Albert N. Shiryaev Steklov Mathematical Institute. On sharp maximal inequalities for stochastic processes

Transcription

1 Albert N. Shiryaev Steklov Mathematical Institute On sharp maximal inequalities for stochastic processes joint work with Yaroslav Lyulko, Higher School of Economics 1

2 TOPIC I: TOPIC II: Sharp maximal inequalities for continuous time processes Sharp maximal inequalities for discrete time processes 2

3 TOPIC I: Sharp maximal inequalities for continuous time processes 1. Introduction. The main method for obtaining a sharp maximal inequalities 2. Maximal inequalities for standard Brownian motion and it s modulus. Martingale and «Stefan problem» approaches 3. Maximal inequalities for skew Brownian motion. Solution to the corresponding Stefan problem 4. Maximal inequalities for Bessel processes. Solution to the corresponding Stefan problem 5. Doob maximal inequalities 3

4 TOPIC II: Sharp maximal inequalities for discrete time processes 1. Maximal inequalities for modulus of simple symmetric Random walk 2. Maximal inequalities for simple symmetric Random walk 4

5 TOPIC I: Sharp maximal inequalities for continuous time processes 1. Introduction. The main method for obtaining a sharp maximal inequalities Let X = X t t 0 be a process on Ω, F, P with natural filtration F = F t t 0, F t = σx s, s t. For any Markov time τ w.r.t. F t t 0 the inequalities E sup 0 t τ X t C X fegτ are called maximal inequalities for X. Here C X is a constant, f and g are some functions. Markov times τ = τω usually belong to the set M = {τ Markov time w.r.t. F t t 0, Eτ < }. 5

6 The inequality * is called sharp maximal inequality if there exist a non-trivial Markov time τ M such that E sup 0 t τ X t = C X feg τ. Examples of maximal inequalities for some well-known processes include Graversen, Peskir, Shiryaev : for geometric Brownian motion X t = expσb t + µ σ 2 /2t with µ < 0, σ > 0: E max X t 1 σ2 ; 0 t τ 2µ + σ2 2µ exp σ2 2µ 2 2σ 2 Eτ 1 for Ornstein-Uhlenbeck process X t t 0 with dx t = βx t dt + db t, β > 0: C 1 E ln1 + βτ E max X t C 2 E ln1 + βτ, β 0 t τ β where C 1, C 2 > 0 are some universal constants; 6

7 for bang-bang process X t t 0 with dx t = µ sgnx t dt + db t, µ > 0: where G µ x = inf c>0 E max X t 0 t τ cx + 1 2µ ln G µ Eτ, 1 + µ c. 7

8 Assume that X = X t t 0 is the Markov process. For given measurable functions L = Lx and K = Kx we define I t = t 0 LX sds, S t = max 0 s t KX s, t 0. s X t, S t s = x 0 x Consider the following optimal stopping problem: V c = sup τ E FI τ, X τ, S τ c GI τ, X τ, S τ, 1 where F, G are given measurable functions, τ M, c > 0 is a parameter. 8

9 Suppose we solved the problem 1 and found the function V c. Then for any τ and c we have EF I τ, X τ, S τ V c + c EGI τ, X τ, S τ Taking the infimum on both sides by c > 0 we obtain the inequality EFI τ, X τ, S τ HEGI τ, X τ, S τ := inf c>0 V c + c EGI τ, X τ, S τ which is true for any Markov time τ M. If infimum is minimum and it is achieved on some c > 0 then inequality 2 is sharp. 2 The corresponding solution τ c of problem 1 when c = c is a stopping time on which 2 becomes an equality. 9

10 Consider the particular case F x, y, z = z, Gx, y, z = x, Lx = cx, Kx = x. The function c = cx is assumed to be positive and continuous and it is called cost for observations. We obtain the following optimal stopping problem: where V x, s = sup τ E x,s S τ τ 0 cx t dt, 3 E s,x, s x is expectation under the measure P x,s = LawX, S P, X 0 = x, S 0 = s τ is the optimal stopping time such that E x,s τ 0 cx t dt < 10

