Linearly Decreasing Payment Intensities

Introduction

A continuous annuity pays at an intensity that can vary over time. When that intensity declines linearly, the stream is larger near the start of the term and smaller near the end. Its value therefore depends on both the total amount paid and when those payments occur. A declining payment intensity may reach zero at the end, or it may remain positive.

The central task is to specify the intensity over the full payment interval and discount each payment at the time it is made. The formulas below assume a constant force of interest when giving closed forms. The same payment stream can be valued under a time-varying force using the corresponding accumulation function, but the constant-force formulas do not apply unchanged in that setting.

Definition: A linearly decreasing payment intensity over $0\leq t\leq n$, with initial intensity $c_0$ and terminal intensity $c_n$, is $$ c(t)=c_0-\frac{c_0-c_n}{n}t =c_0\left(1-\frac{t}{n}\right)+c_n\frac{t}{n}. $$ Here $n>0$ is the term in the chosen time unit, and the intensities are measured in currency per that time unit. The intensity is nonnegative throughout the term exactly when $c_0\geq 0$ and $c_n\geq 0$.

The condition on the endpoints is sufficient because every value of $c(t)$ is a weighted average of $c_0$ and $c_n$, with weights $1-t/n$ and $t/n$, both between zero and one. Thus, if both endpoints are nonnegative, the intensity cannot become negative between them. If either endpoint is negative, the proposed schedule is not a nonnegative payment stream over the entire stated term.

Present Value Under Constant Force

Assume payments are made continuously over $0\leq t\leq n$, there are no payments outside that interval, and the force of interest is a constant $\delta$. A payment made at time $t$ has present-value factor $e^{-\delta t}$. Consequently, a small payment of amount $c(t)\,dt$ contributes $c(t)e^{-\delta t}\,dt$ to present value.

Theorem: Under a constant force of interest $\delta$, the present value of the linearly decreasing intensity is $$ \operatorname{PV}=\int_0^n c(t)e^{-\delta t}\,dt. $$ For $\delta\ne 0$, define $$ J_0=\frac{1-e^{-\delta n}}{\delta}, \qquad J_1=\frac{1-(1+\delta n)e^{-\delta n}}{\delta^2}. $$ Then $$ \operatorname{PV}=c_0J_0-\frac{c_0-c_n}{n}J_1. $$ If $\delta=0$, the value is the undiscounted total payment, $$ \operatorname{PV}=\frac{n(c_0+c_n)}{2}. $$

The units provide a useful check. If time is measured in years, then $n$ is in years, $\delta$ is per year, and $c_0,c_n$ are currency per year. The integral has units of currency. A force quoted per month must instead be paired with time in months and payment intensities per month, or converted consistently before calculation.

Why the Formula Holds

Substitute the linear intensity into the present-value integral. The two integrals that result are the discounted value of a unit level intensity and the discounted first time moment:

\operatorname{PV} =\int_0^n\left(c_0-\frac{c_0-c_n}{n}t\right)e^{-\delta t}\,dt =c_0\int_0^n e^{-\delta t}\,dt -\frac{c_0-c_n}{n}\int_0^n te^{-\delta t}\,dt.

For $\delta\ne0$, direct integration gives

\int_0^n e^{-\delta t}\,dt =\left[-\frac{e^{-\delta t}}{\delta}\right]_0^n =\frac{1-e^{-\delta n}}{\delta}=J_0.

Integration by parts evaluates the second integral. Take $u=t$ and $dv=e^{-\delta t}dt$, so $du=dt$ and $v=-e^{-\delta t}/\delta$. Then

\begin{aligned} \int_0^n te^{-\delta t}\,dt &=\left[-\frac{te^{-\delta t}}{\delta}\right]_0^n +\frac{1}{\delta}\int_0^n e^{-\delta t}\,dt\\ &=-\frac{ne^{-\delta n}}{\delta} +\frac{1-e^{-\delta n}}{\delta^2}\\ &=\frac{1-(1+\delta n)e^{-\delta n}}{\delta^2}=J_1. \end{aligned}

Putting these two integrals back into the expression for present value proves the stated formula. When $\delta=0$, the discount factor is one, so integrating the straight-line intensity gives the area of a trapezoid with parallel sides $c_0$ and $c_n$ and width $n$. That area is $n(c_0+c_n)/2$.

Worked Valuations

In each example, the intensity is stated in dollars per year, time is measured in years, and the force of interest is constant over the full payment term. Intermediate calculations retain additional digits; final dollar amounts are rounded to the nearest cent.

Worked Example: Intensity Declining to Zero

A hypothetical fund pays continuously for 5 years. Its payment intensity declines linearly from $1,200 per year at time 0 to zero at time 5. The force of interest is $\delta=0.04$ per year. Find the present value.

Method. Use the constant-force integral, with $c_0=1200$, $c_n=0$, $n=5$, and $\delta=0.04$. The intensity is nonnegative because both endpoints are nonnegative.

First calculate the two integral factors, using $e^{-0.2}=0.818730753078$:

J_0=\frac{1-e^{-0.2}}{0.04} =\frac{1-0.818730753078}{0.04} =4.53173117305,

J_1=\frac{1-(1+0.2)e^{-0.2}}{0.04^2} =\frac{1-(1.2)(0.818730753078)}{0.0016} =10.9519351915.

The slope magnitude is $(1200-0)/5=240$ dollars per year squared. Therefore,

\begin{aligned} \operatorname{PV} &=1200J_0-240J_1\\ &=1200(4.53173117305)-240(10.9519351915)\\ &=5438.07740766-2628.46444596\\ &=2809.61296170. \end{aligned}

The present value is $2,809.61 to the nearest cent. As a check, the undiscounted total is the area of a triangle: $1200(5)/2=\$3,000$. A positive force discounts the payments, so a present value below $3,000 is reasonable. This method avoids treating the stream as a level annuity at its average intensity; that shortcut would not account for the timing of the payments.

