Policy-Year Indexing and Off-by-One Errors

Why Policy-Year Indexing Matters

In a discrete life insurance calculation, three events must be matched: the policyholder survives to the start of a policy year, dies during that year, and receives the benefit at the end of that year. Each event has its own time index. Confusing those indices can shift a probability by one year, omit a survival factor, or discount the benefit for the wrong number of years.

The same contract can be described using policy years numbered 1, 2, 3, and so on, or using a curtate-lifetime index that starts at 0. Both descriptions are valid, but their indices must be translated consistently. This tutorial focuses on that translation for discrete death benefits, aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Policy year $k$ is the interval from time $k-1$ to time $k$ after issue. The probability of surviving to its start and then dying during it is $$ {}_{k-1}p_x q_{x+k-1}, $$ where ${}_{k-1}p_x$ is the probability that a life aged $x$ survives $k-1$ years, and $q_{x+k-1}$ is the one-year death probability for a life aged $x+k-1$. If the benefit is paid at the end of the year of death, its payment time is $k$, so its discount factor is $v^k$.

Two Indexing Systems for the Same Event

Let $K_x$ be the curtate future lifetime of a life aged $x$: the number of complete years lived after age $x$. Under the usual annual discrete model, $K_x=k-1$ means death occurs in policy year $k$. In particular, death in the first policy year corresponds to $K_x=0$, not $K_x=1$.

The conditional death probability during policy year $k$, given survival to its start, is $q_{x+k-1}$. That is not yet the probability of dying in that year as measured from issue. The unconditional probability from issue must also include survival to the start:

P(K_x=k-1) = {}_{k-1}p_x q_{x+k-1}.

If the annual mortality rates are $q_x,q_{x+1},\ldots$, then ${}_{k-1}p_x$ is the product of the previous survival probabilities:

{}_{k-1}p_x = \prod_{j=0}^{k-2}(1-q_{x+j}), \qquad P(K_x=k-1) = \left(\prod_{j=0}^{k-2}(1-q_{x+j})\right)q_{x+k-1}.

For $k=1$, the product has no factors and equals 1. Thus the first-year death probability is simply $q_x$. For $k=2$, the probability is $(1-q_x)q_{x+1}$: the life must first survive year 1, then die during year 2.

Policy year	Curtate-lifetime event	Probability from issue	Payment time if paid at year-end	Discount factor
1	$K_x=0$	$q_x$	1	$v$
2	$K_x=1$	${}_1p_xq_{x+1}$	2	$v^2$
3	$K_x=2$	${}_2p_xq_{x+2}$	3	$v^3$

Matching the Benefit Date to the Index

Assume an effective annual interest rate $i$, with $v=1/(1+i)$. A level benefit $B$ payable at the end of the policy year of death has present value $Bv^k$ when death occurs in policy year $k$. Its expected present value for an $n$-year term contract is therefore

B\sum_{k=1}^{n} v^k\,{}_{k-1}p_x q_{x+k-1}.

The probability refers to the event measured from issue; the discount exponent refers to when the payment is made. Those are related but distinct parts of each term. Equivalently, using the curtate-lifetime index $j=k-1$, the same expression is

B\sum_{j=0}^{n-1}v^{j+1}\,{}_jp_x q_{x+j}.

The exponent is $j+1$, not $j$, because $j$ counts complete years lived before death, while the benefit is paid one year later, at the end of the year of death.

Why the term formula holds: The events “death in policy year $k$” for $k=1,\ldots,n$ are mutually exclusive. On the event of death in year $k$, the life survives $k-1$ years with probability ${}_{k-1}p_x$, then dies in the next year with conditional probability $q_{x+k-1}$. The event probability is their product. The benefit $B$ is paid at time $k$, so its present value on that event is $Bv^k$. Summing present value times event probability over the mutually exclusive death years gives the expected present value.

Worked Example: A Three-Year Term Benefit

A hypothetical three-year term policy covers a life aged 40. The annual mortality probabilities are $q_{40}=0.10$, $q_{41}=0.20$, and $q_{42}=0.25$. The effective annual interest rate is $i=5\%$, and the death benefit is $10,000, payable at the end of the policy year of death. Find the expected present value. Assume the stated mortality rates apply to each policy year and there are no other benefits.

Method. For each policy year, calculate the probability of surviving to its start and dying during it, then discount the $10,000 payment to that year's end.

Policy year 1. The life is at issue age at the start of the year, so the death probability is $q_{40}=0.10$. The payment is at time 1, so its discount factor is $v$.

v=\frac{1}{1.05}=0.9523809524, \qquad \text{year 1 contribution} = 10{,}000(0.10)(0.9523809524) = \$952.3809524.

