Piecewise Payment Intensities

Why Piecewise Payment Intensities Need Careful Valuation

A continuous payment stream need not maintain one payment rate throughout its term. A hypothetical project might receive payments at one rate during an initial phase, at a higher rate after a specified date, and no payments during a temporary pause. The payment rate can jump at a boundary without any lump sum being paid there. To value such a stream correctly, identify every interval on which its payment intensity and the force of interest are fixed or have a known expression, then value each interval using the appropriate discount or accumulation factor.

The method is to split the timeline at every change relevant to the calculation. A payment-rate change requires a new segment; so does a change in the force of interest. If the payment intensity is zero on an interval, that interval contributes no value, although it still affects the timing of later payments through accumulated interest.

Definition: A continuous payment intensity $c(t)$ is the payment amount per unit of time at time $t$. If $c(t)=1{,}200$ dollars per year on an interval, the stream pays $1{,}200\,dt$ dollars during a short interval of length $dt$ near time $t$. A piecewise payment intensity is specified separately on consecutive intervals. A jump in $c(t)$ changes the rate; it does not itself create a lump-sum payment.

Throughout the worked examples, time is measured in years and payment intensities are in dollars per year. Assume payments are made continuously as described, there are no additional lump sums, and the force of interest is deterministic and integrable. Let $\delta(s)$ denote the force of interest at time $s$. The discount factor from time $t$ to time 0 is

\exp\left(-\int_0^t \delta(s)\,ds\right).

Thus, when the stream ends at time $T$, its present value is

\operatorname{PV} = \int_0^T c(t)\exp\left(-\int_0^t \delta(s)\,ds\right)\,dt.

For an accumulated value at time $T$, each payment made at time $t$ accumulates from $t$ to $T$. The corresponding formula is

\operatorname{AV}_T = \int_0^T c(t)\exp\left(\int_t^T \delta(s)\,ds\right)\,dt.

When the force is constant at $\delta$, the present-value contribution from a constant intensity $C$ on $[a,b]$ is $C(e^{-\delta a}-e^{-\delta b})/\delta$, for $\delta\ne0$. Its accumulated-value contribution at $T$, where $b\le T$, is $C(e^{\delta(T-a)}-e^{\delta(T-b)})/\delta$, again for $\delta\ne0$. If $\delta=0$, either contribution is simply $C(b-a)$. These formulas apply only when the force is constant over the interval being valued; a force change inside an interval requires a further split.

Why Splitting the Timeline Works

Integration is additive over adjacent intervals. If $a=u_0<u_1<\cdots<u_n=b$, then the integral over $[a,b]$ is the sum of the integrals over $[u_j,u_{j+1}]$. In a piecewise valuation, choose the boundaries so that each interval has a single applicable payment-intensity expression and a single applicable force expression. On each interval, insert those expressions into the present-value or accumulated-value integral.

A finite set of boundary points has zero length. Therefore, assigning a boundary such as $t=2$ to either adjacent interval does not change a continuous-stream integral. This is why a rate can jump at $t=2$ without paying a lump sum at that instant. A separately specified lump sum would need its own discounted or accumulated value; it is not represented by an ordinary finite payment intensity at one point.

Mark every change date.
Include dates where the payment intensity changes, where it becomes zero or resumes, and where the force of interest changes.

Write the intensity and force on each segment.
State explicitly which payment rate and force apply between each pair of consecutive dates.

Value each segment at the requested date.
Use the accumulated force from time 0 to each payment date for present value, or from each payment date to the target date for accumulated value.

Add the segment values and check coverage.
Confirm that every payment phase is included once, every pause contributes zero, and no interval has been discounted under the wrong force.

Worked Example: Present Value With a Payment Jump and a Pause

A hypothetical fund receives continuous payments for 8 years. Its payment intensity is $1{,}200$ dollars per year from time 0 to 2, $1{,}800$ dollars per year from time 2 to 5, zero from time 5 to 6, and $900$ dollars per year from time 6 to 8. The force of interest is a constant $0.05$ per year. Find the present value at time 0, to the nearest cent.

