Temporary Continuous Life-Annuity Values

Introduction

A temporary continuous life annuity pays at a constant rate while a person is alive, but only up to a specified time. Its value therefore depends on both the discounting of future payments and the probability that those payments will be made. The fixed term matters: payments stop at the earlier of death and the end of the term.

The central calculation is an integral over time. At each instant, the payment rate is weighted by the probability of survival to that instant and by the discount factor for that time. This tutorial develops that integral, explains why it gives the expected present value, and applies it to several mortality assumptions.

Definition: Let $T_x$ be the future lifetime of a life aged $x$. A temporary continuous life annuity paying at rate 1 per year for at most $n$ years has present value $\int_0^{\min(T_x,n)} v^t\,dt$, where $v^t$ is the discount factor for a payment at time $t$. Its expected present value is denoted here by $\bar{a}_{x:\overline{n}|}$. For a payment rate $c$ per year, the expected present value is $c\bar{a}_{x:\overline{n}|}$.

The Survival-Weighted Integral

Write ${}_tp_x=P(T_x>t)$ for the probability that a life aged $x$ survives at least $t$ more years. Let $D(t)$ be the discount factor from time $t$ to time 0. Under a constant force of interest $\delta$, $D(t)=e^{-\delta t}$. More generally, if the effective annual interest rate is $i$, then $D(t)=(1+i)^{-t}$, with $\delta=\ln(1+i)$ for an equivalent constant force.

Assume payments accrue continuously at rate 1 while the life is alive, and cease at death or at time $n$, whichever occurs first. The present value for a particular lifetime is the integral of discounted payments actually made. Taking its expectation gives the temporary annuity value:

\bar{a}_{x:\overline{n}|}=\int_0^n D(t)\,{}_tp_x\,dt.

The survival probability appears at every time $t$, not just at the end of the term. If the payment rate is $c$, multiplying the integrand by $c$ gives the expected present value $c\bar{a}_{x:\overline{n}|}$. The formula assumes the stated payment rate continues without interruption while the life is alive and that the term is measured in the same time units as the interest and mortality models.

Why the Integral Gives the Expected Present Value

The payment stream can be described using an indicator: at time $t$, a payment is made only if $T_x>t$ and $t<n$. Thus the present value for a payment rate of 1 is the integral of $D(t)\mathbf{1}_{\{T_x>t\}}$ from 0 to $n$. Taking expectations moves the expectation inside the integral because the integrand is nonnegative. The expected value of the indicator is the probability of survival to time $t$.

Write the present value for one lifetime.
$\displaystyle Z=\int_0^n D(t)\mathbf{1}_{\{T_x>t\}}\,dt$. The indicator is 1 when the life is alive at $t$, and 0 otherwise.

Take the expectation.
$\displaystyle E[Z]=\int_0^n D(t)E[\mathbf{1}_{\{T_x>t\}}]\,dt$, since the payment and discount factors are nonnegative.

Replace the expected indicator by a survival probability.
$\displaystyle E[\mathbf{1}_{\{T_x>t\}}]=P(T_x>t)={}_tp_x$. Therefore $\displaystyle E[Z]=\int_0^n D(t){}_tp_x\,dt$.

This justification also identifies the assumptions behind the valuation: the payment is conditional on being alive at the time it accrues, and discounting is applied to each payment at its own time. It does not value a guarantee to pay the full term after death. Such a guarantee would require adding the value of payments that continue regardless of survival.

Constant Force of Mortality

Under a constant force of mortality $\mu$ over the term, ${}_tp_x=e^{-\mu t}$. If the force of interest is also constant at $\delta$, the survival-weighted discount factor is $e^{-(\delta+\mu)t}$. For $\delta+\mu>0$, integration gives:

\bar{a}_{x:\overline{n}|} =\int_0^n e^{-(\delta+\mu)t}\,dt =\frac{1-e^{-(\delta+\mu)n}}{\delta+\mu}.

