Benefit Timing in Insurance-Annuity Identities

Why Benefit Timing Changes the Identity

An insurance-annuity identity links the present value of a death benefit with the present value of payments made while the insured is alive. The link depends on when the benefit is paid. A benefit paid at the end of the year of death is discounted by whole years; a benefit paid at the moment of death is discounted by the exact time of death. Consequently, the discrete identity uses the effective discount rate \(d\), whereas the continuous identity uses the force of interest \(\delta\).

The distinction is easy to miss because both identities have the same broad form: insurance value equals one minus an interest-rate factor times an annuity value. But the annuities are paid on different schedules, and the factors are not interchangeable. The results below assume a unit benefit, no expenses, and present values calculated at issue. For the whole-life identities, assume positive interest and that death occurs eventually.

Definition: In the discrete model, \(K_x\) is the curtate future lifetime, the number of complete years lived after age \(x\). A unit whole-life insurance benefit paid at the end of the year of death has present value \(v^{K_x+1}\), and its expected present value is \(A_x\). The whole-life annuity-due value \(\ddot{a}_x\) is the expected present value of unit payments at times \(0,1,2,\ldots\), for as long as the insured is alive at each payment time.

Definition: In the continuous model, \(T_x\) is the complete future lifetime. A unit whole-life insurance benefit paid at the moment of death has present value \(e^{-\delta T_x}\), and its expected present value is \(\bar{A}_x\). The continuous whole-life annuity value \(\bar{a}_x\) is the expected present value of a payment stream at rate one per year while the insured is alive.

The Discrete Identity: End of the Year of Death

Let \(i\) be the effective annual interest rate, \(v=1/(1+i)\), and \(d=i/(1+i)=1-v\) the effective annual discount rate. For a particular lifetime with \(K_x=k\), the annuity-due payments occur at times \(0,1,\ldots,k\). Its discounted value is \(1+v+\cdots+v^k\). The insurance benefit is paid at time \(k+1\), so its discounted value is \(v^{k+1}\).

The finite geometric sum gives the key pointwise relation. Taking expectations then gives the actuarial identity; this is why the factor is \(d\), not \(i\) or \(\delta\).

d(1+v+\cdots+v^k) =(1-v)\frac{1-v^{k+1}}{1-v} =1-v^{k+1}.

A_x=1-d\ddot{a}_x.

The identity matches a specific pair: end-of-year-of-death insurance and an annuity-due with annual payments. The annuity’s first payment is at issue, at time zero. The relation does not assert that the same factor applies to a continuous annuity, or to a benefit paid immediately upon death.

The Continuous Identity: Benefit at the Moment of Death

In the continuous model, the force of interest \(\delta\) gives the discount factor \(e^{-\delta t}\) at time \(t\). For a fixed death time \(T_x\), the value of a continuous annuity paid at rate one until death is \(\int_0^{T_x}e^{-\delta t}\,dt\). Integrating the discount function shows why \(\delta\) appears in this identity:

\delta\int_0^{T_x} e^{-\delta t}\,dt =1-e^{-\delta T_x}.

The right-hand side is one minus the present value of the unit benefit paid at death. Taking expectations gives the continuous whole-life identity.

\bar{A}_x=1-\delta\bar{a}_x.

This is the continuous counterpart of the discrete identity, not a substitution rule for its interest factor. In particular, replacing \(d\) by \(\delta\) in \(A_x=1-d\ddot{a}_x\) is generally wrong: that equation describes annual annuity-due payments and a benefit paid at the end of the year of death.

Worked Example: Comparing Both Whole-Life Timings

A hypothetical life has a constant force of mortality \(\mu=0.08\) per year. The effective annual interest rate for the discrete model is \(i=e^{0.04}-1\), so the force of interest for the continuous model is \(\delta=0.04\). Find (a) the discrete whole-life insurance value and its annuity-due value, and (b) the continuous whole-life insurance value and its annuity value. Use the appropriate identity in each case.

Method and assumptions. Under constant force mortality, \({}_tp_x=e^{-0.08t}\), so one-year survival is \(p=e^{-0.08}\) and one-year death probability is \(q=1-p\). In the discrete model, \(v=e^{-0.04}\). Annual benefits are paid at the end of the year of death; the annuity-due payments are annual. In the continuous model, the benefit is paid at death and the annuity is paid continuously.

(a) Discrete values. The combined one-year discount-and-survival factor is \(vp=e^{-0.12}\). The geometric series for the annuity-due has ratio \(vp\), so

\ddot{a}_x=\frac{1}{1-vp} =\frac{1}{1-e^{-0.12}} =\frac{1}{0.1130796} =8.8433.

The effective discount rate is

d=1-v=1-e^{-0.04}=0.0392106.

Applying the discrete identity,

A_x=1-d\ddot{a}_x =1-(0.0392106)(8.8433) =0.6532.

