{Exp}}=\log(0.25)-3\approx -4.3863\).
For a Weibull distribution, \(S_X(t)=e^{-(t/\eta)^k}\) and \(f_X(t)=(k/\eta)(t/\eta)^{k-1}e^{-(t/\eta)^k}\). With \(k=2\) and \(\eta=2\), the event densities at \(1\) and \(2\) are \(0.5e^{-0.25}\) and \(e^{-1}\); the censoring contribution at \(3\) is \(e^{-2.25}\). Therefore,
Its log-likelihood is \(\log(0.5)-3.5\approx -4.1931\). Since \(0.0151>0.0124\), the specified Weibull model has the higher likelihood on these records. This is a likelihood comparison for these parameter values, not a claim about which model would have the higher maximized likelihood.
Common mistake avoided: Treating the censoring time of \(3\) as an event would replace the survival term with a density. The record says only that the event time exceeds \(3\), so its contribution must be \(S_X(3)\).
Comparing Maximized Likelihoods
When parameters are estimated, maximize each candidate model’s observed-data likelihood using the same event and censoring records. A useful closed-form case is the exponential model. If \(D=\sum_i\Delta_i\) events are observed and \(T_\bullet=\sum_iT_i\) is the sum of all observed times, then
For \(D>0\), differentiating the log-likelihood gives \(D/\lambda-T_\bullet\). Setting this equal to zero yields \(\widehat{\lambda}=D/T_\bullet\). The censored follow-up times are included in \(T_\bullet\) because their survival contributions contain \(e^{-\lambda T_i}\).
For comparison, consider a Weibull model with a fixed shape \(k>0\) and estimated scale \(\eta\). For event times \(t_i\) and censored times \(c_j\), define \(Q=\sum_i t_i^k+\sum_j c_j^k\). The log-likelihood’s scale-dependent terms are \(-Dk\log\eta-Q/\eta^k\). Differentiating with respect to \(\eta\) and setting the derivative to zero gives \(\widehat{\eta}^{\,k}=Q/D\), provided \(D>0\). These estimates illustrate why censored records affect both the fitted model and the comparison.
Worked Example: Fitting Exponential and Weibull Models
A hypothetical reliability study observes an event at time \(1\), a right-censored record at time \(2\), and an event at time \(3\). Fit an exponential model and a Weibull model whose shape is fixed at \(k=2\); estimate the rate or scale by maximum likelihood, then compare their maximized likelihoods.
Method. There are \(D=2\) events. The sum of all observed times is \(T_\bullet=1+2+3=6\). Thus the exponential estimate is \(\widehat{\lambda}=2/6=1/3\). Its maximized log-likelihood is
For the Weibull model, \(Q=1^2+2^2+3^2=14\), including the censored time. The scale estimate satisfies \(\widehat{\eta}^{\,2}=14/2=7\), so \(\widehat{\eta}=\sqrt{7}\) years. The event-density product has factors \(2(1)/\eta^2\) and \(2(3)/\eta^2\), while the censored record contributes \(e^{-2^2/\eta^2}\). Together the log-likelihood is
The Weibull maximized log-likelihood exceeds the exponential value by approximately \(0.7903\), and the likelihood ratio is
Thus, among these two fitted models, the Weibull model has the greater maximized likelihood on this dataset. The check is consistent: \(\log(108/49)\approx0.7903\), matching the log-likelihood difference.
Common mistake avoided: Excluding the censored time from \(Q\) would give \(1^2+3^2=10\) and an incorrect Weibull scale estimate. The censoring record supplies the survival term \(S_X(2)\), so its squared time belongs in \(Q\).
Worked Example: How Omitting Censoring Can Reverse a Choice
A hypothetical study has an event at time \(2\), an event at time \(4\), and a right-censored record at time \(3\). Compare two fully specified exponential models, \(\lambda=0.2\) and \(\lambda=0.4\), with rates per time unit.
Method. The correct likelihood has two event densities and one censoring survival contribution. The sum of observed times is \(2+4+3=9\), so \(L(\lambda)=\lambda^2e^{-9\lambda}\). For \(\lambda=0.2\), \(\ell=2\log(0.2)-1.8\approx-5.0189\). For \(\lambda=0.4\), \(\ell=2\log(0.4)-3.6\approx-5.4326\). The \(0.2\) model has the higher likelihood.
If the censored contribution is mistakenly omitted, the likelihood becomes \(\lambda^2e^{-6\lambda}\), based only on the two event densities. The corresponding log-likelihoods are \(2\log(0.2)-1.2\approx-4.4189\) and \(2\log(0.4)-2.4\approx-4.2326\). This incorrect calculation selects the \(0.4\) model instead.
Common mistake avoided: A censored record is not “no data.” Its survival contribution can be important enough to change the ranking of candidate models.
Conditions and Exam Traps
A valid comparison requires both models to describe the same observed records: same times, same event indicators, and same time units. Do not compare a likelihood calculated from events alone under one model with a likelihood that includes censored records under another. If different censoring distributions are being modeled, or censoring depends on the event time, the censoring mechanism may need to enter the joint likelihood rather than being dropped.
A larger maximized likelihood means a better fit in the likelihood sense for the observed data. It does not, by itself, establish that the model is adequate or that it should be preferred after accounting for complexity. Nor does a likelihood ratio automatically have a standard chi-square interpretation: that depends on the relationship between the models and the relevant assumptions for calibration. State whether parameters are fixed or maximized, and avoid attaching a significance conclusion that the comparison alone does not justify.
For continuous event-time models, a left-censored record \(X\le c\) instead contributes \(F_X(c)\); an interval-censored record \(a<X\le b\) contributes \(F_X(b)-F_X(a)\). These are different observation mechanisms. The right-censoring survival contribution applies when the information is \(X>c\), not when the event is known to lie in an interval or below a threshold.
Check Your Understanding
For each problem, use the stated event and censoring status when forming the likelihood.
- A hypothetical lifetime study has an event at time \(1\) and right-censoring at time \(2\). Under an exponential model with rate \(\lambda=0.25\) per time unit, write the likelihood and its numerical value. (Answer: \(L=0.25e^{-0.75}\approx0.1181\).)
- Two specified exponential models have rates \(0.1\) and \(0.2\). The data are an event at time \(3\) and right-censoring at time \(4\). Which has the larger likelihood? (Answer: \(\lambda=0.1\), with log-likelihood \(\log(0.1)-0.7\approx-3.0026\), versus \(\log(0.2)-1.4\approx-3.0094\).)
- Under a Weibull model with shape \(k=2\) and scale \(\eta=3\), a hypothetical observation is an event at time \(1\), followed by a separate right-censored observation at time \(2\). What is their joint likelihood contribution? (Answer: \((2/9)e^{-1/9}e^{-4/9}=(2/9)e^{-5/9}\approx0.1275\).)
- A study records an event at time \(2\), right-censoring at time \(4\), and an event at time \(6\). Find the maximum-likelihood estimate and maximized log-likelihood for an exponential model. (Answer: \(\widehat{\lambda}=2/(2+4+6)=1/6\); \(\ell=2\log(1/6)-2\approx-5.5835\).)
- Two specified exponential models, with rates \(0.25\) and \(0.5\), are compared using an event at time \(1\), an event at time \(2\), and right-censoring at time \(5\). Which model has the larger likelihood? (Answer: \(\lambda=0.25\); the log-likelihoods are \(2\log(0.25)-2\approx-4.7726\) and \(2\log(0.5)-4\approx-5.3863\).)