A Barrier Version of the Russian Option

Transcription

1 A Barrier Version of the Russian Option L. A. Shepp, A. N. Shiryaev, A. Sulem Rutgers University; Steklov Mathematical Institute; INRIA- Rocquencourt; March 5, 2002 Abstract For geometrical Brownian motion we consider the problem of finding the optimal stopping time and the value function for a Russion (put) option, assuming that the decision about stopping should be taken before the process of prices reaches dangerous a barrier on the level ε > 0. 1 Introduction 1. In the stochastic financial mathematics there are two classical models which form a basis for description of dynamics of the financial indexes (say, stocks). The first one, the Bachelier model, assumes that the stocks prices X = (X t ) t 0 are described as follows: X t = x + µt + σb t, (1) where x, µ R, σ R + and B = (B t ) t 0 is a standard Brownian motion. The second one, more realistic, (Samuelson, Black, Merton, Scholes) assumes that the process of prices X = (X t ) t 0 is a geometrical (economical) Brownian motion: } X t = x exp {(µ )t 12 σ2 + σb t. (2) 1

2 By Ito s formula or, symbolically, X t = x + t 0 µx s ds + t 0 σx s db s (3) dx t = X t (µ dt + σ db t ), X 0 = x. (4) Note that in the Bachelier model (1) the values X t R, but in the more realistic model (2) the values of X t R In [1-[3, some problems were considered in connection with options of American type. These papers give the explicit formulae for fair (rational) price of the Russian options (put and call) and also expressions for the optimal time of the decision about buying or selling the corresponding options. For the Bachelier model in the case of the put option the problem is to find the fair (rational) price V (x) = sup E τ M [ max X u cτ, (5) u τ where M is the class of all stopping times with respect to the Brownian filtration (Ft B) t 0, Ft B = σ(b s, s t), and to find also the corresponding optimal stopping time τ. The financial meaning of the expression max u τ X u cτ is clear: if the buyer of the option takes a decision at time τ, then he obtains the maximum of the values of X u on the time interval [0, τ (that is, max u τ X u ) taking into account a linear inflation ( cτ). The problem dual to (5) is to find the fair (rational) price of the call option [ V (x) = inf E min X u + cτ, (6) τ M u τ and the corresponding optimal stopping time τ. An interpretation of the expression min u τ X u + cτ is also clear: the buyer has a right to buy the stock at minimum past price (min u τ X u ) but with linear discount (cτ). Let us indicate the connection between formulations (5) and (6) and the standard (X K) + - and (K X) + -formulations. 2

3 Consider the American call option with payoff function f τ = ( X τ K (X, τ) ) +, where K (X, τ) = min u τ X u + rτ ( minimum price in discounted dollars ). If τ is such that X τ K (X, τ) 0 then (if EB τ = 0) sup τ E(X τ K (X, τ)) + = sup τ E(X τ [ K (X, τ)) = x + sup τ E σb τ + µτ min u τ X u rτ = x + sup τ E [(µ r)τ min u τ X u [ = x inf τ E min u τ X u + cτ, where c = r µ (r > 0 if and only if c > µ). Similarly, for a put option with K (X, τ) = max u τ X u rτ we find that ( sup K (X, τ) X τ ) ) [ + = x + sup E max X u cτ, τ τ u τ where c = r + µ (r > 0 if and only if c > µ). So, we see that problems (5) and (6) can be rewritten in the standard (K X) + - and X K) + -forms. From [1, it follows that if c > µ then for problem (5) an optimal stopping time is { } τ = inf t : max X u = X t + θ, (7) u t where θ is some positive constant. (For the case c µ see [1). For problem (6) in the case c > µ we have from [3 that { } τ = inf t : min X u = X t θ, (8) u t where θ is a positive constant. (For the case c µ see again [3.) 3. Let us denote T a = inf{t : X t = a}, a R, (9) and put [ Vε (x) = sup E max X u cτ, (10) τ M ε u τ 3

