Constant-Force Interpolation Across Integer Boundaries

Why Integer Boundaries Matter

When an interval of future lifetime stays within one year of age, a single constant force of mortality can describe survival over that interval. When the interval crosses an integer age, the force may change. The survival probability must then account for each age-year separately. The key is to split the interval at every integer age it crosses and multiply the conditional survival probabilities for the resulting segments.

For example, survival from age 30.25 to age 32.75 includes part of the year of age 30, all of the year of age 31, and part of the year of age 32. Applying the force for age 30 to the whole interval would ignore the specified changes at later ages.

Definition: Suppose the force of mortality is constant at $\mu_j$ during the year of age $[j,j+1)$. For an interval from age $x$ to age $x+t$, split the interval at each integer age it crosses. If the lengths of the resulting segments are $\Delta_j$, then the conditional survival probability is $$ {}_tp_x=\prod_j e^{-\mu_j\Delta_j} =\exp\left(-\sum_j\mu_j\Delta_j\right). $$ Each $\Delta_j$ is measured in years, and the force $\mu_j$ used for a segment is the force during that segment’s year of age.

The formula assumes a piecewise constant force over the stated age-years. The force can differ from one year to the next. A segment of length zero contributes a factor of 1, so an integer endpoint does not require an extra segment beyond that endpoint. If a force is given per year, segment lengths must be expressed in years.

Why the Segment Probabilities Multiply

Within a segment of length $\Delta_j$ with constant force $\mu_j$, the probability of surviving the segment, conditional on reaching its beginning, is $e^{-\mu_j\Delta_j}$. To survive the full interval, a life must survive the first segment, then the next, and so on. Applying the multiplication rule for conditional probabilities gives the product of these segment survival probabilities.

The same result follows by adding the integrated force over the interval. The cumulative force is the sum of each segment’s force multiplied by its duration; survival is the exponential of the negative cumulative force.

{}_tp_x =\prod_j e^{-\mu_j\Delta_j} =e^{-\sum_j\mu_j\Delta_j}.

The product form is often easiest when the interval contains only a few segments. The exponential-of-a-sum form is useful for checking the arithmetic: each force-duration product is a contribution to cumulative force, and all contributions must be included.

Worked Calculations Across Several Ages

A reliable calculation begins by identifying the starting and ending ages, listing each age-year crossed, and recording the length of time spent in it. Only then should the appropriate force or annual survival probability be applied.

Worked Example: Survival from One Fractional Age to Another

A hypothetical insured is currently age 40.6. Assume the force of mortality is constant at $0.08$ during ages 40 to 41, $0.12$ during ages 41 to 42, $0.20$ during ages 42 to 43, and $0.10$ during ages 43 to 44. Calculate the probability of survival to age 43.2.

Method. Split the interval at ages 41, 42, and 43. From age 40.6 to 41 is $0.4$ years; ages 41 to 42 and 42 to 43 each contribute one full year; age 43 to 43.2 contributes $0.2$ years.

The cumulative force over the interval is $$ (0.08)(0.4)+(0.12)(1)+(0.20)(1)+(0.10)(0.2) =0.032+0.12+0.20+0.02 =0.372. $$

Thus, $$ {}_{2.6}p_{40.6} =e^{-0.372} \approx 0.6894. $$ The interval length is $43.2-40.6=2.6$ years, consistent with the segment lengths $0.4+1+1+0.2=2.6$. The survival probability is between 0 and 1, as required.

Common mistake avoided: using the age-40 force $0.08$ for all 2.6 years would give $e^{-(0.08)(2.6)}$, which does not reflect the later age-specific forces. The correct calculation uses each force only for the time spent in its age-year.

For some questions, the annual death probabilities $q_j$, rather than the forces, are given. Under a constant force during age $j$ to $j+1$, annual survival is $p_j=1-q_j=e^{-\mu_j}$. Therefore survival through a fraction $r$ of that age-year is $e^{-r\mu_j}=(1-q_j)^r$.

Result: If $q_j$ is the probability of death between integer ages $j$ and $j+1$, and the force is constant within that year, then a segment of length $r$ in that year has survival probability $(1-q_j)^r$, for $0\le r\le 1$. This follows from $e^{-\mu_j}=1-q_j$. For a multi-year interval, multiply the appropriate fractional or full-year factors. The conversion assumes $0\le q_j<1$, so that the corresponding force $\mu_j=-\ln(1-q_j)$ is finite.

Worked Example: Combining Fractional and Full Years from Annual Probabilities

For a hypothetical life, the annual death probabilities are $q_{50}=0.19$, $q_{51}=0.36$, and $q_{52}=0.51$. Assume a constant force within each year of age. Calculate the probability of survival from age 50.5 to age 52.5.

