From Installments to Continuous Payments
A continuous payment stream does not mean that a finite amount is paid at every instant. It means that payments accumulate at a specified rate per unit of time. One way to understand its present value is to divide the payment period into equal intervals, pay the amount attributable to each interval at its end, and let the interval length shrink. The limiting value is the present value of the continuous stream.
This limit connects discrete and continuous valuation in a precise way. It also clarifies an important distinction: an installment is an amount of money, while a payment intensity is an amount of money per unit of time. The installment amount must therefore shrink with the interval length if the total amount paid over a fixed term is to remain stable.
The assumption of a constant force means that a dollar payable at time \(t\) has present value \(e^{-\delta t}\) dollars. The model here uses end-of-interval payments, with each amount equal to the intensity at that endpoint multiplied by the interval length. Other choices, such as beginning- or midpoint-of-interval payments, can also give the same limit under the stated continuity condition, but they produce different finite approximations.
This tutorial uses finite terms. A continuous perpetuity requires an additional convergence condition on the payment intensity and discounting; a finite-term formula should not be applied to an infinite horizon without checking that the integral exists.
Why the Limit Is an Integral
For an interval of length \(h\), the installment \(r(kh)h\) represents the payments attributed to that interval. Discounting this installment to time 0 multiplies it by \(e^{-\delta kh}\). Thus the present value contribution is \(r(kh)e^{-\delta kh}h\). Summing these contributions gives a right-end Riemann sum for the function \(r(t)e^{-\delta t}\). Continuity makes that function integrable, so the sums converge to its integral.
For a constant payment intensity \(c\), the integral can be evaluated explicitly. When \(\delta\ne 0\), the finite-term present value factor is \((1-e^{-\delta T})/\delta\). The discrete geometric sum gives the same result in the limit, as shown below.
Let \(h=T/n\). Each interval's payment is \(ch\), and the \(k\)th payment is made at time \(kh\).
Discount each payment under the constant force \(\delta\): \(P_n=ch\sum_{k=1}^{n}e^{-\delta kh}\).
For \(\delta\ne0\), \(P_n=ch e^{-\delta h}(1-e^{-\delta T})/(1-e^{-\delta h})\).
Since \(h e^{-\delta h}/(1-e^{-\delta h})=h/(e^{\delta h}-1)\) and \(\lim_{h\to0}h/(e^{\delta h}-1)=1/\delta\), the limiting present value is \(c(1-e^{-\delta T})/\delta\).
The limit in Step 4 follows, for example, from the standard limit \(\lim_{x\to0}(e^x-1)/x=1\), with \(x=\delta h\). The same result follows directly by integrating the discounted intensity:
If \(\delta=0\), there is no discounting. The discrete sum is \(ch n=cT\), and the integral is also \(cT\). This is the limiting case of the factor as \(\delta\) approaches zero.
Worked Examples
Worked Example: Quarterly Installments Approaching a Continuous Stream
A hypothetical fund pays continuously at an intensity of $800 per year for 2 years. The force of interest is constant at \(\delta=0.06\) per year. First find the present value of the continuous stream. Then approximate it using equal quarterly installments paid at the ends of the quarters. Round dollar amounts to the nearest cent.
Method. Use the continuous finite-term factor for the limiting value. For the quarterly approximation, use eight installments: there are four quarters per year for two years. Each installment is the intensity multiplied by one quarter of a year.
The continuous value is $$ P=800\frac{1-e^{-0.06(2)}}{0.06} =800\frac{1-e^{-0.12}}{0.06}. $$ Using \(e^{-0.12}=0.88692044\), $$ P=800\frac{1-0.88692044}{0.06} =800\left(\frac{0.11307956}{0.06}\right) =800(1.88465933) =\$1{,}507.73. $$
For quarterly installments, \(h=0.25\) year, so each payment is $$ 800(0.25)=\$200. $$ There are \(2/0.25=8\) payments, at times \(0.25,0.50,\ldots,2.00\). Their present value is $$ P_8=200\sum_{k=1}^{8}e^{-0.06(0.25k)} =200\sum_{k=1}^{8}e^{-0.015k}. $$ This is a geometric sum: $$ P_8=200\frac{e^{-0.015}(1-e^{-0.12})}{1-e^{-0.015}} =200\frac{1-e^{-0.12}}{e^{0.015}-1}. $$ Using \(e^{0.015}-1=0.01511306\), $$ P_8=200\frac{0.11307956}{0.01511306} =200(7.48224) =\$1{,}496.45. $$
The quarterly approximation is $11.28 below the continuous value, and the installment value approaches the integral as the interval length shrinks. Common mistake avoided: four quarterly installments cover only one year, not two. The correct two-year quarterly schedule has eight installments.
Worked Example: A Linearly Varying Payment Intensity
A hypothetical grant pays at intensity \(r(t)=100+20t\) dollars per year for 2 years, where \(t\) is measured in years. The constant force is \(\delta=0.10\) per year. Find the continuous present value, then approximate it with four equal end-of-interval installments. Round to the nearest cent.
Method. Discount the payment intensity inside the integral. For the finite approximation, use \(h=2/4=0.5\) years and pay \(r(kh)h\) at each right endpoint.
