Effects of Interest and Mortality on Continuous Values

Why Interest and Mortality Both Matter

A continuous life insurance benefit is paid at the moment of death. Its present value depends on two features of the future lifetime: when death occurs and how strongly a payment at that time is discounted. A change in either the force of mortality or the force of interest can therefore change the expected present value.

This tutorial studies those effects with the timing and benefit amount held fixed. The central comparisons are straightforward for whole-life insurance: a higher force of interest lowers the value, while mortality that makes death earlier raises it when interest is positive. Derivatives make these directions precise and quantify local sensitivity.

Definition: For a unit continuous whole-life insurance on a life aged $x$, the present value at issue is $Z=e^{-\delta T_x}$, where $T_x$ is the future lifetime in years and $\delta$ is a constant force of interest. Its actuarial present value is $\bar{A}_x=E[Z]$. If the survival function is ${}_tp_x$ and the force of mortality at duration $t$ is $\mu_{x+t}$, then the density of $T_x$ is ${}_tp_x\mu_{x+t}$, and $$ \bar{A}_x=\int_0^\infty e^{-\delta t}{}_tp_x\mu_{x+t}\,dt. $$ This integral assumes a continuous death benefit of one unit, payable immediately at death, with no other benefits or expenses. The discount factor $e^{-\delta t}$ uses a constant force of interest.

The integral separates the two influences: the mortality model determines the probability density of the payment time, and the force of interest determines the discount applied at each possible time. When comparing assumptions, keep the benefit and payment timing the same; otherwise a change in value cannot be attributed to interest or mortality alone.

Effect of the Force of Interest

For a fixed lifetime distribution, differentiate the whole-life value with respect to the force of interest. Provided differentiation can be passed under the integral—for example, when the relevant discounted moments are finite—the result is

\frac{\partial \bar{A}_x}{\partial \delta} =-\int_0^\infty t e^{-\delta t}{}_tp_x\mu_{x+t}\,dt =-E[T_xe^{-\delta T_x}].

The integrand is nonnegative, so this derivative is nonpositive. If death occurs at a positive time with probability one, the derivative is strictly negative. The reason is also visible without calculus: for every fixed $t>0$, increasing $\delta$ decreases $e^{-\delta t}$, so it decreases the present value of a payment at that time.

The magnitude of the derivative is a local sensitivity: it estimates the change in $\bar{A}_x$ for a small change in $\delta$. It is not generally the exact change for a large rate adjustment. A second differentiation gives

\frac{\partial^2 \bar{A}_x}{\partial \delta^2} =E[T_x^2e^{-\delta T_x}]\geq 0,

when this expectation exists. Thus the value is decreasing and convex as a function of the force of interest: its slope is negative, but becomes less negative as $\delta$ increases.

Effect of Mortality and Survival Ordering

A mortality comparison is meaningful only when it specifies how the survival probabilities change. Suppose two models have future lifetimes $T_1$ and $T_2$, and the second model has ${}_tp_2\leq{}_tp_1$ for every $t\geq0$. The second model has a stochastically shorter lifetime: death occurs earlier in the survival-order sense. For a whole-life benefit and $\delta>0$, the discount function $g(t)=e^{-\delta t}$ decreases with $t$. Therefore, earlier death increases the expected discounted payment.

This comparison presumes death eventually occurs under both models. If $\delta=0$, the value is one under either model when death is certain, because the unit benefit is undiscounted. If the models do not have an ordering of their survival probabilities, there is no general conclusion from the phrase “higher mortality” alone; the time pattern of mortality changes matters.

Compare survival.
Check whether one model has a smaller survival probability at every duration, not merely a larger force of mortality at one selected age or time.

Identify the payment pattern.
For whole-life insurance, a death payment occurs eventually under the stated death-certainty assumption. Confirm that the benefit amount and payment timing are unchanged.

Apply the ordering to discounted value.
With positive interest, an earlier payment has a larger discount factor. The model with stochastically shorter lifetime consequently has the larger whole-life value.

This reasoning also explains why mortality changes should not be judged by looking only at the probability of death in one year. A whole-life value reflects the timing distribution across all future durations, with earlier payments weighted more heavily.

Constant-Force Model: A Useful Sensitivity Formula

Suppose the force of mortality is the constant $\mu>0$ at every duration, and the force of interest is the constant $\delta\geq0$, both measured per year. Then ${}_tp_x=e^{-\mu t}$, and the density of the future lifetime is $\mu e^{-\mu t}$. For unit continuous whole-life insurance,

\bar{A}_x =\int_0^\infty e^{-\delta t}\mu e^{-\mu t}\,dt =\mu\int_0^\infty e^{-(\mu+\delta)t}\,dt =\frac{\mu}{\mu+\delta}.

