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Solution of Derivative

1 question 2212 when using equation 2252 for the

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Unformatted text preview: 2 ) = 1. Finally, if at time T , S T < H or ST > K we must check the probability is zero. If ST > K the barrier must have been hit and the probability is N (−d2 ) = 0. If the barrier has not been hit, i.e. S T < H , then ST < H < K . In this case d2 = d6 and −d8 = −∞ implying the probability equals zero. Question 22.6. Equivalently, we could use x0 Q0 = 200 as the stock price and set the dividend yield equal to δQ + ρsσQ + r − rf . This gives BSCall(200, 195, .15, .08, 1, .02 + .2 × .1 × .15 + .08 − .04) = $15.319. Question 22.8. The two $ traded assets are Y and x . Applying Proposition 20.4 to these ratio of these two yields Ft,T Y x = Ft,T (Y ) Ft,T 1 −ρσY s(T −t) . e x (1) Note that ρ is the correlation between Y and 1/x which implies ρσY s is the covariance given by equation (22.28). Since the forward price for Y is Yt e(r −δQ )(T −t) and the forward price for (1/x) is 2 (1/xt ) e rf −r +s (T −t) we have Ft,T Y x = Qt e r −δQ +rf −r +s 2 −ρsσQ −s 2 (T −t) = Qt e(rf −δQ −ρsσQ )(T −t) . (2) The prepaid forward price is therefore Qt e(rf −δQ −ρsσQ −r )(T −t) . Question 22.10. The value of the option is going to depend upon the probability that the stock at expiration will be greater than K , conditional on it having exceeded H1 without ever having exceeded H2 . One way to value an option like this is to use Monte Carlo simulation. By simulating the path of the stock price, we can isolate those paths along which we hit H1 without hitting H2 . However, we can also view this relatively complicated option as a spread of barrier options. Consider the following strategy: •Buy an ordinary knock-in call with strike K and barrier H1 137 Part 5 Advanced Pricing Theory •Write an ordinary knock-in call with strike K and barrier H2 Now consider the payoffs given the different possible combinations of the stock hitting or not hitting the two barriers: Payoff H1 not hit H1 hit H2 not hit H2 hit Purchased knock-in 0 Max(0, ST − K) Max(0, ST − K) Written knock-in 0 0 −Max(0, ST − K) Total 0 Max(0, ST − K) 0 By entering into a knock-in spread, we are able to replicate the payoff to the knock-in, knock-out.1 Question 22.12. When using equation (22.52) for the lookback call, St = S t and ω = 1. a) If S → 0 then S must also go to zero. The option should have no value. The formula given in equation (22.52) is the difference between two terms. The first term must be equal to zero as the term in brackets is bounded and S → 0. The second term can be rewritten as −S e −r(T −t) e−r(T −t) σ 2 S N (d6 ) + 2 (r − δ) S S 2(r −δ)/σ 2 N (d8 ) (3) which must also go to zero for S ≤ S and both go to zero. b) At maturity if ST > S T then d5 → ∞ which implies both N d5 and N d6 → 1 and N −d5 → 0. Similarly d8 → −∞, implying N d8 → 0. Hence equation (22.52) becomes ST − S T . (4) If we happen to have ST = S T , then each di = 0 and equation (22.52) becomes ST σ2 1 − − ST 2 4 (r − δ) 1 The σ2 1 − 2 4 (r − δ) = 0. (5) easy pricing formula in this example relied on the fact that the knock-out price exceeded the knock-in price. There are also knock-in knock-out options for which the stock price is between the knock-in and knock-out prices. For example if the current stock price is 100 we might have an option which knocks in at 80 and then knocks out at 120. These are not priced so easily, since we might hit the knock-out price before the option has knocked in. Numerical methods are required to price these options. 138 Chapter 22 Exotic Options: II Question 22.14. A shout option allows the holder to lock in the moneyness of the option (i.e. the intrinsic value). If you have only one shout, once you “shout” and lock in an intrinsic value, you lose the chance to lock in a higher amount if the option goes deeper in the money. a) If you never shout you have a call option with payoff max [0, S − K ]. Consider the strategy of shouting at a predetermined value G > K ; e.g. the first time the option is $1 in the money. Since max [0, G − K, S − K ] ≥ max [0, S − K ] and there is a positive probability of reaching G, shouting at G will have a higher value than never shouting. b) The lookback pays max [0, Smax − K ], which is always at least as great as max [0, S − K, H − K ]. Thus the lookback is more expensive. One can think of a lookback option as a shout option with perfect foresight. The ladder option pays max [0, S − K, L − K ] for arbitrarily chosen L. If you could choose L optimally, you would have a shout option and would do no worse than a ladder option. Thus the shout is at least as expensive as the ladder. c) To value a shout option, consider what happens at the moment you shout. If you shout at H , you have a guaranteed payoff of H − K , plus the payoff max [0, S − H ], i.e. if S > H at expiration, you also get that difference. Thus at the moment you shout you receive P V (H − K) plu...
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