Sensitivity of Continuous-Annuity Values to Force

Why Sensitivity Matters

A present value depends on the force of interest used to discount future payments. Knowing the value at one force is useful, but knowing how that value responds to a small change in the force is also important. It measures the immediate effect of a changed valuation assumption and shows which payment times contribute most to that effect.

Let $c(t)$ denote a payment intensity: the amount paid per unit of time at time $t$. Under a constant force of interest $\delta$, a payment at time $t$ is discounted by $e^{-\delta t}$. The present value over a finite term is therefore an integral. This tutorial differentiates that integral while holding the payment schedule fixed; it does not change the payments when the force changes.

Definition: For a nonnegative payment intensity $c(t)$ over $0\leq t\leq T$, its present value under constant force $\delta$ is $$ V(\delta)=\int_0^T c(t)e^{-\delta t}\,dt. $$ The force sensitivity is $V'(\delta)$, the derivative of present value with respect to the force. It measures the change in value per unit change in $\delta$, with $c(t)$ held fixed.

The First Derivative and Its Interpretation

For a finite term, suppose $c(t)$ is integrable and does not depend on $\delta$. Differentiating the discount factor gives $\frac{\partial}{\partial\delta}e^{-\delta t}=-t e^{-\delta t}$. Thus,

V'(\delta)=-\int_0^T t\,c(t)e^{-\delta t}\,dt.

The factor $t$ means that, dollar for dollar of discounted payment, a later payment contributes more to the magnitude of the sensitivity. If $c(t)\geq 0$, then $V'(\delta)\leq 0$: increasing the force cannot increase the present value of nonnegative future payments.

When $V(\delta)>0$, define the duration-like weighted payment time by

D(\delta)=\frac{\int_0^T t\,c(t)e^{-\delta t}\,dt}{\int_0^T c(t)e^{-\delta t}\,dt} =-\frac{V'(\delta)}{V(\delta)}.

The weights in this average are the discounted payments $c(t)e^{-\delta t}\,dt$, divided by their total $V(\delta)$. Consequently, $D(\delta)$ is between the earliest and latest times at which positive payments occur. It is a time, not a dollar amount. The sensitivity itself is $V'(\delta)=-V(\delta)D(\delta)$, in dollars per unit of force when values are in dollars and time is measured in years.

Important distinction: A change of 0.01 in a force is one percentage point of force, not a 1% relative change. For a small change $\Delta\delta$, the first-order estimate is $$ V(\delta+\Delta\delta)\approx V(\delta)+V'(\delta)\Delta\delta. $$ This is a local approximation; it is not generally the exact value at the new force.

Why the Derivative Formula Holds

For a finite term, differentiation under the integral is justified, for example, when $c(t)$ is integrable and the force is restricted to a bounded neighborhood of the value being considered. On the finite interval, the derivative of the integrand is $-t c(t)e^{-\delta t}$, which is integrable under these conditions. The derivative of the integral is therefore the integral of that derivative.

Write the value using the force.
For each payment at time $t$, the present-value contribution is $c(t)e^{-\delta t}\,dt$, so integrate these contributions over the payment period.

Differentiate the discount factor.
The payment schedule is fixed, and $\frac{\partial}{\partial\delta}e^{-\delta t}=-t e^{-\delta t}$. The time $t$ appears because it is the exponent’s multiplier on the force.

Integrate the resulting contributions.
Each discounted payment contributes $-t c(t)e^{-\delta t}\,dt$ to the derivative. Adding those contributions gives $V'(\delta)=-\int_0^T t c(t)e^{-\delta t}\,dt$.

The same reasoning gives a second derivative when differentiation under the integral remains valid:

V''(\delta)=\int_0^T t^2c(t)e^{-\delta t}\,dt.

For nonnegative payments, $V''(\delta)\geq 0$. The present-value function is therefore decreasing and convex in the force: its slope is nonpositive, and the slope becomes less negative as the force rises. This curvature is one reason a first-order sensitivity estimate should not be treated as an exact revaluation for a large change in force.

