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Exam ALTAM · Risk and Solvency Analysis · Tutorial 887 of 999

Worked Mortality and Lapse Stress Case

Trace how mortality and lapse stresses change expected cash flows, reserves, and projected profits for a two-year protection portfolio.

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What You'll Learn

  • Set out a consistent annual timing convention for death, lapse, premiums, and benefits.
  • Calculate expected in-force counts and claims by duration under baseline and stressed decrements.
  • Distinguish a fixed premium rate from the total premium volume collected.
  • Calculate prospective reserves on baseline and stressed assumptions.
  • Project expected profit using baseline reserves when experience follows stressed assumptions.
  • Check whether the direction and size of each stress effect are consistent with the product mechanics.

Setting Up a Mortality and Lapse Stress Case

A mortality-and-lapse stress changes both the probability of a claim and the number of policies exposed to future claims. The effects can point in different directions: higher mortality increases the claim rate, while higher lapses usually reduce later in-force exposure. A sound projection keeps these effects separate, applies them in the stated order, and uses a consistent reserve basis.

This tutorial uses a hypothetical two-year protection product. It focuses on expected annual cash flows and reserves, not on statistical variation around those expectations. The calculations are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Model and timing convention: At the start of each policy year, each in-force policy pays the annual premium. During the year, death occurs with conditional probability \(q_t\), and a death benefit is paid at year-end. A policy that survives the year lapses at year-end with conditional probability \(w_t\). Thus lapse is conditional on survival, and the probability of remaining in force into the next year is \((1-q_t)(1-w_t)\). There is no lapse refund, expense, tax, or benefit other than the death benefit. The effective annual interest rate is \(i\).

The order matters. Mortality is applied to the policies in force at the start of the year. Lapse is then applied only to survivors. The expected number continuing is therefore not obtained by subtracting \(q_t+w_t\) from one unless the interaction term is also accounted for.

Worked Case: Expected Exposure, Claims, and Premium Volume

Suppose a hypothetical insurer has 1,000 two-year policies in force at the start of year 1. Each pays an annual premium of $100 at the start of each year in which it is in force. The death benefit is $10,000. The annual effective interest rate is 5%. The baseline assumptions are \(q_1=0.010\), \(w_1=0.050\), \(q_2=0.020\), and \(w_2=0.040\). Under stress, they are \(q_1=0.015\), \(w_1=0.080\), \(q_2=0.025\), and \(w_2=0.060\). Coverage continues through year 2; it does not expire after year 1.

Worked Example: Projecting Claims and Premiums Under Both Assumption Sets

Method. Project the year-start exposure recursively. Multiply that exposure by the mortality probability to obtain expected deaths, and apply the benefit to expected deaths to obtain expected claims. Avoid treating lapse as an exit before mortality in the same year.

Baseline year 1. There are 1,000 policies in force. Expected deaths are \(1{,}000(0.010)=10\), so expected claims are \(10(\$10{,}000)=\$100{,}000\). Expected survivors are \(1{,}000(0.990)=990\). Of these, \(990(0.050)=49.5\) are expected to lapse, leaving \(990(0.950)=940.5\) policies in force at the start of year 2. Premium volume in year 1 is \(1{,}000(\$100)=\$100{,}000\).

Stressed year 1. Expected deaths are \(1{,}000(0.015)=15\), giving claims of \(15(\$10{,}000)=\$150{,}000\). Expected survivors are \(1{,}000(0.985)=985\); expected lapses are \(985(0.080)=78.8\). Thus year-2 exposure is \(985(0.920)=906.2\), and year-1 premium volume remains \(1{,}000(\$100)=\$100{,}000\).

Baseline year 2. Expected deaths are \(940.5(0.020)=18.81\), so claims are \(18.81(\$10{,}000)=\$188{,}100\). Premium volume is \(940.5(\$100)=\$94{,}050\).

Stressed year 2. Expected deaths are \(906.2(0.025)=22.655\), so claims are \(22.655(\$10{,}000)=\$226{,}550\). Premium volume is \(906.2(\$100)=\$90{,}620\).

Check. Year-2 exposure is lower under stress, but the year-2 mortality rate is higher. Claims nevertheless rise from $188,100 to $226,550. Over both years, expected claims rise from $288,100 to $376,550, while premium volume falls from $194,050 to $190,620. The annual premium rate did not change; the reduction in total premium volume comes from lower year-2 exposure.

Prospective Reserves on a Specified Basis

A reserve calculation must specify its valuation assumptions and timing. Let \(V_t\) be the reserve at the start of year \(t\), immediately before that year’s premium is collected, for a policy then in force. Let \(V_{t+1}\) be the corresponding reserve at the next year-start for a policy that survives and does not lapse. The final reserve after the last covered year is zero.

