Exponentially Varying Payment Intensities

Why Exponential Payment Intensities Matter

A continuous cash flow need not arrive at a constant rate. A payment intensity may increase by a fixed percentage over time, as in a hypothetical benefit stream indexed to a growth assumption, or decrease exponentially as a hypothetical service obligation winds down. When the force of interest is constant, exponential intensity has a particularly direct valuation: discounting the cash flow changes its exponential rate, so the value can be found from a single integral.

The important comparison is between the rate of change in the payment intensity and the force of interest. If the intensity grows more slowly than the force, later payments receive a smaller present-value weight. If the rates are equal, discounting exactly offsets the growth, leaving a constant discounted intensity. The formula must therefore include a separate case for equal rates.

Definition: Over the interval from time 0 to time $n$, an exponentially varying payment intensity is $c(t)=C e^{gt}$, where $C$ is the intensity at time 0, $g$ is the continuous growth rate (which may be negative), and $t$ is measured in years. Under a constant force of interest $\delta$, a payment made at time $t$ is discounted to time 0 by $e^{-\delta t}$. The present value is the integral of the discounted payment intensity over the interval.

Here $C$ is measured in currency per unit of time, while $g$ and $\delta$ are measured per the same time unit. Thus, if $t$ is in years, both rates must be annual continuous rates. The model assumes that payments are made continuously at the stated intensity throughout the specified interval, the force of interest is constant on that interval, and there are no additional fees, interruptions, or contingencies affecting the payments.

Present Value Formula and Its Derivation

At time $t$, the payment intensity is $C e^{gt}$. A small payment over an interval of length $dt$ is approximately $C e^{gt}dt$. Its present value is that amount multiplied by $e^{-\delta t}$. Summing these discounted amounts over the full payment period gives the integral below.

\operatorname{PV}_0 =\int_0^n C e^{gt}e^{-\delta t}\,dt =C\int_0^n e^{(g-\delta)t}\,dt

Combine the exponential factors.
At time $t$, growth contributes $e^{gt}$ and discounting contributes $e^{-\delta t}$, so their product is $e^{(g-\delta)t}$.

Integrate when the rates differ.
If $g-\delta\ne 0$, an antiderivative of $e^{(g-\delta)t}$ is $e^{(g-\delta)t}/(g-\delta)$. Evaluating at the endpoints gives the quotient formula.

Evaluate the equal-rate case directly.
If $g=\delta$, the integrand is $C$, so the present value is $Cn$, not a quotient with a zero denominator.

Theorem: For $c(t)=C e^{gt}$ over $0\le t\le n$, under a constant force $\delta$, the present value at time 0 is

\operatorname{PV}_0= \begin{cases} C\dfrac{e^{(g-\delta)n}-1}{g-\delta}, & g\ne\delta,\\[6pt] Cn, & g=\delta. \end{cases}

The payment intensity is defined over a finite interval in this formula. The formula does not, by itself, value payments after time $n$; any additional payments require a separate valuation.

The equal-rate result can also be understood as the limit of the unequal-rate formula. As $g-\delta$ approaches zero, the exponent $e^{(g-\delta)n}$ approaches 1, and the quotient approaches $n$. Direct integration is the clearest way to compute the equal-rate case and avoids dividing by zero.

\lim_{x\to 0}\frac{e^{xn}-1}{x}=n \qquad\Longrightarrow\qquad \lim_{g\to\delta}C\frac{e^{(g-\delta)n}-1}{g-\delta}=Cn

Worked Valuations

Worked Example: A Growing Intensity That Is Outpaced by Interest

A hypothetical continuous payment stream has intensity $C=\$1{,}200$ per year at time 0 and grows at a continuous rate $g=0.04$ per year for 5 years. The constant force of interest is $\delta=0.06$ per year. Find the present value at time 0.

Method. Use the finite-interval exponential-intensity formula. Since $g-\delta=0.04-0.06=-0.02$, the payment intensity grows more slowly than the force of interest.

Substitute the given values:

\begin{aligned} \operatorname{PV}_0 &=1200\frac{e^{(0.04-0.06)5}-1}{0.04-0.06}\\ &=1200\frac{e^{-0.10}-1}{-0.02}\\ &=1200\frac{0.9048374180-1}{-0.02}\\ &=1200\frac{-0.0951625820}{-0.02}\\ &=1200(4.75812910)\\ &\approx \$5{,}709.75. \end{aligned}

The present-value factor is positive because both the numerator and denominator in the quotient are negative. As a check, the discounted intensity is $1200e^{-0.02t}$, which is positive and decreasing; its average over five years is below its initial value of $\$1{,}200$ per year. A present value of about $\$5{,}710$ is therefore reasonable, below the undiscounted total of a payment intensity that begins at $\$1{,}200$ per year and increases.