11 In addition we assume that X = X t t 0 is a diffusion process and it is a solution of stochastic differential equation dx t = bx t dt + σx t db t, X 0 = 0, where B = B t t 0 is the Brownian motion on Ω, F, P. Diffusion coefficient σ = σx > 0 and drift coefficient b = bx are continuous. We need to know a scale function R = Rx and a speed measure m = mx in order to obtain a solution of the problem 3. It is well known that in the case of diffusion process X we have Rx = x exp y 2bu σ 2 u du dy, x R, mdx = 2dx R xσ 2 x 11

12 From the general optimal stopping theory we may decompose the state space E = {x, s R 2 : x s, s 0} of the process X, S, S t = max X u s by u t where E = C D, C = {x, s E : V x, s > s} is a continuation set. If x, s C we need to continue our observations; D = {x, s E : V x, s = s} is a stopping set. If x, s D we need to stop our observations Therefore if we start in C we need to stop at the first time when the process X, S reaches D. In other words, τ = inf{t 0 : X t, S t D }. Proposition 1. The diagonal {x, s E : x = s} does not belong to the continuation set C. 12

13 Reduce the problem V x, s = sup E x,s S τ τ cx t dt to the optimal τ 0 stopping problem in standard formulation. Consider the process t A t = a + cx udu, a 0 0 and observe that Z t = A t, X t, S t, t > 0, Z 0 = a, x, s is a Markov process. Define the function Ga, x, s = s a and observe that the initial problem takes the form Ṽ a, x, s = sup τ E a,x,s GZ τ, where times τ are such that EA τ <. However since Ṽ a, x, s = V x, s a it is sufficient to find the function V x, s i.e. to solve 2-dimensional optimal stopping problem for X, S. The infinitesimal operator of the process Z = Z t t 0 equals L Z = cx a + L X = cx a + bx x + σ2 x 2 x 2 if x < s. 13 2

14 Since the cost for observations cx is positive we should not allow the process X to decrease too fast when s is fixed. It means that for given s there exist a point gs such that we should stop the observations when X, S achieves a point gs, s. In other words τ = inf{t > 0 : X t gs t } The unknown function g = gs is called the boundary of the stopping set D. The function V x, s, gs < x s is a solution of the system L X V x, s = cx if gs < x < s, 4 V x, s = 0 normal reflection, 5 s x=s V x, s x=gs+ = s instanteneous stopping, 6 V x, s = 0 smooth fit, 7 x x=gs+ which is called a Stefan problem with moving boundary g = gs. 14

15 Explain the meaning of each equation 4-7. According to the general optimal stopping theory, L Z V x, s = 0 when x, s C. Thus we get the equation 4; The instanteneous stopping condition 6 follows from the fact that V x, s = s when x, s D ; The smooth fit condition 7 means that the derivative of the function V x, s is contintinuous on the boundary of C and D ; Clarify the normal reflection condition. Applying the Ito formula for semimartingales dfx t, S t = f xx t, S t dx t + f sx t, S t ds t f xxx t, S t d X, X t to the process fx t, S t t 0, taking the expectation E s,s on both sides and multiplying on t 1 we get 15

16 E s,s fx t, S t fs, s E s,s 1 t t 0 t = E s,s 1 t t 0 L X fx u, S u du f s X u, S u ds u L X fs, s + f s, s s + lim t 0 E s,s S t s t as t 0. Since the diffusion coefficient σ > 0 as t 0 we have 1 t E s,ss t s Therefore the condition f ss, s = 0 assures us that the limit L X fs, s+ f E s,s S t s s, s lim is finite. s t 0 t 16

17 Find the functions V x, s and gs the solutions of system 4-7. Denote τ g = inf{t > 0: X t gs t }, τ gs,s = inf{t > 0: X t / gs, s} and consider the function V g x, s = E x,s S τg τ g 0 cx t dt Using the strong Markov property of X w.r.t. time τ gs,s when x gs, s we have V g x, s = s P x,s X τgs,s = gs + V g s, sp x,s X τgs,s = s. E x,s τ gs,s 0 cx t dt = 17

