Barrier options. In options only come into being if S t reaches B for some 0 t T, at which point they become an ordinary option.

Transcription

1 Barrier options A typical barrier option contract changes if the asset hits a specified level, the barrier. Barrier options are therefore path-dependent. Out options expire worthless if S t reaches the barrier value B before expiry, T. For a down-and-out call option, if S t B for some 0 t T then the option dies, if S t > B for 0 t T it survives and has the usual payoff max(s T K, 0). In options only come into being if S t reaches B for some 0 t T, at which point they become an ordinary option. This is equivalent to a payoff max(s T K, 0) 1 {mt >B}, where m T = min 0 t T {S t}. 1

2 Price paths with a down-and-out barrier S in K out S 0 B out t T 2

3 The down-and-out call If S t falls to B, the option dies and becomes worthless. If S t stays above B, it has the usual payoff max(s T K, 0). For the moment we ll assume that B < K. While the option remains in its price function, C do (S, t), satisfies the Black-Scholes equation, C do + 1 t 2 σ2 S 2 2 C do S 2 + (r y)s C do S r C do = 0. This only holds if the option hasn t knocked out, i.e., only if S u > B for all 0 u t. 3

4 If S t = B at any time 0 t T the option knocks out and becomes worthless, so C do (B, t) = 0. If the option is still alive at expiry then C do (S, T ) = max(s K, 0). Note that the barrier feature manifests itself as a boundary condition. If the option has already knocked out then C do (S t, t) = 0 no matter what the value of S t. 4

5 A binomial method can incorporate this path dependence: S S in in S 0 S 0 B B out out t T t T The barrier may hit grid points, which is good. Usually, however, it misses them all which is less good. 5

6 A reflection result for Black Scholes Suppose V (S, t) is a solution of the Black Scholes equation, V t σ2 S 2 2 V V + (r y)s S2 S for all S > 0 and t < T. r V = 0, (1) If B > 0 is a constant and then α = 1 2 (r y)/σ2 W (S, t) = (S/B) 2α V ( ) B 2 /S, t also satisfies the Black Scholes equation (1) for S > 0 and t < T. 6

7 There is a derivation of this result in the introductory notes on the Black Scholes equation, reproduced along with two alternative proofs in the Appendix of these notes. In general, B may be any constant, but in what follows we will take it to be the level of our barrier. If V (S, t) satisfies the Black Scholes equation for all S > 0 then so too does W (S, t) and so, by linearity, V (S, t) + A W (S, t) is also a solution of the Black Scholes equation for all S > 0 for any constant A. 7

8 An explicit solution for a down-and-out call The down-and-out call s price function, C do (S, t), satisfies the Black Scholes equation C do + 1 t 2 σ2 S 2 2 C do S 2 + (r y) S C do S r C do = 0, (2) for t < T and S > B, has payoff and vanishes on the barrier B, C do (S, T ) = max(s K, 0), S B (3) C do (B, t) = 0, t T. (4) Recall that we only need to find C do (S, t) for S > B. Let us assume that B < K. 8

9 Let C bs be the price of a vanilla version of our down-and-out call, i.e., a vanilla call on the same asset with the same strike and expiry. Its price function, C bs (S, t), is necessarily a solution of the Black Scholes equation (2), for all S > 0 and t < T. We look for a solution of the form C do (S, t) = C bs (S, t) W (S, t). As the Black Scholes equation is linear, W (S, t) must also be a solution of (2). The barrier condition (4) is satisfied if W (B, t) = C bs (B, t). 9

10 The obvious candidate is W (S, t) = (S/B) 2α C bs ( B 2 /S, t ). The reflection principle assures us that this satisfies (2) if C bs (S, t) does, and setting S = B shows W (B, t) = C bs (B, t). The only remaining condition is (3), the payoff condition C do (S, T ) = C bs (S, T ) W (S, t) = max(s K, 0), and this requires that W (S, T ) = 0, S > B. 10

11 To see that this is indeed the case we argue as follows. If 0 < B < S then 0 < 1/S < 1/B. Multiplying by B 2 > 0 shows B 2 /S < B. Our assumption that B < K shows that S > B = B 2 /S < B < K. Thus if S > B then (S/B) 2α C bs (B 2 /S, T ) = (S/B) 2α max ( B 2 /S K, 0 ) = 0, since B 2 /S K < 0 and so max(b 2 /S K, 0) = 0. This establishes that the payoff condition (3) is indeed satisfied. 11

