The Whole-Life Discrete Insurance-Annuity Identity

Why the Insurance-Annuity Identity Matters

A whole-life insurance benefit and a whole-life annuity-due make payments at different times. The insurance pays at the end of the year of death; the annuity-due pays at the beginning of each year, provided the life is then alive. Despite this difference, their present values satisfy a simple relationship. It is useful both for interpreting the two values together and for checking calculations.

The timing convention is essential. An annuity-due has its first payment immediately, at time 0. A discrete whole-life insurance has no payment at time 0: if death occurs in the first year, its benefit is paid at time 1. The identity below depends on those specific payment patterns.

Definition: Assume a life aged \(x\), a positive effective annual interest rate \(i\), and a unit benefit. Let \(v=1/(1+i)\) be the annual discount factor and \(d=1-v=i/(1+i)\) be the annual discount rate. The discrete whole-life insurance value \(A_x\) is the expected present value of 1 paid at the end of the year of death. The whole-life annuity-due value \(\ddot{a}_x\) is the expected present value of 1 paid at time 0 and at the beginning of each later year in which the life is alive.

Write \({}_kp_x\) for the probability that a life aged \(x\) survives at least \(k\) years, and \(q_{x+k}\) for the probability that a life aged \(x+k\) dies within one year. Thus, the probability of surviving \(k\) years and then dying in the next year is \({}_kp_xq_{x+k}\). No constant-mortality assumption is needed: the annual death probabilities may vary with age.

A_x=\sum_{k=0}^{\infty}v^{k+1}{}_kp_xq_{x+k}, \qquad \ddot{a}_x=\sum_{k=0}^{\infty}v^k{}_kp_x.

The first sum discounts each insurance payment to the end of its death year. The second discounts each annuity payment at the beginning of a year, weighted by the probability the life is alive then. In particular, the \(k=0\) annuity term is 1: the life is alive at time 0 with probability 1, and the first payment is immediate.

Deriving the Identity

The key probability relation is that, among lives alive at time \(k\), each either survives to time \(k+1\) or dies during the following year. Therefore \({}_kp_xq_{x+k}={}_kp_x-{}_{k+1}p_x\). Substituting this difference into the insurance value lets adjacent terms cancel.

Express each death-year probability as a survival difference.
The probability of death in year \(k+1\) is \({}_kp_x-{}_{k+1}p_x\).

Substitute into the insurance sum.
Each term becomes the difference between two discounted survival probabilities.

Collect adjacent terms.
The coefficient of each later survival probability is the difference between two consecutive discount factors.

Use \(d=1-v\).
The resulting expression is \(A_x=1-d\ddot{a}_x\), which gives the stated identity.

For completeness, collect the terms through year \(n\). With \({}_0p_x=1\), the partial sum is

\begin{aligned} \sum_{k=0}^{n-1}v^{k+1}({}_kp_x-{}_{k+1}p_x) &=v+\sum_{j=1}^{n-1}(v^{j+1}-v^j){}_jp_x-v^n{}_np_x\\ &=v-d\sum_{j=1}^{n-1}v^j{}_jp_x-v^n{}_np_x. \end{aligned}

Since \(i>0\), \(0<v<1\), and \(0\le {}_np_x\le1\), the boundary term satisfies \(0\le v^n{}_np_x\le v^n\), so it tends to zero. Taking the limit gives

\begin{aligned} A_x &=v-d\sum_{j=1}^{\infty}v^j{}_jp_x\\ &=v-d(\ddot{a}_x-1)\\ &=1-d\ddot{a}_x. \end{aligned} \qquad\text{Hence}\qquad A_x+d\ddot{a}_x=1.

This is a relation between values under the same interest and mortality assumptions, not an assumption that mortality follows a particular law. It applies to the unit-benefit, annual-payment pattern defined above. If the benefit is \(B\) rather than 1 and every annuity payment is also scaled by \(B\), both present values scale by \(B\), so the corresponding relation is \(BA_x+d(B\ddot{a}_x)=B\).

