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Tutorial_2_Solutions

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Name: userid: ACTSC 232 – Winter 2011 Tutorial #2 – Jan 17, 2011 1. You are given the following information: 3 p 51 = 0 . 9126 , 2 q 50 = 0 . 0298 , q 52 = 0 . 0300 , 2 p 52 = 0 . 9312 . (a) Find the numerical values of the following quantities (and also make sure you understand the interpretation of each): i. p 52 Answer: p 52 = 1 - q 52 = 1 - 0 . 0300 = 0 . 9700 . ii. 4 p 50 Answer: 4 p 50 = 2 p 50 2 p 52 = (1 - 1 p 50 )( 2 p 52 ) = (1 - 0 . 0298)(0 . 9312) = 0 . 9305 iii. p 51 Answer: p 51 2 p 52 = 3 p 51 so that p 51 = 3 p 51 2 p 52 = 0 . 9126 0 . 9312 = 0 . 9800 iv. 2 | 2 q 50 Answer: 2 | 2 q 50 = 2 p 50 2 q 52 = (1 - 2 q 50 )(1 - 2 p 52 ) = (1 - 0 . 0298)(1 - 0 . 9312) = 0 . 0667 1

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v. 3 | q 50 Answer: 3 | q 50 = 3 p 50 q 53 = ( 2 p 50 p 52 )( q 53 ) Now in order to ﬁnd q 53 we note that p 52 p 53 = 2 p 52 so that p 53 = 2 p 52 /p 53 = 0 . 9312 / 0 . 9700 = 0 . 9600 . Then q 53 = 1 - p 53 = 1 - 0 . 9600 = 0 . 0400 . Finally, 3 | q 50 = ( 2 p 50 p 52 )( q 53 ) = (1 - 2 p 50 )( p 52 )( q 53 ) = (1 - 0 . 0298)(0 . 97)(1 - 0 . 9600) = 0 . 0376 . (b) Using the above information, ﬁll in the blank entries of the following life table: x x d x 50 10 , 000 . 00 100 . 00 51 9 , 900 . 00 198 . 00 52 9 , 702 . 00 291 . 06 53 9 , 410 . 94 376 . 44 54 9,034.50 . . . . . . . . . Answer: Answers are given in bold in the table above. In general, perhaps the
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Unformatted text preview: simplest way to go about this is to work backwards one year at a time, ﬁrst ﬁnding p 53 (or q 53 ) and using this to obtain ‘ 53 and d 53 , and so on. 2 2. Show that d dx t p x = t p x ( μ x-μ x + t ). Answer: d dx t p x = d dx S x ( t ) = d dx S ( x + t ) S ( x ) = d dx S ( x + t )[ S ( x )]-1 =-S ( x + t )[ S ( x )]-2 S ( x ) + [ S ( x )]-1 S ( x + t ) =-S ( x ) S ( x ) · S ( x + t ) S ( x ) + S ( x + t ) S ( x ) = μ x t p x + S ( x + t ) S ( x ) = μ x t p x + S ( x + t ) S ( x + t ) · S ( x + t ) S ( x ) = μ x t p x--S ( x + t ) S ( x + t ) · S ( x + t ) S ( x ) = μ x t p x-μ x + t t p x = t p x ( μ x-μ x + t ) 3...
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