Unformatted text preview: ACTSC 445: AssetLiability Management Department of Statistics and Actuarial Science, University of Waterloo Solutions for Unit 4 Exercises 1. To find s 5 , we first find z 5 by solving 103 . 49 = 5 . 5((1 + z 1 ) 1 + (1 + z 2 ) 2 + (1 + z 3 ) 3 + (1 + z 4 ) 4 ) + 105 . 5(1 + z 5 ) 5 . Therefore (105 . 5 / (103 . 49 5 . 5((1 + z 1 ) 1 + (1 + z 2 ) 2 + (1 + z 3 ) 3 + (1 + z 4 ) 4 ))) 1 / 5 1 = 0 . 047342 , and so s 5 = 0 . 094684. Similarly, we find z 6 by solving 99 . 49 = 4 . 75((1 + z 1 ) 1 + (1 + z 2 ) 2 + (1 + z 3 ) 3 + (1 + z 4 ) 4 + (1 + z 5 ) 5 ) + 104 . 75(1 + z 6 ) 6 , and get z 6 = 0 . 048935, and so s 6 = 0 . 09787. 2. We use the formula P = 25((1 + z 1 ) 1 + (1 + z 2 ) 2 + (1 + z 3 ) 3 + (1 + z 4 ) 4 + (1 + z 5 ) 5 + (1 + z 6 ) 6 ) + 1000(1 + z 6 ) 6 , where the z i ’s have been found in Question 1. We thus get P = 879 . 2609. (Note: all the spot rates are higher than the coupon rate, which is why we have a discount bond (i.e., whose price is smaller thanhigher than the coupon rate, which is why we have a discount bond (i....
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This note was uploaded on 10/24/2010 for the course ACTSC 445 taught by Professor Christianelemieux during the Fall '09 term at Waterloo.
 Fall '09
 ChristianeLemieux

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