Introduction
A continuous payment stream may change its payment rate over time while the force of interest also changes. The present value must account for both changes at once: the amount paid near time \(t\) is determined by the payment intensity at that time, while its discount factor depends on the force accumulated from time 0 to time \(t\). Treating either function as constant when it is not can materially change the value.
This tutorial develops a single valuation integral for a finite continuous stream, explains why that integral is correct, and applies it to exponential payment intensities and piecewise-linear forces. When the resulting integral has no convenient elementary antiderivative, numerical integration provides a systematic estimate.
Why Both Functions Appear in the Integral
For a short interval from \(t\) to \(t+\Delta t\), the stream pays approximately \(c(t)\Delta t\). If the accumulation function from time 0 is \(a(t)\), then the discount factor to time 0 is \(1/a(t)\). Under a force of interest \(\delta(t)\), the accumulation function satisfies \(a(t)=\exp(\int_0^t\delta(s)\,ds)\). Thus the approximate present value of the payment in that short interval is \(c(t)\Delta t/a(t)\). Summing these discounted payments over finer and finer intervals gives the integral.
This argument assumes the payment intensity and force are integrable over the term, and that the stream is paid continuously according to its stated intensity. It does not describe a separate set of payments made only at selected dates; those payments require their own cash-flow valuation. The force in the discount factor is accumulated over the whole interval \([0,t]\), not simply evaluated at the payment time.
Express the rate of payment \(c(t)\) in dollars per unit time, including any time dependence.
Find \(\int_0^t\delta(s)\,ds\), accounting for any changes in the force over the term.
Multiply the intensity by the discount factor and integrate over the payment period. Use an exact antiderivative when available; otherwise specify and apply a numerical method.
Exponential Payments Under Constant Force
When the payment intensity grows or declines exponentially and the force is constant, the integral has a simple closed form. Suppose \(c(t)=C e^{gt}\), where \(C\) is the initial intensity and \(g\) is its continuous growth rate, and suppose \(\delta(t)=\delta\). Then the discounted intensity is \(C e^{(g-\delta)t}\). If \(g\ne\delta\), integration gives the following result.
If \(g=\delta\), the discounted intensity is constant at \(C\), so the value is \(Cn\). This equality case is important: the displayed quotient has a zero denominator when \(g=\delta\), but the present value itself remains finite.
Worked Example: Exponentially Growing Payments Under a Constant Force
A hypothetical service contract pays continuously for 4 years at an intensity of \(200e^{0.03t}\) dollars per year. The force of interest is constant at \(\delta=0.05\) per year. Find the present value at time 0.
Method. Discount the payment intensity using the accumulated force, then integrate. This avoids the mistake of discounting only the initial payment rate while ignoring its growth.
The accumulated force to time \(t\) is \(\int_0^t0.05\,ds=0.05t\), so the discount factor is \(e^{-0.05t}\). The discounted intensity is therefore \(200e^{0.03t}e^{-0.05t}=200e^{-0.02t}\).
The present value is $$ \operatorname{PV} =200\int_0^4 e^{-0.02t}\,dt =200\left[\frac{e^{-0.02t}}{-0.02}\right]_0^4 =200\frac{1-e^{-0.08}}{0.02}. $$
Using \(e^{-0.08}=0.9231163\), the value is \(200(0.0768837/0.02)=200(3.844185)=768.837\). Rounded only at the final step to the nearest cent, the present value is $768.84.
Check. The payment intensity grows at 3% per year while discounting acts at a 5% force, so the discounted intensity declines over time. A present value less than \(200\times4=$800\), the value if the discounted intensity stayed at its initial level, is consistent with that decline.
When the Force Varies Linearly
If \(\delta(t)=\alpha+\beta t\) on an interval starting at time 0, its accumulated force is \(\alpha t+\frac12\beta t^2\). An exponential payment intensity \(C e^{gt}\) then produces a discounted intensity with a quadratic exponent: \(C\exp((g-\alpha)t-\frac12\beta t^2)\). In general, this integral does not have an elementary antiderivative. The valuation setup is still exact; a numerical method can be used to estimate the integral.