18 Rs Rx = s Rs Rgs + V Rx Rgs gs, s Rs Rgs s gs G gs,s x, ycymdy, where G a,b x, y is the Green function of X on the segment [a, b]: Rb RxRy Ra if a y x, Rb Ra G a,b x, y = Rb RyRx Ra if x y b. Rb Ra Rewrite the expression for V g x, s in the following form: V g s, s s = Rs Rgs Rx Rgs V gx, s s + s gs G gs,s x, ycymdy 18

19 Suppose that V g x, s satisfies the smooth fit condition. Then lim x gs V g x, s s Rx Rgs = 1 R gs V g x x, s = 0, x=gs+ lim x gs s gs Therefore we have Rs Rgs Rx Rgs s gs Rs Rycymdy. V g s, s = s + s gs G gs,s x, ycymdy = Rs Rycymdy, 19

20 Finally we obtain V g x, s = s + x gs Rx Rycymdy, 8 for all gs x s. Now suppose that the function V g x, s is given by 8. Then it is easy to show that V g x, s is a solution of Stefan problem 4-7 if and only if the boundary g = gs belongs to C 1 and satisfies the equation g s = σ 2 gsr gs 2cgsRs Rgs. 9 20

21 Observe that the equation 9 has a whole family of solutions. We need to specify the criteria which enables us to choose the solution g = g s a boundary of the stopping set D. We call the solution gs of the equation 9 an admissible solution if gs < s for all s 0. Theorem [maximality principle]. The boundary g = g s of the stopping set D in the problem V x, s = sup τ E x,s S τ τ 0 cx t dt is a maximal admissible solution of the differential equation 9. 21

22 Theorem. Consider the stopping problem * for diffusion process X = X t t 0 such that dx t = bx t dt+σx t db t. Supremum is taken by all Markov times τ such that E x,s τ 0 cx t dt <. 10 Assume that there exist the maximal admissible solution g s of 9. Then 1 The value function V x, s in problem * is finite and can be determined on E by V x, s = s, if x g s, s + x g s Rx Rycymdy, if g s x s. 2 The Markov time τ = inf{t > 0: X t g S t } is optimal in problem * if it satisfies the condition 10; 22

23 3 If there exist an optimal stopping time σ in problem * such σ that E x,s cx t dt < then P x,s τ σ = 1 for all x, s and 0 time τ is also optimal in problem *. If the equation 9 doesn t have a maximal admissible solution then V x, s = + for all x, s and there is no optimal stopping time in problem *. Theorem [verification theorem]. Assume that for the solution V = V x, s of Stefan problem 4-7 the following statements are true: i V x, s s, x, s E; ii V x, s = E x,s Sτg τ g 0 cx tdt, x, s E for some Markov time τ g = inf{t 0: X t gs t } satisfying 10; iii V x, s E x,s V X τ, S τ for any Markov time τ satisfying 10. Then V x, s coincides with the value function V x, s in problem * and τ g is optimal. 23

24 2. Maximal inequalities for standard Brownian motion and it s modulus. Martingale and «Stefan problem» approaches Consider the standard Brownian motion B = B t t 0, B 0 = 0. This was the first process for which sharp maximal inequalities were established. square root inequality E max 0 t τ B t square root of two inequality E max B t 0 t τ Eτ 2Eτ Inequalities 11 and 12 are also called Dubins-Jacka-Schwarz- Shiryaev inequalities. 24

25 Denote S t B = max 0 u t B u and S t B = max 0 u t B u. Martingale approach. First proof the inequality 11. Consider a stochastic process Z t = cs t B B t 2 t + 1 4c, t 0 when c > 0. Due to Levy theorem LawSB B = Law B and the process B 2 t t is a martingale. Therefore Z t t 0 is also martingale w.r.t. natural filtration of B. It is easy to see that cx 1/2 c 2 0. From this inequality it follows that x ct cx 2 t + 1/4c for all x R. Thus for any τ M we get ES τ t B cτ t = ES τ t B B τ t cτ t EZ τ t = EZ 0 = 1 4c Taking the limit as t from Doob s optional sampling theorem we have ES τ B ceτ +1/4c. Taking an infimum on c > 0 on both sides we obtain

