Solving for an Unknown Force of Mortality

Why an Insurance Value Can Reveal Mortality

A continuous insurance value reflects both the chance that death occurs at each time and the discount applied to a benefit paid at that time. If the interest assumption is known and mortality follows a constant force, a specified insurance value can therefore be used to infer the unknown mortality force. Whole-life insurance leads to a direct algebraic solution; term insurance generally requires solving an equation.

The inversion is meaningful only when the product and assumptions match the formula. In particular, the benefit must be paid at the moment of death, and the mortality force and force of interest must be constant over the covered period. The examples below use unit-benefit values first, then show how a specified benefit amount enters.

Definition: Under a constant force of mortality $\mu$, the future lifetime $T$ has survival probability ${}_tp_x=e^{-\mu t}$ and density $f_T(t)=\mu e^{-\mu t}$, for $t\geq 0$. With constant force of interest $\delta$, a unit benefit paid at time $t$ has present value $e^{-\delta t}$. A continuous term insurance with term $n$ pays one unit at death if death occurs before time $n$, and pays nothing otherwise.

Valuation Formulas to Invert

For a unit benefit payable continuously at death, the present value is $e^{-\delta T}$ if death occurs within the coverage period. Taking its expectation gives the valuation integral. For whole life, the integral extends over all future times; for an $n$-year term, it extends only to $n$.

\bar{A}_x =\int_0^\infty e^{-\delta t}\mu e^{-\mu t}\,dt =\frac{\mu}{\mu+\delta}, \qquad \mu+\delta>0,

\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), \qquad \mu+\delta>0.

The bar indicates that the death benefit is paid at the moment of death. These are unit-benefit values. If the benefit is a constant amount $B$, its expected present value is $B\bar{A}_x$ for whole life or $B\bar{A}_{x:\overline{n}|}$ for term insurance. Divide the observed expected present value by $B$ before using a unit-benefit formula.

Derivation: At time $t$, the probability density of death is $\mu e^{-\mu t}$, and the discount factor for a benefit paid then is $e^{-\delta t}$. Their product is the discounted expected payment density, $\mu e^{-(\mu+\delta)t}$. Integrating it over the times at which the policy pays gives the expected present value. For whole life, $\int_0^\infty e^{-(\mu+\delta)t}\,dt=1/(\mu+\delta)$; for term insurance, $\int_0^n e^{-(\mu+\delta)t}\,dt=(1-e^{-(\mu+\delta)n})/(\mu+\delta)$. Multiplication by $\mu$ yields the stated formulas.

Solving the Whole-Life Equation Directly

If the unit whole-life value $A$ and the force of interest $\delta$ are known, set $A=\bar{A}_x$ and rearrange. For $\delta>0$ and $0<A<1$, the solution is unique and positive:

A=\frac{\mu}{\mu+\delta} \quad\Longrightarrow\quad A(\mu+\delta)=\mu \quad\Longrightarrow\quad \mu=\frac{A\delta}{1-A}.

The expression requires a force of interest, not an effective annual interest rate. If the given rate is an effective annual rate $i$, convert it using $\delta=\ln(1+i)$. An effective annual rate of 5%, for example, corresponds to $\delta=\ln(1.05)$, not $0.05$.

Worked Example: Recovering Mortality From a Whole-Life Value

A hypothetical continuous whole-life policy pays $250,000 at the moment of death. Its expected present value is $150,000. Assume a constant force of interest $\delta=0.04$ and a constant force of mortality $\mu$. Find $\mu$, then verify the value.

Method. First convert the dollar expected present value to a unit-benefit value: $A=150{,}000/250{,}000=0.6$. Use the whole-life inversion, then substitute the result in the original formula.

Solve. The known force of interest is $0.04$, so $\mu=A\delta/(1-A)=0.6(0.04)/(1-0.6)=0.024/0.4=0.06$. Thus the inferred force of mortality is $0.06$ per year.