Common mistake avoided: replacing the denominator with $\delta-g$ while leaving the numerator as $e^{(g-\delta)n}-1$ would reverse the sign and produce an impossible negative value. Keep the numerator and denominator paired as shown, or change the signs in both.

Worked Example: Growth Exactly Matches the Force

A hypothetical fund pays continuously for 4 years at intensity $900e^{0.05t}$ dollars per year. The fund is valued using a constant force of interest $\delta=0.05$ per year. Find both the present value at time 0 and the accumulated value at time 4.

Method. The growth rate $g=0.05$ equals the force $\delta=0.05$, so the discounted payment intensity is constant. Use the special-case formula for present value.

\begin{aligned} \operatorname{PV}_0 &=\int_0^4 900e^{0.05t}e^{-0.05t}\,dt\\ &=\int_0^4 900\,dt\\ &=900(4)\\ &=\$3{,}600.00. \end{aligned}

To accumulate this value from time 0 to time 4, multiply by the accumulation factor under the same constant force:

\begin{aligned} \operatorname{AV}_4 &=\operatorname{PV}_0 e^{\delta(4)}\\ &=3600e^{0.05(4)}\\ &=3600e^{0.20}\\ &=3600(1.221402758)\\ &\approx \$4{,}397.05. \end{aligned}

The result can be checked by accumulating each payment directly: an amount paid at time $t$ grows to time 4 by the factor $e^{0.05(4-t)}$. The accumulated-value integrand is therefore $900e^{0.05t}e^{0.05(4-t)}=900e^{0.20}$, constant over the interval. Integrating it over 4 years gives $3600e^{0.20}$, the same result.

Common mistake avoided: substituting $g=\delta$ into the unequal-rate quotient gives $0/0$, which is undefined. The correct present value is obtained from the integral, or equivalently from the equal-rate case $Cn$.

Worked Example: A Declining Intensity and Its Accumulated Value

A hypothetical obligation is paid continuously for 6 years, starting at an intensity of $\$2{,}000$ per year and declining at a continuous rate of 3% per year. The constant force of interest is $\delta=0.04$ per year. Find its present value at time 0 and its accumulated value at time 6.

Method. A decline of 3% per year corresponds to $g=-0.03$, so the intensity is $2000e^{-0.03t}$. Since $g-\delta=-0.03-0.04=-0.07$, use the unequal-rate formula.

\begin{aligned} \operatorname{PV}_0 &=2000\frac{e^{(-0.03-0.04)6}-1}{-0.03-0.04}\\ &=2000\frac{e^{-0.42}-1}{-0.07}\\ &=2000\frac{0.6570468198-1}{-0.07}\\ &=2000\frac{-0.3429531802}{-0.07}\\ &=2000(4.899331146)\\ &\approx \$9{,}798.66. \end{aligned}

Accumulate to time 6 using $e^{\delta n}=e^{0.04(6)}=e^{0.24}$:

\begin{aligned} \operatorname{AV}_6 &=9798.662292e^{0.24}\\ &\approx 9798.662292(1.271249150)\\ &\approx \$12{,}456.54. \end{aligned}

As a check, discounting the accumulated value for 6 years returns the present value: $12{,}456.5411e^{-0.24}\approx 9{,}798.6623$. The positive values are consistent with a positive payment intensity throughout the term.

Common mistake avoided: a decline is represented by a negative $g$, but it does not make the payment intensity negative. The intensity remains positive because $2000e^{-0.03t}>0$ for every time $t$.

Accumulated Value and Other Timing Intervals

For a stream paid from time 0 to time $n$, the accumulated value at time $n$ can be found by accumulating each payment separately. A payment made at time $t$ earns interest for $n-t$ years, so its accumulation factor is $e^{\delta(n-t)}$. Therefore,

\begin{aligned} \operatorname{AV}_n &=\int_0^n C e^{gt}e^{\delta(n-t)}\,dt\\ &=e^{\delta n}\operatorname{PV}_0\\ &= \begin{cases} C\dfrac{e^{gn}-e^{\delta n}}{g-\delta}, & g\ne\delta,\\[6pt] Cn e^{\delta n}, & g=\delta. \end{cases} \end{aligned}

The last expression follows by multiplying the present-value formula by $e^{\delta n}$. It is also obtained directly by integrating the accumulated payment intensity. In the equal-rate case, the accumulated-value integrand is $Ce^{gt}e^{\delta(n-t)}=Ce^{\delta n}$, constant in $t$, giving $Cn e^{\delta n}$.

