Whole-Life Insurance From Life-Table Counts

Why Life-Table Counts Give a Whole-Life Value

A life table can describe mortality using counts rather than probabilities. The count $l_x$ represents the number of lives surviving to exact age $x$, while $d_x$ represents the number of deaths between ages $x$ and $x+1$. To value a discrete whole-life insurance benefit, convert those counts into the probability that a life aged $x$ dies in each future year, then discount the benefit paid at the end of that year.

The key denominator is the number alive at the policy’s starting age. For example, $d_{x+2}$ counts deaths in the third year, but the probability of that year’s death for someone insured at age $x$ is $d_{x+2}/l_x$, not $d_{x+2}/l_{x+2}$. The latter is a one-year death probability conditional on having reached age $x+2$; it does not account for the chance of surviving the earlier years.

Definition: In a life table, $l_y$ is the number of lives surviving to exact age $y$, and $d_y$ is the number of those lives who die between ages $y$ and $y+1$. For a consistent table, $l_{y+1}=l_y-d_y$. For an individual selected from the lives surviving to age $x$, the probability of dying between ages $x+k$ and $x+k+1$ is ${}_k|q_x=d_{x+k}/l_x$, provided $l_x>0$.

Here, ${}_k|q_x$ denotes the probability that a life aged $x$ survives $k$ full years and then dies in the following year. The life-table counts may use any convenient starting number of lives, often called a radix. They need not be actual counts of insured people: the ratios determine the probabilities.

The Discrete Whole-Life Insurance Formula

Assume a level benefit of $B$ dollars is paid at the end of the policy year of death, with no payment if the insured never dies. Let the effective annual interest rate be $i$, and let $v=1/(1+i)$. If death occurs in the first policy year, the benefit is paid at time 1 and is discounted by $v$. If death occurs in policy year $k+1$, it is paid at time $k+1$ and is discounted by $v^{k+1}$.

Theorem: Given a life table for ages $x,x+1,\ldots$, the actuarial present value of a discrete whole-life insurance with benefit $B$ paid at the end of the year of death is $$ A_x^{(B)}=B\sum_{k=0}^{\infty}v^{k+1}\frac{d_{x+k}}{l_x}. $$ For a unit benefit, $B=1$. If the table ends at a finite ultimate age, the sum includes each death interval represented by the table, through the interval in which the remaining lives die.

This formula requires that the benefit is paid at the end of the year of death and that the annual effective interest rate is constant. It uses the life table to specify the death probabilities for the insured at age $x$. A benefit paid at a different time would require a different discount factor; the formula is not a continuous-payment formula.

Why the Formula Holds

The possible death years form mutually exclusive events: an insured dies in the first policy year, the second policy year, and so on. The present value of the benefit is $Bv^{k+1}$ on the event that death occurs in policy year $k+1$, and zero for that event otherwise. Taking the expectation over these events gives the formula.

\begin{aligned} \operatorname{EPV} &=\sum_{k=0}^{\infty} \bigl(\text{present value if death is in year }k+1\bigr) \bigl(\text{probability of death in year }k+1\bigr)\\ &=\sum_{k=0}^{\infty} Bv^{k+1}\,{}_k|q_x\\ &=B\sum_{k=0}^{\infty}v^{k+1}\frac{d_{x+k}}{l_x}. \end{aligned}

The probability $d_{x+k}/l_x$ can also be understood by following the life-table cohort. Out of the $l_x$ lives at the starting age, $d_{x+k}$ die in the specified interval. Thus the fraction of the original age-$x$ cohort whose death occurs in that interval is exactly the required death-year probability. The annual death probabilities sum to 1 if the table accounts for the eventual death of every life in the cohort.

Worked Calculations

Worked Example: A Four-Interval Life Table

A hypothetical insurer issues a discrete whole-life policy to a life aged 50. The benefit is $10,000, paid at the end of the year of death. The effective annual interest rate is 5%. A mortality table gives $l_{50}=1{,}000$, with death counts $d_{50}=100$, $d_{51}=180$, $d_{52}=240$, and $d_{53}=480$, after which the cohort is exhausted. Find the actuarial present value.

