Discrete and Continuous Benefit-Timing Comparisons

Why Benefit Timing Changes the Identity

Two life insurance contracts can promise the same amount when death occurs but have different present values because they pay at different times. A discrete whole-life policy pays at the end of the year of death; a continuous whole-life policy pays at the moment of death. The corresponding insurance-annuity identities look similar, but their discount quantities are different: the discrete identity uses \(d\), while the continuous identity uses \(\delta\).

For the discrete model, take \(i\) to be a fixed effective annual interest rate, \(v=1/(1+i)\), and \(d=i/(1+i)=1-v\). Let \(K_x\) be the curtate future lifetime, so death occurs between times \(K_x\) and \(K_x+1\), and let \({}_kp_x\) be the probability of surviving \(k\) years. The discrete whole-life insurance \(A_x\) pays 1 at time \(K_x+1\). The annuity-due value \(\ddot a_x\) is the expected present value of unit payments at times \(0,1,2,\ldots\), conditional on being alive at each payment time.

For the continuous model, let \(T_x\) be the complete future lifetime and let the force of interest \(\delta\) be constant, so a payment at time \(t\) has discount factor \(e^{-\delta t}\). The continuous whole-life insurance \(\bar A_x\) pays 1 at time \(T_x\), and the continuous life annuity value \(\bar a_x\) is the expected present value of a unit payment stream at rate 1 while the life is alive.

Definition: \(A_x\) denotes a unit death benefit paid at the end of the year of death; \(\bar A_x\) denotes a unit death benefit paid at the moment of death. Their matching annuity values are \(\ddot a_x\), for payments at integer times beginning immediately, and \(\bar a_x\), for a continuous payment stream. The symbols \(d\) and \(\delta\) are not interchangeable: \(d=1-v\) is an effective annual discount rate, while \(\delta\) is a force of interest.

The Discrete-Time Identity

The discrete identity follows by considering each year in which the policyholder is alive. In a year, the annuity-due value accrues a discount difference of \(v^k-v^{k+1}=dv^k\); the insurance value accounts for the probability-weighted payment at the end of that year. Summing across years makes the intermediate terms cancel.

A_x=1-d\ddot a_x.

To see the cancellation explicitly, write \(p_k={}_kp_x\), with \(p_0=1\). The probability of dying in year \(k+1\) is \(p_k-p_{k+1}\). Therefore

A_x=\sum_{k=0}^{\infty}v^{k+1}(p_k-p_{k+1}), \qquad \ddot a_x=\sum_{k=0}^{\infty}v^kp_k.

Using \(d=1-v\), subtract \(d\ddot a_x\) from 1 and expand the partial sums through year \(n-1\):

1-d\sum_{k=0}^{n-1}v^kp_k = \sum_{k=0}^{n-1}v^{k+1}(p_k-p_{k+1})+v^np_n.

The final term is the discounted probability of surviving to time \(n\). If the discounted survival term tends to zero as \(n\) tends to infinity, the infinite sums converge and the identity \(A_x=1-d\ddot a_x\) follows. This condition holds, for example, if \(i>0\), since \(v^n\) tends to zero and \(0\le p_n\le1\). If the condition does not hold, a nonzero limiting term must be retained.

The same relation can be rearranged using the annuity-immediate value \(a_x=\ddot a_x-1\): \(A_x=v-da_x\). The \(v\) term matters; replacing it with 1 would lose the discount on a payment made at the end of the first year.

Worked Example: Discrete Whole-Life Values Under Constant Mortality

A hypothetical policy pays 1 at the end of the year of death. Assume an effective annual interest rate \(i=25\%\), and assume independent annual mortality with a constant one-year death probability \(q=0.20\) at every age. Find \(A_x\) and \(\ddot a_x\), then verify their identity.

Method. Under this model, \(p=1-q=0.80\), \(v=1/(1.25)=0.80\), and \(d=1-v=0.20\). Survival for \(k\) years has probability \(p^k\), so the annuity-due series is geometric:

\ddot a_x=\sum_{k=0}^{\infty}(vp)^k =\frac{1}{1-vp} =\frac{1}{1-(0.80)(0.80)} =\frac{1}{0.36} =\frac{25}{9}.

The probability of dying in year \(k+1\) is \(p^kq\), and the benefit is discounted for \(k+1\) years. Thus

A_x=\sum_{k=0}^{\infty}v^{k+1}p^kq =\frac{vq}{1-vp} =\frac{(0.80)(0.20)}{0.36} =\frac{4}{9}.

The identity gives \(1-d\ddot a_x=1-(0.20)(25/9)=1-5/9=4/9\), agreeing with the direct insurance calculation. No rounding is needed. A common mistake is to discount the death benefit for \(k\) rather than \(k+1\) years; that would omit one year of interest discount for every possible death year.

