Why Fractional-Age Interpolation Matters
A decrement table commonly gives probabilities over whole years of age, while a calculation may begin or end partway through a year. Under the uniform distribution of decrements (UDD) assumption, exits are spread uniformly through each age year. This gives a direct way to interpolate survival and cause-specific exits between the table’s integer-age points.
The calculation must be conditional on being active at the actual starting age. In particular, an exit probability for a fraction of a year is not generally just that fraction multiplied by the annual probability when the person is already partway through the year. The denominator must reflect the probability of having survived to the fractional starting age.
These statements concern a cohort active at exact age \(x\). Conditional probabilities for a life already active at \(x+s\) are obtained by dividing by the proportion \(1-sq_x\) that reaches that starting age. UDD is applied separately within each integer-age interval; when a calculation crosses an integer age, the intervals must be handled as separate segments.
Deriving the Fractional-Age Formulas
Take two ages \(x+s\) and \(x+t\) within the same age year, where \(0\leq s\leq t\leq 1\). Relative to the cohort active at age \(x\), the proportions still active at these ages are \(1-sq_x\) and \(1-tq_x\). Dividing the latter by the former gives conditional survival from \(x+s\) to \(x+t\). For cause \(j\), the proportion of the original cohort exiting in the interval is \((t-s)q_x^{(j)}\); dividing by the proportion active at \(x+s\) gives the conditional cause-specific exit probability.
The notation \({}_{t-s}q_{x+s}^{(j)}\) here means the probability, conditional on being active at age \(x+s\), of exiting from cause \(j\) before age \(x+t\). It is not a separate annual table entry. The formulas require \(1-sq_x>0\), so that survival to the starting age has positive probability.
The formulas are consistent because the conditional survival and all conditional cause-specific exits must account for the entire interval:
This identity is also a practical check on arithmetic. The cause-specific exit probabilities retain their relative proportions within the year: conditional on an exit during the fractional interval, cause \(j\) has probability \(q_x^{(j)}/q_x\), provided \(q_x>0\).
Calculate the total annual exit probability by summing the cause-specific probabilities.
Use the fractional-age formulas if both ages lie in the same age year. The fractions are measured from that year’s integer age.
Divide fractional exits and survival proportions by the proportion active at the fractional starting age.
Calculate each segment conditionally, then weight later-segment outcomes by the probability of surviving the earlier segment.
Conditional survival plus the conditional probabilities of all causes must equal one, subject to rounding.
Worked Calculations Within One Age Year
Worked Example: Conditional Survival and Cause Exits Between Two Fractions
A hypothetical decrement table has two causes during the year of age \(x\): \(q_x^{(1)}=0.09\) and \(q_x^{(2)}=0.06\). Under UDD, find the probabilities, conditional on being active at age \(x+0.2\), of surviving to age \(x+0.7\), exiting from cause 1 before \(x+0.7\), and exiting from cause 2 before \(x+0.7\). Give probabilities to four decimal places.
Method. Apply the fractional-age formulas within one age year. First, the total annual exit probability is \(q_x=0.09+0.06=0.15\). The proportion of the age-\(x\) cohort active at \(x+0.2\) is \(1-0.2(0.15)=0.97\); at \(x+0.7\), it is \(1-0.7(0.15)=0.895\).
Conditional survival is \(0.895/0.97=0.9226804\), which rounds to \(0.9227\). The cause 1 exit probability over the interval, conditional on reaching \(x+0.2\), is \((0.7-0.2)(0.09)/0.97=0.045/0.97=0.0463918\), or \(0.0464\). The corresponding cause 2 probability is \((0.7-0.2)(0.06)/0.97=0.030/0.97=0.0309278\), or \(0.0309\).
Check and trap avoided. The unrounded probabilities sum to \(0.9226804+0.0463918+0.0309278=1.0000000\). A common mistake is to calculate each exit as the interval length times its annual probability, omitting the conditioning denominator \(0.97\). That would give probabilities for the original age-\(x\) cohort, not for a life known to be active at \(x+0.2\).
Worked Example: Starting Partway Through a Year
For a hypothetical population, the annual cause-specific probabilities in the year of age \(w\) are \(q_w^{(1)}=0.05\) and \(q_w^{(2)}=0.05\). Find, conditional on being active at \(w+0.5\), the probability of surviving to \(w+1\) and the probabilities of exiting from each cause before \(w+1\). Give probabilities to four decimal places.
Method. Here \(q_w=0.10\). The proportion active at \(w+0.5\), relative to those active at \(w\), is \(1-0.5(0.10)=0.95\). The proportion active at \(w+1\) is \(1-0.10=0.90\). Thus conditional survival is \(0.90/0.95=0.9473684\), or \(0.9474\).
Each cause has probability \(0.5q_w^{(j)}/0.95\) of occurring after \(w+0.5\) and by \(w+1\), conditional on being active at \(w+0.5\). For either cause this is \(0.5(0.05)/0.95=0.025/0.95=0.0263158\), or \(0.0263\).
Check and trap avoided. The unrounded results sum to \(0.9473684+0.0263158+0.0263158=1.0000000\). Simply assigning \(0.5(0.10)=0.05\) as the conditional total exit probability would ignore that only \(0.95\) of the original cohort reaches \(w+0.5\). The correct conditional total exit probability is \(0.05/0.95=0.0526316\).
