Discrete Payments as Riemann Approximations

From Payment Intensity to Installments

A payment intensity describes a flow of money per unit of time. A discrete installment, by contrast, is a dollar amount paid at a particular time. To approximate a continuous stream by frequent installments, first multiply the intensity by the length of the payment interval, then discount each resulting amount at its payment time. This distinction is central to valuing frequent payments and is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: Let $c(t)$ be a payment intensity, measured in dollars per unit of time, over $0\leq t\leq T$. For a partition $0=t_0<t_1<\cdots<t_n=T$, write $\Delta_k=t_k-t_{k-1}$. An installment representing the interval $[t_{k-1},t_k]$ has amount $c(\xi_k)\Delta_k$, where $\xi_k$ is a chosen point in that interval. If it is paid at time $\tau_k$, its present value at time 0 is $c(\xi_k)\Delta_kD(\tau_k)$, where $D(t)$ is the discount factor from time $t$ to time 0.

In this tutorial, time is measured in years unless stated otherwise. Thus, an intensity of $1{,}200$ dollars per year over a quarter-year interval corresponds to an installment of $1{,}200(0.25)=300$ dollars. The intensity is not itself the installment.

Discounting Each Installment

Suppose the force of interest at time $t$ is $\delta(t)$. Under the standard accumulation model, the accumulation from time 0 to time $t$ is $\exp(\int_0^t\delta(s)\,ds)$, so the discount factor is its reciprocal:

D(t)=\exp\left(-\int_0^t\delta(s)\,ds\right).

This formula applies when the force is specified as a function of time and the financial model accumulates continuously according to that force. With a constant force $\delta$, it reduces to $D(t)=e^{-\delta t}$. The present value of $n$ installments is the sum of their individual present values:

\operatorname{PV} = \sum_{k=1}^{n}c(\xi_k)\Delta_kD(\tau_k).

A common schedule uses equal intervals of length $h=T/n$, with an installment paid at the end of each interval. Then $t_k=kh$, $\Delta_k=h$, and $\tau_k=t_k$. Choosing the right endpoint to determine the intensity gives

\operatorname{PV}_n = \sum_{k=1}^{n}c(t_k)hD(t_k).

The time used to measure the intensity and the time used for discounting have different jobs. In this right-endpoint convention, $c(t_k)h$ specifies the installment amount for the interval, while $D(t_k)$ discounts it from its payment date. Other conventions, such as using a midpoint intensity, are also possible; the convention must be stated so the installment schedule is unambiguous.

Why the Sum Has a Riemann-Sum Limit

A Riemann sum approximates the integral of a function by adding function values multiplied by interval widths. Here the relevant function is the payment intensity multiplied by the discount factor. This gives a useful interpretation of a frequent-installment schedule.

Result: If $c(t)$ and $\delta(t)$ are continuous on $[0,T]$, then as the largest partition interval tends to zero, the present value sums converge to $\int_0^T c(t)D(t)\,dt$. The same conclusion holds for bounded, piecewise-continuous inputs, because the resulting integrand is Riemann integrable.

Identify the interval amount.
For interval $k$, the installment amount is $c(\xi_k)\Delta_k$. The interval width supplies the time unit needed to turn an intensity into dollars.

Discount at the payment time.
Multiply that amount by $D(\tau_k)$, using the actual time $\tau_k$ at which the installment is paid.

Refine the partition.
For fine intervals, the intensity and discount factor vary little within each interval. The sum therefore takes the form of a Riemann sum for $c(t)D(t)$.

For a more explicit justification, let $\xi_k$ be the same point used to evaluate the intensity and the discount factor. If an installment is paid at $\tau_k$ in its interval, then $D(\tau_k)$ approaches $D(\xi_k)$ uniformly as the largest interval shrinks, because $D$ is continuous on the closed interval $[0,T]$. Since $c$ is bounded, replacing each $D(\tau_k)$ by $D(\xi_k)$ changes the sum by an amount tending to zero. The resulting sum is

\sum_{k=1}^{n}c(\xi_k)D(\xi_k)\Delta_k,

which is a Riemann sum for the continuous function $c(t)D(t)$. Its limit is therefore $\int_0^T c(t)D(t)\,dt$. This argument explains why both the interval width and the payment-time discount factor belong in the discrete calculation.

