Deferred Continuous Annuities

Introduction

A continuous payment stream can begin well after the valuation date and stop at a specified future time. Its value depends on both the length of the payment period and how far into the future that period begins. Keeping those two timing details separate is essential: the deferral period locates the start, while the duration determines how long payments continue.

This tutorial values a level continuous stream under a constant force of interest. The method is to discount each payment from its payment time to the valuation date and integrate over the actual interval during which payments are made.

Definition: A deferred level continuous annuity pays continuously at a constant rate $R$ dollars per year from time $m$ to time $n$, where $0\leq m<n$. The deferral period is $m$, the payment duration is $n-m$, and the payment intensity is $R$. The endpoint convention does not affect the value: a single instant has zero length in a continuous stream.

Present Value Under a Constant Force

Assume a constant force of interest $\delta>0$ per year over the whole interval from the valuation date to the end of the stream. A payment made at time $t$ has time-0 value $e^{-\delta t}$ times its amount. Since the stream pays $R$ dollars per year, its payment over a short interval of length $dt$ is $R\,dt$. Adding the discounted values of these payments gives the present value.

\operatorname{PV}_0 =\int_m^n R e^{-\delta t}\,dt =\frac{R}{\delta}\left(e^{-\delta m}-e^{-\delta n}\right) =R e^{-\delta m}\frac{1-e^{-\delta(n-m)}}{\delta}.

The last form separates the two timing effects. The factor $e^{-\delta m}$ discounts from the stream’s start time $m$ to time 0. The remaining factor values a level continuous annuity of duration $n-m$ at its start time. Thus the stream’s value at time $m$, immediately before it begins, is

\operatorname{PV}_m =\int_m^n R e^{-\delta(t-m)}\,dt =R\frac{1-e^{-\delta(n-m)}}{\delta}.

Discounting this start-date value by $e^{-\delta m}$ gives the time-0 value. This two-stage view is often useful, but it does not change the payment interval: the duration remains $n-m$, not $n$.

Condition: These formulas assume a level payment intensity and a constant force over the entire relevant interval. If the force or payment intensity varies with time, the integral must use that time-varying quantity; the closed form above does not apply unchanged. If $\delta=0$, there is no discounting and the value is $R(n-m)$, the limit of the formula as $\delta$ approaches zero.

Why the Formula Works

The formula follows directly from the definition of present value as the sum of discounted payments. Divide the payment interval into short pieces. On a piece of length $dt$ near time $t$, the amount paid is approximately $R\,dt$, and its present value is approximately $R e^{-\delta t}dt$. As the pieces become arbitrarily short, the sum becomes the integral over the interval from $m$ to $n$.

Identify the actual payment interval.
Record the start time $m$ and the end time $n$. The payment duration is $n-m$.

Discount a payment at time $t$.
Under a constant force $\delta$, a payment at time $t$ is multiplied by $e^{-\delta t}$ to obtain its time-0 value.

Integrate over the payment interval.
Integrate $R e^{-\delta t}$ from $m$ to $n$, then evaluate and round the final dollar amount.

Evaluating the integral gives the stated formula because

\int_m^n R e^{-\delta t}\,dt =R\left[-\frac{1}{\delta}e^{-\delta t}\right]_m^n =\frac{R}{\delta}\left(e^{-\delta m}-e^{-\delta n}\right).

Worked Valuations

Worked Example: A Stream Deferred for Two Years

A hypothetical contract pays continuously at $R=\$1{,}000$ per year from time 2 through time 5. The constant force of interest is $\delta=0.06$ per year. Find the value at time 0 and check it by first valuing the stream at time 2.

Method. The payment interval is $2\leq t\leq5$, so $m=2$, $n=5$, and the duration is $5-2=3$ years. Apply the deferred continuous-annuity formula:

\operatorname{PV}_0 =1{,}000e^{-0.06(2)} \frac{1-e^{-0.06(3)}}{0.06}.

The required exponential values are $e^{-0.12}\approx0.886920437$ and $e^{-0.18}\approx0.835270211$. Equivalently, the value can be calculated as

\operatorname{PV}_0 =1{,}000\frac{e^{-0.12}-e^{-0.30}}{0.06} =1{,}000\frac{0.886920437-0.740818221}{0.06} \approx \$2{,}435.04.

Check at time 2. At the stream’s start, the value of its remaining three years of payments is

\operatorname{PV}_2 =1{,}000\frac{1-e^{-0.18}}{0.06} =1{,}000\frac{1-0.835270211}{0.06} \approx \$2{,}745.50.

Discounting to time 0 gives $\$2{,}745.4965e^{-0.12}\approx\$2{,}435.04$, agreeing with the direct calculation. The check confirms that the three-year duration and two-year deferral have both been used correctly.

Worked Example: Do Not Confuse the End Time with the Duration

A hypothetical fund pays continuously at $\$1{,}200$ per year from time 2 to time 5. The constant force is $\delta=0.08$ per year. Find the value at time 0 and the value at time 2.

Method. The duration is $5-2=3$ years. At time 2, value the three years of payments; then discount that amount two years to time 0.

\operatorname{PV}_2 =1{,}200\frac{1-e^{-0.08(3)}}{0.08} =1{,}200\frac{1-e^{-0.24}}{0.08}.

