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Time-To-Lapse Models With Survival Analysis

Learn to construct and interpret time-to-lapse models when some policies have not lapsed by the end of observation.

Advanced 12 min read

What You'll Learn

  • Define the time origin, lapse event, follow-up time, and censoring indicator
  • Interpret survival and hazard functions for policy retention
  • Calculate Kaplan–Meier survival estimates from event-time risk sets
  • Fit and interpret a constant-hazard exponential model with censored observations
  • Recognize independent-censoring and proportional-hazards assumptions
  • Avoid treating survival estimates as fixed-horizon renewal probabilities

Why Model Time to Lapse?

A fixed-horizon renewal model answers whether a policy renews by a specified date. A time-to-lapse model asks a different question: how long does it take for a defined lapse event to occur? It uses the timing of observed lapses and can retain information from policies that are still active when observation ends. That distinction matters when policies enter the data at different times or have different follow-up periods.

Before estimating anything, the analyst must choose a time origin and define the event. Time might begin at a renewal offer or at policy inception; those choices describe different durations and should not be mixed without a clear modeling reason. The event might be lapse within a particular observation system. A policy still active at the end of observation has not necessarily demonstrated long-term retention; its lapse time is simply not yet observed.

Definition: Let \(T\) be the nonnegative time from a specified origin to the defined lapse event. The survival function is \(S(t)=P(T>t)\): the probability that the policy has not lapsed by elapsed time \(t\). The cumulative probability of lapse by \(t\) is \(1-S(t)\). The hazard \(h(t)\) describes the instantaneous lapse rate at time \(t\), conditional on having remained event-free up to \(t\); it is a rate, not a probability.

For continuous event times, the hazard can be written as \(h(t)=f_T(t)/S(t)\), where \(f_T\) is the event-time density. Equivalently, for a small interval \(\Delta t\), \(h(t)\Delta t\) approximates the conditional probability of a lapse in that interval given survival to its start. A hazard ratio compares hazards; it does not directly compare the probabilities of lapse by a selected horizon.

Event Times and Right Censoring

A common observation pattern is right censoring: the policy has not lapsed when its observation ends, so its exact event time is unknown but exceeds the observed follow-up. Let \(C\) denote the time observation ends. The recorded follow-up time and event indicator are

$$ U=\min(T,C), \qquad \delta=I(T\le C). $$

Here, \(\delta=1\) means a lapse was observed at time \(U\); \(\delta=0\) means the policy was censored at \(U=C\). A censored policy contributes its known event-free follow-up, but it must not be recorded as a lapse at the censoring time or discarded as though it had provided no information.

Administrative censoring occurs when observation ends by design, such as at a study cutoff. It is often more defensible as independent censoring than a censoring time driven by policyholder behavior, but it is not automatically independent. Standard survival estimation requires censoring to be independent of the lapse time, at least conditional on covariates or strata used in the analysis. If another type of policy termination prevents the defined lapse from occurring, treating that termination as ordinary censoring requires careful justification; otherwise a single-event model may not answer the intended question.

Worked Example: Defining Follow-Up From a Renewal Offer

A hypothetical insurer starts follow-up at each policy’s renewal offer and records duration in months using its established duration convention. The data are: Policy A, lapse at 2.5 months; Policy B, still active at the 4-month administrative cutoff; Policy C, lapse at 5 months; Policy D, last confirmed active at 6 months and then lost to observation; Policy E, lapse at 7 months. Construct \(U\) and \(\delta\) for each policy. State the interpretation of the observations for a 4-month horizon.

Method. Use elapsed time from the offer as the origin. For an observed lapse, set \(U=T\) and \(\delta=1\). For a policy without an observed lapse, set \(U=C\) and \(\delta=0\).

A has \(U=2.5\) and \(\delta=1\). B has \(U=4\) and \(\delta=0\). C has \(U=5\) and \(\delta=1\). D has \(U=6\) and \(\delta=0\). E has \(U=7\) and \(\delta=1\). The time unit for every value is months.

At 4 months, A has lapsed, while B is known to have remained event-free through 4 months. C and E are observed to remain event-free for at least 4 months before their later lapses. D is known to remain event-free through 4 months, even though observation ends at 6 months. These facts distinguish an event time from a censoring time; the observations alone do not imply that all five policies have known ultimate lapse outcomes.

Common mistake avoided. Recording B as a lapse at month 4 confuses the observation cutoff with the event. B contributes four months of event-free follow-up and a censoring indicator of zero.

