Multi-Period Benefit Amounts in Continuous Insurance

Why Split a Continuous Benefit?

A continuous life insurance benefit can change with duration. For example, a hypothetical policy might pay one amount for death in the first year, a larger amount for death in the next several years, and a different amount thereafter. The payment is made at the time of death, so both the benefit amount and the discount factor depend on that time.

The valuation method is to divide the possible death times into intervals on which the benefit is constant. If mortality also changes by duration, split at those mortality boundaries as well. On each resulting interval, use the benefit for that interval, the probability of surviving to its start, and the discounted death density within it. The mortality force is part of that density and must not be omitted.

Definition: Let $T_x$ be the future lifetime of a life aged $x$, and let $b(t)$ be the benefit paid at the moment of death at duration $t$. With a constant force of interest $\delta$, the expected present value (EPV) of a continuous death benefit is $$ \operatorname{EPV}=\int_0^\infty e^{-\delta t}b(t)\,{}_tp_x\mu_{x+t}\,dt. $$ Here ${}_tp_x$ is the probability of surviving $t$ years from age $x$, and $\mu_{x+t}$ is the force of mortality at duration $t$. This formula assumes the benefit is paid immediately at death and that the death-time distribution has a density. For a term contract, the integral ends at the term; any maturity benefit must be valued separately.

The factor ${}_tp_x\mu_{x+t}$ is the density of death at duration $t$: the life must survive to $t$, then die in the following short interval. The factor $e^{-\delta t}$ discounts a payment at duration $t$ to issue. If $b(t)$ is piecewise constant, each interval contributes a separate integral.

Valuing One Constant-Benefit Interval

Suppose the force of mortality is constant at $\mu_j$ on an interval $[a,b)$, and the benefit on that interval is $B_j$. Assume the force of interest $\delta$ is constant. Conditional on survival to $a$, survival for a further $s$ years within this interval is $e^{-\mu_j s}$. Thus, for $a\le t<b$, ${}_tp_x={}_ap_xe^{-\mu_j(t-a)}$.

\begin{aligned} \operatorname{EPV}_{[a,b)} &=B_j\int_a^b e^{-\delta t}{}_tp_x\mu_j\,dt\\ &=B_j\,{}_ap_xe^{-\delta a}\mu_j \int_0^{b-a}e^{-(\delta+\mu_j)s}\,ds\\ &=B_j\,{}_ap_xe^{-\delta a} \frac{\mu_j}{\delta+\mu_j} \left(1-e^{-(\delta+\mu_j)(b-a)}\right). \end{aligned}

The substitution $s=t-a$ separates the value accumulated before the interval from the valuation within it. The term ${}_ap_xe^{-\delta a}$ accounts for survival and discounting to the start; $\mu_j$ supplies the death density; and the remaining exponential integral covers deaths during the interval. For an interval extending indefinitely, take the limit as $b$ tends to infinity, giving the tail contribution ${B_j{}_ap_xe^{-\delta a}\mu_j}/{(\delta+\mu_j)}$, provided $\delta+\mu_j>0$.

When both benefit and mortality change, create intervals using the union of their boundaries. On each interval, the benefit and mortality force are constant, so the formula applies directly. This is simply an efficient way to evaluate the original integral, not a different valuation principle.

List every boundary.
Include each duration at which the benefit or the mortality force changes. Sort the boundaries from earliest to latest.

Track survival and discounting to each start.
Compute ${}_ap_x$ from the mortality model and use $e^{-\delta a}$ for discounting to duration $a$.

Value each interval with its local force.
Use the benefit $B_j$ and force $\mu_j$ that apply there. Include $\mu_j$ as the multiplier in the death-density integral.

Add all contributions.
If the last benefit continues for life, evaluate the final interval as an infinite tail.

Worked Valuations

Worked Example: Two Benefit Levels Under Constant Forces

A hypothetical whole-life policy pays \$6,000 if death occurs during the first two years and \$9,000 thereafter. Assume a constant force of mortality $\mu=0.10$ and a constant force of interest $\delta=0.05$. Find the EPV, rounded to the nearest cent.

Method. Split at duration 2. Because the forces are constant throughout, ${}_tp_x=e^{-0.10t}$, and the discounted death density is $0.10e^{-0.15t}$. The mortality multiplier is $0.10$, while $0.15=\delta+\mu$.

