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Unformatted text preview: Introduction to derivatives, Fall 2011 BUSI 588, Case 7 solutions Solutions to Topalov Inc: BlackScholes and binomial trees 1. (*) In order to solve this problem, first we compute the riskneutral probabilities. Since we have u = 1 . 0594, d = 0 . 9439 and r = 1 . 0041, the riskneutral probabilities are given by p u = r d u d = 1 . 0041 . 9439 1 . 0594 . 9439 = 0 . 5208 and p d = 1 p u = 0 . 4792. With these, we price the option working backwards through the tree. The tree below sketches my calculations (see case07.xls for details). 1.23 2.01 3.16 4.73 0.40 0.77 1.49 0.00 0.00 0.00 Note that the values are expiration (last column) are simply max( S T 25 , 0), where the values in the previous columns are the future values of the option discounted at the riskfree rate using the riskneutral probabilities, i.e. 3 . 16 = (4 . 73(0 . 52) + 1 . 49(0 . 48)) / 1 . 0041. 2. (*) One could have equivalently found the price of these securities working backwards through out the tree. It turns out that one can more simply compute these prices using a little bit more thinking. (a) In order to the stock to reach 29.73, it must be the case that it went up three times in a row. The riskneutral probability of this happening is p 3 u , so the price of a security that gives you $1 in the stock reaches 29.73 (nothing otherwise) should be p 3 u /r 3 = 0 . 13957. (b) There are three paths that yield a value of 26.49: uud , udu , or duu (the sequence denoting the order in which the stock went up or down). The price of a security that yields $1 if the stock follows the path uud (nothing otherwise) is p 2 u p d /r 3 . Since this is the same price for the other two paths, one concludes that the value of the security that pays off $1 in the stock reaches 26.49 in the threeperiod binomial tree should be 3 p 2 u p d /r 3 = 0 . 3852....
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 Fall '10
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 Derivatives

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