Whole-Life Insurance and Life Annuity-Due Identity

Why These Two Present Values Are Linked

A whole-life insurance benefit and a life annuity-due respond to the same event—how long the insured survives—but their payments occur at different times. For a discrete whole-life policy, the benefit is paid at the end of the year of death. An annuity-due pays at the start of each year, provided the insured is alive then. Comparing those payment dates reveals a useful identity that can convert one actuarial present value into the other.

The identity applies to a particular pairing of benefits and timing. It is not a general equation for every life insurance contract or every annuity: a benefit paid immediately on death, for example, has different timing from a benefit paid at the end of the year of death.

Definition: Let $K$ be the curtate future lifetime of a life aged $x$: death occurs during the $(K,K+1]$ year of age. With effective annual interest rate $i>0$ and $v=1/(1+i)$, $A_x$ is the expected present value of a unit whole-life insurance benefit paid at time $K+1$. The life annuity-due value $\ddot{a}_x$ is the expected present value of unit payments at times $0,1,\ldots,K$, while the life is alive. The effective annual discount rate is $d=1-v=i/(1+i)$.

These definitions use an annual payment schedule: the annuity makes its first payment immediately at time 0, and a payment at time $j$ is made only if the life is alive then. The insurance pays one unit at the end of the year of death. All values are per unit of benefit or payment; a contract with another amount scales the relevant value proportionally.

Deriving the Whole-Life Identity

Consider one possible lifetime, with $K=k$. The insurance pays at time $k+1$, so its present value on this lifetime is $v^{k+1}$. The annuity pays at times $0$ through $k$, so its present value is $1+v+\cdots+v^k$. Since $d=1-v$, multiplying the annuity present value by $d$ gives a telescoping sum:

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

Consequently, for this lifetime, the insurance present value plus $d$ times the annuity present value is exactly 1. Taking expectations over the lifetime distribution gives the actuarial identity.

Theorem: For a unit whole-life insurance benefit payable at the end of the year of death and a unit life annuity-due payable annually while the life is alive, at effective annual interest rate $i>0$, $$ A_x+d\ddot{a}_x=1. $$ Equivalently, $A_x=1-d\ddot{a}_x$ and $\ddot{a}_x=(1-A_x)/d$.

This argument compares the two present values on the same lifetime before taking expectations, so it does not require a particular mortality law. The interest rate and benefit timing do matter. The assumption $i>0$ ensures $v<1$ and $d>0$, so the conversion by division is defined. The result concerns unit benefits; if the insurance benefit is $B$ and the annuity payment is $C$, their respective expected present values are $BA_x$ and $C\ddot{a}_x$.

Using the Identity to Convert Values

When one value is known, the identity supplies the other without requiring a separate calculation of every death-year probability. For an insurance value $A_x$, first calculate $d=i/(1+i)$, then use $\ddot{a}_x=(1-A_x)/d$. For a known annuity value, use $A_x=1-d\ddot{a}_x$. Keeping the discount rate $d$ distinct from the interest rate $i$ is essential.

Worked Example: Converting Values Under a Two-Year Mortality Model

A hypothetical life aged $x$ has probability $1/2$ of dying in the first year and probability $1/2$ of dying in the second year; death by the end of year 2 is certain. A whole-life policy pays \$50,000 at the end of the year of death, and a matching annuity-due pays a level amount annually while the life is alive. The effective annual interest rate is 25%. Find $A_x$, $\ddot{a}_x$, and the level annual payment whose expected present value equals that of the insurance benefit.

Method. Find the unit insurance and annuity values from the stated lifetime probabilities, then use the identity to check the conversion. Here $P(K=0)=1/2$ and $P(K=1)=1/2$; there is no probability of survival beyond year 2. The interest and discount factors are $v=1/1.25=0.8$ and $d=1-v=0.2$.

If $K=0$, the unit insurance pays at time 1 and has present value $0.8$. If $K=1$, it pays at time 2 and has present value $0.8^2=0.64$. Thus $A_x=(1/2)(0.8)+(1/2)(0.64)=0.40+0.32=0.72$.

