Why Guarantee and Persistency Risk Interact
A maturity guarantee may pay a top-up when an investment account falls below a specified amount. But the top-up is generally payable only if the policy remains in force until maturity. The guarantee’s cost therefore depends on two related questions: how often does the market leave a shortfall, and how often are policies still in force on those market paths?
This interaction matters because lapse rates may differ across market outcomes. Policyholders may be more likely to lapse after poor performance, or they may be more likely to retain a policy when its guarantee is valuable. Either pattern can change the insurer’s expected guarantee outgo. A single average lapse rate can conceal this effect.
The calculations below use hypothetical products and assumptions. “Expected guarantee shortfall” means the expected value of the guarantee top-up, not a formal tail-risk measure such as Expected Shortfall used in some capital frameworks. Where charges are included, the net expected deficit is separately defined.
This simple model isolates the interaction between market outcomes and lapse. It assumes that the stated market paths and their probabilities are appropriate for the valuation being performed, that \(r_j\) is conditional on the path, and that a policy which lapses before maturity receives no maturity guarantee top-up. It also assumes the guarantee payment occurs at maturity and ignores other benefits, expenses, taxes, and charges unless stated. If lapse benefits include a guarantee payment, or if persistency depends on additional policyholder information, the cash flows and conditioning must be adjusted.
How to Combine Market and Persistency Assumptions
For each path, first determine whether the guarantee is in the money. A path with \(F_j\geq G\) has no top-up, so its gross guarantee claim is zero even if the policy persists. A path with \(F_j<G\) has a top-up of \(G-F_j\), but that amount contributes to expected outgo only to the extent that policies remain in force.
For each market path, calculate \(\max(G-F_j,0)\). Do not count a guarantee payment on a path where the account already exceeds the guarantee.
Multiply that path’s shortfall by \(r_j\), not by an unrelated overall average persistency rate.
Multiply the conditional amount by \(p_j\) and sum over all paths to obtain expected outgo per issued policy.
Apply the discount factor for the actual payment time, if a present value is required.
The method follows directly from conditioning on the market path. Let \(I\) equal 1 if the policy is in force at maturity and 0 otherwise. On path \(j\), the top-up is \(I\max(G-F_j,0)\). The shortfall is fixed once the path is specified, and the conditional expectation of \(I\) on that path is \(r_j\). Therefore:
Weighting these conditional expectations by the path probabilities gives the stated sum. This also explains why averaging persistency first can give the wrong answer: the relevant persistency is the one attached to each path’s shortfall.
Worked Examples
Worked Example: Uniform Persistency Under Two Market Outcomes
A hypothetical equity-linked policy pays a maturity benefit of at least \(G=\$11{,}000\) at the end of two years. Its account value will be \(\$8{,}000\) with probability 0.40 or \(\$12{,}000\) with probability 0.60. Compare two assumptions: persistency is 0.70 on both paths under the first assumption and 0.90 on both paths under the second. Use an effective annual interest rate of 5% to value the expected top-up at issue. Assume a policy that lapses receives no maturity top-up.
Method. Calculate the top-up on each path, apply the relevant persistency, weight by the market probability, and discount two years. The common mistake to avoid is multiplying the guarantee shortfall by the probability of the good path or treating the guarantee as payable on every issued policy.
Path shortfalls. On the \(\$8{,}000\) path, the shortfall is \(\$11{,}000-\$8{,}000=\$3{,}000\). On the \(\$12{,}000\) path, the shortfall is \(\max(\$11{,}000-\$12{,}000,0)=\$0\).
With persistency 0.70 on both paths, expected undiscounted outgo per issued policy is:
The two-year discount factor is \(v^2=(1.05)^{-2}=1/1.1025=0.907029\). The value at issue is \(\$840/1.1025=\$761.9048\), or \$761.90 to the nearest cent.
