Solution to Homework 4

ACTS 4301
Instructor: Natalia A. Humphreys
SOLUTION TO HOMEWORK 4
Lesson 6. Survival Distributions: Percentiles, Recursions, and Life Table Concepts.
Lesson 7. Survival Distributions: Fractional Ages.
Problem 1
You are given:
1
, 0 ≤ x < 90.
180 − 2x
Calculate the median future lifetime of (30).
µ(x) =
Solution. Note that
1
0.5
=
180 − 2x
90 − x
is the force of mortality for a beta distribution with ω = 90 and α = 0.5.
We need to find t such that t p30 = 0.5. For a beta distribution,
1
60 − t 2
ω−x−t α
; t p30 =
t px =
ω−x
60
1
60 − t 2
60 − t
= 0.5 ⇔
= 0.25 ⇔ t = 45
t p30 = 0.5 ⇔
60
60
µ(x) =
Problem 2
A population has 40% who are smokers with a constant force of mortality 0.3 and 60% who are
non-smokers with a constant force of mortality 0.15.
Calculate the 75th percentile of the distribution of the future lifetime of an individual selected at
random from this population.
(A) 6.2238
(B) 1.3945
(C) 1.2155
(D) 3.1115
(E) 7.1975
Key: E
Solution. For a constant force of mortality µ, t px = e−µt . For an individual I we are given:
−0.3t
e
, p = 0.4
(I)
t px =
e−0.15t , p = 0.6
This is a mixture distribution. For an individual selected at random from this population, t px =
0.4 · e−0.3t + 0.6 · e−0.15t . The 75th percentile of the distribution of the future lifetime is t such that
−0.15t . Then y 2 = e−0.3t and
t px = 1 − 0.75 = 0.25. Let y = e
0.4 · e−0.3t + 0.6 · e−0.15t = 0.25 ⇔ 0.4y 2 + 0.6y − 0.25 = 0 ⇔
√
√
−0.6 + 0.76
−0.6 − 0.76
2
D = 0.6 + 4 · 0.4 · 0.25 = 0.76, y1 =
= 0.3397, y2 =
= −1.8397 < 0
0.8
0.8
e−0.15t = 0.3397 ⇔ −0.15t = ln 0.3397 ⇒ t = 7.1975 ACTS 4301. SP 2015. SOLUTION TO HOMEWORK 4.
Copyright ©Natalia A. Humphreys, 2015
Problem 3
For T , the future lifetime random variable for (0):
(i) ω > 80
(ii) 50 p0 = 0.55
(iii) E[T ] = 93
(iv) E[min(T, t)] = 2t − 0.006t2
Calculate the complete expectation of life at 50.
Solution. We are given ˚
e0 = E[T ] = 93 and ˚
e0:t = E[min(T, t)] = 2t − 0.006t2 . We need to find ˚
e50 .
ex+n ⇒ ˚
e0 = ˚
e0:50 +50 p0˚
e50 ⇒
˚
ex = ˚
ex:n +n px˚
˚
e50 =
˚
e0 −˚
e0:50
93 − 2 · 50 + 0.006 · 502
8
=
=
= 14.5455
p
0.55
0.55
50 0
Problem 4
Let Tx be the total future lifetime of a cohort of lx individuals defined as:
Z ∞
Z ∞
Z ∞
lx+t
ex · lx
· lx dt =
Tx =
lx+t dt =
t px · lx dt = ˚
lx
0
0
0
You are given a survivorship group with 1000 initial members at age zero and the survival function is
x
s(x) = 1 −
, 0 ≤ x ≤ 120,
120
where x is age.
Calculate T55 , the total number of years lived beyond age 55 by the survivorship group.
Solution. T55 = ˚
e55 · l55 . For a uniform distribution,
120 − 55
ω−x
⇒˚
e55 =
= 32.5
˚
ex =
2
2
ω−x−t
lx+t
l55
120 − 55
=
⇒
=
= 0.54167 ⇒ l55 = 541.67 ⇒
t px =
ω−x
lx
l0
120
T55 = 32.5 · 541.67 = 17,604.17 Page 2 of 5
Copyright ©Natalia A. Humphreys, 2015
ACTS 4301. SP 2015. SOLUTION TO HOMEWORK 4.
Problem 5
Let Tx be defined as in Problem 4. You are given:
(i) t px = (0.7)t , t ≥ 0
(ii) lx+2 = 5.6
Calculate Tx+1 .
Solution. By definition, Tx+1 =
0
lx+1+t dt. Since
lx+1+t
⇒ lx+1+t = lx+1 ·t px+1
lx+1
lx+2
lx+2
5.6
=
⇒ lx+1 =
=
= 8 ⇒ lx+1+t = 8(0.7)t
lx+1
px+1
0.7
t px+1
px+1
R∞
=
Hence,
Z
Tx+1 = 8
∞
(0.7)t dt =
0
8
8
· (0.7)t |∞
= 22.4294
0 =−
ln 0.7
ln 0.7
Problem 6
A survival distribution is defined by
1
(i) s(n) = (1+0.02n)
5 for n a non-negative integer.
(ii) Deaths are uniformly distributed between integral ages.
