4 Option Futures and Other Derivatives. A contingent claim is a random variable that represents the time T payo from seller to buyer.
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1 4 Option Futures and Other Derivatives 4.1 Contingent Claims A contingent claim is a random variable that represents the time T payo from seller to buyer. The payo for a European call option with exercise price e expiring at T is C(T )=max[s(t ) e; 0] and for a put option P T = max[e S(T ); 0]. The famous put call parity is C T P T = S(T ) e, so that put and call have the relation C T = S T +P T e. Note that the relation hold for all = 0; 1;:::;T if r = 0. In the general case e=(1+r) T t = S t +P t C t. So that, e.g., C t = S t +P t e=(1+r) T t An Asian option depends on the whole price history such that and Asian call where C(T )=max[¹s e; 0] ¹S = 1 T TX t=1 S(t): A contingent claim X is said to be marketable or attainable if there exist a self- nancing strategy such that V T (!) =X(!). Thus the corresponding strategy H is replicating X. 77
2 It assumed that there exist a risk neutral probability measure Q. And the contingent claim X is assumed to be marketable. Then as in the single period model: (4.1) Risk neutral valuation principle: The time t value of a marketable contingent claim X is equal to V t, the time t value of the portfolio replicating X. Moreover, V t = V t =B t = E Q [X=B T jf t ];t=0; 1;:::;T for all risk neutral probability measures Q. V t is called the arbitrage price of the contingent claim X at time t. It can be easily demonstrated that if (4.1) does not hold there is immediately an arbitrage. Example 4.1! t =0 t =1 t =2! 1 S 0 =5 S 1 =8 S 2 =9! 2 S 0 =5 S 1 =8 S 2 =6! 3 S 0 =5 S 1 =4 S 2 =6! 4 S 0 =5 S 1 =4 S 2 =3 78
3 If r = 0 then (see example 3.3) Q(! 1 )=1=6, Q(! 2 )=1=12, Q(! 3 )=1=4, and Q(! 4 )=1=2. If the (European) call option with the exercise price e = 5 is attainable, then it value Vt C is V C 0 = E Q[X] =E Q (S2 5) + = =1 In states! 1 and! 2 V C 1 = E Q (S2 5) + js 1 =8 = =3 In states! 3 and! 4 V C 1 = E Q (S2 5) + js 1 =4 = =1 3 : From the put call parity we obtain further V P t V P 0 =1 5+5=1 = V C t S t + e V P 1 =3 8+5=0instates! 1;! 2 V P 1 = =11 3 in states! 3;! 4 Replicating trading strategies: Computing the replicating strategy veri es that the contingent claim is indeed attainable. There are several methods. One based on the predictability argument, another on self- nancing and a third using the discounted processes. 79
4 The predictability method: Basically the replicating strategy is solved from the linear equations V t = H 0 (t)b t + NX n=1 where H is predictable process. H n (t)s n (t) Example 4.1 (continued) For t =2 V 2 (! 1 )=4=H 0 (2;! 1 ) 1+H 1 (2;! 1 ) 9 V 2 (! 2 )=1=H 0 (2;! 2 ) 1+H 1 (2;! 2 ) 6 V 2 (! 3 )=1=H 0 (2;! 3 ) 1+H 1 (2;! 3 ) 6 V 2 (! 4 )=0=H 0 (2;! 4 ) 1+H 1 (2;! 4 ) 3 There are four equations and eight unknowns, so we need additional information. H is predictable, so H n (t +1) must be constant on each subset of the time t generating partition P t. Thus H 0 (2;! 1 ) = H 0 (2;! 2 ) H 0 (2;! 3 ) = H 0 (2;! 4 ) H 1 (2;! 1 ) = H 1 (2;! 2 ) H 1 (2;! 3 ) = H 1 (2;! 4 ) With this additional information there are four unknowns and four equations. Solving the equations gives H 0 (2) = 5 and H 1 (2)=1instated! 1 ;! 2,andH 0 (2) = 1 andh 1 (2) = 1=3 for states! 3 ;! 4. InthesamemannerwecansolvevaluesforH n (1). 80
5 The self- nancing approach: If we know only the value X of the contingent claim, then another method is to go backwards in time. Then V T = X, sothat V T = H 0 (T )B T + NX n=1 H n (T )S n (T ) H(T ) can be now solved as above. In the next step the self- nancing is utilized to give V T 1, such that V T 1 = H 0 (T )B T 1 = NX n=1 H n (T )S n (T 1): Having V T 1 we can solve H n (T 1) and H n (T 1) as above, and so on. 81
