Why Conditional Mortality at Fractional Ages Matters
An annual life table gives mortality information over whole years of age, but a question may condition on survival to a fractional age. For example, an insurer might assess the probability of death between ages \(x+0.4\) and \(x+0.9\), given survival to age \(x+0.4\). The relevant probability is not the unconditional probability of dying during that interval: it is the probability of dying after reaching its starting age.
Under the constant force assumption, mortality is constant within each one-year age interval. The force may change from one age-year to the next. Therefore, an interval wholly within one year uses one force, while an interval crossing an integer age must be split into parts and use the appropriate force in each part.
Conditional Survival Within One Year
Suppose \(r+s\leq 1\), so the entire interval lies between ages \(x\) and \(x+1\). Since the force is constant at \(\mu_x\), survival over an additional \(s\) years is
The starting fraction \(r\) does not appear in \(p_x^s\). Under a constant force, once survival to age \(x+r\) is given, the probability of surviving the next \(s\) years depends on the duration \(s\), not on how far into the age-year the life already is. This does not mean the unconditional probability of surviving from age \(x\) to \(x+r+s\) is \(p_x^s\); that probability also includes survival during the first \(r\) years.
Worked Example: Conditional Mortality Within an Age-Year
A hypothetical life table has \(q_x=0.36\). Under constant force within the year of age, find the probability that a life aged \(x+0.2\) dies before age \(x+0.7\), given survival to age \(x+0.2\).
Method. The interval lasts \(0.7-0.2=0.5\) years and stays within \([x,x+1)\), so use the constant force for age \(x\). First, \(p_x=1-q_x=1-0.36=0.64\).
The conditional death probability is \(1-0.8=0.2\). Thus, the probability of death in the stated interval, conditional on reaching age \(x+0.2\), is 0.20.
Check. The one-year force is \(\mu_x=-\ln(0.64)\), so \(e^{-0.5\mu_x}=e^{0.5\ln(0.64)}=0.8\), agreeing with the survival calculation. The common mistake avoided here is using the annual death probability \(0.36\) directly for a half-year interval.
Intervals That Cross an Integer Age
When an interval crosses age \(x+1\), the force after that boundary is generally \(\mu_{x+1}\), not \(\mu_x\). Conditional survival over the full interval is the product of survival over each part. For instance, if \(0\leq r<1\) and \(r+s>1\), the first part lasts \(1-r\) years in the age-\(x\) year. The remaining part begins at age \(x+1\).
If the remaining part also crosses another integer age, split it again. More generally, partition the interval from \(x+r\) to \(x+r+s\) at every integer age it crosses. In each age-year, raise that year's one-year survival probability \(p_{x+j}\) to the number of years spent in that age-year. Multiply those factors to obtain conditional survival; subtract the result from 1 to obtain conditional death.
The ratio form is a direct consequence of conditional probability. If \(\ell_a\) and \(\ell_b\) represent the numbers surviving to ages \(a\) and \(b\) from a common starting cohort, then the proportion of those alive at age \(a\) who remain alive at age \(b\) is \(\ell_b/\ell_a\). Under constant force, fractional-age values are interpolated within each year using \(\ell_{x+r}=\ell_xp_x^r\). Taking the ratio cancels all survival before age \(a\), leaving only survival over the requested interval.
Write the conditional interval as \(a=x+r\) to \(b=x+r+s\), and calculate its duration \(s=b-a\).
Split the interval wherever it enters a new one-year age interval. Use the relevant annual mortality information for each piece.
For a piece of length \(t\) in age-year \(y\), use \(p_y^t=(1-q_y)^t\).
Multiply the piecewise survival factors. The conditional death probability is one minus that product.
Worked Example: Conditional Mortality Across an Integer Age
For a hypothetical insured population, suppose \(q_x=0.19\) and \(q_{x+1}=0.36\). Find the probability that a life aged \(x+0.5\) dies before age \(x+1.5\), given survival to age \(x+0.5\).
Method. The interval has one-half year in \([x,x+1)\) and one-half year in \([x+1,x+2)\). Convert the annual death probabilities to survival probabilities: \(p_x=1-0.19=0.81\) and \(p_{x+1}=1-0.36=0.64\).
Therefore, \({}_{1.0}q_{x+0.5}=1-0.72=0.28\). The requested conditional probability of death is 0.28.