4 where Correspondingly, denote where M ε = {τ M : τ T ε }. (11) V ε (x) = inf E τ M ε [ min X u + cτ, (12) u τ M ε = {τ M : τ T ε }. (13) Comparing (5) and (10) we see that if in (5) there is no restrictions on the class of admissible stopping times τ, in the case of (10) we admit only stopping times τ M ε which are less or equal to the time of the first reach of the barrier on the level ε, that is, τ T ε = inf{t : X t = ε}. Financial and economical meaning of the problems (10) and (12) is clear. In the case of (10) we must take decision before default time T ε after which all economic activity is stopped. Similarly, we may give the corresponding interpretation of problem (12). We formulate now the structure of the optimal stopping times τε and τ ε. In the first case, τε = inf { t : X t gε(s t ) }, where S t = max u t X u and In the second case, where M t = min u t X t and g ε (s) = max{ ε, s θ }. τ ε = inf { t : X t g ε (M t ) }, g ε (m) = min{ε, m + θ }. These results can be obtained by the method which we apply below for a more realistic case of the geometrical Brownian motion. Our main aim here is to consider the barrier put version of the Russian option for ε > 0. For completeness we shall give also the corresponding results for the case (5), when ε = 0. It will make more clear our approach to the corresponding barrier version (ε > 0). 4

5 2 Russian put option. 1. We consider a (R, S)-market, where a bank account R = (R t ) t 0 satisfies dr t = rr t dt, R 0 = 1, that is, R t = e rt, t 0. General Arbitrage theory ([4, Chapters 7 and 8) claims that (R, X)-market with geometrical Brownian motion X = (X t ) t 0, given by (4), is arbitrage free and for finding the rational (fair) price and optimal stopping times we need to make all calculations with respect to the Wiener measure taking µ = r. Let us denote where V (x, s) = sup E x,s e (λ+r)τ S τ, (14) τ M S t = max u t X u s, and E x,s is the expectation under the assumption X 0 = x and S 0 = s, s x > 0. The term e (λ+r)τ in (14) plays the role of the exponential inflation (see [5, where inflation was linear). Theorem 2.1 (see [1,[2). In problem (14), the optimal stopping time is given by where g (s) = s/θ with τ = inf { t : X t g (S t ) }, (15) ( 1 θ 1/γ1 = 1 1/γ 2 ) 1/(γ2 γ 1 ) (16) and γ 1 (< 0) and γ 2 (> 1), γ k = A 2 + ( 1)k (A 2 ) 2 + B, k = 1, 2, (17) are roots of the equation γ 2 Aγ B = 0, 5

6 s s = θ x C(θ ) x = y 0 x Figure 1: Optimal stopping area and continuation region for the Russian put option with A = 1 2r σ 2, B = 2(λ+r). Moreover, the value function (14) is given by σ 2 ( ) γ1 ( ) γ2 x γ 2 γ 1, g (s) x s [ s γ 2 γ 1 x g (s) V (x, s) = g (s) s, x g (s). (18) Proof : From euristic consideration (and also from the general theory of optimal stopping for Markov processes), we may conclude that one should search optimal stopping in the class of stopping times of the form τ(θ) = inf{t : X t S t /θ}, (19) with θ > 1. Hence, the problem of finding an optimal stopping time τ is reduced to the problem of finding the optimal value θ such that sup E x,s e (λ+r)τ(θ) S τ(θ) = E x,s e (λ+r)τ(θ ) S τ(θ ). (20) θ>1 Note that after finding the value θ, it is necessary to check that the time τ(θ ) is optimal in the class M (not only in the class of stopping times {τ(θ), θ > 1}). Below we shall do it by using a standard technique of the verification theorems based on the Ito Meyer formula. Consider the space of states E = {(x, s) : 0 < x s} in which the two-dimensional Markov process (X t, S t ) t 0 takes its values. For θ > 1 we 6

7 represent E as follows: E = C(θ) D(θ), where is the stopping area and D(θ) = {(x, s) E : x s/θ} C(θ) = {(x, s) E : s/θ < x < s} is the continuation region. Suppose that g θ (s) = s/θ, s > 0, is a boundary between C(θ) and D(θ) (assuming that the points (s, g θ (s)) D(θ)). For search of the optimal θ and the value V (x, s) we consider an auxiliary Stefan (free boundary) problem with unknown functions g θ (s) and V = V (x, s): V (x, s) = s, x g θ (s), (21) with L X V (x, s) = (λ + r)v (x, s), g θ (s) < x < s, (22) and additional conditions on the boundary g θ (s): L X = rx x σ2 x 2 2 x 2, (23) lim V (x, s) = s, (24) x g θ (s) V (x, s) lim x g θ (s) x and on the diagonal {(x, s) : 0 < x = s}: V (x, s) lim x s s = 0, (25) = 0. (26) Condition (21) means that in the stopping area D(θ) the value V (x, s) should be equal to s, which is natural since for τ = 0 ( immediate stopping ), 7

