Convergence of the stochastic Euler scheme for locally Lipschitz ...
arXiv:0912.2609v1 [math.NA] 14 Dec 2009
Convergence of the stochastic Euler scheme for locally Lipschitz coefficients Martin Hutzenthaler and Arnulf Jentzen University of Munich (LMU) and Bielefeld University December 14, 2009
Abstract Stochastic differential equations are often simulated with the Monte Carlo Euler method. Convergence of this method is well understood in the case of globally Lipschitz continuous coefficients of the stochastic differential equation. The important case of superlinearly growing coefficients, however, remained an open question for a long time now. The main difficulty is that numerically weak convergence fails to hold in many cases of superlinearly growing coefficients. In this paper we overcome this difficulty and establish convergence of the Monte Carlo Euler method for a large class of one-dimensional stochastic differential equations whose drift functions have at most polynomial growth.
1
Introduction
Many applications require the numerical approximation of moments or expectations of other functionals of the solution of a stochastic differential equation (SDE) whose coefficients are superlinearly growing, see Section 3 for some examples. Moments are often approximated by discretizing time using the stochastic Euler scheme (see e.g. [9, 12] and [8]) (a.k.a. Euler-Maruyama scheme) and by approximating expectations with the Monte Carlo method. This Monte Carlo Euler method has been shown to converge in the case of globally Lipschitz continuous coefficients of the SDE (see e.g. Section 14.1 in [9] and Section 12 in [12]). The important case of superlinearly growing coefficients, however, has been an open problem for a long time now. The main difficulty is that numerically weak convergence fails to hold in many cases of superlinearly growing AMS subject classifications Primary 65C05; Secondary 65C30. Euler scheme, stochastic differential equations, Monte Carlo Euler method, local Lipschitz condition.
1
coefficients, see [6]. In this paper we overcome this difficulty and establish convergence of the Monte Carlo Euler method for a large class of SDEs with at most polynomial growing drift functions and with globally Lipschitz continuous diffusion functions, see Section 2 for the exact statement. For clarity of exposition, we concentrate in this section on the following prominent example. Let T ∈ (0, ∞) be fixed and let (Xt )t∈[0,T ] be the unique strong solution of the one-dimensional SDE dXt = −Xt3 dt + σ ¯ dWt ,
X0 = x0
(1)
for all t ∈ [0, T ] where (Wt )t∈[0,T ] is a one-dimensional standard Brownian motion and where σ ¯ ∈ [0, ∞) and x0 ∈ R are given constants. Our goal is then to solve the weak approximation problem of the SDE (1). More formally, we want to compute moments and, more generally, the deterministic real number h i E f XT (2) for a given smooth function f : R → R whose derivatives have at most polynomial growth. A frequently used scheme for solving this problem is the Monte Carlo Euler method. In this method, time is discretized through the stochastic Euler scheme and expectations are approximated by the Monte Carlo method. More formally, the Euler approximation (YnN )n∈{0,1,...,N } of the solution (Xt )t∈[0,T ] of the SDE (1) is defined recursively through Y0N = x0 and N Yn+1 = YnN −
3 T YnN + σ ¯ · W (n+1)T − W nT N N N
(3)
for every n ∈ {0, 1, . . . , N − 1} and every N ∈ N := {1, 2, . . .}. Moreover, let YnN,m , n ∈ {0, 1, . . . , N }, N ∈ N, for m ∈ N be independent copies of the Euler approximation defined in (3). The Monte Carlo Euler approximation of (2) with N ∈ N time steps and M ∈ N Monte Carlo runs is then the random real number ! M X 1 N,m f YN . (4) M m=1 In order to balance the error due to the Euler method and the error due to Monte Carlo method, it turns out to be optimal to have M increasing at order of N 2 , see [2]. We say that the Monte Carlo Euler method converges h N2 i X 1 N,m lim E f XT − 2 f YN almost surely =0 N →∞ N m=1
the the if
(5)
holds for every smooth function f : R → R whose derivatives have at most polynomial growth.
