Continuous Benefit Valuation With Staged Coverage

Introduction

A continuous death benefit can have more than one stage. A hypothetical policy might provide no cover during an initial waiting period, pay one amount during a first coverage period, and pay a different amount during a later period. To value it correctly, first identify exactly when each amount is payable. Then account for the chance that the insured survives to each interval and for interest accumulation from the time the policy is issued.

The central task is to turn the benefit schedule into covered time intervals. Each interval has its own benefit amount and may have its own mortality and interest assumptions. The contribution from an interval depends not only on what happens within it, but also on survival and discounting from issue up to its start.

Definition: Let $T_x$ be the future lifetime of a life aged $x$ at issue. A continuous death benefit schedule $B(t)$ pays $B(t)$ at the instant of death $t=T_x$, if $T_x$ lies in a covered interval. If the schedule consists of intervals $[a_j,b_j)$ with benefit $B_j$, then $B(t)=B_j$ on each such interval and is zero when no benefit is covered. For a continuous lifetime model and deterministic force of interest $\delta_t$, the expected present value is the expected value of the discounted benefit payable at death.

Mapping the Coverage Before Valuing It

Write each covered interval as a start time and an end time measured from issue. A deferred period is an interval with no benefit, not a period that can be removed from the survival calculation. A change in benefit amount creates a new interval even if coverage is uninterrupted. Likewise, a gap in coverage must remain a gap in the schedule: death in that gap produces no payment.

In the general continuous model, the probability density of death at time $t$ is ${}_tp_x\mu_{x+t}$, where ${}_tp_x$ is the probability of surviving to time $t$, and $\mu_{x+t}$ is the force of mortality at that time. The discount factor to time $t$ is $\exp(-\int_0^t\delta_s\,ds)$. Thus, for a schedule with finitely many covered intervals, the expected present value is the sum of the contributions over those intervals:

\operatorname{EPV} = \sum_j B_j\int_{a_j}^{b_j} \exp\!\left(-\int_0^t\delta_s\,ds\right) {}_tp_x\mu_{x+t}\,dt.

This expression assumes death has a continuous distribution over the covered periods, with density given by the force-of-mortality formula. It also assumes the benefit is paid at the instant of death and that the stated interest accumulation applies to the payment. If a contract instead pays at the end of a year of death or at a fixed maturity date, its payment timing must be valued differently.

A Useful Formula for a Single Covered Interval

Suppose mortality and interest forces are constant within one covered interval $[a,b)$: $\mu_{x+t}=\mu$ and $\delta_t=\delta$ for $a\leq t<b$. Before using the local formula, calculate the accumulated survival-and-discount factor up to time $a$. Define this factor as $\exp(-\int_0^a(\delta_s+\mu_{x+s})\,ds)$. It includes the assumptions that applied before the covered interval began; they need not be the same as the assumptions within the interval.

Theorem: Under the assumptions above, a constant benefit $B$ payable at death only during $[a,b)$ has expected present value $$ B\exp\!\left(-\int_0^a(\delta_s+\mu_{x+s})\,ds\right) \frac{\mu}{\delta+\mu} \left(1-e^{-(\delta+\mu)(b-a)}\right). $$ If the forces have been constant from issue through $b$, the accumulated factor simplifies to $e^{-(\delta+\mu)a}$. Constancy only within $[a,b)$ is not enough to make that simplification.

The interval formula follows directly by separating the accumulation to the interval start from what happens inside it. The derivation also explains why the survival and discount factors from before $a$ cannot be omitted.

Write the death-time value inside the interval.
At time $t$, survival and discounting from issue contribute their accumulated factors. With constant forces locally, the combined factor from $a$ to $t$ is $e^{-(\delta+\mu)(t-a)}$, and the death density conditional on survival to $a$ contributes $\mu\,dt$.

Keep the accumulation to time $a$.
The probability-and-discount factor through time $a$ is $\exp(-\int_0^a(\delta_s+\mu_{x+s})\,ds)$. Multiply it by the benefit and integrate the local death contribution over the interval.

Integrate over the interval length.
Set $u=t-a$. Then the local integral is $\int_0^{b-a}\mu e^{-(\delta+\mu)u}\,du=\frac{\mu}{\delta+\mu}(1-e^{-(\delta+\mu)(b-a)})$, giving the stated formula.

For a schedule with several stages, apply the formula to each covered interval using that interval's benefit and local forces, and add the contributions. For each interval, the factor at its start must reflect all mortality and interest forces from time zero to that start, including any uncovered period.

Worked Valuations

Worked Example: Deferred Cover Followed by a Benefit Increase

A hypothetical policy is issued to a life aged $x$. The force of mortality is constant at $0.04$ per year and the force of interest is constant at $0.06$ per year from issue through year 8. The policy pays $\$100{,}000$ if death occurs during years 2 through 5, and $\$150{,}000$ if death occurs during years 5 through 8. There is no benefit at other times. Find the expected present value.

Method. Map the two covered intervals as $[2,5)$ and $[5,8)$. The total force $\delta+\mu$ is $0.06+0.04=0.10$ per year throughout. Because both forces are constant from issue, the accumulated factor at time $a$ is $e^{-0.10a}$.

For $[2,5)$, the benefit is $\$100{,}000$ and the interval length is $3$ years:

100{,}000\cdot\frac{0.04}{0.10} \left(e^{-0.10(2)}-e^{-0.10(5)}\right) = 40{,}000(0.818730753-0.606530660) = \$8{,}488.00.

For $[5,8)$, the benefit is $\$150{,}000$, again over 3 years:

150{,}000\cdot\frac{0.04}{0.10} \left(e^{-0.10(5)}-e^{-0.10(8)}\right) = 60{,}000(0.606530660-0.449328964) = \$9{,}432.10.