This closed form applies only when both forces are constant over the full term. If mortality changes by duration, the survival probability must reflect all earlier mortality segments; the integral can then be evaluated segment by segment. The general integral remains valid even when mortality is not constant.

Worked Example: Constant Forces and a Level Payment Rate

A hypothetical temporary annuity pays continuously at \$1,200 per year while a life aged $x$ survives, for at most 5 years. Assume a constant force of interest $\delta=0.06$ and a constant force of mortality $\mu=0.04$ over the term. Find the expected present value, rounded to the nearest cent.

Method. Use the constant-force formula. The discount and survival factors combine as $e^{-(0.06+0.04)t}=e^{-0.10t}$. The annuity factor for a unit payment rate is

\bar{a}_{x:\overline{5}|} =\frac{1-e^{-0.10(5)}}{0.10} =\frac{1-e^{-0.5}}{0.10}.

Using $e^{-0.5}=0.6065306597$, the numerator is $1-0.6065306597=0.3934693403$. Dividing by $0.10$ gives a unit-rate value of $3.934693403$. Multiplying by the payment rate gives

$$ 1{,}200(3.934693403)=4{,}721.632084. $$

The expected present value is \$4,721.63, rounded to the nearest cent. As a check, the value is below the undiscounted maximum of $\$1{,}200(5)=\$6{,}000$, as it must be when payments may stop at death and future payments are discounted. This method avoids treating the payment rate as a single payment at the end of the term.

Piecewise-Constant Mortality

When the force of mortality changes at a specified duration, calculate survival up to the start of each segment using all preceding mortality forces. Within a segment, survival from the segment’s start is then governed by that segment’s force. This gives a sum of integrals, with each later segment multiplied by the probability of surviving to its beginning.

Worked Example: Mortality Changes Partway Through the Term

A hypothetical annuity pays continuously at rate 1 per year for at most 8 years. Assume a constant force of interest $\delta=0.03$. The force of mortality is $0.02$ during years 0 through 5 and $0.05$ during years 5 through 8. Find the expected present value, rounded to four decimal places.

Method. Split the integral at time 5. In the first segment, the combined force of interest and mortality is $0.03+0.02=0.05$. Its contribution is

\int_0^5 e^{-0.05t}\,dt =\frac{1-e^{-0.25}}{0.05} =\frac{1-0.7788007831}{0.05} =4.423984338.

At time 5, the accumulated discount-and-survival factor is $e^{-0.25}=0.7788007831$. During the next 3 years, the combined force is $0.03+0.05=0.08$. The second contribution is therefore

e^{-0.25}\int_0^3 e^{-0.08s}\,ds =0.7788007831\left(\frac{1-e^{-0.24}}{0.08}\right).

Using $e^{-0.24}=0.7866278611$, the integral in parentheses is $(1-0.7866278611)/0.08=2.667151736$. The second contribution is $0.7788007831(2.667151736)=2.077179861$. Thus the total is

$$ 4.423984338+2.077179861=6.501164199. $$

The expected present value is 6.5012, rounded to four decimal places. The check is that the second segment’s contribution includes survival and discounting through the first 5 years. Omitting the factor $e^{-0.25}$ would incorrectly value the later segment as though the life had certainly reached time 5 and the payments had not been discounted before then.

Using a Specified Survival Function

The integral is not limited to constant-force mortality. If the survival function is supplied directly, substitute it into $\int_0^n D(t){}_tp_x\,dt$. For example, under De Moivre’s law with limiting age $\omega$, the future lifetime is uniform on $[0,\omega-x]$, so ${}_tp_x=1-t/(\omega-x)$ for $0\leq t\leq\omega-x$. This is a model assumption, not a general property of human lifetimes.

Worked Example: De Moivre Survival Over a Temporary Term

A hypothetical life aged $x$ is modeled by De Moivre’s law with $\omega-x=10$ years. A continuous annuity pays \$900 per year for at most 5 years. Assume a constant force of interest $\delta=0.04$. Find the expected present value, rounded to the nearest cent.

