Fall 2002 - Course 3 Solutions

Fall 2002
Course 3 solutions
Question #1
Answer: E
b g
b g b g
µ 4 = −s' 4 / s 4
d
4
= - - e / 100
4
1-
e / 100
4
e / 100
=
-
4
1 e / 100
4
e
=
100-
e
4
= 1202553 .
i

Question # 2
Answer: A
q
b g
i
x
= q
b g
b g
τ
x
L
N
M
τ µ
= q x ×
µ
ln p′
ln p
big
bτg
( i)
x
bτg
x
O
Q
P
q
P =
b g
τ
x
L
NM
b g b g b g b g
τ 1 2 3
x x x x
ln e
ln
µ = µ + µ + µ = 15.
b g
τ
x
b g
− µ τ
q = 1− e = 1−
e
=0.7769
b g
q x
2
−15
.
e
−µ
-
big
b g
µ τ
2
b07769 . gµ
b g b05 . gb07769
. g
µ
b τ
15 .
= =
= 0.
2590
O
QP

Question # 3
Answer: D
3
2 2
A[ 60] v 2p[ 60] q[ 60]
2
= × × +
B B B +
pay at end live then die
of year 3 2 years in year 3
4
+ v × 3p[ 60]
× q60+
3
pay at end live then die
of year 4 3 years in year 4
1
1
= 1− 0. 09 1− 011 . 013 . + 1−0. 09 1− 011 . 1−
013 . 015 .
3 4
103 .
103 .
b g
b gb gb g
b g
b gb gb gb g
= 019 .
Question # 4
Answer: B
ax = a + E a
x x x+
a
5
x: 5
:5 5 5
b g
−0.
07 5
1 e
= − = 4.
219
0.
07
b g
− 0.
07 5
Ex = e = 0705 .
1
a x+ 5 = =
0.
08
∴ a x = 4. 219 + 0. 705 12. 5 = 13.
03
, where 0. 07 = µ + δ for t < 5
12. 5, where 0. 08 = µ + δ for t ≥ 5
b gb g

Question # 5
Answer: E
The distribution of claims (a gamma mixture of Poissons) is negative binomial.
EbNg = E∧dEcN Λhi = E∧bΛg
= 3
VarbNg = E∧dVarcN Λhi + Var∧d
EcN
Λ hi
= E∧bΛg+ Var∧bΛg = 6
rβ = 3
rβb1+ βg
= 6
+ βg = 6 / 3 = 2 ; β = 1
rβ = 3
r = 3
−r
b g .
p0 = 1+ β = 0125
rβ
p1 = = 01875 .
r+
1
1+
β
g
b g = p +
Prob at most 1 p
= 0.
3125
0 1
Question # 6
Answer: A
EbSg EbNg Eb Xg
= × = 110× 1101 , = 121110 ,
2
b g b g b g b g b g
Var S = E N × Var X + E X × Var N
= 110× 70 + 1101 × 750
=
Std Dev S
909, 689,
750
2 2
b g = 30161 ,
bS
< g = cZ
< b -
= Pr bZ < - 0. 70g
= 0.
242
g
Pr 100, 000 Pr 100, 000 121110 , / 30161 , h where Z has standard normal distribution