6 Discounted prices method: V t = V 0 + G t = V 0 + P N P tu=1 n=1 H n (u) S n (u) = V 0 + P N P t 1 n=1 u=1 H n(u) S n (u) + P N n=1 H n (t) S n(t) = V t 1 + P N n=1 H n (t) S n(t) Thus assuming V T = X=B T is known, and using again the self- nancing property of H, one can solve H and V T 1 from the equations V T 1(!) =V T 1 + NX H n (T;!) S n (T;!);! 2 : n=1 82
7 American Options The option can be exercised at any before T. So it has more rights than the corresponding European option. Consequently its value is at least that of the European. Example 4.8. Y t =(S n (t) e) + American call option. Y t =(e S n (t)) + American put option. Nevertheless, 4.5 Let Y = fy t ; t =0; 1;:::;Tg be an American option, and X thepayo ofthecorresponding European option at T such that X = Y T. If V t denotes the time t value of the European option. If V t Y t for all t and! 2, then,forallt, V t = Z t,the time t value of the American option, and it is optimal to wait until T to execise. 83
8 Example 4.1 (Continued) Y t =(S 5) + with r =0and! k t =0 t =1 t =2 Q! 1 S 0 =5 S 1 =8 S 2 =9 1! 2 S 0 =5 S 2 =8 S 2 =6! 3 S 0 =5 S 3 =4 S 2 =6! 4 S 0 =5 S 4 =4 S 2 = The value process V t for the European call option X = max(0;s 2 5) is as in the following Figure. S 0 =5 Y 0 =0 V 0 =1 µ S 1 =8 Y 1 =3 V 1 =3 R S 1 =4 Y 1 =0 V 1 = 1 3 S 2 =9 ³ ³ ³ ³ ³ ³1 Y 1 =4 V 1 =4 P P P P P Pq S 2 =6 Y 1 =1 V 1 =1 S 2 =6 ³ ³ ³ ³ ³ ³1 Y 1 =1 V 1 =1 P P P P P Pq S 2 =3 Y 1 =0 V 1 =0 Figure. are the same. Values of American and European options 84
9 Suppose next that the stock is paying dividend of $1 between periods 1 and 2. The exercise price is e =4. Then the American option and European call option value processes di er. S 0 =5 Y 0 =1 V 0 =1 Z 0 =1 1 4 µ S 1 =8 Y 1 =4 V 1 =3 Z 1 =4 R S 1 =4 Y 1 =0 V 1 = 1 3 Z 1 = 1 3 * H H H H H Hj * H H H H H Hj S 2 =9 Y 2 =4 V 2 =4 Z 2 =4 S 2 =6 Y 2 =1 V 2 =1 Z 2 =1 S 2 =6 Y 1 =1 V 1 =1 Z 1 =1 S 2 =3 Y 2 =0 V 2 =0 Z 2 =0 Figure. Values of American and European options are di erent. If the option is exercised just prior the exdividend date then its value is (provided the stock is in the money) S t X: If the stock is not exercised the stock price drops to S t D, i.e., by the amount of the dividend, D. 85
10 A lower bound to the option if it is not exercised at time t is then It follows that if i.e., S t D eb t =B T : S t D eb t =B T S t e D e [1 B t =B T ] ; it is not optimal to exercise the option. On the other hand if D>e[1 B t =B T ] ; it is optimal to exercise the option (not proved explicitly here). Example 4.1 (continued) In the example r = 0, i.e., B t =1forallt, so positive dividend makes an early exercise optimal (of course, provided the option has value). Note. In the case of an European option if the dividend is known and the payment period is known. Then the pre-dividend prices should be adjusted by the present value of the dividend in valuing the option. 86
11 Supermartingale Let Q be a martingale measure for the European option such that So V t = B t E Q [X=B T jf t ]: E Q [V t+1 =B t+1 jf t ] = E [E[X=B T jf t+1 ]jf t ] = E[X=B T jf t ]=V t =B t = V t ; i.e., the discounted value process of the European option is a martingale under Q. If Y t is the corresponding American option with Y T = X. Then Z t V t, and it turns out that E Q [Z t+1 =B t+1 jf t ] Z t =B t so that the value process of the American option is a supermartingale. 87
12 Stopping Times A stopping time is a random variable taking values in the set f0; 1;:::; T;1g, such that each event of the form f = tg, t T, is an element in algebra F t. I.e., f = tg = f! : (!) =tg 2F t. Think of as a time at which you decide to quit a gamble: the decision whether or not you quit at time t depends only on the history up to and including t not the future. For example, for a security with S 0 = 10, 1 = minft : S t 20g is a stopping time, whereas 2 =maxft : S t 20g is not, because the value of 2 is never learned until T. The in nity, 1, provides the possibility that the event of interest never occurs (during the observation period). Let ³(s; t) denote the set of random variables that are stopping times and take nite values in the closed interval [s; t]. 88