26 Prove that inequality ES τ B Eτ is sharp. For each a > 0 consider the time τ a = inf{t 0: S t B B t = a} We see that ES τa B = ES τa B B τa = a. Since Lawτ a = Lawinf{t 0: B t = a} from Wald identities we get a 2 = EBτ 2 a = Eτ a. Corollary. For any continuous local martingale M = M t t 0, M 0 = 0 we have E max 0 t T M t E M T, 13 for any T > 0. Here M t t 0 is a quadratic characteristic of M. This inequality follows from 11 and Dambis-Dubins-Schwarz theorem. Indeed, Emax t T M t = Emax t T B M t = Emax t M T B t E M T. 26

27 Prove the inequality ES τ B 2Eτ. Consider a continuous martingale U t = E B τ E B τ F B t τ, t 0 Applying 13 to max t T U t and taking T + we get Emax t 0 U t E B τ E B τ 2. Using this inequality we estimate ES τ B by E max B t = E max B t τ = E max EB τ Ft τ B 0 t τ t 0 t 0 E max t 0 E B τ F B t τ E B τ = max t 0 U t = E Eτ E B τ 2 + E B τ 2Eτ. + E B τ E B τ E B τ 2 + In order to get the last inequality in this series we used a simple inequality A x 2 + x 2A when 0 < x < A. Now show that inequality ES τ B 2Eτ is sharp. Consider the time τ a = inf{t 0: S t B B t = a} It turns out that E τ a = 2a 2 and Emax t τa B t = 2a. 27

28 «Stefan problem» approach. Basically the proof of 11 and 12 is the application of the main theorem of 1 to the problem V x, s = sup τ E x,s S τ τ 0 cx t dt in the case when cx t c > 0, X t = B t or X t = B t. First, prove the inequality ES τ B Eτ. In the case of Brownian motion Rx = x, mdx = 2dx, x R. According to the theorem the equation for boundary is g s = 1 2cs gs The maximal admissible solution of this equation is g s = s 1/2c. 28

29 Therefore the value function V x, s = sup t τ E x,s S τ B cτ when 0 s x 1/2c equals V x, s = s + 2c x x ydy = cx s 2 + x + 1 4c gs Since we need the value V 0, 0 for any τ M we get ES τ B inf c>0 {V 0, 0 + ceτ} = inf c>0 {1/4c + ceτ} = Eτ However we cannot apply directly the method from 1 in the case of X t = B t and obtain the inequality ES τ B 2Eτ. The reason is that we cannot represent X t = B t in the form dx t = bx t dt + σx t db t with continuous b and σ. But we can consider the problem W x, s = sup τ E x,s s max x + B t cτ 0 t τ and reduce it to the Stefan problem. 29

30 Infinitesimal operator of B equals L = 2 1 d2 dx2, x > 0 with endpoint x = 0. Thus Stefan problem in our case is 2 W x, s = 2c, x 0, gs < x s, x2 W 0+, s = 0, s: gs < 0; x W x, s = 0; W x, s s x=gs+ = s; W x, s = 0. x=s x x=gs+ The solution of this system is the function W x, s = s, s x 1 2c, cx s 2 + x + 1 4c, s 1/2c, s x 1/2c, cx c, 0 s 1 2c Since W 0, 0 = 1/2c for each τ inf c>0 {1/2c + ceτ} = 2Eτ. M we have ES τ B 30

31 3. Maximal inequalities for skew Brownian motion. Solution to the corresponding Stefan problem The process X α = X α t t 0 defined on probability space Ω, F, P is called a skew Brownian motion if it satisfies the stochastic equation X α t = X α 0 + B t + 2α 1L 0 t Xα, 14 where L 0 = L 0 t Xα t 0 с L 0 0 Xα = 0 is the local time of X α in zero. The skew Brownian motion with parameter α = 1/2 has the same distribution as standard Brownian motion, with parameter α = 1 as the modulus of standard Brownian motion. Denote by W α = W α t t 0 the unique strong solution of 14 such that W α 0 = 0. 31

