Linearly Increasing Payment Intensities

What a Linearly Increasing Payment Intensity Means

A continuous annuity pays throughout an interval rather than on a sequence of discrete dates. Its payment intensity specifies the rate of payment at each time. If that rate increases linearly, it has both a starting level and a slope: the payment stream need not start at zero.

Let $t$ measure time in years from the beginning of a finite payment period of length $n$. Suppose the payment intensity is $c(t)=c_0+kt$, where $c_0$ is the initial rate and $k$ is the increase in the rate per year. For example, $c_0=\$50$ per year and $k=\$10$ per year per year means the rate is $\$50$ per year at $t=0$, $\$60$ per year at $t=1$, and continues to rise at that slope.

Definition: A finite linearly increasing payment intensity on $0\leq t\leq n$ is $c(t)=c_0+kt$, where $c_0\geq 0$ is the initial payment rate, $k\geq 0$ is the rate's slope, and $n>0$ is the term. The payment made over a short interval of length $dt$ near time $t$ is approximately $c(t)\,dt$. Thus $c(t)$ is a rate, not a lump-sum payment at time $t$.

Assume a constant force of interest $\delta$ over the valuation period. A dollar paid at time $t$ has present value $e^{-\delta t}$ dollars at time 0. The present value of the stream is therefore the integral of its payment intensity multiplied by the appropriate discount factor.

Present Value Under Constant Force

Theorem: For $c(t)=c_0+kt$ over $0\leq t\leq n$, under a constant force of interest $\delta\ne 0$, the present value at time 0 is $$ \operatorname{PV}_0 =\int_0^n(c_0+kt)e^{-\delta t}\,dt =c_0\frac{1-e^{-\delta n}}{\delta} +k\frac{1-(1+\delta n)e^{-\delta n}}{\delta^2}. $$ If $\delta=0$, the value is $c_0n+\frac{1}{2}kn^2$.

The first term is the value of a level payment intensity $c_0$. The second is the value of the additional intensity $kt$, which starts at zero and rises linearly. The two components may be valued separately because integration is linear.

To justify the formula, split the integral into its level and increasing parts. The level component integrates directly. For the increasing component, integration by parts uses $u=t$ and $dv=e^{-\delta t}dt$, so $du=dt$ and $v=-e^{-\delta t}/\delta$.

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

Multiplying this integral by $k$, and adding $c_0\int_0^n e^{-\delta t}dt$, gives the stated result. When $\delta=0$, there is no discounting, so integrate the intensity itself: $\int_0^n(c_0+kt)dt=c_0n+\frac{1}{2}kn^2$. This is also the limit of the formula as $\delta$ approaches zero; the separate terms in the nonzero-force expression should not be evaluated by substituting zero directly.

Worked Example: Immediate Payments with a Nonzero Starting Rate

A hypothetical fund pays continuously for 2 years. Its payment intensity $t$ years after the start is $50+10t$ dollars per year. The force of interest is constant at $\delta=0.04$ per year. Find the present value and the accumulated value at the end of year 2. Round money to the nearest cent.

Method. Use the level-plus-ramp decomposition under constant force. The intensity starts at $\$50$ per year and rises by $\$10$ per year of elapsed time; it does not start at zero.

Here $c_0=50$, $k=10$, $n=2$, and $\delta n=0.08$. First, $e^{-0.08}=0.9231163464$, so $1-e^{-0.08}=0.0768836536$.

The level component is $50(0.0768836536)/0.04=96.1045670$ dollars. For the increasing component, $1.08e^{-0.08}=1.08(0.9231163464)=0.9969656541$, giving $1-1.08e^{-0.08}=0.0030343459$. Therefore the increasing component is $10(0.0030343459)/(0.04)^2=18.9646619$ dollars.

Adding the components gives $\operatorname{PV}_0=96.1045670+18.9646619=115.0692289$, or $\$115.07$.

To accumulate this value to time 2 under the same constant force, multiply by $e^{0.04(2)}=e^{0.08}=1.0832870677$: $115.0692289(1.0832870677)=124.6530$, or $\$124.65$. The check is that accumulation at a positive force must produce a value greater than the present value.

Common mistake avoided: The growth component is not obtained by treating the slope $10$ as a constant payment rate over the whole term. It is the value of the ramp $10t$, so its integral has the factor $1-(1+\delta n)e^{-\delta n}$.

Deferred Payments and Accumulated Values

If the payment period begins later, define elapsed time from the start of the payment period before applying the formula. Suppose payments begin at time $m$, last for $n$ years, and have intensity $c_0+ks$, where $s=t-m$ is elapsed time since payments began. Under a constant force, first calculate the value at time $m$ as though the stream began then, and discount that value back $m$ years.

\operatorname{PV}_0 =e^{-\delta m} \left( c_0\frac{1-e^{-\delta n}}{\delta} +k\frac{1-(1+\delta n)e^{-\delta n}}{\delta^2} \right), \qquad \delta\ne 0.

The accumulated value at the end of the payment period is likewise obtained from the time-0 present value by multiplying by $e^{\delta(m+n)}$. This follows because every payment discounted to time 0 can instead be accumulated to the common endpoint by the same total accumulation factor.

