Hw8_sol

# Hw8_sol - Homework 8 Solutions 1.8(i According to Theorem...

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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 0 (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 0 = ˜ E o h S N - K (1+ r ) N i = ˜ E 0 h S N (1+ r ) N i - K (1+ r ) N = S 0 - K (1+ r ) N . (iv) The amount of money borrowed at time 0 is S 0 - F 0 . At time N this debt is according to (iii) (1 + r ) N ( S 0 - F 0 ) = K . Hence at time N the value of the portfolio is S N - K = F N . 1

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(v) This follows immediately from (ii). (vi) By (ii) if this were true F n = 0 for all n . But clearly this is not always the case since F n = (1 + r ) n ˜ E n S N (1 + r ) N - S 0 .
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