Continuously Increasing Whole-Life Benefits

What Is a Continuously Increasing Whole-Life Benefit?

A whole-life insurance benefit is paid whenever death occurs, with no fixed term after which coverage ends. In a continuously increasing design, the benefit amount depends on the time elapsed since issue: the longer the contract has been in force, the larger the amount paid at death. This tutorial studies the linear design in which the benefit at time $t$ is proportional to $t$.

Let $T_x$ be the complete future lifetime of a life aged $x$, measured in years from issue. If the benefit is $c t$ dollars when death occurs at duration $t$, then the present value at issue is $cT_x e^{-\delta T_x}$. Here $c$ is the benefit increase per year of duration, and $\delta$ is the force of interest. The benefit is zero at duration zero and increases linearly thereafter.

Definition: For a whole-life contract paying $c t$ at the instant of death at duration $t$, the random present value is $Z=cT_x e^{-\delta T_x}$. Its actuarial present value is $$ E[Z]=c\int_0^\infty t e^{-\delta t} f_{T_x}(t)\,dt =c\int_0^\infty t e^{-\delta t} {}_tp_x\mu_{x+t}\,dt. $$ The second expression uses the complete future lifetime density $f_{T_x}(t)={}_tp_x\mu_{x+t}$, where ${}_tp_x$ is the probability of surviving $t$ years and $\mu_{x+t}$ is the force of mortality at attained age $x+t$.

The interest convention matters. If an effective annual interest rate $i$ is given, the matching force of interest is $\delta=\ln(1+i)$, so the discount factor for a payment at time $t$ is $e^{-\delta t}$. Do not discount using $v^t$ with $v=1/(1+i)$ unless the time and interest conventions have been made consistent; continuous-time discounting is expressed using the force.

Why the Valuation Integral Holds

The valuation starts with the payment conditional on the death time. If death occurs at duration $t$, the insurer pays $ct$ immediately, whose value at issue is $ct e^{-\delta t}$. To average that value over all possible death times, multiply by the density that death occurs at duration $t$, then integrate over all $t\geq0$.

Write the value if death occurs at time $t$.
The benefit is $ct$, and its discounted value is $ct e^{-\delta t}$.

Use the complete-lifetime density.
Survival to $t$ followed by death immediately after $t$ has density ${}_tp_x\mu_{x+t}$. Both survival and the instantaneous mortality force are needed.

Take the expectation.
Integrating the discounted benefit times the death-time density gives $c\int_0^\infty t e^{-\delta t}{}_tp_x\mu_{x+t}\,dt$.

This justification also shows why the integral is not simply an integral of the benefit times the discount factor: the density must account for the probability of being alive until the payment time and then dying at that time. The formula assumes death is modeled by a continuous lifetime distribution with density as stated. If a model instead assigns positive probability to death at specific times, those point masses must also be included in the expected present value.

Constant-Force Mortality

Under a constant force of mortality $\mu>0$, survival is ${}_tp_x=e^{-\mu t}$ and the complete-lifetime density is $f_{T_x}(t)=\mu e^{-\mu t}$. With a constant force of interest $\delta\geq0$, the actuarial present value becomes an elementary integral.

Result: Under constant-force mortality $\mu>0$ and constant force of interest $\delta\geq0$, a benefit of $ct$ at death has actuarial present value $$ c\int_0^\infty t e^{-\delta t}\mu e^{-\mu t}\,dt =\frac{c\mu}{(\mu+\delta)^2}. $$

To evaluate the integral, set $r=\mu+\delta$, which is positive. Integration by parts gives $\int_0^\infty t e^{-rt}\,dt=1/r^2$: the boundary term $-t e^{-rt}/r$ tends to zero at both endpoints, leaving $(1/r)\int_0^\infty e^{-rt}\,dt=1/r^2$. Multiplying by $c\mu$ gives the result. If $\mu+\delta$ were not positive, this infinite integral would not converge; the stated assumptions ensure convergence.

