Approximation Error for Discrete Installments

Why installment timing changes an estimate

A continuous payment stream has a present value found by integrating its discounted payment intensity over time. In practice, it can be useful to replace that integral with a finite number of discrete installments. The result depends not only on how many installments are used, but also on where within each interval the installment is placed.

For example, when the discounted payment flow is falling over time, valuing each interval’s installment at its beginning uses a larger flow value than valuing it at the end. This gives a useful upper and lower estimate for the continuous value. A midpoint estimate often improves accuracy, but its error direction cannot be inferred from monotonicity alone: it depends on curvature.

Definition: Let $c(t)$ be a payment intensity at time $t$, and let $D(t)$ be the factor that discounts a payment at time $t$ to time 0. The discounted flow is $g(t)=c(t)D(t)$. The continuous present value over $[0,T]$ is $I=\int_0^T g(t)\,dt$.

For a constant force of interest $\delta$, $D(t)=e^{-\delta t}$, so $g(t)=c(t)e^{-\delta t}$. With a time-varying force $\delta_s$, the corresponding discount factor is $D(t)=\exp(-\int_0^t \delta_s\,ds)$. The endpoint and midpoint methods apply to the discounted flow $g$, not necessarily to the undiscounted intensity $c$. In each example below, the force and payment intensity are specified so that the flow being approximated is clear.

Three installment approximations

Divide the interval $[0,T]$ into $n$ equal subintervals, each of length $h=T/n$. The left-endpoint method uses the discounted flow at the start of each subinterval; the right-endpoint method uses it at the end; and the midpoint method uses it halfway through. Each estimate multiplies the sampled flow by the interval length.

L_n=h\sum_{j=0}^{n-1}g(jh),\qquad R_n=h\sum_{j=1}^{n}g(jh),\qquad M_n=h\sum_{j=0}^{n-1}g\!\left(\left(j+\tfrac12\right)h\right).

These are estimates of the integral $I$, not alternative definitions of the continuous value. Each term represents an installment with amount approximately equal to the sampled payment intensity times $h$, discounted from the time assigned to that installment. If $g$ is continuous, all three estimates converge to $I$ as the intervals become finer.

Key result: If $g$ is increasing on $[0,T]$, then $L_n\le I\le R_n$. If $g$ is decreasing, then $R_n\le I\le L_n$. These inequalities may be equalities when the flow is constant over the relevant intervals.

Why the endpoint error has a predictable direction

On any one subinterval, an increasing function is at least as large at the right endpoint as at every point inside the interval, and at most as large as the right-endpoint value. Thus the rectangle formed using the left-endpoint value lies below the area under the graph, while the rectangle using the right-endpoint value lies above it. Summing those inequalities over all subintervals gives $L_n\le I\le R_n$. For a decreasing function the endpoint inequalities reverse, giving $R_n\le I\le L_n$.

The difference between the two endpoint estimates can also be found without evaluating the integral. Cancel the shared interior terms in their sums:

L_n-R_n=h\bigl(g(0)-g(T)\bigr).

This identity is a useful check on arithmetic and shows that the gap between the endpoint estimates depends only on the flow at the two ends of the full interval. It does not mean that their average is always the exact integral.

Worked Example: Discounting a Constant Payment Intensity

A hypothetical fund pays continuously at intensity $c(t)=100$ dollars per year for 2 years. The force of interest is constant at $\delta=0.10$ per year. Using two equal intervals, calculate the continuous present value and its left-endpoint, right-endpoint, and midpoint estimates. Round dollar values to the nearest cent.

Method. First form the discounted flow, then evaluate the integral and each sum using $T=2$, $n=2$, and $h=1$. This method avoids treating the undiscounted payment intensity as though it were constant in present-value terms.

The discount factor is $D(t)=e^{-0.10t}$, so $g(t)=100e^{-0.10t}$. The exact continuous value is

I=\int_0^2 100e^{-0.10t}\,dt =\frac{100}{0.10}\left(1-e^{-0.20}\right) =1000(1-0.818730753) =181.269247.

The left-endpoint estimate uses $t=0$ and $t=1$:

L_2=1\left(100e^0+100e^{-0.10}\right) =100(1+0.904837418) =190.483742.