Worked Installment Valuations

Worked Example: Quarterly Payments from a Constant Intensity

A hypothetical fund makes four equal quarterly payments over one year to meet a continuous target payment intensity of $1{,}200$ dollars per year. Each payment is made at the end of its quarter. The force of interest is constant at $\delta=0.06$ per year. Find the present value of the four installments.

Method. Convert the annual intensity into a quarterly amount, find each payment time, and discount each amount under the constant force.

There are four equal intervals, so $h=1/4=0.25$ year. Each installment is

1{,}200(0.25)=300\text{ dollars}.

The payment times are $0.25,0.50,0.75,$ and $1.00$ years. For constant force $0.06$, the discount factor at time $t$ is $e^{-0.06t}$. Thus,

\begin{aligned} \operatorname{PV} &=300\left(e^{-0.06(0.25)}+e^{-0.06(0.50)} +e^{-0.06(0.75)}+e^{-0.06(1.00)}\right)\\ &=300\left(e^{-0.015}+e^{-0.030}+e^{-0.045}+e^{-0.060}\right)\\ &=300(0.985112+0.970446+0.955997+0.941765)\\ &=300(3.853320)\\ &=1{,}155.996. \end{aligned}

Rounded to the nearest cent, the present value is $1,156.00. The four payments total $4(300)=1{,}200$ dollars before discounting, as required by the specified intensity and one-year horizon. The calculation avoids the common mistake of treating $1{,}200$ dollars per year as the amount of every quarterly payment.

Worked Example: Changing Intensity and Changing Force

A hypothetical benefit stream has intensity $c(t)=600+200t$ dollars per year for $0\leq t\leq2$, where $t$ is measured in years. The force of interest is $\delta(t)=0.04+0.02t$ per year. Four installments are paid at the ends of equal half-year intervals. Find their present value using the right-endpoint intensity convention.

Method. Integrate the force to obtain the discount factor, evaluate the intensity at each payment time, multiply by the half-year interval, and discount.

The accumulated force through time $t$ is

\int_0^t(0.04+0.02s)\,ds=0.04t+0.01t^2, \qquad D(t)=e^{-(0.04t+0.01t^2)}.

Here $h=0.5$. At the four payment times, the amounts are $c(t_k)h$:

\begin{aligned} t_1=0.5 &: \quad c(0.5)(0.5)=(600+100)(0.5)=350,\\ t_2=1.0 &: \quad c(1)(0.5)=(600+200)(0.5)=400,\\ t_3=1.5 &: \quad c(1.5)(0.5)=(600+300)(0.5)=450,\\ t_4=2.0 &: \quad c(2)(0.5)=(600+400)(0.5)=500. \end{aligned}

The corresponding discount factors are

\begin{aligned} D(0.5)&=e^{-0.0225}=0.977751,\\ D(1)&=e^{-0.0500}=0.951229,\\ D(1.5)&=e^{-0.0825}=0.920811,\\ D(2)&=e^{-0.1200}=0.886920. \end{aligned}

Therefore,

\begin{aligned} \operatorname{PV} &=350(0.977751)+400(0.951229)+450(0.920811)+500(0.886920)\\ &=342.213+380.492+414.365+443.460\\ &=1{,}580.530. \end{aligned}

Rounded to the nearest cent, the installment present value is $1,580.53. As a check on the setup, the installment amounts total $350+400+450+500=1{,}700$ dollars, and each term is discounted from its stated payment time. Using a constant-force discount factor here would not match the specified time-varying force.

Worked Example: A Payment Intensity with a Change at Midyear

A hypothetical two-year payment stream has intensity $800$ dollars per year during the first year and $1{,}200$ dollars per year during the second year. Payments are made at the ends of four half-year intervals. The force of interest is constant at $0.05$ per year. Find the present value, taking the intensity for each installment from the interval that installment represents.

