Numerical Integration for Continuous Annuities

Why Numerical Integration Is Needed

A continuous life annuity pays at a constant rate while the annuitant is alive. Its value depends on survival at every instant during the payment period, not only at integer ages. When survival probabilities are available only at selected durations, the integral defining the value may not be directly calculable. Numerical integration provides a practical approximation from those tabulated values.

The essential quantity to integrate is the product of two factors: the probability the annuitant is alive at time $t$, and the discount factor for a payment at time $t$. The trapezoidal rule approximates the area under this product by adding the areas of trapezoids between consecutive tabulated times. The spacing between those times matters: under regularity conditions, smaller intervals generally reduce the approximation error.

Definition: For a life aged $x$, a temporary continuous life annuity paying at rate 1 per year for at most $n$ years has present value $$ \bar{a}_{x:\overline{n}|}=\int_0^n v^t\,{}_tp_x\,dt, $$ where ${}_tp_x$ is the probability that the life aged $x$ survives at least $t$ years, and $v^t=(1+i)^{-t}$ when the effective annual interest rate is $i$. For a continuous payment rate $R$, the value is $R\bar{a}_{x:\overline{n}|}$. The formula assumes a constant payment rate while alive and a constant effective annual interest rate over the valuation period.

If survival probabilities are tabulated at durations $t_0,t_1,\ldots,t_m$, define the discounted survival function at each node by $g(t_j)=v^{t_j}{}_{t_j}p_x$. Then the target value is the area under $g(t)$ from 0 to $n$. With equal spacing $h=n/m$, the composite trapezoidal estimate is

\bar{a}_{x:\overline{n}|}\ \approx\ \frac{h}{2}\left[g(t_0)+2\sum_{j=1}^{m-1}g(t_j)+g(t_m)\right].

The first and last ordinates have weight 1; each interior ordinate has weight 2. For nonuniform intervals, use the interval width separately for each trapezoid:

\bar{a}_{x:\overline{n}|}\ \approx\ \sum_{j=1}^{m}\frac{t_j-t_{j-1}}{2} \left[g(t_{j-1})+g(t_j)\right].

Why the Trapezoidal Formula Works

On one interval $[t_{j-1},t_j]$, the trapezoidal rule replaces the curve $g(t)$ by the straight line joining its endpoint values. The area under that line is the area of a trapezoid: its parallel sides have lengths $g(t_{j-1})$ and $g(t_j)$, and its width is $t_j-t_{j-1}$. Therefore its area is the interval width multiplied by the average of the endpoint heights. Adding these areas gives the composite formula.

This argument also identifies what the estimate does and does not guarantee. It is exact for the piecewise-linear interpolation of the tabulated discounted survival values. It is not necessarily exact for the true curve between those values. If $g$ is twice differentiable on an interval of length $h$, the composite trapezoidal error has order $h^2$; a standard bound is

\left|\int_0^n g(t)\,dt-T\right| \leq \frac{n h^2}{12}\max_{0\leq t\leq n}|g''(t)|,

where $T$ is the composite trapezoidal estimate. Thus, when the curve is sufficiently smooth and the same range is divided into smaller equal intervals, the error typically falls in proportion to the square of the interval width. The bound requires a bounded second derivative; tabulated data alone do not establish that condition.

Set the valuation range and time units.
Identify the annuity term and make sure durations are measured in years if $i$ is an effective annual rate.

Discount each survival probability.
At every node $t_j$, calculate $g(t_j)=(1+i)^{-t_j}{}_{t_j}p_x$. Do not discount the survival probabilities in annual steps when a node is fractional.

Compute each trapezoid.
Multiply the width of each interval by the average of its two endpoint values of $g$.

Add the areas and apply the payment rate.
The sum estimates the value per unit annual payment rate. Multiply by $R$ only if the stated continuous payment rate is not 1.

