Common Errors in Continuous Annuity Valuation

Why Continuous-Annuity Errors Are Easy to Make

A continuous annuity is valued by adding the discounted or accumulated amounts of payments made throughout an interval. The integral is often short; the harder part is making sure its pieces describe the problem correctly. A payment rate quoted per month must match the time unit of the integration variable, the discounting rate must be expressed in the right form, and the limits must identify when payments actually occur.

The central discipline is to identify four things before integrating: the payment intensity, the payment interval, the interest model, and the valuation date. A correct formula with any one of these entered incorrectly gives a plausible-looking but wrong answer.

Definition: A continuous payment stream has payment intensity $c(t)$ at time $t$, measured as money per unit of time. Under a constant force of interest $\delta$, its present value at time 0 over the interval from $a$ to $b$ is $\int_a^b c(t)e^{-\delta t}\,dt$. Its accumulated value at time $T$, where $T\ge b$, is $\int_a^b c(t)e^{\delta(T-t)}\,dt$. The time unit used for $t$, $a$, $b$, and $T$ must agree with the time unit used for the payment intensity and force.

A Setup That Prevents Most Mistakes

For a constant payment intensity $c$ and constant force $\delta$, the present value at time 0 of payments from time $a$ through time $b$ is

PV_0=\int_a^b ce^{-\delta t}\,dt =\frac{c}{\delta}\left(e^{-\delta a}-e^{-\delta b}\right),\qquad \delta\ne 0.

The accumulated value at a later time $T\ge b$ is

AV_T=\int_a^b ce^{\delta(T-t)}\,dt =\frac{c}{\delta}\left(e^{\delta(T-a)}-e^{\delta(T-b)}\right),\qquad \delta\ne 0.

These expressions assume a deterministic payment intensity and a constant force measured per unit of the integration time. If the force varies with time, the accumulation from $t$ to $T$ is determined by the accumulated force over that interval, rather than by $\delta(T-t)$. If the payment intensity varies, it must remain inside the integral.

Choose the time unit.
For example, if $t$ is measured in years, convert a monthly payment rate to a yearly rate before integrating.

Identify the interest measure.
Use a force directly in exponential discounting. Convert an effective annual rate $i$ to the equivalent constant force $\delta=\ln(1+i)$.

Set the payment limits.
The limits are the actual beginning and end of the payment period, not automatically 0 and the term of the contract.

Use the requested valuation date.
Discount payments to time 0 for present value, or accumulate them to time $T$ for a value at $T$.

Check the direction and scale.
Present value should be below the undiscounted total for positive interest and positive payments. A later accumulated value should exceed the value at an earlier date.

Why the Constant-Force Formula Holds

Under a constant force $\delta$, one dollar at time $t$ has present value $e^{-\delta t}$ at time 0. During a short interval of length $dt$, a stream paying at intensity $c$ pays approximately $c\,dt$. Its present value contribution is therefore approximately $ce^{-\delta t}dt$. Adding these contributions over the payment period gives the integral.

PV_0=\int_a^b ce^{-\delta t}\,dt =c\left[-\frac{1}{\delta}e^{-\delta t}\right]_a^b =\frac{c}{\delta}\left(e^{-\delta a}-e^{-\delta b}\right).

The same reasoning explains the exponent in an accumulated value: a payment made at time $t$ earns interest only from $t$ to the valuation time $T$. Its accumulation factor is $e^{\delta(T-t)}$, not $e^{\delta T}$ or $e^{\delta t}$.

Worked Examples

Worked Example: Converting a Monthly Payment Intensity

A hypothetical fund pays $30 continuously per month from time 0 to time 1.5 years. The effective annual interest rate is 8%. Find the present value at time 0. Round the final answer to the nearest cent.

Method. Use years as the time unit, so first convert the payment intensity to dollars per year. Then convert the effective annual rate to a constant force. This avoids mixing monthly payment units with an annual integration variable.

The annual payment intensity is $c=12(30)=360$ dollars per year. The equivalent constant force per year is

\delta=\ln(1.08)=0.07696104.

The payment interval is $0\le t\le1.5$. Thus

PV_0=\int_0^{1.5}360e^{-0.07696104t}\,dt =\frac{360}{0.07696104}\left(1-e^{-0.07696104(1.5)}\right).

The exponent is $0.07696104(1.5)=0.11544156$, and $e^{-0.11544156}\approx0.89097264$. Therefore,

PV_0\approx 4{,}677.69(1-0.89097264) =4{,}677.69(0.10902736) \approx 510.00.

The present value is $510.00. As a check, the undiscounted total paid is $360(1.5)=$540$, so a value of $510.00 is reasonable under positive interest. A common error is to integrate $30$ over years as if the rate were $30 per year; that understates the stream by a factor of 12.

Worked Example: Discounting a Deferred Stream to Time 0

A hypothetical account receives payments continuously at $300 per year from time 2 to time 4. The force of interest is constant at 0.04 per year. Find the present value at time 0, rounding to the nearest cent.

Method. Use the actual payment limits, 2 and 4, and discount each payment from its own time $t$ to time 0. This handles the deferral directly in the integral.

