General Payment Intensities and Accumulation Functions

From a Payment Intensity to a Value

A continuous cash flow is specified by how quickly money is paid over time. That rate of payment may vary, and the interest accumulation need not follow a constant force. A general accumulation function provides a single framework for both features: discount each payment to the valuation date, or carry it forward to a later date.

The key distinction is between the payment intensity, measured in dollars per unit of time, and the amount paid over an interval, measured in dollars. If the intensity is $c(t)$, then the payment from time $t$ to $t+dt$ is approximately $c(t)\,dt$. Its value depends on the accumulation between the payment time and the date at which it is valued.

Definition: Let $a(t)$ be a deterministic accumulation function with $a(0)=1$ and $a(t)>0$. A dollar invested at time $s$ accumulates to $a(u)/a(s)$ dollars at time $u$, for $u\ge s$. A payment intensity $c(t)$ is an integrable function for which $c(t)\,dt$ represents the cash paid in a short interval around time $t$.

For the formulas below, assume the cash flow and accumulation are deterministic, and that $a(t)$ is positive and continuous on the finite interval being valued. Assume also that the intensity is integrable there. These conditions ensure the value integrals exist. The intensity may be discontinuous at a finite number of points, as happens with stepwise payment rates. The value assigned to an intensity at an isolated endpoint does not affect its integral. A separate lump-sum payment at a date is not represented by an ordinary intensity integral; it must be valued as a separate cash flow.

The accumulation function can describe a changing interest environment. It need not be written as a constant-force model. The payment intensity and the accumulation function are separate inputs: $c(t)$ says how much is paid, while $a(t)$ says how money grows or discounts over time. This framework is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Present and Accumulated Values

A payment made at time $t$ is worth $1/a(t)$ per dollar at time zero. Multiplying that discount factor by the payment $c(t)\,dt$, and adding across the payment period, gives present value. At a later valuation time $T$, each payment is instead multiplied by its accumulation factor from $t$ to $T$.

Theorem: For a continuous payment intensity $c(t)$ over $0\le t\le T$, its present value at time zero and accumulated value at time $T$ are

\operatorname{PV}_0=\int_0^T \frac{c(t)}{a(t)}\,dt, \qquad \operatorname{AV}_T=\int_0^T c(t)\frac{a(T)}{a(t)}\,dt =a(T)\operatorname{PV}_0.

More generally, the value at a time $u$ within or outside the payment period is

V(u)=\int_0^T c(t)\frac{a(u)}{a(t)}\,dt.

For $t>u$, the factor $a(u)/a(t)$ discounts the future payment back to $u$; for $t<u$, it accumulates the earlier payment forward to $u$.

These expressions apply to a finite interval under the stated assumptions. An integral over an infinite interval requires a separate convergence check; finite-interval formulas alone do not establish that a continuous perpetuity has a finite value. If an intensity is nonnegative and bounded by $m\le c(t)\le M$, then its present value also satisfies

m\int_0^T\frac{1}{a(t)}\,dt \le \operatorname{PV}_0 \le M\int_0^T\frac{1}{a(t)}\,dt.

This comparison follows by dividing the pointwise inequalities by the positive $a(t)$ and integrating. It is useful for checking the scale of a result, though it does not replace evaluating the actual intensity.

Why the Value Formulas Hold

The formulas follow by valuing a small payment first and then taking the limit of finer subdivisions of the payment period. This also explains why the correct factor is the ratio of accumulation functions, rather than just the accumulation at the payment date or at the valuation date.

Represent a small payment.
On a short interval of width $\Delta t$ near $t$, the payment is approximately $c(t)\Delta t$.

Value that payment at time $u$.
A dollar paid at $t$ is worth $a(u)/a(t)$ dollars at $u$. Thus this small payment contributes approximately $c(t)\Delta t\,a(u)/a(t)$.

Add the contributions.
Summing over a partition gives a Riemann sum for the integral of $c(t)a(u)/a(t)$. As the partition becomes finer, integrability gives the stated value $V(u)$.