Worked Example: A Positive Terminal Intensity

A hypothetical maintenance contract pays continuously for 4 years. The payment intensity decreases linearly from $1,000 per year to $300 per year. If the constant force of interest is $\delta=0.06$ per year, calculate the present value.

Method. The endpoints are $c_0=1000$ and $c_n=300$, so the intensity remains nonnegative and the decline is $700/4=175$ dollars per year squared. Using $e^{-0.24}=0.786627861067$,

J_0=\frac{1-e^{-0.24}}{0.06} =\frac{1-0.786627861067}{0.06} =3.55620231556,

J_1=\frac{1-(1+0.24)e^{-0.24}}{0.06^2} =\frac{1-(1.24)(0.786627861067)}{0.0036} =6.82818118819.

Apply the formula with the full initial intensity and the decline:

\begin{aligned} \operatorname{PV} &=1000J_0-175J_1\\ &=1000(3.55620231556)-175(6.82818118819)\\ &=3556.20231556-1194.93170793\\ &=2361.27060763. \end{aligned}

The present value is $2,361.27 to the nearest cent. The positive terminal intensity does not mean the annuity continues beyond year 4; the payment stream ends at that boundary. The calculation includes payments only over $0\leq t\leq4$.

Worked Example: Checking Feasibility Before Valuation

A hypothetical four-year service arrangement proposes the intensity $c(t)=800-250t$ dollars per year, with time in years. The force of interest is $\delta=0.03$ per year. Check whether the schedule is nonnegative throughout the four years, and, if so, find its present value.

Feasibility check. The starting intensity is $c(0)=800$. At the stated end time, $c(4)=800-250(4)=-200$. The schedule therefore becomes negative before the contract ends and is not a feasible nonnegative payment intensity over the full four-year term. The zero is reached when $800-250t=0$, or $t=3.2$ years. Thus, the proposed intensity is nonnegative only through year 3.2.

If the arrangement instead ends at year 3, its terminal intensity is $800-250(3)=50$, so that shorter schedule is feasible. For this modified three-year term, use $c_0=800$, $c_n=50$, and decline magnitude $(800-50)/3=250$. With $e^{-0.09}=0.913931185271$,

J_0=\frac{1-e^{-0.09}}{0.03} =\frac{1-0.913931185271}{0.03} =2.86896049096,

J_1=\frac{1-(1.09)e^{-0.09}}{0.03^2} =\frac{1-(1.09)(0.913931185271)}{0.0009} =4.23889783822.

The present value of the modified three-year schedule is

\begin{aligned} \operatorname{PV} &=800J_0-250J_1\\ &=800(2.86896049096)-250(4.23889783822)\\ &=2295.16839277-1059.72445956\\ &=1235.44393321. \end{aligned}

The modified schedule has present value $1,235.44 to the nearest cent. The key step is checking the endpoint before valuing the stated term; mechanically integrating a negative intensity would not represent the intended nonnegative payment stream.

Common Mistakes and Practical Checks

A frequent error is to replace the declining stream with its average intensity and value that as a level annuity. The average intensity gives the correct undiscounted total, but not generally the correct present value, because discount factors differ across payment times. For example, the first example has an average intensity of $600 per year. Its level-annuity shortcut would give $600J_0=600(4.53173117305)=\$2,719.04$, not the correct linearly declining-stream value of $2,809.61. The larger early payments receive less discount, so the declining stream has greater present value than a level stream with the same undiscounted total.

Other useful checks are:

Confirm that $c_0$ and $c_n$ are rates, not total payments. Their units must match the time unit used for $n$ and $\delta$.
Check both endpoints for nonnegativity before evaluating the integral. A straight line between nonnegative endpoints cannot dip below zero.
Use $e^{-\delta t}$ for constant force. An effective annual rate $i$ instead gives the discount factor $(1+i)^{-t}$ for fractional-year time $t$; do not substitute $i$ for $\delta$.
When $\delta=0$, use the trapezoid-area result or the zero-force integral. The displayed expressions for $J_0$ and $J_1$ have zero denominators at $\delta=0$ and should not be evaluated by direct substitution.
Keep the term boundary explicit: a positive intensity at $t=n$ is still not a payment after $n$ unless a later payment period is separately specified.

Key takeaway: Specify the linear intensity from its two endpoint rates, verify nonnegativity over the stated term, and discount the intensity at each payment time. Under a constant force, the resulting value is $c_0J_0-(c_0-c_n)J_1/n$; with zero force, it is the area of the corresponding trapezoid.

Check Your Understanding

For each problem, identify the payment term and use consistent time units. Answers are rounded to the nearest cent where stated.

A hypothetical two-year stream declines linearly from $500 per year to $100 per year. The force of interest is 0.08 per year. Find its present value. (Answer: $564.31)
A hypothetical three-year annuity has initial intensity $900 per year and terminal intensity $300 per year. The force of interest is 0.04 per year. Find its present value. (Answer: $1,713.15)
A four-year continuous payment stream declines linearly from $600 per year to $200 per year. With zero interest, find the total payment. (Answer: $1,600.00)
A proposed intensity is $700-200t$ dollars per year over a four-year term. Find the time at which it first reaches zero and determine whether the full-term schedule is nonnegative. (Answer: It reaches zero at 3.5 years; it is negative during the final 0.5 year, so the four-year schedule is not nonnegative throughout.)
A hypothetical five-year annuity begins at $1,000 per year and declines linearly to $400 per year. With zero interest, find the total payment and state whether the intensity is nonnegative throughout. (Answer: $3,500.00; yes, since both endpoint intensities are nonnegative.)

What You'll Learn