Why a Time-Varying Force Changes Present Value
For a constant force of interest, discounting a payment at time \(t\) requires the force and the time elapsed. If the force itself changes over time, the rate at a single instant is not enough: the discount factor reflects the entire path of the force from time zero to the payment date. This distinction matters whenever a continuous stream of payments is valued under a changing interest environment.
The earlier material on accumulation functions from the force of interest gives the accumulation from time zero to time \(t\) as the exponential of the integral of the force. Present value uses the reciprocal of that accumulation. We will apply this relationship to payment intensities, which specify an amount paid per unit of time.
The force \(\delta(t)\) has units of inverse time; its integral over time is dimensionless. The intensity \(c(t)\) has units of dollars per year, so multiplying it by \(dt\) gives a dollar payment. The formula allows any integrable force and payment intensity. In the examples below, the forces and payment intensities are nonnegative.
Why the Present-Value Integral Works
Imagine dividing the payment period into short intervals of length \(\Delta t\). Around time \(t_j\), the payment over one interval is approximately \(c(t_j)\Delta t\). That payment is made at approximately \(t_j\), so its present value is approximately \(c(t_j)\Delta t D(0,t_j)\). Adding these discounted payments gives a Riemann sum. As the intervals become shorter, the sum approaches the integral.
The key discount factor follows directly from the accumulation function. If the force acts from time zero to \(t\), the accumulation is \(\exp(\int_0^t\delta(s)\,ds)\). Taking its reciprocal gives \(D(0,t)\). Thus each part of the payment stream is discounted according to the force accumulated over all the time before that payment, not just the force at the payment date.
To justify the identity, differentiate the discount factor with respect to time. The chain rule gives \(D'(0,t)=-\delta(t)D(0,t)\). Therefore \(K\delta(t)D(0,t)=-K D'(0,t)\). Integrating from \(0\) to \(T\), and using \(D(0,0)=1\), gives $$ \int_0^T K\delta(t)D(0,t)\,dt =-K\left[D(0,T)-D(0,0)\right] =K\left(1-D(0,T)\right). $$ Substituting the expression for \(D(0,T)\) proves the identity. It is a useful shortcut for a particular relationship between payment intensity and force; it is not a replacement for the general present-value integral.
Worked Examples
Worked Example: A Level Stream Under a Linearly Increasing Force
A fund pays continuously at an intensity of $1,000 per year for 2 years. The force of interest at time \(t\) is \(\delta(t)=0.02+0.01t\), where \(t\) is measured in years. Find the present value to the nearest cent.
Method. Find the accumulated force to each time \(t\), place its negative in the exponential discount factor, and integrate the level payment intensity. The integral does not have an elementary antiderivative, so use composite Simpson's rule with step size \(h=0.25\) years.
First, the accumulated force is $$ \int_0^t(0.02+0.01s)\,ds=0.02t+0.005t^2. $$ Thus $$ \operatorname{PV}=1000\int_0^2 e^{-0.02t-0.005t^2}\,dt. $$ For Simpson's rule, write \(g(t)=e^{-0.02t-0.005t^2}\). The required values, rounded to six decimal places, are:
| \(t\) | 0 | 0.25 | 0.50 | 0.75 | 1.00 | 1.25 | 1.50 | 1.75 | 2.00 |
|---|---|---|---|---|---|---|---|---|---|
| \(g(t)\) | 1.000000 | 0.994702 | 0.988813 | 0.982345 | 0.975310 | 0.967719 | 0.959589 | 0.950932 | 0.941765 |
There are eight subintervals, so composite Simpson's rule gives $$ \begin{aligned} \int_0^2 g(t)\,dt &\approx \frac{0.25}{3}\big[ g(0)+g(2) +4\{g(0.25)+g(0.75)+g(1.25)+g(1.75)\}\\ &\qquad\qquad\qquad +2\{g(0.50)+g(1.00)+g(1.50)\}\big]\\ &\approx \frac{0.25}{3}\left[ 1.941765+4(3.895698)+2(2.923712) \right]\\ &\approx 1.947665. \end{aligned} $$ Therefore the present value is approximately \(1000(1.947665)=\$1{,}947.665\), or $1,947.67 to the nearest cent. The discount factor falls over the period because the force is positive; a value below the undiscounted total of $2,000 is a reasonable check.
Common mistake avoided: using the ending force \(\delta(2)=0.04\) as if it applied throughout the 2 years. The correct discount exponent at time \(t\) is \(0.02t+0.005t^2\), the integral of the changing force up to \(t\).
Worked Example: Payment Intensity Proportional to a Linear Force
A hypothetical account pays continuously for 3 years at an intensity \(c(t)=600\delta(t)\) dollars per year. Its force of interest is \(\delta(t)=0.02+0.01t\). Find the present value to the nearest cent.