2
In the literature, convergence of the Monte Carlo Euler method is usually established by estimating the time discretization error and by estimating the statistical error. More formally, the triangle inequality yields N2 h i 1 X f YNN,m E f XT − 2 N m=1 N2 h h i i h i 1 X N N − 2 ≤ E f XT − E f YN + E f YN f YNN,m N m=1
(6)
for all N ∈ N. The first summand on the right-hand side of (6) is the time discretization error due to approximating the exact solution with Euler’s method. The second summand on the right-hand side of (6) is the statistical error which is due to approximating an expectation with the arithmetic average over independent copies. The time discretization error is usually the more difficult part to estimate. This is why the concept of numerically weak convergence, which concentrates on that error part, has been studied intensively in the literature (see for instance [10, 8, 1, 13, 11, 5, 12, 15] or Part VI in [9]). To give a definition, we say that the stochastic Euler scheme converges in the numerically weak sense (not to be confused with stochastic weak convergence) if the time discretization error converges to zero, i.e., h h i i lim E f XT − E f YNN = 0 (7) N →∞
holds for every smooth function f : R → R whose derivatives have at most polynomial growth. If the coefficients of the SDE are globally Lipschitz continuous, then numerically weak convergence of Euler’s method and convergence of the Monte Carlo Euler method is well-established, see e.g. Theorem 14.1.5 in [9] and Section 12 in [12]. The case of superlinearly growing coefficients is more subtle. The main difficulty in that case is that numerically weak convergence usually fails to hold, see [6] for a large class of examples. More precisely, the sequence E (YNN )2 , N ∈ N, of second moments of the Euler approximations (3) diverges to infinity although the second moment E (XT )2 of the exact solution of the SDE (1) is finite and, hence, we have h i h i lim E (XT )2 − E (YNN )2 = ∞ (8) N →∞
instead of (7). So we should not split up the error of the Monte Carlo Euler method as in (6). To motivate our approach, we first need to understand why Euler’s method does not converge in the sense of numerically weak convergence. In the deterministic case, that is, (1) and (3) with σ ¯ = 0, the Euler approximation diverges if the starting point is sufficiently large. This divergence has been estimated in [6] and turns out to be at least double-exponentially fast. Now in the presence of noise (¯ σ > 0), the Brownian motion has an exponentially small chance to push the Euler approximation outside of [−N, N ]. On this 3
event, the Euler approximation grows at least double-exponentially fast due to the deterministic dynamics. Consequently, as being double-exponentially large over-compensates that the event has an exponentially small probability, the L2 norm of the Euler approximation diverges to infinity and, hence, numerically weak convergence fails to hold. Now we indicate for example (1) with x0 = 0 and σ ¯ = 1 why the Monte Carlo Euler method converges although the stochastic Euler scheme fails to converge in the sense of numerically weak convergence. Consider the event p ΩN := {sup0≤t≤T |Wt | ≤ N/(2T )} and note that the probability of (ΩN )c is exponentially small in N ∈ N. The key step in our proof is to show that the Euler approximation does not diverge on ΩN as N ∈ N goes to infinity. More precisely, one can show that the Euler approximations (3) are uniformly dominated on ΩN by twice the supremum of the Brownian motion, i.e., N sup 1ΩN YN ≤ 2 sup |Wt | (9) N ∈N
0≤t≤T
holds. Consequently, the restricted absolute moments are uniformly bounded p N p p sup E 1ΩN YN ≤2 ·E sup |Wt | 0, we write r− for the convergence order if the convergence order is better than r − ε for every arbitrarily small ε ∈ (0, r). We therefore choose M = N 2 in order to balance the error arising from Euler’s approximation and the error arising from the Monte Carlo approximation. Both error terms are then bounded by a random multiple of N1 . Since O (M · N ) = O N 3 function evaluations, arithmetical operations and random variables are needed to compute the Monte Carlo Euler approximation (4), the Monte Carlo Euler method converges with order 1 3 − with respect to the computational effort in the case of global Lipschitz coefficients of the SDE (see [2]). Theorem 2.1 shows that 13 − is also the convergence order in the case of superlinearly growing coefficients of the SDE. Simulations support this result, see Section 4.
2
Main result
We establish convergence of the Monte Carlo Euler method for more general one-dimensional diffusions than our introductory example (1). More precisely, we pose the following assumptions on the coefficients. The drift function is assumed to be globally one-sided Lipschitz continuous and the diffusion function is assumed to be globally Lipschitz continuous. Additionally, both the drift function and the diffusion function are assumed to have a continuous fourth derivative with at most polynomial growth. We introduce further notation for the formulation of our main result. Fix T ∈ (0, ∞) and let (Ω, F, P) be a probability space with a normal filtration (Ft )t∈[0,T ] . Let ξ (m) : Ω → R, m ∈ N, be a sequence of independent, identih p i cally distributed F0 /B(R)-measurable mappings with E ξ (1) < ∞ for every p ∈ [1, ∞) and let W (m) : [0, T ] × Ω → R, m ∈ N, be a sequence of independent scalar standard (Ft )t∈[0,T ] -Brownian motions with continuous sample paths. Furthermore, let µ, σ : R → R be four times continuously differentiable functions. Generalizing (1), let (Xt )t∈[0,T ] be a one-dimensional diffusion with drift function µ and diffusion function σ. More precisely, let X : [0, T ] × Ω → R be an (up to indistinguishability unique) adapted stochastic process with continuous sample paths which satisfies Z t Z t (1) (1) P Xt = ξ + µ(Xs ) ds + σ(Xs ) dWs =1 (13) 0
0
for every t ∈ [0, T ]. The functions µ and σ 2 are the infinitesimal mean and the infinitesimal variance respectively. Next we introduce independent versions of the Euler approximation. Define F/B(R)-measurable mappings YnN,m : Ω → R, n ∈ {0, 1, . . . , N }, N ∈ N, m ∈ N, 5
by Y0N,m (ω) := ξ (m) (ω) and by T · µ YnN,m (ω) N (m) (m) + σ YnN,m (ω) · W (n+1)T (ω) − W nT (ω)
N,m Yn+1 (ω) := YnN,m (ω) +
N
(14)
N
for every n ∈ {0, 1, . . . , N − 1}, N ∈ N and every m ∈ N. Now we formulate the main result of this article. Theorem 2.1. Suppose that f, µ, σ : R → R are four times continuously differentiable functions with (n) (n) (n) ∀x ∈ R (15) f (x) + µ (x) + σ (x) ≤ L 1 + |x|δ for every n ∈ {0, 1, . . . , 4}, where L ∈ (0, ∞) and δ ∈ (1, ∞) are fixed constants. Moreover, assume that the drift coefficient is globally one-sided Lipschitz continuous 2 (x − y) · (µ(x) − µ(y)) ≤ L (x − y) ∀ x, y ∈ R (16) and that the diffusion coefficient is globally Lipschitz continuous |σ(x) − σ(y)| ≤ L |x − y|
∀ x, y ∈ R.