At $u=0$, $a(0)=1$, so the factor becomes $1/a(t)$, yielding present value. At $u=T$, the factor is $a(T)/a(t)$, yielding accumulated value. Since $a(T)$ is constant with respect to the integration variable, it can be taken outside the integral, which proves $\operatorname{AV}_T=a(T)\operatorname{PV}_0$. This relation is a useful check whenever both values are calculated from the same cash flow.

Worked Examples

Worked Example: Stepwise Payments With a Changing Accumulation Function

A fund receives payments continuously for one year. The payment intensity is \$80 per year on $0\le t<0.5$, and \$160 per year on $0.5\le t\le1$. The accumulation function is $a(t)=1+t$. Find the present value and the accumulated value at time 1.

Method. Integrate the intensity divided by the accumulation function on each interval where the intensity is constant, then multiply the present value by $a(1)$ to obtain the accumulated value.

The present value is

\begin{aligned} \operatorname{PV}_0 &=80\int_0^{0.5}\frac{1}{1+t}\,dt +160\int_{0.5}^{1}\frac{1}{1+t}\,dt\\ &=80\ln(1.5)+160\bigl(\ln 2-\ln 1.5\bigr)\\ &=80\ln(1.5)+160\ln(4/3)\\ &\approx 32.4372+46.0291\\ &\approx \$78.47. \end{aligned}

Because $a(1)=2$, the accumulated value is

\operatorname{AV}_1=2(78.4663\ldots)=\$156.93.

The result checks against the general relationship $\operatorname{AV}_1=a(1)\operatorname{PV}_0$. The common mistake to avoid is incorrectly replacing $160\ln(4/3)$ by $80\ln(4/3)$. That wrong calculation gives $80\ln(1.5)+80\ln(4/3)=80\ln 2\approx \$55.45$, not the present value. The second interval has intensity \$160 per year, so its coefficient must remain 160.

Worked Example: A Smooth Payment Pattern

A hypothetical project pays continuously from time 0 to time 2 at intensity $c(t)=100/(t+2)$ dollars per year. Its accumulation function is $a(t)=1+t$. Find the present value and the accumulated value at time 2.

Method. Substitute both functions into the general present-value integral. The intensity is smooth and bounded on this interval: it decreases from \$50 per year to \$25 per year.

\begin{aligned} \operatorname{PV}_0 &=\int_0^2\frac{100}{(t+2)(t+1)}\,dt\\ &=100\int_0^2\left(\frac{1}{t+1}-\frac{1}{t+2}\right)\,dt\\ &=100\left[\ln(t+1)-\ln(t+2)\right]_0^2\\ &=100\left((\ln 3-\ln 4)-(\ln 1-\ln 2)\right)\\ &=100\ln(3/2)\\ &\approx \$40.55. \end{aligned}

Since $a(2)=3$, the value at time 2 is

\operatorname{AV}_2=3\bigl(100\ln(3/2)\bigr)\approx \$121.64.

As a scale check, the undiscounted total amount paid is $100\int_0^2 1/(t+2)\,dt=100\ln2\approx \$69.31$. The accumulated value exceeds that total because earlier payments earn interest up to time 2. The present value is lower because every payment is discounted to time zero.

Worked Example: Valuing Payments at an Intermediate Date

A fund receives \$60 per year continuously from time 0 to time 1, then \$120 per year from time 1 to time 3. Let $a(t)=1+t$. Find the value of all these payments at time 1, including the payments that will be received after time 1.

Method. Accumulate the payments before time 1 to time 1, and discount the payments after time 1 back to time 1. Equivalently, use the general value formula with $u=1$.

\begin{aligned} V(1) &=60\int_0^1\frac{2}{1+t}\,dt +120\int_1^3\frac{2}{1+t}\,dt\\ &=120\ln 2+240(\ln 4-\ln 2)\\ &=360\ln 2\\ &\approx \$249.53. \end{aligned}

The first term, $120\ln2\approx \$83.18$, is the accumulated value at time 1 of the payments received before that date. The second term, $240\ln2\approx \$166.36$, is the value at time 1 of future payments. Together they give \$249.53. A common timing error is to discount all payments to time zero and report that as the value at time 1; the valuation date determines the accumulation factor for every payment.

