{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Test2-Solutions-F08

Test2-Solutions-F08 - ACTSC 432 Loss Models 2 TEST#2 1(15...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ACTSC 432 - Loss Models 2 TEST #2 1. (15 marks) For a particular line of insurance business, it is believed that 60% of the risks are standard (with risk parameter ° = ° 1 ) and 40% are substandard (with risk parameter ° = ° 2 ). Given that the conditional distribution of the claim amount for these two types of risk are Pr ( X j = 5 j ° = ° 1 ) = 0 : 2 Pr ( X j = 5 j ° = ° 2 ) = 0 : 1 Pr ( X j = 10 j ° = ° 1 ) = 0 : 5 Pr ( X j = 10 j ° = ° 2 ) = 0 : 3 Pr ( X j = 20 j ° = ° 1 ) = 0 : 3 Pr ( X j = 20 j ° = ° 2 ) = 0 : 6 , and that the r.v.°s X 1 ; X 2 ; ::: are independent conditional on the risk parameter ° , (a) ( 2 marks ) determine the hypothetical means for both types of risk Solution : ± ( ° 1 ) = E [ X j j ° = ° 1 ] = (5) (0 : 2) + (10) (0 : 5) + (20) (0 : 3) = 12 , and ± ( ° 2 ) = E [ X j j ° = ° 2 ] = (5) (0 : 1) + (10) (0 : 3) + (20) (0 : 6) = 15 : 5 . (b) ( 2 marks ) ±nd the process variance for both types of risk Solution: v ( ° 1 ) = V ar ( X j j ° = ° 1 ) = E ° X 2 j j ° = ° 1 ± ° ( E [ X j j ° = ° 1 ]) 2 = (5) 2 (0 : 2) + (10) 2 (0 : 5) + (20) 2 (0 : 3) ° (12) 2 = 31 , and v ( ° 2 ) = V ar ( X j j ° = ° 2 ) = E ° X 2 j j ° = ° 2 ± ° ( E [ X j j ° = ° 2 ]) 2 = (5) 2 (0 : 1) + (10) 2 (0 : 3) + (20) 2 (0 : 6) ° (15 : 5) 2 = 32 : 25 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}