Question # 7
Answer: C
This is just the Gambler’s Ruin problem, in units of 5,000 calories.
Each day, up one with p = 0. 45; down 1 with q = 055 .
Will Allosaur ever be up 1 before being down 2?
e
e
b
b
g
g
1−
055 . / 045 . 2
P 2 =
∧
1−
055 . / 0.
45 3
∧
j
j
= 0598 .
Or, by general principles instead of applying a memorized formula:
Let P 1 = probability of ever reaching 3 (15,000 calories) if at 1 (5,000 calories).
Let P 2 = probability of ever reaching 3 (15,000 calories) if at 2 (10,000 calories).
From either, we go up with p = 0. 45, down with q = 055 .
P(reaching 3) = P(up) × P(reaching 3 after up) + P(down) × P(reaching 3 after down)
P2 = 0. 45 × 1+ 0.
55×
P1
P1 = 0. 45× P2 + 0. 55 × 0 = 0.
45 × P2
P2 = 0. 45 + 0. 55× P1 = 0. 45 + 0. 55 × 0. 45× P2 = 0. 45 + 0.
2475P2
P 2 = 045 . / 1− 0. 2475 = 0598 .
b g
Here is another approach, feasible since the number of states is small.
Let states 0,1,2,3 correspond to 0; 5,000; 10,000; ever reached 15,000 calories. For purposes of
this problem, state 3 is absorbing, since once the allosaur reaches 15,000 we don’t care what
happens thereafter.
The transition matrix is
L
N
M
1 0 0 0
0. 55 0 0.
45 0
0 0.55 0 0.45
0 0 0 1
Starting with the allosaur in state 2;
[0 0 1 0] at inception
[0 0.55 0 0.45] after 1
[0.3025 0 0.2475 0.45] after 2
[0.3025 0.1361 0 0.5614] after 3
[0.3774 0 0.0612 0.5614] after 4
[0.3774 0.0337 0 0.5889] after 5
[0.3959 0 0.0152 0.5889] after 6
By this step, if not before, Prob(state 3) must be converging to 0.60. It’s already closer to 0.60
than 0.57, and its maximum is 0.5889 + 0.0152
O
Q
P

Question # 8
Answer: B
b g ≤ .
The investor will receive at least 1.5 if and only if C 5 0
b g
b g
b gb g = ,
C 5 − C 1 is normally distributed with mean 4 0 0
variance = (4)(0.01) = 0.04, standard deviation = 0.2.
c b g b g h c b g b g
Pr dcCb5g
Cb1gh
/ 0. 2 0. 05 / 0.
2i
1 Φ b025 . g 04 .
Pr C 5 ≤ 0C 1 = 0. 05 = Pr C 5 − C 1 ≤ − 005 . =
= − ≤ −
= − =
h

Question # 9
Answer: D
b gb gb g =
Per 10 minutes, find coins worth exactly 10 at Poisson rate 05 . 0.
2 10 1
g
g
b g
b g
b g
g
b g
Per 10 minutes, f 0 = 0. 3679 F 0 = 0.
3679
f
f
f
1 = 0. 3679 F 1 = 0.
7358
2 = 01839 . F 2 = 0.
9197
3 = 00613 . F 3 = 09810 .
Let Period 1 = first 10 minutes; period 2 = next 10.
Method 1, succeed with 3 or more in period 1; or exactly 2, then one or more in period 2
P = 1− F 2 + f 2 1− F 0 = 1− 09197 . + 01839 . 1−0.
3679
c b gh b gc b gh b g b gb g
= 01965 .
g
b2g b0g
b gb g
Method 2: fail in period 1 if < 2; Pr ob = F 1 = 0.
7358
fail in period 2 if exactly 2 in period 1, then 0; Prob = f f
= 01839 . 0. 3679 = 0.
0677
Succeed if fail neither period; Pr ob = 1− 0. 7358 − 0.
0677
= 01965 .
(Method 1 is attacking the problem as a stochastic process model; method 2 attacks it as a ruin
model.)

Question # 10
Answer: C
Which distribution is it from?
0.25 < 0.30, so it is from the exponential.
Given that Y is from the exponential, we want
Pr Y ≤ y = F y = .
b g b g 0 69
-
1- e y θ
= 0.
69
-
1- e y 0. 5
= 069 . since mean = 0.5
− y = ln b1 − 0. 69g
= − 1171 .
0.
5
y = 05855 .
Question # 11
Answer: D
Use Mod to designate values unique to this insured.
b g b g b g b g
60 60 60 b g
a&& = 1− A / d = 1− 0. 36933 / 0. 06 / 106 . = 111418 .
60 60
1000P = 1000A / a&& = 1000 036933 . / 111418 . = 3315 .
d i b gb g
Mod Mod Mod
A60 = v q60 + 1
p60 A61
= 0. 1376 0 8624 0 383 0 44141
106 .
+ . . = .
Mod
Mod
d 60 i b g
a&& = 1− A / d = 1− 044141 . / 0. 06/ 106 . = 9.
8684
Mod Mod Mod
0 60 60 60
E L = 1000 A − P a&&
d
= 1000 0. 44141−
003315 . 98684 .
= 114.
27
i
b
g

