ON THE MOVING BOUNDARY HITTING PROBABILITY FOR THE BROWNIAN MOTION. Dobromir P. Kralchev

Transcription

1

2 Pliska Stud. Math. Bulgar , STUDIA MATHEMATICA BULGARICA ON THE MOVING BOUNDARY HITTING PROBABILITY FOR THE BROWNIAN MOTION Dobromir P. Kralhev Consider the probability that the Brownian motion hits a moving two-sided boundary by a ertain moment. In some speial ases we find formulae for this probability.. Introdution Let B t be the Brownian motion drift = 0, volatility =, B 0 = 0; T > 0; uppert and lowert be two funtions defined at least for t [0; T ], lowert < uppert, t [0; T ], lower0 0 upper0; τ be the first hitting moment, i.e. τ = inf{t [0; T ] : B t = uppert or B t = lowert} τ = T, if the set is empty. We interpret T as a horizon and are interested in the probabilities of the events U = {B τ = upperτ} and L = {B τ = lowerτ}. In 960 T. W. Anderson [] disovered the rossing probabilities for retilinear boundaries with no horizon two straight lines that are parallel or ross to the left of the origin. In 964 A. V. Skorohod [2] found the probability that the motion will get out of the domain through a little door at the horizon; his formula holds for retilinear boundaries. In 967 L. A. Shepp [3] found a formula for the expetation of the first hitting time for a two-sided symmetri square-root boundary with no horizon. In 97 A. A. Novikov [4] solved the same problem 2000 Mathematis Subjet Classifiation: 60J65 Key words: Brownian motion, hitting time, Laplae transformation

3 84 D. P. Kralhev for a one-sided square-root boundary. In 98 A. A. Novikov [5] published a formula for the probability that the motion will get out of the domain through the horizon; it holds for urvilinear boundaries that are lose to eah other. A little later in the same year A. V. Mel nikov and D. I. Had ziev [6] published a solution to a similar problem for Gaussian martingales. In 999 A. Novikov, V. Frishling and N. Kordzakhia [7] found approximate formulae for the rossing probabilities both for a one-sided and a two-sided boundary with a horizon; they were able to derive an exat formula for a one-sided and a two-sided symmetri square-root boundary. In the urrent paper we find formulae for P U and P L for parallel retilinear boundaries as well as for arbitrary square-root boundaries. 2. Calulation of P U We have a random proess starting at time 0 at point 0. Consider a diffusion proess starting at time t at point x instead and let vt, x = P U, 0 t T, lowert x uppert. So we may safely omit the inequality lower0 0 upper0, keeping this one: lowert < uppert, t [0; T ]. Then f. [8], hapter 0 the funtion vt, x satisfies the onditions: v t v vt, x = 0, x lowert ; uppert vt, lowert = 0, t [0; T ] vt, uppert =, t [0; T ] The equation is simple enough, but the boundary is too ompliated. To get a retangular boundary, set ht = uppert lowert > 0, vt, x = v t, Then the funtion v t, x is a solution to the problem: x lowert. ht 2 v t h t. x + lower t ht v T, x = 0, x 0; v t, 0 = 0, t [0; T ] v t, =, t [0; T ] v x + 2 h 2 t 2 v

4 On the Moving Boundary Hitting Probability for the Brownian motion Two parallel urves: ht = = onst. > 0 Therefore h t = 0 and the equation takes this form: v t lower t v x v 2... Two parallel straight lines: lowert = bt + Then uppert = bt + 2, 2 >, = 2 > 0, lower t = b and the equation beomes: v t b v x v Let κ = 2 2 > 0, λ = b. Then we have to solve the problem: v t λ v x + κ 2 v v T, x = 0, x 0; v t, 0 = 0, t [0; T ] v t, =, t [0; T ] We would prefer to have initial onditions rather than final ones, so we set and reformulate the problem as follows: v t, x = v 2 T t, x v 2 t λ v 2 x + κ 2 v 2 v 2 0, x = 0, x 0; v 2 t, 0 = 0, t [0; T ] v 2 t, =, t [0; T ] Now the time horizon T takes part in the intervals only. We an get rid of it by means of the following argument: when the value of T hanges, only the domain of the funtion v 2 t, x hanges, not its values. Sine T an be an arbitrary positive number, we may assume that the funtion v 2 t, x is defined at least for t [0; + : κ 2 v 2 x 2 λ v 2 x v 2 t = 0 v 2 0, x = 0, x 0; v 2 t, 0 = 0, t [0; + v 2 t, =, t [0; +