(17)
Then there are F/B([0, ∞))-measurable mappings Cε : Ω → [0, ∞), ε ∈ (0, 1), ˜ ∈ F with P[Ω] ˜ = 1 such that and a set Ω 2 N X 1 N,m E f (XT ) − 1 ≤ Cε (ω) · f (Y (ω)) (18) N 2 (1−ε) N N m=1 ˜ N ∈ N and every ε ∈ (0, 1). holds for every ω ∈ Ω, The proof is deferred to Section 5. Moreover, note that the assumptions of Theorem 2.1 ensure the existence of an adapted stochastic process X : [0, T ] × Ω → R with continuous sample paths which satisfies (13) and which satisfies p E sup |Xt | < ∞ (19) 0≤t≤T
for all p ∈ [1, ∞) (see Theorem 2.6.4 in [3]). Therefore, the expression E f (XT ) in (18) in Theorem 2.1 is well-defined. Since O N 3 function evaluations, arithmetical and random variP operations N2 N,m 1 (ω)) in (18) for ables are needed to compute the expression N 2 m=1 f (YN N ∈ N, Theorem 2.1 shows that the Monte Carlo Euler method converges under the above assumptions with order 13 − with respect to the computational effort. This is the standard convergence order as in the global Lipschitz case (see e.g. [2]). 6
3
Examples
In order to illustrate Theorem 2.1, we give four examples in this section. First we provide a simple condition for global one-sided Lipschitz continuity of the drift function µ. Let µ : R → R be continuously differentiable and let its derivative be globally bounded above by L ∈ (0, ∞). Then the fundamental theorem of calculus implies Z x (x − y) µ(x) − µ(y) = (x − y) µ0 (s) ds y (20) Z x 2 L ds = L(x − y) ≤ (x − y) y
for all x, y ∈ R. So every continuously differentiable function whose derivative is globally bounded above is globally one-sided Lipschitz continuous. The introductory example We begin with the introductory example (1) of Section 1. More precisely, we consider the SDE dXt = −Xt3 dt + σ ¯ dWt ,
X0 = x0
(21)
for all t ∈ [0, T ] where σ ¯ ∈ [0, ∞) and x0 ∈ R are given constants. Suppose we want to compute the second moment of XT . If we verify the assumptions of Theorem 2.1, then Theorem 2.1 reassures us that the Monte Carlo Euler method converges to the exact value of the second moment of XT . The drift function 0 µ : R → R is here given by µ(x) = −x3 for all x ∈ R. Its derivative µ (x) = −3x2 for all x ∈ R is globally bounded above by zero. So the drift function is globally one-sided Lipschitz continuous. The diffusion function σ : R → R is given by σ(x) = σ ¯ for all x ∈ R. Clearly the diffusion function is globally Lipschitz continuous. The test function at hand is f : R → R given by f (x) = x2 for all x ∈ R. All three functions µ, σ and f are polynomials and are therefore infinitely differentiable with all derivatives being again polynomials. One can check that (15), (16) and (17) hold with δ = 3 and L = 8 + σ ¯ . Hence, the assumptions of Theorem 2.1 are fulfilled in this example. Stochastic Ginzburg-Landau equation The Ginzburg-Landau equation is from the theory of superconductivity. It has been introduced by Ginzburg and Landau (1950) [4] to describe a phase transition. A stochastic version with multiplicative noise is given by the solution (Xt )t∈[0,T ] of 1 2 dXt = (η + σ ¯ )Xt − λXt3 dt + σ ¯ Xt dWt , X0 = x0 (22) 2 for all t ∈ [0, T ] where η ∈ [0, ∞) and λ, σ ¯ , x0 ∈ (0, ∞) are given constants. Its solution is known explicitly (see e.g. Section 4.4 in [9]) and is given by x0 exp (ηt + σ ¯ Wt ) Xt = q R t 1 + 2x20 λ 0 exp (2ηs + 2¯ σ Ws ) ds 7
(23)
for all t ∈ [0, T ]. As in the previous example, we want to approximate the second moment of the exact solution of the SDE (22) at time T . The assumptions of Theorem 2.1 are again easily verified. The drift function is given by µ(x) = ¯ 2 )x − λx3 for all x ∈ R. As its derivative µ0 (x) = (η + 12 σ ¯ 2 ) − 3λx2 , (η + 21 σ 1 2 ¯ , we see that the drift function x ∈ R, is globally bounded above by η + 2 σ is globally one-sided Lipschitz continuous. The diffusion function σ(x) = σ ¯ · x, x ∈ R, is globally Lipschitz continuous. All three functions µ, σ and f : R → R given by f (x) = x2 for all x ∈ R are polynomials and are infinitely differentiable with all derivatives being again polynomials. One can check that (15), (16) and ¯ 2 + 6λ + σ ¯ . Hence, the assumptions of (17) hold with δ = 3 and L = 2 + η + 12 σ Theorem 2.1 are also fulfilled in this example. Stochastic Verhulst equation The Verhulst equation is an ordinary differential equation and is a simple model for a population with competition between individuals. A stochastic version with multiplicative noise is given by the solution (Xt )t∈[0,T ] of 1 2 dXt = (η + σ ¯ )Xt − λXt2 dt + σ ¯ Xt dWt , X0 = x0 (24) 2 for all t ∈ [0, T ] where η ∈ [0, ∞) and λ, σ ¯ , x0 ∈ (0, ∞) are given constants. Its solution is also known explicitly (see e.g. Section 4.4 in [9]) and is given by Xt =
x0 exp (ηt + σ ¯ Wt ) Rt 1 + x0 λ 0 exp (ηs + σ ¯ Ws ) ds
(25)
