Introduction
A mortality model often describes how the rate of death changes with age. To use that description correctly, distinguish the age a person has reached from the time elapsed since a particular observation began. For a person observed at age \(x\), \(t\) years later the attained age is \(x+t\). The force of mortality at that point is written \(\mu_{x+t}\).
The force is an instantaneous rate, not the probability of death over a year or any other fixed interval. This distinction matters when reading a mortality model, comparing rates at different ages, or changing the time unit used to express a rate. The examples below focus on interpreting the force and its age and duration indices, not on converting it into a finite-interval death probability.
Reading the Age and Duration Indices
The subscript in \(\mu_{x+t}\) identifies the attained age at which the rate applies. Here, \(x\) is the age at the start of observation, while \(t\) is elapsed time since that start. Thus \(t\) is not itself the age, and \(x+t\) is not a duration. For instance, if a person is observed at age 48 and 6 years elapse, the person has attained age 54; the relevant age-based force is \(\mu_{54}\).
The same attained age can be reached from different starting ages. Under an age-based model, a force indexed by attained age has the same value whenever the same age is reached. If one person starts at age 48 and is observed for 6 years, while another starts at age 52 and is observed for 2 years, both are age 54 at the comparison point. Both are assigned \(\mu_{54}\) by the same age-based mortality model.
A model can be written as a function of attained age, such as \(\mu_y\), or expressed along a particular life’s time axis as \(\mu_{x+t}\). These are two ways of identifying the same point in an age-based model: \(y=x+t\). If the force varies with age, changing \(t\) changes the force because the attained age changes. If the force is constant over an age range, changing \(t\) within that range need not change its value.
Worked Example: Identifying the Relevant Attained Age
A hypothetical age-based mortality model specifies the following forces: \(\mu_{50}=0.012\), \(\mu_{54}=0.018\), and \(\mu_{60}=0.027\), all per year. Person A is observed at age 50. Person B is observed at age 54. Find the relevant force for A after 4 years and for B after 6 years, and compare them.
Method. Add elapsed time to starting age to find attained age, then read the force indexed by that age. The common mistake to avoid is using elapsed time alone as the age index.
For A, the attained age after 4 years is \(50+4=54\). The corresponding force is \(\mu_{54}=0.018\) per year.
For B, the attained age after 6 years is \(54+6=60\). The corresponding force is \(\mu_{60}=0.027\) per year.
The difference is \(0.027-0.018=0.009\) per year. B’s force at the comparison point is \(0.009\) per year greater than A’s. This comparison concerns the two specified attained ages; it does not treat either force as a fixed-period death probability.
Check. A’s comparison age is 54, matching the subscript of the first selected force; B’s comparison age is 60, matching the second. The units for both are per year.
Units and Rate Interpretation
A force has units of inverse time. If time is measured in years, its unit is \(\text{year}^{-1}\); if time is measured in months, its unit is \(\text{month}^{-1}\). A numerical value without its time unit is incomplete, because the same rate has different numerical values when expressed using different time units.
For example, a force of \(0.024\) per year corresponds to \(0.024/12=0.002\) per month. The division by 12 changes the unit from a yearly rate to a monthly rate; it does not assert that a finite-year death probability is exactly twelve times a finite-month death probability. The force is a rate at an instant, whereas a probability refers to an event over an interval.
A force is also not bounded above by 1. A value greater than 1 per year is possible as a rate; it does not mean that a probability exceeds 1. The number and its unit must be interpreted together. Nor should a force be reported as a percentage probability without specifying a time interval and using an appropriate survival model to calculate that probability.
Worked Example: Converting the Time Unit
A hypothetical model gives a force of \(0.030\) per year at attained age 64. Express this force per month. Then state the force for a person observed at age 59 after 5 years, assuming the model is indexed by attained age.
Method. First convert the rate unit; separately use starting age plus elapsed time to find the attained age. The common mistake to avoid is either multiplying the rate by 12 or indexing the force by the elapsed duration alone.
There are 12 months in a year, so the rate per month is \(0.030/12=0.0025\) per month.
The person observed at age 59 has attained age \(59+5=64\) after 5 years. Therefore the applicable force is \(\mu_{64}=0.030\) per year, or \(0.0025\) per month when expressed in monthly units.
Check. Converting back gives \(0.0025\times12=0.030\) per year. The attained age is 64, not 5, so the model value at age 64 is the relevant one.