Question # 12
Answer: D
The prospective reserve at age 60 per 1 of insurance is A 60 , since there will be no future
premiums. Equating that to the retrospective reserve per 1 of coverage, we have:
&&
A P s 40:
10 Mod
60 = 40 + P50 s&&
− k
E
5010 : 20 40
A
60
10 50
1
A a&&
a&&
A
40 4010 : Mod 50: 10 :
= × + P50
−
a&&
E E E E
40
10 40 10 50
10 50
40 20
20 40
0.
16132 7.
70
7.
57 0.
06
0.
36913 = × + P Mod
50 −
14.
8166 0. 53667 0.
51081 0.
51081 0.
27414
b gb g
0. 36913 = 0. 30582 + 14. 8196 − 0.
21887
1000 P Mod 50 = 19.
04
P Mod
50
Alternatively, you could equate the retrospective and prospective reserves at age 50. Your
equation would be:
1
A a&&
A
Mod
40 4010 : :
A50 - P50 a&&
= · -
50:
10
a&&
E E
1
where A 40 : 10
40
10 40
= A40−10E40 A50
40 10
10 40
b gb g
= 016132 . − 053667 . 0.
24905
= 0.
02766
d ib g
016132 . 7.
70 0.
02766
0. 24905 − P Mod
50 7.
57 = × −
14.
8166 0.
53667 0.
53667
b1000gb014437
1000P Mod . g
50 = = 19.
07
7.
57
Alternatively, you could set the actuarial present value of benefits at age 40 to the actuarial
present value of benefit premiums. The change at age 50 did not change the benefits, only the
pattern of paying for them.
Mod
40 40 4010 : 50 10 40 50:
10
A = P a&&
+ P E a&&
b g d ib gb g
F 016132 .
0.
16132 = 7. 70 50 0. 53667 7.
57
H G I
14.
8166 K J + P Mod
b1000gb0 07748
1000P Mod . g
50 = = 19.
07
4.
0626

Question # 13
Answer: A
b g b g b g
2 2 τ
x x x
d = q × l = 400
b g
d x
1
b2g
b g
= 0.
45 400 = 180
q′ x =
l
p x
2
b g
2
dx
b g 1
x − db g
.
τ
x
b g
. .
′ = 1− 0 488 = 0512
400
= = 0 488
1000−
180
Note: The UDD assumption was not critical except to have all deaths during the year so that
1000 - 180 lives are subject to decrement 2.

Question #14
Answer: D
Use “age” subscripts for years completed in program. E.g., p 0 applies to a person newly hired
(“age” 0).
Let decrement 1 = fail, 2 = resign, 3 = other.
b g b g b g
1 1 1 1 1 1
0 4 1 5 2 3
′
b2g = 1 2
0 5 1 ′ = 1 2
,
b g
3 , 2 ′
b g
= 1 8
′
b3g = 1 3
0 10 1 ′ = 1 3
,
b g
9 , 2 ′
b g
= 1 4
Then q′ = , q′ = , q′ =
q q q
q q q
bτg b gb gb g
bτg b gb gb g
bτg b gb gb g
τ
τ
τ
0 = 1 = =
2
This gives p 0 = 1− 1/ 4 1− 1/ 5 1− 1/ 10 = 054 .
p 1 = 1− 1/ 5 1−1/ 3 1− 1/ 9 = 0.
474
p 2 = 1− 1/ 3 1−1/ 8 1− 1/ 4 = 0438 .
b g b g b g
b g b g
So 1 200, 1 200 054 . 108, and 1 = 108 0474 . = 512 .
b g b g b g b g
1
2
1
2
τ τ
2 2
q = log p′
/ log p q
b g c h b g
= b0. 405/ 0826 . gb0562
. g
q 2
1 2
= log / log 0. 438 1−
0438 .
3
= 0.
276
b1g
b g b1
= τ g
d2
l2 q2
= 512 . 0276 . = 14
b gb g