8 E x,s e (λ+r)τ S τ = s. Condition (22) means that V (x, s) should satisfy the backward Kolmogorov equation (in x) which is also natural because the functionals E x,s e (λ+r)τ(θ) S τ(θ) for g θ (s) < x < s satisfy (in x) equation (22). Condition (24) is simply the continuity condition of the function V (x, s) on the boundary (then x g θ (s)). Condition (25) is the pasting or smooth fit condition. A priori this condition is not very evident, but from an intuitive point of view, it is supported by the idea that the value function should be sufficiently smooth. Finally, condition (26) is a condition of normal reflection on the diagonal {(x, s) E : 0 < x = s} that follows from the behaviour of the two-dimensional process (X t, S t ) t 0 as a Markov process with reflection. 2. Consider equation (22): rx V x σ2 x 2 2 V = (λ + r)v, (27) x2 for a fixed s > 0 and x (g θ (s), s). General solutions of this equation have the form V (x, s) = B 1 (s)x γ 1 + B 2 (s)x γ 2, (28) where B 1 (s) and B 2 (s) are some constants depending on s, and γ 1 and γ 2 are defined in (17). From (24), (25) we find that B 1 (s)( s θ )γ 1 + B 2 (s)( s θ )γ 2 = s, γ 1 B 1 (s)( s θ )γ γ 2 B 2 (s)( s θ )γ 2 1 = 0. and thus B 1 (s) = sγ 2 γ 2 γ 1 B 2 (s) = sγ 1 γ 2 γ 1 ( ) γ1 θ, (29) s ( ) γ2 θ. (30) s Finally, from (26) under assumption of differentiability of B 1 (s) and B 2 (s) we get B 1(s)s γ 1 + B 2(s)s γ 2 = 0. (31) 8

9 From (29), (30) and (31) we obtain the values θ and V = V (x, s) (denote them θ and Ṽ (x, s)) defined by the right-hand sides of (16) and (18). Now we must prove that the solution of the Stefan problem gives, in fact, the solution to the optimal stopping problem (14), that is, V (x, s) = Ṽ (x, s). For that it is sufficient to prove that (a) for each τ M, (b) for the stopping time we have Indeed, E x,s e (λ+r)τ S τ Ṽ (x, s), (32) τ = inf{t 0 : X t g θ(s t )}, (33) E x,s e (λ+r) τ S τ = Ṽ (x, s). (34) V (x, s) = sup E x,s e (λ+r)τ S τ Ṽ (x, s), τ M and because P x,s ( τ < ) = 1 for all (x, s) E, then from (34) we get that τ is an optimal stopping time in M. (From here it is clear, of course, that θ = θ and g θ(s) = g θ (s) = g (s) = s/θ.) The scheme of the proof of the properties (a) and (b) is the following. Let us take the function Ṽ (x, s) defined by the right-hand side of (18). This function has two continuous first derivatives V e x derivative 2 V e x 2 and e V s. The second is also continuous in (x, s) except for the points on the line x = s/ θ, s > 0. Hence, for each s, the function Ṽ (x, s) has continuous second derivative in x except at the point x = s/ θ. For x < s/ θ, the function Ṽ (x, s) = s, and for x > s/ θ this function has a positive derivative 2 V e. So, this function is convex (in x). This property x 2 allows us to apply the Ito Meyer formula (see, for example, [5, Chapter V or [6, Chapter IV) to the process ( e (λ+r)t Ṽ (X t, S t ) ) which gives (with t 0 left-side version of the second derivatives in the area D( θ)) e (λ+r)t Ṽ (X t, S t ) = Ṽ (x, s) + t [L 0 e (λ+r)u X Ṽ (X u, S u ) (λ + r)ṽ (X u, S u ) + t 0 e (λ+r)u V e ds s u + t 0 e (λ+r)u V σx e u db x u. du (35) 9