Setting Up Other Payment Patterns

A payment intensity does not need a convenient closed-form integral for the framework to apply. For a bounded intensity specified by a formula, substitute that formula directly into the integral. For a stepwise intensity, split the integral at each time when the formula changes. For a smooth intensity, integrate the resulting function on the full interval or use an appropriate numerical integration method if no elementary antiderivative is available.

For example, if a bounded payment rate is $c_1$ on $0\le t<t_1$, $c_2$ on $t_1\le t<t_2$, and $c_3$ on $t_2\le t\le T$, set up present value as

\operatorname{PV}_0 =c_1\int_0^{t_1}\frac{1}{a(t)}\,dt +c_2\int_{t_1}^{t_2}\frac{1}{a(t)}\,dt +c_3\int_{t_2}^{T}\frac{1}{a(t)}\,dt.

The same splitting applies to accumulated value, with $a(T)/a(t)$ in place of $1/a(t)$. Splitting is not an approximation: it simply writes the integral over the full period as a sum over the intervals on which the intensity has a particular expression.

Keep time units consistent. If $t$ is measured in years, $c(t)$ must be in dollars per year for $c(t)\,dt$ to be dollars. The accumulation function must use that same time scale. If time is measured in months instead, convert either the payment intensity or the interest model so the integral still represents the same cash flow and accumulation.

Common Mistakes and Checks

Using the wrong accumulation factor. Present value uses $1/a(t)$; value at time $u$ uses $a(u)/a(t)$. Accumulation and discounting reverse direction depending on whether the payment is before or after the valuation date.
Forgetting the payment intensity is a rate. The amount paid over an interval is the integral of $c(t)$, not $c(t)$ at a single time. In valuation, the discount or accumulation factor also belongs inside that integral.
Failing to split at a change in intensity. Each interval must use the payment rate that applies there. A change at one instant does not itself create a lump-sum payment.
Multiplying present value by the wrong factor. To move a value at time zero to time $T$, multiply by $a(T)$, not by $1/a(T)$. The relation $\operatorname{AV}_T=a(T)\operatorname{PV}_0$ checks the direction.
Assuming every integral is finite. The stated result is for an integrable intensity over a finite interval with a positive, well-behaved accumulation function. An infinite payment period needs a convergence argument.

When setting up a problem, first identify the payment interval, time units, valuation date, intensity on each part of the interval, and accumulation function. Then choose the appropriate value ratio, write the integral, and check whether the result has the expected units and direction.

Check Your Understanding

Set up each valuation using the stated accumulation function, and check whether your answer is consistent with the valuation date.

A continuous payment intensity of \$75 per year applies from time 0 to time 1. The accumulation function is $a(t)=1+0.05t$. Find the present value. (Answer: $1500\ln(1.05)\approx \$73.19$)
A fund receives \$80 per year on $0\le t<1$, then \$120 per year on $1\le t\le2$. If $a(t)=e^{0.04t}$, find the present value. (Answer: $\frac{80(1-e^{-0.04})+120(e^{-0.04}-e^{-0.08})}{0.04}\approx \$191.44$)
For $0\le t\le2$, let $c(t)=120/(t+2)$ dollars per year and $a(t)=1+t$. Find the present value. (Answer: $120\ln(3/2)\approx \$48.66$)
A fund receives \$40 per year on $0\le t<1$ and \$90 per year on $1\le t\le2$. With $a(t)=1+t$, find the value at time 1 of all payments over the full period. (Answer: $80\ln2+180\ln(3/2)\approx \$128.44$)

What You'll Learn