The annuity pays at time 0 in either case. When $K=0$, its present value is 1; when $K=1$, its present value is $1+0.8=1.8$. Therefore, $\ddot{a}_x=(1/2)(1)+(1/2)(1.8)=0.5+0.9=1.4$. The identity check is $A_x+d\ddot{a}_x=0.72+(0.2)(1.4)=0.72+0.28=1$.

The expected present value of the \$50,000 insurance benefit is $50{,}000(0.72)=\$36{,}000$. Let $C$ be the level annual annuity payment that has the same expected present value. Then $1.4C=36{,}000$, so $C=36{,}000/1.4=\$25{,}714.2857\ldots$, or \$25,714.29 to the nearest cent. As a check, $1.4(\$25{,}714.2857\ldots)=\$36{,}000$. A common mistake is to use $i=0.25$ in place of $d$; the correct discount rate is $d=0.2$.

Working Directly from Annual Death Probabilities

The identity can also check values calculated from a mortality model. If $q_{x+k}$ is the probability of death during the year after reaching age $x+k$, then the probability of death in policy year $k+1$ is ${}_kp_xq_{x+k}$. Under this discrete model, $A_x=\sum_{k=0}^{\infty}{}_kp_xq_{x+k}v^{k+1}$. For the annuity-due, the payment at time $j$ is made with probability ${}_jp_x$, giving $\ddot{a}_x=\sum_{j=0}^{\infty}{}_jp_xv^j$. These expressions use end-of-year-of-death insurance and beginning-of-year annuity payments, respectively.

Worked Example: Checking Values from a Three-Year Mortality Table

For a hypothetical life aged $x$, suppose $q_x=0.20$, $q_{x+1}=0.25$, and $q_{x+2}=1$. Thus death occurs by the end of year 3. The effective annual interest rate is 10%. Calculate both values from the death probabilities and verify the identity.

Method. First find the probability of death in each year. The probabilities are $P(K=0)=0.20$, $P(K=1)=(1-0.20)(0.25)=0.80(0.25)=0.20$, and $P(K=2)=0.80(1-0.25)(1)=0.80(0.75)=0.60$. They sum to $0.20+0.20+0.60=1$, as required.

Here $v=1/1.10=10/11$, $d=1/11$, $v^2=100/121$, and $v^3=1000/1331$. The insurance value is $A_x=0.20(10/11)+0.20(100/121)+0.60(1000/1331)$. With denominator 1331, this is $242/1331+220/1331+600/1331=1062/1331$, or approximately $0.7979$ to four decimal places.

The annuity present value is 1 if death occurs in year 1, $1+v=21/11$ if death occurs in year 2, and $1+v+v^2=331/121$ if death occurs in year 3. Therefore, $\ddot{a}_x=0.20(1)+0.20(21/11)+0.60(331/121)$. Using denominator 605 gives $121/605+231/605+993/605=1345/605=269/121$, or approximately $2.2231$ to four decimal places.

The identity check, using exact fractions, is $1062/1331+(1/11)(269/121)=1062/1331+269/1331=1$. Thus the independently calculated values agree. The common mistake this check helps catch is to omit the annuity payment at time 0 or to include an additional annuity payment at the end of the year of death.

Application to a Level Net Premium

For a hypothetical policy with a level annual premium payable in advance while the insured is alive, the premium present value is $P\ddot{a}_x$, where $P$ is the annual premium. If the policy pays a unit benefit at the end of the year of death, its expected present value is $A_x$. Equating these values gives the level net premium $P=A_x/\ddot{a}_x$. This calculation uses only the insurance and annuity present values; it does not include expenses or other loading.

Worked Example: Calculating a Level Net Premium

A hypothetical population follows a constant annual mortality model with $q_{x+k}=0.10$ in every year, so $P(K=k)=0.10(0.90)^k$ for $k=0,1,\ldots$. The effective annual interest rate is 5%. A policy pays \$14,000 at the end of the year of death, and premiums are paid at the start of each year while the insured is alive. Calculate the level annual net premium.