10 Observe that in (35) the integral on ds u is equal to zero because on the diagonal {(x, s) E : x = s}, where the process S can only increase, the derivative ev = 0. In the area C( θ) of continuation of observation, we s have L X Ṽ (x, s) (λ + r)ṽ (x, s) = 0, (36) and in the area D( θ) of stopping, we have L X Ṽ (x, s) (λ + r)ṽ (x, s) 0. (37) Denoting by M τ (P x,s -a.s.) the last integral in (35) we find from (35) (37) that e (λ+r)t Ṽ (X t, S t ) Ṽ (x, s) + M t. (38) The process M = ( M t ) t 0 is a local martingale. If (σ n ) n 1 is a localizing sequence and τ is a stopping time of M then E x,s e (λ+r)(τ σn) Ṽ (X τ σn, S τ σn ) Ṽ (x, s) (39) because E x,s Mτ σn = 0. We have also s Ṽ (x, s). Therefore, from (39) E x,s e (λ+r)(τ σn) S τ σn Ṽ (x, s). (40) Taking n and applying Fatou s lemma to (40) we find that E x,s e (λ+r)τ S τ Ṽ (x, s) and hence V (x, s) Ṽ (x, s). Now let us prove the validity of (34). In [1 it was shown that the stopping time τ (= τ ) M. Also from [1 it follows that the family { Mt τ, t 0 } is uniformly integrable. This together with property E x,s Mt σn τ = 0 gives (when t, n ) that E x,s M τ = 0. Taking into account that in the area C(θ) of continuation of observations, equality (36) does hold we get E x,s e (λ+r) τ Ṽ (X τ, S τ ) = Ṽ (x, s). But Ṽ (X τ, S τ ) = S τ. Therefore, equality (34) holds. This completes the proof of the theorem. 10

11 3 Russian Put option with Barrier Described in Section 1, the barrier version has a simple financial and economical interpretation being also of interest for other models. For our case we give explicit solution which shows some new effects which can appear in similar problems. We denote by M ε = {τ : τ T ε } the class of the stopping times less or equal to T ε = inf{t 0 : X t ε}, ε > 0, that is, the class of first times when the process X reaches the barrier on the level ε. We put Vε (x, s) = sup E x,s e (λ+r)τ S τ. (41) τ M ε It is clear that V ε (x, s) can be defined also as From these formulae Vε (x, s) = sup E x,s e (λ+r)τ S τ I(τ T ε ). (42) τ M V ε (x, s) V (x, s). (43) From a financial point of view, the event {X t < ε} may be interpreted as a state of a default in time t. Hence, if we take τ in M ε, it means that a decision on stopping should be taken before time of default. Since T ε = inf{t : X t ε}, it is clear that for all 0 < x < ε, the optimal stopping time is τε = 0 and in this (noninteresting) case, Vε (x, s) = s for all s x. For this reason in the sequel we assume that (x, s) E ε, where Put E ε = {(x, s) E : ε x s}. (44) and define the following four sets (see Figure 2) g ε (s) = max(ε, s/θ ), (45) D I := { (x, s) E ε : x = ε, ε s εθ }, D II := { (x, s) E ε : ε x s/θ, s > εθ } 11

12 and C I := { (x, s) E ε : ε < x s, ε < s < εθ }, C II := { (x, s) E ε : s/θ < x s, s εθ }. Theorem 3.1 An optimal stopping time for problem (41) (for (x, s) E ε ) is or, equivalently, τ ε = inf { t : X t g ε (S t) }, (46) τ ε = inf { t : (X t, S t ) D I D II}. (47) Moreover, the value function (41) is given by where V is given in (18) and with U ε (x, s) = V ε (x, s) = { Uε (x, s), (x, s) C I D I, V (x, s), (x, s) C II D II. (48) ( ) { γ1 x ε [ β log θ } e (y+1)t ε β β log(s/ε) e t 1 dt + θ γ 2 (49) ( ) { γ2 [ x ε β log θ + ε β β log(s/ε) } e yt e t 1 dt + θ γ 1. (50) β = γ 2 γ 1, and y = β 1. (51) Proof : In the area D := DI D II, we have V ε(x, s) = s and τε = 0. It is clear that if x ε and s εθ, the optimal time τ (for problem (14)) satisfies the property τ T ε, that is, τ M ε, and because Vε (x, s) V (x, s) we see that the optimal time τε in M ε can be taken equal to τ. Hence, if (x, s) CII then τ ε = τ. Consider now on the triangular area C I D I = {(x, s) : ε < x < εθ, ε < s < εθ } {(x, s) : x = ε, ε s < εθ }. 12