Worked Examples

Worked Example: A Temporary Annuity From Half-Year Survival Values

A hypothetical life annuity pays continuously at rate 1 per year for at most 2 years. The effective annual interest rate is 5%. The survival probabilities for a life aged $x$ are tabulated as follows: ${}_0p_x=1$, ${}_{0.5}p_x=0.92$, ${}_1p_x=0.82$, ${}_{1.5}p_x=0.70$, and ${}_2p_x=0.55$. Estimate the annuity value using the composite trapezoidal rule.

Method. The tabulated durations are equally spaced by $h=0.5$ years. At each duration, multiply survival by $v^t=(1.05)^{-t}$.

At $t=0$, $g(0)=1$. At $t=0.5$, $g(0.5)=0.92/\sqrt{1.05}=0.8978281$. At $t=1$, $g(1)=0.82/1.05=0.7809524$. At $t=1.5$, $g(1.5)=0.70/(1.05)^{1.5}=0.6506001$. At $t=2$, $g(2)=0.55/(1.05)^2=0.4988662$.

There are four intervals, so the trapezoidal estimate is

\begin{aligned} T &=\frac{0.5}{2}\left[1+2(0.8978281+0.7809524+0.6506001)+0.4988662\right]\\ &=0.25\left[1+2(2.3293806)+0.4988662\right]\\ &=0.25(6.1576274)\\ &=1.5394069. \end{aligned}

The estimated value is 1.5394 per dollar of annual payment rate, rounded to four decimal places. The endpoint check is that the value uses four half-year widths to cover exactly 2 years. A common mistake is to use $0.82$ rather than $0.82/1.05$ at $t=1$, omitting one year of discounting; the corrected discounted ordinate is $0.7809524$.

Worked Example: Assessing the Effect of Interval Width

Assume a constant force of mortality $\mu_x=0.10$ per year and an effective annual interest rate of $i=0\%$. Estimate the value of a unit continuous annuity over 1 year using interval widths of 1, 0.5, and 0.25 years. Compare the estimates with the exact value.

Method. Under the stated constant-force assumption, survival is ${}_tp_x=e^{-0.10t}$. Since $i=0\%$, the discount factor is 1, so $g(t)=e^{-0.10t}$. The exact value, used here as a benchmark, is

\int_0^1 e^{-0.10t}\,dt =\frac{1-e^{-0.10}}{0.10} =0.9516258.

With width $h=1$, the endpoint values are $g(0)=1$ and $g(1)=e^{-0.10}=0.9048374$. Thus

T_1=\frac{1}{2}(1+0.9048374)=0.9524187.

With width $h=0.5$, the three values are $1$, $e^{-0.05}=0.9512294$, and $e^{-0.10}=0.9048374$. Therefore

T_{0.5}=\frac{0.5}{2}\left[1+2(0.9512294)+0.9048374\right] =0.9518241.

With width $h=0.25$, the values at the interior points are $e^{-0.025}=0.9753099$, $e^{-0.05}=0.9512294$, and $e^{-0.075}=0.9277435$. Hence

\begin{aligned} T_{0.25} &=\frac{0.25}{2}\left[1+2(0.9753099+0.9512294+0.9277435)+0.9048374\right]\\ &=0.9516754. \end{aligned}

The absolute errors are $0.0007929$ for width 1, $0.0001983$ for width 0.5, and $0.0000496$ for width 0.25. Each halving of the interval width reduces the error by approximately a factor of 4, consistent with the $h^2$ error behavior for a smooth curve. All three estimates are slightly above the exact value; that direction is consistent here with the convexity of $e^{-0.10t}$, but should not be assumed for an arbitrary tabulated curve. The check is that all estimates and the benchmark are close to 0.952, and that the finer-grid estimates move closer to the exact value.

Worked Example: Unequally Spaced Survival Data

A hypothetical temporary continuous annuity pays at rate $1,200 per year for at most 3 years. The effective annual interest rate is $i=0\%$. The survival probabilities are tabulated at durations 0, 0.5, 1.5, and 3 years as 1.00, 0.94, 0.80, and 0.50, respectively. Estimate the value using the available nodes.