Question #15
Answer: C
Let:
N = number
X = profit
S = aggregate profit
subscripts G = “good”, B = “bad”, AB = “accepted bad”
c hb g
1 1
c 2hc 3hb60g 10 (If you have trouble accepting this, think instead of a heads-tails rule, that
2
λ G =
3
60 = 40
λ AB = =
the application is accepted if the applicant’s government-issued identification number, e.g. U.S.
Social Security Number, is odd. It is not the same as saying he automatically alternates
accepting and rejecting.)
2
b Gg b Gg b Gg b Gg b Gg
= b40gb10, 000g+ b40gd300 i
2 = 4, 000,
000
2
b ABg b ABg b ABg b ABg b ABg
Var S = E N × Var X + Var N × E X
Var S = E N × Var X + Var N × E X
b gb g b gb g 2
= 10 90, 000 + 10 − 100 = 1, 000,
000
S G and S AB are independent, so
b g b g b g
Var S = Var SG
+ Var S AB = 4, 000, 000 + 1, 000,
000
= 5, 000,
000
If you don’t treat it as three streams (“goods”, “accepted bads”, “rejected bads”), you can
compute the mean and variance of the profit per “bad” received.
1
λ B =
3
60 = 20
c hb g
d i b g b g
2
b g
2 B 2
B B
If all “bads” were accepted, we would have E X = Var X + E X
= 90, 000+ − 100 = 100,
000
Since the probability a “bad” will be accepted is only 50%,
E X = Prob accepted · E X accepted + Prob not accepted · E X not accepted
b Bg b g c B h b g c B h
b gb g b gb g
d i b gb g b gb g
E X B
2
Likewise,
= 05 . − 100 + 05 . 0 = −50
= 05 . 100, 000 + 05 . 0 = 50,
000
2
b Bg b Bg b Bg b Bg b Bg
2
= b20gb47, 500g + b20gd50 i = 1000 , , 000
Now Var S = E N × Var X + Var N × E X
S G and S B are independent, so
b g b g b g
Var S = Var SG
+ Var SB
= 4, 000, 000 + 1, 000,
000
= 5, 000,
000

Question # 16
Answer: C
Let N = number of prescriptions then S
b g
E N
E
E
∞
∑
= 4 = 1−
F j
j=
0
c
b gh
n f n
0
1
2
3
bS − 80g = 40 × E N − 2 +
+
= N × 40
Nb g FNbng 1− FNbng
0.2000
0.1600
0.1280
0.1024
∞
b g = 40× ∑c1
− Fb jgh
j=
2
L ∞
1
O
∑
NM
c Fb jgh ∑c
Fb jgh
j=
0
j=
0 QP
b . g . .
= 40 × 1 − − 1 −
= 40 4 − 144 = 40 × 2 56 = 102 40
∞
bS −120g = 40× E N − 3 = 40× 1−
F j
+
b g+
∑c
b gh
j=
3
L ∞
2
O
∑
NM
c Fb jgh ∑c
Fb jgh
j=
0
j=
0 QP
b . g . .
= 40 × 1 − − 1 −
= 40 4 − 1952 = 40 × 2 048 = 8192
Since no values of S between 80 and 120 are possible,
0.2000
0.3600
0.4880
0.5904
0.8000
0.6400
0.5120
0.4096
E
b
g
S − 100 =
+
b120− 100g × E bS − 80g + 100− 80 × E S −120
+
b g b g
+
= 9216 .
120
Alternatively,
∞
b g+
∑b g Nb g Nb g Nb g Nb g
j=
0
E S − 100 = 40j − 100 f j + 100 f 0 + 60 f 1 + 20 f 2
(The correction terms are needed because b40 j −100g would be negative for j = 0, 1, 2; we need
to add back the amount those terms would be negative)
¥
¥
Nb g Nb g b gb g b gb g b gb g
j=
0 j=
0
= 40 j · f j - 100 f j + 100 0. 200 + 016 . 60 + 0128 . 20
b g . .
= 40 E N − 100+ 20+ 9 6+
256
= 160 − 67. 84 = 9216 .

