Hw8_sol - Homework 8: Solutions 1.8 (i) According to...

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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...
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This note was uploaded on 09/29/2011 for the course MATH 4740 taught by Professor Durrett during the Spring '10 term at Cornell.

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Hw8_sol - Homework 8: Solutions 1.8 (i) According to...

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