Method. With zero interest, $v^t=1$, so each discounted survival value equals the survival probability. The interval widths are 0.5, 1.0, and 1.5 years; they are not equal, so calculate a separate trapezoid for each.

From 0 to 0.5 years, the area is $0.5(1.00+0.94)/2=0.485$. From 0.5 to 1.5 years, it is $1.0(0.94+0.80)/2=0.870$. From 1.5 to 3 years, it is $1.5(0.80+0.50)/2=0.975$. The total estimate per unit annual payment rate is

$$ 0.485+0.870+0.975=2.330. $$

Multiplying by the payment rate gives an estimated present value of $1{,}200(2.330)=\$2{,}796.00$. The estimate covers the full 3-year term because the three widths sum to $0.5+1.0+1.5=3$. A common mistake is to apply the equal-spacing formula using an assumed width of 1 year; the unequal-width formula is required here.

Practical Checks and Common Traps

Numerical integration is only as sound as the data and the time indexing used to construct it. Before accepting an estimate, check that survival probabilities are nonincreasing, lie between 0 and 1, and correspond to the durations shown. For a positive effective annual rate, discount factors should also be nonincreasing. Consequently, the discounted survival values should not increase when both factors are nonincreasing.

Discounting at the wrong time. At a node $t=1.5$, use $(1+i)^{-1.5}$, not $(1+i)^{-1}$ and not $(1+i)^{-2}$.
Using the wrong interval weights. In the equal-spacing formula, endpoints receive weight 1 and interior points receive weight 2. For irregular nodes, compute each trapezoid with its own width.
Mixing time units. If $t$ is measured in months, convert it to years when using an annual effective rate; for example, 6 months is $t=0.5$ years.
Forgetting the payment rate. The integral values a payment rate of 1 per year. A stated rate $R$ requires multiplying the result by $R$.
Claiming exactness from tabulated values. The trapezoidal rule is exact for the piecewise-linear approximation between nodes, not necessarily for the underlying survival curve.

When a finer table is available, comparing coarse- and fine-grid estimates is a useful stability check. A large change signals that the coarse grid may not capture the shape of the discounted survival function adequately. However, refinement does not repair incorrect survival data, inconsistent units, or a payment term that has been indexed incorrectly.

Key takeaway: First form the discounted survival values $g(t)=v^t{}_tp_x$. Then integrate those values with the trapezoidal rule, using the actual interval widths. A finer grid often improves accuracy for a smooth underlying function, but the estimate remains an approximation unless the behavior between tabulated points is known.

Check Your Understanding

For each problem, use the stated assumptions and apply the trapezoidal rule. Round numerical annuity values to four decimal places unless a dollar amount is requested.

A unit continuous annuity lasts 1 year. Assume $\mu_x=0.08$ per year is constant, $i=0\%$, and use nodes at 0, 0.5, and 1 year. Estimate its value. (Answer: 0.9612)
A unit continuous annuity lasts 3 years at $i=0\%$. Survival probabilities at years 0, 1, 2, and 3 are 1.00, 0.90, 0.75, and 0.50. Estimate its value. (Answer: 2.4000)
A unit continuous annuity lasts 1 year at an effective annual rate of 5%. Survival probabilities at years 0 and 1 are 1.00 and 0.80. Use one trapezoid to estimate its value. (Answer: 0.8810)
A unit continuous annuity lasts 3 years at $i=0\%$. Survival probabilities are 1.00 at time 0, 0.80 at time 1, and 0.40 at time 3. Use the unequal-interval trapezoidal rule. (Answer: 2.1000)
Assume $\mu_x=0.10$ per year is constant and $i=0\%$. For a unit continuous annuity over 1 year, use nodes at 0, 0.25, 0.5, 0.75, and 1 year. Estimate its value. (Answer: 0.9517)

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