LTAM MC Solutions, Fall 2019 - Society of Actuaries

LTAM MC Solutions, Fall 2019

1. Answer A 15 px 5 px 10 px5 15 px 0.764591 5 px 0.955290 10 px5 0.800

2. Answer D

0.4q4(10)

0.032

q(1) 40

0.032 0.4

0.08

p( ) 40

1

q(1) 40

q(2) 40

0.845

and

0.6q4(20)

0.045

q(2) 40

0.045 0.6

0.075

3. Answer B The central exposed-to-risk is E7c5 0.8 1 0.4 2.2 dx 2 p^ x e2/2.2 0.403 q^x 1 e2/2.2 0.597

4. Answer C log(m(65, 2018)) 65 65K2018 K2018 ~ N (, 2 ) where 9 2(0.35) 9.7 and 0.75( 2) 1.06066 80%ile of K2018 0.842 8.8069 80%ile of log(m(65, 2018)) 2.6 0.04(8.8069) 2.95

5. Answer C

3 px00 e3(0.01) 0.970446

1

p00 x3

e(0.010.120.08)

0.810584

4 px00 0.970446 0.810584 0.78663

6. Answer A

d

dt

t

V

(0

)

V (0)

t

10, 000 65t

t V (1) t V (0) 55t

tV (2) tV (0)

At t 10 :

10V (1) 8, 000 a65 103,133; 10V (2) 8, 000 a75 77,810

d dt

tV

(0)

0.05109, 650

10, 000 0.017552103,133 109, 650

0.00560577,810 109, 650

4224.65

7. Answer B

EPV Death Benefit: 100, 000 A1 65:20

100, 000( A 65:20

20

E65 )

100, 000(0.43371 0.24381) 18,990

EPV Annuity Benefit: 45, 00020E65 a85 45, 000(0.24381)(6.7993) 74,598 EPV Premiums: Pa 11.8920P

65:20

P 7,870

8. Answer D

a(4) 80

a80

3 8

15 1216

80

8.5484

3 8

15 1216

0.030162 ln(1.05)

8.1672

a(4) 90

a90

3 8

15 1216

90

5.1835

3 8

15 1216

0.096590 ln(1.05)

4.7971

a(4) 80:10

8.1672 10 E80

4.7971 6.5385

20, 000 a(4) 130, 770 80:10

9. Answer E 1, 250,000 250,000 Xa55 100,000A55 X 60,802

10. Answer D

EPV 1000 p6001v 2 p6001v2 3 p6001v3 4 p6001v4 5 p6001v5 1000 0.01v 0.03v2 0.04v3 0.05v4 0.07v5

173

11. Answer D

12P 0.95 a(12) 0.6 1, 000, 000A(12)1

50:10

50:10

a(12) 50:10

(12)a 50:10

(12) 110E50 1.00028.0550 0.46651(1 0.60182) 7.87086

A(12)1 i

i 50:10

(12)

A 50:10

E 10 50

1.02271 0.61643 0.60182

0.014942

P 181.05

12. Answer A

Pa5000

30,

000

A01 50

50,

000

A02 50

20,

000

A03 50

P 1193.7

13. Answer A Pn 1000 A60 / a60 33.115; PFPT 1000 A61 / a61 35.110

10V n 10V FPT 1000 A70 Pna70 1000 A70 PFPT a70 PFPT Pn a70 17.09

Or

10V

n

10001

a70 a60

an d

V FPT

10

10001

a70 a61

10V n

10 V FPT

10001

a70 a60

10001

a70 a61

1000

a70 a61

a70 a60

17.13

14. Answer B

P

100, 000( A61

0.5A 1 61:20

)

100, 000

0.30243 0.50.41417 0.28641

1628.43

a61

14.6491

or

P 50, 000( A61 20 E61 A81)

50, 000

0.30243 0.50.41417 (0.28641)(0.60984)

1628.41

a61

14.6491

20V n 100, 000A81 Pa81 47, 641

15. Answer E

1V

100, 000A01 61:4

200, 000 A02 61:4

7000a 61:4

A01 61:4

4

v p t 00 01 t 61 61t

0

dt 4 e0.02te0.06t 0.05 dt 0.05

0

0.08

1 e4(0.08)

0.17116

A02 61:4

4

v p t 00 02 t 61 61t

0

dt

4

e e 0.02t 0.06t 0.01 dt

0

0.01 0.08

1 e4(0.08)

0.03423

a00 61:4

1

v1

p00 61

v

2 2

p00 61

v33

p00 61

1 e0.08 e0.16

e0.24

3.5619

1V 971

16. Answer D

Pr2 1V P E (1 i) q61 100, 000 p61 2V

(325 700 (0.04)(700))(1.06) 100, 000(0.003792) (1 0.003792)(600) 79.9

17. Answer C

States at t =1,2 0, 0

0, 1 1, 0 1, 1

Probability 0.72 0.49 0.70.3 0.21 0.30.6 0.18 0.30.4 0.12

Final Average Salary (100,000+100,000)/2 = 100,000

(100,000+105,000)/2=102,500 (105,000+105,000)/2=105,000 (105,000+110,250)/2=107,625

So the expected final salary is 0.49100,000 0.21102,500 0.18105,000 0.12110, 250 102,340

18. Answer C

AL0 20(55, 000)(0.013) (33, 000)(0.02) 15 p50 v15(13.086) 166, 083 EPV0 of AL1 21(55, 000)(0.013) (35, 200)(0.02) 15 p50 v15(13.086) 179,968

NC 179,968 166, 083 13,885

19. Answer E

(1 k )Ba6(142)

Bvp64

a (12 ) 65

a(12) 64

(12)a64

(12)

(1.0002)(13.8363)

0.46651 13.3726

1 k 0.994712v(13.086) 0.927 13.3726

k 0.073

20. Answer E aB (63, 0) 1 (1 j)cvp63 (1 j)2 c2v2 2 p63(aB (65, 2))

1

(1.04)(1.03)(1.06)1

95, 95,

082.5 534.4

(1.04)2

(1.03)2

(1.06)2

94, 95,

579.7 534.4

(26.708)

29.0

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