Deterministic Survivorship Groups: Practice Problems and Solutions

The following are properties of a deterministic survivorship group, according to Bowers, Gerber, et. al.

1. Initially, the group is comprised of l₀ lives aged 0.

2. "The members of the group are subject, at each age of their lives, to effective annual rates of mortality (decrement) specified by the values of q_x in the life table."
3. "The group is closed. No future entrants are allowed beyond the initial l₀. The only decreases come as a result of the effective annual rates of mortality (decrement)."

The following formulas illustrate how membership in a deterministic survivorship group changes with the passage of time.

l₁ = l₀(1- q₀) = l₀ - d₀

l₂ = l₁(1- q₁) = l₁ - d₁ = l₀ - (d₀ + d₁)

...

l_x = l_x-1(1- q_x-1) = l_x-1 - d_x-1 = l₀ - _y=0^x-1Σd_y = l₀(1 - _xq₀)

The above series of equalities is called the radix. We can also write these equalities as follows:

l₁ = l₀p₀

l₂ = l₁p₁ = l₀p₀p₁

...

l_x = l_x-1p_x-1 = l₀*_xp₀

We can express q_x, the effective annual rate of mortality, as follows:

q_x = (l_x - l_x+1)/l_x

We can express _nq_x, the n-year rate of mortality, as follows:

_nq_x = (l_x - l_x+n)/l_x

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 60-61.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L8-1. For a particular population of yellow-tailed squawkers, ₂q₀ = 0.012, ₄₂q₂ = 0.24, and l₀ = 13616. Find l₄₄. Round down to the nearest whole yellow-tailed squawker.

Solution S3L8-1. We note that l₂ = l₀(1 - ₂q₀) and l₄₄ = l₂(1 - ₄₂q₂). Thus, l₄₄ = l₀(1 - ₂q₀)(1 - ₄₂q₂) = 13616(1 - 0.012)(1 - 0.24) = l₄₄ = 10223.98208 = about 10223 yellow-tailed squawkers.

Problem S3L8-2. For a particular population of coelacanths, ₇p₀ = 0.8735, ₇p₇ = 0.9623, ₇p₁₄ = 0.5567, and l₀ = 23135. Find l₂₁. Round down to the nearest whole coelacanth.

Solution S3L8-2. We note that l₇ = l₀*₇p₀, l₁₄ = l₇*₇p₇, and l₂₁ = l₁₄*₇p₁₄. Thus, l₂₁ = l₀*₇p₀*₇p₇*₇p₁₄ = 23135*0.8735*0.9623*0.5567 = 10825.90272 = about 10825 coelacanths.

Problem S3L8-3. In a particular deterministic survivorship group of minotaurs, 8742 minotaurs of age 9 are present. A year later, only 7235 of these minotaurs are found to have survived to age 10. What is the effective annual rate of mortality for minotaurs during year 9?

Solution S3L8-3. We use the formula q_x = (l_x - l_x+1)/l_x. We know that l₉ = 8742 and l₁₀ = 7235. Thus,

q₉ = (8742 - 7235)/8742 = q₉ = about 0.1723861817.

Problem S3L8-4. In a particular deterministic survivorship group of blue squirrels, 10035 squirrels of age 2 are present. Only 8515 of these squirrels survive to age 6. What is the 4-year rate of mortality for blue squirrels from age 2 to age 6?

Solution S3L8-4. We use the formula _nq_x = (l_x - l_x+n)/l_x. We know that l₂ = 10035 and l₆ = 8515. Thus, ₄q₂ = (10035 - 8515)/10035 = ₄q₂ = 0.154698555.

Problem S3L8-5. Sktrfs always survive to age 5. Upon turning 5, sktrfs in a particular deterministic survivorship group face a 4-year rate of mortality of 0.13, followed by an 8-year rate of mortality of 0.262, followed by a 3-year rate of mortality of 0.416. Given that 40006 sktrfs survived to age 20, how many of them must have originally been in the group? Use ordinary rounding conventions to round to the nearest whole sktrf.

Solution S3L8-5. We know that l₉ = l₅(1 - ₄q₅), l₁₇ = l₉(1 - ₈q₉), and l₂₀ = l₁₇(1 - ₃q₁₇). Thus,

l₂₀ = l₅(1 - ₄q₅)(1 - ₈q₉)(1 - ₃q₁₇) and l₀ = l₅ = l₂₀/[(1 - ₄q₅)(1 - ₈q₉)(1 - ₃q₁₇)] =

40006/[(1 - 0.13)(1-0.262)(1-0.416)] = 106693.1823 = about 106693 sktrfs.

See other sections of The Actuary's Free Study Guide for Exam 3L.