Deriving Continuous Retirement Annuity Values

Why Continuous Retirement Annuity Values Use Survival Integrals

A pension paid continuously after retirement generates payments only while the retired member is alive. Its expected present value therefore depends on two quantities at every future duration: the discounted value of a payment made then, and the probability that the member is still alive to receive it. Integrating their product gives the value at the date of retirement.

This is a conditional value: it assumes the member has reached retirement and is entitled to the pension. Any probability of reaching retirement from an earlier active state belongs to a broader valuation that combines pre-retirement decrements with the post-retirement pension. It is not part of the retirement annuity value itself.

Definition: Suppose a member retires at age $r$. Let ${}_tp_r^{(R)}$ denote the probability that the member survives for at least $t$ years after retirement, on the applicable post-retirement mortality basis. If the pension is paid continuously at rate $b(t)$ dollars per year while the member is alive, and the force of interest is the constant $\delta$, its expected present value at retirement is $$ \operatorname{EPV}_r=\int_0^\infty b(t)e^{-\delta t}{}_tp_r^{(R)}\,dt. $$ The superscript $R$ labels the retired-member survival basis; it does not indicate another decrement. For a level pension with payment rate $b$, set $b(t)=b$.

The rate $b$ is an annual payment rate, not a single annual payment made at the start or end of each year. Over a short interval of length $dt$, the member receives approximately $b\,dt$ if alive. The formula adds the expected discounted value of these small payments over all future durations.

Deriving the Continuous Pension Formula

First consider a level pension rate $b$. At duration $t$, the payment in an interval of length $dt$ is $b\,dt$ if the member is alive at that time, and zero otherwise. Its expected amount is $b\,{}_tp_r^{(R)}dt$. Discounting that expected payment to retirement multiplies it by $e^{-\delta t}$. Summing over small intervals and taking the limit gives the integral below.

\operatorname{EPV}_r =\lim_{\max \Delta t\to 0}\sum_j b\,{}_{t_j}p_r^{(R)}e^{-\delta t_j}\Delta t =\int_0^\infty b\,{}_{t}p_r^{(R)}e^{-\delta t}\,dt.

If the post-retirement force of mortality at duration $s$ is $\mu_{r+s}^{(R)}$, then survival to duration $t$ is ${}_tp_r^{(R)}=\exp\{-\int_0^t\mu_{r+s}^{(R)}\,ds\}$, provided the mortality model is continuous with that force. For a constant force $\mu$, this becomes $e^{-\mu t}$. This survival expression uses mortality after retirement; it does not substitute an active-member multiple-decrement probability.

When both the payment rate $b$, the force of interest $\delta$, and the post-retirement force of mortality $\mu$ are constant, the level whole-life value has a simple closed form. The conditions matter: the pension must continue for life, the payment rate must remain level, and the combined force $\delta+\mu$ must be positive.

\operatorname{EPV}_r =b\int_0^\infty e^{-\delta t}e^{-\mu t}\,dt =b\int_0^\infty e^{-(\delta+\mu)t}\,dt =\frac{b}{\delta+\mu}.

The mortality force appears alongside the interest force because future payments are less valuable both when they are further in the future and when the member may not survive to receive them. For a temporary pension payable only for $n$ years, the upper limit is $n$, not infinity. With constant forces, the value is $b[1-e^{-(\delta+\mu)n}]/(\delta+\mu)$.

Worked Example: Level Whole-Life Pension

A hypothetical plan pays a retired member a continuous pension at the rate of $12,000 per year for life. At retirement, assume a constant force of interest $\delta=0.04$ and a constant post-retirement force of mortality $\mu=0.06$. Find the expected present value at retirement.

Method. Since the payment rate and both forces are constant, use the whole-life integral. Under the stated mortality assumption, survival to duration $t$ is $e^{-0.06t}$, and the discount factor is $e^{-0.04t}$.

\begin{aligned} \operatorname{EPV}_r &=\int_0^\infty 12{,}000e^{-0.04t}e^{-0.06t}\,dt\\ &=12{,}000\int_0^\infty e^{-0.10t}\,dt\\ &=\frac{12{,}000}{0.10}\\ &=120{,}000. \end{aligned}

The expected present value is $120,000.00. The calculation checks because the combined force is $0.04+0.06=0.10$, and $12{,}000/0.10=120{,}000$. A common mistake is to discount the continuous stream but omit survival; that would give $12{,}000/0.04=$300,000$, which values payments as though the member were certain to remain alive.

Temporary and Increasing Continuous Pensions

A pension may be limited to a fixed period, or its payment rate may change with time. The integral definition accommodates both situations directly. For a temporary pension, the payment rate is zero after the guarantee or payment period if no further benefit is due. For an increasing pension, the time-varying rate belongs inside the integral; it should not be replaced by a level rate unless that is an explicitly chosen approximation.

Worked Example: Temporary Pension With Constant Forces

A hypothetical retiree receives $18,000 per year continuously while alive, but payments stop after 10 years. Assume a constant force of interest $\delta=0.05$ and a constant post-retirement force of mortality $\mu=0.03$. Calculate the value at retirement.

Method. Integrate only over the 10-year payment period. The probability of survival to duration $t$ is $e^{-0.03t}$, and discounting is $e^{-0.05t}$, so the combined exponential rate is $0.08$.

\begin{aligned} \operatorname{EPV}_r &=\int_0^{10}18{,}000e^{-0.05t}e^{-0.03t}\,dt\\ &=18{,}000\int_0^{10}e^{-0.08t}\,dt\\ &=18{,}000\left(\frac{1-e^{-0.08(10)}}{0.08}\right)\\ &=18{,}000\left(\frac{1-e^{-0.8}}{0.08}\right). \end{aligned}

Using $e^{-0.8}=0.4493289641$, the factor is $(1-0.4493289641)/0.08=6.883387949$. Thus the value is $18{,}000(6.883387949)=$123,900.98$, rounded to the nearest cent.