With persistency 0.90, expected undiscounted outgo is \(0.40(0.90)(\$3{,}000)=\$1{,}080\). Its value at issue is \(\$1{,}080/1.1025=\$979.5918\), or \$979.59 to the nearest cent. The higher persistency raises expected guarantee outgo because the same fraction of policies remains in force on both paths, including the path where the guarantee pays.
Check. Only the 40% shortfall path contributes a positive amount, and the discounted values are less than their corresponding undiscounted values, as required at a positive interest rate.
Worked Example: Market-Dependent Persistency Changes the Cost
A hypothetical contract has a maturity guarantee of \(\$10{,}000\) in two years. The account value is \(\$7{,}000\) on a poor-market path with probability 0.30 and \(\$13{,}000\) on a strong-market path with probability 0.70. Compare (A) persistency of 0.80 on both paths with (B) persistency of 0.95 on the poor path and 0.65 on the strong path. Use an effective annual rate of 4%. There are no other maturity benefits or charges.
Method. Calculate each path’s contribution separately, then compare the results. Avoid replacing the conditional rates in (B) with their average before considering which path carries the guarantee shortfall.
The poor-market path has a shortfall of \(\$10{,}000-\$7{,}000=\$3{,}000\). The strong-market path has no top-up because \(\$13{,}000>\$10{,}000\).
Under (A), expected outgo before discounting is \(0.30(0.80)(\$3{,}000)=\$720\). Under (B), it is \(0.30(0.95)(\$3{,}000)=\$855\). The strong-market path contributes \(\$0\) to gross guarantee claims under both assumptions.
The two-year discount factor is \(v^2=(1.04)^{-2}=1/1.0816=0.924556\). Thus the issue-date values are \(\$720/1.0816=\$665.6804\), or \$665.68, under (A), and \(\$855/1.0816=\$790.4965\), or \$790.50, under (B).
Under (B), overall persistency averaged across the market paths is \(0.30(0.95)+0.70(0.65)=0.74\), below the 0.80 in (A). Nevertheless, guarantee outgo is higher under (B), because persistency is greater on the path that produces a claim. This is the key reason a single average lapse assumption can misstate guarantee exposure.
Check. The undiscounted expected outgo under (B), \(\$855\), exceeds that under (A), \(\$720\); the discounted results preserve that ordering.
Worked Example: Greater Overall Persistency Can Reduce a Net Expected Deficit
A hypothetical two-year product has a \(\$10{,}000\) maturity guarantee. The account is \(\$6{,}000\) on a poor-market path with probability 0.30 and \(\$12{,}000\) on a strong-market path with probability 0.70. In addition to any guarantee top-up, the insurer receives a net charge of \(\$1{,}500\) at maturity on each policy that persists, regardless of the market path. Treat that charge as a separate cashflow; assume it does not change the specified account values. Compare assumption A, with persistency 0.80 on the poor path and 0.50 on the strong path, with assumption B, with persistency 0.50 on the poor path and 0.90 on the strong path. No discounting is required.
For this example, define the net expected deficit as the greater of zero and expected guarantee outgo less expected charge income per issued policy. This is a simplified expected-cashflow measure, not a tail-capital measure. The guarantee top-up on the poor path is \(\$10{,}000-\$6{,}000=\$4{,}000\); the strong path has no top-up.
Assumption A. Expected guarantee outgo is \(0.30(0.80)(\$4{,}000)=\$960\). The probability of persisting, including both market paths, is \(0.30(0.80)+0.70(0.50)=0.24+0.35=0.59\). Expected charge income is \(0.59(\$1{,}500)=\$885\). The net expected deficit is \(\max(\$960-\$885,0)=\mathbf{\$75}\).
Assumption B. Expected guarantee outgo is \(0.30(0.50)(\$4{,}000)=\$600\). The probability of persisting is \(0.30(0.50)+0.70(0.90)=0.15+0.63=0.78\). Expected charge income is \(0.78(\$1{,}500)=\$1{,}170\). The net expected deficit is \(\max(\$600-\$1{,}170,0)=\mathbf{\$0}\).