Method. Since the payment intensity is 600 times the force, use the identity derived above. First integrate the force over the full payment period: $$ \int_0^3(0.02+0.01t)\,dt =0.02(3)+0.005(3^2) =0.06+0.045 =0.105. $$ The discount factor at year 3 is \(D(0,3)=e^{-0.105}\approx0.900325\). Hence $$ \begin{aligned} \operatorname{PV} &=600(1-e^{-0.105})\\ &\approx600(1-0.900325)\\ &\approx600(0.099675)\\ &\approx \$59.81. \end{aligned} $$ The undiscounted total is \(\int_0^3 600(0.02+0.01t)\,dt=600(0.105)=\$63\), so the present value is appropriately smaller.
Common mistake avoided: multiplying the total undiscounted payments by the discount factor at year 3. Different portions of a continuous stream are paid at different times and receive different discount factors. The identity accounts for that timing through the full integral.
Worked Example: An Exponentially Increasing Force
A hypothetical stream pays at intensity \(c(t)=800\delta(t)\) dollars per year for 4 years. The force is \(\delta(t)=0.03e^{0.2t}\). Find the present value to the nearest cent.
Method. Again, the payment intensity is proportional to the force. Integrate the force, taking care to integrate the exponential with respect to time: $$ \begin{aligned} \int_0^4 0.03e^{0.2t}\,dt &=\frac{0.03}{0.2}(e^{0.2(4)}-1)\\ &=0.15(e^{0.8}-1)\\ &\approx0.15(2.225541-1)\\ &\approx0.183831. \end{aligned} $$ Thus \(D(0,4)=e^{-0.183831}\approx0.832075\). The present value is $$ \begin{aligned} \operatorname{PV} &=800(1-e^{-0.183831})\\ &\approx800(1-0.832075)\\ &\approx800(0.167925)\\ &\approx \$134.34. \end{aligned} $$ The force increases from \(0.03\) at time zero to \(0.03e^{0.8}\approx0.06677\) at year 4. The discount factor of approximately 0.832 is based on the accumulated force over the full period, not on either endpoint force alone.
Common mistake avoided: treating the force as \(0.03e^{0.2(4)}\) for all four years, or forgetting the factor \(1/0.2\) when integrating \(e^{0.2t}\). Both errors change the accumulated force and therefore the valuation.
Setting Up and Checking a Valuation
For a general payment intensity, the main work is usually setting up the nested integral correctly. The inner integral accumulates the force from time zero to the payment time; the outer integral adds the discounted payments over the payment period. A practical sequence is:
Express \(t\), the force, and the payment intensity consistently. If \(t\) is in years, \(\delta(t)\) is per year and \(c(t)\) is dollars per year.
Compute \(J(t)=\int_0^t\delta(s)\,ds\). The dummy variable \(s\) makes clear that the force is integrated up to the payment time \(t\).
Use \(e^{-J(t)}\) for the discount factor and evaluate \(\int_0^T c(t)e^{-J(t)}\,dt\). Use numerical integration if the resulting integral has no convenient elementary form.
Two checks are especially useful. First, if \(\delta(t)\ge0\) throughout the period, then \(0<D(0,t)\le1\), so the present value of a nonnegative stream cannot exceed its undiscounted total. Second, the discount factor must equal 1 at time zero. A formula that gives a different value at \(t=0\) has likely used the force or its integral incorrectly.
Common Exam Trap
Suppose a payment is made at time \(t\). A tempting but incorrect calculation is to discount it using \(e^{-t\delta(t)}\), as though the force at time \(t\) had applied throughout the whole interval. Under a changing force, the correct discount factor is $$ e^{-\int_0^t\delta(s)\,ds}, $$ which generally differs from \(e^{-t\delta(t)}\). They are equal only in special cases, such as when the force is constant over the interval. For a continuous stream, discount each time-\(t\) payment using its own accumulated force before integrating.
The present-value integral is aligned with the SOA Exam FM (Financial Mathematics) syllabus. The essential discipline is to distinguish the instantaneous force from its accumulation: integrate the force first, then use the resulting discount factor in the payment integral.
Check Your Understanding
For each problem, identify the accumulated force before calculating or setting up the value.
- A hypothetical fund pays $4,000 at time 2. The force is \(\delta(t)=0.015+0.005t\) per year. Find the present value to the nearest cent. (Answer: $3,843.16)
- A continuous stream pays at intensity \(100\delta(t)\) dollars per year for 2 years, where \(\delta(t)=0.01+0.01t\). Find its present value to the nearest cent. (Answer: $3.92)
- A stream pays at intensity \(250\delta(t)\) dollars per year for 5 years, where \(\delta(t)=0.04e^{0.1t}\). Find its present value to the nearest cent. (Answer: $57.14)
- A level intensity of $300 per year is paid continuously from time 0 to time 1. The force is \(\delta(t)=0.01+0.01t\). Set up the present-value integral and estimate its value using numerical integration. (Answer: \(\int_0^1 300e^{-0.01t-0.005t^2}\,dt\approx \$298.01\))