Method. Split at times 2, 5, and 6. For a constant force $\delta$, the discount factor at time $t$ is $e^{-\delta t}$. The segment from $a$ to $b$ with intensity $C$ therefore contributes $C(e^{-\delta a}-e^{-\delta b})/\delta$.

The first phase contributes

1{,}200\frac{1-e^{-0.05(2)}}{0.05} = 1{,}200\frac{1-0.90483742}{0.05} = 1{,}200(1.90325164) = 2{,}283.90197.

The second phase contributes

1{,}800\frac{e^{-0.05(2)}-e^{-0.05(5)}}{0.05} = 1{,}800\frac{0.90483742-0.77880078}{0.05} = 1{,}800(2.52073268) = 4{,}537.31882.

The interval from 5 to 6 is a payment pause, so its contribution is zero. The final phase contributes

900\frac{e^{-0.05(6)}-e^{-0.05(8)}}{0.05} = 900\frac{0.74081822-0.67032005}{0.05} = 900(1.40996350) = 1{,}268.96715.

Adding the contributions gives $2{,}283.90197+4{,}537.31882+0+1{,}268.96715=8{,}090.18794$. Rounded only at the final step to two decimal places, the present value is $\$8{,}090.19$. The intervals cover the full term, and the pause adds no payments. The common mistake avoided is treating the rate change at time 2 as a lump sum rather than as a change in the rate paid after that date.

When Payment and Force Changes Do Not Coincide

A payment phase can cross a force-change date. In that case, the payment intensity may stay constant while its discounting rule changes. Split the phase at the force-change date as well as at payment-change dates. The accumulated force up to the split date determines the discount factor there; discounting after that date uses the new force. This separation is essential: payment timing and interest accumulation are distinct parts of the valuation.

Worked Example: Present Value With Different Payment and Force Boundaries

A hypothetical stream pays continuously at $1{,}000$ dollars per year from time 0 to 2, then $1{,}500$ dollars per year from time 2 to 4, pauses from time 4 to 5, and pays $2{,}000$ dollars per year from time 5 to 7. The force is $0.04$ per year up to time 3 and $0.06$ per year thereafter. Find the present value at time 0, to the nearest cent.

Method. Split at 2, 3, 4, 5, and 7. The payment phase from 2 to 4 must be split at time 3 because the force changes there. The cumulative force through time $t$ is $0.04t$ for $t\le3$, and $0.12+0.06(t-3)$ for $t>3$.

For times 0 to 2, the contribution is

1{,}000\frac{1-e^{-0.08}}{0.04} = 1{,}000\frac{1-0.92311635}{0.04} = 1{,}922.09135.

From time 2 to 3, the contribution is

1{,}500\frac{e^{-0.08}-e^{-0.12}}{0.04} = 1{,}500\frac{0.92311635-0.88692044}{0.04} = 1{,}357.34659.

From time 3 to 4, discount first to time 3 under the force $0.04$, then from time 3 onward under $0.06$:

1{,}500e^{-0.12}\frac{1-e^{-0.06}}{0.06} = 1{,}500(0.88692044)\frac{1-0.94176453}{0.06} = 1{,}291.25562.

There are no payments from 4 to 5. From time 5 to 7, the cumulative force at time 5 is $0.12+0.06(5-3)=0.24$, so the contribution is

2{,}000e^{-0.24}\frac{1-e^{-0.12}}{0.06} = 2{,}000(0.78662786)\frac{1-0.88692044}{0.06} = 2{,}965.05098.

The total is $1{,}922.09135+1{,}357.34659+1{,}291.25562+0+2{,}965.05098=7{,}535.74454$. To two decimal places, the present value is $\$7{,}535.74$. The interval from 2 to 4 is not valued using one force: the force changes at time 3 even though the payment rate does not. Applying $0.04$ through time 4, or applying $0.06$ before time 3, would both be incorrect.

Accumulated Value at a Future Date

For an accumulated value, the same segmentation principle applies, but each payment is carried forward from its own date to the target date. A payment made earlier accumulates for longer than a payment made later. A pause still contributes zero directly; it does not erase the interest earned by earlier payments.

