Exposures to Risk From Table Lives

Why Exposed-to-Risk Time Matters

A decrement table records how many lives are present at the beginning and end of an interval, but those endpoint counts do not, by themselves, reveal how long the lives were exposed between those dates. A person who exits early in the interval contributes less exposure than one who remains until the end. To estimate total time at risk, we must specify how decrements are distributed within the interval.

The result is measured in person-years when time is measured in years. It is a total duration across all lives at risk, not a count of lives or exits. This distinction is important in multiple-decrement work: the same people are exposed to the possibility of any of the competing decrements while they remain in the group.

Definition: For a closed cohort with $L(t)$ lives remaining at time $t$, the exposure over an interval of length $h$ is the area under the lives-at-risk function: $$ E=\int_0^h L(t)\,dt. $$ Here $L(0)=L_0$ is the number at the start and $L(h)=L_h$ is the number at the end. If time is measured in years, $E$ is in person-years. Endpoint table lives alone do not determine this integral; an assumption about the timing of decrements is needed.

The endpoint method applies when the counts refer to the same closed group over a common interval, with no entries, re-entry, or other unaccounted additions. In a study with staggered entry or censoring, exposure should instead be assembled from each person's actual observed time at risk. Interest is not involved in this calculation: exposure is time at risk, not a present value.

Uniform Distribution of Decrements

Under a uniform distribution of decrements (UDD) assumption for the interval, the total number of lives declines linearly from $L_0$ to $L_h$. Thus, each equal-length portion of the interval has the same expected number of exits. The assumption concerns timing within the interval; it does not mean that the number of lives remaining stays constant.

Result: If the total lives remaining decline linearly over an interval of length $h$, exposure is the interval length multiplied by the average of the beginning and ending counts: $$ E_{\text{UDD}}=h\frac{L_0+L_h}{2}. $$ For a one-year interval, this reduces to $(L_0+L_h)/2$ person-years.

This is the area of a trapezoid whose parallel sides have lengths $L_0$ and $L_h$, and whose width is $h$. Equivalently, the average number at risk during the interval is $(L_0+L_h)/2$, so multiplying by the duration gives total person-time.

Worked Example: One-Year UDD Exposure With Competing Decrements

A hypothetical multiple-decrement table has 1,200 lives at the start of a one-year interval and 1,080 at its end. During the interval, 72 exits are recorded for one cause and 48 for another. Estimate the total exposure under UDD.

Model and assumptions. The cohort is closed, the interval is one year, and total decrements are assumed uniformly distributed over that year, so the lives remaining decline linearly. There is no interest assumption or present-value calculation.

The two cause counts sum to $72+48=120$, which agrees with the table decrease $1{,}200-1{,}080=120$. Use the beginning and ending total lives: $$ E_{\text{UDD}}=1\cdot\frac{1{,}200+1{,}080}{2} =\frac{2{,}280}{2} =1{,}140\text{ person-years}. $$

Check. The average number at risk is between 1,080 and 1,200, as required, and multiplying that average by one year gives 1,140 person-years. The cause breakdown checks the total decrease but is not needed to calculate total exposure. The method avoids the mistake of assigning a separate exposure amount to each cause and then treating those as different underlying populations.

For an interval shorter or longer than one year, retain the duration $h$ in the formula. For example, a three-month interval has $h=3/12=0.25$ years. Using $h=3$ would incorrectly treat three months as three years.

Worked Example: Scaling UDD Exposure to a Short Interval

A hypothetical closed group has 1,600 lives at the start of a nine-month period and 1,480 at its end. Assume total decrements are uniformly distributed over the period. Find the exposure in person-years.

Model and assumptions. The period is $h=9/12=0.75$ years. The total lives remaining decline linearly over those nine months. No interest rate is needed.

First calculate the average number at risk: $$ \frac{1{,}600+1{,}480}{2} =\frac{3{,}080}{2} =1{,}540. $$ Then multiply by the time in years: $$ E_{\text{UDD}}=0.75(1{,}540)=1{,}155\text{ person-years}. $$

Check. The exposure is less than $1{,}600(0.75)=1{,}200$, the exposure that would result if all lives remained for all nine months, and greater than $1{,}480(0.75)=1{,}110$, the exposure if all lives were at the ending count throughout. The answer is in person-years, not person-months.

Constant Force of Decrement

A different within-interval assumption is that the total force of decrement is constant. If that force is $\mu$ per year, the number remaining declines exponentially rather than linearly. This assumption is appropriate only when the model specifies a constant total force over the interval; it should not be substituted for UDD merely because both use the same endpoint counts.

Result: Under a constant total force of decrement over an interval of length $h$, with $L_0>0$ and $L_h>0$, $$ L(t)=L_0e^{-\mu t},\qquad \mu=\frac{1}{h}\ln\left(\frac{L_0}{L_h}\right). $$ The corresponding exposure is $$ E_{\text{CF}}=\int_0^h L(t)\,dt =\frac{L_0-L_h}{\mu} =h\frac{L_0-L_h}{\ln(L_0/L_h)}. $$ If $L_h=L_0$, there are no decrements under this model and the exposure is $hL_0$.