As a direct check, the annual death benefit value is \(v q/(1-vp)\). Here \(q=1-e^{-0.08}=0.0768837\), so

\frac{vq}{1-vp} =\frac{(e^{-0.04})(0.0768837)}{0.1130796} =0.6532.

(b) Continuous values. Under constant force mortality, the joint density of death at time \(t\) is \(\mu e^{-\mu t}\). Therefore the death benefit value is \(\int_0^\infty e^{-\delta t}\mu e^{-\mu t}\,dt\), while the continuous annuity value is \(\int_0^\infty e^{-\delta t}e^{-\mu t}\,dt\). Evaluating,

\bar{A}_x=\frac{\mu}{\mu+\delta} =\frac{0.08}{0.08+0.04} =0.6667, \qquad \bar{a}_x=\frac{1}{\mu+\delta} =\frac{1}{0.12} =8.3333.

The identity checks: \(1-\delta\bar{a}_x=1-(0.04)(8.3333)=0.6667\). Values are rounded to four decimal places. The direct calculations and the identities agree in each timing model.

Common mistake avoided. Using \(\delta=0.04\) in the discrete identity would give \(1-(0.04)(8.8433)=0.6463\), not \(0.6532\). That combines the discrete annuity-due value with the continuous interest factor, so it does not represent the stated contract.

Temporary Coverage: Include the Survival Benefit

For a term insurance, the benefit is paid only if death occurs during the term. The whole-life identity must therefore account for the possibility that the insured survives the term. In the discrete model, let \(A^1_{x:\overline{n}|}\) denote the expected present value of a unit benefit paid at the end of the year of death if death occurs within \(n\) years. Let \(\ddot{a}_{x:\overline{n}|}\) be the temporary annuity-due value, with payments at times \(0,\ldots,n-1\) if the insured is alive then, and let \({}_nE_x=v^n{}_np_x\) be the pure endowment value.

For death in year \(k+1\), where \(0\leq k<n\), multiplying the annuity payments through time \(k\) by \(d\) gives \(1-v^{k+1}\). For survival through all \(n\) years, multiplying the \(n\) annuity-due payments by \(d\) gives \(1-v^n\); the remaining value is the pure endowment discount factor \(v^n\). Thus the temporary identity is

A^1_{x:\overline{n}|} =1-d\ddot{a}_{x:\overline{n}|}-{}_nE_x.

The continuous version has the same structure but different timing. Write \(\bar{A}^1_{x:\overline{n}|}\) for the value of a unit benefit paid at death if death occurs by time \(n\), and \(\bar{a}_{x:\overline{n}|}\) for the continuous annuity value up to the earlier of death and time \(n\). The pathwise integral to time \(\min(T_x,n)\) leaves a residual \(e^{-\delta T_x}\) if death occurs by \(n\), and \(e^{-\delta n}\) if the insured survives to \(n\). Taking expectations yields

\bar{A}^1_{x:\overline{n}|} =1-\delta\bar{a}_{x:\overline{n}|}-{}_nE_x, \qquad {}_nE_x=e^{-\delta n}{}_np_x \quad\text{in the continuous model.}

The pure endowment term is not optional: it represents the discounted survival outcome not covered by the term insurance. The discrete and continuous forms use the same survival idea, but their annuity schedules and interest factors correspond to different benefit timings.

Worked Example: Two-Year Discrete Term Insurance

A hypothetical group has a constant annual probability \(p=0.80\) of surviving each year, independently of earlier years. The effective annual interest rate is \(i=0.25\). A unit benefit is paid at the end of the year of death if death occurs within two years. Find the insurance value using the temporary identity.

Method and assumptions. Here \(v=1/(1.25)=0.80\), \(d=1-v=0.20\), and the two-year survival probability is \({}_2p_x=(0.80)^2=0.64\). The temporary annuity-due has a payment at time zero and a second payment at time one only if alive then.

First calculate the annuity value:

\ddot{a}_{x:\overline{2}|} =1+vp =1+(0.80)(0.80) =1.64.

The pure endowment value is

{}_2E_x=v^2{}_2p_x =(0.80)^2(0.64) =(0.64)(0.64) =0.4096.

Now apply the temporary identity:

A^1_{x:\overline{2}|} =1-(0.20)(1.64)-0.4096 =1-0.3280-0.4096 =0.2624.

Check by summing death-year benefits. Death in year one has probability \(0.20\) and discount factor \(0.80\), giving \(0.1600\). Death in year two has probability \((0.80)(0.20)=0.16\) and discount factor \(0.80^2=0.64\), giving \(0.1024\). Their sum is \(0.1600+0.1024=0.2624\), as required. The value is exact; no rounding is needed.