Worked Example: A Deferred Increasing Payment Stream

A hypothetical grant begins making continuous payments at time 1 and finishes at time 4. At elapsed time $s$ years after payments begin, its intensity is $12+3s$ dollars per year. The force of interest is constant at $0.06$ per year. Find the present value at time 0, rounded to the nearest cent.

Method. Find the value at the start of payments for a 3-year stream, then discount it for the 1-year deferral. The slope is measured against elapsed payment time $s$, not calendar time from time 0.

For the 3-year stream, $c_0=12$, $k=3$, $n=3$, and $\delta n=0.18$. Since $e^{-0.18}=0.8352702114$, the level component at time 1 is $12(1-0.8352702114)/0.06=32.9459577$ dollars.

For the ramp, $1.18(0.8352702114)=0.9856188495$, so the numerator is $1-0.9856188495=0.0143811505$. The ramp component at time 1 is $3(0.0143811505)/(0.06)^2=11.9842921$ dollars. Thus the value at time 1 is $32.9459577+11.9842921=44.9302498$ dollars.

Discount for the one-year deferral: $44.9302498e^{-0.06}=44.9302498(0.9417645336)=42.3137$. The present value at time 0 is therefore $\$42.31$.

Check. The payment rates range from $\$12$ to $\$21$ per year during the payment period. A value of $\$42.31$ at time 0 is plausible for three years of payments of that scale, discounted for both the payment timing and the one-year deferral.

Common mistake avoided: Using $n=4$ in the annuity formula would count the deferral year as a payment year. The payment term is 3 years; the separate factor $e^{-0.06}$ accounts for the deferral.

Reading the Parameters and Checking the Result

The initial intensity $c_0$ and slope $k$ have different units. If time is measured in years and currency in dollars, $c_0$ is dollars per year, while $k$ is dollars per year squared: multiplying $k$ by elapsed years gives a payment intensity in dollars per year. The total undiscounted payments are $c_0n+\frac{1}{2}kn^2$, which provides a useful check when $\delta=0$, or a benchmark when discounting is present.

Quantity	Interpretation	Useful check
$c_0$	Payment intensity at the start of the payment period	The intensity at $t=0$ must equal $c_0$
$k$	Increase in intensity per unit of elapsed time	The intensity at $t=n$ is $c_0+kn$
$n$	Length of the actual payment period	A deferral is not included in $n$
$\delta$	Constant force of interest during the valuation interval	For $\delta>0$, the present value is below total undiscounted payments

The formula assumes that the stated intensity applies throughout the payment interval and that the force of interest is constant over the period being valued. It does not describe discrete payments made at selected dates, nor does it apply unchanged if the payment rate or force follows a different pattern. A negative slope would describe a decreasing intensity, which is a different case.

A common exam trap is to confuse a linearly increasing intensity with a sequence of annual payments that increases by a fixed amount. Here, the cash paid in an interval is the integral of the intensity over that interval. Also check the time origin: for a deferred stream, the starting rate is the intensity when payments begin if the problem defines it that way. Always express the intensity in terms of elapsed payment time before applying the formula.

Key takeaway: For a finite stream with intensity $c_0+kt$, value the nonzero starting level and the linear ramp as separate components. Under constant force, discount each instant's payment by $e^{-\delta t}$; for a deferred stream, value the payment-period annuity first and then apply the deferral factor.

Check Your Understanding

For each problem, use the stated time origin and force of interest. Round dollar values to the nearest cent unless an exact expression is requested.

A continuous stream pays for 2 years with intensity $10+2t$ dollars per year, where $t$ is years from the start. The force is constant at $0.05$. Find the present value at time 0. (Answer: $\$22.78$)
A 3-year continuous payment stream has intensity $6+1.5t$ dollars per year, with $t$ measured from the start of payments. Under constant force $0.08$, find its present value at the start. (Answer: $\$21.76$)
A 4-year stream has intensity $3+2t$ dollars per year and no interest. Find the total payment. (Answer: $\$28.00$)
A stream begins at time 2 and ends at time 5. Its intensity is $5+s$ dollars per year, where $s$ is elapsed time since it began. The force is constant at $0.03$. Find its present value at time 0. (Answer: $e^{-0.06}\left[5(1-e^{-0.09})/0.03+\{1-(1.09)e^{-0.09}\}/0.0009\right]\approx \$17.50$)
A 2-year stream has intensity $14+3t$ dollars per year and constant force $0.04$. Find its accumulated value at time 2. (Answer: $\$35.31$)

Quantity	Interpretation	Useful check
\(c_0\)	Payment intensity at the start of the payment period	The intensity at \(t=0\) must equal \(c_0\)
\(k\)	Increase in intensity per unit of elapsed time	The intensity at \(t=n\) is \(c_0+kn\)
\(n\)	Length of the actual payment period	A deferral is not included in \(n\)
\(\delta\)	Constant force of interest during the valuation interval	For \(\delta>0\), the present value is below total undiscounted payments

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