# sol4 - ACTSC 445 Asset-Liability Management Department of...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACTSC 445: Asset-Liability 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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online