The final equality uses $\int_0^\infty e^{-ct}\,dt=1/c$ for $c>0$. Differentiating this expression shows the effects of each force while holding the other fixed:

\frac{\partial \bar{A}_x}{\partial\delta} =-\frac{\mu}{(\mu+\delta)^2}<0, \qquad \frac{\partial \bar{A}_x}{\partial\mu} =\frac{\delta}{(\mu+\delta)^2}\geq0.

If $\delta>0$, increasing the constant force of mortality increases the whole-life value. If $\delta=0$, the value is one for every $\mu>0$: a certain unit payment is made at death, and there is no discounting. These derivatives are rates of change per unit change in the corresponding force; they do not mean a percentage change.

Worked Example: Raising the Force of Interest

A hypothetical insurer considers a unit continuous whole-life benefit. Assume a constant force of mortality $\mu=0.03$ per year and a constant force of interest initially equal to $\delta=0.04$ per year. Calculate the value, then find the value if the force of interest rises to $0.06$. Compare the exact change with the local sensitivity at the original assumption.

Method. Under constant mortality and interest, use $\bar{A}_x=\mu/(\mu+\delta)$. The benefit, mortality assumption, and payment timing remain fixed; only $\delta$ changes.

At the original assumption, $\mu+\delta=0.03+0.04=0.07$, so $$ \bar{A}_x=\frac{0.03}{0.07}=0.428571\ldots. $$

At the higher force of interest, $\mu+\delta=0.03+0.06=0.09$, giving $$ \bar{A}_x=\frac{0.03}{0.09}=0.333333\ldots. $$

The exact change is $0.333333\ldots-0.428571\ldots=-0.095238\ldots$, a decrease. At the original assumption, the interest sensitivity is $$ \frac{\partial\bar{A}_x}{\partial\delta} =-\frac{0.03}{(0.07)^2} =-\frac{0.03}{0.0049} =-6.122449\ldots. $$

The force increases by $0.02$, so the first-order sensitivity estimate is $(-6.122449\ldots)(0.02)=-0.122449\ldots$. This is only a local estimate; it differs from the exact decrease because the change in force is not infinitesimal. Rounded to four decimal places, the original and revised values are $0.4286$ and $0.3333$, and the exact change is $-0.0952$. The decrease confirms the direction predicted by the derivative. The common mistake avoided is treating a derivative-based estimate as an exact finite change.

Worked Example: Changing the Force of Mortality

For a hypothetical unit continuous whole-life benefit, assume a constant force of interest $\delta=0.03$ per year. Compare a constant force of mortality $\mu=0.02$ with a revised constant force $\mu=0.05$. Find both values and use the mortality derivative to interpret the direction of change.

Method. Apply $\bar{A}_x=\mu/(\mu+\delta)$ separately under each mortality assumption. The same constant force of interest and unit benefit apply in both cases.

With $\mu=0.02$, the denominator is $0.02+0.03=0.05$, so $\bar{A}_x=0.02/0.05=0.400000$. With $\mu=0.05$, the denominator is $0.05+0.03=0.08$, so $\bar{A}_x=0.05/0.08=0.625000$. The revised value is higher by $0.625000-0.400000=0.225000$.

At the original assumption, the derivative with respect to mortality is $$ \frac{\partial\bar{A}_x}{\partial\mu} =\frac{0.03}{(0.02+0.03)^2} =\frac{0.03}{0.0025} =12. $$

The change in force is $0.05-0.02=0.03$, so the local linear estimate of the change is $12(0.03)=0.36$. The exact increase, $0.225000$, is smaller; the derivative is a local measure, not an exact finite-change formula. Rounded to four decimal places, the values are $0.4000$ and $0.6250$, with an increase of $0.2250$. This result is consistent with a mortality increase making death earlier under the constant-force model. The common mistake avoided is applying the derivative at the original point as if the value were linear in mortality.

Term Insurance and the Role of Duration

The same assumptions give a useful term-insurance calculation. For a unit continuous $n$-year term benefit—paid at death only if death occurs by time $n$, with no payment at time $n$ for survival—the value under constant $\mu$ and $\delta$ is

\bar{A}_{x:\overline{n}|} =\int_0^n e^{-\delta t}\mu e^{-\mu t}\,dt =\frac{\mu}{\mu+\delta}\left(1-e^{-(\mu+\delta)n}\right).

The upper limit is $n$ because a death after the term produces no benefit. This expression also shows why the whole-life formula cannot simply be used for a term contract: the term value includes only the part of the death-time distribution within the coverage period.

