Tutorials › CAS Exam 8 - Advanced Ratemaking › Quantifying Uncertainty In Retention Forecasts

CAS Exam 8 · Retention and Churn Modeling · Tutorial 862 of 989

Quantifying Uncertainty In Retention Forecasts

Learn how policy-level probabilities, shared uncertainty, and future portfolio mix each affect the uncertainty range around an aggregate renewal forecast.

Advanced 10 min read

What You'll Learn

  • Calculate the mean and variance of aggregate renewals from policy-level probabilities.
  • Distinguish independent policy outcomes from uncertainty shared across a cohort.
  • Apply the law of total variance to build a forecast variance without omitting between-scenario variation.
  • Separate renewal-outcome uncertainty from uncertainty in future segment composition.
  • Interpret predictive ranges and recognize what they do—and do not—measure.

Why a Renewal Forecast Needs an Uncertainty Range

An expected renewal count is a useful summary, but it is not a promise that exactly that many policies will renew. If a portfolio forecast is 700 renewals, decision-makers also need to understand whether counts near 680 or 720 are plausible, and what assumptions drive that range. This tutorial develops uncertainty calculations for aggregate renewal forecasts, aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

There are several distinct sources of variation. Individual policyholders may renew or lapse even when their renewal probabilities are known. A cohort may also share an uncertain renewal environment, so that a market or model scenario changes many probabilities together. Finally, the future portfolio may contain a different mix of segments than expected. These sources should be identified before combining them; otherwise a forecast can omit important variation or count the same uncertainty twice.

Definition: Let \(R_i=1\) if policy \(i\) renews and \(R_i=0\) otherwise. The aggregate renewal count is \(M=\sum_{i=1}^{N} R_i\). A predictive range describes plausible values of this future count under a specified model. It includes the randomness of future outcomes and may also include uncertainty in probabilities or portfolio composition, if those are modeled.

Policy-Level Variation When Probabilities Are Fixed

Suppose the portfolio contains \(N\) policies and policy \(i\) has a known renewal probability \(p_i=P(R_i=1)\). If renewal outcomes are conditionally independent and the probabilities are treated as fixed, then \(R_i\) is Bernoulli with mean \(p_i\) and variance \(p_i(1-p_i)\). Linearity of expectation gives the expected renewal count; independence allows the individual variances to be added.

$$ E[M]=\sum_{i=1}^{N}p_i,\qquad \operatorname{Var}(M)=\sum_{i=1}^{N}p_i(1-p_i). $$

The distribution of \(M\) in this case is a Poisson-binomial distribution, because the independent Bernoulli outcomes can have different probabilities. Replacing all policy probabilities with their average does not generally preserve the variance. If every policy has the same probability \(p\), then \(M\) is binomial and its variance is \(Np(1-p)\). But with varying probabilities, the individual variances should be summed.

Worked Example: A Predictive Range From Policy Probabilities

A hypothetical renewal book has 60 policies with renewal probability 0.80 and 40 policies with renewal probability 0.55. Assume each policy's outcome is independent of the others and these probabilities are fixed. Find the expected number of renewals, its standard deviation, and an approximate central 90% predictive range.

Method. Add the policy-level expected counts and variances. The mean is \(60(0.80)+40(0.55)=48+22=70\) renewals.

For the first group, the variance contribution is \(60(0.80)(1-0.80)=60(0.16)=9.60\). For the second group, it is \(40(0.55)(1-0.55)=40(0.2475)=9.90\). Therefore, \(\operatorname{Var}(M)=9.60+9.90=19.50\), and the standard deviation is \(\sqrt{19.50}=4.4159\) renewals.

A normal approximation gives a central 90% range of approximately \(70\pm1.645(4.4159)\). The margin is \(7.2641\), so the approximate range is 62.74 to 77.26 renewals, or about 63 to 77 when expressed as whole counts. The approximation is reasonable here: each group has expected renewals and nonrenewals comfortably above 10. The exact predictive distribution is discrete, so an exact 90% range would be obtained from its Poisson-binomial distribution.

Check and common mistake. The group means sum to 70, and the variance contributions sum to 19.50. Treating all 100 policies as though they had probability 0.70 would give variance \(100(0.70)(0.30)=21\), which is not the correct variance for these fixed, heterogeneous probabilities.

Shared Uncertainty and the Law of Total Variance

A different situation arises when a probability is uncertain for the whole cohort. For example, suppose renewal outcomes are independent conditional on a market scenario, but the scenario determines a common probability for all policies. The outcomes are then not independent unconditionally: observing a favorable scenario makes many renewals more likely at once.