12 Call payoff and its reflection Payoff Reflection < B Reflection > B payoffs B 2 /K B K S 12

13 Thus, if the option has not yet knocked out its value for S > B is given by C do (S, t) = C bs (S, t) (S/B) 2α C bs (B 2 /S, t). For S B the option has knocked out and C do (S, t) = 0. Provided the option is alive and S > B, its delta is found using do (S, t) = bs (S, t) + (S/B) 2α 2 bs (B 2 /S, t) where 2α (S 2α 1 /B 2α ) C bs (B 2 /S, t), bs = C bs S. 13

14 While the option is alive and S > B, its gamma is given by Γ do (S, t) = Γ bs (S, t) (S/B) 2α 4 Γ bs (B 2 /S, t) + 2 (2α 1) (S 2α 3 /B 2α 2 ) bs (B 2 /S, t) where 2α (2α 1) (S 2α 2 /B 2α ) C bs (B 2 /S, t), Γ bs = 2 C bs S 2. Similar results apply to the other greeks. 14

15 C d/o against S with B = 80, K = 100, T t = 1.25, r = 0.05, q = 0 and σ =

16 d/o against S with B = 80, K = 100, T t = 1.25, r = 0.05, q = 0 and σ =

17 Γ d/o against S with B = 80, K = 100, T t = 1.25, r = 0.05, q = 0 and σ =

18 ϕ d/o against S with B = 80, K = 100, T t = 1.25, r = 0.05, q = 0 and σ =

19 Down-and-out calls: barrier above the strike Call payoff and its reflection Payoff Reflection < B Reflection > B payoffs K B B 2 /K S 19

20 If the barrier lies above the strike, B > K our trick above fails, because there are some values of S > B where C bs (B 2 /S, T ) 0. We want the reflected solution to vanish at expiry for S > B. So before reflecting, value an option with a truncated call payoff which is zero for S < B we don t care about the payoff for S < B because we aren t pricing the option there. This is the payoff for a vanilla call struck at B plus (B K) standard digital calls, each paying either 0 or 1, and also struck at B. More generally if we find the reflected vanilla option price with payoff P o (S) does not result in the correct payoff for the barrier option, the trick is to work with the price of a new vanilla option with the truncated payoff P o (S) 1 {S>B}. 20

21 truncated payoff strike B call digital call payoffs B K K B S 21

22 Modified call payoff and its reflection Payoff Reflection < B Reflection > B payoffs K B B 2 /K S 22

23 The Black Scholes value of a down-and-out call which hasn t already knocked out is therefore C do (S, t) = C bs (S, t; B) + (B K)C d (S, t; B) for S > B. ( S B ) 2α ( Cbs(B 2 /S, t; B) + (B K)C d (B 2 /S, t; B) ) The first two terms are the unreflected solution, which takes care of the payoff. The second two are the reflection and take care of the barrier condition. From this we may compute formulae for the other Greeks; they are lengthy. 23

24 C d/o as a function of S with K = 100, B = 120, r = 0.05, q = 0, σ = 0.3 and T t = 1.25 (left) and T t = (right). 24

25 d/o as a function of S with K = 100, B = 120, r = 0.05, q = 0, σ = 0.3 and T t = 1.25 (left) and T t = (right). 25

26 Γ d/o as a function of S with K = 100, B = 120, r = 0.05, q = 0, σ = 0.3 and T t = 1.25 (left) and T t = (right). 26

27 vega d/o as a function of S with K = 100, B = 120, r = 0.05, q = 0, σ = 0.3 and T t = 1.25 (left) and T t = (right). 27

28 When B > K we find that do becomes singular as t T and S B. This makes delta-hedging the option impractical in this limit. 28

29 Other barrier options Up-and-out call: this is superficially similar to a down-and-out call, except that it knocks out when the asset rises to B. Its risk characteristics, however, are completely different as its value increases with S but then falls sharply as the barrier is approached. Down-and-in call : this expires worthless unless S t falls to B, in which case it turns into a vanilla call with strike K. The latter is worth C bs (B, t; K) at exchange. The corresponding put options have obvious definitions. 30

30 Up-and-out call To deal with this option, note that we must have B > K or the option is worthless. The up-and-out option has the same payoff as the vanilla option for S < B, but we have to truncate this payoff for S > B if we wish to use a reflection argument. Therefore, let V bs (S, t) be the price of an option with the payoff V bs (S, T ) = This payoff may be written as 0 if S K, S K if K < S B, 0 if S > B. max(s K, 0) max(s B, 0) (B K) 1 {B>S}. 31

31 Payoff to reflect for an up-and-out call up-and-out cal l strike K call strike B call digital calls payoffs B K K B S 32