Intervals That Cross an Integer Age
UDD is specified within each age year. If an interval crosses an integer age, use the first year’s probabilities up to that boundary and the next year’s probabilities after it. Outcomes in the later segment are conditional on surviving the first segment, so they must be multiplied by that survival probability when expressed relative to the original starting population.
For example, if \(S_1\) is survival from the fractional starting age to the intervening integer age, and \(Q_2^{(j)}\) is the conditional probability of cause \(j\) in the next segment, the overall cause \(j\) probability includes the first-segment cause \(j\) probability plus \(S_1Q_2^{(j)}\). This multiplication is essential even when the annual probabilities in the two years are similar.
Worked Example: Fractional Interval Across an Integer Age
In a hypothetical decrement table, the year of age \(w\) has \(q_w^{(1)}=0.05\) and \(q_w^{(2)}=0.05\). The year of age \(w+1\) has \(q_{w+1}^{(1)}=0.10\) and \(q_{w+1}^{(2)}=0.05\). Under UDD in each year, calculate conditional survival from \(w+0.5\) to \(w+1.5\) and the two cause-specific exit probabilities. Give results to four decimal places.
Method. Split the interval at \(w+1\). In the first segment, from \(w+0.5\) to \(w+1\), the starting active proportion relative to age \(w\) is \(1-0.5(0.10)=0.95\), and the endpoint active proportion is \(0.90\). Conditional survival in this segment is \(0.90/0.95=18/19=0.9473684\). Each cause-specific exit probability in this segment is \(0.5(0.05)/0.95=0.0263158\).
In the second segment, from \(w+1\) to \(w+1.5\), the total annual exit probability is \(0.10+0.05=0.15\). Conditional survival is \(1-0.5(0.15)=0.925\). The segment cause 1 probability is \(0.5(0.10)=0.05\), and the segment cause 2 probability is \(0.5(0.05)=0.025\).
Overall conditional survival is \((18/19)(0.925)=0.8763158\), or \(0.8763\). Overall cause 1 exit probability is \(0.0263158+(18/19)(0.05)=0.0263158+0.0473684=0.0736842\), or \(0.0737\). Overall cause 2 exit probability is \(0.0263158+(18/19)(0.025)=0.0263158+0.0236842=0.0500000\), or \(0.0500\).
Check and trap avoided. The results sum to \(0.8763158+0.0736842+0.0500000=1.0000000\). A common error is to add the second segment’s cause probabilities without weighting them by survival to \(w+1\). Those exits apply only to lives that reached the second segment; the required weight is \(18/19\), not 1.
Further Checks and Limits of the Method
The UDD interpolation is based on straight-line change in the number of active lives within each year. It does not assert that the instantaneous force of decrement is constant. The constant-force assumption gives a different fractional-age model, so the two approaches should not be interchanged. Here the annual probabilities are the inputs; no interest-rate assumption is needed because the quantities being calculated are probabilities rather than present values.
Before accepting a result, check that the starting and ending fractions are measured from the correct integer age, and that all cause probabilities refer to the same conditioning event. In a single year, the conditional outcomes must sum to one. Across years, calculate the next segment’s outcomes conditionally at its own start and weight them by survival to that start. These checks catch the most common errors: using the wrong fractional interval, omitting the starting-age denominator, or failing to condition later-segment exits on survival.
Check Your Understanding
For each problem, apply UDD separately within each stated age year. Give probabilities to four decimal places.
- During the year of age \(a\), \(q_a^{(1)}=0.08\) and \(q_a^{(2)}=0.12\). Find conditional survival from \(a+0.25\) to \(a+0.75\), and the two cause-specific exit probabilities. (Answer: survival \(0.8947\); cause 1 \(0.0421\); cause 2 \(0.0632\).)
- During the year of age \(b\), \(q_b^{(1)}=0.09\) and \(q_b^{(2)}=0.06\). Conditional on being active at \(b+0.2\), find survival to \(b+0.7\) and the probability of exit from each cause. (Answer: survival \(0.9227\); cause 1 \(0.0464\); cause 2 \(0.0309\).)
- During the year of age \(c\), \(q_c^{(1)}=0.04\) and \(q_c^{(2)}=0.06\). Conditional on being active at \(c+0.5\), find survival to \(c+1\) and the two cause-specific exit probabilities. (Answer: survival \(0.9474\); cause 1 \(0.0211\); cause 2 \(0.0316\).)
- In the year of age \(d\), \(q_d^{(1)}=0.08\) and \(q_d^{(2)}=0.02\). In the year of age \(d+1\), \(q_{d+1}^{(1)}=0.04\) and \(q_{d+1}^{(2)}=0.06\). Conditional on being active at \(d+0.4\), find survival to \(d+1.2\) and the cause-specific exit probabilities. (Answer: survival \(0.9188\); cause 1 \(0.0575\); cause 2 \(0.0238\).)
- In the year of age \(e\), \(q_e^{(1)}=0.12\) and \(q_e^{(2)}=0.08\). Conditional on being active at \(e+0.3\), calculate survival to \(e+0.8\) and the two cause-specific exit probabilities. (Answer: survival \(0.8936\); cause 1 \(0.0638\); cause 2 \(0.0426\).)