The Progress of Liberty

The Actuary’s Free Study Guide for Exam 3L - Section 40

This section of sample problems and solutions is a part of The Actuary’s Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 40 of the Study Guide. See an index of all sections by following the link in this paragraph.

The actuarial accumulated value at the end of the term of an n-year temporary life annuity of 1 per year payable continuously while (x) survives is denoted by the symbol s^-_x:n¬(with a straight line drawn directly over the “s” whenever circumstances permit) and can be found as follows.

s^-_x:n¬= ā_x:n¬ /_nE_x

s^-_x:n¬= ₀ⁿ∫(1/_n-tE_x+t)dt

We recall that _nE_x = vⁿ*_np_x.

It is possible to calculate the derivatives of actuarial present values of continuous life annuities through the following formulas.

d(ā_x)/dx = [μ(x) + δ]ā_x - 1

∂(ā_x:n¬)/∂x = [μ(x) + δ]ā_x:n¬- (1 - _nE_x)

∂(_n│ā_x)/∂n = -vⁿ*_np_x

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 140-141.

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

Problem S3L40-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is currently 0.06. Desiderius the Triceratops has a continuous 4-year temporary life annuity. At the end of four years, if Desiderius survives, what will be the actuarial accumulated value s^-_x:4¬ of Desiderius’s annuity?

Solution S3L40-1. We use the formula š_x:n¬= ā_x:n¬ /_nE_x. To find ā_x:4¬, we use the formula

ā_x:n¬ = ₀ⁿ∫v^t*_tp_x*dt. Since the lifetimes of triceratops are exponentially distributed, _tp_x = e^-0.34t for all x. Moreover, v^t = e^-0.06t. Thus,

ā_x:4¬  = ₀⁴∫e^-0.06t*e^-0.34t*dt = ₀⁴∫e^-0.4t*dt = (-5/2)e^-0.4t│₀⁴ = (5/2)(1 - e^-1.6) = ā_x:4¬  = about 1.99525871.

Now we find ₄E_x = v⁴*₄p_x = e^-0.24e^-1.36 = e^-1.6

Thus, s^-_x:4¬= 1.99525871/e^-1.6 = s^-_x:4¬= about 9.882581086.

Problem S3L40-2. The lives of blue bears exhibit a survival function s(x) = 1 - x/67 for 0 ≤ x ≤ 67. Carlos the Blue Bear is 34 years old and has a continuous 6-year temporary life annuity. The annual force of interest is currently 0.035. At the end of six years, if Carlos survives, what will be the actuarial accumulated value s_34:6¬ of his annuity? Set up the appropriate integral and use any calculator to evaluate it.

Solution S3L40-2. We use the formula s^-_x:n¬= ₀ⁿ∫(1/_n-tE_x+t)dt.

First, we try to find _6-tE_34+t = v^6-t*_6-tp_34+t.

Since we have a constant force of interest of 0.035, v^6-t = e^-0.035(6-t) = e^-0.21e^0.035t.

_6-tp_34+t = s(40)/s(34+t) = (27/67)/((33 – t)/67) = _6-tp_34+t = 27/(33-t).

Thus, s^-_34:6¬= ₀⁶∫[e^0.21e^-0.035t(33-t)/27]dt

s^-_34:6¬= about 7.444296275.

Problem S3L40-3. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is currently 0.06. Find Let f(n) = ∂(_n│ā_x)/∂n for triceratopses. Find f(7).

Solution S3L40-3. We use the formula ∂(_n│ā_x)/∂n = -vⁿ*_np_x. For triceratopses, _np_x = e^-0.34n and vⁿ = e^-0.06n

Thus, f(n) = -vⁿ*_np_x = -e^-0.4n and f(7) = -e^-0.4*7 = -e^-2.8 = about -0.0608100626.

Problem S3L40-4. For orange-spotted flying hippopotami, you are given that ā_6:3¬= 2.55, ₃E₆ = 0.88, and the force of mortality for 6-year-old orange-spotted flying hippopotami is 0.06. Let g(x) = ∂(ā_x:3¬)/∂x. The annual force of interest is 0.09. Find g(6).

Solution S3L40-4. We use the formula ∂(ā_x:n¬)/∂x = [μ(x) + δ]ā_x:n¬ - (1 - _nE_x).

Thus, g(6) = [μ(6) + δ]ā_6:3¬ - (1 - ₃E₆) = [0.06 + 0.09]2.55 – (1 – 0.88) = g(6) = 0.2625.

Problem S3L40-5. A 15-year-old yaac (short for Yet Another Absurdist Creature) can get a continuous whole life annuity paying 1*dt at every instant for a price of 6.64 Golden Hexagons (GH). Yaac force of mortality follows the equation μ(x) = 0.1x^1/8 for all x between 12 and 16, thereafter resorting to a constant force of mortality. Define h(x) = d(ā_x)/dx. You are given h(15) = 0.2. Find the annual force of interest.

Solution S3L40-5. We use the formula d(ā_x)/dx = [μ(x) + δ]ā_x – 1, rearranging it thus:

(d(ā_x)/dx + 1)/ā_x= μ(x) + δ

δ = (d(ā_x)/dx + 1)/ā_x- μ(x)

δ = (h(x) + 1)/ā_x- μ(x)

δ = (h(15) + 1)/ā₁₅ – μ(15)

δ = (0.2 + 1)/6.64 – 0.1*15^1/8

δ = about 0.0404378364

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

Trackback URI | Comments RSS