Kaplan–Meier Estimation

The Kaplan–Meier estimator is a nonparametric estimate of \(S(t)\) from observed event times and right-censored follow-up. At each distinct lapse time \(t_j\), let \(n_j\) be the number of policies still under observation and event-free immediately before \(t_j\), and let \(d_j\) be the number of lapses at that time. The estimator is

$$ \widehat{S}(t)=\prod_{t_j\le t}\left(1-\frac{d_j}{n_j}\right). $$

Censoring reduces the risk set for later event times but does not itself make the survival estimate drop. When a lapse and censoring occur at the same recorded time, a common convention is to include the censored policies in that time’s risk set and remove them afterward. The data convention and event-time resolution should be stated when ties affect the calculation.

The product form follows from conditional survival. The probability of surviving beyond \(t_j\), conditional on having survived to just before \(t_j\), is estimated by the fraction that do not lapse at \(t_j\), namely \((n_j-d_j)/n_j\). Survival through \(t\) requires surviving each event time up to \(t\). Multiplying these conditional fractions gives the product above. Censored observations are included in the risk sets only while their follow-up is known.

Worked Example: Kaplan–Meier Survival at Several Horizons

In a hypothetical cohort of 10 policies, two lapse at month 1. At month 2, one lapses and one is censored; apply the convention that the censoring occurs after the month-2 lapse risk set is formed. Two lapse at month 3. One is censored at month 4, and the remaining three are censored at month 5. Calculate the Kaplan–Meier estimates at months 1, 2, 3, and 4, and interpret the estimate at month 3.

Method. At month 1, all 10 policies are at risk and two lapse. Thus \(n_1=10\), \(d_1=2\), and the first survival factor is \(1-2/10=8/10=0.8\).

After those two lapses, eight policies remain at risk at month 2. One lapses, so the month-2 factor is \(1-1/8=7/8\). The estimate is \(\widehat{S}(2)=0.8(7/8)=0.7\). The one policy censored at month 2 is removed after it contributes to that risk set.

Six policies remain at risk just before month 3: eight after month 1, less the month-2 lapse and the month-2 censoring. Two lapse at month 3, so the factor is \(1-2/6=4/6\). Therefore, \(\widehat{S}(3)=0.7(4/6)=7/15=0.4667\), rounded to four decimal places. The month-4 censoring does not change the estimate, so \(\widehat{S}(4)=0.4667\).

The estimate at month 3 means that the estimated probability of remaining without a recorded lapse beyond month 3 is about 0.4667 under the model’s assumptions. The estimated probability of lapse by month 3 is \(1-0.4667=0.5333\). This is a cohort survival estimate, not a claim that exactly that fraction of every future cohort will remain active.

Check. The survival estimate cannot increase as time passes: \(0.8\), \(0.7\), and \(0.4667\) are nonincreasing. Censoring at month 4 leaves the month-4 estimate unchanged.

Common mistake avoided. Multiplying by a factor for each censoring time would make the estimate fall even though no lapse occurred. Censoring changes later risk sets; it is not an event.

Parametric and Regression Models

A parametric model specifies a distribution for lapse time. For example, the exponential model assumes a constant hazard \(\lambda>0\) over the time scale being modeled. Under this assumption, \(S(t)=e^{-\lambda t}\), and the probability of lapse by \(t\) is \(1-e^{-\lambda t}\). The constant-hazard assumption is restrictive: it does not allow lapse risk to rise or fall with tenure unless the model is extended.

With independent right censoring and exponential event times, each observed lapse contributes a density \(\lambda e^{-\lambda U_i}\) to the likelihood, and each censored policy contributes a survival probability \(e^{-\lambda U_i}\). If \(D=\sum_i\delta_i\) is the number of observed lapses and \(E=\sum_i U_i\) is total exposure time, the likelihood, apart from factors not involving \(\lambda\), is

$$ L(\lambda)=\lambda^D e^{-\lambda E}. $$

Taking logarithms gives \(\ell(\lambda)=D\log(\lambda)-\lambda E\). Differentiating gives \(d\ell/d\lambda=D/\lambda-E\). Setting this equal to zero yields the maximum-likelihood estimate \(\widehat{\lambda}=D/E\). This derivation explains why censored follow-up belongs in the exposure total even though it contributes no lapse event.

Worked Example: Estimating a Constant Lapse Hazard

A hypothetical portfolio has 8 observed lapses and 200 policy-years of total observed follow-up, including follow-up from policies censored while still active. Assume lapse times follow an exponential distribution with a constant hazard and censoring is independent of lapse time. Estimate the annual hazard, the probability of remaining event-free for one year, and the probability of lapse by three years.

