W12Tute 1001

W12Tute 1001 - 2) 1 X 2 ] = E [( X-1) 1 X 1 ]-P ( X...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Examples 1. Aggregate claims have the distribution X Geometric(0 . 6). 1.1 Suppose that D is a positive integer. Show that E [( X - ( D + 1)) 1 X D +1 ] = E [( X - D ) 1 X D ] - P ( X > D ) 1.2 Hence, or otherwise, show that E [( X - 3) 1 X 3 ] = 8 75 2. Motor claims Y have distribution given by P ( Y k ) = k 500 for 0 k 500. An insurer sells a policy on these motor claims so that there is a deductible of $50, and any loss amount above $300 (including the deductible) is covered by reinsurance. (Thus, after the deductible and reinsurance are accounted for, the insurer’s loss on any individual policy is limited to $250.) Calculate the insurer’s expected loss. UNSW Week 12 ACTL1001 Tutorial
Background image of page 1

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

View Full DocumentRight Arrow Icon
Examples Q1 - Recursive Relationship E [( X - ( D + 1)) 1 X D +1 ] = X x = D +1 ( x - ( D + 1)) f ( x ) = " X x = D ( x - ( D + 1)) f ( x ) # - ( D - ( D + 1)) f ( D ) = " X x = D ( x - D ) f ( x ) # - " X x = D f ( x ) # + f ( D ) = " X x = D ( x - D ) f ( x ) # - " X x = D +1 f ( x ) # = E [( X - D ) 1 X D ] - P ( X > D ) as required. UNSW Week 12 ACTL1001 Tutorial
Background image of page 2
Examples Q1 - Recursive Relationship Note that P ( X > D ) = X x = D +1 0 . 6(0 . 4) x - 1 | {z } Geometric Series = 0 . 6(0 . 4) D 1 - 0 . 4 = 0 . 4 D Then E [( X - 0) 1 X 0 ] = E [ X ] = 1 0 . 6 = 5 3 E [( X - 1) 1 X 1 ] = E [( X - 0) 1 X 0 ] - P ( X > 0) = 5 3 - 1 = 2 3 E [( X -
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2) 1 X 2 ] = E [( X-1) 1 X 1 ]-P ( X > 1) = 2 3-. 4 = 4 15 E [( X-3) 1 X 3 ] = E [( X-2) 1 X 2 ]-P ( X > 2) = 4 15-. 4 2 = 8 75 Alternatively, let x-3 = u so X x =3 ( x-3)(0 . 6)(0 . 4) x-1 = 0 . 4 3 X u =0 u (0 . 6)(0 . 4) u-1 = 0 . 4 3 E [ X ] UNSW Week 12 ACTL1001 Tutorial Examples Q2 - Deductible and Reinsurance The insurers loss is given by if Y 50; Y-50 if 50 < Y 300; 300-50 = 250 if 300 < Y 500 . and the density function of Y is f ( y ) = d dy P ( Y y ) = 1 500 . Hence, the insurers expected loss is Z 50 (0) 1 500 dy + Z 300 50 ( y-50) 1 500 dy + Z 500 300 250 1 500 dy = 0 + ( y-50) 2 1000 300 50 + 100 = 250 2 1000 + 100 = 325 2 = 162 . 5 UNSW Week 12 ACTL1001 Tutorial...
View Full Document

Page1 / 4

W12Tute 1001 - 2) 1 X 2 ] = E [( X-1) 1 X 1 ]-P ( X...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online