Overall persistency rises from 0.59 to 0.78, yet the defined net expected deficit falls from \(\$75\) to zero. There are two reasons: persistency falls on the poor path, where the guarantee pays, and rises on the strong path, where it increases charge income without increasing gross guarantee claims. This does not show that raising persistency on every path reduces gross guarantee exposure; under fixed market outcomes and fixed benefits, greater persistency on a shortfall path increases expected gross claims.
Check. Under B, expected charges exceed expected guarantee outgo by \(\$570\). The defined deficit is zero because it is floored at zero, not negative \(\$570\).
Interpreting the Comparison
For gross guarantee claims, higher conditional persistency on a path with \(F_j<G\) increases that path’s contribution, provided its market probability and shortfall remain unchanged. Higher persistency on a path with \(F_j\geq G\) does not increase gross top-ups in this model because the top-up remains zero. These statements concern expected guarantee payments, not the product’s full profit or capital position.
A broader insurer deficit may also reflect charges, premiums, expenses, surrender payments, and other benefits. In that broader calculation, persistency affects more than the guarantee payment. Greater persistence may increase future premium or charge income, but it may also keep an expensive guarantee in force. The net effect depends on the cashflows included, their timing, and their relationship to market outcomes. State the measure being compared before drawing a conclusion.
Market and lapse assumptions should therefore be examined jointly when persistency varies by market path. In particular, identify whether the scenarios with high guarantee shortfalls are also the scenarios with high persistency. Two assumptions with the same average persistency can imply different expected claims if their conditional rates differ across paths.
Common Exam Trap
Suppose a guarantee has a \(\$3{,}000\) shortfall with probability 0.30 and no shortfall otherwise. Persistency is 0.95 on the shortfall path and 0.65 on the other path. A tempting but incorrect calculation uses average persistency, \(0.30(0.95)+0.70(0.65)=0.74\), and then calculates \(0.30(0.74)(\$3{,}000)=\$666\). That calculation applies average persistency to a path where it does not belong.
The correct expected gross claim is \(0.30(0.95)(\$3{,}000)=\$855\). The no-shortfall path contributes zero, regardless of its persistency. The general correction is to multiply each path’s shortfall by its own conditional persistency and probability before summing.
Check Your Understanding
For each problem, calculate per issued policy unless the question asks for a different quantity. Assume lapse before maturity eliminates the specified maturity top-up.
- A one-year guarantee is \(\$9{,}000\). The account is \(\$6{,}500\) with probability 0.20 and at least \(\$9{,}000\) otherwise. Persistency on the shortfall path is 0.75. With an effective annual rate of 3%, find the issue-date expected guarantee outgo. (Answer: \(\$364.08\))
- A two-year guarantee has a shortfall of \(\$2{,}000\) with probability 0.40 and no shortfall otherwise. Compare undiscounted expected guarantee outgo when persistency on the shortfall path is 0.90 versus 0.60. (Answer: \(\$720\) versus \(\$480\) per issued policy)
- A two-year guarantee is \(\$12{,}000\). Terminal accounts are \(\$9{,}000\), \(\$11{,}000\), and \(\$14{,}000\), with probabilities 0.25, 0.35, and 0.40. Their respective persistencies are 0.80, 0.70, and 0.90. At an effective annual rate of 4%, find the issue-date expected top-up value. (Answer: \(\$781.25\))
- A one-year guarantee has a \(\$4{,}000\) shortfall on a poor-market path with probability 0.30 and no shortfall on a strong-market path with probability 0.70. Each persisting policy produces a net charge of \(\$1{,}500\), on either path. Using the net expected deficit definition in the third worked example, calculate the deficit for (A) persistency 0.80 poor and 0.50 strong, and (B) persistency 0.50 poor and 0.90 strong. (Answer: A, \(\$75\); B, \(\$0\))