{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Test1-Solutions-S11

Test1-Solutions-S11 - ACTSC 331 Life Contingencies I Test 1...

This preview shows pages 1–3. Sign up to view the full content.

ACTSC 331 - Life Contingencies I Test 1 1. (18 marks) Consider a special fully discrete whole life insurance with: b K +1 = 400 e 0 : 02( K +1) , K = 0 ; 1 ;::: premium k payable at time k (if survival): k = , k = 0 ; 1 ; 2 ;::: A force of interest of 4% is assumed. Given that ± A @ x A @ x +15 0 : 02 0 : 669 0 : 726 0 : 04 0 : 492 0 : 564 0 : 06 0 : 382 0 : 456 0 : 08 0 : 308 0 : 380 0 : 1 0 : 256 0 : 324 (a) (3 marks) Solution: Under the equivalence principle, EPV @0 ( ) = EPV @0 (future premiums), where EPV @0 ( ) = 1 X k =0 b k +1 v k +1 k j q x = 1 X k =0 400 e 0 : 02( k +1) e & 0 : 04( k +1) k j q x = 400 1 X k =0 e & 0 : 02( k +1) k j q x = 400 A @ =2% x , and EPV @0 (future premiums) = 1 X k =0 v k k p x = a @ =4% x = 1 ± A @ =4% x d = 1 ± A @ =4% x 1 ± e & 0 : 04 . 1

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

View Full Document
One concludes that = 400 A @ =2% x 1 A @ =4% x 1 e & 0 : 04 = 400 (0 : 669) 1 & 0 : 492 1 e & 0 : 04 = 20 : 655 . (b) (3 marks) calculate the time- 15 policy value. Solution: Using the prospective formula, we have 15 V = EPV @15 ( ) EPV @15 ( future premiums ) = 1 X k =0 b 15+ k +1 v k +1 k j q x +15 1 X k =0 k k p x +15 = 400 e 0 : 02(15) 1 X k =0 e 0 : 02( k +1) e & 0 : 04( k +1) k j q x +15 a @ =4% x +15 = 400 e 0 : 02(15) A @ =2% x +15 1 A @ =4% x +15 1 e & 0 : 04 = 400 e 0 : 02(15) (0 : 726) (20 : 655) 1 & 0 : 564 1 e & 0 : 04 = 162 : 33 . (c)
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}