Worked Calculations

Worked Example: Varying Death Probabilities over Four Years

A hypothetical policy pays 1 at the end of the year of death. The effective annual interest rate is \(i=25\%\). The one-year death probabilities are \(q_x=0.20\), \(q_{x+1}=0.10\), \(q_{x+2}=0.25\), and \(q_{x+3}=1\). Calculate \(A_x\) and \(\ddot{a}_x\), and verify the identity. Use the method of first finding the unconditional death-year probabilities; this avoids applying each \(q\) to the original group rather than to survivors.

The discount factor and annual discount rate are

v=\frac{1}{1.25}=0.8, \qquad d=1-v=0.2.

The probability of death in each year is the probability of reaching that year multiplied by its conditional death probability. The successive survival probabilities at the starts of the years are \(1\), \(0.8\), \(0.8(0.9)=0.72\), and \(0.72(0.75)=0.54\). Thus the death-year probabilities are

0.20,\qquad 0.80(0.10)=0.08,\qquad 0.72(0.25)=0.18,\qquad 0.54(1)=0.54.

They sum to \(0.20+0.08+0.18+0.54=1\), as every life in this model dies by the end of year 4. Discount each payment by its payment time:

\begin{aligned} A_x &=0.20(0.8)+0.08(0.8^2)+0.18(0.8^3)+0.54(0.8^4)\\ &=0.160000+0.051200+0.092160+0.221184\\ &=0.524544. \end{aligned}

The annuity pays at times 0, 1, 2, and 3 only if the life is alive at each payment time. The probabilities of being alive at those times are \(1\), \(0.8\), \(0.72\), and \(0.54\), respectively. Therefore

\begin{aligned} \ddot{a}_x &=1+0.8(0.8)+0.72(0.8^2)+0.54(0.8^3)\\ &=1+0.640000+0.460800+0.276480\\ &=2.377280. \end{aligned}

The check is \(d\ddot{a}_x=0.2(2.377280)=0.475456\), so \(A_x+d\ddot{a}_x=0.524544+0.475456=1.000000\). The identity holds.

Worked Example: Constant Annual Mortality

For a hypothetical population, suppose the conditional probability of death in each year is \(q_{x+k}=0.25\) for every \(k\ge0\), and the effective annual interest rate is \(i=25\%\). Find both values and check their relationship. Use geometric series because the probability of surviving each year is the same. The important trap is to use \(d=1-v\), not \(i\), in the identity.

Here \(p_{x+k}=1-0.25=0.75\), \(v=1/1.25=0.8\), and \(d=1-0.8=0.2\). The probability of death in year \(k+1\) is \(0.75^k(0.25)\), so

\begin{aligned} A_x &=\sum_{k=0}^{\infty}0.8^{k+1}(0.75^k)(0.25)\\ &=(0.8)(0.25)\sum_{k=0}^{\infty}(0.8\cdot0.75)^k\\ &=\frac{0.20}{1-0.60} =0.50. \end{aligned}

At time \(k\), the probability of being alive is \(0.75^k\). Thus the annuity terms form another geometric series:

\ddot{a}_x =\sum_{k=0}^{\infty}(0.8\cdot0.75)^k =\frac{1}{1-0.60} =2.50.

The check is \(A_x+d\ddot{a}_x=0.50+0.20(2.50)=0.50+0.50=1\). Using \(i=0.25\) in place of \(d=0.20\) would incorrectly give \(0.50+0.25(2.50)=1.125\).

Worked Example: A Life That Cannot Survive Beyond Two Years

A hypothetical life has \(q_x=0.50\), \(q_{x+1}=1\), and no possibility of survival beyond age \(x+2\). The effective annual interest rate is \(i=10\%\). Calculate the discrete whole-life insurance and annuity-due values, then verify the identity. Keep the annuity payment at time 0: omitting it is a common timing error.

The interest quantities are

v=\frac{1}{1.10}=\frac{10}{11}, \qquad d=1-v=\frac{1}{11}.