Verify. The unit whole-life value from the assumed model is $\bar{A}_x=0.06/(0.06+0.04)=0.06/0.10=0.6$. Multiplying by the benefit gives $250{,}000(0.6)=\$150{,}000$, as required.

Mistake avoided. Using the dollar amount $150{,}000$ as if it were the unit-benefit value would be invalid. The formula takes a value per dollar of benefit; scaling by the $250,000 benefit must be removed first.

Solving the Term Equation

For a finite term, the unknown force appears both in the fraction and in the exponential. In general, the equation does not simplify to a direct algebraic formula for $\mu$. Instead, define a function from the valuation formula and solve for the force at which that function equals the observed unit value.

For fixed $\delta\geq 0$ and $n>0$, the term value increases with $\mu$. When $\delta>0$, differentiating the term formula gives

\frac{\partial \bar{A}_{x:\overline{n}|}}{\partial\mu} = \frac{\delta}{(\mu+\delta)^2} \left(1-e^{-(\mu+\delta)n}\right) + \frac{\mu}{\mu+\delta}n e^{-(\mu+\delta)n}.

Both terms are nonnegative, and the derivative is positive for $\mu\geq0$, $n>0$, and $\delta>0$. The value starts at zero when $\mu=0$ and approaches one as $\mu$ increases without bound. Consequently, each observed unit term value strictly between zero and one identifies exactly one nonnegative $\mu$ under these assumptions. A numerical solver can use bisection, which keeps a bracket around the solution, or Newton's method, which uses the derivative to refine a trial value.

Worked Example: Numerically Solving a Six-Year Term Value

A hypothetical six-year continuous term policy pays one unit at death before the end of six years. The force of interest is $\delta=0.04$, and the observed unit value is $0.270713018$. Assume a constant force of mortality. Infer $\mu$ and verify the result.

Method. Set $g(\mu)=\frac{\mu}{\mu+0.04}(1-e^{-6(\mu+0.04)})$ and solve $g(\mu)=0.270713018$. The derivative is $g'(\mu)=\frac{0.04}{(\mu+0.04)^2}(1-e^{-6(\mu+0.04)})+\frac{\mu}{\mu+0.04}6e^{-6(\mu+0.04)}$. Because the term value is increasing, a bracket or Newton iteration is appropriate.

Bracket the solution. At $\mu=0.05$, $\mu+\delta=0.09$, $e^{-6(0.09)}=e^{-0.54}=0.582748252$, and $g(0.05)=(0.05/0.09)(1-0.582748252)=0.231806527$. This is below the target. At $\mu=0.06$, $\mu+\delta=0.10$, and $g(0.06)=(0.06/0.10)(1-e^{-0.60})=0.6(1-0.548811636)=0.270713018$. The target is therefore attained at $\mu=0.06$.

Show a numerical refinement. Starting Newton's method at $0.05$, the derivative formula gives $g'(0.05)\approx4.002995$. The first update is $\mu_1=0.05+(0.270713018-0.231806527)/4.002995\approx0.059719$. Evaluating at this trial value gives $g(0.059719)\approx0.269654$ and $g'(0.059719)\approx3.785$. The next update is approximately $\mu_2=0.059719+(0.270713018-0.269654)/3.785\approx0.059999$. The unrounded solution is $0.060000$ to six decimal places.

Verify. Substituting $\mu=0.06$ into the term formula gives $0.6(1-e^{-0.6})=0.270713018$, matching the stated value. The inferred force is $0.0600$ per year to four significant figures.

Mistake avoided. The whole-life inversion $\mu=A\delta/(1-A)$ cannot be applied to this term value. It omits the factor $1-e^{-(\mu+\delta)n}$, which accounts for the finite coverage period.

Effective Interest and a Benefit-Amount Check

In applications, an interest assumption may be stated as an effective annual rate, while an insurance value may be stated in dollars. Both must be converted to the form required by the unit-value equation. The next example makes those steps explicit.