for all t ∈ [0, T ]. Note that the unbounded interval (0, ∞) is the state space in this example. Both the drift function µ(x) = (η + 12 σ ¯ 2 )x − λx2 and the diffusion function σ(x) = σ ¯ · x are defined for x ∈ (0, ∞). In order to apply Theorem 2.1, we need to extend µ and σ to the whole real line such that µ and σ satisfy the assumptions of Theorem 2.1. One can check that this indeed is possible. Of course, the exact solution (Xt )t∈[0,T ] is not changed by such an extension of µ and σ. An SDE with non-polynomial coefficients All previous examples have polynomials as coefficients. Now we give an example with irrational drift function and rational diffusion function. Let (Xt )t∈[0,T ] be the solution of 1 + Xt dXt = sin (Xt ) − Xt5 dt + dWt , 1 + Xt2
X0 = x0
(26)
for all t ∈ [0, T ] where x0 ∈ R is a given constant. The drift function and 1+x the diffusion function are here given by µ(x) = sin(x) − x5 and σ(x) = 1+x 2 respectively for all x ∈ R. Both functions are infinitely differentiable and one can check that all derivatives have polynomial growth. The derivative of the drift function is given by µ0 (x) = cos(x) − 5x4 for all x ∈ R and is globally bounded above by 1. The derivative of the diffusion function is given by σ 0 (x) = 8
N = 20 1.1379 N = 25 0.4641
N = 21 0.9118 N = 26 0.4663
N = 22 0.4258 N = 27 0.4859
N = 23 0.2942 N = 28 0.4904
N = 24 0.4386 N = 29 0.4935
Table 1: Monte Carlo Euler approximations (29) of E (X1 )2 of the SDE (27). ˆ
1+x2 −(1+x)·2x (1+x2 )2
˜
2
= 1−2x−x (1+x2 )2 for all x ∈ R and is globally bounded. Hence, the drift function is globally one-sided Lipschitz continuous and the diffusion function is globally Lipschitz continuous. Suppose we want to compute the fourth moment of the exact solution of the SDE (26) at time T . Then the test function is the monomial f : R → R given by f (x) = x4 for all x ∈ R. Hence, the assumptions of Theorem 2.1 are fulfilled in this example.
4
Simulations
In this section, we simulate the second moment of two stochastic differential equations. First we simulate the stochastic Ginzburg-Landau equation, which we choose as there exists an explicit solution for this SDE. Let (Xt )t∈[0,T ] be the solution of (22) with the parameters being set to T = 1, η = 0, σ ¯ = 1, λ = 1 and x0 = 1. The SDE (22) then reads as 1 3 Xt − Xt dt + Xt dWt , X0 = 1 (27) dXt = 2 for all t ∈ [0, 1]. Moreover, the exact solution at time 1 is given by X1 = q
x0 exp (W1 ) . R1 1 + 2 0 exp (2Ws ) ds
(28)
The exact value of the second moment E (XT )2 = E (X1 )2 is not known. Instead we use the exact solution (28) at time 1 to approximate the second moment. For this, we approximate the Lebesgue integral in the denominator of (28) with a Riemann sum with 3 · 103 summands. Moreover, we approximate the second moment at time 1 by a Monte Carlo simulation with 107 independent approximations of X1 . This results in the approximate value E (X1 )2 ≈ 0.4945. Next we approximate the second moment at time 1 with the Monte Carlo Euler method. We will sample one random ω ∈ Ω and calculate the Monte Carlo Euler approximations for this ω ∈ Ω for different discretization step sizes. More precisely, Table 1 shows the Monte Carlo Euler approximation N2 1 X N,m 2 (ω) Y N 2 m=1 N
9
(29)
of the second moment at time 1 of the SDE (27) for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. In Figure 1, the approximation error of these Monte
Figure ˆ ˜1:
Approximation error (30) of the Monte Carlo Euler approximations (29) of E (X1 )2 of the SDE (27).
Carlo Euler approximations, i.e., the quantity N2 2 X 1 0.4945 − Y N,m (ω) , N 2 m=1 N
(30)
is plotted against N 3 for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. Note that N 3 is the computational effort up to a constant. The three order lines in Figure 1 correspond to the convergence orders 61 , 13 and 21 . Hence, Figure 1 indicates that the Monte Carlo Euler method converges in the case of the stochastic Ginzburg-Landau equation (27) with its theoretically predicted order 31 −. Next we simulate our introductory example. Let (Xt )t∈[0,T ] be the solution of the SDE (21) with T = 1, σ ¯ = 1 and x0 = 0. The SDE (21) thus reads as dXt = −Xt3 dt + dWt ,
X0 = 0
(31)
for all t ∈ [0, 1]. Here there exists no explicit expression for the or its solution second moments. As an approximation of the exact value E (X1 )2 , we now take a Monte Carlo Euler approximation with a larger N . We choose N = 212 10
N = 20 1.4516 N = 25 0.4452
N = 21 0.5166 N = 26 0.4602
N = 22 0.4329 N = 27 0.4517
N = 23 0.5308 N = 28 0.4548
N = 24 0.4285 N = 29 0.4537
Table 2: Monte Carlo Euler approximations (32) of E (X1 )2 of the SDE (31). ˆ
˜
and obtain as approximation the value E (X1 )2 ≈ 0.4529. Table 2 shows the value of the Monte Carlo Euler approximation N2 1 X N,m 2 Y (ω) N 2 m=1 N
(32)
of the second moment at time 1 of the SDE (31) for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. In Figure 2, the approximation error of these Monte
Figure ˆ ˜2:
Approximation error (33) of the Monte Carlo Euler approximations (32) of E (X1 )2 of the SDE (31).