Prospective reserve recursion: For a level annual premium \(P\), death benefit \(B\), annual mortality probability \(q_t\), annual lapse probability \(w_t\) conditional on survival, and discount factor \(v=1/(1+i)\), $$ V_t=v\left[q_tB+(1-q_t)(1-w_t)V_{t+1}\right]-P. $$ This is a reserve immediately before the premium at the start of year \(t\). It assumes the stated premium is received then, the death benefit is paid at year-end, and only policies still in force at the next year-start carry \(V_{t+1}\). It does not apply unchanged if premiums, benefits, or payment timing differ.

The recursion follows by valuing future outgo and future reserves at the start of the year. Expected death outgo at year-end is \(q_tB\); its present value is \(vq_tB\). The probability of continuing to the next year is \((1-q_t)(1-w_t)\), so the present value of the next reserve is \(v(1-q_t)(1-w_t)V_{t+1}\). Subtract the premium received immediately to obtain the reserve.

Worked Example: Comparing Baseline and Stressed Reserves

Method. Apply the prospective recursion backward, first calculating the final-year reserve, then the initial reserve. Use \(P=\$100\), \(B=\$10{,}000\), and \(v=1/1.05\). The calculation below treats each assumption set as a complete valuation basis; it does not yet project stressed experience against baseline reserves.

Baseline year 2. Since \(V_3=0\), $$ V_2=v(0.020)(\$10{,}000)-\$100 =\frac{\$200}{1.05}-\$100 =\$90.4762. $$

Baseline year 1. The continuation probability is \((1-0.010)(1-0.050)=0.9405\). Therefore, $$ V_1=\frac{(0.010)(\$10{,}000)+(0.9405)(\$90.4762)}{1.05}-\$100 =\frac{\$100+\$85.0929}{1.05}-\$100 =\$76.2789. $$

Stressed year 2. On the stressed basis, $$ V_2^{\text{stress}}=\frac{(0.025)(\$10{,}000)}{1.05}-\$100 =\frac{\$250}{1.05}-\$100 =\$138.0952. $$

Stressed year 1. The stressed continuation probability is \((1-0.015)(1-0.080)=0.9062\). Thus, $$ V_1^{\text{stress}}=\frac{\$150+(0.9062)(\$138.0952)}{1.05}-\$100 =\frac{\$150+\$125.1429}{1.05}-\$100 =\$162.0400. $$

Check. The stressed-basis reserves exceed the baseline-basis reserves at both year starts: by \(\$138.0952-\$90.4762=\$47.6190\) at the start of year 2 and by \(\$162.0400-\$76.2789=\$85.7611\) at the start of year 1. Higher mortality raises expected claims; the lapse change also affects the probability that a future reserve is required.

Projected Profit When Experience Is Stressed

A stress projection often asks a different question from a stressed-basis reserve valuation: what profit emerges if experience follows stressed assumptions but reserves remain on the baseline basis? This distinction separates the change in cash flows from a reserve revaluation. For the projection below, use the baseline reserves \(V_1=\$76.2789\) and \(V_2=\$90.4762\), and apply stressed mortality and lapse to expected claims and continuing exposure.

For one policy in force at the start of year \(t\), expected profit at the year-end is defined here as accumulated premium and opening reserve, less expected claims and the expected reserve held for policies continuing to the next year:

$$ \Pi_t=(P+V_t)(1+i)-q_t^{\text{actual}}B -(1-q_t^{\text{actual}})(1-w_t^{\text{actual}})V_{t+1}. $$

This is a one-year expected profit measure under the stated timing. In a portfolio projection, multiply it by the number of policies in force at the start of the year. The formula assumes that opening reserves earn the annual interest rate, claims are paid at year-end, and no expenses or other cash flows occur.

Worked Example: Projecting Duration Profits Against Baseline Reserves

Method. Use the profit equation with baseline reserves and stressed experience. Calculate the portfolio profit separately by duration; do not multiply a year-2 per-policy profit by the original year-1 cohort size.

Year 1. The opening reserve is \(V_1=\$76.2789\), and the stressed continuation probability is \(0.9062\). Expected year-end reserve for continuing policies, per year-1 policy, is \(0.9062(\$90.4762)=\$81.9895\). Accumulated premium and reserve are \((\$100+\$76.2789)(1.05)=\$185.0929\). Expected claims are \((0.015)(\$10{,}000)=\$150\). Hence, $$ \Pi_1=\$185.0929-\$150-\$81.9895=-\$46.8967 $$ per policy in force at the start of year 1. For 1,000 policies, expected year-1 profit is \(-\$46{,}896.67\).

