Why Partial-Year Death Probabilities Need Care
A life table commonly gives an annual probability of death, but a calculation may concern only part of an age year: for example, death between ages \(x+0.2\) and \(x+0.5\). Under the constant-force assumption, mortality is modeled continuously within the year. The key distinction is whether the requested probability is measured from age \(x\), or conditional on being alive at the beginning of the specified interval.
This tutorial focuses on probabilities within one integer age year. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The annual death probability alone does not specify the timing of deaths within the year; the constant-force assumption supplies that interpolation rule.
From Annual Probability to a Partial-Year Probability
The earlier material on converting annual \(q_x\) to a fractional-age force gives the force for this model as \(\mu=-\ln(1-q_x)\). Equivalently, if \(p_x=1-q_x\), then the survival probability for \(t\) years within the age year is \(p_x^t\). The same annual survival probability is therefore matched at \(t=1\), while the constant force determines survival at fractional times.
For a subinterval from age \(x+s\) to age \(x+t\), where \(0\le s<t\le1\), there are two useful probabilities. The unconditional probability is measured from age \(x\): the life must survive to \(x+s\) and then die before \(x+t\). The conditional probability assumes the life is already alive at \(x+s\).
Here \(T_x\) denotes the future lifetime of a life aged \(x\). The conditional expression is also the probability of death over the next \(t-s\) years for a life alive at age \(x+s\), written \({}_{t-s}q_{x+s}\). These formulas apply only while both endpoints are in the same age year under the stated constant-force assumption. If an interval crosses an integer age, the force for the next age year must also be taken into account.
Why the Interval Formulas Hold
The derivation follows from the defining relationship between force of mortality and survival. Within the year, if \(S(t)={}_tp_x\), the force is \(\mu=-\frac{d}{dt}\ln S(t)\). When \(\mu\) is constant, integrating from 0 to \(t\) gives \(\ln S(t)-\ln S(0)=-\mu t\). Since \(S(0)=1\), it follows that \(S(t)=e^{-\mu t}\). At \(t=1\), \(S(1)=p_x\), so \(e^{-\mu}=p_x\) and \(S(t)=p_x^t\).
To die in \((s,t]\), a life must be alive at \(s\) but not survive to \(t\). Subtracting the probability of survival to \(t\) from the probability of survival to \(s\) gives \(S(s)-S(t)\). For the conditional probability, divide this difference by \(S(s)\), the probability of being alive at \(s\). This yields \(1-S(t)/S(s)=1-p_x^{t-s}\). The argument also shows why the conditional probability depends on the interval length \(t-s\), while the unconditional probability additionally depends on the chance of reaching its start.
Set \(p_x=1-q_x\) for the age year in question.
Use \(p_x^s-p_x^t\) for an unconditional interval probability from age \(x\); use \(1-p_x^{t-s}\) if survival to age \(x+s\) is given.
Confirm \(0\le s<t\le1\) and verify the result is between 0 and 1.
Worked Calculations
Worked Example: Death Between Two Fractional Ages
A hypothetical life table has \(q_x=0.19\). Assume constant force within the year of age \(x\). Find (a) the unconditional probability that a life aged \(x\) dies between ages \(x+0.2\) and \(x+0.5\), and (b) the probability of death in that interval conditional on being alive at \(x+0.2\).
Method. Convert the annual death probability to annual survival, then use the difference of survival probabilities for the unconditional answer and the interval-length formula for the conditional answer. This avoids treating an interval probability as though it were automatically conditional.
The annual survival probability is \(p_x=1-0.19=0.81\). The corresponding force is \(\mu=-\ln(0.81)=0.210721\) per year. Under constant force, survival to \(x+0.2\) is
Survival to \(x+0.5\) is \(0.81^{0.5}=0.90\). Therefore, the unconditional probability of dying in the specified interval is
Given survival to \(x+0.2\), the interval length is \(0.5-0.2=0.3\) years. Thus the conditional probability is
Check. Multiplying the conditional probability by the probability of reaching the interval start gives approximately \(0.958731516(0.061260)\approx0.0587\), consistent with the unconditional result. Both probabilities are reported to four decimal places.
Worked Example: Conditional Probability From an Interval Length
For a hypothetical age year, let \(q_x=0.20\) and assume constant force. Find the unconditional probability of death between \(x+0.25\) and \(x+0.75\), then find the probability conditional on survival to \(x+0.25\).
Method. First calculate survival to each endpoint. For the conditional result, use only the length of the interval, not the time already survived.