Method and table check. First verify the survivor counts implied by the death counts: $1{,}000-100=900$, $900-180=720$, $720-240=480$, and $480-480=0$. The counts are consistent. The four death-year probabilities for a life aged 50 are each death count divided by the original $l_{50}=1{,}000$:

\frac{100}{1{,}000}=0.10,\quad \frac{180}{1{,}000}=0.18,\quad \frac{240}{1{,}000}=0.24,\quad \frac{480}{1{,}000}=0.48.

They sum to $0.10+0.18+0.24+0.48=1$, as expected when the cohort is exhausted. The annual discount factor is $v=1/1.05=0.95238095$. The required powers, rounded here for display, are $v^2=0.90702948$, $v^3=0.86383760$, and $v^4=0.82270247$.

Discount and weight each payment. The benefit for a first-year death is discounted one year; a fourth-year death benefit is discounted four years:

\begin{aligned} A_{50}^{(10{,}000)} &=10{,}000\left[ (0.10)(0.95238095) +(0.18)(0.90702948) +(0.24)(0.86383760) +(0.48)(0.82270247) \right]\\ &=10{,}000\left[ 0.09523810+0.16326531+0.20732102+0.39489719 \right]\\ &=10{,}000(0.86072162)\\ &=\$8{,}607.22. \end{aligned}

The actuarial present value is $8,607.22, rounded to the nearest cent. The method avoids treating the counts as probabilities without first conditioning on the number alive at age 50.

Worked Example: Converting Survivor Counts to Death Counts

A hypothetical policy covers a life aged 60 for a $12,000 whole-life benefit paid at the end of the year of death. The effective annual interest rate is 6%. The life table gives $l_{60}=800$, $l_{61}=720$, $l_{62}=540$, $l_{63}=270$, and $l_{64}=0$. Calculate the actuarial present value.

Method. Since the table supplies survivor counts, obtain each annual death count by subtracting consecutive survivor counts:

d_{60}=800-720=80,\quad d_{61}=720-540=180,\quad d_{62}=540-270=270,\quad d_{63}=270-0=270.

Use $l_{60}=800$ as the denominator for all four death-year probabilities. They are $80/800=0.10$, $180/800=0.225$, $270/800=0.3375$, and $270/800=0.3375$. These add to 1. The discount factor is $v=1/1.06=0.94339623$, and the powers needed are $v^2=0.88999644$, $v^3=0.83961928$, and $v^4=0.79209366$.

Apply the end-of-year discounting. The first death interval receives $v$, not 1, because its benefit is paid at the end of year 1:

\begin{aligned} A_{60}^{(12{,}000)} &=12{,}000\left[ (0.10)(0.94339623) +(0.225)(0.88999644) +(0.3375)(0.83961928) +(0.3375)(0.79209366) \right]\\ &=12{,}000\left[ 0.09433962+0.20024920+0.28337151+0.26733161 \right]\\ &=12{,}000(0.84529194)\\ &=\$10{,}143.50. \end{aligned}

The result is $10,143.50, rounded to the nearest cent. The calculation uses differences in survivor counts to recover deaths, then divides each death count by the starting-age survivor count.

Worked Example: A Larger Benefit and a Different Radix

A hypothetical whole-life contract pays $25,000 at the end of the year of death for a life aged 45. The effective annual interest rate is 4%. A life table has $l_{45}=400$ and death counts $d_{45}=40$, $d_{46}=60$, $d_{47}=100$, and $d_{48}=200$, with no survivors after the final interval. Find the actuarial present value.

Method. The cohort is consistent because its successive survivor counts are $400-40=360$, $360-60=300$, $300-100=200$, and $200-200=0$. Divide every death count by $l_{45}=400$, giving death-year probabilities $0.10$, $0.15$, $0.25$, and $0.50$. These sum to 1.

