Deriving Continuous Annuity Values from Survival Functions

Why Continuous Annuity Values Are Integrals

A continuous life annuity pays an amount gradually over time rather than as a single payment at the end of each year. To value it, consider a small payment made at time $t$: it is discounted back to the valuation date, and it is paid only if the annuitant is alive then. The present value is therefore built by adding discounted payment rates across time, weighted by the probability of being alive at each time.

The key is to distinguish the payment rate from the total payment. A rate of $b$ dollars per year paid continuously for a full year produces $b$ dollars in total. If death occurs during that year, only the portion paid while the annuitant is alive is received.

Definition: Let $T_x$ be the future lifetime of a life aged $x$, and let ${}_tp_x=P(T_x>t)$ be the probability that the life survives at least $t$ years. Under a constant force of interest $\delta$, a continuous life annuity paying at rate 1 per year while the life is alive has expected present value \[ \bar{a}_x=\int_0^\infty e^{-\delta t}{}_tp_x\,dt. \] For a temporary annuity payable for at most $n$ years, the corresponding value is \[ \bar{a}_{x:\overline{n}|}=\int_0^n e^{-\delta t}{}_tp_x\,dt. \]

Here $\bar{a}_x$ and $\bar{a}_{x:\overline{n}|}$ are values per unit of annual payment rate. If the rate is a constant $b$, multiply the appropriate integral by $b$. If the rate varies with time, place the rate $b(t)$ inside the integral. These formulas assume the stated interest force applies throughout the payment period and that payments stop when the life dies or, for a temporary annuity, when the term ends.

Deriving the Survival-Weighted Integral

The integral follows directly from the present value of the actual payments. It is useful to make that random present value explicit before taking its expectation.

Represent whether a payment is made.
At time $t$, let $\mathbf{1}_{\{T_x>t\}}$ equal 1 if the annuitant is alive and 0 otherwise. A payment rate $b(t)$ therefore contributes $b(t)e^{-\delta t}\mathbf{1}_{\{T_x>t\}}dt$ to present value over a short interval of length $dt$.

Write the random present value.
If payments are due over an interval $I$, the present value for a particular lifetime is $Z=\int_I b(t)e^{-\delta t}\mathbf{1}_{\{T_x>t\}}dt$.

Take expectations.
The integrand is nonnegative, so the expectation can be interchanged with the integral. Since $E[\mathbf{1}_{\{T_x>t\}}]=P(T_x>t)={}_tp_x$, it follows that $E[Z]=\int_I b(t)e^{-\delta t}{}_tp_x\,dt$.

Thus the integral is not an arbitrary valuation rule: it is the expected present value of the stream of payments. Each time-$t$ payment is discounted by $e^{-\delta t}$, then weighted by the probability it is actually made. For a continuous annuity deferred $m$ years and payable for at most $n$ years, the payment interval is $[m,m+n]$, so its value is $\int_m^{m+n} b(t)e^{-\delta t}{}_tp_x\,dt$. For a deferred whole-life annuity, the upper limit is infinity.

The survival probability must match the life and the valuation date. In the notation of the survival function $S_0$, if $S_0(x)>0$, then \[ {}_tp_x=\frac{S_0(x+t)}{S_0(x)}. \] This is conditional survival: it describes a life already known to have reached age $x$. It is not generally correct to use $S_0(x+t)$ alone when valuing a contract issued to a life aged $x$.

Worked Applications

Worked Example: A Temporary Annuity with Stepwise Survival

A hypothetical two-year continuous annuity pays at a rate of $1,000 per year while the annuitant is alive. Its future lifetime $T$ has survival function $P(T>t)=1$ for $0\leq t<1$, $0.6$ for $1\leq t<2$, and 0 for $t\geq2$. The effective annual interest rate is $i=100\%$, so the constant force is $\delta=\ln(1+i)=\ln 2$. Find the expected present value at time 0.

Method. Use the temporary-annuity integral with the survival probability on each time interval. The discount factor is $e^{-\delta t}=2^{-t}$. The survival probability is 1 during the first year and 0.6 during the second.

\begin{aligned} E[Z] &=1000\left(\int_0^1 2^{-t}\,dt+0.6\int_1^2 2^{-t}\,dt\right)\\ &=1000\left(\frac{1-0.5}{\ln 2} +0.6\frac{0.5-0.25}{\ln 2}\right)\\ &=1000\left(\frac{0.5+0.15}{\ln 2}\right) =\frac{650}{\ln 2} \approx \$937.75. \end{aligned}

The annuity factor per dollar of annual payment rate is $0.65/\ln 2\approx0.93775$. Multiplying by $1,000 gives $937.75, rounded to the nearest cent. The common mistake avoided here is to weight both years by survival probability 1: after the first year, only 0.6 of the hypothetical population remains alive.

Worked Example: Conditioning Survival at the Issue Age

A hypothetical life annuity is issued to a person aged $x$. The survival function satisfies $S_0(x)=0.8$, $S_0(x+t)=0.8$ for $0\leq t<1$, and $S_0(x+t)=0.4$ for $1\leq t<2$. The annuity pays continuously at a rate of $1,500 per year for at most two years. The effective annual interest rate is $i=100\%$, so $\delta=\ln 2$. Find its expected present value at age $x$.

