Continuous Term Insurance Values

Why a Fixed Term Changes the Valuation

A continuous whole-life insurance pays a benefit whenever death occurs, whenever that may be. A continuous term insurance pays only if death occurs within a specified period. Its value therefore depends both on the time value of money and on the probability that death occurs before the term ends. In particular, the relevant probability at each instant is a death density, not merely the probability of being alive at that instant.

This tutorial considers a unit benefit, payable immediately at death if death occurs within \(n\) years. There is no payment if the insured survives to the end of the term. The benefit amount can be scaled after valuing the unit benefit.

Definition: Let \(T_x\) be the future lifetime of a life aged \(x\), and let the force of interest be the constant \(\delta\). The present value random variable for a unit continuous term insurance of \(n\) years is \(e^{-\delta T_x}\) if \(0<T_x\le n\), and \(0\) otherwise. Its actuarial present value is denoted here by \(\bar A^1_{x:\overline{n}|}\).

Write \({}_tp_x=P(T_x>t)\) for the probability that a life aged \(x\) survives at least \(t\) more years, and \(\mu_{x+t}\) for the force of mortality at age \(x+t\). When the future lifetime has a density, the density of death at time \(t\) is \({}_tp_x\mu_{x+t}\). Thus, to value the insurance, discount each possible death-time payment to time zero and average over the times at which death can occur.

The Continuous Term Insurance Formula

Assume a constant force of interest \(\delta\) and a future lifetime with a density over the term. At time \(t\), the probability density of death is \({}_tp_x\mu_{x+t}\), and a unit payment at that time has present value \(e^{-\delta t}\). Therefore,

\bar A^1_{x:\overline{n}|} =\int_0^n e^{-\delta t}\,{}_tp_x\mu_{x+t}\,dt.

This integral ends at \(n\) because deaths after the term do not generate a benefit. Under a continuous lifetime model, the probability of death at exactly \(n\) is zero, so writing the upper endpoint as \(n\) includes deaths within the term without creating a boundary payment.

Key result: A continuous term insurance value is the discounted integral of the death-time density over the coverage period. The death-time density is \({}_tp_x\mu_{x+t}\); the survival probability alone is not the density of a payment.

The formula also follows directly from the expected present value. The present value is nonzero only when death occurs within the term. For a small interval from \(t\) to \(t+dt\), the probability of surviving to \(t\) and then dying during that interval is approximately \({}_tp_x\mu_{x+t}dt\). The discounted expected contribution from that interval is consequently approximately \(e^{-\delta t}{}_tp_x\mu_{x+t}dt\). Summing these contributions over the term and taking the limit gives the integral.

For a benefit amount \(b\), payable at death under the same assumptions, multiply the unit-benefit value by \(b\). This linear scaling applies because the present value at every possible death time is multiplied by the same constant benefit amount.

Constant Force of Mortality

A particularly useful model assumes the force of mortality is constant over the term: \(\mu_{x+t}=\mu\) for \(0\le t\le n\). Under this assumption, the survival probability is \({}_tp_x=e^{-\mu t}\). Substitution into the general integral gives a closed-form expression.

\begin{aligned} \bar A^1_{x:\overline{n}|} &=\int_0^n e^{-\delta t}e^{-\mu t}\mu\,dt\\ &=\mu\int_0^n e^{-(\mu+\delta)t}\,dt\\ &=\frac{\mu}{\mu+\delta} \left(1-e^{-(\mu+\delta)n}\right). \end{aligned}

The constant-force formula requires constant mortality and interest forces over the covered period. If mortality changes by duration or age, the general integral still applies, but the integrand must reflect the appropriate survival probability and force at each time. If the interest force is not constant, the discount factor \(e^{-\delta t}\) must also be replaced by the appropriate accumulation-based discount factor.

Worked Example: Constant Mortality and Interest Forces

A hypothetical three-year contract pays a unit benefit immediately at death if death occurs within the three-year term. The force of mortality is constant at \(\mu=0.12\) per year, and the force of interest is constant at \(\delta=0.03\) per year. Calculate the actuarial present value to four decimal places.