Policy year 2. The life must survive year 1 and then die during year 2:

P(\text{death in year 2}) = (1-0.10)(0.20) = 0.90(0.20) = 0.18.

The payment is at time 2. Since $v^2=0.9070294785$, the contribution is

10{,}000(0.18)(0.9070294785) = \$1{,}632.653061.

Policy year 3. The life must survive both previous years before dying in year 3:

P(\text{death in year 3}) = (0.90)(0.80)(0.25) = 0.18.

The payment is at time 3. Since $v^3=0.8638375985$, the contribution is

10{,}000(0.18)(0.8638375985) = \$1{,}554.907677.

Adding the three contributions gives $952.3809524+1{,}632.653061+1{,}554.907677=4{,}139.941690$. Rounded to the nearest cent, the expected present value is $4,139.94. As a check, the three death probabilities sum to $0.10+0.18+0.18=0.46$, the probability of death during the three-year term.

Common mistake avoided: The year-3 probability is not $q_{42}=0.25$ from issue. That rate is conditional on surviving to age 42; from issue, the probability is $0.90(0.80)(0.25)=0.18$. The benefit is also discounted three years, not two.

Using Life-Table Counts

When a life table gives $\ell_x$, the number of lives surviving to exact age $x$, the number dying between ages $x+k-1$ and $x+k$ is $\ell_{x+k-1}-\ell_{x+k}$. Dividing by the original cohort size $\ell_x$ gives the probability from issue of dying during policy year $k$:

P(\text{death in policy year }k) = \frac{\ell_{x+k-1}-\ell_{x+k}}{\ell_x}.

This is the same as ${}_{k-1}p_xq_{x+k-1}$, since survival to the start of the year is $\ell_{x+k-1}/\ell_x$, and the conditional one-year death probability is $(\ell_{x+k-1}-\ell_{x+k})/\ell_{x+k-1}$. Their product reduces to the life-table expression above.

Worked Example: Reading Policy Years From Life-Table Counts

A hypothetical life table has $\ell_{60}=1{,}000$, $\ell_{61}=900$, and $\ell_{62}=720$. A two-year term insurance pays $5,000 at the end of the policy year of death. The effective annual interest rate is $i=4\%$. Find the expected present value, assuming the table counts represent the relevant annual mortality experience.

Method. Divide deaths in each age interval by the initial cohort size $\ell_{60}$, then apply the discount factor for the end of the corresponding policy year.

Policy year 1, ages 60 to 61. There are $1{,}000-900=100$ deaths, so the probability from issue is $100/1{,}000=0.10$. The payment occurs at time 1.

Policy year 2, ages 61 to 62. There are $900-720=180$ deaths, so the probability from issue is $180/1{,}000=0.18$. This is not $180/900=0.20$: $0.20$ is conditional on reaching age 61, whereas the present-value calculation from issue needs the unconditional probability. The payment occurs at time 2.

v=\frac{1}{1.04}=0.9615384615, \qquad v^2=\frac{1}{1.04^2}=0.9245562130.

The year-1 contribution is $5{,}000(0.10)(0.9615384615)=\$480.7692308$. The year-2 contribution is $5{,}000(0.18)(0.9245562130)=\$832.1005917$. Their sum is $\$1{,}312.8698225$, so the expected present value rounded to the nearest cent is $1,312.87.

Check. The probability of dying in the two-year term is $(1{,}000-720)/1{,}000=0.28$, which agrees with $0.10+0.18$. The conditional second-year death probability is $180/900=0.20$; multiplying by the $0.90$ probability of surviving the first year recovers $0.18$.

Benefit Schedules and the Off-by-One Trap

The indexing rule applies whether the benefit is level or varies by policy year. If a benefit $B_k$ is payable at the end of policy year $k$, then its expected present value over an $n$-year term is

\sum_{k=1}^{n} B_k v^k\,{}_{k-1}p_x q_{x+k-1}.

The payment amount $B_k$ can change from year to year; the survival, death, and payment-time indices do not. Before substituting values, write a row for each policy year and identify four items: the conditional mortality rate, survival to the year's start, death probability from issue, and payment time.

Number the policy years.
For a term of $n$ years, use $k=1,\ldots,n$. Policy year $k$ runs from time $k-1$ to time $k$.

Find the event probability.
Use ${}_{k-1}p_xq_{x+k-1}$. If using life-table counts, use $(\ell_{x+k-1}-\ell_{x+k})/\ell_x$ for the probability from issue.

Set the payment time.
An end-of-year-$k$ payment has discount factor $v^k$. In curtate notation, $K_x=k-1$, so the factor is $v^{K_x+1}$.

Multiply and sum.
For each year, multiply its benefit, event probability, and discount factor; then add the mutually exclusive year-by-year contributions.