5 86 D. P. Kralhev After the Laplae transformation V p, x = L[v 2 t, x] we get the problem: 3 κ. V λ. V p. V = 0 V 0 = 0 V = p V is onsidered a funtion of x, and p is just a parameter. The harateristi equation is κν 2 λν p = 0 D = λ 2 + 4κp > 0, beause p > 0, so ν, 2 = λ ± λ 2 + 4κp and λ V x = C. osh 2 + 4κp. x λ + C 2. sinh 2 + 4κp. x λx exp From V 0 = 0 it follows that C = 0. Then from V = p exp λ C 2 = λ p. sinh 2 +4κp Finally, we obtain the equality: we get V x = sinh p. sinh λ 2 +4κp. x λ 2 +4κp λx exp So we have just proved the following theorem where L stands for the reversed Laplae transformation: Theorem. If lowert = bt +, uppert = bt + 2, = 2 > 0, x bt then P U = vt, x, vt, x = v t,, v t, x = v 2 T t, x, v 2 t, x = L [V p, x], V p, x = sinh λ 2 +4κp. x λx exp, λ p. sinh 2 +4κp where κ = 2 2, λ = b.

6 On the Moving Boundary Hitting Probability for the Brownian motion 87 Fortunately, the funtion V p, x an be expliitly transformed to the funtion v 2 t, x and then bak to vt, x. Aording to [9], we have: = sinh λx sinh λ v 2 t, x = L [V p, x] = p n + π n = res p n V p, x expp t = n. n. sinnπx. exp κn 2 π 2 + λ2 4κ t κn 2 π 2 + λ2 4κ exp λx The value of the single addend must be onsidered equal to x, when λ = 0. This addend omes from the residuum at p 0 = 0. The n-th addend in the sum omes from the residuum at p n = κn 2 π 2 λ2 4κ, n N. Then = sinh λx sinh λ + π n = v t, x = v 2 T t, x = n. n. sinnπx. exp κn 2 π 2 + λ2 4κ t T κn 2 π 2 + Substituting κ and λ in the last expression, we get sinhbx + 2π sinhb n = λ2 4κ n. n. sinnπx. exp n 2 π 2 + b 2 2 t T n 2 π 2 + b 2 2 Finally, the probability we are looking for is equal to P U = vt, x = v t, and an be found from the expression above. 2 2 x bt exp λx exp {bx } Theorem 2. If lowert = bt +, uppert = bt + 2, = 2 > 0, then P U = vt, x = e bx bt π n = sinh{bx bt } + sinhb n. n. sin nπ x bt n 2 π 2 + b 2 2 where the single addend is equal to x, when b = 0.. exp n 2 π 2 + b 2 2 t T 2 2,

7 88 D. P. Kralhev To hek, let T + ; we an do this, beause the series is onvergent uniformly with respet to T [t; + ; thus we obtain the formula for the ase, when there is no horizon: Corollary. If lowert = bt +, uppert = bt + 2, = 2 > 0 and there is no horizon, then e 2bx bt for b 0 P U = vt, x = e 2b x for b = 0 The formula for b = 0 is well known from the martingale theory. The formula for b 0 an be found in different denotation in [] as Theorem 4. on page 75. Again, let in Corollary ; thus we get the solution for a one-sided boundary with no horizon: Corollary 2. If uppert = bt + 2 horizon, then and there is no lower boundary and no P U = vt, x = { e 2bx bt 2 for b > 0 for b 0 This result an be verified by means of Kendall s famous formula. Unfortunately, this tehnique does not work for Theorem 2: the series is not uniformly onvergent with respet to ; x 0 ], x 0 x bt Square-root boundaries Let lowert = a t + γ + 0, uppert = b t + γ + 0, b > a. Then the problem 2 takes the form: x + v t 2t + γ v T, x = 0, x 0; v t, 0 = 0, t [0; T ] v t, =, t [0; T ] a b a v x + 2 b a 2 t + γ 2 v Remark: The number γ must be positive, beause vt, x is defined for t [0; T ] and uppert > lowert, t [0; T ]. If the domain of the funtion hanges, the domain of γ will hange respetively. It is essential that the multiplier t + γ is raised to the same power in both denominators; this happens for square-root boundaries only. That is why, by

