Convergence of Continuous Perpetuities Under Variable Force

Why Convergence Matters

A finite-term continuous annuity has a finite valuation interval. A perpetuity has no final payment time, so its present value is an improper integral: the value is the limit of the present values over increasingly long finite intervals. That limit may be finite or infinite. The answer depends on the cumulative discount over time, not just on the force of interest at one particular time.

This distinction matters especially when the force varies. A force that declines toward zero can still produce enough cumulative discount for a perpetuity to have finite value. Conversely, merely knowing that the force is positive at every finite time does not guarantee a finite present value. The test is whether the discounted payment intensity has a finite integral over the entire timeline.

Definition: Let $t$ be time measured in years, let $\delta(t)$ be a locally integrable force of interest, and let $c(t)\geq 0$ be a locally integrable payment intensity in dollars per year. Define the cumulative force and discount function by $$ G(t)=\int_0^t \delta(s)\,ds, \qquad D(t)=e^{-G(t)}. $$ The present value of payments made continuously from time 0 onward is the improper integral $$ V=\int_0^\infty c(t)D(t)\,dt =\lim_{T\to\infty}\int_0^T c(t)e^{-G(t)}\,dt. $$ The perpetuity has finite value exactly when this limit is finite.

The assumptions ensure that the force and payment intensity are integrable on every finite interval, so the finite-time valuation is well defined. Since $c(t)$ is nonnegative, the partial present values cannot decrease as $T$ increases. Thus the improper integral either approaches a finite limit or grows without bound. If a model has signed cash flows, positive and negative amounts can cancel; the convergence test here concerns a nonnegative payment stream.

Why the Discount Function Uses Cumulative Force

Over a short time interval of length $dt$, the accumulation factor is approximately $1+\delta(t)dt$. Accumulating from time 0 to time $t$ therefore gives the accumulation function $e^{G(t)}$, where $G(t)$ is the integral of the force. Discounting a payment at time $t$ back to time 0 uses its reciprocal.

D(t)=\frac{1}{e^{G(t)}}=e^{-G(t)}.

A continuous payment intensity $c(t)$ produces an amount approximately $c(t)dt$ over a short interval near time $t$. Its present value is approximately $c(t)D(t)dt$. Adding these contributions over a finite interval and then extending the endpoint to infinity gives the improper-integral formula for $V$.

Theorem: Under the stated nonnegativity and local-integrability assumptions, a continuous perpetuity has finite present value if and only if $\int_0^\infty c(t)e^{-G(t)}dt$ converges. It is the long-run behavior of the product $c(t)e^{-G(t)}$, rather than the force or payment intensity considered alone, that determines convergence.

A Comparison Test for Eventual Discounting

A useful sufficient condition compares the payment intensity with exponential growth and the cumulative force with linear growth. The time at which a force bound begins matters: a lower bound that holds only after some time $t_0$ must retain the discount accumulated before $t_0$.

State eventual bounds.
Suppose that for some $t_0$, $\delta(t)\geq\varepsilon$ for $t\geq t_0$, and $c(t)\leq C e^{gt}$ there, where $C>0$ and $\varepsilon>g$.

Bound the cumulative force.
For $t\geq t_0$, integrate the force bound from $t_0$, not from 0: $$ G(t)=G(t_0)+\int_{t_0}^t\delta(s)\,ds \geq G(t_0)+\varepsilon(t-t_0). $$

Bound the discounted intensity.
Exponentiating the negative cumulative-force bound and applying the payment bound gives $$ c(t)D(t)\leq C e^{\varepsilon t_0-G(t_0)}e^{-(\varepsilon-g)t}. $$

Conclude convergence.
The right side has a finite integral on $[t_0,\infty)$ because $\varepsilon-g>0$. The integral on $[0,t_0]$ is finite by local integrability, so the full present value is finite.