Worked Example: Converting Interest Before Inverting

A hypothetical continuous whole-life policy pays $80,000 at death. Its expected present value is $36,000. The effective annual interest rate is $i=0.05$. Assuming constant forces of interest and mortality, find the mortality force and check the value.

Convert to a unit value. The observed unit value is $A=36{,}000/80{,}000=0.45$.

Convert the interest rate. The equivalent force of interest is $\delta=\ln(1+i)=\ln(1.05)=0.0487902$, to seven decimal places.

Infer mortality. $\mu=A\delta/(1-A)=0.45(0.0487902)/0.55=0.0399192$ per year, to seven decimal places.

Verify. The denominator in the whole-life formula is $\mu+\delta=0.0399192+0.0487902=0.0887094$. Thus $\mu/(\mu+\delta)\approx0.0399192/0.0887094=0.4500$, to four decimal places. The dollar expected present value is $80{,}000(0.4500)=\$36{,}000$, agreeing with the data.

Mistake avoided. Substituting $0.05$ for $\delta$ would treat an effective annual rate as a force. The correct conversion is the logarithm; the small numerical difference changes the inferred mortality force.

Limits, Checks, and Common Traps

An inferred force is conditional on the model. A value alone does not establish that mortality is constant; constancy is an assumption used to make the inversion. Before accepting a result, check the product type, benefit timing, term length, benefit scale, and interest convention.

Check the benefit timing. These formulas are for a benefit paid at the moment of death. A benefit paid at the end of the year of death has a different present-value formula.
Check the term boundary. A term contract pays only when death occurs during its coverage period. The upper integration limit is $n$, not infinity.
Check the value scale. Divide an expected dollar value by the constant benefit amount to obtain the unit value.
Check the rate convention. Use $\delta$ in the continuous discount factor. If the data give effective $i$, first calculate $\ln(1+i)$.
Check feasibility. Under positive force of interest, a whole-life unit value must lie strictly between zero and one. A term unit value must also lie between zero and one.

There is one important identifiability exception. If $\delta=0$, a whole-life benefit paid at eventual death has value one for any $\mu>0$, because death occurs eventually with probability one under a positive constant force. That value cannot identify the force. In contrast, with zero interest and a finite term, the value is $1-e^{-\mu n}$, which still identifies $\mu$ for a value strictly between zero and one.

Key takeaway: For continuous whole-life insurance with $\delta>0$, invert directly using $\mu=A\delta/(1-A)$. For continuous term insurance, retain the finite-term exponential factor, solve the resulting increasing equation numerically when needed, and substitute the inferred force back into the original valuation formula.

Check Your Understanding

For each problem, assume a constant force of mortality and continuous payment at the moment of death. For whole-life problems, use the specified force of interest unless an effective rate is stated.

A hypothetical whole-life policy has a unit-benefit value of $0.4$ under force of interest $\delta=0.03$. Find the force of mortality and verify it in the whole-life formula. (Answer: $\mu=0.02$ per year; $0.02/(0.02+0.03)=0.4$.)
A continuous whole-life policy has a unit-benefit value of $0.6$, and the force of interest is $\delta=0.05$. Infer the force of mortality. (Answer: $\mu=0.075$ per year.)
With zero force of interest, a hypothetical eight-year continuous term policy has unit value $0.5$. Infer $\mu$, using the finite-term formula. (Answer: $\mu=\ln(2)/8\approx0.0866$ per year.)
A five-year continuous term policy has $\delta=0.02$ and $\mu=0.08$. Calculate its unit value, showing the exponent and the leading fraction. (Answer: $(0.08/0.10)(1-e^{-0.50})\approx0.3148$.)
A hypothetical four-year continuous term policy has $\delta=0.03$ and observed unit value $0.230776$. Use the term formula to infer the force of mortality and verify by substitution. (Answer: $\mu\approx0.0700$ per year; $(0.07/0.10)(1-e^{-0.40})\approx0.230776$.)

What You'll Learn