Method. Both forces are constant, so use the closed-form continuous term insurance formula. The combined exponent rate is \(\mu+\delta=0.12+0.03=0.15\).

Substitute the parameters and term:

\begin{aligned} \bar A^1_{x:\overline{3}|} &=\frac{0.12}{0.15}\left(1-e^{-0.15(3)}\right)\\ &=0.8\left(1-e^{-0.45}\right). \end{aligned}

Using \(e^{-0.45}\approx0.6376282\), we obtain \(1-e^{-0.45}\approx0.3623718\), and \(0.8(0.3623718)\approx0.2898975\). Rounded to four decimal places, the actuarial present value is 0.2899.

Check. The probability of death within three years is \(1-e^{-0.12(3)}=1-e^{-0.36}\approx0.3023\). The present value must be less than this probability because the payment is discounted; \(0.2899<0.3023\), as expected. The common mistake to avoid is discounting the entire term-end death probability to year three, even though each benefit is paid at the actual time of death.

Changing Mortality During the Term

When the force of mortality changes at a specified duration, divide the integral into intervals. On each interval, use that interval’s mortality force and the survival probability accumulated from the start of the term to the time of death. Discounting also continues from time zero, not from the beginning of the current interval.

Worked Example: A Piecewise Constant Mortality Force

A hypothetical two-year term contract pays a unit benefit immediately at death during the term. The force of interest is constant at \(\delta=0.10\) per year. The force of mortality is \(\mu=0.10\) during the first year and \(\mu=0.20\) during the second year. Calculate the actuarial present value to six decimal places.

Method. Evaluate the death-time integral separately over the two mortality intervals. On the first interval, survival to time \(t\) is \(e^{-0.10t}\). On the second interval, survival to time \(t\) is \(e^{-0.10}e^{-0.20(t-1)}\): the life must survive the first year and then survive for \(t-1\) years under the second force.

For the first year, combine the discount and survival factors: \(e^{-0.10t}e^{-0.10t}=e^{-0.20t}\). The first-year contribution is

\begin{aligned} C_1 &=\int_0^1 e^{-0.10t}e^{-0.10t}(0.10)\,dt\\ &=\frac{0.10}{0.20}\left(1-e^{-0.20}\right)\\ &=0.5(1-0.8187308)\\ &=0.0906346. \end{aligned}

For the second year, set \(s=t-1\), so \(0\le s\le1\). Survival to time \(1+s\) is \(e^{-0.10}e^{-0.20s}\), and discounting to time zero is \(e^{-0.10(1+s)}\). Their product is \(e^{-0.20}e^{-0.30s}\). Therefore the second-year contribution is

\begin{aligned} C_2 &=\int_0^1 e^{-0.20}e^{-0.30s}(0.20)\,ds\\ &=e^{-0.20}\frac{0.20}{0.30}\left(1-e^{-0.30}\right)\\ &=(0.8187308)(0.6666667)(1-0.7408182)\\ &=(0.8187308)(0.6666667)(0.2591818)\\ &\approx0.1414667. \end{aligned}

Add the interval contributions: \(0.0906346+0.1414667=0.2321013\). The actuarial present value is 0.232101 to six decimal places.

Check. The total probability of death within two years is \(1-e^{-(0.10)(1)-(0.20)(1)}=1-e^{-0.30}\approx0.2592\). The calculated value is positive and below that probability, as it must be under positive interest. The common mistake to avoid is restarting survival or discounting at time one: both factors must account for the full time elapsed since time zero.

Using a Survival Model Directly

A mortality force does not need to be specified as a constant. If a survival model is supplied, its derivative gives the death-time density wherever the survival function is differentiable. In particular, if the survival probability is \({}_tp_x\), then the density is \(-\frac{d}{dt}{}_tp_x\), which equals \({}_tp_x\mu_{x+t}\) when the force of mortality is defined.

Worked Example: A Uniform Future Lifetime Model

In a hypothetical model, a life aged \(x\) has future lifetime uniformly distributed from 0 to 10 years. Thus \({}_tp_x=1-t/10\) for \(0\le t\le10\). A continuous term insurance pays a unit benefit immediately at death during the first four years. The force of interest is constant at \(\delta=0.05\) per year. Find the actuarial present value to four decimal places.