Composite Simpson’s rule is one such method. For an interval divided into an even number \(m\) of equal subintervals of width \(h\), let \(f(t)\) denote the discounted intensity. The rule estimates the integral as $$ \int_0^n f(t)\,dt\ \approx\ \frac{h}{3}\left[f(t_0)+f(t_m) +4\sum_{\substack{j=1\\j\text{ odd}}}^{m-1}f(t_j) +2\sum_{\substack{j=2\\j\text{ even}}}^{m-2}f(t_j)\right]. $$
The alternating weights apply to the values of the entire discounted intensity, not to the payment intensity or discount factor separately. Simpson’s rule gives an approximation, not an exact value; the number of subintervals and the smoothness of the integrand affect its accuracy.
Worked Example: Exponential Payments with an Increasing Linear Force
A hypothetical fund pays continuously for 2 years at an intensity of \(100e^{0.02t}\) dollars per year. Its force of interest is \(\delta(t)=0.04+0.02t\) per year. Estimate the present value using composite Simpson’s rule with four equal subintervals.
Method. Integrate the force first, form the full discounted intensity, and then apply Simpson’s rule. This avoids substituting the force at time \(t\) for the accumulated force from 0 to \(t\).
The accumulated force is \(\int_0^t(0.04+0.02s)\,ds=0.04t+0.01t^2\). Thus \(f(t)=100\exp(0.02t-0.04t-0.01t^2)=100\exp(-0.02t-0.01t^2)\). For Simpson’s rule, use the factor 100 after estimating the integral of the exponential part. The width is \(h=2/4=0.5\).
The exponential part at the five points is: \(f(0)/100=1\); \(f(0.5)/100=e^{-0.0125}=0.987578\); \(f(1)/100=e^{-0.03}=0.970446\); \(f(1.5)/100=e^{-0.0525}=0.948854\); and \(f(2)/100=e^{-0.08}=0.923116\).
Applying the Simpson weights gives the estimated integral of the exponential part: $$ \frac{0.5}{3}\left[1+0.923116 +4(0.987578+0.948854)+2(0.970446)\right] =\frac{11.609736}{6} =1.934956. $$
Multiplying by the intensity scale gives an estimated present value of \(100(1.934956)=$193.4956\), or $193.50 to the nearest cent. The answer is a numerical estimate from the specified four-subinterval Simpson calculation.
Check. The initial discounted intensity is $100 per year and then decreases, since the force exceeds the payment growth rate throughout the term. A value below $200, the initial intensity multiplied by 2 years without discounting or growth effects, is reasonable.
Piecewise-Linear Forces
A force can be linear on each of several intervals, with its formula changing at specified times. The accumulated force must then be carried forward from one interval to the next. On a segment beginning at time \(r\), if the force is \(\delta(t)=A+B(t-r)\), the accumulated force at \(t\ge r\) is the accumulated force at \(r\), plus \(A(t-r)+\frac12B(t-r)^2\). This preserves the contribution from earlier time periods and ensures the discount factor is continuous at the breakpoint.
A useful valuation strategy is to write the accumulated force segment by segment and integrate the discounted intensity over each segment. If the resulting integrals are not elementary, numerical integration may be applied separately on each interval. Breakpoints should be included as integration endpoints, particularly when the slope changes there.
Worked Example: Constant Payments with a Piecewise-Linear Force
A hypothetical grant pays continuously at a constant intensity of $80 per year for 3 years. The force is \(\delta(t)=0.02+0.01t\) for \(0\le t\le2\), and \(\delta(t)=0.04-0.02(t-2)\) for \(2\le t\le3\). Estimate its present value using composite Simpson’s rule with width \(h=0.5\) on each segment.
Method. Accumulate the force over the first segment, then add its total to the accumulated force on the second segment. This avoids restarting the discount factor at the breakpoint \(t=2\).
For \(0\le t\le2\), the accumulated force is \(0.02t+0.005t^2\). At \(t=2\) this equals \(0.04+0.02=0.06\). For \(2\le t\le3\), set \(s=t-2\). The force is \(0.04-0.02s\), so the accumulated force is \(0.06+0.04s-0.01s^2\).