Worked Example: Constant-Force Mortality and a Scaled Benefit

A hypothetical whole-life contract pays $1{,}000t$ dollars at death $t$ years after issue. Assume a constant force of mortality $\mu=0.10$ per year and a constant force of interest $\delta=0.20$ per year. Find the actuarial present value.

Method. Use the constant-force result with benefit scale $c=\$1{,}000$ per year of duration. Under the mortality assumption, $f_{T_x}(t)=0.10e^{-0.10t}$, and discounting contributes $e^{-0.20t}$.

Combining the exponents gives the integral $$ 1{,}000\int_0^\infty t(0.10)e^{-(0.10+0.20)t}\,dt =1{,}000(0.10)\int_0^\infty t e^{-0.30t}\,dt. $$

The remaining integral is $1/(0.30)^2=1/0.09=11.111111$. Therefore, $$ 1{,}000(0.10)(11.111111)=\$1{,}111.1111. $$

Rounded to the nearest cent, the actuarial present value is \$1,111.11. The method uses the full death-time density, not just the mortality force or the discount factor. The result is consistent with the expected payment being made at a random future time: the benefit grows with duration, while mortality and interest both reduce the weight assigned to long durations.

Piecewise Mortality Forces

A mortality force need not be constant over the entire lifetime. If it changes at a specified duration, survival to that duration carries forward into every later interval. In particular, the time $t$ in the benefit $ct$ remains total time since issue; it does not restart when the force changes.

For a constant mortality force $\mu_j$ on an interval $[a,b)$, suppose the survival probability at its start is $S(a)={}_ap_x$. On that interval, ${}_tp_x=S(a)e^{-\mu_j(t-a)}$, so the density is $S(a)\mu_j e^{-\mu_j(t-a)}$. Including discounting, the interval's contribution is $$ cS(a)\mu_j e^{-\delta a}\int_a^b t e^{-(\mu_j+\delta)(t-a)}\,dt. $$

Worked Example: A Change in Mortality Force at Two Years

A hypothetical policy pays $100t$ dollars at death, where $t$ is measured from issue in years. The mortality force is $0.10$ per year for the first two years and $0.20$ per year thereafter. The force of interest is $\delta=0.05$ per year. Find the actuarial present value, rounding to the nearest cent.

Method. Split the density integral at $t=2$. For the first interval, survival is $e^{-0.10t}$; for the second, survival to year 2 is $e^{-0.20}$, followed by survival $e^{-0.20(t-2)}$. Retain the total duration $t$ in the benefit.

For $0\leq t<2$, the integral before applying the \$100 scale is $$ 0.10\int_0^2 t e^{-0.15t}\,dt =\frac{0.10}{(0.15)^2}\left[1-e^{-0.30}(1+0.30)\right]. $$

Using $e^{-0.30}=0.74081822$, the bracket is $1-(0.74081822)(1.30)=0.03693631$. Since $(0.15)^2=0.0225$, this interval contributes $$ \frac{0.10(0.03693631)}{0.0225}=0.1641614 $$ before the benefit scale, or $100(0.1641614)=\$16.4161$.

For the interval $t\geq2$, write $t=2+s$, where $s\geq0$. Survival through the first two years is $e^{-0.20}$, and the combined discount and post-year-2 mortality factor for $s$ is $e^{-0.25s}$. The contribution before the benefit scale is $$ 0.20e^{-0.30}\int_0^\infty (2+s)e^{-0.25s}\,ds =0.20e^{-0.30}\left(\frac{2}{0.25}+\frac{1}{(0.25)^2}\right). $$

The two terms in parentheses are $8$ and $16$, respectively, so the contribution is $0.20(0.74081822)(24)=3.5559275$, or $100(3.5559275)=\$355.5927$. Adding the intervals gives $\$16.4161+\$355.5927=\$372.0088$. The actuarial present value, to the nearest cent, is \$372.01.

The interval contributions are positive and the later interval carries survival to year 2. Those are useful checks on the calculation. In particular, the benefit at a death after year 2 is based on $2+s$, not merely on $s$.