Question #17
Answer: B
10 30:
40 10 30 10 40
10
10
10 10
d 10 30 id 10 40 ib1
g
10
b10E30gb 10 E40gb1
ig
E = p p v = p v p v + i
= +
= b054733 . gb053667 . gb179085
. g
= 052604 .
10
The above is only one of many possible ways to evaluate 10 p 30 10 p 40 v , all of which should
give 0.52604
a = a − E a
30: 4010 : 30: 40 10 30: 40 30+ 10:
40+
10
= ba&& 30: 40 −1g −b0. 52604gba&&
40:
50 −1g
b g b gb g
= 13. 2068 − 0. 52604 114784 .
= 71687 .
Question #18
Answer: A
Equivalence Principle, where π is annual benefit premium, gives
d
b g
1000 35 35
π =
d
A + IA × π = a&&
x π
i
1000A35
1000·
042898 .
=
a&&
- IA ( 1199143 . - 616761 . )
b g
35 35
We obtained a&&
35 from
a&&
35
i
42898 .
=
582382 .
= 7366 .
1 A35
1 0.
42898
= − = − = 11.
99143
d 0.
047619

Question #19
Answer: D
Low random ⇒ early deaths, so we want t q = 0. 42 or l + t = 0.
58l
Using the Illustrative Life Table, l80+ t = b 058 . g 3 , 914 , 365
= 2, 270,
332
l > l > l
80+ 5 80+ t 80+
6
so K = curtate future lifetime = 5
80 80 80
30
L = 1000
1
v k + – (Contract premium) a&&
k +1
= 705 − 20 5.
2124
= 601
b gb
g
Question #20
Answer: C
Time until arrival = waiting time plus travel time.
Waiting time is exponentially distributed with mean 1 λ
. The time you may already have been
waiting is irrelevant: exponential is memoryless.
You: E (wait) =
20 1 hour = 3 minutes
E (travel) = b0. 25gb16g + b0.
75gb28g
= 25 minutes
E (total) = 28 minutes
Co-worker:
E (wait) = 1 5
hour = 12 minutes
E (travel) = 16 minutes
E (total) = 28 minutes

Question #21
Answer: B
Bankrupt has 3 states 0, 1, 2, corresponding to surplus = 0, 1, 2
Transition matrix is
M =
L
NM
1 0 0
0. 1 0.
9 0
0. 01 0. 09 0.
9
O
QP
Initial vector t 0 = 0 0 1
t M = t =
0 1
t M = t =
1 2
t M = t =
2 3
0. 01 0. 09 090 .
0028 . 0162 . 081 .
00523 . 0. 2187 0729 .
At end of 2 months, probability ruined = 0.0285%
Question #22
Answer: D
This is equivalent to a compound Poisson surplus process. Water from stream is like premiums.
Deer arriving to drink is like claims occurring. Water drunk is like claim size.
E[water drunk in a day] = (1.5)(250) = 375
b1
g
+ θ =
500
375
Prob(surplus level at time t is less than initial surplus, for some t)
1 375
= Ψb0g
= = = 75%
1 + θ 500

Question #23
Answer: C
Time of first claim is T 1 = − 1/ 3log 05 . = 023 .
0.
23
b g . Size of claim = =
Time of second claim is T 2 = 0. 23− 1/ 3log 0. 2 = 0.
77
Time of third claim is T 3 = 077 . − 1/ 3log 01 . = 1.
54
10 1.
7
0.
77
b g . Size of claim = =
1.
54
b g . Size of claim = =
Since initial surplus = 5> first claim, the first claim does not determine c
0 z.
77
4
Test at T 2 : Cumulative assets = 5 + ct dt = 5 + 0 . 054c
Cumulative claims = 1.7 + 5.9 = 7.6
Assets ≥ claims for c ≥ 4815 .
Test at T 3 : Cumulative claims = 1.7 + 5.9 + 34.7 = 42.3
Cumulative assets = 5+ ct dt = 5+
1732 . c
0
z. 154
0
Assets ≥ claims for c ≥ 2154 .
If c < 4815 . , insolvent at T 2 .
If c > 4815 . , solvent throughout.
49 is smallest choice > 48.15.
4
10 5.
9
10 34.
7
Question #24
Answer: C
µ xy = µ x + µ y = 014 .
µ
Ax
= Ay
=
+ = 007 .
= 05833 .
µ δ 007 . + 005 .
µ xy
Axy
=
axy
+ = 0.
14 014 .
1
= = 0 7368 =
µ δ 0 14 + 0 05 019
µ + δ
= 1
. and
. . .
014 . + 0.
05
xy
P
A xy A + A − A
= =
a a
xy
x y xy
xy
b g
2 0. 5833 −0.
7368
=
= 0.
0817
52632 .
xy
= 5.
2632