Method. Over the 5-year term, ${}_tp_x=1-t/10$. The unit-rate value is

\int_0^5 e^{-0.04t}\left(1-\frac{t}{10}\right)\,dt = \int_0^5 e^{-0.04t}\,dt -\frac{1}{10}\int_0^5 t e^{-0.04t}\,dt.

The first integral is $(1-e^{-0.2})/0.04=4.531731172$, using $e^{-0.2}=0.8187307531$. Integration by parts gives $\int_0^n te^{-kt}\,dt=[1-e^{-kn}(1+kn)]/k^2$. Here, the second integral is

\frac{1-e^{-0.2}(1+0.2)}{0.04^2} =\frac{1-(0.8187307531)(1.2)}{0.0016} =\frac{0.0175230963}{0.0016} =10.95193519.

Therefore the unit-rate value is $4.531731172-(10.95193519/10)=3.436537653$. Multiplying by \$900 per year gives $900(3.436537653)=\$3,092.883888$, so the expected present value is \$3,092.88. This result is below the corresponding certain 5-year annuity value because survival is less than 1 for positive durations. The method avoids replacing the specified survival function with a constant force assumption.

Common Mistakes and Interpretation

A temporary continuous life annuity has a payment rate, a fixed term, a survival model, and an interest model. Each plays a distinct role. The term sets the upper integration limit; survival weights whether the payment is made; and the discount factor converts each payment to time 0.

Using only the survival probability at the end of the term. Payments at earlier times require survival only to those times. The correct integrand uses ${}_tp_x$ at each $t$.
Forgetting accumulated survival in a later mortality segment. A later segment’s contribution must include the probability of surviving through every earlier segment.
Using a discrete-payment formula. A continuous stream is valued by integrating over time, not by counting annual payments at integer times.
Confusing a temporary annuity with a guaranteed annuity. A temporary life annuity stops at death. A guaranteed payment period continues according to its guarantee even if death occurs.
Mixing time units. If $t$ is measured in years, the force of interest and mortality must be per year. Converting a monthly rate or duration incorrectly changes the exponent and the value.

A useful check is that the value cannot exceed the present value of a certain continuous annuity paying at the same rate for the full term, when the discount factor is nonnegative and survival probabilities are at most 1. In symbols, ${}_tp_x\leq 1$ implies $\int_0^n D(t){}_tp_x\,dt\leq\int_0^n D(t)\,dt$. This bound can reveal a misplaced survival factor or a term that was inadvertently extended beyond $n$.

Key takeaway: For a temporary continuous life annuity, value each instant’s payment by its discount factor and the probability the life survives to that instant, then integrate only over the fixed term: $\bar{a}_{x:\overline{n}|}=\int_0^n D(t){}_tp_x\,dt$.

Check Your Understanding

For each problem, use the stated survival and interest assumptions. Round monetary values to the nearest cent and unit annuity values to four decimal places.

A unit-rate continuous annuity is payable for 5 years. The force of interest is $0.04$, and the force of mortality is $0.06$, both constant. Find its expected present value. (Answer: 3.9347)
A hypothetical annuity pays \$500 per year continuously for at most 4 years. Assume constant forces $\delta=0.02$ and $\mu=0.08$. Find its expected present value. (Answer: \$1,648.40)
A unit-rate continuous annuity has a 5-year term. The force of interest is $0.02$; the force of mortality is $0.03$ for the first 2 years and $0.08$ for the next 3 years. Find its expected present value. (Answer: 4.2484)
Under De Moivre’s law, $\omega-x=8$. A unit-rate continuous annuity is payable for at most 4 years, and $\delta=0.05$. Find its expected present value. (Answer: 2.7492)
A hypothetical continuous annuity pays \$800 per year for at most 6 years. Assume constant forces $\delta=0.03$ and $\mu=0.07$. Find its expected present value. (Answer: \$3,609.51)

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