Numerical Evaluation of Continuous Insurance Integrals

Why Numerical Evaluation Is Needed

A continuous whole-life insurance pays a benefit at the moment of death. When the force of mortality or the force of interest changes with time, the present value may not reduce to a simple constant-force formula. Numerical integration provides a practical approximation, provided the integrand is constructed correctly, the integration interval is chosen sensibly, and the remaining tail is handled rather than ignored.

Let $T_x$ be the future lifetime of a life aged $x$. Write ${}_tp_x=P(T_x>t)$, and let $\mu_{x+t}$ be the force of mortality at duration $t$. Suppose the force of interest at duration $s$ is $\delta(s)$, so the discount factor to time 0 for a payment at time $t$ is $\exp(-\int_0^t \delta(s)\,ds)$. For a unit benefit payable at death, the continuous whole-life insurance APV is $\bar{A}_x$.

Definition: The integrand for a unit continuous whole-life death benefit is $$ g(t)={}_tp_x\,\mu_{x+t}\exp\left(-\int_0^t\delta(s)\,ds\right). $$ Thus $$ \bar{A}_x=\int_0^\infty g(t)\,dt. $$ Here $g(t)\,dt$ is the discounted probability of death in the small interval from $t$ to $t+dt$, to first order in $dt$.

The mortality and interest assumptions must be specified before evaluating this integral. In particular, the survival probability is determined by mortality alone: ${}_tp_x=\exp(-\int_0^t\mu_{x+s}\,ds)$. Interest contributes a separate discount factor. Combining these gives

g(t)=\mu_{x+t}\exp\left(-\int_0^t\bigl(\mu_{x+s}+\delta(s)\bigr)\,ds\right).

This form is convenient for numerical work: calculate the force of mortality at each integration point, then multiply by the accumulated survival-discount factor. If the forces are discontinuous or change formula at known durations, split the integral at each change point. A quadrature interval should not straddle a point where the integrand’s formula changes unless the method explicitly accounts for that change.

Composite Simpson’s Rule and the Tail

For a finite interval $[0,T]$, composite Simpson’s rule uses an even number $n$ of equal subintervals, with width $h=T/n$. It gives the approximation

\int_0^T g(t)\,dt\ \approx\ \frac{h}{3}\left[g(t_0)+g(t_n)+4\sum_{\substack{j=1\\j\ \mathrm{odd}}}^{n-1}g(t_j)+2\sum_{\substack{j=2\\j\ \mathrm{even}}}^{n-2}g(t_j)\right], \qquad t_j=jh.

Simpson’s rule approximates the integrand by quadratic curves across pairs of subintervals. For a sufficiently smooth integrand, its composite error generally decreases on the order of $h^4$ as the step size is reduced. This is not a guarantee of a particular error for every mortality model: a kink, jump, or unaccounted change point can undermine the approximation. An odd number of subintervals does not meet the standard composite Simpson requirement; use an even number or apply another method to the leftover interval.

The whole-life integral has an infinite upper limit, so a finite-interval estimate must be combined with a tail estimate or a justified truncation error. If mortality and interest forces are constant after duration $T$, with respective values $\mu$ and $\delta$, then

\int_T^\infty g(t)\,dt ={}_Tp_x\exp\left(-\int_0^T\delta(s)\,ds\right) \int_0^\infty \mu e^{-(\mu+\delta)u}\,du =C(T)\frac{\mu}{\mu+\delta},

where $C(T)={}_Tp_x\exp(-\int_0^T\delta(s)\,ds)$ and $u=t-T$. The last equality follows because the integral of $e^{-(\mu+\delta)u}$ over nonnegative $u$ is $1/(\mu+\delta)$, when $\mu+\delta>0$. This tail formula does not apply if the post-$T$ forces are not constant; in that case, use an appropriate tail model or integrate farther with a defensible approximation.

Specify the model.
Write the mortality and interest forces, including every duration at which either changes.

Build the integrand.
Calculate the mortality force and the accumulated survival-discount factor at each quadrature node.