Method. The interval contains half a year at age 50, one full year at age 51, and half a year at age 52. The annual survival probabilities are $$ 1-q_{50}=0.81,\qquad 1-q_{51}=0.64,\qquad 1-q_{52}=0.49. $$

Apply the fractional-year rule to the two half-year segments and use the full annual survival probability for age 51: $$ {}_{2}p_{50.5} =(0.81)^{0.5}(0.64)(0.49)^{0.5} =(0.9)(0.64)(0.7) =0.4032. $$

The segment lengths sum to $0.5+1+0.5=2$ years, matching $52.5-50.5=2$. Each segment survival factor is no greater than 1, and their product is $0.4032$. Reporting to four decimal places, the required probability is $0.4032$.

Common mistake avoided: taking the square root of the age-51 survival probability as well would incorrectly treat that full year as a half-year. The age-51 segment has length one, so its factor is $0.64^1=0.64$.

Worked Example: Starting and Ending Within Different Age-Years

A hypothetical population has piecewise constant forces of $0.20$ during ages 30 to 31, $0.40$ during ages 31 to 32, and $0.10$ during ages 32 to 33. Calculate the probability that a life aged 30.25 survives to age 32.75.

Method. The life spends $31-30.25=0.75$ years in the first age-year, one full year at ages 31 to 32, and $32.75-32=0.75$ years in the last age-year.

The cumulative force is $$ (0.20)(0.75)+(0.40)(1)+(0.10)(0.75) =0.15+0.40+0.075 =0.625. $$

Therefore, $$ {}_{2.5}p_{30.25} =e^{-0.625} \approx 0.5353. $$ The interval is $32.75-30.25=2.5$ years, and the segment lengths add to $0.75+1+0.75=2.5$. Rounded to four decimal places, the survival probability is $0.5353$.

Common mistake avoided: using $0.75$ years for the middle segment would omit one quarter of the year at age 31. The middle segment runs from age 31 to age 32, so its length is exactly one year.

A Boundary-Checking Procedure

Most errors in these calculations come from assigning a segment the wrong length or using the force for the wrong age-year. A short boundary check before exponentiating can prevent both. In particular, the segment starting at an integer age belongs to the year beginning at that age, and an interval ending exactly at an integer age has no positive-length segment after that endpoint.

Write the endpoints.
Record the starting age $x$ and the ending age $x+t$. Confirm that their difference is the requested duration $t$.

Mark every integer boundary.
Split the interval wherever the applicable age-year changes. For example, an interval from age 18.4 to 20.2 crosses ages 19 and 20.

Assign lengths and assumptions.
For each segment, subtract its beginning age from its ending age and use the force or annual survival probability specified for that age-year.

Combine and check.
Multiply the segment survival factors, or add the force-duration products and exponentiate their negative. Check that segment lengths sum to $t$, and that the result lies between 0 and 1.

The length check is especially useful when both endpoints are fractional. If the segment lengths do not sum to the difference between the endpoints, a boundary has likely been missed or counted twice. A second check is to verify that every survival factor uses the force or annual probability for the age-year in which its segment lies.

Common Exam Traps

Using one force over the whole interval. This is valid only when the force is constant over the entire interval. If the force changes at integer ages, split there.
Confusing an age with a duration. From age 30.25 to age 32.75 is $2.5$ years, not $32.75$ years. The age-year segment lengths, not the endpoint labels, determine the exponents.
Applying a full-year probability to a fractional segment. Under a constant force, a segment of length $r$ uses $(1-q_j)^r$, not $1-rq_j$.
Misreading the boundary year. A segment from age 31 to age 32 uses the assumptions for the year of age 31, while a segment from age 32 to age 32.5 uses those for the year of age 32.
Rounding too early. Keep the segment factors or cumulative force unrounded until the final probability is calculated.

Key takeaway: Under a force that is constant within each integer year of age, survival across multiple age-years is the product of the survival factors for those years. Split at every integer boundary, use the correct segment length and age-year assumption, and verify that the lengths add to the total interval.

Check Your Understanding

For each problem, assume the force is constant within each stated year of age. Use the indicated rounding only for the final answer.

A life aged 24.2 faces forces $0.05$ from age 24 to 25 and $0.15$ from age 25 to 26. Find the probability of survival to age 25.7. (Answer: 0.8650)
For a hypothetical life, $q_{60}=0.19$ and $q_{61}=0.36$. Find the probability of survival from age 60.5 to age 61.5 under constant force within each year. (Answer: 0.7200)
For a hypothetical life aged 70.25, the forces during ages 70, 71, and 72 are $0.10$, $0.20$, and $0.30$, respectively. Find the probability of survival to age 72.5, to four decimal places. (Answer: 0.6538)
A hypothetical life aged 40.5 has annual death probabilities $q_{40}=0.36$, $q_{41}=0.19$, and $q_{42}=0.51$. Assuming constant force within each age-year, find the probability of survival to age 42.5. (Answer: 0.4536)
For a hypothetical life aged 35.4, the forces during ages 35, 36, and 37 are $0.10$, $0.20$, and $0.05$, respectively. Calculate the probability of survival to age 37.6, to four decimal places. (Answer: 0.7483)

What You'll Learn