T1-S-232-2010W

T1-S-232-2010W - SOLUTIONS TO TUTORIAL 1 - ACTSC 232,...

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SOLUTIONS TO TUTORIAL 1 - ACTSC 232, WINTER 2010 1. (a) S 0 ( x ) = e - R x 0 μ t dt = (1 - x 100 ) 2 for 0 x < 100 and S 0 ( x ) = 0 for x 100. (b) The probability is Pr { 10 < T 30 < 30 } = S 30 (10) - S 30 (30) = S 0 (40) - S 0 (60) S 0 (30) = 0 . 4082. (c) We have t p x = S 0 ( x + t ) S 0 ( x ) = ( 100 - x - t 100 - x ) 2 for 0 t 100 - x and t p x = 0 for t > 100 - x . (d) f 30 ( t ) = t p 30 μ 30+ t = (70 - t ) / 2450 , 0 t < 70. (e) E [ T 30 ] = R 0 t p 30 dt = R 70 0 ( 70 - t 70 ) 2 dt = 70 / 3 or E [ T 30 ] = R 0 tf 30 ( t ) dt = R 70 0 t (70 - t ) 2450 dt = 70 / 3. (f) We have E [ K 30 ] = k =1 k p 30 = 70 k =1 (1 - k 70 ) 2 = 22 . 84 . (g) Pr { K 30 = 20 } = 20 p 30 - 21 p 30 = 0 . 0202 . (h) Pr { K 30 < 30 } = Pr { T 30 < 30 } = 30 q 30 = 0 . 6735 . (i) Pr { K 30 40 } = Pr { T 30 < 41 } = 41 q 30 = 0 . 8284 . 2. We are given that 0 . 02 = q x = 1 - p x = 1 - e - R 1 0 μ x + t dt and 0 . 04 = q x = 1 - p x = 1 - e - R 1 0 ( μ x + t + c ) dt . Hence, c = ln 98 96 . 3. Note that 0
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T1-S-232-2010W - SOLUTIONS TO TUTORIAL 1 - ACTSC 232,...

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