Method. Convert the unconditional survival-function values into probabilities conditional on being alive at age $x$. During the first year, ${}_tp_x=0.8/0.8=1$. During the second year, ${}_tp_x=0.4/0.8=0.5$.

\begin{aligned} E[Z] &=1500\left(\int_0^1 2^{-t}\,dt +0.5\int_1^2 2^{-t}\,dt\right)\\ &=1500\left(\frac{0.5}{\ln 2} +0.5\frac{0.25}{\ln 2}\right)\\ &=\frac{1500(0.625)}{\ln 2} =\frac{937.5}{\ln 2} \approx \$1,352.53. \end{aligned}

The result is rounded to the nearest cent. A common error is to use the unconditional values $0.8$ and $0.4$ directly as the probabilities for the life aged $x$. That would give $1500(0.8(0.5)+0.4(0.25))/\ln 2=750/\ln 2\approx\$1,082.02$, which is too low. The contract is valued conditional on the person having reached age $x$, so the correct probabilities are the ratios $S_0(x+t)/S_0(x)$.

Worked Example: A Deferred Annuity with Changing Payment Rates

A hypothetical annuity pays continuously at a rate of $800 per year from time 0 to 1, $1,200 per year from time 1 to 2, and $400 per year from time 2 to 3, provided the annuitant is alive. The future-lifetime survival probabilities are 1 on $[0,1)$, 0.5 on $[1,2)$, and 0.25 on $[2,3)$, and 0 from time 3 onward. The effective annual interest rate is $i=100\%$, hence $\delta=\ln 2$. Find the expected present value.

Method. Split the integral at each point where either the payment rate or survival probability changes. On each interval, both are constant, while the discount factor remains $2^{-t}$.

\begin{aligned} E[Z] &=800\int_0^1 2^{-t}\,dt +1200(0.5)\int_1^2 2^{-t}\,dt +400(0.25)\int_2^3 2^{-t}\,dt\\ &=\frac{800(0.5)+1200(0.5)(0.25)+400(0.25)(0.125)}{\ln 2}\\ &=\frac{400+150+12.5}{\ln 2} =\frac{562.5}{\ln 2} \approx \$811.52. \end{aligned}

The expected present value is $811.52, rounded to the nearest cent. Each term uses the payment rate for that interval, the probability of being alive during it, and the discount integral over that same interval. This setup avoids applying one payment rate or one survival probability to the entire three-year period.

Common Mistakes and Checks

The integral makes the timing and survival assumptions visible, but several errors can still arise when setting it up. Before calculating, identify the valuation age, the payment interval, the payment rate on each part of that interval, and the relevant survival probability.

Using annual-payment timing. A continuous annuity is not valued by summing payments at integer times. Its payments occur throughout the interval, so the payment rate is integrated over time.
Forgetting the probability weight. Discounting alone values a guaranteed stream. A life-contingent stream also requires the factor ${}_tp_x$.
Using unconditional survival at age $x$. Use $S_0(x+t)/S_0(x)$ when the life is already known to be age $x$ at the valuation date.
Starting a deferred annuity too early. If payments are deferred $m$ years, the integral starts at $m$, not at 0. Survival to each payment time is still included through ${}_tp_x$.
Mixing up payment rate and total payment. A rate measured in dollars per year is integrated over time. Do not multiply it by the term a second time after evaluating the integral.

A quick reasonableness check is to compare a life annuity with a guaranteed annuity having the same payment rate and term. Since $0\leq{}_tp_x\leq1$, multiplying by survival probabilities cannot increase the value: $\int_I b(t)e^{-\delta t}{}_tp_x\,dt\leq\int_I b(t)e^{-\delta t}\,dt$ for nonnegative $b(t)$. This check applies when both streams have the same payment interval, rates, and interest basis.

Key takeaway: The expected present value of continuous life-annuity payments is the integral of payment rate times discount factor times probability of survival to the payment time. First choose the correct time interval; then use survival conditional on the valuation age.

Check Your Understanding

For each hypothetical contract, set up the survival-weighted integral and check the numerical answer.

A two-year continuous annuity pays $1,000 per year while alive. The survival probability is 1 on $[0,1)$ and 0.4 on $[1,2)$, and $\delta=\ln 2$. Find the expected present value. (Answer: $865.62)
For a life aged $x$, $S_0(x)=0.9$, $S_0(x+t)=0.9$ on $[0,1)$, and $S_0(x+t)=0.45$ on $[1,2)$. A two-year annuity pays continuously at $800 per year and $\delta=\ln 2$. Find its value at age $x$. (Answer: $721.35)
A continuous annuity is deferred one year and then pays $1,000 per year through time 3, provided the annuitant is alive. Survival is 1 on $[0,1)$, 0.8 on $[1,2)$, and 0.4 on $[2,3)$; $\delta=\ln 2$. Find the expected present value at time 0. (Answer: $360.67)
A three-year continuous annuity pays $600 per year. Survival probabilities are 0.9 on $[0,1)$, 0.6 on $[1,2)$, and 0.3 on $[2,3)$, and $\delta=\ln 2$. Find its expected present value. (Answer: $551.83)
A two-year annuity pays at a rate of $1,000 per year during year 1 and $500 per year during year 2, while the annuitant is alive. Survival is 1 on $[0,1)$ and 0.5 on $[1,2)$, and $\delta=\ln 2$. Find its expected present value. (Answer: $811.52)

What You'll Learn