Since $e^{-0.24}\approx0.786627861$, this is

\operatorname{PV}_2 =1{,}200\frac{0.213372139}{0.08} \approx \$3{,}200.58.

Now discount to time 0. Using $e^{-0.16}\approx0.852143789$,

\operatorname{PV}_0 =3{,}200.5821e^{-0.16} \approx \$2{,}727.36.

The direct calculation agrees:

\operatorname{PV}_0 =1{,}200\frac{e^{-0.16}-e^{-0.40}}{0.08} =1{,}200\frac{0.852143789-0.670320046}{0.08} \approx \$2{,}727.36.

Common mistake avoided. Treating “ends at time 5” as “lasts five years” incorrectly gives a five-year duration and extends payments to time 7. That wrong calculation is $1{,}200e^{-0.16}(1-e^{-0.40})/0.08\approx\$4{,}214.02$. It does not value the stated contract. The check is that the time-0 value must use the interval from time 2 to time 5, not from time 2 to time 7.

Valuing at a Date After Payments Have Begun

If the valuation date lies inside the payment interval, it is important to specify whether the question asks for the value of the remaining payments or the value at that date of the entire stream. The remaining-payment value excludes payments already made. The entire-stream value includes those earlier payments accumulated to the valuation date as well as the future payments discounted to it.

Worked Example: Remaining Payments Versus the Entire Stream

A hypothetical stream pays continuously at $\$700$ per year from time 1 to time 4. The constant force is $\delta=0.08$ per year. Find (a) the time-2 value of the remaining payments, (b) the time-0 value of the full stream, and (c) a check by accumulating the full stream to time 2 and discounting that value to time 0.

(a) Remaining payments at time 2. Only payments from time 2 to time 4 are included:

V_{\text{remaining},2} =700\int_2^4 e^{-0.08(t-2)}\,dt =700\frac{1-e^{-0.16}}{0.08} =700\frac{0.147856211}{0.08} \approx \$1{,}293.74.

(b) Full stream at time 0. The original payment interval is time 1 to time 4, a duration of three years:

\operatorname{PV}_0 =700\frac{e^{-0.08}-e^{-0.32}}{0.08} =700\frac{0.923116346-0.726149037}{0.08} \approx \$1{,}723.46.

(c) Check using the full time-2 value. Payments made from time 1 to time 2 accumulate to time 2. Their accumulated value is

V_{\text{past},2} =700\int_1^2 e^{0.08(2-t)}\,dt =700\frac{e^{0.08}-1}{0.08} =700\frac{1.083287068-1}{0.08} \approx \$728.76.

Thus the value at time 2 of the entire stream is $\$728.76+\$1{,}293.74=\$2{,}022.50$. Discounting that total by two years gives

2{,}022.5037e^{-0.16}\approx \$1{,}723.46,

which agrees with the direct time-0 value. Discounting only the remaining-payment value at time 2 would omit the payments from time 1 to time 2, so it would not verify the value of the full stream.

Common Timing Traps

The integral makes timing explicit, but errors can still arise when translating the wording of a problem into its limits. Before calculating, mark the valuation date, the first payment time, and the final payment time on a timeline. Then check the formula’s exponents against those dates.

Using the end time as the duration. If payments run from time $m$ to time $n$, the duration is $n-m$, not $n$.
Discounting from the wrong date. A value calculated at time $m$ must be discounted $m$ years, using $e^{-\delta m}$, to obtain its time-0 value.
Mixing remaining value with full-stream value. At a date inside the payment interval, the remaining annuity value does not include payments already made.
Applying an annual-payment factor without checking the model. A continuous stream is valued by integrating its payment intensity. A discrete-payment formula does not automatically value payments made continuously.
Using a constant-force factor when the force varies. The simple exponential discount factor applies here because the force is constant; a time-varying force requires a different accumulation model.

Key takeaway: For a level continuous stream with intensity $R$, constant force $\delta$, and payment interval from $m$ to $n$, its time-0 value is $R(e^{-\delta m}-e^{-\delta n})/\delta$. The start and end times determine the integral limits; the duration is their difference.

Check Your Understanding

For each problem, identify the payment interval before calculating. Round final dollar amounts to the nearest cent.

A hypothetical trust pays continuously at $\$450$ per year from time 2 to time 4. The constant force is $0.03$ per year. Find its time-0 value. (Answer: $\$822.66$)
A hypothetical grant pays continuously at $\$1{,}200$ per year from time 0.5 to time 3.5, with constant force $0.05$. Find its time-0 value. (Answer: $\$3{,}260.47$)
A hypothetical fund pays $\$800$ per year continuously from time 1 to time 4, with constant force $0.08$. Find both its time-0 value and the time-2 value of the remaining payments. (Answer: Time 0, $\$1{,}969.67$; remaining value at time 2, $\$1{,}478.56$)
A hypothetical contract pays continuously from time 2 to time 5 under a constant force of $0.06$. Its time-0 value is $\$5{,}000$. Find the payment intensity and the contract’s value at time 2. (Answer: Payment intensity $\$2{,}053.36$ per year; value at time 2, $\$5{,}637.48$)

What You'll Learn