De Moivre Model Insurance Valuations

Introduction

De Moivre’s law is a simple survival model with an important actuarial feature: the future lifetime of a life aged $x$ is uniformly distributed between zero and the number of years remaining to a fixed limiting age. That makes continuous insurance values straightforward to integrate, provided the survival model and the end of the coverage period are represented correctly.

This tutorial develops continuous whole-life and term insurance values under De Moivre’s law. The central calculation is the expected discounted benefit at the moment of death. The finite endpoint matters: a term longer than the time remaining to the limiting age cannot cover deaths after that age.

De Moivre Survival and Death-Time Distributions

Assume a limiting age $\omega$, and consider a life currently aged $x$, where $0\leq x<\omega$. Let $m=\omega-x$ be the remaining lifetime to the limiting age. Under De Moivre’s law, deaths are distributed uniformly by age between age 0 and $\omega$. Consequently, the remaining future lifetime $T_x$ is uniform on $[0,m]$. The model assigns probability zero to death at the single endpoint $m$.

Definition: Under De Moivre’s law, the survival function from age 0 is $S_0(y)=1-y/\omega$ for $0\leq y\leq\omega$. For a life aged $x<\omega$, the conditional probability of surviving a further $t$ years is ${}_tp_x=S_0(x+t)/S_0(x)=(m-t)/m$ for $0\leq t\leq m$. It is zero for $t\geq m$. Thus ${}_tq_x=1-{}_tp_x=t/m$ for $0\leq t\leq m$.

The density of $T_x$ is $f_{T_x}(t)=1/m$ for $0<t<m$. It is constant because the future lifetime is uniform. The force of mortality is not constant: at duration $t<m$, it is $\mu_{x+t}=1/(m-t)$, which increases as the life approaches the limiting age. These are different quantities: the density describes the unconditional probability per unit time of death at duration $t$, while the force of mortality is conditional on survival to that duration.

Continuous Whole-Life Insurance

Suppose a unit benefit is payable immediately at death, with a constant force of interest $\delta\geq0$. The present value is $e^{-\delta T_x}$ if death occurs, and its expected value is the continuous whole-life insurance value $\bar A_x$. Under De Moivre’s law, all possible death times lie between 0 and $m$, so the expectation is an integral over that finite interval.

\bar A_x =E[e^{-\delta T_x}] =\int_0^m e^{-\delta t}\frac{1}{m}\,dt =\frac{1-e^{-\delta m}}{\delta m},\qquad \delta>0.

If $\delta=0$, there is no discounting and the unit benefit is paid with probability 1, so $\bar A_x=1$. Equivalently, this is the limit of the displayed expression as $\delta$ approaches zero. A benefit of $B$ dollars has expected present value $B\bar A_x$.

Worked Example: Continuous Whole-Life Cover

A hypothetical life aged 40 follows De Moivre’s law with limiting age 100. A whole-life policy pays $ \$120{,}000$ immediately at death. The force of interest is $\delta=0.04$. Find the expected present value and the probability that a benefit is paid.

Method and model parameters. The remaining lifetime is $m=100-40=60$ years. The death-time density is $1/60$ over $0<t<60$. The force of interest is constant at $0.04$ per year.

Unit insurance value. Here $\delta m=0.04(60)=2.4$, so $\bar A_{40}=(1-e^{-2.4})/2.4$. Using $e^{-2.4}=0.0907179533$, this gives $\bar A_{40}=(1-0.0907179533)/2.4=0.9092820467/2.4=0.3788675195$.

Expected present value. Multiply the unrounded unit value by the benefit: $ \$120{,}000(1-e^{-2.4})/2.4=\$45{,}464.1023\ldots$, or $\$45{,}464.10$ to the nearest cent.

Probability check. The policy covers death at any future time before the limiting age, and the life dies before that age with probability 1. Therefore, the probability of a benefit payment is 1. The expected present value is less than the undiscounted benefit of $\$120{,}000$, as it should be when $\delta>0$.