Carlo Euler approximations, i.e., the quantity N2 X 1 N,m 2 0.4529 − (YN (ω)) , 2 N m=1 11
(33)
is plotted against N 3 for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. Note that N 3 is the computational effort up to a constant. The three order lines in Figure 2 correspond to the convergence orders 16 , 31 and 12 . Therefore, Figure 2 suggests that the Monte Carlo Euler method converges in the case of the SDE (31) with its theoretically predicted order 31 −.
5
Proof of Theorem 2.1
First we introduce more notation. Recall the Brownian motion W (1) : [0, T ] × Ω → R and the Euler approximations YnN,1 : Ω → R, n ∈ {0, 1, . . . , N }, N ∈ N, from Section 2. Throughout this section, we use the stochastic process W : [0, T ] × Ω → R and the F/B(R)-measurable mappings YnN : Ω → R, n ∈ {0, 1, . . . , N }, N ∈ N, given by (1)
Wt (ω) := Wt (ω)
and
YnN (ω) := YnN,1 (ω)
(34)
for every t ∈ [0, T ], n ∈ {0, 1, . . . , N }, ω ∈ Ω and every N ∈ N.
5.1
Outline
For general drift and diffusion functions, our proof of Theorem 2.1 somewhat buries the main new ideas. To explain these ideas, we give now an outline. The main step will be to establish uniform boundedness of the restricted absolute moments of the Euler approximations h i sup E YNN 1ΩN < ∞ (35) N ∈N
where (ΩN )N ∈N is a sequence of events whose probabilities converge to 1 sufficiently fast. From here, one can then adapt the arguments of the global Lipschitz case to derive the modified numerically weak convergence (12) and to obtain Theorem 2.1. The idea behind (35) is now explained on the example of negative cubic drift and multiplicative noise. Formally, we consider µ(x) = −x3 , σ(x) = x for all x ∈ R and ξ (1) (ω) = 1 for all ω ∈ Ω. The Euler approximation (34) is then given by Y0N = 1 and N Yn+1 = YnN −
3 T YnN + YnN · W (n+1)T − W nT N N N
(36)
for all n ∈ {0, 1, . . . , N − 1} and all N ∈ N. Fix N ∈ N and assume that the Euler approximation (YkN )k∈{0,1,...,N } does not change sign until and including the n-th approximation step for some n ∈ {0, 1, . . . , N } which we fix for now, that is, YkN ≥ 0 for all k ∈ {0, 1, . . . , n}. Then, using 1 + x ≤ exp(x) for all
12
x ∈ R, we have 3 T N N N YkN = Yk−1 − Yk−1 + Yk−1 · W kT − W (k−1)T N N N N ≤ Yk−1 1 + W kT − W (k−1)T N N N ≤ Yk−1 exp W kT − W (k−1)T N
(37)
N
and iterating this inequality shows N YkN ≤ Yk−2 exp W (k−1)T − W (k−2)T exp W kT − W (k−1)T N N N N N = Yk−2 exp W kT − W (k−2)T N N N ˜ kN ≤ . . . ≤ Y0 exp W kT =: D
(38)
N
for all k ∈ {0, 1, . . . , n}. Thus the Euler approximation (YkN )k∈{0,1,...,n} is ˜ N )k∈{0,1,...,n} . This dominatbounded above by the dominating process (D k ing process has uniformly bounded absolute moments. So the absolute moments of the Euler approximation can only be unbounded if the Euler approximation changes its sign. Now if Yk happens to be very large for one k ∈ {0, 1, . . . , N − 1}, then Yk+1 is negative with absolute value being very large because of the negative cubic drift. Through a sequence of changes in sign, it could happen that the absolute value of the Euler approximation increases more and more. To avoid this, we restrict the Euler approximation to an event ΩN on which the drift alone cannot change the sign of the Euler approximation. On ΩN , the Euler approximation changes sign only due to the diffusive part. As the diffusion function is at most linearly growing, these changes of sign can be controlled. In between consecutive changes of sign, the Euler approximation is again bounded by a dominating process as above. Through this pathwise comparison with a dominating process, we will establish the uniform boundedness (35) of the restricted absolute moments. For the details, we refer to Lemma 5.2, which is the key result in our proof of Theorem 2.1.