32 We can write this as V bs (S, T ) = C bs (S, T ; K) C bs (S, T ; B) (B K) C d (S, T ; B), so we have V bs (S, t) = C bs (S, t; K) C bs (S, t; B) (B K) C d (S, t; B). If S < B then B 2 /S > B, hence V bs (B 2 /S, T ) = 0, and, as usual, at S = B ( S B ) 2α V bs(b 2 /S, t) = V bs(b, t). 33

33 The value of an up-and-out call which has not yet knocked out is C uo (S, t) = V bs (S, t) ( S B ) 2α Vbs (B 2 /S, t) for S < B. = C bs (S, t; K) ( ) ( ) S 2α B Cbs B 2 S, t; K (C bs (S, t; B) ( ) ( )) S 2α B Cbs B 2 S, t; B (B K) (C d (S, t; B) ( ) ( S 2α B Cd B 2 S ),, t; B ) The singularity in near S = B as t T means it is impractical to delta-hedge the option in this limit. 34

34 payoff T t =0.10 T t =0.50 T t =1.00 Up-and-out call prices 140 C u/ o K = B = r =0.10 y =0.05 σ = S 35

35 Out-in parity Without rebates, for any barrier options with just one barrier, (down-and-out) + (down-and-in) = vanilla, because whether the asset hits the barrier or not, one and only one of the barrier options is exercised, the other expiring worthless. So V di (S, t) + V do (S, t) = V bs (S, t). This applies to up options, with the obvious modification, as well. 36

36 down-out cal l down-in call vanilla call Out-in parity for barrier call prices price K = B = r =0.10 y =0.00 σ =0.25 T = S 37

37 Forward-start barrier options Many barrier options have a forward-start condition. At time t = 0, it is agreed that at an intermediate time T 1 the holder will receive a barrier contract with later expiry T > T 1. The strike and barrier are set by reference to the asset value at T 1. For example, a six-month down-and-out call, starting in three months, with the strike to be set at-the-money (i.e., at the spot price three months from initiation) and the barrier to be set at 90% of the strike. Here T 1 is three months and T is nine months. There is no barrier for the first three months; for the following six months it is a regular down-and-out call with B = 0.9 K. 38

38 These options are very like forward-start vanilla options. To value them, first work back to the intermediate date T 1, then use the values at that time as an intermediate payoff. We ll assume that the barrier is continuously monitored. So, with the example above, the barrier option is at-the-money at time T 1, and the barrier is set to K = S(T 1 ), B = c S(T 1 ), where 0 < c < 1 in our case c =

39 At time T 1, set K = S and B = c S in the earlier formula for a downand-out call, C fs do (S, T 1) = C do (S, T 1 ; K = S, B = c S, T ) = C bs (S, T 1 ; S, T ) (S/c S) 2α C bs ((c S) 2 /S, T 1 ; S, T ) = A do (T 1, T ) S where A do (T 1, T ) is easily worked out and does not depend on S or t. For earlier times, t < T 1, C fs do(s, t) = A do (T 1, T ) S e y(t 1 t). Note that this procedure also works if different constant volatilities are used for each leg of the contract, as is sometimes done for forwardstart vanilla options. 40

40 Appendix: three proofs of the reflection principle We present three proofs of the reflection principle, which asserts that if V (S, t) is a solution of the Black Scholes equation then so too is where α = 1 2 (r y)/σ2. W (S, t) = (S/B) 2α V (B 2 /S, t), We also note that this change of variables is its own inverse, V (S, t) = (S/B) 2α W (B 2 /S, t), which is one reason for the term reflection principle. The proof of this fact is left as an exercise. We now proceed with the proofs of the principle. 41

41 Proof by reduction to a diffusion equation We start with the Black Scholes equation V t σ2 S 2 2 V V + (r y) S S2 S r V = 0 and introduce the change of variables x = log(s/b), τ = σ 2 (T t), V (S, t) = e αx βτ u(x, τ), where α = 1 2 (r y)/σ2 and β = 1 2 α2 r/σ 2. Note that ˆL τ = e αw τ βτ = e rt e αw τ α 2 τ/2 = e rt L τ is essentially a discounted change of measure. 42

42 We find that u(x, τ) satisfies the diffusion equation u τ = 1 2 u 2 x 2, which has the property that if u(x, τ) is a solution then so too is u( x, τ). The map x x is a reflection in the origin, which is another reason for the term reflection principle. Therefore, we get two solutions of the Black Scholes equation for the price of one! These are V (S, t) = e αx βτ u(x, τ), W (S, t) = e αx βτ u( x, τ). This is easily verified by setting z = x and using the chain rule. 43

43 Now observe that and that W (S, t) = e 2αx ( e αx βτ u( x, τ) ), e αx βτ u( x, τ) is simply the expression for V (S, t) with all of the x s replaced by x. Since x = log(s/b), x = log (B/S) = log ( ) B 2/ B and so, to get from x to x, all we need do is replace S by B 2 /S in the expression x = log(s/b). S 44

