Continuous Whole-Life Insurance and Annuity Identity

Why the Continuous Identity Is Useful

A life insurance benefit payable at the moment of death and an annuity paid continuously while the same person is alive have different payment patterns. Yet, when both are valued using the same interest and mortality assumptions, their values are linked by a simple identity. The identity is useful for checking a calculation and for finding one value when the other is known.

The key is to compare the two present values for a single lifetime before taking expectations. At every possible death time, the discounted insurance payment complements the discounted stream of annuity payments. This argument does not require a particular mortality law; the values will depend on the mortality model, but the identity holds across models under the stated contract and interest assumptions. This relationship is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let \(T_x\) be the future lifetime of a life aged \(x\), measured in years. Assume a constant force of interest \(\delta>0\). The value \(\bar{A}_x\) is the actuarial present value of a unit whole-life insurance benefit payable at the instant of death, so \(\bar{A}_x=E[e^{-\delta T_x}]\). The value \(\bar{a}_x\) is the actuarial present value of a continuous whole-life annuity paying at rate 1 per year while the life is alive, so \(\bar{a}_x=E[\int_0^{T_x}e^{-\delta t}\,dt]\). Both values are calculated at the same time, for the same life, and on the same interest and mortality basis.

The interest assumption must be interpreted consistently. With an effective annual interest rate \(i\), the corresponding constant force is \(\delta=\ln(1+i)\), and \(e^{-\delta t}=(1+i)^{-t}\). The continuous cash flows are discounted over the fraction of a year at which they occur. Substituting \(i\) for \(\delta\) in an identity involving a continuous annuity would generally be incorrect.

Deriving the Identity

First consider one individual whose future lifetime is a particular finite value \(T_x=t\). The present value of the unit death benefit is \(e^{-\delta t}\). The present value of the unit-rate annuity payments received before death is the integral of the discount factor from time 0 to time \(t\). Evaluating this integral gives the pathwise relationship.

\int_0^{t} e^{-\delta s}\,ds =\frac{1-e^{-\delta t}}{\delta} \quad\Longrightarrow\quad e^{-\delta t}+\delta\int_0^{t}e^{-\delta s}\,ds=1.

This equality holds for each finite death time \(t\), not merely on average. Taking expectations over \(T_x\) therefore gives the actuarial present value identity. The expectation of the death-benefit present value is \(\bar{A}_x\), and the expectation of the annuity present value is \(\bar{a}_x\).

\boxed{\bar{A}_x+\delta\bar{a}_x=1}

Theorem: For unit whole-life insurance payable at the instant of death and a unit-rate continuous whole-life annuity payable while the same life is alive, valued at constant force of interest \(\delta>0\) on the same mortality basis, \(\bar{A}_x+\delta\bar{a}_x=1\). Equivalently, \(\bar{A}_x=1-\delta\bar{a}_x\) and \(\bar{a}_x=(1-\bar{A}_x)/\delta\).

The identity does not assert that either value is fixed independently of the mortality model. Rather, mortality determines how the present-value amounts are distributed across possible lifetimes; for each lifetime, the discounted amounts satisfy the same complement relationship. The formulas also rely on the specific benefit timing and amounts in the definition: a unit benefit paid at death and an annuity paid continuously at rate 1 while alive.

Worked Applications

Worked Example: Constant-Force Mortality

A hypothetical life has a future lifetime with constant force of mortality \(\mu=0.03\) per year. The force of interest is \(\delta=0.04\) per year. Find \(\bar{A}_x\) and \(\bar{a}_x\) for the contracts in the definition, and verify the identity. The corresponding effective annual interest rate is \(i=e^{0.04}-1\), approximately 4.0811%.

Method and model. Under constant force of mortality, \(T_x\) has an exponential distribution with density \(f_{T_x}(t)=\mu e^{-\mu t}\) for \(t\geq 0\). The death benefit is paid at death, and the annuity is paid continuously while alive. We calculate each expected present value under that model.

For the insurance, multiply the discounted benefit \(e^{-\delta t}\) by the lifetime density and integrate:

\bar{A}_x =\int_0^\infty e^{-0.04t}(0.03e^{-0.03t})\,dt =0.03\int_0^\infty e^{-0.07t}\,dt =\frac{0.03}{0.07} =\frac{3}{7} \approx 0.4286.