Here \(p_x=1-0.20=0.80\). The endpoint survival probabilities are
Subtracting gives the unconditional probability from age \(x\):
The interval lasts \(0.75-0.25=0.50\) years. Conditional on being alive at its start, the death probability is
Check. The conditional probability exceeds the unconditional probability because the latter also requires survival through the first \(0.25\) years. Multiplying \(0.945741609\) by \(0.105572809\) gives approximately \(0.099844598\), as required. Probabilities are rounded to four decimal places.
Worked Example: Expected Deaths in a Partial-Year Interval
A hypothetical group has 500 lives, each aged \(x\), and \(q_x=0.36\). Assume each life follows the same constant-force model within the year. Find the expected number of deaths between ages \(x+0.25\) and \(x+0.75\). Also find the expected number of deaths in the next half-year if exactly 400 lives are known to be alive at age \(x+0.25\).
Method. Calculate the unconditional interval probability for the original group and multiply by 500. For the specified survivors, use the conditional probability and multiply by 400. Expected counts follow by summing individual death indicators; independence is not needed for these expectations.
The annual survival probability is \(p_x=1-0.36=0.64\). Survival to the two endpoints is
Thus the unconditional probability of death in the interval is
The expected number of deaths among the original 500 lives is \(500(0.178885438)=89.442719\), or \(89.44\) lives to two decimal places. For a life alive at \(x+0.25\), the remaining interval is \(0.50\) years. Its conditional death probability is
Among 400 lives actually alive at \(x+0.25\), the expected deaths before \(x+0.75\) are \(400(0.20)=80\). Check. The second calculation conditions on the 400 survivors; it does not apply the original-cohort probability \(0.178885438\) to them. The first expected count is rounded to two decimal places, and the second is exact.
Unconditional and Conditional Probabilities Compared
| Quantity requested | Condition supplied | Formula within age year \(x\) |
|---|---|---|
| Probability of death in \((s,t]\), measured from age \(x\) | Life is aged \(x\) at the start | \(p_x^s-p_x^t\) |
| Probability of death in the interval, given alive at \(x+s\) | Survival to the interval start is known | \(1-p_x^{t-s}\) |
| Probability of survival from \(x+s\) to \(x+t\), given alive at \(x+s\) | Survival to the interval start is known | \(p_x^{t-s}\) |
An important consistency identity is
The probability of dying in the interval from age \(x\) is the probability of reaching its start multiplied by the probability of dying during the interval given that the life reached its start. This identity is useful both as a calculation method and as a check.
Common Exam Trap
Suppose \(q_x=0.19\), and the requested event is death between \(x+0.2\) and \(x+0.5\) for a life aged \(x\). A common error is to report \(1-0.81^{0.3}\approx0.0613\) as the answer. That is the probability conditional on survival to \(x+0.2\), not the unconditional probability from age \(x\). The unconditional probability is
The wording of the condition matters. “A life aged \(x\)” calls for the unconditional calculation. “Given that the life is alive at age \(x+0.2\)” calls for the conditional calculation. Also check the time unit: \(t-s\) is measured in years, so an interval from \(x+0.2\) to \(x+0.5\) has length \(0.3\), not \(0.5\).
Check Your Understanding
For each question, use constant force within the stated age year. Give probabilities to four decimal places unless another quantity is requested.
- A hypothetical age year has \(q_x=0.19\). Find the unconditional probability of death between \(x+0.2\) and \(x+0.5\), and the conditional probability given survival to \(x+0.2\). (Answer: \(0.0587\) unconditional; \(0.0613\) conditional.)
- If \(q_x=0.20\), find the unconditional probability of death from \(x+0.25\) to \(x+0.75\), and the probability conditional on survival to \(x+0.25\). (Answer: \(0.0998\) unconditional; \(0.1056\) conditional.)
- For \(q_x=0.36\), find the unconditional probability of death between \(x+0.25\) and \(x+0.75\). Then find the expected number of deaths in that interval among 250 lives initially aged \(x\). (Answer: probability \(0.1789\); expected deaths \(44.72\).)
- Suppose \(q_x=0.10\). Given survival to \(x+0.3\), find the probability of death before \(x+0.7\). (Answer: \(1-0.9^{0.4}\approx0.0413\).)
- For \(q_x=0.19\), a life is alive at \(x+0.2\). How long, in years, must the following interval be for the conditional probability of death during it to equal \(0.05\)? (Answer: \(\ln(0.95)/\ln(0.81)\approx0.2434\) years.)