Continuous Life-Annuity Cash Flows

Introduction

A life annuity pays while a specified person is alive. In a discrete annuity, payments occur at specified dates; in a continuous life annuity, payments accrue continuously over time. The distinction affects the present value because each portion of the payment stream is discounted according to when it is paid, and payment stops when the annuitant dies.

The central valuation question is therefore: at each future time, what is the chance that the annuitant is still alive to receive the payment then accruing? The answer is expressed directly using survival probabilities. This tutorial develops that expected-present-value formula and illustrates how to use it under different survival models.

Definition and Assumptions

Definition: Let $T_x$ be the future lifetime, in years, of a person currently aged $x$, and let $c$ be a level payment rate in dollars per year. A continuous life annuity pays at rate $c$ for as long as the person is alive. Its present value at age $x$, under a constant force of interest $\delta$, is $$ Z=c\int_0^{T_x} e^{-\delta t}\,dt. $$ The standard whole-life continuous annuity factor is denoted $\bar{a}_x$. It is the expected present value when the payment rate is one dollar per year.

The integral describes an accumulating stream, not a single payment at the end of each year. For example, a payment rate of $c=1{,}200$ dollars per year means that $100$ dollars accrues during each month the annuitant is alive, or equivalently that payments accrue at a uniform rate throughout the year. The total paid in a particular year depends on how much of that year the annuitant survives.

Unless a problem states otherwise, the valuation here uses a fixed effective annual interest rate $i$, expressed as a force of interest $\delta=\ln(1+i)$. Thus, the present-value discount factor for a payment at time $t$ is $e^{-\delta t}$. The model also assumes that the payment rate is fixed, payments cease at death, and no expenses, guarantees, or other benefits are included.

Expected present value: With ${}_tp_x=P(T_x>t)$, the unit-rate whole-life continuous life-annuity value is $$ \bar{a}_x=\int_0^\infty e^{-\delta t}\,{}_tp_x\,dt. $$ At payment rate $c$, the expected present value is $c\bar{a}_x$. The integral requires the survival probability at every future time, not just at integer-year payment dates.

Why Survival Probabilities Enter the Formula

The cash flow at time $t$ occurs only if the annuitant is alive at that time. Its expected contribution is therefore the payment rate multiplied by the discount factor and the probability of being alive. Adding those expected contributions over time gives the expected present value.

Express the cash flow using an indicator.
At time $t$, the payment rate applies if $T_x>t$, and does not apply otherwise. Write this condition as the indicator $\mathbf{1}(T_x>t)$.

Discount the stream.
The random present value is $Z=c\int_0^\infty e^{-\delta t}\mathbf{1}(T_x>t)\,dt$. This integral stops contributing after death because its indicator is then zero.

Take expectations.
The integrand is nonnegative, so expectation and integration can be interchanged. Since $E[\mathbf{1}(T_x>t)]={}_tp_x$, the result is $E[Z]=c\int_0^\infty e^{-\delta t}{}_tp_x\,dt=c\bar{a}_x$.

This derivation explains why survival probabilities, rather than probabilities of death during a year, appear in the integrand: the payment accruing at time $t$ depends on being alive at time $t$. The formula is general for a level payment rate and constant force of interest; it does not require a constant force of mortality. A model that supplies survival probabilities only at integer durations is not, by itself, enough to determine this continuous value unless an assumption is made about survival between those durations.

Worked Valuations

Worked Example: Constant Mortality and a Level Payment Rate

A hypothetical whole-life continuous annuity pays at a rate of $1{,}000$ dollars per year to a person aged $x$ while alive. Assume a constant force of mortality $\mu=0.07$ per year and an effective annual interest rate $i= e^{0.03}-1$, so the force of interest is $\delta=0.03$. Find the expected present value.

Method. Use the survival function implied by the constant mortality force, then integrate the discounted survival-weighted payment rate. The constant force assumption means ${}_tp_x=e^{-\mu t}$ for $t\geq 0$.

Step 1: Combine interest and survival. At time $t$, the discounted probability-weighted payment rate is $1{,}000e^{-0.03t}e^{-0.07t}=1{,}000e^{-0.10t}$ dollars per year.

Step 2: Integrate over the lifetime.

E[Z]=1{,}000\int_0^\infty e^{-0.10t}\,dt =1{,}000\left[\frac{-e^{-0.10t}}{0.10}\right]_0^\infty =1{,}000\left(\frac{1}{0.10}\right).

As $t$ tends to infinity, $e^{-0.10t}$ tends to zero, which gives the stated integral. Therefore, $E[Z]=\$10{,}000.00$. The unit-rate annuity factor is $\bar{a}_x=10$ years, and multiplying it by $1{,}000$ dollars per year gives dollars. This calculation uses the full continuous stream, not annual payments at year-end.

Check. Both interest and mortality reduce the expected contribution at time $t$, so their exponential factors multiply. The resulting combined rate, $0.10$, is greater than either $0.03$ or $0.07$, as expected.

Worked Example: Using a Non-Exponential Survival Function

A hypothetical annuity pays continuously at a rate of $600$ dollars per year. The force of interest is $\delta=0.06$, and the future lifetime has survival function ${}_tp_x=0.8e^{-0.04t}+0.2e^{-0.16t}$ for $t\geq0$. Find the expected present value.

Method. Substitute the given survival function directly into the expected-present-value integral. No constant-mortality assumption is needed. The stated survival function is valid at time zero because $0.8+0.2=1$, and it decreases to zero as $t$ increases.

Step 1: Form the discounted survival-weighted rate.

600e^{-0.06t}{}_tp_x =600\left(0.8e^{-0.10t}+0.2e^{-0.22t}\right).