Calculate
0.3 p31.4 .
Solution.
0.3q31
, q31 = 1 − p31
1 − 0.4q31
S(x + t)
S(n + 1)
S(32)
1 + 0.02 · 31 5
1.62 5
, pn =
⇒ p31 =
=
=
= 0.9405
t px =
S(x)
S(n)
S(31)
1 + 0.02 · 32
1.64
0.3 · 0.0595
= 0.01829 ⇒0.3 p31.4 = 1 − 0.01829 = 0.9817
q31 = 1 − 0.9405 = 0.0595, 0.3 q31.4 =
1 − 0.4 · 0.0595
0.3 p31.4
= 1 −0.3 q31.4 ,
0.3 q31.4
=
Problem 7
Deaths are uniformly distributed over each year of age. You are given:
x
45
46 47 48 49
lx 100 98 95 93 86
Calculate:
a) 1|2 q46
b) 0.25 q48.4
Solution.
a) 1|2 q46 = p46 −3 p46
p46 =
l47
95
l49
86
95 − 86
9
= , 3 p46 =
=
⇒ q46 =
=
= 0.0918
l46
98
l46
98 1|2
98
98
Page 3 of 5
Copyright ©Natalia A. Humphreys, 2015
ACTS 4301. SP 2015. SOLUTION TO HOMEWORK 4.
b)
0.25 q48.4
0.25 q48.4
0.25q48
86
7
, q48 = 1 −
=
⇒
1 − 0.4q48
93
93
7
0.25 · 93
=
7 = 0.0194
1 − 0.4 · 93
=
Problem 8
You are given:
(i) q58 = 0.03
(ii) q59 = 0.032
(iii) Deaths are uniformly distributed over each year of age.
Calculate ˚
e58:1.6 .
(A) 1.5614
(B) 1.5764
(C) 1.5776
(D) 1.5577
(E) 1.5558
Key: A
Solution. Since the mortality changes at age 59, we must split the group of (58) into those who die
before 59, those who die between ages 59 and 59.6, and those who survive beyond age 59.6.
Those who die before 59 live 0.5 years on average with probability q58 = 0.03.
Those who die between ages 59 and 59.6 live 1.3 years on average with probability
1 |0.6 q58
= p58 ·0.6 q59 = (1 − q58 ) · 0.6q59 = 0.97 · 0.0192 = 0.018624
Those who survive beyond age 59.6 live 1.6 years with probability
1.6 p58
= p58 ·0.6 p59 = 0.97 · (1 − 0.6 · 0.032) = 0.97 · 0.9808 = 0.951376
Hence,
˚
e58:1.6 = 0.5 · 0.03 + 1.3 · 0.018624 + 1.6 · 0.951376 = 1.5614128
Problem 9
For a 4-year medical school, you are given the following probabilities for overall coursework failure
from all causes:
q0 = 0.17, q1 = 0.11, q2 = 0.07, q3 = 0.02
Failure is uniformly distributed over each year.
Compute the temporary 2.5-year complete expected medical school lifetime for a student entering
the second semester (˚
e0.5:2.5 ).
Solution. Since the mortality changes at age 1 and at age 2, we must split the group of (0.5) into
those who die before 1, those who die between ages 1 and 2, those who die between ages 2 and 3, and
those who survive beyond age 3.
Those who die before 1 live 0.25 years on average with probability
0.5q0
0.5 · 0.17
=
= 0.0929
0.5 q0.5 =
1 − 0.5q0
1 − 0.5 · 0.17
Those who die between ages 1 and 2 live 1 year on average with probability
0.5 |q0.5
=0.5 p0.5 · q1 = 0.9071 · 0.11 = 0.09978
Those who die between ages 2 and 3 live 2 years on average with probability
1.5 |q0.5
=1.5 p0.5 · q2 =0.5 p0.5 · p1 · q2 = 0.9071 · 0.89 · 0.07 = 0.05651
Those who survive beyond age 3 live 2.5 years with probability
2.5 p0.5
=0.5 p0.5 · p1 · p2 = 0.9071 · 0.89 · 0.93 = 0.7508
Page 4 of 5
Copyright ©Natalia A. Humphreys, 2015
ACTS 4301. SP 2015. SOLUTION TO HOMEWORK 4.
Hence,
˚
e0.5:2.5 = 0.25 · 0.0929 + 1 · 0.09978 + 2 · 0.05651 + 2.5 · 0.7508 = 2.113055
Problem 10
You are given:
(i) ˚
ex+0.4:0.5 = 0.42
(ii) Deaths are uniformly distributed between integral ages.
Calculate px .
Solution. Note that px = 1 − qx . By definition,
Z 0.5
Z 0.5
Z
˚
ex+0.4:0.5 =
(1 −t qx+0.4 ) dt =
t px+0.4 dt =
0
= 0.5 −
0.52
2
·
0
0
0.5
(1 −
tqx
) dt =
1 − 0.4qx
qx
qx
= 0.42 ⇔
= 0.64 ⇔ qx = 0.50955 ⇒ px = 0.4904
1 − 0.4qx
1 − 0.4qx
Page 5 of 5