For deaths during the first two years:

\begin{aligned} \operatorname{EPV}_{[0,2)} &=6000\int_0^2 0.10e^{-0.15t}\,dt\\ &=6000\left(\frac{0.10}{0.15}\right)(1-e^{-0.30})\\ &=6000\left(\frac{2}{3}\right)(1-0.7408182)\\ &=6000(0.1727879)\\ &=\$1{,}036.73. \end{aligned}

For deaths at duration 2 or later, the payment is \$9,000. The contribution is the infinite tail:

\begin{aligned} \operatorname{EPV}_{[2,\infty)} &=9000\int_2^\infty 0.10e^{-0.15t}\,dt\\ &=9000\left(\frac{0.10}{0.15}\right)e^{-0.30}\\ &=9000\left(\frac{2}{3}\right)(0.7408182)\\ &=9000(0.4938788)\\ &=\$4{,}444.91. \end{aligned}

Adding the contributions gives $\$1{,}036.73+\$4{,}444.91=\boxed{\$5{,}481.64}$, rounded to the nearest cent. The two discounted death-density coefficients sum to $0.1727879+0.4938788=0.6666667$, which is $\mu/(\delta+\mu)=0.10/0.15$, as expected for the whole lifetime under these constant forces.

Worked Example: Benefit and Mortality Boundaries Both Matter

A hypothetical whole-life policy pays \$5,000 during the first year, \$9,000 from duration 1 to 4, and \$3,000 thereafter. The force of mortality is $0.08$ for durations 0 to 2 and $0.12$ after duration 2. The force of interest is $\delta=0.02$ throughout. Find the EPV, rounded to the nearest cent.

Method. Split at durations 1, 2, and 4. The first boundary changes the benefit, the second changes both benefit-interval calculations and mortality, and the third changes the benefit to its whole-life tail. Survival and discounting to duration $t$ together give $e^{-0.10t}$ up to duration 2, since $0.08+0.02=0.10$. After duration 2, the combined force is $0.12+0.02=0.14$.

For duration 0 to 1, the benefit is \$5,000 and the mortality force is $0.08$:

\begin{aligned} \operatorname{EPV}_{[0,1)} &=5000\left(\frac{0.08}{0.10}\right)(1-e^{-0.10})\\ &=5000(0.8)(1-0.9048374)\\ &=5000(0.0761301)=\$380.65. \end{aligned}

For duration 1 to 2, the benefit is \$9,000. The accumulated survival-discount factor is $e^{-0.10}$ at duration 1 and $e^{-0.20}$ at duration 2:

\begin{aligned} \operatorname{EPV}_{[1,2)} &=9000\left(\frac{0.08}{0.10}\right)(e^{-0.10}-e^{-0.20})\\ &=9000(0.8)(0.9048374-0.8187308)\\ &=9000(0.0688853)=\$619.97. \end{aligned}

For duration 2 to 4, the force is now $0.12$, and the combined force is $0.14$. The accumulated survival-discount factor at duration 2 is $e^{-0.20}$; at duration 4 it is $e^{-0.48}$, because the combined force acts for two more years:

\begin{aligned} \operatorname{EPV}_{[2,4)} &=9000\left(\frac{0.12}{0.14}\right)(e^{-0.20}-e^{-0.48})\\ &=9000\left(\frac{6}{7}\right)(0.8187308-0.6187834)\\ &=9000(0.1713835)=\$1{,}542.45. \end{aligned}

Finally, the benefit is \$3,000 from duration 4 onward. The accumulated survival-discount factor at duration 4 is $e^{-0.48}$:

\begin{aligned} \operatorname{EPV}_{[4,\infty)} &=3000\left(\frac{0.12}{0.14}\right)e^{-0.48}\\ &=3000\left(\frac{6}{7}\right)(0.6187834)\\ &=3000(0.5303858)=\$1{,}591.16. \end{aligned}

Therefore the total is $\$380.65+\$619.97+\$1{,}542.45+\$1{,}591.16=\boxed{\$4{,}134.23}$, rounded to the nearest cent. The result uses the local mortality force in each segment and carries the accumulated survival and discount factor forward to every boundary.

Worked Example: A Whole-Life Tail After Several Benefit Changes

A hypothetical policy pays \$10,000 for death during the first year, \$14,000 from duration 1 to 3, and \$8,000 thereafter. Assume a constant force of interest $\delta=0.05$. The force of mortality is $0.05$ through duration 2 and $0.10$ thereafter. Find the EPV, rounded to the nearest cent.