44 Thus e αx βτ u( x, τ) = V (B 2 /S, t), and hence we have the reflection principle, W (S, t) = e 2αx V (B 2 /S, t) = ( S B ) 2α V (B 2 /S, t). We leave it as an exercise to interpret this in terms of the risk-neutral processes ds t S t = (r y)dt + σ dw t and dx t = dw t. [Hint: it comes down to showing the two processes x + t and x t with dx + t = dw t, dx t = dw t and x + 0 = x 0 = x 0 have the same terminal distributions.] 45

45 Proof by brute force Suppose that Set V t σ2 S 2 2 V V + (r y) S S2 S r V = 0. W (S, t) = S β V (ξ, t), ξ = B2 S, where β is an as-yet-undetermined constant. The chain rule shows that S = ξ S ξ. 46

46 Using the chain and product rules, we find that W t = S β V t, W S = Sβ 1 2 W S 2 = S β 2 ( ( β V ξ V ξ ), β(β 1) V 2(β 1) ξ V ξ + ξ2 2 V ξ 2 ). We now try to choose β so that W satisfies the same Black Scholes equation as V, W t σ2 S 2 2 W S 2 + (r y) S 2 W S 2 r W = 0. 47

47 This implies that we must have V t σ2 ξ 2 2 V ξ 2 + ( ) (β 1)σ 2 + r y ξ V ξ ( 1 2 β(β 1)σ 2 + β(r y) r which will be the case if we can find β so that ) V = 0, (β 1)σ 2 r y = r y, 1 2 β(β 1)σ2 + β(r y) r = r, as this reduces the expression above to which is certainly true. V t σ2 ξ 2 2 V V + (r y) ξ ξ2 ξ r V = 0 48

48 Fortunately, these two equations for β are not independent. both reduce to They which implies that (β 1)σ 2 = 2(r y), 2(r y) β = 1 σ 2. Setting β = 2α gives the reflection formula that we have used in the main text of these lectures, namely, with W (S, t) = S 2α V (B 2 /S, t) α = 1 2 (r y)/σ2. 49

49 Proof by probability The Gaussian shift theorem says that if Z is drawn from an N(0, 1) normal distribution, F is a function and c is a constant, then E[ e cz F (Z) ] = 1 2π = ec2 /2 2π = ec2 /2 2π F (Z) e Z2 /2+cZ dz F (Z) e (Z c)2 /2 dz F (Y + c) e Y 2 /2 dy = e c2 /2 E[ F (Z + c) ], provided, of course, that the expectations involved exist. 50 (5)

50 Under the usual Black Scholes assumptions we have V (S, t) = e rτ E Q [ t Po (S T ) S t = S ], where S T = S t exp ( ξ + η Z ), (6) Note that τ = T t, ξ = (r y 1 2 σ2 ) τ, η = σ τ, Z N(0, 1). β = 2α = 1 2(r y)/σ 2 = 2ξ/η 2. (7) We may write (6) in the form V (S, t) = e rτ E[ Po ( S e ξ+η Z ) ]. 51

51 Therefore, we can write W (S, t) = ( ) S βv ( B 2 ) B S, t = [ ( ) e rτ S βpo ( B 2 E B = [ ( e rτ e βξ E e βηz Se ξ ηz B S eξ+ηz ) ] ) βpo ( B 2 = e rτ+β(ξ+βη2 /2) E [ ( Se ξ η(z+βη) = e rτ E B [ ( Se ξ βη 2 ηz) ) βpo ( B 2 B S eξ+ηz ) ] ) βpo ( B 2 S S eξ+βη2 +η Z since β = 2ξ/η 2 implies that β (ξ + β η 2 /2) = 0. eξ+η (Z+βη) ) ], ) ] 52

52 Furthermore, β = 2ξ/η 2 also implies that ξ βη 2 = ξ, ξ + βη 2 = ξ and, since Z N(0, 1) if and only if Z N(0, 1), we have W (S, t) = [ ( e rτ S e ξ+ηz E B = [ ( e rτ S e ξ+ηz E B which is equivalent to W (S, t) = e rτ E Q t ) βpo ( B 2 ) βpo ( [ ( ) ST βpo ( B 2 B S T S e ξ ηz B 2 S e ξ+ηz ) S t = S ) ], ]. ) ] 53

53 Therefore we have W (S, t) = e rτ E Q t [ F (ST ) S t = S ] which satisfies the Black Scholes equation and where the the payoff function is F (S T ) = (S T /B) β P o ( B 2 /S T ). 54