Accumulated Value Under a Time-Varying Force

Why Each Payment Has Its Own Accumulation Period

For a continuous stream of payments, money arriving at different times does not earn interest for the same length of time. A payment made near the terminal date has little time to accumulate, while an earlier payment accumulates for longer. Under a time-varying force of interest, the rate also changes during each payment’s accumulation period. The correct terminal value therefore uses the force over the interval from each payment time to the terminal date.

Assume interest is deterministic and described by a continuous force of interest $\delta(t)$, measured per unit time. Let $c(t)$ be the payment intensity at time $t$, in dollars per unit time, and let $T$ be the terminal date. Payments are made continuously over the specified interval, with no uncertainty or transaction costs. We seek the accumulated value at time $T$.

Definition: The accumulation factor from time $s$ to time $t$, for $s\leq t$, is $$ A(s,t)=\exp\left(\int_s^t \delta(u)\,du\right). $$ A continuous payment intensity $c(t)$ over $0\leq t\leq T$ has terminal accumulated value $$ \operatorname{AV}(T)=\int_0^T c(t)A(t,T)\,dt =\int_0^T c(t)\exp\left(\int_t^T\delta(u)\,du\right)dt. $$

The lower limit of the inner integral is the payment time $t$, not the start of the entire contract. If payments begin later than time 0, the outer integral begins at that start time. An initial lump sum $P$ at time 0 is accumulated separately as $P A(0,T)$, then added to the accumulated value of the continuous payments.

Why the Terminal-Value Formula Holds

The accumulation factor follows from the defining differential equation for a force of interest. If one unit is invested at time $s$, its value $B(t)$ at time $t$ satisfies $dB(t)/dt=\delta(t)B(t)$ and $B(s)=1$. Separating variables and integrating gives $\ln B(t)=\int_s^t\delta(u)\,du$, so $B(t)=A(s,t)$.

Now divide the payment stream into short time intervals. In an interval of length $\Delta t$ near time $t_j$, the payment is approximately $c(t_j)\Delta t$. It accumulates from $t_j$ to $T$ by the factor $A(t_j,T)$. Adding these accumulated amounts gives a Riemann sum. As the intervals become arbitrarily short, the sum converges to the integral below.

\sum_j c(t_j)\Delta t\,A(t_j,T) \longrightarrow \int_0^T c(t)A(t,T)\,dt.

This reasoning also explains an important distinction: $A(0,T)$ is the factor for a lump sum invested at time 0. It is not the factor for every payment in a continuous stream.

Worked Examples with a Linear Force

Suppose the force varies linearly with time, $\delta(t)=\alpha+\beta t$. For a payment made at time $t$ and accumulated to time $T$, its accumulation factor is

A(t,T)=\exp\left(\alpha(T-t)+\frac{\beta}{2}(T^2-t^2)\right).

The formula follows by integrating $\alpha+\beta u$ from $t$ to $T$. Even though the force is linear, integrating a constant payment intensity times this exponential will not generally produce an elementary antiderivative. A numerical integral or an equivalent special-function expression may be used when needed.

Worked Example: A Level Continuous Payment Under a Linear Force

A hypothetical fund receives payments continuously at an intensity of $1,000 per year from time 0 to time 1. The force is $\delta(t)=0.02+0.02t$ per year. Find the accumulated value at time 1. Use Simpson’s rule with eight equal subintervals, then round to the nearest cent.

Method. A payment at time $t$ accumulates for the force from $t$ to 1. Thus

\operatorname{AV}(1) =1000\int_0^1\exp\left(\int_t^1(0.02+0.02u)\,du\right)dt =1000\int_0^1 e^{0.03-0.02t-0.01t^2}\,dt.

For Simpson’s rule, the step size is $h=1/8=0.125$. The integrand values at $t=0,0.125,\ldots,1$, rounded here to seven decimals, are $1.0304545,\ 1.0277209,\ 1.0246744,\ 1.0213179,\ 1.0176540,\ 1.0136866,\ 1.0094191,\ 1.0048555,\ 1.0000000$. Simpson’s rule gives

\begin{aligned} \int_0^1 e^{0.03-0.02t-0.01t^2}\,dt &\approx \frac{0.125}{3}\big[ 1.0304545+1.0000000\\ &\qquad+4(1.0277209+1.0213179+1.0136866+1.0048555)\\ &\qquad+2(1.0246744+1.0176540+1.0094191)\big]\\ &\approx 1.0168447. \end{aligned}

Therefore, the terminal value is approximately $1000(1.0168447)=\$1,016.8447$, or $1,016.84 to the nearest cent. As a check, the value is greater than the $1,000 total deposited and less than $1000e^{0.03}\approx \$1,030.45$, the value that would result if all deposits were made at time 0.

Common mistake avoided: multiplying the total deposits by the full-term factor $e^{0.03}$ assumes every payment was made at time 0. Later payments earn less interest, so that calculation overstates the accumulated value.

