Present Value Under Constant Force

Present Value of a Continuous Payment Stream

A continuous payment stream does not arrive as one payment at the end of each year. Instead, payments flow throughout the term at a specified rate per unit of time. To find the present value, value each small payment at the time it is made, discount it to time zero, and add the resulting values across the whole interval.

This tutorial considers a constant payment intensity over a finite term and a constant force of interest. The payment intensity and force must use a consistent time unit. For example, if the force is stated per year, measure the term in years and express the payment intensity in dollars per year.

Definition: A constant payment intensity $c$ over the interval from time 0 to time $n$ means that the payment made during a short interval of length $dt$ is $c\,dt$. If $c$ is measured in dollars per year, then $dt$ is measured in years and $c\,dt$ is measured in dollars. The present value is measured at time 0.

Under a constant force of interest $\delta$, a payment made at time $t$ has discount factor $e^{-\delta t}$ when valued at time 0. Thus a small payment $c\,dt$ at time $t$ contributes $c e^{-\delta t}\,dt$ to the present value. Adding those contributions over the payment interval gives the valuation integral.

PV_0=\int_0^n c e^{-\delta t}\,dt

Deriving the Finite-Term Formula

The formula follows directly by integrating the discounted payment intensity. The calculation assumes that the force $\delta$ and payment intensity $c$ remain constant throughout the term, and that payments occur continuously from time 0 through time $n$. The exact treatment of a payment at either endpoint does not affect the integral: a single instant has zero duration and therefore contributes no separate payment.

Write the value of a small payment.
During an interval of length $dt$ at time $t$, the payment is $c\,dt$. Its value at time 0 is $c e^{-\delta t}\,dt$.

Add values across the term.
Integrate the discounted contributions from $t=0$ to $t=n$: $PV_0=\int_0^n c e^{-\delta t}\,dt$.

Evaluate the integral.
For $\delta\ne 0$, an antiderivative of $e^{-\delta t}$ is $-e^{-\delta t}/\delta$, so $PV_0=\frac{c}{\delta}(1-e^{-\delta n})$.

Theorem: For a constant payment intensity $c$ over $0\le t\le n$, under a constant force of interest $\delta$, $$ PV_0= \begin{cases} \displaystyle \frac{c}{\delta}\left(1-e^{-\delta n}\right), & \delta\ne 0,\\[6pt] cn, & \delta=0. \end{cases} $$ The $\delta=0$ case means there is no discounting: the present value is the total payment $cn$. The formula allows a constant real $\delta$; in the usual positive-interest case, $\delta>0$.

The result also has a useful interpretation. If $\delta>0$, then $e^{-\delta t}$ is less than 1 for $t>0$, so later portions of the stream receive more discounting than earlier portions. Consequently, the present value is less than the undiscounted total $cn$, provided $c>0$.

Timelines and Inputs

Before substituting numbers, identify the valuation date, the payment interval, the intensity, and the force. The timeline for this topic is always read from the present value date at time 0 to the end of the term at time $n$.

Quantity	Meaning	Unit example
$t$	Time at which a small payment is made	Years after time 0
$n$	Length of the payment term	Years
$c$	Constant payment intensity	Dollars per year
$\delta$	Constant force of interest	Per year
$c e^{-\delta t}\,dt$	Time-0 value of the payment made during $dt$ at time $t$	Dollars

If a problem gives an effective annual interest rate $i$, first convert it to a force of interest for the same annual time unit:

\delta=\ln(1+i)

This conversion follows because one year of accumulation under a constant force is $e^\delta$, which must equal $1+i$, the one-year accumulation under effective annual rate $i$. The conversion is valid when $i>-1$. Do not substitute an effective annual rate directly for $\delta$ in an exponential discount factor.

Worked Example: Annual Intensity Over Four Years

A hypothetical fund pays continuously at an intensity of $1{,}200$ dollars per year for four years. The force of interest is $0.05$ per year. Find the present value at time 0, rounded to the nearest cent.

Method. The payment intensity is constant, the force is constant, and the stream runs from $t=0$ to $t=4$ years. The formula applies directly:

PV_0=\frac{c}{\delta}\left(1-e^{-\delta n}\right)

Substitute $c=1{,}200$, $\delta=0.05$, and $n=4$:

\begin{aligned} PV_0 &=\frac{1{,}200}{0.05}\left(1-e^{-(0.05)(4)}\right)\\ &=24{,}000\left(1-e^{-0.2}\right)\\ &=24{,}000(1-0.818730753)\\ &=24{,}000(0.181269247)\\ &=4{,}350.461928 \end{aligned}

Rounded to the nearest cent, the present value is $\$4{,}350.46$. As a check, the undiscounted total is $1{,}200(4)=\$4{,}800$. The answer is positive and less than that total, as expected with a positive force.

Common mistake avoided: Do not discount the entire $\$4{,}800$ as though it were paid at year 4. That would give $4{,}800e^{-0.2}=\$3{,}929.91$, which treats all payments as a single end-of-term payment rather than a continuous stream.