The discount factors \(e^{-\int_0^t\delta(u)\,du}\) at \(t=0,0.5,1,1.5,2,2.5,3\), rounded here to six decimals for display, are respectively \(1.000000,\ 0.988813,\ 0.975310,\ 0.959589,\ 0.941765,\ 0.925425,\ 0.913931\). For example, at \(t=2.5\), \(s=0.5\), the accumulated force is \(0.06+0.04(0.5)-0.01(0.5)^2=0.0775\), giving the discount factor \(e^{-0.0775}=0.925425\).
Apply Simpson’s rule on \(0\le t\le2\): $$ \frac{0.5}{3}\left[1+0.941765+ 4(0.988813+0.959589)+2(0.975310)\right] \approx 1.947666. $$ On \(2\le t\le3\), use the values at 2, 2.5, and 3: $$ \frac{0.5}{3}\left[0.941765+0.913931+4(0.925425)\right] \approx 0.926233. $$
The estimated integral of the discount factor over the full term is \(1.947666+0.926233=2.873899\). Multiplying by the constant intensity, the present value is approximately \(80(2.873899)=$229.9119\), or $229.91 to the nearest cent.
Check. The force is positive throughout, so each payment is discounted. The discount factor falls from 1 to about 0.914 over the term, making the result less than the undiscounted total of \(80(3)=$240\), as expected.
Common Mistakes and a Practical Check
The most important distinction is between a force at a particular time and the accumulated force up to that time. For the increasing force in the second example, \(\delta(2)=0.08\), but the accumulated force through time 2 is \(\int_0^2(0.04+0.02s)\,ds=0.12\). Using \(e^{-\delta(2)\cdot2}=e^{-0.16}\) as the discount factor at time 2 is wrong; the correct factor is \(e^{-0.12}\). In general, multiplying the force at the endpoint by elapsed time is valid only when the force is constant over that entire elapsed period.
- Forgetting payment growth. A payment intensity \(Ce^{gt}\) must remain inside the present-value integral. Discounting a constant \(C\) instead values a different stream.
- Using the wrong sign. Positive force reduces present value, so its accumulated value appears with a negative sign in the exponential discount factor.
- Resetting accumulation at a breakpoint. The accumulated force on a later segment includes all earlier segments, even when the formula for the force changes.
- Presenting a numerical estimate as exact. State the numerical method and its step size when using an approximation such as Simpson’s rule.
A final reasonableness check is to compare the answer with the undiscounted total \(\int_0^n c(t)\,dt\). If the force is positive throughout and the intensity is nonnegative, each payment is discounted, so the present value must be below that total. If the force is zero throughout, the present value equals the undiscounted total.
Check Your Understanding
For numerical questions, use the stated integration method and round money only at the final step to the nearest cent.
- A stream pays at intensity \(150e^{0.02t}\) dollars per year for 3 years under a constant force \(\delta=0.06\). Find its present value. (Answer: $424.05)
- A stream pays at a constant intensity of $90 per year for 2 years under \(\delta(t)=0.03+0.01t\). Estimate the present value using Simpson’s rule with two equal subintervals. (Answer: approximately $173.57)
- A stream pays at intensity \(50e^{0.04t}\) dollars per year for 5 years under a constant force \(\delta=0.07\). Find its present value. (Answer: $232.15)
- A stream pays at a constant intensity of $100 per year for 2 years. The force is 0.02 for \(0\le t\le1\), then \(\delta(t)=0.02+0.02(t-1)\) for \(1\le t\le2\). Estimate the present value by applying Simpson’s rule with width 0.5 over the full interval. (Answer: approximately $195.73)
- A stream pays at intensity \(120e^{0.01t}\) dollars per year over \(0\le t\le2\), and \(\delta(t)=0.03+0.01t\). Write the exact present-value integral and state the accumulated force at \(t=2\). (Answer: \(\int_0^2 120e^{0.01t}e^{-(0.03t+0.005t^2)}\,dt\); accumulated force at \(t=2\) is 0.08.)