The continuous value is $$ P=\int_0^2(100+20t)e^{-0.1t}\,dt. $$ For the constant part, $$ 100\int_0^2 e^{-0.1t}\,dt =100\frac{1-e^{-0.2}}{0.1} =100\frac{1-0.81873075}{0.1} =\$181.26925. $$ For the part involving \(t\), integration by parts gives $$ \int_0^T t e^{-\delta t}\,dt =\frac{1-(1+\delta T)e^{-\delta T}}{\delta^2}. $$ Therefore, with \(T=2\) and \(\delta=0.1\), the varying part is $$ 20\frac{1-(1+0.1(2))e^{-0.2}}{0.1^2} =20\frac{1-1.2(0.81873075)}{0.01} =20\frac{0.01752310}{0.01} =\$35.04620. $$ Thus \(P=181.26925+35.04620=\$216.32\), to the nearest cent.
For four intervals, the payment times are \(0.5,1.0,1.5,2.0\). The intensities there are \(110,120,130,140\) dollars per year, so the installments are \(55,60,65,70\) dollars. Their discounted value is $$ P_4=55e^{-0.05}+60e^{-0.10}+65e^{-0.15}+70e^{-0.20}. $$ Using the discount factors \(0.95122942,0.90483742,0.86070798,0.81873075\), respectively, $$ P_4=52.31762+54.29025+55.94602+57.31115 =\$219.87. $$
This is a finite right-end approximation, not the limiting value; in particular, the installment amounts are based on endpoint intensities. Common mistake avoided: using \(r(kh)\) as the installment itself would omit the factor \(h\) and confuse dollars per year with dollars.
Worked Example: A Deferred Continuous Payment Stream
A hypothetical trust pays continuously at a constant intensity of $500 per year from time 1 through time 4. The constant force is \(\delta=0.05\) per year. Find its value at time 0, and compare it with six equal semiannual installments, each paid at the end of its interval. Round to the nearest cent.
Method. Discount the stream over its actual payment period, \([1,4]\). For the installment approximation, there are \(3/0.5=6\) intervals; the payments occur at \(1.5,2.0,\ldots,4.0\).
The continuous present value is $$ P=\int_1^4 500e^{-0.05t}\,dt =500e^{-0.05}\frac{1-e^{-0.05(3)}}{0.05}. $$ Using \(e^{-0.05}=0.95122942\) and \(e^{-0.15}=0.86070798\), $$ P=500(0.95122942)\frac{1-0.86070798}{0.05} =500(0.95122942)(2.78584040) =\$1{,}324.99. $$
Each semiannual installment is \(500(0.5)=\$250\). Thus $$ P_6=250\sum_{j=0}^{5}e^{-0.05(1.5+0.5j)} =250e^{-0.075}\frac{1-e^{-0.15}}{1-e^{-0.025}}. $$ Using \(e^{-0.075}=0.92774349\), \(1-e^{-0.15}=0.13929202\), and \(1-e^{-0.025}=0.02469009\), $$ P_6=250(0.92774349)\left(\frac{0.13929202}{0.02469009}\right) =250(5.23397) =\$1{,}308.49. $$
The deferred start is accounted for by the factor \(e^{-0.05}\) in the continuous value. Common mistake avoided: valuing the stream as though it began at time 0 would overstate its present value; the first payment is not made until time 1.
Conditions, Interpretation, and Exam Traps
The continuous limit is useful because it gives a clean valuation method without requiring an arbitrary payment frequency. Its assumptions should still be checked: the payment intensity must be expressed per unit of time, the interval length must use that same unit, and the force must be stated on a compatible time basis. The Riemann-sum result applies when the discounted intensity is continuous on the finite payment interval; more general integrable intensities can also be treated, but continuity is a simple sufficient condition.
A finite installment approximation need not equal the continuous value. With right-end payments and a constant positive intensity, the discounted value decreases through time, so each right-end rectangle uses the lower discount factor for its interval. The approximation is consequently below the continuous value. This observation concerns this particular setup; it is not a license to assume every discrete payment pattern is below its continuous counterpart.
- Check the interval count. A two-year term divided into quarters has eight intervals, not four.
- Multiply intensity by interval length. An intensity of \(c\) dollars per year produces an installment of \(ch\) dollars over an interval of \(h\) years.
- Discount at the actual payment time. For right-end payments, the \(k\)th payment is discounted to time 0 from \(kh\).
- Keep the payment period distinct from the valuation date. A stream beginning after time 0 must be discounted from its actual start.
- Do not use a finite-term factor for an infinite term without checking convergence.
The method is aligned with the SOA Exam FM (Financial Mathematics) syllabus. In a problem, write down the intensity, term, force, time units, and installment timing before forming either the finite sum or its limiting integral.
Check Your Understanding
For each problem, use a constant force of interest and state the present value to the nearest cent where a dollar answer is requested.
- A hypothetical stream pays continuously at $600 per year for 1.5 years. The force is \(\delta=0.03\) per year. Find its present value. (Answer: $880.05)
- A hypothetical project pays continuously at $900 per year for 2 years under \(\delta=0.08\) per year. Find its present value. (Answer: $1,663.38)
- A hypothetical fund has intensity \(r(t)=200+50t\) dollars per year for 2 years, with \(\delta=0.05\) per year. Find its continuous present value. (Answer: $474.23)
- A hypothetical scholarship pays continuously at $300 per year from time 2 through time 5. The force is \(\delta=0.04\) per year. Find its time-0 value. (Answer: $782.89)
- A hypothetical one-year stream pays continuously at $1,000 per year under \(\delta=0.04\) per year. Find both its continuous present value and its approximation using four equal quarterly installments paid at quarter-end. (Answer: continuous $980.26; quarterly approximation $975.37)