Competing Risks For Policy Termination

Why Policy Termination Has Competing Causes

A policy can stop being active for different reasons. A customer may choose not to renew, an insurer may decide not to offer renewal, or another event may end the policy before renewal, such as a midterm cancellation. These outcomes matter for retention decisions, but they are not interchangeable: a customer nonrenewal is not evidence that the insurer declined the risk.

Competing-risks methods model the first termination event and its cause. They are useful when the question is not merely whether a policy terminates, but how likely each specific termination outcome is over a stated period. The event causes are mutually exclusive for a given policy at its first termination time. Administrative end of observation, by contrast, is censoring: the policy’s eventual termination cause is not observed during the study window.

Definition: Let \(T\) be the time from a defined starting point to the first policy termination, and let \(J\) identify its cause. For example, \(J=C\) may denote customer nonrenewal, \(J=I\) insurer nonrenewal, and \(J=O\) another termination. A competing-risks outcome is the pair \((T,J)\). A policy still active when observation ends is censored, not classified as having a termination cause.

The starting point and event definitions must be operationally clear. For a renewal cohort, the time origin might be a specified period before expiration, and customer or insurer nonrenewal might be recorded at the renewal decision. If a policy instead cancels midterm, that may be a separate cause, depending on the study design. A termination code that combines customer and insurer decisions cannot support a reliable distinction between those causes.

Cause-Specific Hazards and Cumulative Incidence

For cause \(k\), the cause-specific hazard describes the instantaneous rate of that cause among policies that have not yet experienced any termination. In continuous time, for small \(dt\),

\lambda_k(t) = \lim_{dt\to 0} \frac{P(t\leq T<t+dt,\ J=k\mid T\geq t)}{dt}.

The condition \(T\geq t\) is essential: a policy that already terminated for any cause is no longer at risk of terminating for another cause. The cause-specific hazard is a rate, not the unconditional probability of that cause by time \(t\). With several causes, the overall hazard is \(\lambda(t)=\sum_k\lambda_k(t)\), and the probability of remaining active is \(S(t)=P(T>t)\).

The cumulative incidence function (CIF) for cause \(k\) is the probability that termination from that cause has occurred by time \(t\):

F_k(t)=P(T\leq t,\ J=k) =\int_0^t S(u-)\lambda_k(u)\,du.

Here \(S(u-)\) is the probability of being event-free just before time \(u\). To experience cause \(k\) at time \(u\), a policy must first remain free of every termination cause until then. The CIF therefore combines the cause-specific rate with the chance of still being exposed to that rate.

Key result: When the listed causes cover all first terminations, the probabilities satisfy \(S(t)+\sum_k F_k(t)=1\). A cause-specific hazard describes the rate among policies still at risk; its CIF describes the unconditional probability of that cause by time \(t\).

Discrete-Interval Calculation

For data divided into intervals, define \(h_{kj}\) as the probability of termination from cause \(k\) during interval \(j\), conditional on being active at its start. The interval hazards must be defined over the same interval and at the same conditioning point. The total termination probability in interval \(j\) is \(\sum_k h_{kj}\), which cannot exceed one. The conditional probability of surviving the interval is \(1-\sum_k h_{kj}\).

S_j=S_{j-1}\left(1-\sum_k h_{kj}\right), \qquad F_{kj}=F_{k,j-1}+S_{j-1}h_{kj}, \qquad S_0=1,\quad F_{k0}=0.

The CIF increment is the probability of reaching the interval without a prior termination, multiplied by the conditional probability of cause \(k\) in that interval. This is the discrete counterpart of the continuous-time integral.

Worked Example: Two Intervals With Three Termination Causes

Hypothetical scenario. A renewal cohort is followed for two successive intervals. Conditional on being active at the start of an interval, the interval 1 probabilities are \(h_{C1}=0.10\), \(h_{I1}=0.05\), and \(h_{O1}=0.05\). For policies still active at the start of interval 2, they are \(h_{C2}=0.20\), \(h_{I2}=0.10\), and \(h_{O2}=0.05\). Find each cause’s cumulative incidence and the active probability at the end of interval 2.

Method. Apply the discrete competing-risks recursion. In interval 1, the total termination probability is \(0.10+0.05+0.05=0.20\), so \(S_1=1(1-0.20)=0.80\). The cumulative incidences are \(F_{C1}=0.10\), \(F_{I1}=0.05\), and \(F_{O1}=0.05\).