Theorem: For any random variables \(M\) and \(\Theta\) with finite variance, \(\operatorname{Var}(M)=E[\operatorname{Var}(M\mid\Theta)]+\operatorname{Var}(E[M\mid\Theta])\). The first term is average within-scenario outcome variation. The second is variation between scenario-specific expected counts.

The identity follows by writing \(M-E[M]\) as \([M-E(M\mid\Theta)]+[E(M\mid\Theta)-E(M)]\). When the square is expanded and its expectation taken, the cross-product has expectation zero: conditional on \(\Theta\), the first bracket has mean zero, while the second is fixed. The two remaining squared terms give the conditional variance averaged over scenarios and the variance of the conditional mean. Thus scenario variation cannot be discarded just because the overall average probability is known.

$$ \operatorname{Var}(M) = E[\operatorname{Var}(M\mid\Theta)] + \operatorname{Var}(E[M\mid\Theta]). $$

Worked Example: A Shared Renewal Scenario

A hypothetical cohort contains 40 policies. With probability one-half, a shared market scenario produces renewal probability 0.40 for every policy; with probability one-half, it produces probability 0.80 for every policy. Conditional on the scenario, policy outcomes are independent. Find the expected renewal count and its total variance. Compare that variance with a calculation using only the average probability.

Method. Let \(\Theta\) denote the scenario's common renewal probability. The average probability is \(E[\Theta]=(0.40+0.80)/2=0.60\). Therefore the expected count is \(40(0.60)=24\).

In the 0.40 scenario, the conditional mean is \(40(0.40)=16\) and the conditional variance is \(40(0.40)(0.60)=9.6\). In the 0.80 scenario, the conditional mean is \(40(0.80)=32\) and the conditional variance is \(40(0.80)(0.20)=6.4\). The expected conditional variance is \((9.6+6.4)/2=8\).

The scenario-specific means are 16 and 32, each occurring with probability one-half. Their average is 24, so the between-scenario variance is \(\big[(16-24)^2+(32-24)^2\big]/2=(64+64)/2=64\). The total variance is \(8+64=72\).

Using only the average probability as if the 40 outcomes were independent Bernoulli variables with fixed probability 0.60 gives \(40(0.60)(0.40)=9.6\). That is not the total variance under the stated shared-scenario model. It omits the between-scenario variance of 64. The correct check is \(8+64=72\), not 152: the squared deviations from 24 are 64 in each scenario, and averaging them still gives 64.

More generally, if all \(N\) policies share a random probability \(\Theta\), and outcomes are independent conditional on \(\Theta\), then \(E[M\mid\Theta]=N\Theta\) and \(\operatorname{Var}(M\mid\Theta)=N\Theta(1-\Theta)\). Substitution into the total-variance identity gives \(\operatorname{Var}(M)=N E[\Theta(1-\Theta)]+N^2\operatorname{Var}(\Theta)\). Equivalently, this is \(N\mu(1-\mu)+N(N-1)\operatorname{Var}(\Theta)\), where \(\mu=E[\Theta]\). The additional term reflects positive dependence induced by the shared probability. This formula does not apply if policies have independent, policy-specific uncertainty rather than a common random probability.

Uncertainty in Future Portfolio Composition

A renewal forecast can also be uncertain because the future portfolio's segment counts are not fixed. Suppose there are segments \(j=1,\ldots,K\), with renewal probability \(p_j\) for a policy in segment \(j\). Let \(N_j\) be the future number of policies in segment \(j\), so the aggregate count is \(M=\sum_j M_j\). Conditional on the segment counts, and assuming independent renewal outcomes with fixed segment probabilities, the conditional mean and variance are

$$ E[M\mid N_1,\ldots,N_K]=\sum_{j=1}^{K}N_jp_j,\qquad \operatorname{Var}(M\mid N_1,\ldots,N_K)=\sum_{j=1}^{K}N_jp_j(1-p_j). $$

If segment counts themselves are uncertain, apply the law of total variance again: \(\operatorname{Var}(M)=E[\operatorname{Var}(M\mid N_1,\ldots,N_K)]+\operatorname{Var}(E[M\mid N_1,\ldots,N_K])\). The second term is the variability of the expected renewal count as the portfolio mix changes. It is zero when the segment counts are fixed. If segment counts move together, their covariances also affect this term; treating them as independent without justification can misstate the range.