For the annuity, use its present value conditional on lifetime \(t\), then average over the exponential lifetime:

\bar{a}_x =\int_0^\infty \frac{1-e^{-0.04t}}{0.04}(0.03e^{-0.03t})\,dt =\frac{1}{0.04}\left(1-\frac{0.03}{0.07}\right) =\frac{1}{0.07} \approx 14.2857.

Check. Using unrounded values, \(3/7+0.04(100/7)=3/7+4/7=1\). Rounded to four decimals, \(0.4286+0.04(14.2857)\) is approximately \(1.0000\); the slight displayed discrepancy, if any, is due to rounding.

Common mistake avoided. The mortality force and interest force play different roles. The denominator \(0.07\) comes from adding them, while the insurance numerator is the mortality force \(0.03\). Using the interest force as the numerator would not calculate the death-benefit value.

Worked Example: A Uniformly Distributed Future Lifetime

In a hypothetical model, \(T_x\) is uniformly distributed from 0 to 20 years. The force of interest is \(\delta=0.05\) per year. Calculate \(\bar{A}_x\) directly, calculate \(\bar{a}_x\) both directly and using the identity, and check the results. This model has a finite maximum future lifetime and does not have constant force of mortality.

Method and model. The density is \(1/20\) on \(0\leq t\leq20\), and zero elsewhere. For the insurance, average the discounted death benefit over that density:

\bar{A}_x =\frac{1}{20}\int_0^{20} e^{-0.05t}\,dt =\frac{1-e^{-1}}{20(0.05)} =1-e^{-1} \approx 0.6321.

Here \(e^{-1}\approx0.3678794\), so the numerator \(1-e^{-1}\) is approximately \(0.6321206\), and the denominator \(20(0.05)\) is 1.

For a direct annuity calculation, the probability of surviving to time \(t\), for \(0\leq t\leq20\), is \(1-t/20\). Thus the expected present value of the continuous annuity can be integrated over the payment times:

\bar{a}_x =\int_0^{20}e^{-0.05t}\left(1-\frac{t}{20}\right)\,dt =20-\frac{1-e^{-1}}{20(0.05)^2} =20-\frac{0.6321206}{0.05} \approx 7.3576.

The identity gives the same result: \((1-\bar{A}_x)/0.05=(1-0.6321206)/0.05=0.3678794/0.05\approx7.3576\). Check: \(0.6321206+0.05(7.357588)=1.000000\) before rounding.

Common mistake avoided. A finite maximum lifetime does not make the annuity certain to pay for the full 20 years. The direct calculation weights each payment time by the probability of surviving to that time, \(1-t/20\), rather than integrating the discount factor alone.

Worked Example: Two Latent Mortality Groups

A hypothetical population consists of two equally likely mortality groups. Conditional on group 1, future lifetime has constant force of mortality \(\mu_1=0.01\) per year; conditional on group 2, it has constant force \(\mu_2=0.05\) per year. The force of interest is \(\delta=0.04\) per year for both groups. Find the population values \(\bar{A}_x\) and \(\bar{a}_x\), and verify the identity.

Method and model. Calculate each expected present value conditional on the group, then average those values using the group probabilities. For constant mortality force \(\mu\), the formulas are \(\bar{A}_x=\mu/(\mu+\delta)\) and \(\bar{a}_x=1/(\mu+\delta)\), as follows by the exponential-density integrations in the previous example.

For group 1, the denominator is \(0.01+0.04=0.05\). Therefore \(\bar{A}_{x,1}=0.01/0.05=0.2\) and \(\bar{a}_{x,1}=1/0.05=20\).

For group 2, the denominator is \(0.05+0.04=0.09\). Therefore \(\bar{A}_{x,2}=0.05/0.09=5/9\approx0.5555556\), and \(\bar{a}_{x,2}=1/0.09=100/9\approx11.1111111\).

Average the two insurance values and the two annuity values separately:

\bar{A}_x =\tfrac12(0.2)+\tfrac12(5/9) =\frac{17}{45} \approx0.3778, \qquad \bar{a}_x =\tfrac12(20)+\tfrac12(100/9) =\frac{70}{9} \approx15.5556.