Method. The exponential maximum-likelihood estimate is the observed event count divided by total exposure: \(\widehat{\lambda}=8/200=0.04\) per policy-year. The estimated one-year survival is \(e^{-0.04(1)}=e^{-0.04}=0.9608\), rounded to four decimal places.

For three years, \(\widehat{S}(3)=e^{-0.04(3)}=e^{-0.12}=0.8869\), rounded to four decimal places. The estimated probability of lapse by three years is \(1-\widehat{S}(3)=1-0.8869=0.1131\), or 11.31%.

Check. The three-year survival estimate is lower than the one-year estimate, and its complementary lapse probability is \(0.1131\). The annual hazard is a rate per policy-year, not an annual lapse probability.

Common mistake avoided. Using \(8\) divided by the number of policies would ignore differences in follow-up. Here the likelihood uses 200 policy-years, so a policy censored after substantial event-free observation still contributes exposure.

Regression models allow the lapse hazard to vary with policy characteristics. A standard proportional-hazards specification is \(h(t\mid x)=h_0(t)\exp(\beta^\mathsf{T}x)\), where \(h_0(t)\) is an unspecified baseline hazard and \(x\) is a vector of covariates. Holding other covariates fixed, a one-unit increase in covariate \(x_k\) multiplies the hazard by \(\exp(\beta_k)\). The proportional-hazards assumption means this ratio does not vary with time. It does not mean the lapse probability changes by a constant amount, and it does not by itself provide an individual survival probability without the baseline survival information.

Covariates and their measurement times must be appropriate to the chosen time origin. A predictor measured only after the policy has already remained active for some time cannot simply be treated as known at inception. Doing so can use future information to explain earlier survival. If a predictor changes over time, the analysis must represent that timing rather than silently treating its later value as a baseline value.

Interpretation and Practical Checks

Survival estimates are useful for questions such as “what fraction is expected to remain lapse-free through six months?” or “how does the lapse timing pattern differ across defined groups?” Their interpretation depends on the endpoint, the time origin, the observation process, and the model assumptions. Report the time unit and the horizon with every estimate; a survival probability at six months is not interchangeable with one at six years.

  • Define the event precisely. State what operational outcome counts as lapse and when its event time is recorded.
  • Keep the origin consistent. Time since offer and time since inception answer different questions.
  • Check censoring. Policies censored before a horizon still contribute information before that time, but estimates at later horizons rely on the remaining risk set and the censoring assumption.
  • Separate rates from probabilities. A hazard or hazard ratio is not a probability of lapse over a fixed period.
  • Check model fit and scope. The Kaplan–Meier estimator avoids a parametric event-time distribution, while the exponential model imposes a constant hazard. Neither repairs a poorly defined endpoint or informative censoring.
Key takeaway: A time-to-lapse analysis records both how long each policy is observed and whether lapse actually occurred. Correctly handling censoring preserves the information in incomplete follow-up; the resulting survival probabilities remain conditional on a clear endpoint and defensible assumptions. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, state the key assumption where relevant and keep time units explicit.

  1. A policy is followed from its renewal offer for 9 months and remains active when a planned study cutoff occurs. Give its observed follow-up and event indicator. (Answer: \(U=9\) months and \(\delta=0\).)
  2. Under an exponential model, 6 lapses occur over 150 policy-years of total exposure. Find the estimated hazard and estimated probability of remaining lapse-free for 2 years. (Answer: \(\widehat{\lambda}=0.04\) per policy-year; \(\widehat{S}(2)=e^{-0.08}=0.9231\).)
  3. Of 8 policies at risk at month 1, one lapses. At month 2, five policies are at risk and two lapse; a censoring occurs at month 2 after those lapses are counted. Find the Kaplan–Meier estimate through month 2. (Answer: \((7/8)(3/5)=21/40=0.5250\).)
  4. A proportional-hazards model estimates a coefficient of \(0.25\) for a one-unit increase in a price-change variable. Interpret its hazard ratio, holding other covariates fixed. (Answer: \(e^{0.25}=1.2840\); the modeled lapse hazard is about 28.40% higher, not the lapse probability.)
  5. An exponential model is fitted with 9 observed lapses over 300 policy-years of exposure. Find the estimated probability of lapse within 6 months, using years as the model’s time unit. (Answer: \(\widehat{\lambda}=9/300=0.03\) per year; \(t=0.5\) years, so \(1-e^{-0.03(0.5)}=1-e^{-0.015}=0.0149\), or 1.49%.)
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