Lecture 11

Lecture 11 - One way of solving problems of option...

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One way of solving problems of option valuation is to split the problem into tiny steps: first, suppose that the stock would either rise or fall by some discrete amount. Of course this is a terrible assumption if we're talking about a period of time like 3 months, but it's not a bad assumption is we're talking about a period of time like a second. And if we can solve it, then we can model the first second, then the next and the next, up to three months! (Or, we might not have that many steps, but at least we can have enough to get a decent approximation.) The full model is called a tree, because each branch splits into more branches and so on. But first we have to solve the 2-step tree. Suppose the stock is worth 100. Either it goes up to 110 or down to 90 after 6 months. This tree would be drawn like this: The upper node has a value of 110 and the lower node has value of 90. Suppose I have a short position in a call with strike of 100 (ATM) and enough stock to make the total portfolio riskless. How much stock does it take? Call it " ". If the stock rises to 110 then I would have 110* worth of shares minus 10 for the short call, for total 110 - 10. If the stock went down to 90 then I would have a portfolio worth 90 . What delta would give the same payoff in each state? Set 110 - 10 = 90 so =10/20 = 0.5. So a portfolio with a half share of stock and one short call would always have the same payoff (either 55 10 if the stock goes to the up node or 45 if the stock goes to the down node) so its value must grow at the risk-free rate interest rate or else there would be arbitrage opportunities, if there were 2 portfolios with the same (riskless) payoff but different values. So the present value of this payoff in 6 months had better be 45e -rT = 45e -0.5r . If the current risk-free six-month zero rate is 4.5% then this is 45e -0.045*0.5 = 45e -0.0225 = 45* 0.97775 = 43.9988. So the current value of this
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This note was uploaded on 04/04/2012 for the course FIN 420 taught by Professor Poniachek during the Spring '12 term at Rutgers.

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Lecture 11 - One way of solving problems of option...

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