Question #25
Answer: E
b gb g b g
b gb i g b g
0545 1− 0022 + 0022
b1
+ ig b . gb . g .
=
V + P 1+ i − q 1− V = V
20 20 20 40 21 20 21 20
0. 49 + 0. 01 1+ − 0. 022 1− 0545 . = 0545 .
= 111 .
050 .
b gb g b g
b gb g q 41b g
V + P 1+ i − q 1− V = V
21 20 20 41 22 20 22 20
0545 . + . 01 111 . − 1− 0. 605 = 0.
605
0. 61605 − 0.
605
q 41 =
0.
395
= 0.
028
Question #26
Answer: E
1000 P = 1000 A / a&&
60 60 60
b 60 60 61g b 60 61g
= 1000 v q + p A / 1+
p v a&&
b 60 60 61g b 60 61g
= 1000 q + p A / 1. 06 + p a&&
c b gb gh c b gb gh
= 15+ 0. 985 382. 79 / 106 . + 0. 985 10. 9041 = 3322 .

Question #27
Answer: E
Method 1:
In each round,
N = result of first roll, to see how many dice you will roll
X = result of for one of the N dice you roll
S = sum of X for the N dice
b g = EbNg
= 35 .
b g = VarbNg
= 2.
9167
b g = b g* b g = 12.
25
b g = b g b g + b g b g
2
b35 . gb2. 9167g b2. 9167gb35
. g
E X
Var X
E S E N E X
Var S E N Var X Var N E X
= +
= 45.
938
Let S 1000 = the sum of the winnings after 1000 rounds
b g = =
b g b g
E S 1000 1000 * 12. 25 12,
250
Stddev S1000 = sqrt 1000* 45. 938 = 214.
33
2
After 1000 rounds, you have your initial 15,000, less payments of 12,500, plus winnings
of S 1000 .
Since actual possible outcomes are discrete, the solution tests for continuous outcomes greater
than 15000-0.5. In this problem, that continuity correction has negligible impact.
b
Pr 15000 - 12500 + S > 14999.
5 =
c
1000
b g b g
= Pr S 1000 - 12250 / 214. 33> 149995 . - 2500 - 12250 / 214.
33 =
= 1− Φ 117 . = 012 .
Method 2
b g
g
Realize that you are going to determine N 1000 times and roll the sum of those 1000 N’s dice,
adding the numbers showing.
Let N 1000 = sum of those N’s
h

g b g b gb g
E 1000E N 1000 3 5 3500
Varb = N1000g = . =
= 1000Var( N ) = 2916.
7
E S1000 = E N1000 E X = 3500 3. 5 = 12.
250
b g b g b g b gb g
b 1000g = b 1000g b g + b 1000g b g
2
= b gb . g + b . gb . g = .
Var S E N Var X Var N E X
3500 2 9167 29167 35 45938
2
StddevbS 1000 g = 214.
33
Now that you have the mean and standard deviation of S 1000 (same values as method 1), use the
normal approximation as shown with method 1.
Question #28
Answer: B
p
k
F bI = a +
k K J p k−1
HG
b g
F bI = +
HG
2 K J F bI a × ⇒ −
HG K J × =
0. 25 = a + b × 0.
25⇒ a + b = 1
0. 1875 0. 25 1 0. 25 0.
1875
2
b = 0.
5
a = 05.
F I
HG K J × =
p 3 0 5 0 .
= . +
5
3
01875 . 0125 .

Question #29
Answer: C
Limiting probabilities satisfy (where B = Bad = Poor):
P = 0. 95P + 015 . S
S = 0. 04P + 080 . S + 0.
25B
B = 001 . P + 005 . S + 0.
75B
P + S + B = 100 .
Solving, P = 0.
694
Question #30
Answer: B
Transform these scenarios into a four-state Markov chain, where the final disposition of rates in
any scenario is that they decrease, rather than if rates increase, as what is given.
2
00
P
2
01
from year t – 3 to from year t – 2 to Probability that year t will
State
year t – 2 year t – 1 decrease from year t - 1
0 Decrease Decrease 0.8
1 Increase Decrease 0.6
2 Decrease Increase 0.75
3 Increase Increase 0.9
Transition matrix is
+ P = 0. 8* 0. 8 + 0. 2 * 0. 75 = 0.
79
L
N
M
0. 80 0. 00 0. 20 0.
00
0. 60 0. 00 0. 40 0.
00
0. 00 0. 75 0. 00 0.
25
0. 00 0. 90 0. 00 0.
10
For this problem, you don’t need the full transition matrix. There are two cases to consider.
Case 1: decrease in 2003, then decrease in 2004; Case 2: increase in 2003, then decrease in 2004.
For Case 1: decrease in 2003 (following 2 decreases) is 0.8; decrease in 2004 (following 2
decreases is 0.8. Prob(both) = 0.8 × 0.8 = 0.64
For Case 2: increase in 2003 (following 2 decreases) is 0.2; decrease in 2004 (following a
decrease, then increase) is 0.75. Prob(both) = 0.2 × 0.75 = 0.15
Combined probability of Case 1 and Case 2 is 0.64 + 0.15 = 0.79
O
Q
P