Worked Example: A Benefit Payable Only in Policy Year 2

A hypothetical three-year term policy covers a life aged 35. The annual death probabilities are $q_{35}=0.04$, $q_{36}=0.05$, and $q_{37}=0.06$. The effective annual interest rate is $i=4\%$. The policy pays $10,000 at the end of policy year 2 if death occurs during that year, and pays nothing for deaths in years 1 and 3. Find the expected present value. Assume the annual mortality rates apply to the corresponding policy years.

Method. Only the year-2 death event produces a benefit. Its probability must include survival through year 1, and its payment is discounted to time 2.

The probability of surviving policy year 1 is $1-q_{35}=0.96$. Conditional on that survival, the probability of death in policy year 2 is $q_{36}=0.05$. Thus the probability from issue of death in year 2 is

P(\text{death in year 2}) = 0.96(0.05) = 0.048.

The payment is at the end of year 2, so $v^2=1/(1.04)^2=0.9245562130$. The expected present value is

10{,}000(0.048)(0.9245562130) = \$443.7869822.

Rounded to the nearest cent, the answer is $443.79. The year-2 death event probability is checked by its components: $0.96$ is survival to the start of policy year 2, and $0.05$ is death during that year conditional on reaching it. The year-2 benefit is discounted exactly two years.

Common mistake avoided: Using $10{,}000(0.05)v$ would make two errors: it would omit the probability of surviving the first year, and it would discount the year-2 payment as though it were paid at time 1.

Common Indexing Errors

A reliable calculation keeps conditional rates, unconditional event probabilities, and payment dates separate until each is correctly identified. These errors are especially easy to make when switching between policy-year numbering and curtate-lifetime notation.

Using the wrong year’s mortality rate. Policy year $k$ uses $q_{x+k-1}$, because the life is age $x+k-1$ at the start of that year.
Omitting survival to the year’s start. The conditional rate $q_{x+k-1}$ alone is not the probability from issue for death in policy year $k$. Multiply by ${}_{k-1}p_x$.
Discounting by the curtate index. If $K_x=k-1$, an end-of-year-of-death payment is discounted by $v^{K_x+1}$, not $v^{K_x}$.
Using a conditional life-table denominator. For a probability from issue, divide deaths in the age interval by $\ell_x$, not by the number alive at the start of that interval.
Starting policy years at zero without changing the meaning. If an index starts at 0, state whether it counts completed years before death or numbers payment times. Do not use a zero-based index as if it were a policy-year number.

A quick check is to examine the first year. There is no prior year of survival to include, so the first-year death probability is $q_x$; if the benefit is payable at the end of that year, its discount factor is $v$. A formula that gives either a survival factor before the first year or a time-0 payment should be reviewed.

Key takeaway: For death in policy year $k$, use probability ${}_{k-1}p_xq_{x+k-1}$ and, for an end-of-year benefit, discount by $v^k$. In curtate notation the same event is $K_x=k-1$, and the discount factor is $v^{K_x+1}$.

Check Your Understanding

For each problem, identify the relevant policy year before calculating; all policies below use end-of-year benefit payments unless stated otherwise.

A two-year term policy has annual death probabilities $q_x=0.10$ and $q_{x+1}=0.20$, with interest rate $i=0$. The benefit is $1,000 in either year of death. Find its expected present value. (Answer: $280.00)
For a life aged $x$, $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$. Find the probability, from issue, of death in policy year 3. (Answer: 0.1800)
A two-year term policy has $q_x=0.05$, $q_{x+1}=0.10$, an effective annual interest rate of $5\%$, and a $8,000 benefit payable at the end of the year of death. Find the expected present value, rounded to the nearest cent. (Answer: $1,070.29)
A life table gives $\ell_{70}=500$, $\ell_{71}=450$, and $\ell_{72}=360$. Find the probability from age 70 of death during policy year 2, and state the conditional probability of death during that year given survival to its start. (Answer: 0.1800 from issue; 0.2000 conditional on reaching age 71)
A three-year term policy has $q_x=0.04$, $q_{x+1}=0.05$, and $q_{x+2}=0.06$. It pays $10,000 only for death in policy year 2, at an effective annual interest rate of $4\%$. Find its expected present value, rounded to the nearest cent. (Answer: $443.79)

Policy year	Curtate-lifetime event	Probability from issue	Payment time if paid at year-end	Discount factor
1	\(K_x=0\)	\(q_x\)	1	\(v\)
2	\(K_x=1\)	\({}_1p_xq_{x+1}\)	2	\(v^2\)
3	\(K_x=2\)	\({}_2p_xq_{x+2}\)	3	\(v^3\)

What You'll Learn