13 (4.6) Suppose here exists a risk neutral probability measure Q and de ne the adapted process Z = fz t ; t =0; 1;:::Tg by (7) Z t = max E Q[B t Y =B jf t ]: 2³(t;T ) Then the process Z=B is the smallest Q-supemaringale satisfying (8) Z t Y t ; for all t and!: Moreover, the stopping time (9) (t) =minfs t : Z s = Y s g maximizes the right hand side of (7) for t =0; 1;:::;T. We will return to the meaning of this result later. 89
14 The value process of the American option is solved backwards, and we use induction to verify (4.6). If t = T then (7), (8) and (9) clearly hold (since trivially ³(T;T) =[T; T], so Z T =max 2³(T;T) E Q [B T Y T =B T jf T ]=Y T, and hence (T )=minfs T : Z s = Y s g = T ). The next step is to compute Z T 1 by (10) Z T 1 =maxfy T 1 ;E Q [B T 1 Z T =B T jf T 1 ]g But, as observed at the previous Z T will be equal to Y T. Furthermore, at this stage, it will be optimal to take = T 1 if and only if Z T 1 = Y T 1 (otherwise it is optimal to exercise at T with Z T = Y T, and = T ). So (9) holds, and (7) can be rewritten for t = T 1 Z T 1 =maxfy T 1 ;E Q [B T 1 Y T =B T ]jf T 1 ]g; so that again (8) holds. Next apply induction, and assume statement (4.6) is true for Z t. Then for t 1 (11) Z t 1 =maxfy t 1 ;E Q [B t 1 Z t =B t jf t 1 ]g: 90
15 We see that (8) is true, because by the law of iterated expectations, we obtain Z t 1 = maxfy t 1 ; E Q [B t 1 max 2³(t;T) E Q [B t Y =B jf t ]=B t jf t 1 ]g (12) maxfy t 1 ;E Q [B t 1 E Q [B t Y =B jf t ]jf t 1 ]g = maxfy t 1 ;E Q [B t 1 Y =B jf t 1 ]g for all 2 ³(t; T ). So Z t 1 Y t 1 for all! implying (8) for t 1. On the other hand taking the stopping time in (9) for time t 1, we see that (12) becomes an equality. It follows that Z t 1 as has been computed satis es (7) for time t and that the stopping time given by (9) for time t 1 is the one that maximizes the right hand side of (7) for time t 1. 91
16 Example 4.1 (Continued) Starting with Z 2 = Y 2, for states! 1 and! 2 i.e., E Q [Z 2 jf 1 ] = E Q [Y 2 jf 1 ] = Y 2 (! 1 )Q(f! 1 gjf! 1 ;! 2 g) +Y 2 (! 2 )Q(f! 2 gjf! 1 ;! 2 g) E Q [Z 2 jf 1 ]=E Q [Y 2 jf 1 ]= =3 Hence Z 1 =maxfy 1 ;E Q [Z 2 jf 1 ]g =maxf4; 3g =4for! 1 and! 2. For! 3 and! 4 E Q [Z 2 jf 1 ]=1=3, so that Z 1 =maxf0; 1=3g =1=3. Finally, E Q [Z 1 jf 0 ]=E Q [Z 1 ]= =11 4 : Thus Z 0 =maxf1; g =
17 An European option X is marketable if there exist a self- nancing replicating portfolio V t such that V T = X. The time t value of X is V t. The American option Y is marketable or attainable if for each stopping time T there exists a self- nancing trading strategy with portfolio value V such that V = Y. Such a strategy is called replicating or hedging trading strategy. Example 4.1 (Continued) The stopping times of the form (9), i.e., (t) = minfs t : Y s = Z s g, to be checked are ½ 1;! =!1 ;! 2 (0;!)= (1;!)= 2;! =! 3 ;! 4 and (2;!)=2forall! 2. The trading strategies for each of the optimal stopping times can be checked the same manner as was done with the European option earlier. 93
18 Example 4.1 (continued) Now Y (2) = Y 2 is the same as the European call option X that was found marketable already earlier, and hence there is a trading strategy such that V (t) = Y (t) for t =2. For (0) = (1): 8 < Y = : 4;! =! 1 ;! 2 1;! =! 3 0;! =! 4 To check that there is a trading strategy such that V (t) = Y (t) for t = 0 and 1. Beginning with the largest value of H 0 (2) + 6H 1 (2) = Y 2 (! 3 )=1 H 0 (2) + 3H 1 (2) = Y 2 (! 4 )=0 which has the solution H 0 (2) = 1 andh 1 (2) = 1=3. The trading strategy is self- nancing, which implies V 1 =1=3 for! 3 ;! 4. Because (t) =minfs t : Z s = Y s g = 1 the values H 0 (2) and H 1 (2) do not really matter. Next we seek H 0 (1) and H 1 (1): H 0 (1) + 8H 1 (1) = Y 1 (!) =4;! 1 ;! 2 H 0 (1) + 4H 1 (1) = V 1 (!) =1=3;! 3 ;! 4 Solving these gives H 0 (1) = 10=3 and H 1 (1) = 11=12. Thus there exist a trading strategy satisfying V (t) = Y (t) for t = 0 or 1, and Y is a marketable American option. Note V 0 = 10=3 +(11=12) 5= = Z 0. 94