In particular, if the payment intensity is bounded above eventually, take $g=0$. A force bounded below by a positive constant eventually is then sufficient for convergence. It is not necessary: a force can decline toward zero and still give a finite value if its cumulative effect grows fast enough.

Worked Example: Constant Force and Level Payment Intensity

A hypothetical continuous scholarship fund pays at an intensity of $1{,}200$ dollars per year forever. The force of interest is constant at $\delta(t)=0.04$ per year. Find the present value and check whether the perpetuity converges.

Method. Form the cumulative force, obtain the discount function, and evaluate the resulting improper integral. This avoids treating a continuous payment intensity as a single payment at the end of each year.

The cumulative force is $$ G(t)=\int_0^t 0.04\,ds=0.04t. $$ Therefore, the discount function is $D(t)=e^{-0.04t}$, and the present value is $$ V=\int_0^\infty 1{,}200e^{-0.04t}\,dt =1{,}200\left[-\frac{1}{0.04}e^{-0.04t}\right]_0^\infty. $$

As $t\to\infty$, $e^{-0.04t}\to 0$. At $t=0$, $e^0=1$. Hence $$ V=1{,}200\left(0-\left(-\frac{1}{0.04}\right)\right) =1{,}200(25)=30{,}000. $$ The integral is finite, so the present value is $30{,}000$ dollars. The check is consistent with the continuous-perpetuity factor $1/\delta=25$ under a constant force: $1{,}200$ dollars per year multiplied by 25 years of present-value factor gives $30{,}000$ dollars.

Increasing Forces Can Produce Finite Value

When the force increases over time, the cumulative force can grow faster than linearly. This can make the discount function shrink rapidly, but an increasing force should still be integrated before drawing a conclusion. The next example has a force that increases toward a positive limit; its discount function and value can both be found exactly.

Worked Example: An Increasing Force with an Exact Value

A hypothetical research grant pays continuously at $600$ dollars per year forever. Its force of interest is $\delta(t)=t/(1+t)$ per year, with $t$ in years. Determine the present value.

Method. Integrate the time-varying force to find cumulative discount, then integrate the discounted payment intensity. A common mistake is to replace the varying force by its limiting value of 1; that would incorrectly treat the force as constant from time 0.

First, the force is increasing because $$ \delta'(t)=\frac{1}{(1+t)^2}>0 $$ for $t\geq0$. Its cumulative force is $$ G(t)=\int_0^t\frac{s}{1+s}\,ds =\int_0^t\left(1-\frac{1}{1+s}\right)ds =t-\ln(1+t). $$ Thus $$ D(t)=e^{-G(t)} =e^{-t+\ln(1+t)} =(1+t)e^{-t}. $$ The present value becomes $$ V=600\int_0^\infty(1+t)e^{-t}\,dt =600\left(\int_0^\infty e^{-t}\,dt+\int_0^\infty te^{-t}\,dt\right). $$

The first integral is $$ \int_0^\infty e^{-t}\,dt=1. $$ For the second, integration by parts gives $$ \int_0^\infty te^{-t}\,dt =\left[-te^{-t}\right]_0^\infty+\int_0^\infty e^{-t}\,dt =0+1=1, $$ since $te^{-t}\to0$ as $t\to\infty$. Therefore $$ V=600(1+1)=1{,}200. $$ The present value is finite and equals $1{,}200$ dollars. This result uses the complete force path: substituting the limiting force 1 into a constant-force perpetuity formula would give $600$, not the correct value.

Declining Forces and the Borderline Case

For a declining force, it is particularly important to calculate the cumulative force. Consider the family $\delta(t)=k/(1+t)$, where $k>0$. Although the force approaches zero, its integral grows logarithmically. The resulting discount function is a power of $1+t$, and convergence depends on that power.

G(t)=\int_0^t\frac{k}{1+s}\,ds=k\ln(1+t), \qquad D(t)=e^{-G(t)}=(1+t)^{-k}.