5.2
Notation and auxiliary lemmas
In order to show Theorem 2.1, the following objects are needed. First of all, nT define tN n := N for every n ∈ {0, 1, . . . , N } and every N ∈ N and let the B(R)/B(R)-measurable mapping σ ˜ : R → R be given by ( (σ(x)−σ(0)) : x 6= 0 x (39) σ ˜ (x) := 0 :x=0 for all x ∈ R. Moreover, let the F/B(R)-measurable mappings αnN,m : Ω → R and βnN,m : Ω → R be given by TL (m) (m) αnN,m (ω) := +σ ˜ (YnN,m (ω)) · WtN (ω) − WtN (ω) (40) n n+1 N 13
and
T µ(0) (m) (m) + σ(0) · WtN (ω) − WtN (ω) (41) n n+1 N for every ω ∈ Ω, n ∈ {0, 1, . . . , N − 1} and every N, m ∈ N. For simplicity we also use αnN , βnN : Ω → R given by αnN (ω) := αnN,1 (ω) and βnN (ω) := βnN,1 (ω) for every ω ∈ Ω, n ∈ {0, 1, . . . , N − 1} and every N ∈ N. Using these ingredients, we now define the dominating process. Let the F/B(R)-measurable mapping N,m Dv,w : Ω → R be given by Pw−1 N,m N,m Dv,w (ω) := e( l=v αl (ω)) T |µ(0)| + |σ(0)| + ξ (m) (ω) + 1 w−1 Pw−1 X N,m + sgn(YkN,m (ω)) e( l=k+1 αl (ω)) βkN,m (ω) (42) βnN,m (ω) :=
k=v
for every ω ∈ Ω, v, w ∈ {0, 1, . . . , N } and every N, m ∈ N, where sgn : R → {−1, 1} is given by sgn(x) := 1 for every x ∈ [0, ∞) and sgn(x) := −1 for N,m : Ω → R only depends on the Brownian every x ∈ (−∞, 0). Note that Dv,w (m) motion W : [0, T ] × Ω → R and the initial random variable ξ (m) : Ω → R N,m , m ∈ N, for every v, w ∈ {0, 1, . . . , N } and every N, m ∈ N. Therefore, Dv,w is a sequence of independent random variables for every v, w ∈ {0, 1, . . . , N } and every N ∈ N. We will show that the Euler approximation is dominated by the dominating process since the last change of sign. More formally, let the F/P ({0, 1, . . . , n})-measurable mapping τnN : Ω → {0, 1, . . . , n} be given by N N N τn (ω) := max {0} ∪ k ∈ {1, 2, . . . , n} sgn(Yk−1 (ω)) 6= sgn(Yk (ω)) (43) for every ω ∈ Ω, n ∈ {0, 1, . . . , N } and every N ∈ N. The random time τnN : Ω → {0, 1, . . . , n} is the last time of a change of sign of YkN , k ∈ {0, 1, . . . , n}, for every n ∈ {0, 1, . . . , N } and every N ∈ N. Lemma 5.2 below shows that |YnN | is on a certain event ΩN,n for every n ∈ {0, 1, . . . , N } and every bounded by DτN,1 N n ,n N ∈ N. Next we define these events ΩN,n , n ∈ {0, 1, . . . , N }, N ∈ N, such that the drift alone cannot cause a change of sign and such that the increment of the Brownian motion is not extraordinary large. Let the real number rN ∈ [0, ∞) be given by 1 (δ−1) 1 N N 4 − 1 , max 0, rN := min (44) L 0 T sup |µ (s)| + 3L s∈[−1,1]
˜ N,n , ΩN,n , Ωm ∈ F and sets ΩN ∈ F by for every N ∈ N. Now define sets Ω N − 41 ˜ N,n := ω ∈ Ω N N Ω sup , (45) (ω) − W (ω) ≤ N Wtk+1 tk k∈{0,1,...,n−1}
by ( ΩN,n :=
ω ∈ Ω
) N,1 Dv,w (ω) ≤ rN
sup v,w∈{0,1,...,n}
14
˜ N,n , ∩Ω
(46)
by
Ωm N
:=
ω ∈ Ω
sup v,w∈{0,1,...,N }
N,m Dv,w (ω) ≤ rN ,
sup n∈{0,1,...,N −1}
(m) (m) − 41 WtN (ω) − WtN (ω) ≤ N n+1
(47)
n
and by ΩN := ΩN,N = Ω1N
(48)
˜ ∈ F by for every n ∈ {0, 1, . . . , N } and every N, m ∈ N. Finally, define Ω ˜ := Ω
2
[
∞ M \ \
Ωm M
N ∈N M =N m=1
P 2 N (ω)·f (YNN,m (ω)) − E 1ΩN f (YNN ) m=1 1Ωm \ N < ∞ . ω ∈ Ω sup N (1+ε) N ∈N
ε>0
(49) p
1
˜ is indeed in F. Moreover, we write kZk p := (E [|Z| ]) p ∈ [0, ∞] Note that Ω L for all p ∈ [1, ∞) and all F/B(R)-measurable mappings Z : Ω → R. Our proof of Theorem 2.1 uses the following lemmas. Lemma 5.1 (Burkholder-Davis-Gundy inequality). Let N ∈ N and let Z1 , . . . , ZN : Ω → R be F/B(R)-measurable mappings with E|Zn |2 < ∞ for all n ∈ {1, . . . , N } and with E [Zn+1 |Z1 , . . . , Zn ] = 0 for all n ∈ {1, . . . , N − 1}. Then we obtain 12 2 2 kZ1 + . . . + ZN kLp ≤ Kp · kZ1 kLp + . . . + kZN kLp
(50)
for every p ∈ [2, ∞), where Kp , p ∈ [2, ∞), are universal constants. The following lemma is the key result in our proof of Theorem 2.1. N,1 Lemma 5.2 (Dominator Lemma). Let YnN : Ω → R, Dn,m : Ω → R, τnN : Ω → R and ΩN,n ∈ F for n, m ∈ {0, 1, . . . , N } and N ∈ N be given by (34), (42), (43) and (46). Then we have N Yn (ω) ≤ DN,1 (ω) (51) τ N (ω),n n
for every ω ∈ ΩN,n , n ∈ {0, 1, . . . , N } and every N ∈ N. The domination (51) might not look helpful at first view since DτN,1 depends N n ,n N on the Euler approximation and since τn is in general not a stopping time for all n ∈ {1, . . . , N } and all N ∈ N. However, the dependence of the dominating 15
process on the Euler approximation can be controlled as σ ˜ is bounded and the N dependence of DτN,1 on τ for all n ∈ {0, 1, . . . , N } and all N ∈ N is no problem N n n ,n N,1 as Dv,w can be controlled uniformly in v, w ∈ {0, 1, . . . , N } and N ∈ N. This is subject of the following lemma. Lemma 5.3 (Bounded moments of the dominator). N,1 ˜ N,n ∈ F and Dn,m Let Ω : Ω → [0, ∞) for n, m ∈ {0, 1, . . . , N } and N ∈ N be given by (45) and (42). Then we have " !# N,1 p Dv,w sup sup E 1Ω˜ N,n sup Yn+1 (ω) ≥ YnN (ω) · 1 + (ω) +σ ˜ YnN (ω) · WtN (ω) − W tn n+1 N T µ(0) + (ω) − WtN (ω) + σ(0) · WtN n n+1 N „
TL σ N +˜
„
(YnN (ω))·
«« WtN
(ω)−WtN (ω)
n n+1 ≥ YnN (ω) · e T µ(0) N + + σ(0) · WtN (ω) − W (ω) tn n+1 N N
= YnN (ω) · eαn (ω) + βnN (ω) N = − YnN (ω) · eαn (ω) + sgn(YnN (ω)) · βnN (ω) .