Approximate the finite part.
Apply Simpson’s rule with an even number of subintervals on smooth pieces, or choose a method appropriate to the integrand.

Complete and check the APV.
Add a tail estimate, compare interval widths where feasible, and retain sufficient precision until the final rounding.

Worked Example: Simpson’s Rule with a Constant-Force Tail

A hypothetical unit whole-life benefit is payable continuously at death. For durations $0\le t\le2$, suppose the force of mortality is $\mu_{x+t}=0.02+0.01t$ and the force of interest is $\delta(t)=0.03+0.005t$. After duration 2, both forces remain constant at their duration-2 values. Approximate $\bar{A}_x$ by composite Simpson’s rule with width $h=0.5$, then add the tail.

Method. The sum of the forces over the first two years is $0.05+0.015t$, so its integral from 0 to $t$ is $0.05t+0.0075t^2$. Therefore

g(t)=(0.02+0.01t)e^{-(0.05t+0.0075t^2)}.

There are four subintervals, an even number. Evaluating the integrand gives, to seven decimal places,

\begin{array}{c|ccccc} t&0&0.5&1&1.5&2\\ g(t)&0.0200000&0.0243371&0.0283237&0.0319277&0.0351238 \end{array}

The Simpson estimate on $[0,2]$ is

\frac{0.5}{3}\left[0.0200000+0.0351238+ 4(0.0243371+0.0319277)+2(0.0283237)\right] =\frac{0.3368304}{6}\approx0.0561384.

For the tail, the accumulated survival-discount factor at duration 2 is $C(2)=e^{-(0.05(2)+0.0075(2^2))}=e^{-0.13}\approx0.8780954$. The post-2 mortality and interest forces are both $0.04$, so the tail is $0.8780954(0.04/(0.04+0.04))=0.4390477$.

Adding the finite part and tail gives $0.0561384+0.4390477\approx0.4951861$. Thus the approximate APV is 0.4952 to four decimal places. The result is positive and below 1, as a discounted unit benefit should be. The calculation also avoids the common error of treating the finite-interval integral as the whole-life value.

Interval Choice and Rounding

A finer interval generally improves a smooth-integrand Simpson approximation, but it also requires more function evaluations. A useful practical check is to calculate estimates with two interval widths and compare them. Agreement is evidence of stability, not a proof that the answer is exact; both calculations can still share a modeling or tail error.

Rounding at intermediate steps can matter, especially when an accumulated factor multiplies a substantial tail. Retain several guard digits in the forces, exponentials, node values, and tail calculations. Round the reported APV only at the end. A table rounded for display need not be the table used for the underlying calculation.

Worked Example: Checking the Integration Interval

Use the same hypothetical mortality and interest model as in the preceding example. Compare composite Simpson estimates on $[0,2]$ using $h=1$ and $h=0.5$, then compare the resulting whole-life values. Use the unrounded integrand in the calculations.

With $h=1$, there are two subintervals and the nodes are $0,1,2$. The estimate is

\frac{1}{3}\left[g(0)+4g(1)+g(2)\right] \approx\frac{1}{3}\left[0.0200000+4(0.0283237)+0.0351238\right] \approx0.0561395.

The $h=0.5$ estimate from the previous calculation is approximately $0.0561384$. Their absolute difference is approximately $0.0000011$. Adding the same tail, $0.4390477$, gives whole-life estimates of approximately $0.4951872$ and $0.4951861$, respectively. They both round to $0.4952$ to four decimal places.

The finer-grid estimate is not automatically the exact value, but its close agreement with the coarser-grid estimate supports the reported precision. The common mistake is to compare just the finite integrals while overlooking that both estimates require the same tail. Here the tail is the largest part of the APV, so omitting it would make the comparison misleading.

Worked Example: Rounding the Accumulated Factor Too Early

A hypothetical unit continuous whole-life benefit has force of mortality $0.03$ and force of interest $0.01$ during the first year. After year 1, the respective forces are constant at $0.04$ and $0.02$. Approximate the first-year integral by Simpson’s rule using nodes $0,0.5,1$, calculate the tail, and assess rounding the accumulated factor prematurely.

