Why Recover Insurance from an Annuity?
Continuous life insurance and continuous life annuities use the same survival and discounting assumptions, but their payments occur at different times. A continuous annuity pays while the insured is alive; continuous insurance pays at death. Their present values are therefore linked. If an annuity value is known or easier to calculate, a complement identity can recover the corresponding insurance value.
The force of interest \(\delta\) is essential to these identities. The discount factor for a payment at time \(t\) is \(e^{-\delta t}\). The term-insurance identity also requires attention to what happens at the end of the term: a pure endowment is the discounted value of a payment made at that date only if the insured survives.
These are expected present values for unit benefits or unit annuity payments. The continuous annuity pays at rate one per year while the insured is alive, up to the stated term if temporary. The insurance benefit is paid immediately at the moment of death. The term contract described here has no benefit payable at maturity; its pure-endowment value is a separate component in the identity, not an additional benefit of the term policy.
The Complement Identities
For whole life, exactly one of two events occurs: death occurs at a finite future time, or the annuity continues through each finite time while the insured is alive. Their discounted values give a direct complement.
For a term, the corresponding complement has three parts: the term insurance payment if death occurs before \(n\), the stream of temporary annuity payments while alive before \(n\), and the pure endowment value if the insured survives to \(n\). The last part is needed to complete the identity.
Why the Term Identity Holds
Assume the future lifetime has an absolutely continuous distribution, so survival probability changes according to \(\frac{d}{dt}{}_tp_x=-{}_tp_x\mu_{x+t}\). This is the usual continuous mortality model. The term insurance value is the discounted death-payment integral, while the temporary annuity is the integral of discounted survival. The following derivation shows why their values, together with the pure endowment, sum to one.
\(\bar A^1_{x:\overline{n}|}=\int_0^n e^{-\delta t}{}_tp_x\mu_{x+t}\,dt\), and \(\bar a_{x:\overline{n}|}=\int_0^n e^{-\delta t}{}_tp_x\,dt\).
By the product rule, \(\frac{d}{dt}(e^{-\delta t}{}_tp_x)=-e^{-\delta t}{}_tp_x(\delta+\mu_{x+t})\).
Rearranging and integrating gives \(\int_0^n e^{-\delta t}{}_tp_x(\delta+\mu_{x+t})\,dt=1-e^{-\delta n}{}_np_x\), since \({}_0p_x=1\).
The mortality part of the integral is \(\bar A^1_{x:\overline{n}|}\), the interest part is \(\delta\bar a_{x:\overline{n}|}\), and the remaining boundary term is \({}_nE_x\). Thus their sum is one.
Letting the term boundary extend to infinity gives the whole-life identity when discounted survival tends to zero. The proof also identifies a key limitation: these integral formulas presume a continuous lifetime distribution. If the model allows an atom of probability at the term boundary or at a particular death time, the boundary or death-payment convention must be handled explicitly.
Worked Applications
When a mortality model has constant force \(\mu\), survival is \({}_tp_x=e^{-\mu t}\), and the death density is \(\mu e^{-\mu t}\). With force of interest \(\delta\), the whole-life annuity integral is \(1/(\delta+\mu)\). For an \(n\)-year term, the temporary annuity and pure endowment values are \((1-e^{-(\delta+\mu)n})/(\delta+\mu)\) and \(e^{-(\delta+\mu)n}\), respectively. These formulas follow by integrating the discounted survival function.
Worked Example: Recovering Whole-Life Insurance
A hypothetical whole-life contract is valued at force of interest \(\delta=0.05\). The mortality basis assumes a constant force of mortality \(\mu=0.03\) at all future durations. Calculate the continuous whole-life annuity and recover the unit continuous whole-life insurance value.
Method. Use the constant-force survival model, then apply the whole-life complement identity.
The combined discount and mortality force is \(\delta+\mu=0.05+0.03=0.08\). Therefore,
The whole-life identity gives
Check. Direct integration of the death-payment value gives \(\int_0^\infty 0.03e^{-0.08t}\,dt=0.03/0.08=0.375\), matching the complement result. The common mistake avoided here is multiplying the annuity by the mortality force: the complement uses the force of interest \(\delta\).
Worked Example: Recovering Five-Year Term Insurance
A hypothetical five-year term contract is valued at force of interest \(\delta=0.04\). Assume constant force of mortality \(\mu=0.06\). Calculate the temporary continuous annuity, the pure-endowment value, and the unit continuous term insurance value.
Method. Calculate the two survival-related values under the stated exponential survival model, then use the term complement identity.