The probability of death in the first year is \(0.50\). Half the lives survive to the second year, and all of those die in that year, so the second-year death probability is \(0.50(1)=0.50\). The two probabilities sum to 1. The insurance value is

\begin{aligned} A_x &=0.50\left(\frac{10}{11}\right) +0.50\left(\frac{10}{11}\right)^2\\ &=\frac{5}{11}+\frac{50}{121} =\frac{55}{121}+\frac{50}{121} =\frac{105}{121} \approx0.8678. \end{aligned}

The annuity pays 1 at time 0 with certainty and another 1 at time 1 only for the half who survive the first year. There are no payments at time 2 because no life is alive then. Hence

\ddot{a}_x=1+0.50\left(\frac{10}{11}\right) =1+\frac{5}{11} =\frac{16}{11} \approx1.4545.

Using exact fractions, \(d\ddot{a}_x=(1/11)(16/11)=16/121\). Therefore \(A_x+d\ddot{a}_x=105/121+16/121=121/121=1\). The rounded values are shown to four decimal places; the verification uses exact values.

Common Exam Traps and Interpretation

The identity is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In an exam calculation, it can reveal a mistake in timing, probabilities, or interest conversion even when each individual step appears plausible.

Confusing \(i\) and \(d\). The annual discount rate is \(d=i/(1+i)=1-v\), not \(i\). At \(i=25\%\), \(v=0.8\) and \(d=0.2\). The incorrect check \(0.50+0.25(2.50)=1.125\) in the constant-mortality example uses the wrong rate; the correct check is \(0.50+0.20(2.50)=1\).
Shifting the annuity by one year. The annuity-due begins at time 0, so its first term is 1. Starting the sum at \(v\) describes payments one year later, not an annuity-due.
Discounting the insurance to the wrong time. Death in year \(k+1\) produces a payment at time \(k+1\), discounted by \(v^{k+1}\), not \(v^k\).
Using conditional death probabilities as unconditional probabilities. The probability of dying in a later year must include the probability of surviving to that year. For example, with \(q_x=0.20\) and \(q_{x+1}=0.10\), the second-year death probability is \(0.80(0.10)=0.08\), not \(0.10\).

A useful final check is to state the timing in words before substituting numbers: insurance payments occur at the end of the death year, while annuity-due payments occur at the beginning of each year of life. The identity does not apply unchanged if a benefit is paid immediately at death, if annuity payments are made at year-end, or if payment amounts are not level. Those are different cash-flow patterns and require their own valuation expressions.

Key takeaway: For unit whole-life insurance payable at the end of the year of death and a unit whole-life annuity-due payable annually while alive, under the same mortality and interest assumptions, \(A_x+d\ddot{a}_x=1\), where \(d=1-v\). Check the payment times and discount rate before using the relation.

Check Your Understanding

For each problem, identify the payment timing before calculating. Values are exact unless a decimal approximation is requested.

A hypothetical life has \(q_x=0.40\), \(q_{x+1}=1\), and effective annual interest \(i=10\%\). Find the death-year probabilities, \(A_x\), and \(\ddot{a}_x\), then check the identity. (Answer: death-year probabilities \(0.40,0.60\); \(A_x=104/121\approx0.8595\); \(\ddot{a}_x=17/11\approx1.5455\); \(A_x+(1/11)\ddot{a}_x=1\).)
For a hypothetical life, \(q_{x+k}=0.20\) for every year \(k\ge0\), and \(i=25\%\). Find \(A_x\) and \(\ddot{a}_x\) using geometric series. (Answer: \(A_x=4/9\); \(\ddot{a}_x=25/9\); \(A_x+0.20\ddot{a}_x=1\).)
A hypothetical life has \(q_x=0.30\), \(q_{x+1}=0.20\), and \(q_{x+2}=1\); \(i=25\%\). Find the unconditional death-year probabilities, both present values, and the identity check. (Answer: probabilities \(0.30,0.14,0.56\); \(A_x=0.61632\); \(\ddot{a}_x=1.9184\); \(0.61632+0.20(1.9184)=1\).)
A hypothetical life has constant annual death probability \(0.40\), and the effective annual interest rate is \(20\%\). Calculate the insurance and annuity-due values from their geometric series and verify the identity. (Answer: \(v=5/6\), \(d=1/6\), \(A_x=2/3\), \(\ddot{a}_x=2\), and \(2/3+(1/6)(2)=1\).)

What You'll Learn