Constant Force Within Each Year of Age

Why the Age Year Matters

Life-table information is commonly organized by integer age, while a survival question may begin or end at a fractional age. A model is therefore needed to describe mortality between the integer ages. Under the constant force within each year of age assumption, the force is constant during each age year, but it may take a different value in the next age year.

This is a local assumption: for survival from age \(x+s\) to age \(x+s+t\), first check that both ages lie in the same age year. If they do, use the force assigned to age \(x\), not a force from a different year. The time \(t\) must be measured in years when the force is expressed per year.

Definition: Let \(x\) be an integer age. The constant-force assumption within the age year \([x,x+1)\) states that the force of mortality is constant at \(\mu_x\) throughout that interval: \(\mu_{x+r}=\mu_x\) for \(0\le r<1\). The value may change at the next integer age. For an individual alive at age \(x+s\), where \(0\le s\le s+t\le1\), conditional survival for \(t\) years is denoted \({}_tp_{x+s}\).

The force of mortality is an instantaneous rate. In general, it describes the limiting rate of death over a very short interval, conditional on survival to the start of that interval. A value such as \(\mu_x=0.06\) per year is a rate, not a probability of death over a full year or over a fractional interval.

Survival Within One Age Year

When the force is constant at \(\mu_x\) throughout the relevant interval, survival for \(t\) years is exponential:

{}_tp_{x+s}=\exp(-\mu_x t), \qquad 0\le s\le s+t\le1.

The starting fraction \(s\) identifies the person’s attained age, but does not change the force within that age year. The duration \(t\) determines the exponent. The restriction \(s+t\le1\) is important: it ensures that the full interval uses the same age-year force.

A brief justification follows from the definition of force. If \(R(t)\) is survival for \(t\) years from a point within the age year, then over a short additional interval \(h\), the conditional probability of surviving is approximately \(1-\mu_x h\). Thus \(R(t+h)\) is approximately \(R(t)(1-\mu_x h)\). In the limit, \(R'(t)=-\mu_xR(t)\), with \(R(0)=1\). Solving this differential equation gives \(R(t)=\exp(-\mu_xt)\). The formula applies only while the force remains \(\mu_x\); it does not assert that the force stays unchanged after the next integer age.

Key result: If the interval begins at age \(x+s\) and ends no later than age \(x+1\), use \(\mu_x\) and the elapsed duration \(t\): \({}_tp_{x+s}=\exp(-\mu_xt)\). Do not substitute the starting fractional age for \(x\), and do not use the force for a different integer age year.

Choosing the Correct Force and Duration

The practical calculation has three parts: identify the age year containing the entire interval, determine the elapsed time in years, and apply the exponential expression. The force is selected by the integer age year, while the duration is the difference between the ending and starting ages.

Locate the interval.
Write the starting and ending attained ages and verify that they lie in the same integer age year.

Select the force.
Use the force specified for that integer age year. A force for a later or earlier age year is not interchangeable.

Measure the duration.
Subtract the starting age from the ending age, keeping the result in years if the force is per year.

Evaluate and check.
Calculate the exponential survival probability. It must be between 0 and 1 and must decrease as the duration increases when the force is positive.

For a positive force, longer exposure within the same age year gives lower survival because \(\exp(-\mu_xt)\) decreases as \(t\) increases. The starting fraction \(s\) does not appear in the exponential once the correct year’s force has been selected. It still matters for checking that the interval does not cross an integer-age boundary.

Worked Example: Survival Over a Fraction of an Age Year

In a hypothetical mortality model, the force during the age year from 47 to 48 is \(0.06\) per year. A person is alive at age 47.2. Find the probability that the person survives to age 47.7. Give the probability to four decimal places.

Method. Both ages are in the age year \([47,48)\), so use \(\mu_{47}=0.06\). The elapsed time is \(47.7-47.2=0.5\) years. The interval stays within the year, so the constant-force formula applies.

Substitute the force and duration:

{}_{0.5}p_{47.2} =\exp(-0.06(0.5)) =\exp(-0.03) =0.9704455\ldots

Rounded only at the final step, survival is 0.9704. As a check, the result is less than 1 and close to 1 because the force is modest and the exposure is only half a year. The common mistake to avoid is treating \(0.06\) as a half-year death probability; it is the force per year used in the exponent.

Worked Example: Splitting an Interval Without Changing the Force

A hypothetical model assigns a force of \(0.10\) per year to the age year from 60 to 61. An individual is alive at age 60.2. Calculate the probability of surviving to age 60.7. Verify the result by splitting the interval at age 60.4. Give the probability to four decimal places.