To justify the formula, constant force means that $L(t)$ solves $dL(t)/dt=-\mu L(t)$, with initial value $L(0)=L_0$. Solving gives $L(t)=L_0e^{-\mu t}$. Applying the endpoint condition $L(h)=L_h$ yields $e^{-\mu h}=L_h/L_0$, and hence $\mu=h^{-1}\ln(L_0/L_h)$. Integrating the exponential lives function gives $L_0(1-e^{-\mu h})/\mu$. Since $L_0e^{-\mu h}=L_h$, this equals $(L_0-L_h)/\mu$, producing the stated exposure formula.

The constant-force formula assumes $L_h>0$: a positive exponential curve cannot reach zero in a finite interval. When $L_h=L_0$, the displayed logarithmic expression has a zero denominator, so use the no-decrement limit $hL_0$ instead. With rounded table counts, either assumption produces an estimate based on those rounded endpoints.

Worked Example: Exposure Under a Constant Force

A hypothetical group has 1,000 lives at the beginning of a one-year interval and 500 at the end. Assume a constant total force of decrement during the year. Find the exposure, rounded to two decimal places.

Model and assumptions. The interval is one year, $L_0=1{,}000$, and $L_h=500$. The force is constant over the year. There is no interest rate or benefit amount in this exposure calculation.

First find the force implied by the endpoints: $$ \mu=\ln\left(\frac{1{,}000}{500}\right) =\ln(2) =0.693147\text{ per year}. $$ Then calculate exposure: $$ E_{\text{CF}} =\frac{1{,}000-500}{\ln(2)} =\frac{500}{0.693147} =721.3475\ldots\text{ person-years}. $$ Rounded to two decimal places, the estimated exposure is $721.35$ person-years.

Check. The exposure must lie between 500 and 1,000 person-years, since the number at risk stays between those endpoints for a one-year interval. The result satisfies this check. The constant-force lives curve is not linear, so using the UDD trapezoid would give a different estimate.

Combining Intervals and Choosing the Assumption

If a table provides intermediate counts, calculate exposure separately on each interval using the assumption stated for that interval, then add the results. This preserves the information in the intermediate lives. Applying an assumption across a longer combined interval can give a different answer, even if the beginning and final counts are unchanged.

Worked Example: Adding Annual Exposures From a Table

A hypothetical closed group has 800 lives at time 0, 720 at the end of year 1, and 600 at the end of year 2. Assume UDD separately in each one-year interval. Estimate the two-year exposure.

Model and assumptions. Both intervals have length one year, and the lives remaining decline linearly within each year. The intermediate count of 720 is part of the table and must be used.

For year 1: $$ E_1=\frac{800+720}{2}=760\text{ person-years}. $$ For year 2: $$ E_2=\frac{720+600}{2}=660\text{ person-years}. $$ Therefore, $$ E_{\text{total}}=E_1+E_2=760+660=1{,}420\text{ person-years}. $$

Check. The year 1 exposure lies between 720 and 800 person-years, and the year 2 exposure lies between 600 and 720 person-years. Their sum is therefore consistent with the two intervals. Applying a single straight line from 800 to 600 over the full two years would give $2(800+600)/2=1{,}400$, not 1,420. That calculation imposes a different path and ignores the supplied intermediate count.

Common Exam Traps

A frequent error is to count every initial life as exposed for the full interval. That overstates exposure when decrements occur. For example, if 1,000 lives fall to 800 over one year under UDD, the incorrect calculation $1{,}000(1)=1{,}000$ person-years assumes that no one exits during the year. The correct calculation is $$ E_{\text{UDD}}=1\cdot\frac{1{,}000+800}{2}=900\text{ person-years}. $$ The difference of 100 person-years is half the 200 decrements, as expected when exits are spread uniformly across the year.

Other checks help prevent errors:

Keep the time unit consistent. Convert months to years before reporting person-years.
Use the specified timing assumption. UDD and constant force generally produce different exposures from the same endpoints.
Use the total lives at risk. Cause-specific exit counts do not create separate exposure populations when the causes compete within the same group.
Respect interval boundaries. Add interval exposures when intermediate table lives are given; do not discard them without a stated reason.
Check the range. For a closed cohort with no entries, exposure over $h$ years must be between $hL_h$ and $hL_0$.

Key takeaway: Exposure from table lives is the area under the lives-at-risk curve. Under UDD, use the trapezoid area; under a constant force, integrate the exponential curve. State the assumption, use the correct interval length, and add exposures across intervals when the table supplies intermediate lives. This treatment is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the stated timing assumption and keep exposure in person-years.

A closed group falls from 2,400 to 2,280 lives over one year. Under UDD, calculate its exposure. (Answer: 2,340 person-years)
A group falls from 1,600 to 1,480 over nine months. Under UDD, convert the interval to years and calculate exposure. (Answer: 1,155 person-years)
A group falls from 640 to 512 lives over two years under a constant total force. Use $E=h(L_0-L_h)/\ln(L_0/L_h)$ and round to two decimals. (Answer: 1,147.24 person-years)
A table records 1,000 lives, 940 after year 1, and 846 after year 2. Assuming UDD separately in each year, calculate each year's exposure and the total. (Answer: 970 and 893 person-years; total 1,863 person-years)
Two hypothetical groups are observed over one year. Group A has 900 lives at the start and 810 at the end, with UDD. Group B has 600 lives at the start and 480 at the end, with a constant total force. Calculate the combined exposure, rounding to two decimals. (Answer: 1,392.77 person-years)

What You'll Learn