Piecewise Benefits for Discrete Term Insurance

Why Group Policy Years?

A term insurance benefit need not stay level throughout the coverage period. A hypothetical policy might pay a larger amount during its first two years and a smaller amount afterward, or it might increase once and then decrease. The actuarial present value (APV) accounts for both the probability that death occurs in each policy year and the benefit payable in that year.

When the benefit is constant within blocks of years, there is no need to treat every year as a separate benefit amount. Instead, calculate the discounted death probabilities for each block, add them, and multiply by that block’s benefit. This is an efficient way to organize the same year-by-year APV.

Definition: Consider an $n$-year discrete term insurance issued to a life aged $x$. Let $B_j$ be the benefit payable if death occurs during policy year $j$, with payment at the end of that year. For $j=1,\ldots,n$, the probability of death in policy year $j$ is ${}_{j-1}p_x q_{x+j-1}$. With a constant effective annual interest rate $i$, the APV is $$ \operatorname{APV}=\sum_{j=1}^{n} B_j v^j\,{}_{j-1}p_x q_{x+j-1}, \qquad v=\frac{1}{1+i}. $$

The notation ${}_{j-1}p_x q_{x+j-1}$ separates the two events needed for death in year $j$: the insured survives the first $j-1$ years, then dies during the next year. The factor $v^j$ discounts a payment made at the end of policy year $j$ back $j$ years.

Grouping Years With Equal Benefits

Suppose a block of policy years $r$ through $s$ has the same benefit $C$. The contribution of this block to the APV is $$ C\sum_{j=r}^{s} v^j\,{}_{j-1}p_x q_{x+j-1}. $$

If the schedule has several blocks, each policy year must belong to exactly one block. The total APV is the sum of the block contributions. Grouping changes the bookkeeping, not the valuation: every year’s benefit is still multiplied by that year’s discounted probability of death.

Theorem: If the policy years $1,\ldots,n$ are partitioned into nonoverlapping blocks $I_1,\ldots,I_m$, and the benefit is $C_h$ in block $I_h$, then $$ \operatorname{APV}=\sum_{h=1}^{m} C_h \left(\sum_{j\in I_h}v^j\,{}_{j-1}p_x q_{x+j-1}\right). $$ Each block includes its stated first and last policy years. A year omitted from or counted in two blocks would give an incorrect benefit schedule.

This grouping formula follows directly by collecting terms in the year-by-year sum. For any block $I_h$, $B_j=C_h$ for every $j\in I_h$. Therefore, $$ \sum_{j=1}^{n}B_jv^j\,{}_{j-1}p_xq_{x+j-1} = \sum_{h=1}^{m}\sum_{j\in I_h}C_hv^j\,{}_{j-1}p_xq_{x+j-1} = \sum_{h=1}^{m}C_h\sum_{j\in I_h}v^j\,{}_{j-1}p_xq_{x+j-1}. $$

The first equality just separates the full sum according to the blocks. The second takes the constant benefit $C_h$ outside the inner sum for each block. No assumption of constant mortality is needed; the annual survival and death probabilities can vary.

Worked Example: Two Benefit Blocks With Constant Mortality

A hypothetical four-year discrete term policy pays $10,000 for death in policy years 1 or 2, and $5,000 for death in years 3 or 4. Assume a constant annual death probability $q=0.10$, so $p=0.90$, and an effective annual interest rate $i=0.05$. Find the APV.

Method. Compute the probability of death in each policy year, discount it to time 0, and group the first two and last two years by their benefits. This avoids treating the benefit blocks as if they changed the timing of payment.

The death probabilities in years 1 through 4 are $q,\ pq,\ p^2q,\ p^3q$, respectively:

0.10,\quad 0.90(0.10)=0.09,\quad 0.90^2(0.10)=0.081,\quad 0.90^3(0.10)=0.0729.

The discount factor is $v=1/1.05=0.9523809524$. Thus the first block contributes

10{,}000\left(0.10v+0.09v^2\right) = 10{,}000\left(0.0952380952+0.0816326531\right) = 1{,}768.71.

The second block contributes

5{,}000\left(0.081v^3+0.0729v^4\right) = 5{,}000\left(0.0699708510+0.0599750122\right) = 649.73.

Adding the block contributions gives an APV of $1{,}768.71+649.73=$ $2,418.44. As a check, the death probabilities sum to $0.3439$, which is $1-0.90^4$, the probability of death within the four-year term.

Changing Mortality From Year to Year

The block method applies just as well when the mortality probability changes by policy year. In that case, build each year’s death probability by multiplying the probability of reaching the start of that year by the year-specific death probability. Do not use a single $q$ for all years unless the model states that mortality is constant.

Worked Example: Three Blocks With Varying Annual Mortality

A hypothetical five-year term policy pays $20,000 in year 1, $12,000 in years 2 and 3, and $6,000 in years 4 and 5. The annual death probabilities for a life at the start of successive policy years are $0.05, 0.10, 0.15, 0.20,$ and $0.25$. Assume an effective annual interest rate of $i=0.04$. Calculate the APV.

Method. First find each year’s probability of death from the start of the term. Then discount the death probabilities and add them within their benefit blocks. This avoids applying a year’s mortality rate before the life has survived to that year.

The probabilities of reaching the starts of years 1 through 5 are $1,\ 0.95,\ 0.855,\ 0.72675,$ and $0.5814$. Multiplying each by that year’s death probability gives

0.05,\quad 0.95(0.10)=0.095,\quad 0.855(0.15)=0.12825,\quad 0.72675(0.20)=0.14535,\quad 0.5814(0.25)=0.14535.