24
Hence, the induction hypothesis yields N Yn+1 (ω) ≤ DN,1 N =e
N
τn (ω),n “P ” n N α (ω) l=τ N (ω) l
(T |µ(0)| + |σ(0)| + |ξ(ω)| + 1)
n
n−1 X
+
N (ω) k=τn “P n
Pn
sgn(YkN (ω)) e(
N (ω) l=τn
=e
n X
+
(ω) · eαn (ω) + sgn(YnN (ω)) · βnN (ω)
” αN l (ω)
l=k+1
αN l (ω)) β N (ω) k
+ sgn(YnN (ω)) · βnN (ω)
(T |µ(0)| + |σ(0)| + |ξ(ω)| + 1) Pn
sgn(YkN (ω)) e(
l=k+1
αN l (ω)) β N (ω) k
N (ω) k=τn
N,1 = DτN,1 N (ω),n+1 (ω) = Dτ N
n+1 (ω),n+1
n
(ω),
which shows that (51) also holds in this case for ω and n + 1. N (ω) < 0 holds. Then 3.) In the third case assume that YnN (ω) ≥ 0 and that Yn+1 N we obtain τn+1 (ω) = n + 1. Additionally, note that N N,1 Yn (ω) ≤ sup (69) Dk,l (ω) ≤ rN k,l∈{0,1,...,n}
holds due to the induction hypothesis and since ω ∈ ΩN,n+1 ⊂ ΩN,n . Hence, (64) yields T µ(0) T ≥0 (70) YnN (ω) + µ YnN (ω) − N N and therefore N 0 ≥ Yn+1 (ω)
T N (ω) (ω) − W = YnN (ω) + µ YnN (ω) + σ YnN (ω) · WtN tn n+1 N T T µ(0) = YnN (ω) + µ YnN (ω) − N N T µ(0) + (ω) − WtN (ω) + σ YnN (ω) · WtN n n+1 N T µ(0) N (ω) + σ YnN (ω) · WtN (ω) − W ≥ t n n+1 N and T µ(0) − σ YnN (ω) · WtN (ω) − WtN (ω) 0≥ ≥ − n n+1 N N N ≥ −T |µ(0)| − L · Yn (ω) + |σ(0)| · WtN (ω) − W (ω) . tn n+1 N Yn+1 (ω)
25
(71)
This implies N N N (ω) − W (ω) 0 ≥ Yn+1 (ω) ≥ −T |µ(0)|− L · DτN,1 (ω) + |σ(0)| · W N tn tn+1 n (ω),n (ω) − WtN (ω) ≥ −T |µ(0)| − (L · rN + |σ(0)|) · WtN n n+1 ! 1 N4 N (ω) (ω) − W + |σ(0)| · WtN ≥ −T |µ(0)| − L · t n n+1 L and 1 1 N 0 ≥ Yn+1 (ω) ≥ −T |µ(0)| − |σ(0)| + N 4 · 1 N4 |σ(0)| = −T |µ(0)| − − 1 ≥ −T |µ(0)| − |σ(0)| − 1 1 N4 and finally N Yn+1 (ω) ≤ T |µ(0)| + |σ(0)| + 1 + |ξ(ω)| Pn
= e( +
l=n+1
n X
αN l (ω))
sgn(YkN (ω)) e(
k=n+1 „ Pn
=e +
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|)
l=τ N (ω) n+1
n X
αN l (ω)
Pn
l=k+1
αN l (ω)) β N (ω) k
«
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|) Pn
sgn(YkN (ω)) e(
l=k+1
αN l (ω)) β N (ω) k
N (ω) k=τn+1
= DτN,1 N
n+1 (ω),n+1
(ω),
which shows that (51) also holds in this case for ω and n + 1. N 4.) In the last case assume that YnN (ω) < 0 and that Yn+1 (ω) ≥ 0 holds. Then N we also obtain τn+1 (ω) = n + 1. Note that N N,1 Yn (ω) ≤ sup (72) Dk,l (ω) ≤ rN k,l∈{0,1,...,n}
holds due to the induction hypothesis and since ω ∈ ΩN,n+1 ⊂ ΩN,n . Therefore, (65) implies T µ(0) T YnN (ω) + µ YnN (ω) − ≤0 (73) N N
26
and hence N 0 ≤ Yn+1 (ω)
T N (ω) = YnN (ω) + µ YnN (ω) + σ YnN (ω) · WtN (ω) − W t n n+1 N T µ(0) T = YnN (ω) + µ YnN (ω) − N N T µ(0) N (ω) + + σ YnN (ω) · WtN (ω) − W t n n+1 N T µ(0) N (ω) + σ YnN (ω) · WtN (ω) − W ≤ t n n+1 N
(74)
and T µ(0) N + σ YnN (ω) · WtN (ω) − WtN (ω) 0 ≤ Yn+1 (ω) ≤ n n+1 N N N (ω) ≤ T |µ(0)| + L · Yn (ω) + |σ(0)| · WtN (ω) − W t n n+1 N N (ω) + |σ(0)| · (ω) − W (ω) ≤ T |µ(0)| + L · DτN,1 W . N (ω),n tn tn+1
(75)
n