In interval 2, the total conditional termination probability is \(0.20+0.10+0.05=0.35\). The CIF increments use \(S_1=0.80\), not the original cohort size alone: customer nonrenewal adds \(0.80(0.20)=0.16\), insurer nonrenewal adds \(0.80(0.10)=0.08\), and other termination adds \(0.80(0.05)=0.04\).

Thus \(F_C=0.10+0.16=0.26\), \(F_I=0.05+0.08=0.13\), \(F_O=0.05+0.04=0.09\), and \(S_2=0.80(1-0.35)=0.52\). Check: \(0.26+0.13+0.09+0.52=1.00\). The common mistake is to add the interval 2 probabilities directly to the interval 1 CIFs without weighting them by the probability still active at interval 2.

Constant Cause-Specific Hazards

A useful continuous-time model assumes each cause-specific hazard is constant over the prediction period. If the rates are \(\lambda_1,\ldots,\lambda_K\), the total rate is \(\lambda=\sum_k\lambda_k\). Under this assumption, the probability of remaining event-free through \(t\) is \(S(t)=e^{-\lambda t}\). For \(\lambda>0\), integrating the CIF formula gives:

F_k(t)=\int_0^t e^{-\lambda u}\lambda_k\,du =\frac{\lambda_k}{\lambda}\left(1-e^{-\lambda t}\right).

The share \(\lambda_k/\lambda\) is the fraction of the total instantaneous rate attributable to cause \(k\); \(1-e^{-\lambda t}\) is the probability of any termination by time \(t\). Their product is the cause-specific termination probability by \(t\). This result requires constant hazards over the period. If hazards vary over time, the general integral must be used instead.

Worked Example: One-Year Incidence Under Constant Hazards

Hypothetical scenario. Suppose the annual cause-specific hazards for customer nonrenewal, insurer nonrenewal, and other termination are constant at \(0.12\), \(0.06\), and \(0.02\) per year. Calculate each one-year cumulative incidence and the probability of remaining active. The model uses continuous time, constant hazards, and one year as the prediction horizon.

Step 1: Total hazard. \(\lambda=0.12+0.06+0.02=0.20\) per year. At \(t=1\), \(S(1)=e^{-0.20}=0.8187308\).

Step 2: Probability of any termination. \(1-S(1)=1-0.8187308=0.1812692\).

Step 3: Allocate that probability by cause. For customer nonrenewal, \(\lambda_C/\lambda=0.12/0.20=0.60\), so \(F_C(1)=0.60(0.1812692)=0.1087615\). For insurer nonrenewal, the rate share is \(0.06/0.20=0.30\), so \(F_I(1)=0.30(0.1812692)=0.0543808\). For other termination, the rate share is \(0.02/0.20=0.10\), so \(F_O(1)=0.10(0.1812692)=0.0181269\).

Rounded to four decimal places, the probabilities are \(0.1088\), \(0.0544\), \(0.0181\), and \(0.8187\) for customer nonrenewal, insurer nonrenewal, other termination, and remaining active, respectively. Check: the unrounded cause probabilities sum to \(0.1812692\), and adding \(S(1)\) gives \(1.0000000\). The common mistake is to report \(0.12\) as the one-year probability of customer nonrenewal: it is a hazard rate, not the CIF.

Estimating Incidence From Observed Event Counts

When event times and causes are observed, the Aalen-Johansen estimator estimates the probability of remaining event-free and each cause’s cumulative incidence. At event time \(t_j\), let \(Y_j\) be the number at risk immediately beforehand, \(d_j\) the number of all-cause terminations, and \(d_{kj}\) the number of terminations from cause \(k\). The estimator updates event-free probability by \(1-d_j/Y_j\); the increment for cause \(k\) is the event-free probability just before \(t_j\), multiplied by \(d_{kj}/Y_j\).

\widehat S(t_j)=\widehat S(t_{j-1})\left(1-\frac{d_j}{Y_j}\right), \qquad \widehat F_k(t_j)=\widehat F_k(t_{j-1}) +\widehat S(t_{j-1})\frac{d_{kj}}{Y_j}.

The risk set at each event time includes policies that have not yet terminated or been censored. Under the usual independent-censoring assumption, censoring removes a policy from later risk sets but does not count as a termination. The estimator accounts for the fact that competing events reduce the number of policies still able to experience the cause of interest.

