# T4-Answer-232-2010W - a 70 = 147 . 152 . (b) We have V ar [...

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ANSWERS TO TUTORIAL 4 – ACTSC 232, WINTER 2010 1. It follows from ¨ a 50: 10 = 1 - A 50: 10 d and A 50: 10 = A 1 50: 10 + v 10 10 p 50 that ¨ a 50: 10 = 7 . 12468. 2. It follows from A 1 x : 10 = 1 - d ¨ a x : 10 - 10 E x , ¨ a x : 10 = ¨ a x - 10 | ¨ a x , and ¨ a x : 10 = 10 E x ¨ s x : 10 that A 1 x : 10 = 0 . 24. 3. (a) The present value is 300 ¨ a 20 + 500 ± ¨ a K x +1 - ¨ a ( K x +1) 20 ² . (b) The APV of the annuity is 300 ¨ a 20 + 500 20 | ¨ a x = 10440 . 80. (c) It follows from π ¯ a x : 20 = 300 ¨ a 20 + 500 20 | ¨ a x that π = 755 . 507. (d) The number of the payments made in the guaranteed life annuity is N = max { 20 , K x + 1 } . We have for 0 k 19, Pr { N > k } = 1 and for k 20, Pr { N > k } = Pr { K x k } = Pr { T x > k } = e - 0 . 01 k . Thus, E ( N ) = k =0 Pr { N > k } = 102 . 28 . 4. (a) The variance is 1000 2 V ar a T x ] = 1000 2 2 ¯ A x - ( ¯ A x ) 2 δ 2 = 4 . 44444 × 10 7 . (b) The probability is Pr { 4000 < 1000¯ a T x < 8000 } = Pr { 4 . 3588 < T x < 9 . 6416 } = 0 . 16 . (c) The probability = Pr { ¯ a T x > ¨ a x : 30 } = Pr { 1 - v Tx δ > 15 . 9291 } = 0 . 77612 . 5. (a) Y = 20 ¯ a T 70 . Thus, E [ Y ] = 20 ¯
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Unformatted text preview: a 70 = 147 . 152 . (b) We have V ar [ Y ] = 20 2 2 A 70-( A 70 ) 2 2 = 5240 . 81 . (c) The probability is . 02 = Pr { v T 70 > 20 a T 70 } = 1 20 log( 20 + 20 ) . Thus, = 8 . 08 . 6. (a) Y = 2 a T 40 + 10 v T 40 = 33 . 33-23 . 33 v T 40 . (b) E ( Y ) = 33 . 33-23 . 33 A 40 = 27 . 50. (c) V ar ( Y ) = 23 . 33 2 ( 2 A 40-( A 40 ) 2 ) = 43 . 7. 7. We have E ( a T x n ) = Z a t n f T x ( t ) dt = Z n Z t v s ds f T x ( t ) dt + a n Z n f T x ( t ) dt = Z n ( s p x-n p x ) v s ds + a n n p x = Z n s p x v s ds. 1...
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## This note was uploaded on 03/20/2011 for the course ACTSC 232 taught by Professor Matthewtill during the Summer '08 term at Waterloo.

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