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Unformatted text preview: 03 with probability 0 . 4 with probability 0 . 6 and . 02 + 0 . 03 X = ( . 05 with probability 0 . 4 . 02 with probability 0 . 6 UNSW Week 13 ACTL1001 Tutorial Examples 1.3 E [ S 18+ t ] = E [ S 18 e 1 e 2 . . . e t ] = 20000 E [ e 1 ] E [ e 2 ] . . . E [ e t ] due to independence = 20000( E [ e 1 ]) t since the deltas are identically distributed = 20000(0 . 4 e . 05 + 0 . 6 e . 02 ) t = 20000 e . 02 t (0 . 4 e . 03 + 0 . 6) t UNSW Week 13 ACTL1001 Tutorial Examples 1.4 We want . 01 10 X i =1 E [ S 18+ t ] t ( ap ) 18 1 + i t = 0 . 01 10 X i =1 20000 e . 02 t (0 . 4 e . 03 + 0 . 6) t e. 02 t (1 + i ) t = 200 10 X i =1 . 4 e . 03 + 0 . 6 1 . 1 t = 200 101 1 = 1302 . 0336 where = . 4 e . 03 + 0 . 6 1 . 1 UNSW Week 13 ACTL1001 Tutorial...
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This note was uploaded on 06/12/2011 for the course ASB 1001,2522, taught by Professor Nicole during the One '09 term at University of New South Wales.
 One '09
 Nicole

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