Chapter 8
Chapter 8
1. Mayfawny purchases a whole life insurance policy.
There are three ways that Mayfawny's policy can terminate: a. Death (1) b. Diagnosis of a critical illness (2); and c. Lapse (3).
The policy pays a death benefit of 10,000 at the moment of death. The policy will also pay a critical illness benefit of 20,000 if Mayfawny is diagnosed with a critical illness. Only one benefit will be paid.
There is no benefit paid upon lapse.
You are also given:
i.
(1) x
0.01
ii.
(2) x
0.015
iii.
(3) x
0.06
iv. 0.035
Mayfawny pays a net premium continuously for her lifetime as long as the policy is inforce. The net premium is determined using the equivalence principle.
Calculate the net premium paid by Mayfawny.
February 17, 2015 Copyright Jeffrey Beckley 2012
Solution:
PVP PVB Pax00 10, 000 Ax01 20, 000 Ax02 0 Ax03
t
t
( x
) s
ds
(0.010.0150.06)ds
ax00 vt t px( ) dt et e 0
dt e0.035t e 0
dt
0
0
0
e0.035t
0
e0.085t
dt
e0.12t
0
dt
e0.12t
0.12
0
1 0.12
Ax01
vt t
0
px(
)
(1) x
dt
e0.035t e0.085t 0.01 dt
0
0.01 0.12
Ax02
vt t
0
px(
)
(2) x
dt
e0.035t e0.085t 0.015 dt
0
0.015 0.12
Pax00
10, 000 Ax01
20, 000 Ax02
0 Ax03
P
10, 000 Ax01
20, 000 Ax02 ax00
0 Ax03
10,
000
0.01 0.12
20,
000
0.015 0.12
0
400
1
0.12
February 17, 2015 Copyright Jeffrey Beckley 2012
2. Jeff is receiving a salary paid continuously for as long as he is in employed at Purdue. Jeff can leave employment through death (1), retirement (2), or disability (3). Once Jeff leaves employment, the salary stops.
You are given:
i. The salary pays at an annual rate of 70,000 per year.
ii. 0.05
iii. Jeff is currently age 59.
iv. Jeff will retire at age 65 if he is still teaching. He will not retire prior to age 65.
v.
(1) 59t
0.01 0.001t
vi.
(3) 59t
0.03 0.001t
Calculate the present value of Jeff's future earnings while employed at Purdue.
Solution:
6
PV 70, 000 vt t p5(9) dt The limits on the integral are determined by Jeff's retirement date. 0
t
t
t
t
5(9)s ds
( 5(19)s 5(92)s )ds
(0.010.001t 0.030.001t )ds
0.04ds
vt
et
e0.05t ;t
p( ) 59
e
0
e 0
e 0
e 0
e0.04t
6
6
6
PV=70,000
vt
t
p( ) 59
dt
70, 000
e0.05t e0.04t dt 70, 000
e0.09t dt
0
0
0
e0.09t 6
e0.09(6) 1
70,
000
0.09
0
70, 000
0.09
324,529.14
February 17, 2015 Copyright Jeffrey Beckley 2012
3. You are given the following table where decrement (1) is death, decrement (2) is lapse, and decrement (3) is diagnosis of critical illness:
x
q(1) x
q(2) x
q(3) x
qx( )
px( )
lx( )
d (1) x
d(2) x
d (3) x
55 0.02 0.15 0.010
56 0.03 0.06 0.015
57 0.04 0.04 0.020
58 0.05 0.03 0.025
59 0.06 0.02 0.030
a. Complete the table using a radix of 10,000.
See answers at end.
b. Calculate:
i.
p( )
3 55
Solution:
3
p( ) 55
l( )
58
l( )
55
6605.1 10, 000
0.66051
ii.
q(2)
2 56
Solution:
q(2)
2 56
d(2) 56
d(2) 57
l( )
56
492 293.56 8200
0.0958
iii.
q(3)
1|2 55
Solution:
q(3)
1|2 55
d(3) 56
d(3) 57
l( )
55
123 146.78 10, 000
0.026978
February 17, 2015 Copyright Jeffrey Beckley 2012
iv. The probability that a person age 55 will decrement from death or critical illness before age 60.
Solution:
d (1) 55
d (1) 56
...
d (1) 59
d (3) 55
d (3) 56
...
d (3) 59
l ( )
55
2136.763305 0.21368 10, 000
c. Assuming uniform distribution of each decrement between integer ages, calculate:
i.
q(2)
0.25 55
Solution:
0.25
q(2) 55
(0.25)(q5(52) )
(0.25)
1500 10, 000
0.0375
ii.
p( )
0.5 56
Solution:
0.5
p( ) 56
1 0.5
q( ) 56
1 (0.5)(q5(6) )
1 (0.5)(0.105)
0.9475
iii.
p( )
0.5 56.8
Solution:
p( )
0.5 56.8
l( )
57.3
l( )
56.8
(0.7)l5(7 ) (0.2)l5(6 )
(0.3)l5(8) (0.8)l5(7)
(0.7)(7339) (0.3)(6605.1) (0.2)(8200) (0.8)(7339)
0.94776
iv.
q(1)
0.5 55.6
Solution:
q(1)
0.5 55.6
d d (1)
(1)
0.4 55.6 0.1 56
l( )
55.6
(0.4)(200) (0.1)(246) (0.4)(10,000) (0.6)(8200)
0.01173
February 17, 2015 Copyright Jeffrey Beckley 2012
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