The first exponent is $0.06+0.04=0.10$; the second is $0.06+0.16=0.22$.

Step 2: Integrate each term separately. For a positive constant $k$, $\int_0^\infty e^{-kt}\,dt=1/k$. Hence

E[Z] =600\left(0.8\frac{1}{0.10}+0.2\frac{1}{0.22}\right) =600\left(8+\frac{10}{11}\right) =600\left(\frac{98}{11}\right) =\$5{,}345.45.

The displayed dollar value is rounded to the nearest cent; the unrounded value is $5{,}345.4545\ldots$. The unit-rate factor is $98/11$, and multiplying by $600$ dollars per year gives the expected present value in dollars.

Check. The survival function is a weighted sum of two exponential terms, so each term can be integrated after adding the force of interest to its exponent. Replacing the given survival function with a single exponential would impose an unsupported constant-mortality model.

Worked Example: Mortality Force Changes After Three Years

A hypothetical whole-life continuous annuity pays at a rate of $800$ dollars per year. The force of interest is $\delta=0.05$. The force of mortality is $0.10$ per year for the first three years and $0.20$ per year thereafter. Find the expected present value.

Method. Split the integral where the mortality assumption changes. For $0\leq t\leq3$, ${}_tp_x=e^{-0.10t}$. For $t>3$, survival must include both survival through the first three years and survival in the period after year three: ${}_tp_x=e^{-0.30}e^{-0.20(t-3)}$.

Step 1: Value the first three years of accrual.

800\int_0^3 e^{-0.05t}e^{-0.10t}\,dt =800\int_0^3 e^{-0.15t}\,dt =800\frac{1-e^{-0.45}}{0.15}.

Using $e^{-0.45}=0.63762815$, this contribution is $800(0.36237185/0.15)=\$1{,}932.65$.

Step 2: Value accrual after year three. Set $s=t-3$, so $s$ starts at zero at the change in mortality force. At time $3+s$, the discounted survival factor is $e^{-0.05(3+s)}e^{-0.30}e^{-0.20s}=e^{-0.45}e^{-0.25s}$. Thus the tail contribution is

800e^{-0.45}\int_0^\infty e^{-0.25s}\,ds =800\frac{e^{-0.45}}{0.25} =800\frac{0.63762815}{0.25} =\$2{,}040.41.

Step 3: Add the contributions. The expected present value is $\$1{,}932.65+\$2{,}040.41=\$3{,}973.06$, rounded to the nearest cent.

Check. At the start of the tail, the combined discount-and-survival factor is $e^{-0.45}$: $e^{-0.15}$ from interest over three years and $e^{-0.30}$ from survival over those years. The tail contribution therefore begins from the correctly accumulated first-three-year factors.

Common Mistakes and Exam Traps

Continuous annuity calculations are especially sensitive to the time origin and to what a payment rate means. A small error in either can change the whole tail of an integral.

Using a death probability instead of a survival probability. A payment at time $t$ is made if the annuitant is alive then, so its expected contribution uses ${}_tp_x$, not ${}_tq_x$. The two probabilities describe different events.
Treating a payment rate as a payment at each integer year. A rate of $800$ dollars per year accrues throughout the year. It is not an $800$-dollar payment made only at year-end.
Forgetting survival before a change in mortality force. After year three in the preceding example, survival is $e^{-0.30}e^{-0.20(t-3)}$, not merely $e^{-0.20(t-3)}$. The latter incorrectly assumes the annuitant is certain to have survived to year three.
Changing the time origin without changing the discount factor. In the tail, writing $s=t-3$ makes the integral begin at zero, but the factor $e^{-0.45}$ must remain outside it. Omitting it values the future stream as if the annuitant were alive at time zero and no discount had occurred before year three.

A related algebra trap is to rewrite the tail using $t$ while dropping a factor. In the piecewise-mortality example, $e^{-0.05t}e^{-0.30}e^{-0.20(t-3)}=e^{0.30}e^{-0.25t}$, not $e^{-0.30}e^{-0.25t}$. The factor $e^{0.30}$ appears because $e^{-0.20(t-3)}=e^{0.60}e^{-0.20t}$. The shifted-time form $e^{-0.45}e^{-0.25s}$, with $s=t-3$, is often less error-prone.

Key takeaway: For a level continuous payment rate $c$, discount the rate at each time and weight it by the probability the annuitant is alive at that time. Under constant force of interest, the expected present value is $c\int_0^\infty e^{-\delta t}{}_tp_x\,dt$. At a change in mortality assumptions, preserve all survival and discount factors accumulated before the change.

Check Your Understanding

For each problem, use the specified payment rate, interest force, and survival model. Report dollar answers to the nearest cent.

A hypothetical whole-life continuous annuity pays $500$ dollars per year. The force of interest is $0.04$, and ${}_tp_x=e^{-0.06t}$. Find its expected present value. (Answer: $\$5{,}000.00$)
A continuous annuity pays $900$ dollars per year while the annuitant is alive. Assume $\delta=0.03$ and constant mortality force $\mu=0.02$. Derive the discounted survival-weighted rate and find the expected present value. (Answer: $\$18{,}000.00$)
A hypothetical annuity pays $1{,}000$ dollars per year. The force of interest is $0.02$; mortality force is $0.04$ for the first two years and $0.10$ thereafter. Split the valuation at year two and find the expected present value. (Answer: $\$9{,}275.66$)
A continuous annuity pays $800$ dollars per year, with $\delta=0.05$ and survival function ${}_tp_x=0.6e^{-0.03t}+0.4e^{-0.13t}$. Integrate the two discounted terms separately to find the expected present value. (Answer: $\$7{,}777.78$)

What You'll Learn