Worked Example: Aalen-Johansen Estimates From a Small Cohort

Hypothetical scenario. Ten policies are followed from the same starting point, with no censoring before the two event times. At \(t_1\), one customer nonrenewal and one insurer nonrenewal occur. At \(t_2\), among the eight policies then at risk, two customer nonrenewals, one insurer nonrenewal, and one other termination occur. The remaining four policies are censored after \(t_2\). Estimate the cause-specific CIFs and event-free probability after \(t_2\).

At \(t_1\). The risk set is \(Y_1=10\), and the total number of terminations is \(d_1=2\). Therefore \(\widehat S(t_1)=1(1-2/10)=0.80\). The CIF increments are \(1(1/10)=0.10\) for customer nonrenewal, \(1(1/10)=0.10\) for insurer nonrenewal, and \(0\) for other termination.

At \(t_2\). The risk set is \(Y_2=8\) and \(d_2=4\). Event-free probability becomes \(\widehat S(t_2)=0.80(1-4/8)=0.40\). The customer CIF increases by \(0.80(2/8)=0.20\), the insurer CIF by \(0.80(1/8)=0.10\), and the other-termination CIF by \(0.80(1/8)=0.10\).

The final estimates are \(\widehat F_C=0.10+0.20=0.30\), \(\widehat F_I=0.10+0.10=0.20\), \(\widehat F_O=0+0.10=0.10\), and \(\widehat S=0.40\). Check: \(0.30+0.20+0.10+0.40=1.00\). The common mistake is to use the original risk set of ten at \(t_2\); four policies had already terminated and are no longer at risk.

Why Competing Events Must Stay in the Calculation

A common alternative is to estimate a cause-specific survival curve by treating competing terminations as censored, then report one minus that curve as the probability of the cause of interest. This generally does not estimate the CIF. It answers a hypothetical net-risk question in which competing causes are removed from the event process, rather than the real-world probability that the specified cause occurs first while the other causes remain possible.

The distinction matters for interpretation. A higher insurer-nonrenewal hazard can reduce the number of policies that remain exposed to customer nonrenewal, and therefore can reduce customer nonrenewal’s CIF even if its own cause-specific hazard does not change. Cause-specific hazards are conditional rates among policies still active; CIFs are unconditional probabilities that incorporate all competing causes. Neither should be substituted for the other.

In practice, define the cohort, time origin, termination causes, and censoring rule before fitting or interpreting a model. State the prediction horizon and report probabilities in a way that makes their conditioning clear. These distinctions are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Cause-specific hazard: rate of a particular termination cause among policies still event-free.
Cumulative incidence: probability of that cause occurring first by a specified time, accounting for all competing causes.
Censoring: incomplete observation, not a termination cause; its treatment relies on an appropriate independent-censoring assumption.

Check Your Understanding

For each problem, treat the stated causes as mutually exclusive first terminations. Give probabilities to four decimal places where needed.

In one discrete interval, conditional on being active at its start, customer nonrenewal has probability \(0.08\), insurer nonrenewal has probability \(0.03\), and other termination has probability \(0.04\). Find the end-of-interval probability of each termination cause and of remaining active. (Answer: customer \(0.0800\), insurer \(0.0300\), other \(0.0400\), active \(0.8500\))
In interval 1, conditional termination probabilities are \(0.10\) for customer nonrenewal, \(0.10\) for insurer nonrenewal, and \(0.05\) for other termination. In interval 2, among policies still active, they are \(0.20\), \(0.05\), and \(0.05\), respectively. Find all three cumulative incidences and the probability still active after interval 2. (Answer: customer \(0.2500\), insurer \(0.1375\), other \(0.0875\), active \(0.5250\))
Constant annual cause-specific hazards are \(0.09\) for customer nonrenewal, \(0.03\) for insurer nonrenewal, and \(0.03\) for other termination. Find the one-year customer nonrenewal CIF. (Answer: \(0.0836\))
At \(t_1\), from \(Y_1=20\), there are two customer nonrenewals, two insurer nonrenewals, and one other termination. At \(t_2\), \(Y_2=15\), there are three customer nonrenewals, no insurer nonrenewals, and three other terminations. Use Aalen-Johansen updates to find the final CIFs and event-free probability. (Answer: customer \(0.2500\), insurer \(0.1000\), other \(0.2000\), event-free \(0.4500\))
Suppose customer nonrenewal and insurer nonrenewal have constant hazards of \(0.10\) and \(0.10\) per year, with no other cause. Find the one-year customer CIF and compare it with one minus the net cause-specific survival \(e^{-0.10}\). (Answer: customer CIF \(0.0906\); net-risk complement \(0.0952\), which is not the CIF)

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