sols29_41 - Mich 2010 Mathematical Finance: Short Solutions...

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Unformatted text preview: Mich 2010 Mathematical Finance: Short Solutions 4 29. We have K = 42, S (0) = 40, T = 1 / 3, = 0 . 15 and = 0 . 24 so - 2 / 2 = 0 . 1212. We have to calculate P ( S (1 / 3) > K ). We know that W log( S (1 / 3) /S (0)) N (0 . 0404 , . 0192) and consequently P ( S (1 / 3) > K ) = = P (log S (1 / 3) /S (0) > log K/S (0)) = P ( W > log 42 / 40) . 476. 30. The value of is necessary but not included in the list. When using the risk neutral drift to calculate call option prices is not needed but otherwise it is. 31. (a) C ( K,T ) = 4 . 015. (b) Use drift to get P ( S (1 / 4) K ) = 0 . 631. (c) Use independence of the two conditions to get C = e- . 04 50 . 1557 = 7 . 478. 32. Either directly or by finding C and using put-call parity we get P = 4 . 41. 33. If K = 0 then C = S (0). Also lim T C ( K,T ) = S (0). 34. Using drift , P ( S (1 / 2) < . 9 S ) = P (log S (1 / 2) /S < log 0 . 9) . 036....
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This note was uploaded on 02/04/2011 for the course MATH 3301 taught by Professor Dri.m.macphee during the Spring '10 term at Durham.

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