Common mistake avoided. The integral ends at 60, not infinity: the model gives no probability of survival beyond the limiting age.

Continuous Term Insurance and the Integration Limit

For a continuous $n$-year term policy, the benefit is paid only if death occurs during the first $n$ years. If $n\leq m$, the coverage ends before or at the limiting age, and the integral ends at $n$. If $n>m$, the life cannot survive to the end of the term under the model; the benefit can only be paid before duration $m$. In both cases, the upper integration endpoint is $\min(n,m)$.

Definition: For a unit continuous term insurance benefit payable immediately at death during the term, write its value as $\bar A^1_{x:\overline{n}|}$. Under De Moivre’s law and force of interest $\delta>0$, $\bar A^1_{x:\overline{n}|}=\int_0^{\min(n,m)}e^{-\delta t}(1/m)\,dt=[1-e^{-\delta\min(n,m)}]/(\delta m)$. For $\delta=0$, the value is $\min(n,m)/m$, the probability of a covered death.

The probability that death occurs during the term is ${}_{\min(n,m)}q_x=\min(n,m)/m$. This probability is not the same as the insurance value when interest is positive: the value also weights each possible death time by its discount factor. A benefit of $B$ dollars has expected present value $B\bar A^1_{x:\overline{n}|}$.

Worked Example: Term Ends Before the Limiting Age

A hypothetical life aged 30 follows De Moivre’s law with limiting age 90. A 20-year continuous term policy pays $ \$200{,}000$ immediately at death during the term. The force of interest is $\delta=0.05$. Find the unit insurance value, expected present value, and probability of a covered death.

Method and model parameters. The remaining lifetime is $m=90-30=60$ years. Since $n=20<60$, the benefit can be paid only during durations $0\leq t\leq20$; the death-time density is $1/60$.

Unit insurance value. The exponent at the term endpoint is $\delta n=0.05(20)=1$, and the denominator is $\delta m=0.05(60)=3$. Therefore, $\bar A^1_{30:\overline{20}|}=[1-e^{-1}]/3$. Using $e^{-1}=0.3678794412$, the numerator is $1-0.3678794412=0.6321205588$, and the unit value is $0.6321205588/3=0.2107068529$.

Expected present value. Using the unrounded unit value, $\$200{,}000(1-e^{-1})/3=\$42{,}141.3706\ldots$, or $\$42{,}141.37$ to the nearest cent.

Probability check. The probability of death within 20 years is $20/60=1/3=0.3333$. The unit insurance value $0.2107$ is lower than this probability because covered benefits are discounted.

Common mistake avoided. The denominator uses the full remaining lifetime $m=60$, because the death-time density is $1/m$; it is not $1/n$. The term endpoint changes the integration limit, not the density.

Worked Example: Term Extends Beyond the Limiting Age

A hypothetical life aged 55 follows De Moivre’s law with limiting age 80. A 30-year continuous term policy pays $ \$80{,}000$ immediately at death during the term. The force of interest is $\delta=0.02$. Find the expected present value and probability of a covered death.

Method and model parameters. The remaining lifetime is $m=80-55=25$ years. The stated term is $n=30$, but no death can occur after duration 25 under this model. Thus the effective integration limit is $\min(30,25)=25$, and the density before that endpoint is $1/25$.

Unit insurance value. The exponent at the effective endpoint is $0.02(25)=0.5$, and $\delta m=0.02(25)=0.5$. Hence $\bar A^1_{55:\overline{30}|}=[1-e^{-0.5}]/0.5$. Using $e^{-0.5}=0.6065306597$, the numerator is $0.3934693403$, giving $0.3934693403/0.5=0.7869386806$.

Expected present value. The unrounded calculation is $\$80{,}000(1-e^{-0.5})/0.5=\$62{,}955.0944\ldots$, or $\$62{,}955.09$ to the nearest cent.

Probability check. The probability of death before the limiting age is 1, and the 30-year term includes all possible future death times under the model. Thus the probability of a covered death is 1.

