Continuous Death-Benefit Present Values

Introduction

For a death benefit paid at the moment of death, the payment time is not restricted to the end of a year. If death occurs at time $t$, the benefit is discounted for exactly $t$ years. The present value is therefore a random variable determined by the future lifetime and the benefit payable at that time.

The key calculation is an expectation over possible death times. This tutorial develops that calculation for a general death-benefit function and a general deterministic discount function, then applies it to hypothetical term contracts. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For a life aged $x$, the complete future lifetime, denoted $T_x$, is the random time from age $x$ until death, measured in years. Its survival function is ${}_tp_x=P(T_x>t)$. If this survival function is differentiable and the lifetime has a density, the density of $T_x$ is $f_{T_x}(t)=-\frac{d}{dt}{}_tp_x$. Where the force of mortality exists, this is equivalently $f_{T_x}(t)={}_tp_x\mu_{x+t}$.

Present Value at the Moment of Death

Let $D(t)$ be the deterministic discount factor for a payment made $t$ years from now, with $D(0)=1$. At a constant effective annual interest rate $i$, $D(t)=v^t=(1+i)^{-t}$. At a constant force of interest $\delta$, $D(t)=e^{-\delta t}$. These are alternative ways to specify discounting; use the one consistent with the stated interest assumption.

Suppose the contract pays a benefit $b(t)$ if death occurs at time $t$, and pays nothing if death occurs outside the coverage period $[0,n]$. The present-value random variable is

Z=b(T_x)D(T_x)\mathbf{1}_{\{T_x\le n\}}.

Here $\mathbf{1}_{\{T_x\le n\}}$ equals 1 when death occurs during the coverage period and 0 otherwise. For whole-life coverage, there is no finite coverage cutoff, so the indicator is omitted. A benefit that depends on time can be represented by $b(t)$; a level benefit $B$ is the special case $b(t)=B$.

Theorem: If $T_x$ has density $f_{T_x}$, then the expected present value of a death benefit paid at the moment of death is $$ E[Z]=\int_0^n b(t)D(t)f_{T_x}(t)\,dt. $$ For coverage without a finite term, integrate over all possible death times, provided the integral exists.

The density in this formula is unconditional: $f_{T_x}(t)\,dt$ is the probability of dying between $t$ and $t+dt$ from the starting age. It already accounts for the chance of surviving to time $t$. Consequently, do not multiply the integrand by a separate survival probability a second time.

Why the Density Integral Gives the Expected Present Value

For a continuous random variable $T_x$, the expectation of a function $g(T_x)$ is the integral of $g(t)$ weighted by the density $f_{T_x}(t)$. In this contract, the function is the discounted benefit when death occurs during coverage and zero otherwise:

g(t)= \begin{cases} b(t)D(t), & 0\le t\le n,\\ 0, & t>n. \end{cases}

Substituting this function into the density formula for expectation gives the result:

\begin{aligned} E[Z] &=\int_0^\infty g(t)f_{T_x}(t)\,dt\\ &=\int_0^n b(t)D(t)f_{T_x}(t)\,dt. \end{aligned}

The upper limit is $n$ because the present value is zero for deaths after coverage ends. Using $f_{T_x}(t)={}_tp_x\mu_{x+t}$ gives an equivalent form:

E[Z]=\int_0^n b(t)D(t){}_tp_x\mu_{x+t}\,dt.

This form separates the ingredients: survival to time $t$, the instantaneous force of mortality at that time, the benefit amount, and discounting. It applies when the lifetime is continuously distributed over the relevant range. If the lifetime model includes point masses, their probability-weighted present values must also be included; the density integral alone would then be incomplete.

Worked Applications

Worked Example: Level Benefit with a Uniform Death-Time Model

A hypothetical two-year contract pays $12,000 if the insured dies during the term, with payment made at the moment of death. For this model, $T_x$ is uniform on $[0,2]$, so its density is $f_{T_x}(t)=1/2$ for $0\le t\le2$. The force of interest is $\delta=\ln 2$, so $D(t)=e^{-t\ln 2}=2^{-t}$. Find the expected present value.

Method. Insert the level benefit, discount factor, and density into the death-time integral. The common mistake avoided here is discounting the benefit to the end of the year of death instead of to the actual death time.

For this contract, $b(t)=12{,}000$ and $n=2$. Therefore,

\begin{aligned} E[Z] &=\int_0^2 12{,}000\cdot 2^{-t}\cdot \frac12\,dt\\ &=6{,}000\int_0^2 e^{-t\ln 2}\,dt\\ &=6{,}000\left[\frac{-e^{-t\ln 2}}{\ln 2}\right]_0^2\\ &=6{,}000\frac{1-2^{-2}}{\ln 2}\\ &=\frac{4{,}500}{\ln 2}\\ &\approx \$6{,}492.13. \end{aligned}

The model assigns probability 1 to death by time 2, and the discount factor ranges from 1 down to $1/4$. The expected present value is between $12{,}000(1/4)=\$3{,}000$ and $12,000, as the result should be. The final amount is rounded to the nearest cent.

Worked Example: Benefit Increasing with Time

A hypothetical one-year contract pays $b(t)=\$4{,}000(1+t)$ at the moment of death if death occurs during the first year. Assume $T_x$ is uniform on $[0,2]$, with density $1/2$ on that interval, and the force of interest is $\delta=\ln 2$. Find the expected present value.