Worked Example: Random Mix Across Two Segments

A hypothetical future portfolio has 100 policies. Each policy independently belongs to segment A with probability 0.60 and to segment B with probability 0.40. Renewal probabilities are 0.80 in A and 0.50 in B. Conditional on segment assignment, renewal outcomes are independent. Find the mean and variance of total renewals, separating outcome variation from mix variation.

Let \(N_A\) be the number in A; then \(N_A\) has mean \(100(0.60)=60\) and variance \(100(0.60)(0.40)=24\). Since \(N_B=100-N_A\), the conditional expected count is \(0.80N_A+0.50(100-N_A)=50+0.30N_A\). Its expectation is \(50+0.30(60)=68\).

Given the mix, the renewal-outcome variance is \(0.80(0.20)N_A+0.50(0.50)(100-N_A)=0.16N_A+0.25(100-N_A)\). Its expectation is \(0.16(60)+0.25(40)=9.6+10=19.6\).

The variance of the conditional expected count due to mix is \(\operatorname{Var}(50+0.30N_A)=0.30^2(24)=2.16\). Thus total variance is \(19.6+2.16=21.76\), and the standard deviation is \(\sqrt{21.76}=4.6648\). The result checks against the equivalent view that each independently assigned policy renews with marginal probability \(0.60(0.80)+0.40(0.50)=0.68\), giving variance \(100(0.68)(0.32)=21.76\).

Common mistake. If one freezes the mix at its expected counts, 60 in A and 40 in B, the outcome variance is 19.6, not 21.76. That calculation is conditional on a fixed mix and omits the 2.16 variance contribution from random composition.

Interpreting and Communicating a Forecast Range

A predictive range is conditional on its assumptions. It is not automatically a confidence interval for an average renewal probability, nor does it guarantee that the observed count will fall inside it. A confidence interval for a model parameter addresses uncertainty in that parameter; a predictive range for renewals additionally includes future Bernoulli outcomes. If model-parameter uncertainty is included, it should be propagated into the predictive distribution rather than added informally.

For an applied forecast, state which uncertainties are included. A calculation based on fixed policy probabilities includes policy-level outcome variation. A scenario model can add shared uncertainty in the probabilities. A simulation with future segment assignments can add composition variation. If the same uncertainty is represented both in scenario-specific probabilities and in a separate variance adjustment, it may be counted twice.

  • Fixed mix, fixed probabilities: sum policy-level variances when outcomes are independent.
  • Shared scenario: include both average conditional variance and variance between scenario-specific expected counts.
  • Random composition: account for variation in the expected count across possible segment mixes.
  • Approximate ranges: check that a normal approximation is plausible; otherwise use an appropriate discrete distribution or simulation.

The practical value of this decomposition is diagnostic as well as numerical. A wide range driven mainly by outcome variation may respond to a larger portfolio, while a range driven by a shared market scenario or uncertain mix requires better scenario or composition assumptions. Reporting only a point forecast conceals that distinction.

Key takeaway: First define what is random and what is held fixed. Then calculate the conditional renewal variation and add the variation in conditional expected renewals when probabilities or portfolio composition are uncertain. The total-variance identity makes those components explicit.

Check Your Understanding

For each problem, identify the uncertainty being modeled before calculating.

  1. A cohort has 30 independent policies, each with fixed renewal probability 0.70. Find the expected renewal count and variance. (Answer: Mean 21; variance 6.30.)
  2. For 25 policies, a shared scenario gives probability 0.20 or 0.60, each with probability one-half. Outcomes are independent conditional on the scenario. Find the mean and total variance, and compare with the variance using only the average probability. (Answer: Mean 10; total variance 30; average-probability-only variance 6.)
  3. A future book contains 80 policies. Each independently belongs to segment A with probability 0.25 and B otherwise. Renewal probabilities are 0.90 in A and 0.40 in B. Find the total mean and variance, including composition uncertainty. (Answer: Mean 42; variance 19.95.)
  4. A portfolio has fixed counts of 40 policies with renewal probability 0.50 and 60 with probability 0.75; all outcomes are independent. Find the mean, variance, and standard deviation of renewals. (Answer: Mean 65; variance 21.25; standard deviation 4.61.)
  5. For 100 policies, a common scenario gives renewal probability 0.30 or 0.70, with equal probability, and outcomes are conditionally independent. Find the total variance and explain how much is between-scenario variance. (Answer: Total variance 421; between-scenario variance 400.)
Clinical Biostats