Check. Using unrounded values, \(17/45+0.04(70/9)=17/45+14/45=31/45\) would not equal 1—so this indicates an arithmetic error. Rechecking the annuity average: \(\tfrac12(20)+\tfrac12(100/9)=10+50/9=140/9\), not \(70/9\). Thus the correct population annuity value is \(140/9\approx15.5556\), and \(17/45+0.04(140/9)=17/45+28/45=1\).

Rounded to four decimals, the correct pair is \(\bar{A}_x=0.3778\) and \(\bar{a}_x=15.5556\). Common mistake avoided. Do not replace the two mortality forces by their average, \(0.03\), and then treat the population as one constant-force group. The conditional values must be calculated within each group and averaged with the group probabilities.

Using the Relationship Carefully

For a calculation, the identity can be used in either direction. If the insurance value is available, the annuity value is \((1-\bar{A}_x)/\delta\). If the annuity value is available, the insurance value is \(1-\delta\bar{a}_x\). These are especially useful as independent checks on values obtained from an integral, a life table or a mortality model. Work with unrounded values when checking the identity, since rounding intermediate results can leave a small apparent discrepancy.

Match the contracts.
Confirm the insurance benefit is 1 at the instant of death and the annuity pays continuously at rate 1 while the same life is alive.

Match the valuation basis.
Use the same interest assumption, mortality basis, valuation age and time of valuation for both values.

Use the force of interest.
Substitute \(\delta\), not the effective annual rate \(i\), into \(\bar{A}_x+\delta\bar{a}_x=1\).

Check the result.
Calculate the left side using unrounded values. A value other than 1 signals an inconsistency in the calculation or in the assumptions.

A frequent exam trap is to use an annual effective rate where the identity requires a force. For example, if \(i=0.04\), then \(\delta=\ln(1.04)\), not \(0.04\). Another trap is to apply the identity to benefits with different amounts or timing without adjusting the relationship. The theorem stated here is for the particular continuous contracts defined above; alternative payment timing and benefit amounts require their own setup.

The mortality model can also affect how values should be averaged. In a mixed population, the unconditional values are probability-weighted averages of the conditional values, as in the two-group example. The identity still holds because each group satisfies it on the same interest basis, and averaging preserves the equality. But substituting an average mortality force into a constant-force formula generally does not produce those population averages.

Key takeaway: For unit continuous whole-life insurance payable at death and a unit-rate continuous whole-life annuity payable while alive, on the same basis, \(\bar{A}_x+\delta\bar{a}_x=1\). The identity follows by adding the two discounted values for each possible lifetime, then taking expectations.

Check Your Understanding

For each problem, identify the model and use consistent continuous-time discounting. Answers are rounded to four decimals where appropriate.

A hypothetical life has constant mortality force \(\mu=0.02\) and force of interest \(\delta=0.06\), both per year. Find \(\bar{A}_x\) and \(\bar{a}_x\), then check the identity. (Answer: \(\bar{A}_x=0.2500,\ \bar{a}_x=12.5000\).)
A future lifetime is uniformly distributed from 0 to 10 years, and \(\delta=0.05\). Find the insurance and annuity values. (Answer: \(\bar{A}_x=0.7869,\ \bar{a}_x=4.2612\).)
A population has two equally likely groups with constant mortality forces \(0.02\) and \(0.08\) per year. If \(\delta=0.04\), find the population values by conditioning on group. (Answer: \(\bar{A}_x=0.5000,\ \bar{a}_x=12.5000\).)
For a hypothetical life, \(\bar{A}_x=0.36\) and the force of interest is \(\delta=0.04\). Find the value of the matching continuous whole-life annuity and verify the identity. (Answer: \(\bar{a}_x=16.0000\); \(0.36+0.04(16)=1\).)
A hypothetical life has mortality force \(0.02\) per year during the first five years and \(0.06\) per year thereafter. The force of interest is \(0.04\) per year. Find \(\bar{A}_x\) by splitting deaths into the first five years and the period after year five, then use the identity for \(\bar{a}_x\). (Answer: \(\bar{A}_x=0.5309,\ \bar{a}_x=11.7279\).)

What You'll Learn