Select Life Tables

Problem Preview: Two 43-year-olds apply for insurance. Life A was selected at 40 (duration 3, select period ended). Life B was just selected at 43 (duration 0). Life A uses $q_{43}=0.0080$. Life B uses $q_{[43]}=0.0048$. 40% lower. Same age, very different mortality. The exam tests whether you pick the right rate.

HIGH-FREQUENCY: Reading a select-and-ultimate table correctly (identifying which column to use for each year) is tested on nearly every FAM sitting.

When an insurer underwrites a policy, it screens the applicant. Those who pass have lower mortality than the general population. This advantage fades over time.

Select period (length $d$): mortality depends on both selection age and duration since selection. Ultimate period ($t\ge d$): mortality depends only on attained age.

NOTE: Two lives of the same attained age but different selection ages have different mortality during the select period. A 43-year-old selected at 40 uses $q_{43}$ (ultimate), while one selected at 43 uses $q_{[43]}$ (select), which is significantly lower.

Common mistakes

• Using the select rate from the wrong row for ultimate lookups. After the select period, ultimate rates depend on attained age only. $q_{44}$ is found where $x+3=44$, which is the $[41]$ row. Trap: looking in the $[44]$ row, which gives $q_{47}$.
• Forgetting to transition from select to ultimate. A 5-year survival for a life selected at 40 in a 3-year select table uses 3 select rates and 2 ultimate rates. Using 5 select rates or 5 ultimate rates is wrong.
• Treating two lives of the same age as identical. A 43-year-old selected at 40 and a 43-year-old selected at 43 have different mortality during the select period.

Bottom line

• Selection effect: recently underwritten lives have lower mortality, which fades over the select period, so select rates rise with duration
• During select period ($t<d$): mortality depends on both selection age $[x]$ and duration $t$
• After select period ($t\ge d$): mortality depends only on attained age, drop the brackets
• Chain survival: multiply year-by-year $(1-q_{[x]+k})$, using select rates when $k<d$ and switching to ultimate rates at $k=d$

Exam shortcut

Begin every select-and-ultimate problem by writing a timeline with one box per year. In each box, write the exact notation for the applicable rate. Circle the select-ultimate boundary. Then read each rate from the correct row and column. This takes 30 seconds and saves the problem. "Row for select, column for ultimate." Select reads horizontally from the selection-age row. Ultimate reads vertically in the last column by attained age.

The full lesson (about 1,864 words, 12 min read) adds 3 worked examples, all 5 common mistakes, a self-check, free in the app.

Learning objectives

• 8e

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