Question #31
Answer: B
lx = ω − x = 105 − x
⇒ t P45 = l45+
t / l45 = 60 − t / 60
Let K be the curtate future lifetime of (45). Then the sum of the payments is 0 if K ≤ 19 and is
K – 19 if K ≥ 20 .
&&a
20 45
=
b
60
= 1·
K=
20
F 60 - KI 1
HG
60 K J ·
g b40gb41g
2b60g
40+ 39+ ... + 1 = = 1366 .
60
Hence,
c h c h
bK
33g
K
b g
Prob K − 19 > 1366 . = Prob K > 32 . 66
= Prob ≥ since is an integer
= Prob T ≥ 33
l78
27
= 33p45
= =
l 60
= 0.
450
45

Question #32
Answer: C
2
A x =
µ
µ + 2δ
= 0. 25→ µ = 0.
04
A x =
µ
µ + δ
= 0.
4
dIAi = ¥
A ds
x z0 s x
z
∞
0
s x x
E A ds
d
= ¥ -0.
1s
e
z0
=
F e
b0.
4g HG
ib04
. g ds
-0.
1s
-
01 .
I
¥
04 .
= = 4
KJ
01 .
0
Alternatively, using a more fundamental formula but requiring more difficult integration.
c h
IA = t p µ t e dt
x 0
t x x
=
z
z
¥
¥
0
= 0.
04
t e 0.
04 e dt
t e
b g
b
-δ
t
-0. 04t
-0.
06t
z¥
0
-0.
1t
dt
(integration by parts, not shown)
F -t
1
=
HG -
I -0.
1 t
¥
004 . 01 K J e
. 001 .
0
0.
04
= = 4
0.
01
g

Question #33
Answer: E
Subscripts A and B here just distinguish between the tools and do not represent ages.
ο
We have to find e AB
ο
A
e
F I
HG K J = - = - =
z10
2
10
t t
= 1-
dt t 10 5 5
0 10 20
F I
7
HG K J = − = 49 − 49 = 35 .
0 14
F tI t
dt
οHG K JF
I
-
HG K J z
7
7F
= - - +
0 HG
t
e
ο t
= z7
2
−
B 1 dt t
0 7 14
e
ο
AB
z
t t t
= 1-
1 1 7 10 10 7 70
2 2 3
t t t
= t − − +
20 14 210
49 49 343
= 7 − − + =
20 14 210
7
0
2.
683
0
2
I
KJ
dt
ο ο ο ο
AB A B AB
e = e + e − e
= 5+ 3. 5− 2. 683 = 5817 .

Question #34
Answer: A
b g b g = + =
µ τ x t
0100 . 0004 . 0104 .
t
p
b g =
-0.
104
τ
x
e
t
Actuarial present value (APV) = APV for cause 1 + APV for cause 2.
z5
0
z
−0. 04 t −0 . 104 t 5
−0. 04 t −0.
104 t
0
z5
−0.
144t
. , .
0
b g b g
2000 e e 0100 . dt + 500, 000 e e 0400 . dt
c b g b gh e dt
= 2000 010 + 500 000 0 004
e
b gj
2200 −
= − =
0 144 1 0.
144 5
e
.
7841
Question #35
Answer: A
R = 1 − p = q
x
x
S = 1− p × e since e
0
= e
x
b g
z
1
1
1
−k − µ x + − µ x −
0
0
So S = 0. 75R ⇒ 1− p × e = 0.
75q
x
e
e
−k
k
−k
t k dt t dt k dt
= e
−
z
z
d b g i b g
x
q
= 1−
0.
75
p
x
x
1
0
µ
x
b g
t dt
px
1−
qx
= =
1 − 0. 75q
1−
0.
75q
L
NM
1−qx
k = ln
1−0.
75q
x
x
O
QP
x
e
z
−z
1
0
k dt