Worked Example: Deferred Payments Proportional to the Force

A hypothetical account receives payments from time 1 to time 3 at intensity $c(t)=600\delta(t)$ dollars per year, where $\delta(t)=0.03+0.01t$ per year. Find the value of these payments at time 3, to the nearest cent.

Method. The payments are deferred until time 1, so only payments with $1\leq t\leq3$ are included. The value is

\operatorname{AV}(3) =600\int_1^3\delta(t)\exp\left(\int_t^3\delta(u)\,du\right)dt.

To evaluate this integral, let $G(t)=\exp(\int_t^3\delta(u)\,du)$. Differentiating gives $G'(t)=-\delta(t)G(t)$. Consequently, the integrand $\delta(t)G(t)$ is $-G'(t)$, and

\begin{aligned} \operatorname{AV}(3) &=600\big[G(1)-G(3)\big]\\ &=600\left[\exp\left(\int_1^3(0.03+0.01u)\,du\right)-1\right]\\ &=600\left[e^{0.10}-1\right]\\ &=600(1.1051702-1)\\ &\approx \$63.10. \end{aligned}

The force integral is $0.03(3-1)+0.005(3^2-1^2)=0.06+0.04=0.10$. Also, $G(3)=1$, since a payment made exactly at the terminal date has no time to accumulate. The result is positive, as expected for positive payments and a positive force throughout the payment period.

Common mistake avoided: starting the payment integral at time 0 would add payments that do not exist. The deferral changes the outer integration limits, even though the terminal date remains 3.

Worked Example: Combining an Initial Deposit and Continuous Payments

At time 0, a hypothetical fund receives a lump sum of $2,500. It also receives continuous payments at intensity $400\delta(t)$ dollars per year for $0\leq t\leq2$, where $\delta(t)=0.04+0.01t$ per year. Find the total value at time 2, rounded to the nearest cent.

Method. Accumulate the lump sum for the full two years. Value the continuous payments separately, using their own accumulation periods. The total force integral is

\int_0^2(0.04+0.01u)\,du=0.04(2)+0.005(2^2)=0.10.

The initial deposit accumulates to $2500e^{0.10}=\$2,762.9255$. For the continuous payments, the same differentiation argument as in the previous example gives

\begin{aligned} \operatorname{AV}_{\text{payments}}(2) &=400\int_0^2\delta(t)\exp\left(\int_t^2\delta(u)\,du\right)dt\\ &=400(e^{0.10}-1)\\ &=400(0.1051702)\\ &=\$42.0681. \end{aligned}

Adding the two components gives $\$2,762.9255+\$42.0681=\$2,804.9936$, or $2,804.99 to the nearest cent. The payment value is not $400e^{0.10}$: the continuous payments are made over time, not deposited as a single amount at time 0.

Reading the Limits and Checking the Result

A reliable setup identifies three items before any calculation: when the payments occur, when the value is required, and which force applies between each payment time and that valuation date. For payments from $r$ to $T$, use

\operatorname{AV}(T)=\int_r^T c(t)\exp\left(\int_t^T\delta(u)\,du\right)dt.

If the force is positive throughout the period, every payment made before $T$ accumulates to more than its original amount. A useful check is therefore that the terminal value exceeds total deposits, except for payments made only at the terminal instant, which contribute no interest. If the force is negative, that comparison need not hold. When $c(t)\geq0$, the accumulated value must also be nonnegative.

For a level payment intensity $c$, the total deposited over $[r,T]$ is $c(T-r)$. That total is not itself the accumulated value unless there is no interest. Each infinitesimal deposit $c(t)\,dt$ has its own accumulation factor, so a full-term factor applied to the total is generally incorrect. This time-specific factor is the central feature of accumulated value under a time-varying force.

Check Your Understanding

Set up the accumulation interval carefully in each problem. Round money to the nearest cent.

A hypothetical fund receives payments at intensity $400\delta(t)$ dollars per year from time 0 to time 1, with $\delta(t)=0.02+0.01t$. Find the value at time 1. (Answer: $400(e^{0.025}-1)=\$10.13$)
Payments arrive at intensity $250\delta(t)$ dollars per year over $0\leq t\leq2$, where $\delta(t)=0.04+0.02t$. Find their value at time 2. (Answer: $250(e^{0.12}-1)=\$31.87$)
A payment intensity of $600 per year is paid continuously from time 1 to time 2. The force is $\delta(t)=0.03+0.01t$. Use Simpson’s rule with four equal subintervals to find the value at time 2. (Answer: approximately $614.22)
A $1,500 deposit is made at time 0, and payments arrive at intensity $300\delta(t)$ dollars per year through time 2. If $\delta(t)=0.05+0.01t$, find the total value at time 2. (Answer: $1500e^{0.12}+300(e^{0.12}-1)=\$1,729.49$)
Payments are made at intensity $K\delta(t)$ dollars per year from time 0 to time 2, with $\delta(t)=0.04+0.01t$. What value of $K$ gives a terminal accumulated value of $105.17, to the nearest dollar? (Answer: $K=105.17/(e^{0.10}-1)\approx \$1,000$)

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