Method. Integrate the time-dependent benefit against the discounted death-time density over the coverage period. The common mistake avoided here is using the benefit at the end of the term for every death, rather than the benefit amount specified at the actual death time.

For $0\le t\le1$, the integrand is $4{,}000(1+t)2^{-t}(1/2)$. Thus,

\begin{aligned} E[Z] &=2{,}000\int_0^1(1+t)e^{-t\ln 2}\,dt. \end{aligned}

Write $a=\ln 2$. Evaluating the two parts of the integral gives

\begin{aligned} \int_0^1 e^{-at}\,dt&=\frac{1-e^{-a}}{a}=\frac{1}{2a},\\ \int_0^1 te^{-at}\,dt &=\frac{1-(1+a)e^{-a}}{a^2} =\frac{1-a}{2a^2}. \end{aligned}

The second equality in each line uses $e^{-a}=1/2$. Combining the terms,

\begin{aligned} E[Z] &=2{,}000\left(\frac{1}{2a}+\frac{1-a}{2a^2}\right)\\ &=2{,}000\left(\frac{1}{2a^2}\right)\\ &=\frac{1{,}000}{(\ln 2)^2}\\ &\approx \$2{,}081.37. \end{aligned}

As a check, the probability of death in the first year is $1/2$, and the benefit ranges from $4,000 to $8,000. Discounting reduces the value below the undiscounted expected benefit. The final amount is rounded to the nearest cent.

Worked Example: Term Benefit Under a Constant Mortality Force

A hypothetical three-year term contract pays $25,000 at the moment of death if death occurs within the term. Assume a constant force of mortality $\mu_{x+t}=0.2$ per year during the term and a constant force of interest $\delta=0.1$ per year. Find the expected present value.

Method. First obtain the death-time density from survival and the mortality force, then integrate the discounted benefit over the three-year term. The common mistake avoided here is treating the probability of death by year 3 as though all deaths occurred at time 3.

Under the stated mortality assumption, ${}_tp_x=e^{-0.2t}$. Therefore,

f_{T_x}(t)={}_tp_x\mu_{x+t}=0.2e^{-0.2t},\qquad 0\le t\le3.

The discount factor is $D(t)=e^{-0.1t}$. Substituting both into the expected present value formula gives

\begin{aligned} E[Z] &=25{,}000\int_0^3 e^{-0.1t}\left(0.2e^{-0.2t}\right)\,dt\\ &=5{,}000\int_0^3 e^{-0.3t}\,dt\\ &=5{,}000\left[\frac{-e^{-0.3t}}{0.3}\right]_0^3\\ &=\frac{5{,}000}{0.3}(1-e^{-0.9})\\ &\approx \$9{,}890.51. \end{aligned}

The probability of death during the term is $1-e^{-0.6}\approx0.4512$, so the expected benefit before discounting is $\$25{,}000(0.4512)$, approximately $11,280. Discounting must reduce that amount, and the calculated value is lower. The final amount is rounded to the nearest cent.

Common Traps and Practical Checks

The present-value random variable is tied to the actual payment time. A curtate future lifetime counts completed years, so it is appropriate for benefits paid at a policy-year boundary, not for a benefit paid at the moment of death. Replacing $T_x$ with a completed-year count changes both the discount period and, potentially, the benefit amount.

Several checks help keep the calculation consistent:

Use the coverage interval. A term benefit paid only on death by time $n$ has an integral ending at $n$, not infinity.
Use the correct density. The density $f_{T_x}(t)={}_tp_x\mu_{x+t}$ includes survival to time $t$. Do not multiply it by ${}_tp_x$ again.
Match discounting to the interest model. Use $v^t$ for a constant effective annual rate, or the appropriate discount function for the stated force of interest.
Check units. The time variable, mortality force, and interest force must all use the same time unit, such as years.
Check bounds. For a nonnegative benefit and a discount factor no greater than 1, the expected present value cannot exceed the undiscounted expected benefit.

The integral does not by itself assume a particular mortality law or a particular benefit pattern. Those enter through $f_{T_x}(t)$ and $b(t)$. Keeping the model, coverage condition, payment time, and discount function explicit is the safest way to set up a continuous death-benefit valuation.

Check Your Understanding

For each problem, identify the relevant death-time density and coverage interval before calculating.

A hypothetical one-year contract pays $8,000 at death. The complete future lifetime is uniform on $[0,2]$, and interest is zero. Find the expected present value. (Answer: $4,000.00)
A hypothetical two-year term contract pays $10,000 at death. The force of mortality is $0.1$ per year and the force of interest is $0.1$ per year throughout the term. Find the expected present value. (Answer: $1,648.40)
A hypothetical three-year contract has $T_x$ uniform on $[0,3]$, with density $1/3$. It pays $9,000 for death during the first year and $15,000 for death after the first year but by time 3. The force of interest is $\ln 2$. Find the expected present value. (Answer: $4,869.11)
A hypothetical two-year term contract pays $10{,}000(1+t)$ dollars at death at time $t$. Assume a constant force of mortality $0.2$ per year and a constant force of interest $0.1$ per year. Find the expected present value. (Answer: $5,716.84)

What You'll Learn