Method. Calculate $A_x$ from the geometric death-year probabilities, obtain $\ddot{a}_x$ from the identity, then equate the expected present values of benefits and premiums. The discount factor is $v=1/1.05=20/21$, and the discount rate is $d=1/21$.

By the geometric-series formula, $A_x=\sum_{k=0}^{\infty}0.10(0.90)^kv^{k+1}=0.10v/[1-0.90v]$. The numerator is $0.10(20/21)=2/21$, and the denominator is $1-0.90(20/21)=1-18/21=3/21$. Hence $A_x=(2/21)/(3/21)=2/3$.

The identity gives $\ddot{a}_x=(1-2/3)/(1/21)=(1/3)(21)=7$. If $P$ is the annual net premium, equivalence of expected present values requires $7P=14{,}000(2/3)=\$9{,}333.3333\ldots$. Thus $P=\$9{,}333.3333\ldots/7=\$1{,}333.3333\ldots$, or \$1,333.33 to the nearest cent. Substitution checks the result: $7(\$1{,}333.3333\ldots)=\$9{,}333.3333\ldots$, equal to the benefit present value before rounding. The timing trap is to treat the premium as payable at year-end; the stated premium is due at the start of each year, so its value uses $\ddot{a}_x$.

Common Timing and Rate Traps

The most reliable check is to put both cash flows on one timeline before applying the formula. The annuity-due begins at time 0; the insurance benefit is paid at time $K+1$. For a lifetime with $K=k$, there are $k+1$ annuity payments, not $k$, and the insurance discount exponent is $k+1$, not $k$.

Using $i$ instead of $d$. The identity uses $d=1-v=i/(1+i)$. Substituting $i$ changes the result.
Changing the benefit timing. A benefit payable immediately on death is not discounted to the end of the year of death, so this identity does not directly apply to it.
Changing the annuity timing. An annuity-immediate does not make a payment at time 0. Its value is not $\ddot{a}_x$, so the stated identity cannot be used without adjusting the payment timing.
Forgetting payment amounts. $A_x+d\ddot{a}_x=1$ compares unit amounts. Scale the insurance and annuity values by their respective benefit and payment amounts before equating present values.

The identity is useful both as a conversion and as a consistency check. When an insurance value and annuity-due value come from the same mortality and interest assumptions, their weighted sum $A_x+d\ddot{a}_x$ should equal 1. A failure to do so can indicate a mistaken discount rate, survival probability, or payment date. This relationship is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated payment timing and effective annual interest rate. Answers are shown at the end of each question for checking.

A hypothetical life has constant annual mortality $q_{x+k}=0.25$, and a unit whole-life benefit is paid at the end of the year of death. At an effective annual interest rate of 25%, find $A_x$ and $\ddot{a}_x$. (Answer: $A_x=0.5000,\ \ddot{a}_x=2.5000$)
A hypothetical life has probability $0.40$ of dying in year 1 and probability $0.60$ of dying in year 2, with no survival beyond year 2. At an effective annual interest rate of 20%, find $A_x$ and $\ddot{a}_x$. (Answer: $A_x=0.7500,\ \ddot{a}_x=1.5000$)
At an effective annual interest rate of 4%, a whole-life value is $A_x=0.42$. Find $\ddot{a}_x$. (Answer: $\ddot{a}_x=15.08$)
A hypothetical life has constant annual mortality $q_{x+k}=0.20$. At an effective annual interest rate of 10%, find $A_x$ and $\ddot{a}_x$. (Answer: $A_x=2/3,\ \ddot{a}_x=11/3$)
A hypothetical policy pays \$30,000 at the end of the year of death. At an effective annual interest rate of 5%, $A_x=0.35$, and premiums are payable at the start of each year while the insured is alive. Find the level annual net premium, to the nearest cent. (Answer: \$769.23)

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