Why Convert an Annual Probability to a Force?
A life table commonly gives mortality in annual steps: \(q_x\) is the probability that a life aged \(x\) dies before reaching age \(x+1\). But a calculation may require survival for only part of that year. A within-year mortality model lets us translate the annual table entry into a continuous-time measure and use it consistently for fractional ages.
The key distinction is that \(q_x\) is a probability over a full year, while the force of mortality is an instantaneous rate. They are not interchangeable. Under the constant force assumption for the age interval from \(x\) to \(x+1\), the annual probability determines a unique force for that interval. The force may differ in the next age year.
The force has units of inverse years. It is nonnegative, and a larger force means lower survival over a fixed duration. When the annual life-table probability is supplied as a decimal, the conversion is direct. For example, \(q_x=0.08\) means an 8% conditional probability of death during the age year, not a force of 0.08.
Deriving the Conversion
For a constant force \(\mu\), survival through duration \(t\) is \(e^{-\mu t}\). This follows because the survival function satisfies the differential equation \(S'(t)=-\mu S(t)\), with \(S(0)=1\); solving gives \(S(t)=e^{-\mu t}\). Over a full year, this gives \(p_x=e^{-\mu}\). Since \(p_x=1-q_x\), solving for \(\mu\) gives the conversion.
Conversely, if the within-year force \(\mu\) is given, then the probability of surviving the year is \(e^{-\mu}\), so the annual death probability is
These equations assume \(0\le q_x<1\), so that the full-year survival probability is positive. If \(q_x=0\), then \(\mu=0\). If \(q_x=1\), no finite constant force produces zero survival at exactly one year: the limiting force is infinite. In ordinary life-table calculations, \(q_x<1\) is the relevant case.
Read \(q_x\) as a conditional one-year probability and set the one-year survival probability to \(1-q_x\).
Use \(1-q_x=e^{-\mu}\) for this age year. Do not treat \(q_x\) itself as the force.
Calculate \(\mu=-\ln(1-q_x)\), or \(q_x=1-e^{-\mu}\) when starting from the force.
Verify that \(e^{-\mu}=1-q_x\). This catches sign errors and incorrect use of the logarithm.
Using the Force Within the Age Year
Once \(\mu\) has been obtained from \(q_x\), the same constant-force model gives survival for any duration \(t\) within that age year. In the notation for fractional-age survival, for \(0\le t\le1\),
The second equality follows by substituting \(e^{-\mu}=1-q_x\). This is a useful direct form when the annual \(q_x\) is known: raise the annual survival probability to the fractional duration. The assumption applies only within the age year whose \(q_x\) was used. To continue across an integer-age boundary, use the mortality information and model for each subsequent age year rather than extending one force automatically.
Worked Example: Converting an Annual Probability and Finding Fractional Survival
A hypothetical life table gives \(q_{40}=0.08\). Assume a constant force throughout the age year from 40 to 41. Find the force and the probability of surviving 0.6 years from exact age 40. Give the force to five significant figures and the probability to four decimal places.
Method. Convert the annual death probability to annual survival, take its negative logarithm to obtain the force, and then use the force for the requested duration.
The one-year survival probability is
Therefore, the constant force for this age year is
For \(t=0.6\), fractional-year survival is
Rounded as requested, \(\mu=0.083382\) per year and \({}_{0.6}p_{40}=0.9512\). Check the conversion over a full year: \(e^{-0.0833816\ldots}=0.92\), which agrees with \(1-q_{40}\). The common mistake this method avoids is setting \(\mu=q_{40}=0.08\); that would give the wrong annual survival, since \(e^{-0.08}\ne0.92\).
Worked Example: Recovering Annual Mortality from a Force
For a hypothetical insured population, assume the force of mortality is constant at \(0.03\) per year from age 55 to 56. Find the annual \(q_{55}\) and the probability of surviving 0.25 years from age 55. Give both probabilities to four decimal places.
Method. First exponentiate the negative force over one year to obtain annual survival; subtract that result from 1 to obtain \(q_{55}\). Then use the same force over one quarter of a year.
