TO SOLUTIONS ASSIGNMENT #1
A 1. (a) By the EP, b A40 = 230 a40 = 230 1- 40 . We have = ln1.06 and
A40 =
0
v t t p40 40 (t) dt =
60 0
e-t dt = 0.2773594. 60 [5 marks] [5 marks]
Hence, b = 10, 284.20. (b) The benefit reserve = b A60 - 230 a60 = 1565.07. (c) Given T (40) > 20,
20 L
= b v T (40)-20 - 230 aT (40)-20 . Thus,
2
230 V ar(20 L|T (40) > 20) = b +
( 2 A60 - (A60 )2 ) = 12651925.25. [5 marks]
2. (a) By the EP, a50:10 = A50:20 . Thus, = 0.0555. (b) Given T (50) > 15,
15 L 15 L
[5 marks]
= v T (50)-15 , =v ,
5
if T (50) 20, if T (50) > 20. [4 marks]
(c) A65:5 . (d)
1
15 E 50
[3 marks]
( a50:10 - A1 ). [3 marks] 50:15 (e) By (a), a50:10 - A1 = A50:20 - A1 . Hence, the formulas in (c) and (d) are 50:15 50:15 equal since A50:20| = A1 + A65:5 15 E 50 . [5 marks] 50:15 [5 marks] (f) By (c), the benefit reserve is A65:5 = 1 - a65:5 = 0.7840.
20 L
3. (a) Given K(65) 20,
= aK(65)-19 .
[3 marks] [4 marks]
(b) E [20 L | K(65) 20] = a85 (c) V ar L [20 | K(65) 20] =
1 d2
( 2 A85 - (A85 )2 ) .
12 E68
[5 marks] - P (15| a65 ) a68: 12 [4 marks] [4 marks]
(d) The prospective formula: 3 V (15| a65 ) = a80 (e) The retrospective formula: 3 V (15| a65 ) =
1
3 E65
P (15| a65 ) a65: 3 .
1
4.
5 V60
= P60 s60: 5 - 5 k60 s60: 5 = 15.227 = P60:15 s60: 5 - 5 k 60 = 0.5053. [5 marks]
5 V60:15
5. (a)
1 V100:3|
=1-
a101:2| , a100:3|
v = e- ,
a101:2| = 1 + v p101 = 1 + v a100:3|
1 V100:3|
l102 = 1.411767, l101 = 1 + v p100 + v 2 2 p100 = 1.9494875, = 0.275827. [5 marks]
(b)
1 1 V100: 3 1 = A1 2 - P100: 3 a101: 2 , 101:
1 P100: 3 ...