Mixtures of Mortality Models

A population may be modeled as a mixture of subgroups, each with its own constant force of mortality. Suppose a life belongs to subgroup $j$ with probability $w_j$, where the weights are nonnegative and sum to 1. If subgroup $j$ has constant force $\mu_j$, then the unconditional lifetime density is the weighted density $f_{T_x}(t)=\sum_j w_j\mu_j e^{-\mu_jt}$. The increasing-benefit present value is found by integrating this density, or equivalently by averaging the subgroup-specific present values.

Worked Example: Averaging Across Two Mortality Subgroups

A hypothetical insured population consists of two fixed subgroups. At issue, the probability of belonging to subgroup 1 is $0.25$, with constant force of mortality $0.10$; the probability of subgroup 2 is $0.75$, with constant force of mortality $0.30$. A whole-life contract pays $100t$ dollars at death, and the force of interest is $\delta=0.10$. Find the actuarial present value.

Method. Condition on subgroup membership. Each subgroup has an exponential complete-lifetime density, and the present value for subgroup $j$ is $100\mu_j/(\mu_j+\delta)^2$. Weight each value by its subgroup probability.

For subgroup 1, the present value is $$ 100\frac{0.10}{(0.10+0.10)^2} =100\frac{0.10}{0.04} =\$250.00. $$ Its weighted contribution is $0.25(\$250.00)=\$62.50$.

For subgroup 2, the present value is $$ 100\frac{0.30}{(0.30+0.10)^2} =100\frac{0.30}{0.16} =\$187.50. $$ Its weighted contribution is $0.75(\$187.50)=\$140.625$. Thus the total is $\$62.50+\$140.625=\$203.125$, or \$203.13 to the nearest cent. The subgroup weights sum to $1$, and the answer lies between the two subgroup-specific present values, as a weighted average should.

Common Mistakes and Checks

Omitting survival from the density. The density is ${}_tp_x\mu_{x+t}$, not just $\mu_{x+t}$. For example, in the first worked example, omitting survival would give $1{,}000(0.10)/(0.20)^2=\$2{,}500$, not $\$1{,}111.11$. That incorrect expression does not represent a death-time density.
Resetting duration when mortality changes. A benefit $ct$ is based on time since issue. After a change at year 2, a death at year 3 receives $3c$, not $c$.
Applying the scale twice or not at all. If an integral has already been multiplied by $c$, its result is already in dollars. If the integral was calculated without $c$, multiply by $c$ once at the end.
Using a discrete benefit convention. This contract pays at the instant of death, so its present value uses $e^{-\delta t}$ and the complete-lifetime density. A year-end payment model is a different contract and requires different timing.

A practical final check is to verify that the integrand includes three elements: the benefit amount at duration $t$, the continuous discount factor, and the density of death at $t$. For piecewise models, also verify that the survival probability accumulated before each interval is retained.

Key takeaway: For a benefit of $ct$ paid at death, value the payment at its actual death time and average over the complete future lifetime density. Under constant forces this reduces to $c\mu/(\mu+\delta)^2$; under changing or mixed mortality, use the appropriate density and preserve duration from issue.

Check Your Understanding

For each problem, state or use the specified mortality and interest assumptions, and keep time measured from issue.

A whole-life benefit is $200t$ dollars. Under constant mortality force $0.05$ and interest force $0.03$, find its actuarial present value. (Answer: \$1,562.50)
A contract pays $75t$ dollars at death. Mortality force is $0.12$ and interest force is $0.08$, both constant. Find the actuarial present value. (Answer: \$225.00)
A population has two fixed subgroups: weight $0.40$ with constant mortality force $0.10$, and weight $0.60$ with constant mortality force $0.20$. For benefit $100t$ and interest force $0.10$, find the population actuarial present value. (Answer: \$233.33)
A benefit of $10t$ dollars is paid at death. The mortality force is $0.10$ for the first year and $0.20$ thereafter, and the interest force is zero. Find the actuarial present value to the nearest cent. (Answer: \$54.76)
Under constant mortality force $0.10$, a benefit of $t$ dollars has actuarial present value $2$. Find the nonnegative force of interest $\delta$. (Answer: $\delta=\sqrt{0.05}-0.10\approx0.1236$ per year, or 12.36%)

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