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Unformatted text preview: ANSWERS TO TUTORIAL 2 – ACTSC 232, WINTER 2010 1. (a) 2 p [95]+1 = l [95]+3 l [95]+1 = 0 . 294 . (b) 2 | 3 q [95] = l [95]+2- l [95]+5 l [95] = 0 . 54545 . (c) 2 d [97]+1 = l [97]+1- l [97]+3 = 40. (d) 6 d [95] = l [95]- l [95]+6 = l [95]- l 101 = 217. (e) e [96] = ∑ ∞ k =1 k p [96] = ∑ 5 k =1 l [96]+ k l [96] = 1 . 095. 2. (a) Z = 1000 v T 30 , < T 30 ≤ 20 , 2000 v T 30 , 20 < T 30 ≤ 40 , 3000 v T 30 , T 30 > 40 . (b) Given De Moivre’s law with ω = 110, we know that the pdf of T 30 is f 30 ( t ) = 1 80 for 0 < t < 80. Thus, the net premium of the insurance is E ( Z ) = Z ∞ b t v t f 30 ( t ) dt = Z 20 1000 v t 1 80 dt + Z 40 20 2000 v t 1 80 dt + Z 80 40 3000 v t 1 80 dt = 946 . 225 . (c) We have E ( Z 2 ) = Z ∞ ( b t ) 2 v 2 t f 30 ( t ) dt = Z 20 1000 2 v 2 t 1 80 dt + Z 40 20 2000 2 v 2 t 1 80 dt + Z 80 40 3000 2 v 2 t 1 80 dt = 934565 . 515 . Thus, V ar ( Z ) = 39223 . 76....
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