For a continuous perpetuity, the same formulas apply only when the relevant integrals converge and the derivative may be passed under the integral. A sufficient practical condition is that the payment intensity and its time-weighted versions are integrable after discounting in a neighborhood of the force being used. If the value or its derivative diverges, the corresponding sensitivity formula is not applicable.

Worked Sensitivity Calculations

Worked Example: Level Payments Over a Finite Term

A hypothetical fund pays continuously at an intensity of \$200 per year for 2 years. The valuation force is $\delta=0.3$ per year. Find the present value, its force sensitivity, and the duration-like weighted payment time. State values to the nearest cent and the weighted time to four decimal places.

Method. Use the present-value integral for a level intensity and its differentiated form. The common mistake to avoid is omitting the time factor in the derivative or reversing its sign.

Here $c(t)=200$, $T=2$, and $\delta=0.3$. The present value is

V(0.3)=200\int_0^2 e^{-0.3t}\,dt =200\left[\frac{1-e^{-0.3t}}{0.3}\right]_0^2 =\frac{200(1-e^{-0.6})}{0.3}.

Using $e^{-0.6}\approx0.5488116361$, the remaining values are

1-e^{-0.6}\approx0.4511883639,\qquad V(0.3)\approx\frac{200(0.4511883639)}{0.3} =\$300.7922426\approx \$300.79.

For the sensitivity, integrate $t e^{-0.3t}$. An antiderivative is $-e^{-0.3t}(t/0.3+1/0.09)$, giving

V'(0.3)=-200\int_0^2t e^{-0.3t}\,dt =-\frac{200[1-(1+0.3\cdot2)e^{-0.6}]}{(0.3)^2}.

The bracket is $1-1.6(0.5488116361)=1-0.8780986178=0.1219013822$. Hence

V'(0.3)\approx-\frac{200(0.1219013822)}{0.09} =-\$270.8920\text{ per unit of force} \approx-\$270.89\text{ per unit of force}.

The weighted payment time is

D(0.3)=-\frac{V'(0.3)}{V(0.3)} \approx\frac{270.8920}{300.7922} \approx0.9006\text{ years}.

Check. The result is between 0 and 2 years, the payment period. The negative derivative also agrees with the fact that a higher force reduces the value of positive future payments.

Worked Example: A Continuously Increasing Payment Intensity

A hypothetical grant pays at an intensity $c(t)=30t$ dollars per year during the first 2 years, with no payments outside that period. The valuation force is zero. Find its present value, sensitivity at zero force, and duration-like weighted payment time. Report the value and sensitivity to the nearest cent and the weighted time to four decimal places.

Method. Integrate the payment intensity for value and multiply it by payment time before integrating for sensitivity. The trap is to use the value integral alone for the derivative.

At $\delta=0$, the discount factor is 1. Therefore,

V(0)=\int_0^2 30t\,dt =30\left[\frac{t^2}{2}\right]_0^2 =30\cdot2 =\$60.00.

The force sensitivity is

V'(0)=-\int_0^2 t(30t)\,dt =-30\left[\frac{t^3}{3}\right]_0^2 =-30\cdot\frac{8}{3} =-\$80.00\text{ per unit of force}.

Consequently,

D(0)=-\frac{V'(0)}{V(0)} =\frac{80}{60} =\frac{4}{3} \approx1.3333\text{ years}.

Check. The intensity rises over time, so the weighted average payment time should be later than the midpoint of the interval. The result, $1.3333$ years, is later than 1 year and lies within the 2-year term.

Worked Example: A Continuous Perpetuity With Declining Payments

A hypothetical perpetual benefit has payment intensity $c(t)=50e^{-0.2t}$ dollars per year, where $t$ is in years. The valuation force is $\delta=0.3$ per year. Find the value, sensitivity, and duration-like weighted payment time. Report the value and sensitivity to the nearest cent and the weighted time to four decimal places.

Method. Combine the force discounting with the decay in the payment intensity, then evaluate the convergent integrals. The common mistake is to ignore the decay rate and use the valuation force alone as the exponential rate.

The discounted payment intensity is $50e^{-0.2t}e^{-0.3t}=50e^{-0.5t}$. Since $0.5>0$, the value integral converges:

V(0.3)=\int_0^\infty50e^{-0.5t}\,dt =50\left[\frac{-e^{-0.5t}}{0.5}\right]_0^\infty =\frac{50}{0.5} =\$100.00.