The right-endpoint estimate uses $t=1$ and $t=2$:

R_2=1\left(100e^{-0.10}+100e^{-0.20}\right) =100(0.904837418+0.818730753) =172.356817.

The midpoint times are $0.5$ and $1.5$:

M_2=1\left(100e^{-0.05}+100e^{-0.15}\right) =100(0.951229425+0.860707976) =181.193740.

The discounted flow is decreasing, so the left estimate should be above $I$ and the right estimate below it. Indeed, $172.356817<181.269247<190.483742$. The midpoint estimate is also below $I$, consistent with the positive curvature of this exponential flow. Rounded to cents, the values are $I=\$181.27$, $L_2=\$190.48$, $R_2=\$172.36$, and $M_2=\$181.19$. The check confirms both endpoint directions and the midpoint’s smaller error.

Midpoint errors depend on curvature

A midpoint rectangle may fall below or above the area under the graph. If $g$ is convex, its graph lies below the straight chord joining the endpoints of an interval. The integral over that interval is therefore at least the area of the midpoint rectangle, so the composite midpoint estimate is no greater than the integral. If $g$ is concave, the inequality reverses. These conclusions require convexity or concavity across the intervals; monotonicity by itself is not enough.

For a twice-differentiable flow, convexity means $g''(t)\ge 0$, and concavity means $g''(t)\le 0$. A useful quantitative bound follows from a Taylor expansion about the midpoint $m$ of one interval. The linear term integrates to zero because it is proportional to $t-m$. If $\lvert g''(t)\rvert\le K$, the remaining quadratic error on that interval has absolute value no more than $Kh^3/24$. Adding the bound over $n$ intervals gives the composite midpoint bound.

\lvert I-M_n\rvert\le \frac{KTh^2}{24}, \qquad \text{if } \lvert g''(t)\rvert\le K \text{ on }[0,T].

The bound is useful when a question asks for a guaranteed error tolerance. It is a maximum-error bound, not a prediction that the actual error equals its right-hand side. It also requires the stated second-derivative bound; without such a condition, the bound cannot be assumed.

Worked Example: An Increasing Linear Discounted Flow

A hypothetical continuous payment stream has constant force of interest $\delta=0.10$ per year over $[0,2]$. Its payment intensity is $c(t)=40t e^{0.10t}$ dollars per year, so its discounted flow is $g(t)=40t$. Using two equal intervals, calculate the exact value and all three approximations. Round dollar values to the nearest cent.

Method. Approximate the discounted flow, rather than the growing undiscounted intensity. This avoids applying a monotonicity rule to the wrong function.

The model gives $g(t)=c(t)e^{-0.10t}=40t$, with $T=2$, $n=2$, and $h=1$. Integrating,

I=\int_0^2 40t\,dt=20t^2\bigg|_0^2=80.

The left-endpoint estimate uses times $0$ and $1$:

L_2=1\bigl(40(0)+40(1)\bigr)=0+40=40.

The right-endpoint estimate uses times $1$ and $2$:

R_2=1\bigl(40(1)+40(2)\bigr)=40+80=120.

The midpoint times are $0.5$ and $1.5$:

M_2=1\bigl(40(0.5)+40(1.5)\bigr)=20+60=80.

The increasing flow requires $L_2\le I\le R_2$, which holds: $40\le80\le120$. The midpoint is exact because $g(t)=40t$ is linear on each interval, not merely because it is increasing. To the nearest cent, the four values are $I=\$80.00$, $L_2=\$40.00$, $R_2=\$120.00$, and $M_2=\$80.00$. This check also shows why the left-endpoint sum uses the values at 0 and 1, not at 0 and 2.

Worked Example: A Concave Flow and a Midpoint Overestimate

A hypothetical payment stream has constant force of interest $\delta=0.05$ per year over $[0,4]$. Its payment intensity is $c(t)=(100+20t-2t^2)e^{0.05t}$ dollars per year, giving discounted flow $g(t)=100+20t-2t^2$. Using two equal intervals, find the exact value and the three estimates. Round dollar values to the nearest cent.

Method. Work with the discounted flow and check its first and second derivatives. This avoids assuming that a midpoint estimate must be an underestimate.