Method. Split at durations 1, 2, and 3. Up to duration 2, the combined force is $0.05+0.05=0.10$. After duration 2, it is $0.10+0.05=0.15$. The survival-discount factors at durations 1, 2, and 3 are therefore $e^{-0.10}$, $e^{-0.20}$, and $e^{-0.35}$.

For duration 0 to 1, use benefit \$10,000 and mortality force $0.05$:

\begin{aligned} \operatorname{EPV}_{[0,1)} &=10000\left(\frac{0.05}{0.10}\right)(1-e^{-0.10})\\ &=10000(0.5)(1-0.9048374)\\ &=10000(0.0475813)=\$475.81. \end{aligned}

For duration 1 to 2, use benefit \$14,000 and the same local forces:

\begin{aligned} \operatorname{EPV}_{[1,2)} &=14000\left(\frac{0.05}{0.10}\right)(e^{-0.10}-e^{-0.20})\\ &=14000(0.5)(0.9048374-0.8187308)\\ &=14000(0.0430533)=\$602.75. \end{aligned}

For duration 2 to 3, the local mortality force is $0.10$, and the combined force is $0.15$:

\begin{aligned} \operatorname{EPV}_{[2,3)} &=14000\left(\frac{0.10}{0.15}\right)(e^{-0.20}-e^{-0.35})\\ &=14000\left(\frac{2}{3}\right)(0.8187308-0.7046881)\\ &=14000(0.0760284)=\$1{,}064.40. \end{aligned}

For the whole-life tail from duration 3, the benefit remains \$8,000:

\begin{aligned} \operatorname{EPV}_{[3,\infty)} &=8000\left(\frac{0.10}{0.15}\right)e^{-0.35}\\ &=8000\left(\frac{2}{3}\right)(0.7046881)\\ &=8000(0.4697921)=\$3{,}758.34. \end{aligned}

Adding the four unrounded contributions gives an EPV of $\boxed{\$5{,}901.29}$, rounded to the nearest cent. As a check, the interval death-density coefficients are $0.0475813$, $0.0430533$, $0.0760284$, and $0.4697921$. Their sum is $0.6364551$, which is less than 1 because discounting reduces the total relative to the undiscounted probability of eventual death.

Common Trap: Omitting the Mortality Force

A frequent error is to integrate $e^{-(\delta+\mu)t}$ and treat the result as a discounted probability of death. It is not: it is missing the mortality-force multiplier. Under constant forces, the discounted death density is $\mu e^{-(\delta+\mu)t}$, not just $e^{-(\delta+\mu)t}$.

For the first interval of Worked Example 1, the incorrect calculation would be $6000(1-e^{-0.30})/0.15=\$10{,}367.27$. That omits the factor $0.10$. The corrected calculation is $6000(0.10/0.15)(1-e^{-0.30})=\$1{,}036.73$. The wrong result is ten times the correct segment value. Always identify the full death density before integrating, especially when the mortality force changes between intervals.

Key takeaway: For each constant-benefit, constant-mortality interval, multiply the survival-discount factor at the interval start by the benefit and by $\mu_j/(\delta+\mu_j)$, then apply the interval-length factor $1-e^{-(\delta+\mu_j)(b-a)}$. Include the mortality-force multiplier and add every segment, including any infinite tail.

Check Your Understanding

For each problem, use the force of mortality and force of interest stated, split at every relevant boundary, and round money to the nearest cent.

A hypothetical whole-life policy pays \$10,000 for death during the first two years and \$15,000 thereafter. Assume constant forces $\mu=0.04$ and $\delta=0.06$. Find the EPV. (Answer: \$5,637.46)
A hypothetical whole-life policy pays \$7,000 during the first year and \$11,000 thereafter. Assume constant forces $\mu=0.08$ and $\delta=0.02$. Find the EPV. (Answer: \$8,495.48)
A hypothetical whole-life policy pays \$6,000 for death during the first two years and \$9,000 thereafter. The force of mortality is $0.05$ for the first year and $0.10$ after duration 1; $\delta=0.05$ throughout. Find the EPV. (Answer: \$5,462.44)
A candidate values a \$4,000 benefit over durations 0 to 1 with constant forces $\mu=0.05$ and $\delta=0.05$, using $4000(1-e^{-0.10})/0.10$. What factor is missing, and what is the correct segment EPV? (Answer: The mortality-force multiplier $0.05$ is missing; the correct EPV is $\$190.33$.)

What You'll Learn