For the sensitivity, $\int_0^\infty t e^{-at}\,dt=1/a^2$ for $a>0$, as follows by integration by parts. Using $a=0.5$,

V'(0.3)=-50\int_0^\infty t e^{-0.5t}\,dt =-\frac{50}{(0.5)^2} =-\$200.00\text{ per unit of force}.

Thus,

D(0.3)=-\frac{V'(0.3)}{V(0.3)} =\frac{200}{100} =2.0000\text{ years}.

Check. The decay in the payment intensity makes the discounted integrals finite. The weighted time is positive, and the negative sensitivity is consistent with discounting positive payments more heavily when the force increases.

Using Sensitivity Without Overstating It

Sensitivity provides a first-order estimate around the force at which the derivative is calculated. For example, if a present value is \$500, its duration-like weighted payment time is 2.4 years, and the force rises by 0.01, then $V'=-500(2.4)=-\$1{,}200$ per unit of force. The first-order change is approximately $-\$1{,}200(0.01)=-\$12$. This is an estimate, not the exact value change.

For nonnegative payment intensities, the nonnegative second derivative means the value curve is convex. Accordingly, over a positive force increase, the local tangent-line estimate generally lies below the actual value at the new force; over a force decrease, the direction of the approximation error reverses. A large change should be evaluated by recalculating the present-value integral rather than relying only on its derivative.

The duration-like quantity here is defined directly as a weighted average of continuous payment times. It is useful for interpreting the derivative of this continuous-annuity value, but it should not be confused with other duration measures that may be defined for different cash flows or valuation settings.

Common Mistakes

Using the wrong sign. Differentiating $e^{-\delta t}$ introduces $-t$, so nonnegative payments have nonpositive sensitivity.
Forgetting the payment-time factor. The derivative is not just the negative of the value integral. Each contribution is multiplied by its time $t$.
Changing the payment schedule while differentiating. The derivative formula holds $c(t)$ fixed. If the payments themselves depend on $\delta$, their derivative must also be included.
Using the wrong exponential rate for a perpetuity. If the payment intensity declines exponentially, combine that decline with the force discounting before evaluating the integral.
Confusing a local estimate with an exact revaluation. The derivative estimates a small change. For a substantial change in force, evaluate the present value again at the new force.
Misreading the units. If time is in years, the derivative is in dollars per unit of annual force. Multiply it by the force change in decimal form.

Key takeaway: For a fixed, nonnegative continuous payment intensity, $V'(\delta)=-\int t c(t)e^{-\delta t}\,dt$. Its magnitude is the present value multiplied by the discounted weighted average payment time, and the value is decreasing and convex in the force when the required integrals exist.

Check Your Understanding

For each problem, calculate the requested sensitivity using the stated force and time units.

A continuous payment intensity of \$80 per year is paid for 3 years. At $\delta=0.2$ per year, find $V$ and $V'$, to the nearest cent. (Answer: $V=\$180.48$; $V'=-\$243.80$ per unit of force.)
A payment intensity is $c(t)=25t$ dollars per year for $0\leq t\leq2$, and the force is zero. Find $V$, $V'$, and $D$. (Answer: $V=\$50.00$; $V'=-\$66.67$ per unit of force; $D=1.3333$ years.)
A 4-year payment stream has fixed intensity $c(t)=100e^{0.05t}$ dollars per year. At $\delta=0.05$, find its value, sensitivity, and weighted payment time. (Answer: $V=\$400.00$; $V'=-\$800.00$ per unit of force; $D=2.0000$ years.)
A perpetual payment intensity is $c(t)=60e^{-0.1t}$ dollars per year, valued at $\delta=0.2$. Find the present value and sensitivity, to the nearest cent. (Answer: $V=\$200.00$; $V'=-\$666.67$ per unit of force.)
A continuous annuity has value \$500 and duration-like weighted payment time 2.4 years at its current force. Estimate the value change for a small increase of 0.01 in force using a first-order approximation. (Answer: approximately $-\$12.00$.)

What You'll Learn