Here $h=4/2=2$. The flow is increasing because $g'(t)=20-4t\ge4>0$ on $[0,4]$, and concave because $g''(t)=-4<0$. Its exact value is

I=\int_0^4(100+20t-2t^2)\,dt =\left[100t+10t^2-\frac{2}{3}t^3\right]_0^4 =400+160-\frac{128}{3} =\frac{1552}{3} =517.333333.

The left endpoints are 0 and 2, where $g(0)=100$ and $g(2)=100+40-8=132$:

$$ L_2=2(100+132)=464. $$

The right endpoints are 2 and 4, where $g(2)=132$ and $g(4)=100+80-32=148$:

$$ R_2=2(132+148)=560. $$

The midpoints are 1 and 3, where $g(1)=118$ and $g(3)=142$:

$$ M_2=2(118+142)=520. $$

Since the flow is increasing, $L_2\le I\le R_2$, as $464<517.333333<560$. Since it is concave, $M_2\ge I$; indeed, $520>517.333333$. Rounded to cents, the values are $I=\$517.33$, $L_2=\$464.00$, $R_2=\$560.00$, and $M_2=\$520.00$. The midpoint estimate is high by $520-517.333333=2.666667$, or $\$2.67$ to the nearest cent. The example shows that knowing the flow is increasing determines the endpoint directions but does not determine the midpoint direction.

Common traps and a practical check

The most frequent conceptual mistake is to infer a midpoint error direction from whether the discounted flow is increasing or decreasing. That information orders the left and right sums around the integral, but it does not reveal whether the midpoint sum is high or low. Check curvature instead: convex flow gives $M_n\le I$, while concave flow gives $M_n\ge I$. If the flow changes curvature within the interval, neither overall midpoint direction follows from these simple conditions.

A second trap is to approximate $c(t)$ instead of $g(t)=c(t)D(t)$. Even when the payment intensity rises, discounting may make the discounted flow fall. Always form the discounted flow before applying an error-direction rule. Finally, use the interval width $h=T/n$ consistently: the sum of sampled flow values must be multiplied by $h$, not by $T$ unless there is only one interval.

Form the discounted flow.
Write $g(t)=c(t)D(t)$, using the stated force of interest and time units.

Set the partition.
Calculate $h=T/n$, then list the left, right, or midpoint sample times explicitly.

Calculate and check.
Multiply each sampled-flow sum by $h$, compare with the exact integral if available, and verify the endpoint inequalities from monotonicity.

Endpoint estimates are especially useful when a problem asks for bounds rather than a highly accurate value. Midpoint estimates can be accurate with fewer intervals, but use curvature or an error bound when the direction or maximum error matters. These distinctions are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the relevant shape of the discounted flow as well as computing the requested values. Give dollar results to the nearest cent where applicable.

A hypothetical stream has discounted flow $g(t)=12+3t$ dollars per year on $[0,2]$. With $n=2$, calculate $I$, $L_2$, $R_2$, and $M_2$. (Answer: $I=30$, $L_2=27$, $R_2=33$, and $M_2=30$.)
A hypothetical constant payment intensity of $\$50$ per year is discounted at constant force $\delta=0.20$ over one year. With $n=2$, calculate the exact continuous value and all three estimates. (Answer: $I=\$45.32$, $L_2=\$47.62$, $R_2=\$43.09$, and $M_2=\$45.30$, to the nearest cent.)
A hypothetical discounted flow is $g(t)=8+4t-t^2$ on $[0,2]$. With $n=2$, calculate $I$, $L_2$, $R_2$, and $M_2$, and state the midpoint error direction. (Answer: $I=64/3$, $L_2=19$, $R_2=23$, $M_2=21.5$; the midpoint estimate is an overestimate because the flow is concave.)
A twice-differentiable discounted flow on $[0,3]$ satisfies $\lvert g''(t)\rvert\le6$. Using $n=3$, give a guaranteed upper bound on the absolute composite midpoint error. (Answer: $\lvert I-M_3\rvert\le0.75$, in the corresponding value units.)
A discounted flow is increasing and convex throughout the valuation period. State the direction of error for $L_n$, $R_n$, and $M_n$. (Answer: $L_n\le I\le R_n$, and $M_n\le I$.)

What You'll Learn