Worked Example: Comparing Interest Assumptions for a Term Benefit

A hypothetical policy pays one unit continuously at death if death occurs within 10 years. Assume a constant force of mortality $\mu=0.03$ per year. Compare the value at $\delta=0.05$ with the value at $\delta=0.07$. There is no survival payment at year 10.

Method. Use the constant-force term formula with $n=10$. First calculate $\mu+\delta$, then evaluate the exponential term for each interest assumption.

At $\delta=0.05$, $\mu+\delta=0.08$ and $(\mu+\delta)n=0.8$. Since $e^{-0.8}=0.449329\ldots$, $$ \bar{A}_{x:\overline{10}|} =\frac{0.03}{0.08}(1-e^{-0.8}) =0.375(1-0.449329\ldots) =0.375(0.550671\ldots) =0.206502\ldots. $$

At $\delta=0.07$, $\mu+\delta=0.10$ and $(\mu+\delta)n=1$. Since $e^{-1}=0.367879\ldots$, $$ \bar{A}_{x:\overline{10}|} =\frac{0.03}{0.10}(1-e^{-1}) =0.3(1-0.367879\ldots) =0.189636\ldots. $$

The value decreases by $0.189636\ldots-0.206502\ldots=-0.016866\ldots$. Rounded to four decimal places, the two values are $0.2065$ and $0.1896$, a decrease of $0.0169$. This is consistent with the fact that every covered death payment is discounted more heavily at the higher force. The common mistake avoided is including a payment for survival at year 10 when the contract specifies death benefits only.

Interest-Rate Units and Exam Traps

A force of interest is not the same as an effective annual interest rate. If the effective annual rate is $i$, the corresponding constant force is $\delta=\ln(1+i)$, since $e^\delta=1+i$. The conversion matters because continuous benefits are discounted with $e^{-\delta t}$, not by substituting an annual effective rate directly for $\delta$.

For example, if $i=0.05$, then $\delta=\ln(1.05)=0.048790\ldots$, not $0.05$. With constant mortality $\mu=0.03$, the whole-life value is $0.03/(0.03+0.048790\ldots)=0.38076\ldots$, or $0.3808$ to four decimal places. Substituting $0.05$ as the force would give $0.03/(0.03+0.05)=0.3750$, which is not the value under a 5% effective annual rate.

Keep the time unit consistent. If $t$ is measured in years, both forces must be annual. A force stated monthly must be converted before combining it with an annual force.
Check what is held fixed. An interest sensitivity assumes mortality and benefit terms do not change; a mortality comparison assumes the discount basis and contract do not change.
Do not overstate the mortality result. A higher force at one duration alone does not establish survival ordering at every duration. Use the full survival pattern for a general comparison.
Distinguish term from whole life. The term integral stops at the end of coverage and has no maturity payment unless one is explicitly specified.

Key takeaway: For continuous whole-life insurance, increasing the force of interest decreases the value. If positive interest applies and one mortality model has lower survival at every duration, that model produces a higher value. Under constant forces, $\bar{A}_x=\mu/(\mu+\delta)$, with partial derivatives that quantify the local effects.

Check Your Understanding

Use the stated constant-force assumptions, keep all forces in annual units, and round final values to four decimal places where needed.

A unit continuous whole-life benefit has constant $\mu=0.04$ and $\delta=0.06$. Calculate its value. Then calculate the value if $\delta$ rises to $0.08$, and give the decrease. (Answer: $0.4000$, then $0.3333$; decrease $0.0667$.)
A unit continuous whole-life benefit has $\delta=0.03$. Find its value at constant mortality $\mu=0.03$ and at $\mu=0.06$. Give the increase. (Answer: $0.5000$, then $0.6667$; increase $0.1667$.)
A unit 5-year continuous term benefit has constant $\mu=0.02$ and $\delta=0.03$, with no survival benefit. Calculate its present value using the term formula. (Answer: $0.0885$.)
A unit 10-year continuous term benefit has constant $\mu=0.04$ and $\delta=0.06$. Calculate its present value, stating the exponent used in the formula. (Answer: exponent $1.0$; present value $0.2528$.)
For constant forces $\mu=0.02$ and $\delta=0.04$, calculate the whole-life value and both partial derivatives with respect to $\delta$ and $\mu$. Interpret the signs. (Answer: value $0.3333$; $\partial\bar{A}_x/\partial\delta=-5.5556$, $\partial\bar{A}_x/\partial\mu=11.1111$; higher interest lowers the value and higher mortality raises it locally.)

These sensitivity comparisons and valuation methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

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