13 εθ s D I D II C I s = θ x CII x = y 0 ε x Figure 2: Optimal stopping area and continuation region for the Russian put option with barrier If the random walk (X t, S t ) t 0 starts in a point (x, s) CI, then there are two possibilities: 1) the random walk is terminated (in time T ε ) on the boundary DI ; or 2) after reaching the point (εθ, εθ ) and penetrating in the area CII, it will be terminated when the random walk reaches the stopping boundary {(x, s) : x = s/θ } E ε separating areas DII and C II. If (x, s) D I, then T ε = 0 and for such points the stopping time τε = 0 is optimal in M ε. Consider now more carefully the case when the initial point (x, s) CI. Let τ (εθ, εθ ) = inf { t 0 : (X t, S t ) = (εθ, εθ ) } be { the time of the } first reaching of the corner point (εθ, εθ ). The event Tε τ (εθ, εθ ) means that the random walk leaves the triangular area through the boundary DI. The event { } τ (εθ, εθ ) < T ε means that departure from CI takes place through the corner point (εθ, εθ ) and then the time τε will be the sum of the time τ (εθ, εθ ) and the time of exit from the point (εθ, εθ ) to the boundary {(x, s) : x = s/θ } E ε. These considerations show how to define the function V ε (x, s) = E x,s e (λ+r)τ ε Sτ ε for the points (x, s) in CI. Consider the following auxiliary problem: find the function U ε = U ε (x, s) such that U ε (x, s) = s, if x = s, ε < s < εθ, (52) 13

14 L X U ε (x, s) = (λ + r)u ε (x, s), if ε < x < εθ, (53) U ε lim x s s (x, s) = 0, if ε < x < εθ. (54) Additionally to (52) (54), we consider also the condition U ε (x, s) = V (x, s), if ε x θ, s = εθ, (55) where V (x, s) is the function defined by (18). The general solution to (53) can be written as (see (28)) ( ) γ1 ( ) γ2 x x U ε (x, s) = A 1 (s) + A 2 (s). (56) ε ε To find A 1 (s) and A 2 (s), we use conditions (52) and (54). Then, assuming that A 1 (s) and A 2 (s) are differentiable we get A 1 (s) + A 2 (s) = s, (57) From (57), we get ( ) γ1 ( ) γ2 s s A 1(s) + A ε 2(s) = 0. (58) ε A 1(s) + A 2(s) = 1. Taking into account (58), we find that from which we get A 1 (s) = (s/ε) γ 2 (s/ε) γ 2 (s/ε) γ 1 = A 2(s) = A 2 (s) (s/ε) γ 1 (s/ε) γ 1 (s/ε) γ 2 = sθ = s = ε β β log θ β log(s/ε) 1 1 (s/ε) γ 1 γ 2, (59) 1 1 (s/ε) γ 2 γ 1, (60) du 1 (u/ε) γ 2 γ 1 + C 2 14 e yt e t 1 dt + C 2, (61)