Worked Example: Accumulated Value Across Multiple Payment Phases

A hypothetical account receives continuous payments at $500$ dollars per year from time 0 to 1, makes no payments from time 1 to 2, receives $1{,}000$ dollars per year from time 2 to 4, and receives $700$ dollars per year from time 4 to 6. The force of interest is a constant $0.03$ per year. Find the accumulated value at time 6, to the nearest cent.

Method. For a constant force $\delta$, a payment at time $t$ accumulates by the factor $e^{\delta(6-t)}$. For an interval $[a,b]$ with intensity $C$, integrate that factor over the interval.

C\int_a^b e^{0.03(6-t)}\,dt = C\frac{e^{0.03(6-a)}-e^{0.03(6-b)}}{0.03}.

The phase from 0 to 1 accumulates to

500\frac{e^{0.18}-e^{0.15}}{0.03} = 500\frac{1.19721736-1.16183424}{0.03} = 589.71867.

There are no payments from 1 to 2. The phase from 2 to 4 accumulates to

1{,}000\frac{e^{0.12}-e^{0.06}}{0.03} = 1{,}000\frac{1.12749685-1.06183655}{0.03} = 2{,}188.67683.

The phase from 4 to 6 accumulates to

700\frac{e^{0.06}-1}{0.03} = 700\frac{1.06183655-1}{0.03} = 1{,}442.85277.

The total accumulated value is $589.71867+0+2{,}188.67683+1{,}442.85277=4{,}221.24827$. Rounded to two decimal places, the accumulated value at time 6 is $\$4{,}221.25$. The early phase has the largest accumulation period, but its smaller payment intensity means it does not necessarily have the largest accumulated contribution.

Common Exam Traps

A reliable valuation depends on matching each payment interval to the correct interest accumulation. In particular, do not use a phase’s payment rate outside its stated dates, and do not discount every payment using only the force that applies at the payment date when the force changes over time.

For example, in the first worked example, an incorrect calculation would use $1{,}800$ dollars per year from time 0 to 5:

1{,}800\frac{1-e^{-0.25}}{0.05} = 1{,}800(4.42398432) = 7{,}963.17178.

That calculation ignores the actual $1{,}200$-dollar rate from time 0 to 2, so it is not the value of the stated stream. The correction is to split at time 2, apply $1{,}200$ on $[0,2]$ and $1{,}800$ on $[2,5]$, and include the later phase after the pause. The correct total is $\$8{,}090.19$, not $\$7{,}963.17$.

Missing a force boundary. If the force changes inside a payment phase, split that phase even if its payment intensity stays constant.
Counting a pause as a payment. An interval with zero intensity contributes zero, not the rate from the preceding phase.
Inventing a lump sum at a jump. A rate change at a single date is not a separate payment unless a lump sum is explicitly specified.
Using the wrong accumulation direction. Present value discounts from each payment date to time 0; accumulated value carries each payment forward to the target date.

Key takeaway: Split a piecewise continuous stream at every payment-intensity change, pause, and force change. Value each segment with the discount or accumulation factor implied by the force over the full path to the valuation date, then add the segment values.

Check Your Understanding

For each question, use years as the time unit and round dollar values to the nearest cent.

A stream pays $800$ dollars per year from time 0 to 3 and $1{,}200$ dollars per year from time 3 to 5. The force is $0.04$ per year throughout. Find the present value at time 0. (Answer: $\$4{,}307.28$)
A stream pays $1{,}000$ dollars per year from time 0 to 2, makes no payments from time 2 to 3, and pays $1{,}600$ dollars per year from time 3 to 4.5. The force is $0.05$ per year throughout. Find the present value at time 0. (Answer: $\$3{,}893.39$)
A stream pays $500$ dollars per year from time 0 to 1, $900$ dollars per year from time 1 to 3, pauses from time 3 to 3.5, and pays $700$ dollars per year from time 3.5 to 4. The force is $0.02$ per year throughout. Find the accumulated value at time 4. (Answer: $\$2{,}761.60$)
A stream pays $700$ dollars per year from time 0 to 2, then $1{,}000$ dollars per year from time 2 to 4. It pauses from time 4 to 5, then pays $1{,}300$ dollars per year from time 5 to 6. The force is $0.04$ per year up to time 2 and $0.06$ per year thereafter. Find the present value at time 0. (Answer: $\$4{,}058.11$)

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