Check. The two parts have conditional survival probabilities \(0.9\) and \(0.8\), so their product cannot exceed either factor; \(0.72\) is consistent. Using \(q_x\) for the entire interval would ignore the mortality assumption for the second half-year.
Derivation: Why the Factors Multiply
Let \(a\) be the starting age and let an interval be divided at integer-age boundaries into successive durations \(t_1,\ldots,t_m\). Write \(y_j\) for the integer age at the start of segment \(j\). For each segment, constant force gives conditional survival \(p_{y_j}^{t_j}\). By the multiplication rule for conditional probabilities, survival over all segments is the product of the successive conditional survival probabilities:
where \(b=a+\sum_{j=1}^{m}t_j\), and \(T\) denotes age at death. Each segment's survival probability is conditional on reaching its beginning, so multiplying the factors accounts for the full path from \(a\) to \(b\) without including survival before \(a\). The death probability is the complement \(1-\prod_{j=1}^{m}p_{y_j}^{t_j}\).
Worked Example: Several Age-Years and a Fractional Endpoint
A hypothetical table gives \(q_x=0.5904\) and \(q_{x+1}=0.36\). Calculate the probability of death between ages \(x+0.5\) and \(x+2\), conditional on survival to age \(x+0.5\).
Method. The interval covers \(0.5\) years in the age-\(x\) year and one full year in the age-\(x+1\) year. The endpoint \(x+2\) is exactly an integer age, so there is no additional exposure in the age-\(x+2\) year. Convert the given values: \(p_x=1-0.5904=0.4096\), and \(p_{x+1}=1-0.36=0.64\).
The conditional death probability is \(1-0.4096=0.5904\). Thus, the answer is 0.5904.
Check. The square root of \(0.4096\) is \(0.64\), and the second factor is also \(0.64\), so the survival product is \(0.4096\). The endpoint matters: the life has spent one full year in the age-\(x+1\) interval, not a fractional year in the next interval.
Common Exam Traps
A frequent error is to treat an annual probability as though it applied unchanged over a fraction of a year. Under constant force, a duration \(t\) uses \(p_y^t\), not \(p_y\) and not \(1-tq_y\). The expression \(1-tq_y\) is associated with a different fractional-age assumption and is not the constant-force formula.
For example, if \(q_x=0.36\) and the conditional interval lasts half a year, the incorrect linear calculation gives survival \(1-0.5(0.36)=0.82\), hence death probability \(0.18\). The constant-force calculation is \(1-(1-0.36)^{0.5}=1-0.8=0.20\). The latter is the correct result under the stated assumption.
- Conditioning error: Use the duration after the stated starting age. Do not include survival from age \(x\) to age \(x+r\) when the question already conditions on reaching \(x+r\).
- Boundary error: When an interval crosses an integer age, use the mortality assumption for each age-year rather than extending the first year's force across the whole interval.
- Endpoint error: Count exposure carefully. An interval ending exactly at an integer age has zero duration in the age-year that begins at that endpoint.
- Complement error: The product of fractional survival factors is a survival probability. Subtract it from 1 only when the question asks for conditional death.
These calculations are useful whenever a life-table problem specifies a fractional starting age or asks for mortality over a partial year. The central discipline is to identify the conditional interval first, then apply the constant-force model separately to every age-year it intersects. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use constant force within each integer age-year and state the conditional probability requested.
- A hypothetical table has \(q_x=0.36\). Find the probability of death from age \(x+0.2\) to \(x+0.5\), conditional on survival to \(x+0.2\). (Answer: 0.1253)
- Suppose \(q_x=0.19\) and \(q_{x+1}=0.36\). Find the probability of death from age \(x+0.4\) to \(x+1.2\), conditional on survival to \(x+0.4\). (Answer: 0.1940)
- A hypothetical life table gives \(p_x=0.64\) and \(p_{x+1}=0.81\). Find the conditional probability of death from age \(x+0.5\) to age \(x+2\). (Answer: 0.3520)
- For a hypothetical table, \(p_x=0.4096\) and \(p_{x+1}=0.64\). Find the conditional probability of death from age \(x+0.25\) to \(x+1.5\). (Answer: 0.5904)
- A hypothetical table has \(q_x=0.19\) and \(q_{x+1}=0.36\). Find the conditional probability of death from age \(x+0.25\) to \(x+1.5\). Give the probability to four decimals. (Answer: 0.3169)