32 Consider the optimal stopping problem V x, s = sup τ E x,s s max x + W t α cτ 0 t τ 15 with constant cost for observations c > 0. We cannot directly apply the methods from 1 since X t = x + Wt α cannot be represented in the form dx t = bx t dt + σx t db t with continuous b and σ. However we can write the analogue of Stefan problem 4-7 in the case of optimal stopping problem. The infinitesimal operator for X equals L = 2 1 d2 functions dx 2 and defined for {f : f exists on R \ {0}, f 0+ = f 0, lim x fx = 0 and αf 0+ = 1 αf 0 } 32

33 Therefore we get the Stefan problem for value function 2 V x, s = 2c, x 0, gs < x s, x2 α V 0+, s = 1 α V 0, s, s: gs < 0; x x V x, s = 0; V x, s s x=gs+ = s; V x, s x=s x The solution of this system is given in the following x=gs+ = 0 Theorem 1. The optimal stopping time τ c in the problem 15 exists and equals τ = inf{t 0 : X t gs t } 33

34 The mapping g = gs, s 0 is given by s = parameter β = 1 α/α. g + 1/2c, if g 0, β 2 1 2cβ 2 e2cβg + g β + 1 2cβ2, if g < 0, s The boundary s = sg of the stopping set when c = 1 and α = 0.1, 0.2,..., 0.9. g 34

35 If we consider the sets D = {x, s E : x gs}, C = E \ D then the value function equals V x, s = s + cx gs 2, x, s C, x 0, s 1 2c or x < 0, s < 1 2c, s + cx gs 2 + 2c1 βxgs, x, s C, x 0, s < 1 2c, s, x, s D The proof of the theorem is based on finding the solution to Stefan problem. Particularly the equation for boundary g = gs is g s = 1, 2cs gs s: gs 0, 1, 2cβs gs s: gs < 0 The general solution of this equation is sg = a 0 e 2cg + g + 1/2c when g 0 and sg = b 0 e 2cβg + g/β + 1/2cβ 2 when g < 0. 35

36 In order to prove that the solution of Stefan problem V x, s coincides with the value function V x, s = sup τ E x,s s max0 t τ x + W α t cτ we use the following analogue of Ito formula: V X t, S t = V X 0, S 0 + 2α 1 2 t 0 t 0 V xx u, S u db u + t 0 V sx u, S u ds u + V x0+, S u + V x0, S u dl 0 u V x0, S u dl 0 u t 0 t V xxx u, S u IX u 0du 0 V x0+, S u 36

37 Once we know the value V 0, 0 it is possible to obtain the maximal inequalities. Theorem 2 Lyulko For any Markov time τ M and for any α 0, 1 the following inequality holds: E max 0 t τ W α t M α Eτ, 16 where M α = α1 + A α /1 α and A α is the unique solution of the equation such that A α > 1. A α e A α+1 = 1 2α α 2, The inequality 16 is sharp i.e. for any T > 0 there exist a Markov time τ with Eτ = T such that E max 0 t τ W α t = M α Eτ. 37

38 The inequalities like 16 can be obtained not only for maximum max. Thus in [Zhitlukhin 2012] there were stated the following 0 t τ W α t inequalities for range of skew Brownian motion: E max 0 t τ W α t min 0 t τ W α t K α Eτ, where K α = C α + C 1 α, C α = α αd 2 0 α 1 α 1 α 2D αx + α 1 α 2α D α 2α 1e x α dx and D α is the unique negative solution of the equation 2α 1α 2 e D α 1 = D α M α 2 1 K α α 2 0 1/2 1α 38

39 4. Maximal inequalities for Bessel processes. Solution to the corresponding Stefan problem A continuous nonnegative Markov process X = X t x t 0, x 0 is called a Bessel process of dimension γ R X Bes γ x if it s infinitesimal operator equals L X = 1 2 γ 1 x d dx + d2 dx 2 The endpoint x = 0 is called trap if γ 0, instantaneously reflecting if γ 0, 2 and entrance if γ 2. 39

40 In the case α = n N the Bessel process can be realized as a radial part of n-dimensional Brownian motion X t x = n i=1 B i t + a i 2 1/2 where a = a 1, a 2,..., a n is a vector in R n with norm x = a a2 n. B 1, B 2,..., B n are independent Brownian motions starting from zero. The Bessel process of dimension γ = 1 is a modulus of standard Brownian motion x + B t., Consider the optimal stopping problem V x, s = sup τ where Markov times τ M. E x,s s max 0 t τ X tx cτ 40