Common mistake avoided. Omitting the pure endowment gives \(1-d\ddot{a}_{x:\overline{2}|}=0.6720\), which is not the term insurance value. That expression fails to subtract the value associated with survival through the coverage period.

Worked Example: Two-Year Continuous Term Insurance

A hypothetical life has constant force of mortality \(\mu=0.10\) per year and force of interest \(\delta=0.05\). A unit benefit is paid immediately at death if death occurs within two years. Find the insurance value using the continuous temporary identity, and verify it directly.

Method and assumptions. Survival to time \(t\) is \(e^{-0.10t}\). The temporary continuous annuity has value \(\int_0^2e^{-(0.10+0.05)t}\,dt\). The pure endowment value is discounted survival to time two.

The annuity value is

\bar{a}_{x:\overline{2}|} =\frac{1-e^{-0.15(2)}}{0.15} =\frac{1-e^{-0.30}}{0.15} =\frac{1-0.740818}{0.15} =1.727879.

The pure endowment value is

{}_2E_x=e^{-0.05(2)}e^{-0.10(2)} =e^{-0.30} =0.740818.

Using the continuous identity,

\bar{A}^1_{x:\overline{2}|} =1-(0.05)(1.727879)-0.740818 =1-0.086394-0.740818 =0.172788.

For a direct check, the density of death at time \(t\) is \(0.10e^{-0.10t}\). Discounting and integrating over the coverage period gives

\bar{A}^1_{x:\overline{2}|} =\int_0^2 0.10e^{-0.15t}\,dt =\frac{0.10}{0.15}(1-e^{-0.30}) =0.172788.

The methods agree. Values are rounded to six decimal places. A frequent timing error is to use \(d\) with this continuous annuity; the payments are not annual annuity-due payments, so the derivation calls for \(\delta\).

Choosing the Correct Identity

Before applying an identity, match the benefit schedule, annuity schedule, and interest measure. The following comparison summarizes the pairings.

Insurance benefit timing	Matching annuity timing	Interest factor	Whole-life identity
End of year of death	Annual payments in advance while alive	Effective discount rate \(d=1-v\)	\(A_x=1-d\ddot{a}_x\)
At the moment of death	Continuous payment stream while alive	Force of interest \(\delta\)	\(\bar{A}_x=1-\delta\bar{a}_x\)

The temporary forms add a pure endowment because survival to the end of the term is an outcome that the death benefit does not pay. The identities are useful both for calculating an unknown value and for checking a value obtained by another method. They also provide a diagnostic: if a calculation uses the correct annuity value but the wrong interest factor, the mismatch usually signals a benefit-timing error.

Key takeaway: Use \(d\) with an annual annuity-due and a benefit payable at the end of the year of death. Use \(\delta\) with a continuous annuity and a benefit payable at the moment of death. For temporary coverage, subtract the pure endowment value in addition to the annuity term.

Check Your Understanding

For each problem, identify the timing model before applying an identity. Values stated as decimals are rounded to four decimal places where needed.

A hypothetical discrete whole-life contract has one-year survival probability \(p=0.75\) in every year and effective annual interest \(i=0.25\). Find \(v\), \(d\), \(\ddot{a}_x\), and \(A_x\), and check the identity. (Answer: \(v=0.8,\ d=0.2,\ \ddot{a}_x=2.5,\ A_x=0.5\).)
A hypothetical life has constant force of mortality \(\mu=0.06\) and force of interest \(\delta=0.03\). For a continuous whole-life benefit paid at death, find \(\bar{A}_x\) and \(\bar{a}_x\). (Answer: \(\bar{A}_x=0.6667,\ \bar{a}_x=11.1111\).)
A hypothetical discrete model has constant annual survival probability \(p=0.80\), effective annual interest \(i=0.25\), and a unit end-of-year-of-death benefit for three years. Find the temporary insurance value using the annuity and pure endowment identity. (Answer: \(\ddot{a}_{x:\overline{3}|}=2.0496,\ {}_3E_x=0.262144,\ A^1_{x:\overline{3}|}=0.327936\).)
A hypothetical life has constant force of mortality \(\mu=0.10\), force of interest \(\delta=0.05\), and a continuous-at-death unit benefit during a two-year term. Find the temporary annuity value, pure endowment value, and insurance value. (Answer: \(\bar{a}_{x:\overline{2}|}=1.727879,\ {}_2E_x=0.740818,\ \bar{A}^1_{x:\overline{2}|}=0.172788\).)
A hypothetical discrete whole-life contract has \(i=e^{0.04}-1\) and one-year survival probability \(p=e^{-0.08}\). The benefit is paid at the end of the year of death, and the annuity is annual and in advance. Find the insurance value and state which interest factor belongs in the identity. (Answer: \(d=1-e^{-0.04}=0.0392106\), \(A_x=0.6532\); use \(d\), not \(\delta=0.04\).)

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