This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 8: Solutions 1.8 (i) According to Theorem 1.2.2 we can find v n (n=0,1,2,3) recursively by using the formula v n ( s,y ) = 1 1 + r (˜ pv n +1 (2 s,y + 2 s ) + (1 ˜ p ) v n +1 (0 . 5 s,y + 0 . 5 s )) = 2 5 ( v n +1 (2 s,y + 2 s ) + v n +1 (0 . 5 s,y + 0 . 5 s )) , and that v 3 ( s,y ) = 1 4 y 4 + . (ii) v (4 , 4) = 1 . 216 . (iii) Again by Theorem 1.2.2 δ n ( s,y ) = v n +1 (2 s,y + 2 s ) v n +1 (0 . 5 s,y + 0 . 5 s ) 1 . 5 s , with v 4 as in part (i). 2.11 (i) This a consequence of the following identity ( x K ) + ( K x ) + = x K, for any real numbers x,K . (ii) We have by linearity of the conditional expectation and Theorem 2.4.7 F n = ˜ E n F N (1 + r ) N n = ˜ E n C N P N (1 + r ) N n = ˜ E n C N (1 + r ) N n ˜ E n P N (1 + r ) N n = C n P n . (iii) F = ˜ E o h S N K (1+ r ) N i = ˜ E h S N (1+ r ) N i K (1+ r ) N = S K (1+ r ) N . (iv) The amount of money borrowed at time 0 is S F . At time N this debt is according to (iii) (1 + r ) N ( S F ) = K...
View
Full
Document
This note was uploaded on 09/29/2011 for the course MATH 4740 taught by Professor Durrett during the Spring '10 term at Cornell.
 Spring '10
 DURRETT

Click to edit the document details