41 Theorem 3. Let X Bes γ x where the dimension γ R and c > 0. The optimal stopping time τ in problem * exists and equals τ = inf{t 0: X t, S t D } with X t = X t x, S t = S t x, s = s max X u and stopping set 0 u t D = {x, s: s s, x g s} where g = g s is the unique nonnegative solution of the equation 2c γ 2 g sgs 1 such that gs s when s 0 and lim s g s gs s = 1, γ 2 = 1 17 s and s is the root of the equation g s = 0. When γ = 2 the equation 17 has the form 2cg sgs lns/g = 1. 41

42 Moreover if we denote C 1 = {x, s R + R + : s > s, g s < x s}, C 2 = {x, s R + R + : 0 x s s } and define a continuation set by C = C 1 C 2 then depending on the value of parameter γ the value function V x, s equals if α > 0 V x, s = s, s + c γ x2 g 2 s + 2cg2 s γγ 2 c γ x2 + s, g s x γ 2 1 x, s D,, x, s C 1, x, s C 2 ; 42

43 if α = 0 if α < 0 V x, s = V x, s = s, s, x, s D, s + c 2 g2 s x 2 + cx 2 ln x g s, x, s C ; s + c γ x2 g 2 s + 2cg2 s γγ 2 g s x γ 2 1 x, s D,, x, s C. 43

44 Using this theorem we can obtain the maximal inequalities for Bessel processes. if γ 0 then the point x = 0 is a trap. Therefore X t x 0 if t 0 and maximal inequalities do not make sense if γ > 0 then from theorem it follows that V 0, 0 = s. Denote V x, s = V γ c x, s, s = s c γ Since Bessel processes are self-similar LawX t x, t 0 = Lawc 1/2 X ct c 1/2 x the value function Vc γ x, s is also self-similar, i.e. cvc γ x, s = V γ 1 Hence s c γ = s 1 γ/c. Therefore we get the inequalties E max X t0 0 t τ inf {s 1γ/c + ceτ} = c>0 inf {V 0, 0 + ceτ} = c>0 4s 1 γeτ cx, cs. 44

45 Theorem 4 Dubins-Shepp-Shiryaev Let X Bes γ 0, γ > 0. Then for any Markov time τ M the following sharp maximal inequality holds: E max X t0 0 t τ 4s 1 γeτ, where s 1 γ is the root of equation g s = 0 such that as γ. s 1 γ γ 1 4 Observe that in the case γ = 1 we have s 1 1 = 1/2 and therefore we get the maximal inequality for modulus of standard Brownian motion E max B t 0 t τ 2Eτ. 45

46 5. Doob maximal inequalities Theorem 5. Let M = M t t 0 be a local martingale on a filtered probability space Ω, F t t 0, P. Then for any p > 0 there exist a universal constants c p и C p such that c p E[M] p/2 τ E max M t p 0 t τ C p E[M] p/2 τ, where [M] t t 0 is called a quadratic variation of M. 18 The inequalities 18 are called Burkholder-Davis-Gundy inequalities. In the case when M t = B t is standard Brownian motion we get c p Eτ p/2 E max B t p/2 0 t τ C p Eτ p/2, 19 Note that if p 2 the exact values of the constants c p and C p when inequalities 19 become sharp are still not known. 46

47 Some particular cases of Burkholder-Davis-Gundy inequalities: Davis inequalities p = 1: c 1 E τ E max B t 0 t τ C 1 E τ Doob inequalities p = 2: c 2 Eτ E max 0 t τ B2 t C 2 Eτ Consider the case p = 1. One of the possible ways to obtain the exact values of c 1, C 1 is to solve the optimal stopping problem V c = sup τ E max B t c τ 0 t τ where c > 0, τ is the Markov time such that E τ <., 20 47