At 4%, $v=1/1.04=0.96153846$, $v^2=0.92455621$, $v^3=0.88899636$, and $v^4=0.85480419$. Weight the benefit by the probability and discount factor for each death year:

\begin{aligned} A_{45}^{(25{,}000)} &=25{,}000\left[ (0.10)(0.96153846) +(0.15)(0.92455621) +(0.25)(0.88899636) +(0.50)(0.85480419) \right]\\ &=25{,}000\left[ 0.09615385+0.13868343+0.22224909+0.42740210 \right]\\ &=25{,}000(0.88448847)\\ &=\$22{,}112.21. \end{aligned}

The actuarial present value is $22,112.21, rounded to the nearest cent. The cohort radix of 400 affects the intermediate counts but not the result once the counts are converted to probabilities.

Checks, Interpretation, and Common Traps

A life-table calculation is easier to audit when the table and the probability weights are checked before discounting. The following steps keep the age intervals, denominator, and payment times aligned.

Check the table.
For each age, confirm that $l_{y+1}=l_y-d_y$. If the table provides only survivor counts, calculate each $d_y$ by subtraction.

Set the starting denominator.
For a policy issued at age $x$, divide every future death count by $l_x$, not by the survivor count at that later age.

Match each year to its discount power.
Deaths between ages $x+k$ and $x+k+1$ generate a payment at time $k+1$, so use $v^{k+1}$.

Check the probability weights and value.
If the table covers all deaths, the death-year probabilities sum to 1. The value for a positive benefit is nonnegative and, for positive interest, below the undiscounted benefit.

A frequent exam trap is to discount by $v^k$ instead of $v^{k+1}$. That mistakenly treats the benefit for a first-year death as paid immediately. In the first worked example, this incorrect indexing would give

10{,}000\left[ 0.10+0.18v+0.24v^2+0.48v^3 \right] =10{,}000(0.90375769) =\$9{,}037.58,

rather than the correct $8,607.22. The correction is to associate the first death interval with $v^1$, the second with $v^2$, and so on. Another common error is to use $d_{x+k}/l_{x+k}$ for every interval. That ratio is conditional on reaching age $x+k$; the whole-life present-value sum over death years for someone initially aged $x$ needs unconditional probabilities from age $x$, namely $d_{x+k}/l_x$.

The value is proportional to the benefit: changing a $10,000 benefit to $20,000 doubles every discounted payment and therefore doubles the actuarial present value. Changing the radix of the life table does not change the result if every survivor and death count is scaled by the same factor. These are useful checks, but they do not replace checking that each probability uses the correct starting age and that each discount power matches the actual payment time.

Key takeaway: For a discrete whole-life benefit paid at the end of the year of death, each table death count $d_{x+k}$ contributes $Bv^{k+1}d_{x+k}/l_x$. The death count identifies the interval; the starting survivor count gives the probability; and the discount power reflects the end-of-year payment.

Check Your Understanding

For each hypothetical policy, use the supplied life-table counts and annual effective interest rate. Give the actuarial present value to the nearest cent.

A $5,000 whole-life benefit is paid at the end of the year of death for a life aged 30. The table has $l_{30}=1{,}000$, followed by death counts $d_{30}=100$, $d_{31}=200$, and $d_{32}=700$, after which the cohort is exhausted. The annual effective interest rate is 10%. (Answer: $3,910.59)
A $10,000 benefit covers a life aged 55, with end-of-year payment at death. Survivor counts are $l_{55}=900$, $l_{56}=810$, $l_{57}=630$, $l_{58}=360$, and $l_{59}=0$. The annual effective interest rate is 5%. Find the value, deriving the death counts from the survivor counts. (Answer: $8,648.76)
A hypothetical insurer issues a $8,000 whole-life benefit at age 40, payable at the end of the year of death. The table gives $l_{40}=2{,}000$ and death counts $160, 240, 400,$ and $1{,}200$ in the four successive years, with no lives remaining after that. At 4% effective annual interest, find the actuarial present value. (Answer: $7,028.41)
Two hypothetical life tables describe the same mortality pattern at age 70. Table A has $l_{70}=500$ and successive death counts $50,100,150,200$; Table B has $l_{70}=1{,}000$ and successive death counts $100,200,300,400$. Each cohort is exhausted after four years. For a $10,000 end-of-year whole-life benefit and a 5% effective annual interest rate, verify whether the two tables give the same value. (Answer: Yes; each gives $8,648.76.)

What You'll Learn