19 The following results give the basic properties of American options. (4:14) Suppose (i) there exists a risk neutral probability measure Q, (ii) the process Z is as in (7), (iii) the American option is marketable. Then Z is the value (price) process of Y,andthe optimal early exercise strategy is given by (0). (4:15) If Y is a marketable American option and Y=B is a Q-submartingale, then = T is always an optimal exercise strategy, and the price coincides with the price of the European option X = Y T. (4:16) In a world of non-negative interest rate and no dividends, an American call option written on individual risky security should not be exercised early. 95
20 Algorithm for calculating the value of an American option. At the rst step we calculate the values of the American option at each node starting from the m end values, m being determined by the number of intermediate steps in the time period from zero to exercise date Y m (j) =maxfe S 0 u j d m j ; 0g; j =0; 1;:::;m: For the following steps k = m 1;m 2;:::;0, Y k (j) = maxfe S 0 u j d k j ; qy k+1 (j)=b +(1 q) Y k+1 (j +1)=Bg; j =0; 1;:::;k Again there are several ready programmed procedures for example for Excel (e.g., Benniga, S. (1997). Financial Modeling. The MIT press, London, England). 96
21 Exercise 4.8 r = 0, American put option with e =6, Y t =(6 S t ) +, V t = E Q [Y t+1 jf t ]and Z t =maxfy t ;E Q [Z t+1 jf t ]g. Thus obviously at time t = 2 the option has only value when! =! 4. At t = 1 the put option has potential value only for f! 3 ;! 4 g. In that case E Q [Z 2 jf 1 ](!) = 0 if! 2 f! 1 ;! 2 g and E Q [Z 2 jf 1 g](!) =0 Q(f! 3 jf! 3 ;! 4 g)+3 Q(f! 4 jf! 3 ;! 4 g)=3 2 3 =2if! 2f! 3;! 4 g. In this case Z 1 =maxfy 1 ;E Q [Z 2 jf 1 ]g =maxf2; 2g =2. At t =0: Y 0 =6 5=1. V 0 = E Q [(6 S 2 ) + ]= 0 Q(! 1 )+0 Q(! 2 )+0 Q(! 3 )+3 Q(! 4 )= 1 2 3= E Q[Z 1 jf 0 ]=E Q [Z 1 ]=0 Q(f! 1 ;! 2 g)+2 Q(f! 3 ;! 4 g)=2 3 4 =11 2. Thus the optimal exercise time is =1. S 0 =5 Y 0 =1 V 0 =1 1 2 Z 0 =1 1 2 µ R S 1 =8 Y 1 =0 V 1 =0 Z 1 =0 S 1 =4 Y 1 =2 V 1 =2 Z 1 =2 * H H H H H Hj * H H H H H Hj S 2 =9 Y 2 =0 V 2 =0 Z 2 =0 S 2 =6 Y 2 =0 V 2 =0 Z 2 =0 S 2 =6 Y 1 =0 V 1 =0 Z 1 =0 S 2 =3 Y 2 =3 V 2 =3 Z 2 =3 97
22 Hedging: V = Y where V is the hedging portfolio. (0) = (1): H 0 (1) + 8H 1 (1) = Y 1 (! 1 )=Y 1 (! 2 )=0 H 0 (1) + 4H 1 (1) = Y 1 (! 3 )=Y 1 (! 4 )=2 from which H 0 (1) = 4 and H 1 (1) = 1=2, which gives the hedge portfolio. The value of the portfolio in the beginning is H 0 (1) +5H 1 (1) = 4+( 1=2) 5 = = Z 0. 98
23 Complete and Incomplete Markets The model is said to be complete if every contingent claim is marketable; otherwise, the model is incomplete. Here are just the main results: (4:17) The multiperiod model is complete if and only if every single period model is complete. (4:18) The multiperiod model is complete if and only if the risk neutral probability Q measure is unique. (4:19) The contingent claim X is marketable (i.e., attainable) if and only if E Q [X=B T ] isthesamefor every Q 2 IM, the set of all risk neutral probability measures (note that in incomplete markets there are several risk neutral probability measures). (4:20) If the model is complete then every American option is marketable. (4:21) The American option is marketable if and only if, for each stopping time in (9), E Q [Y =B ] takes the same value for all Q 2 IM. 99
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