Method. Divide the schedule at the change in intensity, assign each interval its own intensity and amount, and discount at the payment date. The partition boundary at year 1 ensures no interval straddles the change.

The payment times are $0.5,1.0,1.5,$ and $2.0$ years. Each interval is $0.5$ year long. The first two installments are $800(0.5)=400$ dollars each; the last two are $1{,}200(0.5)=600$ dollars each. With $D(t)=e^{-0.05t}$,

\begin{aligned} \operatorname{PV} &=400e^{-0.05(0.5)}+400e^{-0.05(1)} +600e^{-0.05(1.5)}+600e^{-0.05(2)}\\ &=400(0.975310)+400(0.951229) +600(0.927743)+600(0.904837)\\ &=390.124+380.492+556.646+542.902\\ &=1{,}870.164. \end{aligned}

Rounded to the nearest cent, the present value is $1,870.16. The installments total $400+400+600+600=2{,}000$ dollars, matching the first-year and second-year intensities over their respective one-year periods. Using $800$ dollars per year for the interval from year 1 to year 1.5 would incorrectly extend the first-year rate past its stated boundary.

Common Timing and Unit Traps

A correct sum depends on more than choosing the right discount factors. Each installment amount must represent the intended interval, and each discount factor must correspond to the payment date. The following errors are especially easy to make:

Using intensity as installment amount. An intensity of $c$ dollars per year over an interval of length $h$ gives an installment of $ch$ dollars, not $c$ dollars.
Mixing time units. If the intensity is per year, convert a quarter-year interval to $0.25$ year before multiplying. If time is measured in months instead, the force and intensity units must be consistent with months.
Discounting from the wrong date. A payment at the end of an interval uses $D(t_k)$, not the factor for the beginning $D(t_{k-1})$. A different payment date requires a different discount factor.
Ignoring a change in the schedule. When an intensity changes, identify which rate applies to each installment interval. If necessary, place a partition point at the change.

For instance, the incorrect quarterly calculation in the first worked example would use $1{,}200$ dollars for each of four payments, giving $1{,}200\sum_{k=1}^{4}e^{-0.06k/4}$, or approximately $4{,}623.98$ dollars. That calculation treats a dollars-per-year intensity as a dollar amount and therefore assigns four times the intended total cash flow. The correction is to multiply the intensity by $0.25$ year, making each payment $300$ dollars.

Key takeaway: Write each installment as intensity multiplied by interval length, attach the correct payment time, and discount at that time. As the intervals become finer, the discounted sum has the Riemann-sum form for the payment intensity times its discount factor.

Check Your Understanding

For each problem, state the installment construction and use the specified payment dates and discount model.

A hypothetical one-year payment stream has constant intensity $900$ dollars per year and constant force $\delta=0.08$. Four equal quarterly installments are paid at quarter-end. Find their present value. (Answer: $856.32)
A hypothetical one-year stream has intensity $c(t)=500+100t$ dollars per year and constant force $\delta=0.04$. Two equal installments are paid at the ends of the half-year intervals, using the right-endpoint intensity. Find their present value. (Answer: $557.79)
A hypothetical one-year stream has constant intensity $1{,}000$ dollars per year and force $\delta(t)=0.02+0.01t$. Four equal quarterly installments are paid at quarter-end. Find their present value to the nearest cent. (Answer: $985.29)
A hypothetical one-year stream pays at intensity $700$ dollars per year in the first half-year and $1{,}100$ dollars per year in the second half-year. Four equal quarterly installments are paid at quarter-end under constant force $\delta=0.06$, using each interval's applicable intensity. Find their present value. (Answer: $864.11)
A hypothetical stream has constant intensity $1{,}000$ dollars per year over two years and constant force $\delta=0.05$. If equal intervals become arbitrarily fine and each installment is paid at its interval's end, what integral gives the limiting present value, and what is its value to the nearest cent? (Answer: $\int_0^2 1{,}000e^{-0.05t}\,dt=1{,}903.25$ dollars.)

What You'll Learn