48 The problem 20 can be formulated in a standard way for 3- dimensional Markov process Z t = t, X t, S t, X t = B t, S t = max u t B u But this problem is nonlinear and we cannot decrease it s dimensionality. The same situation happens when p 2. In the case p = 2 the corresponding optimal stopping problem sup τ Emax t τ B2 t cτ is linear and we can get the solution explicitly. As a consequence we obtain the Doob maximal inequalities Eτ E max 0 t τ B2 t 4Eτ, 21 where τ is the Markov time such that Eτ <. 48

49 Prove the inequality 21 and show that it is sharp. Denote S t B 2 = max 0 u t B2 u. The lower bound for ES τ B 2 follows from the Wald identity: ES τ B 2 EB 2 τ = Eτ To show that this inequality is sharp it is enough to consider the time τ T = inf{t 0: B t = T }. Then Eτ T = EB 2 τ T = T and ES τ T B2 = T. In order to prove the upper bound E sequence of stopping times max 0 t τ B2 t 4Eτ consider the σ λ,ε = inf{t > 0: max 0 s t B s λ B t ε}, where λ, ε > 0. It is known that Eσ λ,ε p/2 < if and only if λ < p/p 1. 49

50 Therefore if λ 0, 2 we have E max 0 t σ λ,ε B 2 t = λ 2 E B σλ,ε 2 + 2λεE B σλ,ε + ε 2 KE B σλ,ε 2 22 for some constant K > 0. Divide the both sides of 22 on E B σλ,ε 2 and take λ 2. Since E B σλ,ε 2 = Eσ λ,ε and E B σλ,ε /E B σλ,ε 2 1/ Eσ λ,ε 0 if λ 2 then from 22 we get K λ 2 + 2λε E B σ λ,ε EB σλ,ε 2 + ε2 4. E B σλ,ε 2 Therefore K = 4 is the best possible constant in the upper bound for ES τ B 2. 50

51 TOPIC II: Sharp maximal inequalities for discrete time processes 1. Maximal inequalities for modulus of simple symmetric Random walk In this section time t will take discrete values i.e. t = n = 0, 1, 2,... Consider the simple symmetric Random walk X n = S n = ξ ξ n, X 0 = S 0 = 0, where ξ 1,..., ξ n,... are i.i.d. random variables, Pξ 1 = 1 = Pξ 1 = 1 = 1/2 Denote the current maximums of X and X by M n S = max 0 k n S k and M n S = max 0 k n S k. 51

52 In order to obtain the maximal inequalities for S n n 0 and S n n 0 consider the following optimal stopping problems: and V c = sup E τ M W c = sup E τ M max S k cτ 0 k τ max S k cτ 0 k τ For any nonnegative integer l define the stopping times τ l = σ l = inf{k > n : M k S S k = l}, if m s < l, n, if m s l inf{k > n : S k 0, M k S S k = l}, if m s < l, n, if m s l 52

53 and a function Q l = Q l n, s, m, c such that Q l n, s, m, c = sup τ M l E s,m M τ S cτ, where the set of stopping times equals M l = {τ l, σ l : l Z + }. If the conditions 1 Q l n, s, m, c m cn, 2 Q l n, s, m, c EQ l n+1, s+ξ n+1, max{m, s+ξ n+1 }, c excessivity are satisfied then Q l n, s, m, c = sup τ n E s,m M τ S cτ i.e. the supremum on all stopping times is achieved on the stopping times of the special form τ l and σ l. Namely if l [1/2c 1/2, 1/2c] then supremum is achieved on τ l. If l [1/2c 1, 1/2c 1/2] then supremum is achieved on σ l. 53

54 Take an arbitrary l N and compute Eτ l and EM τl S. Represent τ l as a sum τ l = τ 1 + τ 2 where τ 1 = inf{k 0 : S k = l}, τ 2 = inf{k 0 : max 0 i k S i+τ 1 S τ 1 S k+τ 1 S τ 1 = l} Due to Wald identities for Random walk we have Eτ 1 = ES 2 τ 1 = l 2. Also note that the distribution law of τ 2 coincides with distribution law of the time inf{k 0 : M k S S k = l}. This Markov time can be represented as a sum of M τ 2S + 1 i.i.d. random variables with distribution of τ l,1 = inf{k 0 : S k = l or S k = 1}. Therefore since EM τ 2S = EM τ 2S S τ 2 = l we get Eτ 2 = EM τ 2 + 1Eτ l,1 = ll + 1 Here we used Wald identities ES τ l,1 = 0, ES 2 τ l,1 = Eτ l,1 in order to prove that Eτ l,1 = l. 54