Method. The full interval has duration \(60.7-60.2=0.5\) years and remains within \([60,61)\). Direct calculation gives:

{}_{0.5}p_{60.2} =\exp(-0.10(0.5)) =\exp(-0.05) =0.9512294\ldots

For the check, the first segment lasts \(60.4-60.2=0.2\) years, and the second lasts \(60.7-60.4=0.3\) years. Both use \(\mu_{60}=0.10\):

\exp(-0.10(0.2))\exp(-0.10(0.3)) =\exp(-0.02)\exp(-0.03) =\exp(-0.05) =0.9512294\ldots

Both methods give survival of 0.9512 to four decimal places. The split calculation checks that consecutive segments in the same age year combine to the same total duration. The common mistake is to change the force merely because the calculation has been split into two pieces; no integer-age boundary was crossed.

Worked Example: Use the Force for the Current Age Year

A hypothetical model has force \(0.04\) per year during ages 50 to 51 and force \(0.12\) per year during ages 51 to 52. A person is alive at age 50.65. Calculate the probability of surviving to age 50.95. Give the probability to four decimal places.

Method. The interval begins and ends before age 51, so it lies entirely in \([50,51)\). Its duration is \(50.95-50.65=0.30\) years. Therefore the relevant force is \(\mu_{50}=0.04\), not \(\mu_{51}=0.12\).

{}_{0.30}p_{50.65} =\exp(-0.04(0.30)) =\exp(-0.012) =0.9880717\ldots

Rounded to four decimal places, the survival probability is 0.9881. This is consistent with a short interval and a relatively small force. The common mistake is selecting the force from the nearest listed age without checking the interval: age 51 is not reached, so the age-51 force does not apply.

Assumptions, Limits, and Exam Checks

The constant-force-within-year model does not mean mortality is constant at all ages. It permits a different force in each integer age year, so the model is piecewise constant by age. For this tutorial’s fractional-interval formula, the relevant segment must remain within one such year. If an interval crosses an integer age, the force may change there, and the one-force formula cannot be applied to the entire interval without checking the model for the additional age year.

The model also assumes the force is expressed in the same time unit used for the duration. If a force is stated per year, a duration of six months is \(0.5\) years, not 6. If a duration is supplied in months, convert it to years before using a per-year force. Do not round the duration or the exponential result partway through a calculation.

Question to check	Correct interpretation
Which force applies?	The force for the integer age year containing the interval.
What is the duration?	Ending attained age minus starting attained age, in years for an annual force.
Does the interval stay in that year?	Check both endpoints; the formula here requires \(s+t\le1\).
Is the force itself a probability?	No. It is a rate used in the exponent, not a probability over the interval.

A reliable final check is to read the expression in context: \(\mu_x\) identifies the age-year force and \(t\) identifies how long the individual is exposed to it. Then confirm the result is a probability and that increasing \(t\), with the same positive force, would reduce survival. These checks catch age-year selection and time-unit errors before they affect later calculations.

Key takeaway: Within the age year \([x,x+1)\), a constant force \(\mu_x\) gives conditional survival \({}_tp_{x+s}=\exp(-\mu_xt)\) for an interval that remains in that year. The force is chosen by the integer age year, and the duration is measured in years when the force is annual.

Check Your Understanding

For each problem, verify the age year and time unit before evaluating the survival probability.

A hypothetical model has force \(0.03\) per year from age 44 to 45. An individual alive at age 44.15 is observed to age 44.75. Find the conditional survival probability to four decimal places. (Answer: 0.9822)
In a hypothetical model, the force from age 68 to 69 is \(0.12\) per year. Calculate survival from age 68.30 to age 68.80, and state the duration used. (Answer: duration \(0.50\) years; survival probability 0.9418)
A model specifies force \(0.04\) per year for ages 39 to 40 and \(0.20\) per year for ages 40 to 41. Find the probability that someone alive at age 39.70 survives to age 39.95. State which force applies. (Answer: use \(\mu_{39}=0.04\); survival probability 0.9900)
A hypothetical age-year force is \(0.05\) per year from age 55 to 56. Find survival from age 55.2 to 55.5, then from age 55.5 to 55.9, and verify the combined survival from age 55.2 to 55.9. Give probabilities to four decimal places. (Answer: first segment 0.9851; second segment 0.9802; combined survival 0.9656)
A model has force \(0.08\) per year from age 63 to 64. Compare survival over \(0.25\) years with survival over \(0.50\) years, both starting at age 63.2. Give both probabilities to four decimal places and identify which is larger. (Answer: \(0.9802\) over 0.25 years; \(0.9608\) over 0.50 years; survival over 0.25 years is larger)

What You'll Learn