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# L5 - Lecture 5 American Options An American option is a...

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Lecture 5 American Options An American option is a contract between two parties made at a certain time t such that the buyer of the contract has the right, but not the obligation, to exercise the option at any time τ with t τ T . If the option is exercised at τ , then the seller pays the buyer an amount Y ( τ ) 0. For instance, Y ( τ ) = ( S ( τ ) - c ) + for an American call option and Y ( τ ) = ( c - S ( τ )) + for an American put option based on a single stock. One can identify an American option by its payoff process Y A = { Y ( t ) , t = 0 , 1 , . . . , T } . American options enjoy the additional flexibility — possibility of exercising earlier than T — compared to their European option counterparts. What is the value V A ( t ) of an American option? 5.1 A special case: “American = European” Since the holder of an American option can always choose not to exercise the option until time T , V A ( t ) V ( t ) where V ( t ) is the time t value of the European option with the payoff Y = Y ( T ). Nevertheless, there are situations where the two value processes coincide. Proposition 5.1 Consider an American option Y A and the corresponding European option with time T value Y = Y ( T ) . If V ( t ) Y ( t ) for all t , then V ( t ) = V A ( t ) for all t , and it is optimal to wait until time T to exercise. Proof For the holder of an American option, exercising at t only ends up with payoff Y ( t ), while selling the corresponding European option (or shorting the portfolio which replicates the European option) would guarantee you a time t payoff V ( t ). Hence the option should not be exercised at t . Since t is arbitrary, it is optimal to wait until T to decide whether to exercise. Consider the American call option with Y ( t ) = ( S ( t ) - c ) + at each t where c = 2 . 05 in the example given in Lecture 2. Proposition 5.1 applies to this case. See Figure 5.1. Note: “American calls = European calls” Yes, the quote is really true, i.e. American calls have the same values as their European counterparts in the simple set-up given in this section, thus there should be no earlier exercises. This result will be proved in Section 5.2 following Theorem 5.1. Briefly, it is based on the fact that { Y * ( τ ) } is a submartingale. A stochastic process X = { X ( t ) , t = 0 , 1 , . . . , T } is called a Q - supermartingale under a proba- bility measure Q on Ω and with respect to FF , if the conditional expectation E Q ( X ( t ) | F t - 1 ) X ( t - 1) t = 1 , . . . , T ; 1

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V (0) = 0 . 28 Y (0) = 0 . 00 ' ' ' ' ' ' '* H H H H H H Hj V (1) = 0 . 31 Y (1) = 0 . 09 ' ' ' ' ' ' '* H H H H H H Hj V (1) = 0 . 04 Y (1) = 0 . 00 ' ' ' ' ' ' '* H H H H H H Hj V (2) = 0 . 36 Y (2) = 0 . 24 ' ' ' ' ' ' '* H H H H H H Hj V (2) = 0 . 05 Y (2) = 0 . 00 ' ' ' ' ' ' '* H H H H H H Hj V (2) = 0 . 00 Y (2) = 0 . 00 ' ' ' ' ' ' '* H H H H H H Hj V (3) = 0 . 40 Y (3) = 0 . 40 V (3) = 0 . 06 Y (3) = 0 . 06 V (3) = 0 . 00 Y (3) = 0 . 00 V (3) = 0 . 00 Y (3) = 0 . 00 Figure 5.1: Exercise at t or T ?
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