55 Finally we have Eτ l = Eτ 1 + Eτ 2 = l 2 + ll + 1 = l2l + 1 and EM τl S = E max S k + E max S k = 2l i.e. 0 k τ 1 0 k τ 2 Eτ l = l2l + 1, EM τl S = 2l From this system we find that EM τl S = 8Eτ l + 1 1/2. Theorem 6 Dubins-Schwarz For any Markov time τ M the following sharp maximal inequality holds: E max S n 0 n τ If we consider the Markov time 8Eτ τ = inf{n 0 : max 0 k n S k S n = N} for any N N then 23 becomes an equality

56 2. Maximal inequalities for simple symmetric Random walk Consider the optimal stopping problem V c = sup E τ M max S k cτ 0 k τ Theorem 7. The optimal stopping time τ c and value function V c in problem equal inf{k 0 : S k 1 2 = 1 2c }, if 1 2c c, τ c = inf{k 0 : S k 1 2 = 1 2c }, if 1 2c < 1 2c. 1 V c = 2c c 2 2c c 1 1, if c c, 1 2c c 2 2c c , if 2c + 1 < 1 2 2c, where x is the integer part of x. 56

57 Proof. According to the discrete version of Levy theorem [Fujita, Mischenko] Law max S S, max S = Law S , LS, where LS = L n S n 0, L n S is the number of crossings of the level 1/2 by Random walk on [0, n]. Rewriting the problem and using Wald identities we have EM τ S cτ = EM τ S S τ ces 2 τ = E S τ 1/2 1/2 cs 2 τ 1/2 Since S 2 τ = S τ 1/2 2 + S τ 1/4 we can rewrite the last expression E S τ 1/2 cs 2 τ 1/2 = E S τ 1/2 c S τ 1/2 2 + c/4 1/2 24 Observe that the resulting expression does not depend on τ explicitly, there is only dependence on S τ 1/2. That s why the method we use is called the method of space change. 57

58 Consider the function fx = x cx 2, x 0. It attains a maximum at the point c 0 = 1/2c and therefore x cx 2 f 2c 1 = 1/4c. Hence from 24 we get sup E τ M max S n cτ 0 n τ 1 4c + c However this inequality can be not sharp if 1 does not belong to 2c the values set E = {k + 1/2} k 0 of the process S 1/2. fx 1 4c V c i i c 1 c x 58

59 Nevertheless it is clear that the maximum of S τ 1/2 c S τ 1/2 2 is attained at the closest point to 1/2c i.e. at the point i 0 = c 1. The values of optimal stopping time τ c and value function V c depend on the relation between 2 distances 1 = 1/2c i 0 + 1/2 and 2 = i 0 + 1/2 1/2c: τ c = inf{k 0 : inf{k 0 : S k 1 2 = i }, if 1 2, S k 1 2 = i }, if 1 > 2 V c = fi c 4 1 2, if 1 2, fi c 4 1 2, if 1 > 2 59

60 Theorem 8. For any Markov time τ M the following inequality holds: 4Eτ E max S n 0 n τ 2 25 If for any N N we consider the Markov time τ = inf{n 0 : max 0 k n S k S n = N} then 25 becomes an equality. Proof. Use the inequality 24 which we already proved: { E max S n inf c Eτ n τ c>0 4 4c 1 } 4Eτ = 2 2 which gives us exactly

61 Now show that 25 is sharp. Due to the discrete version of Levy theorem the time τ = inf{n 0 : max S k S n = N} coincides by 0 k n distribution with inf{n 0 : S n 1/2 1/2 = N} = inf{n 0 : S n = N or S n = N + 1} = τ N,N+1 Using Wald identities we can check that Eτ = Eτ N,N+1 = NN + 1 On the other hand EM τ = EM τ S τ = N = 4NN

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