STT 455-6: Actuarial Models

437 Pages·2017·1.62 MB·English

by  Albert Cohen

STT 455-6: Actuarial Models
Albert Cohen
ActuarialSciencesProgram
DepartmentofMathematics
DepartmentofStatisticsandProbability
C336WellsHall
MichiganStateUniversity
EastLansingMI
48823
[email protected]
[email protected]
AlbertCohen (MSU) STT455-6:ActuarialModels MSU2013-2018 1/348
Copyright Acknowledgment
Many examples and theorem proofs in these slides, and on in class exam
preparation slides, are taken from our textbook ”Actuarial Mathematics for
Life Contingent Risks” by Dickson,Hardy, and Waters.
Please note that Cambridge owns the copyright for that material.
No portion of the Cambridge textbook material may be reproduced
in any part or by any means without the permission of the
publisher. We are very thankful to the publisher for allowing posting
of these notes on our class website.
Also, we will from time-to-time look at problems from released
previous Exams MLC by the SOA. All such questions belong in
copyright to the Society of Actuaries, and we make no claim on
them. It is of course an honor to be able to present analysis of such
examples here.
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Survival Models
An insurance policy is a contract where the policyholder pays a
premium to the insurer in return for a benefit or payment later.
The contract specifies what event the payment is contingent on. This
event may be random in nature
Assume that interest rates are deterministic, for now
Consider the case where an insurance company provides a benefit
upon death of the policyholder. This time is unknown, and so the
issuer requires, at least, a model of of human mortality
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Survival Models
Define (x) as a human at age x. Also, define that person’s future lifetime
as the continuous random variable T . This means that x +T represents
x x
that person’s age at death.
Define the lifetime distribution
F (t) = P[T ≤ t] (1)
x x
the probability that (x) does not survive beyond age x +t years, and it’s
complement, the survival function S (t) = 1−F (t).
x x
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P[x < T ≤ x +t] (2)
0
=
P[T > x]
0
and so
F (x +t)−F (x)
0 0
F (t) =
x
1−F (x)
0
(3)
S (x +t)
0
S (t) =
x
S (x)
0
In general we can extend this to
S (t +u) = S (t)S (u) (4)
x x x+t
Conditional Equivalence
We have an important conditional relationship
P[T ≤ t] = P[T ≤ x +t | T > x]
x 0 0
AlbertCohen (MSU) STT455-6:ActuarialModels MSU2013-2018 5/348
(3)
S (x +t)
0
S (t) =
x
S (x)
0
In general we can extend this to
S (t +u) = S (t)S (u) (4)
x x x+t
Conditional Equivalence
We have an important conditional relationship
P[T ≤ t] = P[T ≤ x +t | T > x]
x 0 0
P[x < T ≤ x +t] (2)
0
=
P[T > x]
0
and so
F (x +t)−F (x)
0 0
F (t) =
x
1−F (x)
0
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Conditional Equivalence
We have an important conditional relationship
P[T ≤ t] = P[T ≤ x +t | T > x]
x 0 0
P[x < T ≤ x +t] (2)
0
=
P[T > x]
0
and so
F (x +t)−F (x)
0 0
F (t) =
x
1−F (x)
0
(3)
S (x +t)
0
S (t) =
x
S (x)
0
In general we can extend this to
S (t +u) = S (t)S (u) (4)
x x x+t
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Conditions and Assumptions
Conditions on S (t)
x
S (0) = 1
x
lim S (t) = 0 for all x ≥ 0
t→∞ x
S (t ) ≥ S (t ) for all t ≤ t and x ≥ 0
x 1 x 2 1 2
Assumptions on S (t)
x
d S (t) exists ∀t ∈ R
dt x +
lim t ·S (t) = 0 for all x ≥ 0
t→∞ x
lim t2·S (t) = 0 for all x ≥ 0
t→∞ x
The last two conditions ensure that E[T ] and E[T2] exist, respectively.
x x
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P[(0) survives beyond age 30] = S (30) = 1−F (30)
0 0
(cid:18) (cid:19)1
30 6
= 1− = 0.9532
120
P[(30) dies before age 50] = F (20)
30 (5)
F (50)−F (30)
0 0
= = 0.0410
1−F (30)
0
S (65)
P[(40) survives beyond age 65] = S (25) = 0 = 0.9395
40
S (40)
0
Example 2.1
Assume that F (t) = 1−(cid:0)1− t (cid:1)16 for 0 ≤ t ≤ 120. Calculate the
0 120
probability that
(0) survives beyond age 30
(30) dies before age 50
(40) survives beyond age 65
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Example 2.1
Assume that F (t) = 1−(cid:0)1− t (cid:1)16 for 0 ≤ t ≤ 120. Calculate the
0 120
probability that
(0) survives beyond age 30
(30) dies before age 50
(40) survives beyond age 65
P[(0) survives beyond age 30] = S (30) = 1−F (30)
0 0
(cid:18) (cid:19)1
30 6
= 1− = 0.9532
120
P[(30) dies before age 50] = F (20)
30 (5)
F (50)−F (30)
0 0
= = 0.0410
1−F (30)
0
S (65)
P[(40) survives beyond age 65] = S (25) = 0 = 0.9395
40
S (40)
0
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