ACTSC 433 A4 SOLN W12

ACTSC 433 A4 SOLN W12 - Solutions to Assignment 4 ACTSC...

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Unformatted text preview: Solutions to Assignment 4 ACTSC 433/833, Winter 2012 1. (a) We have f ( x ) = F ( x ) = x 2 e- /x ,x > 0. Thus, L ( ) = f (96) f (135) f (278) f (396)( F (90)) 6 4 e- . 0906131 . It follows from d ln L ( ) d = 0 that = 44 . 1437. (b) It follows from F ( x ) = 0 . 5 that the maximum likelihood estimate of the median of a loss is- ln0 . 5 = 63 . 6859. 2. (a) We have F ( x ) = 1- 90 90+ x , x > 0 and L ( ) = f ( 20 . 8 + 15) f ( 40 . 8 + 15) f ( 52 . 8 + 15) f ( 60 . 8 + 15)(1- F ( 100 . 8 + 15)) 2 . 8 4 (1- F (15)) 6 . Hence, it follows from d d ln L ( ) = 0 that = 1 . 25309. (b) The probability is 1- F ( 50 . 8 +15) 1- F (15) = 105 167 . 5 . Hence, its MLE is 105 167 . 5 = 0 . 557 . (c) The expected cost per payment is E ( Y P ) = R 100 y f ( y . 8 +15) . 8(1- F (15)) dy + 100 1- F ( 100 . 8 +15) 1- F (15) = 84 - 1- 84 - 1 105 230 - 1 . Hence, the MLE for the expected cost per payment is 84 - 1- 84 - 1 105 230 - 1 = 59 . 74 . 3. (a) We have f ( x ) = F ( x ) = 90 ( x +90) +1 , x 0. The distribution of Y is F Y ( y ) = , y < 10 , F ( y )- F (10) 1- F (10) = 1- ( 100 90+ y ) , 10 y < 60 , 1 , y 60 . (b) f Y ( y ) = , y < 10 or y > 60 , f ( y ) 1- F (10) = 100 (90+ y ) +1 , 10 y < 60 , Pr { Y = 60 } = 1- F (60) 1- F (10)...
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ACTSC 433 A4 SOLN W12 - Solutions to Assignment 4 ACTSC...

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