The Continuous-Time Identity

Suppose \(T_x\) has density \(f_x(t)\) and survival function \(S_x(t)={}_tp_x=P(T_x>t)\). Under constant force of interest \(\delta\),

\bar A_x=\int_0^\infty e^{-\delta t}f_x(t)\,dt, \qquad \bar a_x=\int_0^\infty e^{-\delta t}S_x(t)\,dt.

The relationship between these values comes from differentiating discounted survival. Since \(S_x'(t)=-f_x(t)\), the product rule gives

\frac{d}{dt}\left(e^{-\delta t}S_x(t)\right) =-e^{-\delta t}\bigl(\delta S_x(t)+f_x(t)\bigr).

Integrate from 0 to \(b\). Since \(S_x(0)=1\), rearrangement yields

1=e^{-\delta b}S_x(b) +\delta\int_0^b e^{-\delta t}S_x(t)\,dt +\int_0^b e^{-\delta t}f_x(t)\,dt.

If the boundary term \(e^{-\delta b}S_x(b)\) tends to zero as \(b\) tends to infinity, the two integrals converge to the whole-life values and

\bar A_x=1-\delta\bar a_x.

Thus the rate in the continuous identity is \(\delta\), not \(d\). The derivation relies on continuous payments and a benefit paid at the actual death time. It does not describe a contract that waits until the end of the year of death.

Worked Example: Continuous Whole-Life Values Under Exponential Mortality

A hypothetical continuous whole-life policy pays 1 at death. Assume \(T_x\) is exponential with force of mortality \(\mu=0.20\) per year, and assume a constant force of interest \(\delta=0.20\) per year. Find \(\bar A_x\) and \(\bar a_x\), and check the continuous identity.

Method. The exponential lifetime has density \(f_x(t)=0.20e^{-0.20t}\) and survival \(S_x(t)=e^{-0.20t}\). Therefore

\bar A_x=\int_0^\infty e^{-0.20t}(0.20e^{-0.20t})\,dt =0.20\int_0^\infty e^{-0.40t}\,dt =\frac{0.20}{0.40} =0.50.

The continuous annuity value is

\bar a_x=\int_0^\infty e^{-0.20t}e^{-0.20t}\,dt =\int_0^\infty e^{-0.40t}\,dt =\frac{1}{0.40} =2.00.

The check is \(1-\delta\bar a_x=1-(0.20)(2.00)=0.60\), which does not equal \(0.50\). This signals an inconsistency in the displayed calculation: the annuity integral is \(1/0.40=2.50\), not \(2.00\). Using the correctly evaluated value, \(1-(0.20)(2.50)=0.50=\bar A_x\). The corrected values are \(\bar A_x=0.50\) and \(\bar a_x=2.50\), with no rounding needed. This check illustrates why evaluating the integral and then testing the identity are both necessary.

Temporary Insurance: Keep the Survival Benefit

For an \(n\)-year term contract, the death benefit is paid only if death occurs within the \(n\)-year term. The annuity side must therefore be temporary as well, and the identity includes a pure endowment term for surviving to the end of the term. For the discrete model, define \(\ddot a_{x:\overline n|}=\sum_{k=0}^{n-1}v^k{}_kp_x\), the value of payments at times \(0\) through \(n-1\) while alive. Write \(A_{x:\overline n|}^{1}\) for the unit term insurance value and \(v^n{}_np_x\) for the discounted survival probability at time \(n\). Then

A_{x:\overline n|}^{1} =1-d\ddot a_{x:\overline n|}-v^n{}_np_x.

The extra term is necessary because the term insurance pays nothing when the life survives all \(n\) years. In the continuous model, define \(\bar a_{x:\overline n|}=\int_0^n e^{-\delta t}{}_tp_x\,dt\) and let \(\bar A_{x:\overline n|}^{1}\) be the value of a unit benefit paid at death before time \(n\). Integrating the discounted-survival derivative only to \(n\) gives

\bar A_{x:\overline n|}^{1} =1-\delta\bar a_{x:\overline n|}-e^{-\delta n}{}_np_x.

Both formulas separate the initial unit value into three parts: the term insurance value, the annuity value multiplied by its timing-specific discount quantity, and the discounted value of surviving the term.

Worked Example: Comparing Two-Year Term Benefit Timing

Consider two hypothetical unit-benefit contracts covering two years. For both, assume annual survival probabilities match an exponential mortality model with force \(\mu=\ln 2\), so the probability of surviving one year is \(e^{-\mu}=0.50\). Set the continuous force of interest to \(\delta=\ln 2\); the matching effective annual interest rate is \(i=e^\delta-1=100\%\), giving \(v=0.50\) and \(d=0.50\). One contract pays at year-end of death; the other pays at the moment of death. Find both term insurance values.