With a constant payment intensity $c_0>0$, the present value is $c_0\int_0^\infty(1+t)^{-k}dt$. For $k>1$, an antiderivative is $(1+t)^{1-k}/(1-k)$, and $(1+t)^{1-k}\to0$. The integral is then $1/(k-1)$. For $k=1$, the integral is $\ln(1+t)$, which grows without bound. For $0<k<1$, the same power antiderivative grows without bound. Thus the threshold for a constant payment intensity is $k>1$.

Worked Example: A Declining Force Above the Convergence Threshold

A hypothetical trust makes continuous payments at $750$ dollars per year forever. Its force of interest is $\delta(t)=1.5/(1+t)$ per year. Determine whether its present value is finite and, if so, calculate it.

Method. Use the cumulative-force power form and test the resulting improper integral. The mistake to avoid is declaring the value infinite merely because the force tends to zero.

Integrating the force gives $$ G(t)=\int_0^t\frac{1.5}{1+s}\,ds=1.5\ln(1+t), \qquad D(t)=(1+t)^{-1.5}. $$ Therefore, $$ V=750\int_0^\infty(1+t)^{-1.5}\,dt. $$ An antiderivative is $$ \int(1+t)^{-1.5}\,dt =-2(1+t)^{-0.5}, $$ because differentiating the right side gives $(1+t)^{-1.5}$. Hence $$ \int_0^\infty(1+t)^{-1.5}\,dt =\lim_{T\to\infty}\left[-2(1+T)^{-0.5}+2\right] =2. $$ It follows that $V=750(2)=1{,}500$ dollars. The value is finite because $1.5>1$, even though the force declines toward zero. As a boundary check, changing the coefficient to $1$ would instead give $\int_0^\infty(1+t)^{-1}dt=\lim_{T\to\infty}\ln(1+T)=\infty$.

Common Convergence Traps

Using only the limiting force. A force approaching zero does not by itself imply divergence. For example, the declining force $1.5/(1+t)$ gives a finite value for a level payment intensity, while $1/(1+t)$ gives an infinite value.
Using the force instead of cumulative force. The discount function is $e^{-\int_0^t\delta(s)ds}$, not $e^{-\delta(t)t}$ in general. That shortcut is valid when the force is constant, not for an arbitrary time-varying force.
Dropping the initial-time factor in a tail bound. If $\delta(t)\geq\varepsilon$ only for $t\geq t_0$, then the correct bound includes $e^{\varepsilon t_0-G(t_0)}$. One cannot act as though the lower bound held from time 0.
Checking discount without checking payments. Even when $D(t)$ decreases rapidly, a payment intensity that grows rapidly must also be included in $c(t)D(t)$. The integral of that product is the convergence test.

Key takeaway: First compute $G(t)=\int_0^t\delta(s)ds$, then form $D(t)=e^{-G(t)}$, and finally test $\int_0^\infty c(t)D(t)dt$. A continuous perpetuity has finite present value exactly when this discounted-intensity integral converges. This convergence framework is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, decide whether the perpetuity has finite value and calculate the value when it is finite. Time is measured in years, and the stated payment intensities are in dollars per year.

A perpetuity pays continuously at $250$ dollars per year under the constant force $\delta(t)=0.08$. Find its present value. (Answer: $3{,}125$ dollars.)
A perpetuity pays $600$ dollars per year continuously, and its force is $\delta(t)=3/(1+t)$. Find its present value. (Answer: $300$ dollars.)
A perpetuity pays $400$ dollars per year continuously under $\delta(t)=1/(1+t)$. Does it have finite present value? (Answer: No; the present value diverges.)
A perpetuity pays $1{,}000$ dollars per year continuously under the increasing force $\delta(t)=t/(1+t)$. Find its present value. (Answer: $2{,}000$ dollars.)
A perpetuity pays $1$ dollar per year continuously. Its force is zero for $0\leq t<2$ and is $1$ for $t\geq2$. Find its present value, accounting for the discount accumulated before time 2. (Answer: $3$ dollars.)

What You'll Learn