Effective Rates and Consistent Time Units

Rate conversion and unit conversion are frequent sources of errors. A continuous intensity of $100$ dollars per month is equivalent to $1{,}200$ dollars per year. If the force is per year, use $1{,}200$ dollars per year and a term measured in years. Alternatively, all quantities could be expressed using months, but then the force would also have to be converted to a monthly force. Mixing annual and monthly units inside the same exponent is not valid.

Worked Example: Converting an Effective Annual Rate

A hypothetical account pays continuously at an intensity of $\$2{,}400$ per year for two years. Its effective annual interest rate is $5\%$. Find the present value at time 0, rounded to the nearest cent.

Method. First convert the effective annual rate to the constant force. Since one year’s accumulation is $1.05$, the force is $\delta=\ln(1.05)$ per year.

\begin{aligned} \delta&=\ln(1.05)=0.048790164\text{ per year},\\ e^{-2\delta}&=e^{-2\ln(1.05)}=\frac{1}{1.05^2} =\frac{1}{1.1025}=0.907029478. \end{aligned}

Now apply the finite-term formula, keeping the two-year term and annual force in matching units:

\begin{aligned} PV_0 &=\frac{2{,}400}{0.048790164}(1-0.907029478)\\ &=49{,}190.0\ldots(0.092970522)\\ &=4{,}573.24\ldots \end{aligned}

Rounded to the nearest cent, the present value is $\$4{,}573.24$. A check on the conversion is $e^\delta=1.05$, the stated one-year accumulation factor.

Common mistake avoided: Substituting $0.05$ for $\delta$ would incorrectly treat the effective annual rate as a force. The correct force is $\ln(1.05)$, not $0.05$.

Worked Example: Monthly Intensity With an Annual Force

A hypothetical service contract produces a continuous payment intensity of $\$100$ per month for 18 months. The constant force of interest is $0.06$ per year. Find the present value at time 0, rounded to the nearest cent.

Method. Convert the intensity and term to annual units: $\$100$ per month is $\$1{,}200$ per year, and 18 months is $1.5$ years. The force is already annual.

\begin{aligned} PV_0 &=\frac{1{,}200}{0.06}\left(1-e^{-(0.06)(1.5)}\right)\\ &=20{,}000\left(1-e^{-0.09}\right)\\ &=20{,}000(1-0.913931185)\\ &=20{,}000(0.086068815)\\ &=1{,}721.376300 \end{aligned}

Rounded to the nearest cent, the present value is $\$1{,}721.38$. The undiscounted total is $\$100(18)=\$1{,}800$, so the calculated value is below the total, as expected.

Common mistake avoided: Using $18$ as $n$ while retaining a force of $0.06$ per year makes the exponent use inconsistent units. With an annual force, the term in the exponent must be $1.5$ years.

What the Formula Assumes—and What It Does Not

The formula is specific to a constant intensity over a finite interval under a constant force. Its inputs should represent the actual payment pattern and interest model:

The intensity $c$ is the amount paid per unit of time throughout the interval, not the total payment over the term.
The force $\delta$ is constant over the whole term. A force that changes with time requires a different discount expression.
The term $n$ is finite and is measured in the same time unit used for $\delta$ and $c$.
The valuation date is time 0, and the stream starts at that time. A stream that begins after a delay requires an additional timing adjustment.

These assumptions matter in exam problems because a familiar-looking integral can be inappropriate if the payment intensity changes, the force varies, or the stream begins later. Check the timeline and the wording before using the closed form. This topic is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Key takeaway: For constant intensity $c$ from time 0 to $n$ under constant force $\delta$, discount each payment at its own time and integrate: $PV_0=\int_0^n c e^{-\delta t}\,dt$. For $\delta\ne0$, this is $\frac{c}{\delta}(1-e^{-\delta n})$; for zero force, it is $cn$.

Check Your Understanding

For each problem, identify consistent time units before applying the finite-term present-value formula. Round dollar answers to the nearest cent.

A hypothetical payment stream has a constant intensity of $\$300$ per year for two years. The force of interest is $0.03$ per year. Find its present value. (Answer: $\$582.35$)
A hypothetical fund pays continuously at $\$1{,}000$ per year for three years. The effective annual interest rate is $4\%$. Find its present value. (Answer: $\$2{,}830.23$)
A hypothetical contract pays continuously at $\$80$ per month for 15 months. The force is $0.06$ per year. Find its present value. (Answer: $\$1{,}156.10$)
A hypothetical stream pays at $\$1{,}500$ per year continuously for six years under a force of $0.05$ per year. Find its present value. (Answer: $\$7{,}775.45$)
A constant continuous payment intensity of $\$500$ per year is valued at $\$2{,}000$ at time 0 under a force of $0.04$ per year. How long is the payment term? (Answer: $4.3588$ years, approximately)

Quantity	Meaning	Unit example
\(t\)	Time at which a small payment is made	Years after time 0
\(n\)	Length of the payment term	Years
\(c\)	Constant payment intensity	Dollars per year
\(\delta\)	Constant force of interest	Per year
\(c e^{-\delta t}\,dt\)	Time-0 value of the payment made during \(dt\) at time \(t\)	Dollars

What You'll Learn