Method. Differentiate the survival probability to obtain the death-time density. For \(0<t<10\), \(-\frac{d}{dt}(1-t/10)=0.10\). Equivalently, the force is \(\mu_{x+t}=0.10/(1-t/10)\), and multiplying it by survival gives the same density: \((1-t/10)\{0.10/(1-t/10)\}=0.10\).

The density is constant over the four-year coverage period, so the term value is

\begin{aligned} \bar A^1_{x:\overline{4}|} &=\int_0^4 e^{-0.05t}(0.10)\,dt\\ &=\frac{0.10}{0.05}\left(1-e^{-0.05(4)}\right)\\ &=2(1-e^{-0.20})\\ &=2(1-0.8187308)\\ &=2(0.1812692)\\ &=0.3625384. \end{aligned}

The actuarial present value is 0.3625 to four decimal places. The probability of death during the term is \(4/10=0.4\), so the discounted value, \(0.3625\), is smaller than the probability of payment. This check also confirms that the density used in the integral is \(0.10\), not the survival probability \(1-t/10\) by itself.

Common Timing and Term Errors

The most frequent errors come from confusing the time at which death occurs with the end of the coverage period. In a continuous term contract, the benefit is paid at death, not automatically at the term’s end. The contract also pays nothing for survival to the end of the term. These features distinguish the term insurance value from an annuity value or a contract with a survival benefit.

For example, in the first worked example, an incorrect year-three discounting calculation would be \(e^{-0.03(3)}(1-e^{-0.12(3)})\). Numerically this is approximately \((0.9139)(0.3023)=0.2763\), not the correct \(0.2899\). That calculation places every payment at year three. The correct calculation discounts each possible payment by \(e^{-0.03t}\), using the actual death time \(t\).

Do not omit the mortality force. The survival probability \({}_tp_x\) is not the death density. The density is \({}_tp_x\mu_{x+t}\).
Do not include a survival payment. A term insurance pays on death within the term, not on survival to the term’s end.
Do not reset the clock between intervals. A death payment in a later interval is discounted from time zero, and survival must include all earlier intervals.
Do not use the constant-force shortcut outside its assumptions. With changing mortality, split the integral or use the supplied survival model directly.

Key takeaway: For a unit continuous term benefit, integrate the product of the discount factor and the death-time density over the coverage term. Under constant mortality and interest forces, this reduces to \(\frac{\mu}{\mu+\delta}(1-e^{-(\mu+\delta)n})\). The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the relevant mortality model and term before calculating. Unless stated otherwise, benefits are unit amounts paid immediately at death and values should be rounded to four decimal places.

A hypothetical two-year term has constant force of mortality \(\mu=0.06\) and force of interest \(\delta=0.04\). Find its actuarial present value. (Answer: \(0.6(1-e^{-0.20})\approx0.1088\))
Future lifetime is uniform from 0 to 8 years, so \({}_tp_x=1-t/8\) for \(0\le t\le8\). A three-year term has force of interest \(\delta=0.08\). Find the actuarial present value. (Answer: \(\frac{0.125}{0.08}(1-e^{-0.24})\approx0.3334\))
A three-year term has force of interest \(\delta=0.05\), mortality force \(\mu=0.10\) during the first year, and mortality force \(\mu=0.20\) during years two and three. Find the actuarial present value. (Answer: \(\frac{0.10}{0.15}(1-e^{-0.15})+e^{-0.15}\frac{0.20}{0.25}(1-e^{-0.50})\approx0.3638\))
Under constant forces \(\mu=0.08\) and \(\delta=0.02\), a unit continuous term benefit covers five years. Find its actuarial present value and compare it with the probability of death during the term. (Answer: \(0.8(1-e^{-0.50})\approx0.3148\), less than \(1-e^{-0.40}\approx0.3297\))
A model has survival probability \({}_tp_x=1-t/6\) for \(0\le t\le6\). A two-year continuous term benefit is discounted at force \(\delta=0.03\). Find the actuarial present value. (Answer: \(\frac{1}{6}\frac{1-e^{-0.06}}{0.03}\approx0.3235\))

What You'll Learn