During the first year, $g(t)=0.03e^{-0.04t}$. The accumulated survival-discount factor at year 1 is $C(1)=e^{-0.04}\approx0.9607894$, not $0.04$. The three integrand values are $g(0)=0.0300000$, $g(0.5)=0.03e^{-0.02}\approx0.0294060$, and $g(1)=0.03e^{-0.04}\approx0.0288237$.

Simpson’s estimate for the first year is

\frac{0.5}{3}\left[0.0300000+4(0.0294060)+0.0288237\right] \approx0.0294079.

After year 1, the tail is $C(1)(0.04/(0.04+0.02))\approx0.6405263$. The whole-life estimate is therefore $0.0294079+0.6405263\approx0.6699342$, or 0.6699 to four decimal places. As a check, the first-year integral is exactly $(0.03/0.04)(1-e^{-0.04})\approx0.0294079$, consistent with Simpson’s estimate.

If $C(1)$ is prematurely rounded to $0.96$, the tail becomes $0.96(2/3)=0.6400000$, producing a total of $0.6694079$. The tail is understated by approximately $0.0005263$. This example shows why even a seemingly small rounding difference in a factor used for a long tail can change the final reported value.

Checks and Common Traps

A numerical APV is only as reliable as its model and implementation. Before accepting a result, confirm that the death density uses the mortality force at the correct duration, the discount factor accumulates interest through the payment time, every change point is respected, and the tail matches the stated post-interval assumptions.

Confusing a force with an accumulated factor. A force such as $0.04$ is not the accumulated survival-discount factor at time 1; in the preceding example that factor is $e^{-0.04}\approx0.9608$.
Applying the wrong step size. Simpson’s rule uses the actual width $h$, and its standard composite form requires an even number of subintervals.
Rounding each node too aggressively. Keep more digits in calculations than in the displayed table and round the final APV only once.
Ignoring a change point or the tail. Integrate separately across model changes, and do not treat an arbitrary finite endpoint as the end of a whole-life contract.
Assuming a finer grid fixes a wrong model. Repeated quadrature cannot correct an incorrect mortality force, interest force, or benefit timing assumption.

These methods are useful whenever a continuous death-benefit APV does not have a convenient closed form. They are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus, where careful construction and evaluation of life-contingency values matters as much as the final arithmetic.

Check Your Understanding

For each problem, state the assumptions used and retain sufficient precision until the final answer.

A hypothetical unit continuous whole-life benefit has constant force of mortality $0.05$ and constant force of interest $0.03$ for all future durations. Evaluate its APV using the constant-force integral. (Answer: $0.6250$)
A hypothetical contract has mortality force $0.02$ and interest force $0.03$ for the first year. Thereafter, the forces are constant at $0.04$ and $0.02$. Calculate the first-year integral and the tail, then give the APV to four decimals. (Answer: approximately $0.6537$)
For a hypothetical model, the computed integrand values at $t=0,0.5,1,1.5,2$ are $0.0180,0.0201,0.0220,0.0238,0.0255$, respectively. The accumulated survival-discount factor at $t=2$ is $0.85$, and after time 2 the forces are constant at mortality $0.03$ and interest $0.05$. Use composite Simpson’s rule on $[0,2]$, add the tail, and report the APV to four decimals. (Answer: approximately $0.3626$)
For constant mortality $0.04$ and constant interest force $0.06$, approximate the whole-life APV by Simpson’s rule on $[0,1]$ with $h=0.5$, then use the exact constant-force tail after time 1. Give the answer to four decimals. (Answer: $0.4000$)
In the first-year model of Worked Example: Rounding the Accumulated Factor Too Early, a student replaces $C(1)=e^{-0.04}$ by $0.96$ before calculating the tail. How much does this change the tail, and in which direction? (Answer: it understates the tail by approximately $0.0005263$)

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