Adding the unrounded interval contributions gives $\$17{,}920.1055\ldots$, or $\$17{,}920.11$ to the nearest cent. The method avoids treating the benefit increase as a new policy issued at year 5: the second interval still includes survival and discounting from issue through year 5.

Worked Example: A Deferred Interval After Different Early Mortality

A hypothetical policy has force of interest $0.04$ per year throughout. Its force of mortality is $0.05$ per year during $[0,1)$, then $0.10$ per year during $[1,3)$. It pays $\$40{,}000$ at death during $[1,2)$ and $\$70{,}000$ at death during $[2,3)$, with no payment otherwise. Find the expected present value.

Method. The accumulated survival-and-discount factor to year 1 is $e^{-(0.05+0.04)(1)}=e^{-0.09}$. On both covered intervals, the local forces are $\mu=0.10$ and $\delta=0.04$, so the local combined force is $0.14$. For a one-year interval beginning at $a$, its factor before multiplying by the benefit is

e^{-0.09}e^{-0.14(a-1)} \frac{0.10}{0.14}(1-e^{-0.14}).

The factor for $[1,2)$ is $0.0852839877$, so its value is $40{,}000(0.0852839877)=\$3{,}411.36$. The factor for $[2,3)$ has one additional year of local survival and discounting: $0.0852839877e^{-0.14}=0.0741423371$. Its value is $70{,}000(0.0741423371)=\$5{,}189.96$.

Adding the unrounded contributions gives $\$8{,}601.3231\ldots$, or $\$8{,}601.32$ to the nearest cent. A common mistake is to omit the accumulated factor $e^{-0.09}$ for the first covered interval. That incorrect calculation gives $40{,}000\cdot\frac{0.10}{0.14}(1-e^{-0.14})=\$3{,}732.62$ for $[1,2)$, as though the life had certainly survived and the interest clock had only started at year 1. The correct value includes survival and discounting from issue.

Worked Example: A Coverage Gap and a Later Mortality Change

A hypothetical contract pays $\$25{,}000$ at death during $[0,1)$, provides no coverage during $[1,2)$, and pays $\$80{,}000$ at death during $[2,4)$. The force of interest is $0.03$ per year throughout. The force of mortality is $0.05$ per year through year 2 and $0.10$ per year during $[2,4)$. Find the expected present value.

Method. The first interval has local combined force $0.03+0.05=0.08$, so its value is

25{,}000\cdot\frac{0.05}{0.08}(1-e^{-0.08}) = 25{,}000(0.0480522835) = \$1{,}201.31.

There is no benefit for death during $[1,2)$, so that interval contributes zero. To reach year 2, the accumulated survival-and-discount factor is $e^{-2(0.03+0.05)}=e^{-0.16}$. During $[2,4)$, the local combined force is $0.03+0.10=0.13$, and the interval length is 2 years. Thus the later benefit contributes

80{,}000e^{-0.16}\frac{0.10}{0.13}(1-e^{-0.13(2)}) = 80{,}000(0.1500745917) = \$12{,}005.97.

The total is $\$1{,}201.3071\ldots+\$12{,}005.9673\ldots=\$13{,}207.2744\ldots$, or $\$13{,}207.27$ to the nearest cent. The interval map prevents two errors: assigning a benefit to deaths in the gap, and using the mortality force from the first interval to value the later interval.

Checks and Common Pitfalls

Before finalizing a staged valuation, check that every covered period appears exactly once, every gap has zero benefit, and the force assumptions used in each interval match the stated time periods. In particular, local constancy is not the same as constancy from issue. The interval formula's starting factor always reaches back to issue, even when the benefit itself is deferred.

Starting the calculation at the deferment date. This drops the probability of surviving to that date and the discounting accumulated before it. Retain the factor from time zero to the interval start.
Using the wrong local force. If mortality or interest changes at a stage boundary, split the calculation there and use the new force only over the interval where it applies.
Filling a coverage gap. A zero-benefit period contributes nothing, even though the life can survive through it and later become eligible for a benefit.
Rounding too early. Keep unrounded interval values when adding them, then round the total to cents. Rounded components may not sum to the correctly rounded total.

Key takeaway: Map the benefit schedule first. For each covered interval, multiply its local death-payment value by the full survival-and-discount accumulation from issue to the interval's start, then add the interval contributions.

Check Your Understanding

For each problem, identify the covered interval or intervals before calculating. Round money to the nearest cent.

A hypothetical policy pays $\$100{,}000$ at death during $[1,3)$. The force of mortality is $0.04$ per year and the force of interest is $0.06$ per year from issue onward. Find the expected present value. (Answer: $\$6{,}560.77$)
A hypothetical benefit of $\$30{,}000$ is payable at death during $[2,5)$. The force of mortality is $0.08$ per year and the force of interest is $0.02$ per year throughout. Find the expected present value. (Answer: $\$5{,}092.80$)
A hypothetical policy pays $\$40{,}000$ at death during $[1,2)$. The force of interest is $0.03$ throughout; the force of mortality is $0.05$ during $[0,1)$ and $0.10$ during $[1,2)$. Find the expected present value. (Answer: $\$3{,}462.53$)
A hypothetical policy pays $\$20{,}000$ at death during $[0,1)$, nothing during $[1,2)$, and $\$50{,}000$ at death during $[2,3)$. The force of mortality is $0.04$ and the force of interest is $0.06$, both constant from issue onward. Find the expected present value. (Answer: $\$2{,}319.55$)

What You'll Learn