Discrete contract. Each year has \(p=0.50\) and \(q=0.50\). The end-of-year death benefit value is

A_{x:\overline 2|}^{1} =vq+v^2pq =(0.50)(0.50)+(0.50)^2(0.50)(0.50) =0.25+0.0625 =0.3125.

The temporary annuity-due value is \(1+vp=1+(0.50)(0.50)=1.25\), and the discounted survival value is \(v^2p^2=(0.50)^2(0.50)^2=0.0625\). The identity checks: \(1-(0.50)(1.25)-0.0625=0.3125\).

Continuous contract. The death density is \(\mu e^{-\mu t}\), so with \(\delta=\mu=\ln 2\), the term insurance value is

\bar A_{x:\overline 2|}^{1} =\int_0^2 \mu e^{-(\mu+\delta)t}\,dt =\frac{1}{2}\left(1-e^{-4\ln 2}\right) =\frac{1}{2}\left(1-\frac{1}{16}\right) =\frac{15}{32} =0.46875.

For the check, the continuous annuity value is \((1-e^{-4\ln2})/(2\ln2)=15/(32\ln2)\), so \(\delta\bar a_{x:\overline 2|}=15/32\). The discounted survival term is \(e^{-2\delta}{}_2p_x=(1/4)(1/4)=1/16\). Thus \(1-15/32-1/16=15/32\), as required. No rounding is needed. The two values differ even though the models agree on survival at integer years: the continuous contract pays earlier within each year of death. The common mistake is to treat that as a change in mortality rather than a change in payment timing.

Choosing and Checking the Correct Identity

Before substituting values, identify the payment time in the contract. If the benefit is paid at the end of the year of death, use \(A_x\) and the discrete identity with \(d\). If it is paid at the moment of death, use \(\bar A_x\) and the continuous identity with \(\delta\). For a term contract, also include the discounted survival term; omitting it incorrectly treats survival to the end of the term as if it produced a death benefit.

A second check is the interest convention. The effective annual discount rate is \(d=1-v=i/(1+i)\), whereas the force of interest satisfies \(e^{-\delta t}\) as the discount factor over time \(t\). For a fixed effective annual rate, the corresponding constant force is \(\delta=\ln(1+i)\). In general, \(d\ne\delta\). For example, in the discrete worked example \(i=0.25\), so \(d=0.20\), while \(\delta=\ln(1.25)\), approximately \(0.2231\). Incorrectly inserting \(\delta\) into \(1-d\ddot a_x\) would give \(1-(0.2231)(25/9)\approx0.3802\), not the correct \(A_x=4/9\approx0.4444\). The correction is to use \(d=0.20\) for the discrete contract; \(\delta\) belongs to the continuous identity.

Key takeaway: Match the discount quantity to the benefit and annuity timing: \(A_x=1-d\ddot a_x\) for end-of-year benefits, and \(\bar A_x=1-\delta\bar a_x\) for benefits at the moment of death, subject to the required convergence conditions. Temporary identities also subtract the discounted probability of surviving the term.

Check Your Understanding

For each problem, identify the timing convention before applying an identity. Answers are given to check your calculations.

A hypothetical discrete whole-life policy pays 1 at the end of the year of death. The effective annual interest rate is \(50\%\), and the annual death probability is constantly \(0.25\), independently from year to year. Find \(\ddot a_x\) and \(A_x\). (Answer: \(\ddot a_x=2\), \(A_x=1/3\).)
A continuous whole-life contract pays 1 at death. Assume exponential future lifetime with force of mortality \(\mu=0.30\) and constant force of interest \(\delta=0.20\). Find \(\bar A_x\) and verify it using \(\bar a_x\). (Answer: \(\bar a_x=2\), \(\bar A_x=0.6000\).)
A three-year discrete term policy pays 1 at year-end of death. The effective annual interest rate is \(25\%\), and the annual death probability is constantly \(0.20\). Find the term insurance value using the temporary annuity and survival identity. (Answer: \(A_{x:\overline 3|}^{1}=0.327936\).)
A three-year continuous term policy has exponential mortality force \(\mu=0.10\) and interest force \(\delta=0.10\). Find its moment-of-death benefit value and state the discounted survival term used in the identity. Give the benefit value to four decimals. (Answer: \(\bar A_{x:\overline 3|}^{1}=\frac12(1-e^{-0.6})=0.2256\); discounted survival term \(e^{-0.6}=0.5488\).)
A hypothetical population has annual survival probability \(0.90\), and a discrete whole-life contract uses effective annual interest \(i=0.21\). Find its end-of-year death benefit value. Then suppose mortality follows a constant force \(\mu=-\ln(0.90)\) and the matching interest force is \(\delta=\ln(1.21)\). Find the moment-of-death benefit value under that continuous model, to four decimals. (Answer: discrete \(A_x=10/31=0.3226\); continuous \(\bar A_x=\mu/(\mu+\delta)=0.3560\).)

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