PV_0=\int_2^4 300e^{-0.04t}\,dt =\frac{300}{0.04}\left(e^{-0.04(2)}-e^{-0.04(4)}\right).

The two exponents are $0.08$ and $0.16$. Using $e^{-0.08}\approx0.92311635$ and $e^{-0.16}\approx0.85214379$,

PV_0=7{,}500(0.92311635-0.85214379) =7{,}500(0.07097256) \approx 532.29.

The present value is $532.29. The undiscounted payments total $300(4-2)=$600$, so the discounted value is below that total. A frequent error is to use limits 0 and 2, which values a different stream. Another is to calculate a value at time 2 and report it as a value at time 0 without discounting it for the additional two years.

Worked Example: Accumulating Payments to a Later Date

A hypothetical trust pays $200 per year continuously from time 1 to time 3. The constant force is 0.05 per year. Find the accumulated value at time 4, rounding to the nearest cent.

Method. Accumulate each payment from its payment time to time 4. A payment at time $t$ earns interest for $4-t$ years.

AV_4=\int_1^3 200e^{0.05(4-t)}\,dt =\frac{200}{0.05}\left(e^{0.05(4-1)}-e^{0.05(4-3)}\right).

The accumulation exponents are $0.15$ and $0.05$. Using $e^{0.15}\approx1.16183424$ and $e^{0.05}\approx1.05127110$,

AV_4=4{,}000(1.16183424-1.05127110) =4{,}000(0.11056314) \approx 442.25.

The accumulated value at time 4 is $442.25. This is larger than the value at time 3 because the stream earns interest for one additional year. The common mistake is to use $e^{0.05t}$ as the accumulation factor: that would accumulate a payment from time 0 to $t$, not from its payment time $t$ to time 4.

Common Traps and How to Check Them

For an effective annual rate $i$, the equivalent constant force is $\delta=\ln(1+i)$. It is not generally correct to substitute $i$ itself for $\delta$ in an exponential formula. The effective annual accumulation factor is $1+i$; the constant-force factor over one year is $e^\delta$, so equality requires $e^\delta=1+i$. Hence $\delta=\ln(1+i)$.

For example, with an 8% effective annual rate, using $0.08$ as the force makes the one-year exponential accumulation $e^{0.08}$, not $1.08$. The resulting valuation uses a different interest model. This substitution may be a close approximation for a small rate, but it is not an exact conversion.

The payment limits also encode timing. A stream from time 2 through time 4 has two years of payments, but its present value is not the same as a two-year stream beginning at time 0: each payment is made later and receives a larger discount. Similarly, an accumulated value at a date after the final payment includes interest earned after that payment.

Situation	Correct factor for a payment at time $t$	Typical error
Present value at time 0	$e^{-\delta t}$	Using $e^{\delta t}$, which accumulates instead of discounts
Accumulated value at time $T$	$e^{\delta(T-t)}$	Using $e^{\delta t}$ or accumulating every payment for $T$ years
Monthly intensity with time in years	Monthly rate multiplied by 12	Leaving the rate per month while integrating over years
Effective annual rate $i$	$\delta=\ln(1+i)$ for constant-force formulas	Using $i$ as though it were the force

A final reasonableness check is to compare against the undiscounted total, $c(b-a)$, when $c$ is constant. With positive interest, the present value at time 0 is less than this total if payments occur at or after time 0. The accumulated value at $T\ge b$ is at least the total, and is strictly greater if positive interest applies over a positive portion of the payment period or accumulation interval. These checks do not replace the calculation, but they often reveal a sign error or a valuation-date mix-up.

Key takeaway: Write the valuation as an integral over the actual payment interval before simplifying. Keep the payment-rate unit, the time unit, the interest measure, and the requested valuation date consistent. The resulting exponential factor should describe exactly the time from each payment to the valuation date.

Check Your Understanding

For each problem, identify the payment interval and valuation date before calculating.

A stream pays $150 per year continuously from time 0 to time 2 under a constant force of 0.03 per year. Find its present value at time 0, to the nearest cent. (Answer: $291.18)
A stream pays $20 per month continuously from time 0 to time 1.5 years. The constant force is 0.04 per year. Find its present value at time 0, to the nearest cent. (Answer: $349.41)
A stream pays $500 per year continuously from time 2 to time 5. The constant force is 0.05 per year. Find its present value at time 0, to the nearest cent. (Answer: $1,260.37)
A stream pays $100 per month continuously from time 0.5 to time 2. The constant force is 0.04 per year. Find its accumulated value at time 3, to the nearest cent. (Answer: $1,930.78)

Situation	Correct factor for a payment at time \(t\)	Typical error
Present value at time 0	\(e^{-\delta t}\)	Using \(e^{\delta t}\), which accumulates instead of discounts
Accumulated value at time \(T\)	\(e^{\delta(T-t)}\)	Using \(e^{\delta t}\) or accumulating every payment for \(T\) years
Monthly intensity with time in years	Monthly rate multiplied by 12	Leaving the rate per month while integrating over years
Effective annual rate \(i\)	\(\delta=\ln(1+i)\) for constant-force formulas	Using \(i\) as though it were the force

What You'll Learn