8 On the Moving Boundary Hitting Probability for the Brownian motion 89 multiplying the equation by 2t + γ we an ensure that the variable t takes part in only one oeffiient: 2t + γ v t a x + v b a x + b a 2 2 v We an get rid of the multiplier t + γ by means of a suitable substitution. Let v t, x = v 2 2 lnt + γ, x ; then the funtion v 2 t, x is a solution to the following problem: v 2 t x + a v 2 b a x + b a 2 2 v 2 v 2 2 lnt + γ, x = 0, x 0; v 2 t, 0 = 0, t [ 2 ln γ; 2 lnt + γ] v 2 t, =, t [ 2 ln γ; 2 lnt + γ] Finally, let v 2 t, x = v 3 t + 2 lnt + γ, x in order to have an initial ondition instead of a final one. v 3 t x + a v 3 b a x + b a 2 2 v 3 v 3 0, x = 0, x 0; v 3 t, 0 = 0, t [ 0; 2 lnt + γ 2 ln γ] v 3 t, =, t [ 0; 2 lnt + γ 2 ln γ] Again, we would prefer to searh for v 3 t, x defined for t [0; + : v 3 t x + a v 3 b a x + b a 2 2 v 3 v 3 0, x = 0, x 0; v 3 t, 0 = 0, v 3 t, =, b a 2 V t [0; + t [0; + Apply the Laplae transformation: V p, x = L[v 3 t, x]. Then x + V p V = 0 V 0 = 0 a b a V = p

9 90 D. P. Kralhev V is onsidered a funtion of x, and p is just a parameter. This boundary value problem has a unique solution; the solution is an analytial funtion: V x = n x n. The differential equation turns into the following n = 0 equation for the oeffiients of the series: n n + b a 2 n + 2n + n+2 n+2 = a b a n + n+ ab a n + 2 n+ + b a2 n + p n + 2n + n, n N 0 p n = 0, n N 0 ; We have to find 0 and in order to speify the sequene n n = 0. From V 0 = 0 it follows that 0 = 0. Let =, n = α n, n N 0. Then α 0 = 0, α =, α n+2 = ab a n + 2 α n+ + b a2 n + p n + 2n + α n, n N 0 ; V x =. α n x n =. n = 0 α n x n ; the unknown onstant an be found from n = the boundary ondition V = p ; it follows that = We have just p. α n proved the following theorem: Theorem 3. If lowert = a t + γ + 0, uppert = b t + γ + 0, b > a, x then P U = vt, x = v t, 0 a t+γ, v t, x = v 2 2 lnt + γ, x, b a t+γ v 2 t, x = v 3 t + 2 lnt + γ, x, v 3 t, x = L [V p, x], where L stands for the reversed Laplae transformation, V p, x =. α n x n, α 0 = 0, α =, α n+2 = = p. α n n = n = n = ab a n + 2 α n+ + b a2 n + p n + 2n + α n, n N 0 ; When a = 0, we an obtain an expliit formula for α n n = 0. Indeed, α 0 = 0, α =, α n+2 = b 2 n+p n+2n+ α n, n N 0 ; so α 2m = 0, m N 0.

10 On the Moving Boundary Hitting Probability for the Brownian motion 9 Let β m = α 2m+, m N 0 ; then β 0 =, β m+ = b 2 2m + + p 2m + 32m + 2 β m, i.e. β m = b 2 2m + p b 2m m β m, hene β m = 2 m + p. 2m + 2m 2m +! m = That is why, the following statement holds: Corollary 3. If lowert = 0, uppert = b t + γ + 0, b > 0, then x 0 P U = vt, x, vt, x = v t, b, v t, x = v 2 2 t + γ lnt + γ, x, v 2 t, x = v 3 t + 2 lnt + γ, x, v 3 t, x = L [V p, x], where L stands for the reversed Laplae transformation, V p, x =. β m x 2m+, β 0 =, β m = b 2m 2m +! m 2 m + p, m N; = m = p. m = 0 m = 0 β m 3. Calulation of P L The only differene is that 0 and hange plaes in the boundary onditions of. This hange propagates through the sequene of substitutions. 3.. Two parallel straight lines Now 3 hanges in this way: κ. V λ. V p. V = 0 V 0 = p V = 0 After the substitution V x = W x we get the problem: κ. W + λ. W p. W = 0 W 0 = 0 W = p