15 where β and y are defined in (51). Similarly, A 1 (s) = ε β β log θ β log(s/ε) e (y+1)t e t 1 dt + C 1. (62) Now we must find the constants C 1 and C 2. We remark that in the point (x, s) = (ε, εθ ), we have Thus, A 1 (εθ ) + A 2 (εθ ) = εθ. (63) C 1 + C 2 = εθ. (64) In the point (x, s) = (εθ, εθ ), the equality U(x, s) = V (x, s) holds and together with (18) and (56) leads to the relationship A 1 (εθ )(θ ) γ 1 + A 2 (εθ )(θ ) γ 2 = Together with (61), (62) it gives C 1 (θ ) γ 1 + C 2 (θ ) γ 2 = Combining with (64), we obtain [ εθ γ 2 (θ ) γ 1 γ 1 (θ ) γ 2. (65) γ 2 γ 1 [ εθ γ 2 (θ ) γ 1 γ 1 (θ ) γ 2. (66) γ 2 γ 1 C 1 = εθ γ 2 γ 2 γ 1, C 2 = εθ γ 1 γ 2 γ 1. Finally, A 1 (s) = ε β [ β log θ β log(s/ε) e (y+1)t e t 1 dt + θ γ 2, (67) A 2 (s) = ε β [ β log θ β log(s/ε) e yt e t 1 dt + θ γ 1. (68) With these values we find that the solution of the problem (52) (54) for (x, s) in C I is given by (50). 15

16 From previous considerations it is easy to see, using the strong Markov property of the process (X t, S t ) t 0, that the function V ε (x, s) = E x,s e (λ+r)τ ε S τ ε (69) in the area E ε is given by the following expression: { Uε (x, s), (x, s) CI V ε (x, s) = D I, V (x, s), (x, s) CII D II. (70) We claim that V ε (x, s) = V ε (x, s) and τ ε 0 is an optimal stopping time in the class M ε. The idea of the proof is the same as in Theorem 2.1 and is based on the verification of the properties (a) and (b) (see (32),(34)), which requires the use of the Ito Meyer formula. For (x, s) CII, we saw already that τ ε = τ. Consequently, one need to consider only the case when (x, s) CI D I. If (x, s) D I, then T ε = 0 and τε = 0. Now we apply the Ito Meyer formula to the function V ε (x, s), where (x, s) CI. As in (35) t e (λ+r)t V ε (X t, S t ) = V ε (x, s) + e [L (λ+r)u X V ε (X u, S u ) (λ + r)v ε (X u, S u ) + t 0 e (λ+r)u V ε s ds u + t 0 du e (λ+r)u σx u V ε x (X u, S u ) db u. Again the integral on ds u is equal to zero by condition (54). We have L X V ε (X u, S u ) (λ + r)v ε (X u, S u ) 0 P x,s -a.s., (x, s) C I. Hence, as in (38) where M t is the local martingale e (λ+r)t V ε (X t, S t ) V ε (x, s) + M t, (71) M t = t 0 e (λ+r)u σx u V ε x (X u, S u ) db u. (72) From (71) we conclude, as in the end of the proof of Theorem 2.1, that for any stopping time τ M ε, E x,s e ( λ+r)τ S τ V ε (x, s). (73) 16

17 Consequently, V ε (x, s) V ε(x, s). As in [1, we may check that the family {M t τ ε, t 0} is uniformly integrable. This property together with the property that L X V ε (x, s) (λ + r)v ε (x, s) = 0 for all points (x, s) CI C II, leads to E x,s e ( λ+r)τ ε V ε (X τ ε, S τ ε ) V ε (x, s). (74) With P x,s -probability one, V ε (X τ ε, S τ ε ) = S τ ε for all (x, s) E ε. Therefore, the stopping time τε is optimal for problem (41). This completes the proof of Theorem 3.1. References [1 L. A. Shepp, A. N. Shiryaev. The Russian option: Reduced regret. Ann. Appl. Probab V. 3. No. 3. P [2 L. A. Shepp, A. N. Shiryaev. A new look at the Russian option. Theory Probab. Appl V. 39. No. 1. P [3 L. A. Shepp, A. N. Shiryaev. A dual Russian option for selling short. Probability Theory and Mathematical Statistics: Lectures presented at the semester held in St. Peterburg, Russia, March 2 April 23, Ed. by I. A. Ibragimov et al. Amsterdam: Gordon & Breach, 1996, [4 A. N. Shiryaev. Essentials of Stochastic Finance. Singapore: World Scientific, [5 J. Jacod. Calcul Stochastique et Problèmes de Martingales. Berlin etc.: Springer-Verlag, (Lecture Notes in Math. V. 714.) [6 Ph, Protter. Stochastic Integration and Differential Equations: A new approach. Berlin etc.: Springer-Verlag, (Appl. Math. V. 21.) 17