Common mistake avoided. Integrating to 30 would incorrectly include durations at which the survival probability is already zero. The value is the same as whole-life cover because the term extends beyond every possible death time.

Why the Formulas Hold

The valuation formulas follow directly from the expectation of the present value. Under De Moivre’s law, the probability of death in a small interval from $t$ to $t+dt$ is approximately $(1/m)dt$, as long as $0<t<m$. If a unit benefit is paid immediately at death, its present value at duration $t$ is $e^{-\delta t}$. Multiplying these terms gives the contribution to expected present value from deaths in that interval. Summing those contributions over the possible death times gives the integral.

Specify the possible death times.
For whole life, $0<t<m$. For an $n$-year term, $0<t<\min(n,m)$.

Weight each death time.
The density is $1/m$, and the present value of a unit benefit at time $t$ is $e^{-\delta t}$. The expected-value contribution is therefore $e^{-\delta t}(1/m)dt$.

Integrate and multiply by the benefit.
Integrating over the relevant finite interval gives the unit value; multiplying by the stated benefit gives the expected present value in dollars.

For example, evaluating the whole-life integral gives $\int_0^m e^{-\delta t}(1/m)dt=(1/m)[-e^{-\delta t}/\delta]_0^m=(1-e^{-\delta m})/(\delta m)$. Replacing $m$ with $\min(n,m)$ in the exponential endpoint gives the term value, while the density denominator remains $m$. This derivation explains both the formula and the most important distinction between the two parameters in it.

Common Checks and Exam Traps

These values are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. A reliable calculation keeps the mortality model, time interval, and discounting separate until the integral is set up.

Using a constant force of mortality. Under De Moivre’s law, the death-time density is constant, but the force of mortality is $1/(m-t)$, not a constant. Using an exponential survival model changes the answer.
Using $n$ as the density denominator. The density is determined by the full remaining lifetime $m$, even when coverage lasts for only $n<m$ years.
Integrating beyond the limiting age. If $n>m$, use $m$ as the endpoint. The term cannot produce a benefit after the life’s last possible duration.
Confusing probability with present value. The probability of covered death is $\min(n,m)/m$. With positive interest, the insurance value is smaller because the benefit is discounted.
Rounding too early. Retain the exponential and unit value at full precision when multiplying by the benefit; round the final dollar amount to the nearest cent.

A quick reasonableness check is that, for $\delta>0$, the expected present value of a unit benefit cannot exceed the probability that the benefit is paid. It also cannot exceed 1. For whole life under this model, that probability is 1; for term insurance, it is $\min(n,m)/m$.

Check Your Understanding

For each hypothetical policy, use De Moivre’s law and the stated constant force of interest. Keep full precision until the final dollar amount.

A life aged 25 has limiting age 65. Find the unit value of continuous whole-life insurance when $\delta=0.03$. (Answer: $0.5823$)
A life aged 40 has limiting age 90. A 10-year continuous term pays $\$90{,}000$ immediately at death, with $\delta=0.04$. Find the expected present value and the probability of a covered death. (Answer: $\$14{,}835.60$; probability $0.2000$)
A life aged 60 has limiting age 80. A 25-year continuous term pays $\$40{,}000$ immediately at death, with $\delta=0.05$. Find the expected present value and the probability of a covered death. (Answer: $\$25{,}284.82$; probability $1.0000$)
A life aged 48 has limiting age 80. An 8-year continuous term pays $\$100{,}000$ immediately at death, with $\delta=0.025$. Find the expected present value and the probability of a covered death. (Answer: $\$22{,}658.66$; probability $0.2500$)
A life aged 35 has limiting age 65. A 12-year continuous term pays $\$75{,}000$ immediately at death, with $\delta=0.03$. Find the unit value, expected present value, and probability of a covered death. (Answer: unit value $0.3359$; $\$25{,}193.64$; probability $0.4000$)

What You'll Learn