If a stream begins at a later time, make the time origin explicit. For example, suppose the stream starts at time $s$, lasts $n$ years, and has intensity $C e^{g(t-s)}$ for $s\le t\le s+n$. Its value at time $s$ is the same finite-interval formula as a stream beginning at time 0, because the elapsed time is $t-s$. To find its value at time 0, discount that time-$s$ value by $e^{-\delta s}$. This separation helps prevent a common timing error: using the full calendar time in the growth exponent when the stated intensity is defined relative to the stream’s start.

Interpreting the Rates and Avoiding Traps

The difference $g-\delta$ is the rate governing the discounted intensity. If $g<\delta$, the discounted intensity declines over time. If $g>\delta$, it increases over time. If $g=\delta$, it is level. These descriptions concern the discounted intensity, not the original payment intensity: a payment stream can grow while its discounted intensity declines.

A formula with $g-\delta$ in both its exponent and denominator is convenient because it works without sign changes when $g$ is above or below $\delta$. For $g<\delta$, both $e^{(g-\delta)n}-1$ and $g-\delta$ are negative, so the present value remains positive. For $g>\delta$, both are positive. If calculations use rounded values of exponentials, retain sufficient precision in intermediate steps and round the dollar answer only at the end.

Condition	Discounted intensity	Present-value calculation
$g<\delta$	Declines over time	Use the unequal-rate formula; numerator and denominator are both negative.
$g=\delta$	Constant at $C$	Use $Cn$, not the quotient with zero denominator.
$g>\delta$	Grows over time	Use the unequal-rate formula; numerator and denominator are both positive.

Check the time unit. If $n$ is in years, $g$ and $\delta$ must be annual continuous rates. A monthly duration cannot be combined directly with annual rates without converting units.
Check the sign of the growth rate. Exponential decline uses $g<0$, while growth uses $g>0$.
Check the valuation date. A value at time $n$ is not the same as a value at time 0; convert between them using the appropriate accumulation or discount factor.
Check the interval. The formula covers only the specified finite payment period. It does not automatically include deferred payments or a perpetuity.

Key takeaway: Under constant force $\delta$, exponential intensity $Ce^{gt}$ has discounted intensity $Ce^{(g-\delta)t}$. Integrate this over the actual payment interval; when $g=\delta$, use the direct integral $Cn$.

Check Your Understanding

For each problem, identify the growth rate and payment interval before applying a valuation formula. Round dollar answers to the nearest cent.

A hypothetical stream pays continuously for 4 years at intensity $1500e^{0.02t}$ dollars per year. The constant force is $0.05$ per year. Find the present value at time 0. (Answer: $\$5{,}653.98$)
A hypothetical continuous payment stream has intensity $800e^{0.04t}$ dollars per year for 7 years, and the constant force is $0.04$ per year. Find its present value. (Answer: $\$5{,}600.00$)
A hypothetical fund pays for 5 years at intensity $1000e^{0.03t}$ dollars per year under a constant force of $0.03$ per year. Find the accumulated value at time 5. (Answer: $\$5{,}809.17$)
A hypothetical obligation pays for 4 years at intensity $2500e^{-0.02t}$ dollars per year. The constant force is $0.03$ per year. Find its present value at time 0. (Answer: $\$9{,}063.46$)
A hypothetical stream begins at time 2 and continues for 3 years. Its intensity at time $t$ is $1200e^{0.01(t-2)}$ dollars per year for $2\le t\le5$. With constant force $0.04$ per year, find its value at time 0. (Answer: $\$3{,}178.06$)

Condition	Discounted intensity	Present-value calculation
\(g<\delta\)	Declines over time	Use the unequal-rate formula; numerator and denominator are both negative.
\(g=\delta\)	Constant at \(C\)	Use \(Cn\), not the quotient with zero denominator.
\(g>\delta\)	Grows over time	Use the unequal-rate formula; numerator and denominator are both positive.

What You'll Learn