The Uniform Distribution of Deaths Assumption for Fractional Ages: Practice Problems and Solutions

To work with life table functions for fractional ages, it is often useful to adopt the assumption of the uniform distribution of deaths. The assumption results in the following formula:

s(x+t) = (1-t)*s(x) +t*s(x+1), where 0 ≤ t ≤ 1.

Under a uniform distribution of deaths for fractional ages, the following equalities hold:

_tq_x =t*q_x

_tp_x =1 -t*q_x

_yq_x+t = yq_x/(1 - t*q_x), where 0 ≤ t ≤ 1, where 0 ≤ y ≤ 1, y + t ≤ 1.

µ_x+t = q_x/(1 - t*q_x)

_tp_xµ_x+t = q_x

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 67-68.

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

Problem S3L15-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Assuming a uniform distribution of deaths for fractional ages, find s(17.6).

Solution S3L15-1. We use the formula s(x+t) = (1-t)*s(x) +t*s(x+1), where x = 17 and t = 0.6. Then s(17.6) = 0.4*s(17) + 0.6*s(18) = 0.4e^-0.34*17 + 0.6e^-0.34*18 = s(17.6) = 0.0025545597.

Problem S3L15-2. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Assuming a uniform distribution of deaths for fractional ages, find _0.2q₅.

Solution S3L15-2. First, we find q₅ = 1 - s(6)/s(5) = 1 - (e^-0.34*6)/(e^-0.34*5) = 1 - e^-0.34 = 0.2882296772. Now we can use the formula _tq_x = t*q_x. So _0.2q₅ = 0.2q₅ = 0.2*0.2882296772 =

_0.2q₅ = 0.0576459354.

Problem S3L15-3. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Assuming a uniform distribution of deaths for fractional ages, find _0.4q_3.3.

Solution S3L15-3. We use the formula _yq_x+t = yq_x/(1 - t*q_x). Here, x = 3, t = 0.3 and y = 0.4. Moreover, q₃ = 1 - s(4)/s(3) = 1 - (e^-0.34*4)/(e^-0.34*3) = 1 - e^-0.34 = 0.2882296772. Thus,

_0.4q_3.3 = 0.4*0.2882296772/(1 - 0.3*0.2882296772) = _0.4q_3.3 = 0.1262046484.

Problem S3L15-4. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Assuming a uniform distribution of deaths for fractional ages, find µ_1.9.

Solution S3L15-4. We use the formula µ_x+t = q_x/(1 - t*q_x). Here, x = 1, t = 0.9.

Moreover, q₁ = 1 - s(2)/s(1) = 1 - (e^-0.34*2)/(e^-0.34*1) = 1 - e^-0.34 = 0.2882296772.

Thus, µ_1.9 = 0.2882296772/(1 - 0.9*0.2882296772) = µ_1.9 = 0.389187535.

Problem S3L15-5. The lives of a group of π-legged grasshoppers do not follow a neat survival function. However, you are given that deaths for fractional ages are uniformly distributed and that _0.3p₂ = 0.656 and µ_2.3 = 0.448. Find q₂ for this group of π-legged grasshoppers.

Solution S3L15-5. We use the formula _tp_xµ_x+t = q_x. Thus, q₂ = _0.3p₂*µ_2.3 = 0.656*0.448 =

q₂ = 0.293888

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