The combined force is \(0.04+0.06=0.10\), and the five-year discounted survival probability is \(e^{-0.10(5)}=e^{-0.5}=0.606530660\). The temporary annuity value is
The pure-endowment value is \({}_5E_x=e^{-0.04(5)}{}_5p_x=e^{-0.5}=0.606530660\). Thus,
Check. Direct integration gives \(\int_0^5 0.06e^{-0.10t}\,dt=0.06(1-e^{-0.5})/0.10=0.236081604\). The two methods agree. The common mistake avoided here is omitting the pure-endowment term: the term insurance value is not \(1-\delta\bar a_{x:\overline{5}|}\).
Worked Example: A Longer Term and an Independent Check
A hypothetical ten-year term contract uses force of interest \(\delta=0.06\) and constant force of mortality \(\mu=0.04\). Calculate the continuous term insurance value from its temporary annuity value, and verify the result from the death-payment integral.
Method. First calculate the discounted survival probability at the ten-year boundary. Then use the complement identity and check it against direct integration.
The combined force is \(0.06+0.04=0.10\), so \(e^{-0.10(10)}=e^{-1}=0.367879441\). The temporary annuity value is
The pure-endowment value is \({}_{10}E_x=e^{-1}=0.367879441\). Substitution into the identity gives
Check. Direct integration yields \(\int_0^{10}0.04e^{-0.10t}\,dt=0.04(1-e^{-1})/0.10=0.252848224\). The calculation is consistent to the displayed precision. The trap avoided is using the mortality force in the annuity complement term: it is \(\delta\bar a\), not \(\mu\bar a\).
Choosing and Checking the Identity
The whole-life identity has no maturity boundary, while the term identity stops both the annuity and death benefit at time \(n\). The term identity therefore needs the pure-endowment value for survival to that boundary. Keeping the contract definitions visible helps prevent an apparently plausible but incomplete complement calculation.
| Value to recover | Complement identity | Boundary term |
|---|---|---|
| Continuous whole-life insurance | \(\bar A_x=1-\delta\bar a_x\) | None; the contract has no finite term boundary |
| Continuous \(n\)-year term insurance | \(\bar A^1_{x:\overline{n}|}=1-\delta\bar a_{x:\overline{n}|}-{}_nE_x\) | \({}_nE_x=e^{-\delta n}{}_np_x\) |
A useful check is to compare the recovered insurance value with its death-payment integral, \(\int e^{-\delta t}{}_tp_x\mu_{x+t}\,dt\), over the appropriate duration. Under standard continuous mortality, the insurance value must be nonnegative. For term insurance, \(\delta\bar a_{x:\overline{n}|}+{}_nE_x\) cannot exceed one under the stated assumptions. A result outside these checks usually signals a missing pure endowment, a mismatched interest basis, or an incorrect term boundary.
These identities are useful when the annuity value is available from a valuation basis and the insurance integral is inconvenient to evaluate directly. They are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In exam calculations, write down whether the contract is whole life or term before substituting values, and label the pure-endowment term explicitly whenever there is a finite boundary.
Check Your Understanding
For each problem, use the continuous payment convention and the stated force of interest and mortality. Give probabilities or insurance values to four decimal places where requested.
- A hypothetical whole-life contract has \(\delta=0.04\) and constant \(\mu=0.06\). Find \(\bar a_x\) and \(\bar A_x\). (Answer: \(\bar a_x=10\), \(\bar A_x=0.6000\))
- A hypothetical four-year term has \(\delta=0.05\) and constant \(\mu=0.05\). Find \({}_4E_x\) and \(\bar A^1_{x:\overline{4}|}\). (Answer: \({}_4E_x=0.6703\), \(\bar A^1_{x:\overline{4}|}=0.1648\))
- A hypothetical six-year term has \(\delta=0.02\) and constant \(\mu=0.03\). Calculate \(\bar a_{x:\overline{6}|}\), then recover the unit term insurance value. (Answer: \(\bar a_{x:\overline{6}|}=5.1836\), \(\bar A^1_{x:\overline{6}|}=0.1555\))
- A hypothetical five-year term has \(\delta=0.03\) and constant \(\mu=0.07\). Calculate the pure-endowment value and the unit continuous term insurance value. (Answer: \({}_5E_x=0.6065\), \(\bar A^1_{x:\overline{5}|}=0.2754\))
- A hypothetical eight-year term uses \(\delta=0.05\) and constant \(\mu=0.02\). Calculate the temporary annuity value, the pure-endowment value, and the unit term insurance value. (Answer: \(\bar a_{x:\overline{8}|}=6.1256\), \({}_8E_x=0.5712\), \(\bar A^1_{x:\overline{8}|}=0.1225\))