Annual survival is
Thus the annual death probability is
For a quarter of a year, survival is
To four decimal places, \(q_{55}=0.0296\) and \({}_{0.25}p_{55}=0.9925\). As a check, the full-year survival \(1-q_{55}\) equals \(e^{-0.03}\). Do not multiply the force by 0.25 and call the result a survival probability: \(0.03(0.25)=0.0075\) is the exponent magnitude, while survival is \(e^{-0.0075}\).
Worked Example: Obtaining the Force from Life-Table Counts
A hypothetical life table records \(\ell_{62}=32{,}000\) and \(\ell_{63}=30{,}400\). Assume a constant force from age 62 to 63. Find \(q_{62}\), the force, and the probability of surviving 0.7 years from age 62. Give the probabilities to four decimal places and the force to five significant figures.
Method. Use the life-table counts to find the fraction dying during the year, convert that annual probability to a force, and then evaluate fractional-year survival.
The number dying between ages 62 and 63 is
The annual death probability is the number dying divided by the number alive at age 62:
The corresponding annual survival probability is \(1-q_{62}=0.95\), so
For 0.7 years of exposure under this force,
Thus \(q_{62}=0.0500\), \(\mu=0.051293\) per year, and \({}_{0.7}p_{62}=0.9647\). The annual check is \(e^{-\mu}=0.95\), and the life-table check is \(30{,}400/32{,}000=0.95\). The common mistake avoided here is dividing the deaths by \(\ell_{63}\); the annual conditional death probability uses the number alive at the start of the interval as its denominator.
Interpretation and Exam Traps
The logarithmic conversion matters because a probability and an instantaneous rate describe mortality differently. For small \(q_x\), \(-\ln(1-q_x)\) is close to \(q_x\), but they are not equal. The approximation may be adequate for a rough estimate, but it should not replace the exact conversion when the question specifies a constant force.
A second frequent trap is mixing interpolation assumptions. Under the uniform distribution of deaths assumption, survival within the year is linear in elapsed time: \({}_tp_x=1-tq_x\). Under constant force, it is exponential: \({}_tp_x=(1-q_x)^t\). The two formulas agree at \(t=0\) and \(t=1\), but generally differ between those endpoints. Use the assumption stated in the problem; do not substitute a familiar fractional-age formula from a different model.
This conversion is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In a calculation, identify both the mortality assumption and the relevant age interval before applying it; a force inferred from one annual \(q_x\) is not automatically a force for all ages.
Check Your Understanding
For each problem, use the stated model and keep the force in units of inverse years. Probabilities are reported to four decimal places unless otherwise indicated.
- A hypothetical table has \(q_{35}=0.06\). Under constant force in that age year, find the force to five significant figures. (Answer: \(\mu=-\ln(0.94)=0.061875\) per year.)
- For a hypothetical age year, \(q_x=0.15\). Assuming constant force, find the probability of surviving 0.4 years from exact age \(x\). (Answer: \({}_{0.4}p_x=0.85^{0.4}=0.9371\).)
- A model specifies a constant force of \(0.04\) per year from age 70 to 71. Find the annual \(q_{70}\) and the probability of surviving 0.8 years from age 70. (Answer: \(q_{70}=1-e^{-0.04}=0.0392\); \({}_{0.8}p_{70}=e^{-0.032}=0.9685\).)
- A hypothetical life table gives \(\ell_{48}=18{,}000\) and \(\ell_{49}=17{,}100\). Under constant force from age 48 to 49, find \(q_{48}\) and the force to five significant figures. (Answer: \(q_{48}=0.0500\); \(\mu=-\ln(0.95)=0.051293\) per year.)
- A hypothetical age-year mortality probability is \(q_x=0.09\). Under constant force, find the probability of surviving 0.3 years and verify the annual survival probability used. (Answer: \({}_{0.3}p_x=0.91^{0.3}=0.9721\); annual survival \(1-q_x=0.9100\).)