11 92 D. P. Kralhev This is the same problem as 3, only λ has hanged its sign. Then W x = sinh p sinh V x = sinh λ 2 +4κp. x λ 2 +4κp λ 2 +4κp. x λ 2 +4κp p sinh and the following statement holds: λ x exp exp λx Theorem 4. If lowert = bt +, uppert = bt + 2, = 2 > 0, x bt then P L = vt, x, vt, x = v t,, v t, x = v 2 T t, x, v 2 t, x = L [V p, x], V p, x = sinh λ 2 +4κp. x λx exp, λ p sinh 2 +4κp where L stands for the reversed Laplae transformation, κ = 2 2, λ = b. Theorem 5 an be dedued from Theorem 4 as Theorem 2 was dedued from Theorem. Or we may notie that the hanges in the formulae are equivalent to swapping the lower and the upper boundary: we replae x lowert with uppert x and vie versa as well as b with b in Theorem 2. Theorem 5. If lowert = bt +, uppert = bt + 2, = 2 > 0, then P L = vt, x = e bx bt. + 2π n = sinh {bbt+ 2 x} + sinhb n. n. sin nπ bt+ 2 x n 2 π 2 + b 2 2 where the single addend is equal to 2 x, when b = 0.,. exp n 2 π 2 + b 2 2 t T 2 2 Corollary 4. If lowert = bt +, uppert = bt + 2, = 2 > 0 and there is no horizon, then e 2b bt+ 2 x for b 0 P L = vt, x = e 2b 2 x for b = 0,

12 On the Moving Boundary Hitting Probability for the Brownian motion 93 Corollary 5. If lowert = bt + and there is no upper boundary and no horizon, then { e 2bx bt for b < 0 P L = vt, x = for b Square-root boundaries Theorem 6. If lowert = a t + γ + 0, uppert = b t + γ + 0, b > a, x then P L = vt, x = v t, 0 a t+γ, v t, x = v 2 2 lnt + γ, x, b a t+γ v 2 t, x = v 3 t + 2 lnt + γ, x, v 3 t, x = L [V p, x], where V p, x = W p, x, W p, x =. α n x n, = p. α n, α 0 = 0, α =, α n+2 = bb a n + 2 n = n = α n+ + b a2 n + p n + 2n + α n, n N 0. Corollary 6. If lowert = a t + γ + 0, uppert = 0, a < 0, then x P L = vt, x, vt, x = v t, 0 a t+γ, v t, x = v 2 2 lnt + γ, x, a t+γ v 2 t, x = v 3 t + 2 lnt + γ, x, v 3 t, x = L [V p, x], where, V p, x = W p, x, W p, x =. β m x 2m+, = p. β m β 0 =, β m = a 2m 2m +! m = 0 m 2 m + p, m N. m = m = 0 4. Numerial experiments In order to return to the original formulation of the problem B 0 = 0, i.e. starting moment = 0, initial position = 0, one has to put t = 0 and x = 0 in our formulae; for square-root boundaries γ must be positive. The formulae were programmed and tabulated. The results were ompared with the values of the rossing probabilities alulated by means of the Monte Carlo method and dynamial programming. The idea of the last method is to alulate the values of vt, x, beginning from the horizon and moving to the starting moment step by step; for eah t an array of values of vt, x is alulated using the array of the preeding step. The three results agree with one another. So we have a numerial onfirmation of our formulae besides the theoretial one. The algorithm that uses the formulae is the fastest one.

13 94 D. P. Kralhev 5. Conlusion The propositions above give a omprehensive answer to the question about the rossing probabilities in two speial ases: retilinear parallel boundaries and square-root boundaries with a time horizon. The obtained formulae are suitable for programming: numerial alulations based on them are fast enough. R E F E R E N C E S [] T.W. Anderson A modifiation of the sequential probability ratio test to redue the sample size Ann.Math.Statist. 3 Marh 960, in English. [2] A.V. Skorohod Random proesses with independent inrements. Mosow, Nauka, 964 in Russian. [3] L.A. Shepp A first passage problem for the Wiener proess Ann.Math.Statist , in English. [4] A.A. Novikov On stopping times for the Wiener proess Theory Prob.Appl , Mosow, in Russian. [5] A.A. Novikov On small deviations of Gaussian proesses Mathematial notes , USSR Aad. Si., Mosow, Nauka, in Russian. [6] A.V. Mel nikov, D.I. Had ziev Asymptotis of small deviations probability of Gaussian martingales Proeedings of Bulgarian Aad. Si Sofia, in English. [7] Alex Novikov, Volf Frishling, Nino Kordzakhia Approximations of boundary rossing probabilities for a Brownian motion, J.Appl.Prob , Israel, in English. [8] A.D. Wentzel A ourse in the theory of random proesses, Mosow, Nauka, 975 in Russian. [9] M.A. Lavrentiev, B.V. Shabbat Methods of the theory of funtions of a omplex variable, Mosow, Nauka, 987 in Russian. Dobromir Pavlov Kralhev University of Food Tehnologies Dept. of Informatis 26 Maritsa Str Plovdiv, Bulgaria dobromir kralhev@ abv.bg