This result can be checked against the whole-life value under the same forces, $18{,}000/0.08=$225,000$. A 10-year temporary pension must be worth less than that whole-life pension, as it excludes all payments after year 10. A common error is to use an upper limit of infinity and thereby include payments beyond the stated period.

For a rate that increases continuously at force $g$, a common model is $b(t)=b_0e^{gt}$, where $b_0$ is the initial annual payment rate. Under constant $\delta$ and $\mu$, the integrand is $b_0e^{-(\delta+\mu-g)t}$. The whole-life integral is finite only if $\delta+\mu-g>0$. If this condition fails, the mathematical expected present value under the assumed infinite-horizon model is not finite.

Worked Example: Increasing Pension With a Change in Mortality Force

A hypothetical pension starts at a continuous payment rate of $15,000 per year and increases at a continuous rate of $g=0.02$, so its rate at duration $t$ is $15{,}000e^{0.02t}$. Assume a constant force of interest $\delta=0.04$. The post-retirement force of mortality is $0.04$ for the first five years and $0.08$ thereafter. Find the whole-life value at retirement.

Method. Split the integral at year 5, where the mortality force changes. In the first interval, payment growth less interest and mortality gives exponent rate $0.02-0.04-0.04=-0.06$. At year 5, the accumulated exponent is $-0.06(5)=-0.30$. After year 5, the exponent decreases at rate $0.02-0.04-0.08=-0.10$. Thus the post-five-year part begins with the survival-and-discount-and-growth factor $e^{-0.30}$, not a newly reset factor of 1.

\begin{aligned} \operatorname{EPV}_r &=15{,}000\left[ \int_0^5 e^{-0.06t}\,dt +e^{-0.30}\int_5^\infty e^{-0.10(t-5)}\,dt \right]\\ &=15{,}000\left[ \frac{1-e^{-0.30}}{0.06} +\frac{e^{-0.30}}{0.10} \right]. \end{aligned}

Using $e^{-0.30}=0.7408182207$, the first factor is $(1-0.7408182207)/0.06=4.319696322$, and the second is $0.7408182207/0.10=7.408182207$. Their unrounded sum is $11.727878529$. Therefore the value is $15{,}000(11.727878529)=$175,918.18$, rounded to the nearest cent.

The calculation checks by examining the two intervals: the first contributes approximately $64,795.44 and the second approximately $111,122.73, for a total of $175,918.18 when added using the unrounded amounts. The transition factor $e^{-0.30}$ carries the accumulated effect through the first five years. Omitting it would treat the member as newly alive and the accumulated payments as undiscounted at year 5.

Conditional Value, Assumptions and Common Traps

A retirement annuity value is conditional on being alive and entitled to the pension at retirement. The survival probability in the integral starts from the retired member at duration zero, so ${}_0p_r^{(R)}=1$. Do not multiply the conditional value by the probability that an active member reaches retirement unless the quantity being sought is explicitly an unconditional value from an earlier valuation date.

The model also assumes that the stated payment rate is payable while alive and that the stated mortality basis governs survival after retirement. If payment instead continues after death to a survivor, or stops for a reason other than death, the payment-contingency model must reflect that benefit design. The simple life annuity integral does not value such additional contingent payments.

Specify the valuation date and conditioning.
For a value at retirement, set duration zero at retirement and treat entitlement at that date as given.

Write the payment rate.
Use dollars per year as a function of duration, and set it to zero after any stated payment term.

Write post-retirement survival.
Use the retired-member mortality assumption and account for every change in force over time.

Discount and integrate.
Multiply the payment rate by survival and discounting, then integrate over exactly the period in which payments may be made.

A frequent exam trap is to apply an active-member decrement probability to a pension already valued conditional on retirement. Another is to use the annual effective rate $i$ directly as though it were the force of interest: continuous discounting uses $e^{-\delta t}$, with $\delta=\ln(1+i)$ when converting from an effective annual rate. In all cases, state the interest and mortality assumptions before applying a simplified formula.

Key takeaway: The conditional value at retirement is the integral of the continuous payment rate multiplied by post-retirement survival and discounting. Constant-force closed forms are shortcuts derived from that integral; piecewise assumptions require carrying the accumulated survival and discount factor across each interval boundary.

Check Your Understanding

For each problem, calculate a value at retirement unless the question states otherwise. Use the assumptions given and round dollar answers to the nearest cent.

A hypothetical plan pays a level continuous pension of $9,000 per year for life. The force of interest is $0.03$, and the post-retirement force of mortality is $0.07$, both constant. Find the value at retirement. (Answer: $90,000.00)
A hypothetical pension pays $15,000 per year continuously while alive for at most 12 years. Assume constant forces $\delta=0.04$ and $\mu=0.06$. Find its value. (Answer: $104,820.87)
A hypothetical whole-life pension pays continuously at a level rate of $20,000 per year. Assume constant forces $\delta=0.03$ and $\mu=0.02$. Find its value. (Answer: $400,000.00)
A hypothetical whole-life pension has payment rate $10{,}000e^{0.01t}$ dollars per year at duration $t$. The constant forces are $\delta=0.03$ and $\mu=0.04$. Find its value and verify that the integral converges. (Answer: $166,666.67; the net exponent rate is $-0.06$, so it converges.)
A hypothetical pension pays $12,000 per year continuously for life. The force of interest is $0.02$; the mortality force is $0.03$ for the first four years and $0.08$ thereafter. Write the split-integral expression and calculate the value. (Answer: $12{,}000[(1-e^{-0.20})/0.05+e^{-0.20}/0.10]=$141,752.31$)

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