Year 2. The baseline opening reserve is \(V_2=\$90.4762\), and the closing reserve is zero. Accumulated premium and reserve are \((\$100+\$90.4762)(1.05)=\$200\). Expected claims per year-2 policy are \((0.025)(\$10{,}000)=\$250\), so $$ \Pi_2=\$200-\$250=-\$50 $$ per policy in force at the start of year 2. There are 906.2 expected policies at that point, giving expected year-2 profit of \(906.2(-\$50)=-\$45{,}310\).

Check. The total expected profit over the two durations, without discounting duration-2 profit back to issue, is \(-\$46{,}896.67-\$45{,}310=-\$92{,}206.67\). Under the baseline assumptions and baseline reserves, the recursion makes expected profit zero in each year: the reserve is constructed to balance the expected cash flows on that same basis. The negative stressed profits therefore reflect the difference between stressed experience and the valuation basis.

Common Exam Traps and Interpretation Checks

A useful way to verify a stress projection is to check each quantity against the product mechanics, rather than expecting every metric to move in the same direction.

  • Applying lapse before mortality. The model specifies lapse among survivors, so continuation is \((1-q_t)(1-w_t)\), not \(1-q_t-w_t\). In year 1 of the case, the stressed continuation probability is \(0.985(0.920)=0.9062\), not \(1-0.015-0.080=0.905\).
  • Calling lower exposure lower mortality. A higher lapse rate can reduce later exposure, but it does not reduce the conditional mortality probability among policies still in force. Keep the year-start exposure and \(q_t\) as separate factors in the claim calculation.
  • Ending coverage too early. These policies provide two years of coverage. Year-2 claims must be calculated from the policies that remain in force after year 1; they would be zero only if coverage expired after the first year.
  • Changing the premium rate unintentionally. The premium rate in the worked case remains $100. Total premium volume falls in year 2 because fewer policies are in force, not because the per-policy premium changed.
  • Mixing reserve bases. A stressed-basis reserve is not the same result as stressed profit against baseline reserves. State explicitly which basis is used for reserve valuation and which assumptions generate projected experience.

The direction check is also useful. In the worked case, higher mortality increases claims directly, while higher lapse reduces the number of policies reaching year 2. The direct mortality increase outweighs the exposure reduction, so year-2 claims rise. The size and even direction of the net result can vary in other products, so the calculation—not a rule of thumb—determines the outcome.

Key takeaway: A mortality-and-lapse stress case requires four distinct steps: project exposure using conditional decrements, calculate claims and premium volume from that exposure, value reserves on an explicitly stated basis, and calculate profit using reserves consistent with the question. Keeping those steps separate makes the results auditable.

Check Your Understanding

For each problem, use the timing and definitions stated in the question. Give monetary answers to the nearest cent.

  1. A hypothetical two-year policy cohort has 800 policies in force at the start of year 1. In year 1, \(q_1=0.012\) and \(w_1=0.050\), with lapse conditional on survival. The death benefit is $6,000. Find the expected year-2 exposure and year-1 claims. (Answer: 750.88 policies; $57,600.00)
  2. A one-year policy has \(P=\$150\), \(B=\$5{,}000\), \(q=0.030\), and \(i=0.040\); there is no future reserve. Using the reserve convention in this tutorial, calculate the reserve immediately before the premium. (Answer: \(-\$5.77\))
  3. At the start of a one-year period, a policy has premium \(P=\$120\), opening reserve \(V=\$40\), and benefit \(B=\$4{,}000\). The annual effective interest rate is 5%, and stressed mortality is \(q=0.020\). The closing reserve is zero. Calculate expected profit per policy. (Answer: $88.00)
  4. A two-year product has \(B=\$10{,}000\), \(P=\$120\), and \(i=0.040\). On the valuation basis, \(q_1=0.010\), \(w_1=0.050\), \(q_2=0.020\), and \(w_2=0.030\). Calculate the reserve at the start of year 1, immediately before premium, using the stated reserve recursion. (Answer: $41.54)
  5. A hypothetical cohort has 1,200 policies at the start of year 1 and a $5,000 death benefit. Baseline year-1 assumptions are \(q_1=0.010\), \(w_1=0.040\); stressed assumptions are \(q_1=0.015\), \(w_1=0.080\). Year-2 mortality is 0.020 on baseline and 0.025 under stress. Calculate baseline and stressed year-2 expected claims and their difference. (Answer: baseline $114,048.00; stress $135,930.00; increase $21,882.00)
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