Why the Duration Index Becomes \(x+t\)
The relationship between starting age, elapsed time, and attained age follows from the way age is measured. If age increases by one year for each year of elapsed time, then after \(t\) years a person who began at age \(x\) is age \(x+t\). Consequently, a model whose force is indexed by attained age \(y\) is evaluated at \(y=x+t\) along that person’s time axis.
At the start of observation, the person’s age is \(x\).
After \(t\) years, attained age is \(x+t\), because age and elapsed time use the same yearly scale.
The force at that point is the value indexed by the attained age, \(\mu_{x+t}\).
This argument justifies the index, not a formula for calculating survival or death probabilities from the force. Those calculations require additional relationships between a force and a survival model. For this topic, the essential result is that the force at elapsed duration \(t\) is identified by the attained age \(x+t\), with units consistent with the chosen time scale.
Worked Example: Evaluating a Duration-Indexed Force Model
For a hypothetical group observed at age 40, suppose the specified force along the observation time axis is \(\mu_{40+t}=0.010+0.002t\) per year, where \(t\) is measured in years and the model is specified for \(0\leq t\leq 10\). Calculate the force at \(t=3\) and \(t=8\), identify each attained age, and compare the results.
Method. Substitute each duration into the stated function, while also calculating \(40+t\) to interpret the age index. The common mistake to avoid is substituting attained age directly for \(t\).
At \(t=3\), the attained age is \(40+3=43\). The force is \(0.010+0.002(3)=0.010+0.006=0.016\) per year. Thus the value is \(\mu_{43}=0.016\) per year.
At \(t=8\), the attained age is \(40+8=48\). The force is \(0.010+0.002(8)=0.010+0.016=0.026\) per year. Thus the value is \(\mu_{48}=0.026\) per year.
The difference is \(0.026-0.016=0.010\) per year. Over the 5 years between these observation times, the specified force value increases by \(0.010\) per year. This is a comparison of rate values at two instants, not a calculation of the probability of death between those times.
Check. Both durations lie in the stated range. The values increase with \(t\), as the positive coefficient \(0.002\) indicates, and each result is stated per year.
Common Exam Traps
Force-of-mortality questions often test whether the rate’s index and unit have been read carefully. Before using a value, identify the time origin, the duration measured from that origin, the resulting attained age, and the unit attached to the force.
- Using duration as age. If someone is age 57 at observation and \(t=4\), the relevant attained age is 61, not 4. The age-based force is \(\mu_{61}\).
- Ignoring the starting age. A duration such as 4 years does not identify an age-based force until the starting age is known.
- Changing the unit in the wrong direction. To convert a rate per year to a rate per month, divide the numerical value by 12. To convert per month to per year, multiply by 12.
- Calling a force a probability. A force is an instantaneous rate with inverse-time units, not a death probability over a stated period.
- Assuming every age has the same force. The notation allows the force to vary with attained age. A constant value should be used only when the model states that it is constant over the relevant ages.
These distinctions make comparisons meaningful. Two values can be compared directly only when their units are consistent; in an age-based model, the ages being compared must also be identified. If two lives have different starting ages but reach the same attained age, an age-indexed force assigns the same rate to both at that point.
Check Your Understanding
For each problem, identify the relevant age or rate carefully before giving the requested result.
- A hypothetical age-based model has \(\mu_{46}=0.011\) per year and \(\mu_{50}=0.017\) per year. A person is observed at age 46. What force applies after 4 years, and what is its unit? (Answer: \(\mu_{50}=0.017\) per year)
- A model specifies a force of \(0.036\) per year at age 72. Convert it to a rate per month. (Answer: \(0.003\) per month)
- Two people are observed at ages 49 and 53. After 7 years for the first and 3 years for the second, what attained age has each reached? Under the same age-based model, how do their force indices compare? (Answer: Both have attained age 56, so both use \(\mu_{56}\))
- For a hypothetical life observed at age 35, the specified force is \(\mu_{35+t}=0.008+0.0015t\) per year, with \(t\) in years. Find the force after 4 years and state the attained age. (Answer: Age 39; \(0.014\) per year)
- A hypothetical model gives \(\mu_{60}=1.2\) per year. Is this invalid because the numerical value exceeds 1, and is it a 120% probability of death during the next year? (Answer: It is not invalid as a force; it is an instantaneous rate, not a finite-year probability)