Question #36
Answer: E
k k+
1
p k
β = mean = 4; = β / 1+
β
n P N = n
b g
b g x f
b1
g bxg f
b2
g bxg f
b3
g bxg
0 0.2 0 0 0 0
1 0.16 1 0.25 0 0
2 0.128 2 0.25 0.0625 0
3 0.1024 3 0.25 0.125 0.0156
f
b k g b x g = probability that, given exactly k claims occur, that the aggregate amount is x.
b1
g b g
bkg b g
b g
x
k−1
e b gj b g
f x = f x ; the claim amount distribution for a single claim
b g
∑
f x = f j x f x − j
j=
0
x
sb g ∑ b g bkgb g
k = 0
k
f x = P N = k × f x ; upper limit of sum is really ∞ , but here with smallest possible
claim size 1, f
f s 0 02
b g b g = 0 for k
x
> x
g = .
g = . * . = .
= . * . + . * . = .
b g = + + =
b g = . + . + . + . = .
f s 1 016 0 25 0 04
f s 2 016 0 25 0128 00625 0048
f s 3 016 . * 025 . 0128 . * 0125 . 01024 . * 00156 . 0.
0576
F s 3 02 004 0048 0 0576 0346

Question #37
Answer: E
Let L = incurred losses; P = earned premium = 800,000
Bonus
F LI = 0. 15× 0.
60 −
HG K J × P if positive
P
= 015 . × 060 . P − Lg if positive
= 015 . × b480, 000−
Lg if positive
= 015 . · c480, 000-b
L 480,
000gh
E (Bonus) = 0.15 (480,000– Ec
L 480, 000gh
From Appendix A.2.3.1
= 0.15{480,000 – [500,000 × (1 – (500,000 / (480,000+500,000)))]}
= 35,265
Question #38
Answer: D
1
28:
2
z2
-δ
t 1 72
0
2
A = e dt
d i b g
1 - δ
= 1- e = 0. 02622 since δ = ln 106 . = 0.
05827
72δ
F 71
a&& = 1+ v 1.
9303
28:
2 H G I
72 K J =
1
3V = 500, 000 A − 6643 a&&
282 : 282 :
= 287

Question #39
Answer: D
Let A x and a x be calculated with µ x t
* *
b g and δ = 006 .
b g
Let Ax
and ax
be the corresponding values with µ x t
increased by 0.03 and δ decreased by 0.03
a
a
x
*
x
1 Ax
= − 0.
4
= = 6.
667
δ 0.
06
= a
x
L
N
M
t
*
¥ - µ x s + 0. 03 ds -0.
03t
x
0
0
Proof: a = e e dt
=
=
z
z
z
= a
* *
x x x
A = 1− 003 . a = 1−
0.
03a
0
x
¥
¥
0
t
x
0
e e e dt
-
z
-z
t
z0
µ
d
µ
= 1−
003 . 6667 .
= 08 .
x
b g
b g
b g
s ds
s ds
-0. 03t
-0.
03t
-0.
06t
e e dt
b gb g
i

Question #40
Answer: A
bulb ages
year 0 1 2 3 # replaced
0 10000 0 0 0 -
1 1000 9000 0 0 1000
2 100+2700 900 6300 0 2800
3 280+270+3150 3700
The diagonals represent bulbs that don’t burn out.
E.g., of the initial 10,000, (10,000) (1-0.1) = 9000 reach year 1.
(9000) (1-0.3) = 6300 of those reach year 2.
Replacement bulbs are new, so they start at age 0.
At the end of year 1, that’s (10,000) (0.1) = 1000
At the end of 2, it’s (9000) (0.3) + (1000) (0.1) = 2700 + 100
At the end of 3, it’s (2800) (0.1) + (900) (0.3) + (6300) (0.5) = 3700
1000 2800 3700
Actuarial present value = + +
2 3
105 . 1. 05 1.
05
= 6688