Here $v=1/1.04=0.9615384615$. For year 1, the discounted probability is $0.05v=0.0480769231$, so its contribution is $20{,}000(0.0480769231)=\$961.54$.

For years 2 and 3, the discounted probabilities are $0.095v^2=0.0878338261$ and $0.12825v^3=0.1140137820$. Their combined contribution is

12{,}000(0.0878338261+0.1140137820)=\$2{,}422.17.

For years 4 and 5, the discounted probabilities are $0.14535v^4=0.1242457893$ and $0.14535v^5=0.1194671051$. Their combined contribution is

6{,}000(0.1242457893+0.1194671051)=\$1{,}462.28.

The total APV is $\$961.54+\$2{,}422.17+\$1{,}462.28=$ $4,845.99, rounded to the nearest cent. As a check, the undiscounted death probabilities sum to $0.564$, equal to one minus the probability of surviving all five years, $1-(0.95)(0.90)(0.85)(0.80)(0.75)=0.564$.

Using a Life Table and Checking Payment Timing

When a problem supplies life-table counts rather than annual death probabilities, the probability of death in policy year $j$ can be obtained as $(\ell_{x+j-1}-\ell_{x+j})/\ell_x$, provided the counts refer to the same life table and the cohort is followed over the relevant ages. This is the unconditional probability, from issue, of dying in that policy year. It is equivalent to ${}_{j-1}p_xq_{x+j-1}$.

Worked Example: Grouping Benefits From Life-Table Counts

A hypothetical four-year policy is issued at age 40 and pays $10,000 in years 1 and 2, then $4,000 in years 3 and 4. The life-table counts are $\ell_{40}=10{,}000,\ \ell_{41}=9{,}500,\ \ell_{42}=8{,}550,\ \ell_{43}=6{,}840,$ and $\ell_{44}=5{,}472$. Assume a level effective annual interest rate of $i=0.05$. Find the APV.

Method. Convert the decreases in the life-table counts to probabilities from issue, then discount those probabilities by the policy-year payment time. This makes the issue-age denominator explicit and avoids confusing conditional death rates with unconditional probabilities.

The deaths in the four years are $10{,}000-9{,}500=500$, $9{,}500-8{,}550=950$, $8{,}550-6{,}840=1{,}710$, and $6{,}840-5{,}472=1{,}368$. Divide each count by $\ell_{40}=10{,}000$, giving probabilities $0.05,\ 0.095,\ 0.171,$ and $0.1368$.

With $v=1/1.05=0.9523809524$, the first benefit block contributes

10{,}000(0.05v+0.095v^2) = 10{,}000(0.0476190476+0.0861678005) = \$1{,}337.87.

The second block contributes

4{,}000(0.171v^3+0.1368v^4) = 4{,}000(0.1477162254+0.1125456980) = \$1{,}041.05.

The APV is therefore $\$1{,}337.87+\$1{,}041.05=$ $2,378.92, rounded to the nearest cent. The life-table counts also give $(10{,}000-5{,}472)/10{,}000=0.4528$ as the probability of death during the term; the four annual probabilities sum to $0.4528$.

Common Mistake: Discounting From the Wrong Year

If death occurs in policy year $j$, the benefit is paid at the end of that year, so the discount factor is $v^j$. A frequent indexing error is to call the first policy year “year 0” and use $v^{j-1}$ while retaining the probability of death in year $j$. That discounts every payment for one year too little.

For the first worked example, this incorrect approach would give

10{,}000(0.10+0.09v)+5{,}000(0.081v^2+0.0729v^3) =\$2{,}539.36.

The corrected expression uses $v,v^2,v^3,v^4$, respectively, and gives $\$2{,}418.44$. A quick timing check is to match the first-year death probability to a payment one year after issue, not immediately at issue.

Key takeaway: For every policy year, pair its benefit with that year’s probability of death and its end-of-year discount factor. Grouping equal benefits can shorten the calculation, but it does not change these year-by-year pairings.

Check Your Understanding

For each problem, assume benefits are paid at the end of the year of death. Round monetary answers to the nearest cent.

A three-year policy pays $10,000 in year 1 and $5,000 in years 2 and 3. The annual death probability is $0.10$ in each year, and $i=0$. Find the APV. (Answer: $1,855.00)
A four-year policy pays $3,000 in years 1 and 2, then $1,000 in years 3 and 4. Assume constant annual death probability $q=0.20$ and effective annual interest $i=0.25$. Find the APV. (Answer: $894.68)
A three-year policy pays $5,000 in years 1 and 2, then $2,000 in year 3. The annual death probabilities are $0.10,\ 0.20,$ and $0.30$, respectively, and $i=0.05$. Find the APV. (Answer: $1,665.69)
A four-year policy pays $10,000 in years 1 and 2 and $4,000 in years 3 and 4. The life-table counts are $\ell_{40}=10{,}000,\ \ell_{41}=9{,}500,\ \ell_{42}=8{,}550,\ \ell_{43}=6{,}840,\ \ell_{44}=5{,}472$, and $i=0.05$. Find the APV. (Answer: $2,378.92)
A five-year policy pays $2,000 in years 1 through 3 and $5,000 in years 4 and 5. Assume constant annual death probability $q=0.10$ and $i=0$. Find the APV. (Answer: $1,234.55)

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