This shows N Yn+1 (ω) ≤ T |µ(0)| + (L · rN + |σ(0)|) · WtN (ω) − WtN (ω) n n+1 ! 1 N4 ≤ T |µ(0)| + L · + |σ(0)| · WtN (ω) − WtN (ω) n n+1 L 1 1 |σ(0)| ≤ T |µ(0)| + |σ(0)| + N 4 · 1 = T |µ(0)| + +1 1 4 N N4
(76)
and finally N Yn+1 (ω) ≤ T |µ(0)| + |σ(0)| + 1 + |ξ(ω)| Pn
= e( +
l=n+1
n X
αN l (ω))
sgn(YkN (ω)) e(
k=n+1 „ Pn
=e +
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|)
l=τ N (ω) n+1
n X
αN l (ω)
Pn
l=k+1
αN l (ω)) β N (ω) k
«
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|) Pn
sgn(YkN (ω)) e(
l=k+1
N (ω) k=τn+1
αN l (ω)) β N (ω) k
= DτN,1 N
n+1 (ω),n+1
(ω),
which shows that (51) holds in this case for ω and n + 1 and which finally yields (51) for every ω ∈ ΩN,n , n ∈ {0, 1, 2, . . . , N } and every N ∈ N by induction.
27
5.5
Proof of Lemma 5.3
In order to bound the restricted moments of the dominating process, we need to know the exponential moments of a standard normal distributed random variable and we also need to estimate the p-th moment of a normal distributed random variable for every p ∈ [1, ∞). Lemma 5.7. Let Y : Ω → R be a standard normal distributed F/B(R)-measurable mapping. Then we obtain that c2 (77) E ecY = e 2 holds for every c ∈ R. Proof. By completing the square, we obtain Z Z 2 1 x2 1 1 √ e− 2 (x −2cx) dx E ecY = ecx √ e− 2 dx = 2π 2π R ZR Z 2 c2 1 1 − 12 (x2 −2cx+c2 ) c2 1 √ e √ e− 2 (x−c) dx = e 2 dx = e 2 2π 2π R R Z c2 c2 1 − x2 √ e 2 dx = e 2 =e2 2π R
(78)
for every c ∈ R. Lemma 5.8. Let Y : Ω → R be a normal distributed F/B(R)-measurable mapping. Then we obtain that kY kLp ≤ p kY kL2
(79)
holds for every p ∈ [1, ∞). Proof. First of all, H¨ older’s inequality implies kY kLp ≤ kY kL2 ≤ p kY kL2
(80)
for every p ∈ [1, 2), which shows (79) in the case p ∈ [1, 2). Denote now the mean r hof Y : Ω i→ R by c := E [Y ] ∈ R and the standard deviation by 2 σ := E (Y − c) ∈ [0, ∞). If Yˆ : Ω → R is a standard normal distributed F/B(R)-measurable mapping, then
Y
Lp
= σ Yˆ + c
Lp
≤ σ Yˆ + c
Ldpe
1 dpe dpe ˆ = E σ Y + c
1 dpe dpe k X dpe k (dpe−k) ≤ E |σ| Yˆ |c| k
k=0
1 dpe dpe k X dpe k (dpe−k) = |σ| E Yˆ |c| k
k=0
28
(81)
k/2 ˆ k holds for for all k ∈ {2, 3, . . . }, qevery p ∈ [2, ∞). Using E|Y | ≤ (k − 1) 2 2 2 E|Yˆ | ≤ E(Yˆ )2 = 1 and (σ + c) ≤ 2(σ + c ) yields 1 dpe dpe X dpe dpe (dpe−k) k ≤ |σ| (dpe − 1) 2 |c| k k=0 p p √ = (dpe − 1) (σ + c) ≤ p (σ + c) ≤ p (σ 2 + c2 ) = p kY kL2
kY kLp
for every p ∈ [2, ∞), which finally shows (79). Proof of Lemma 5.3. Note that the discrete time stochastic integral n−1 X
N z·σ ˜ (YlN ) · WtN − W t l+1 l
(82)
l=0
is a martingale in n ∈ {0, 1, . . . , N } for every z ∈ {−1, 1} and every N ∈ N. Since the exponential function is convex, we obtain that ! n−1 X N exp z·σ ˜ (Yl ) · WtN − WtN (83) l+1 l l=0
is a positive submartingale in n ∈ {0, 1, . . . , N } for every z ∈ {−1, 1} and every N ∈ N. Therefore, Doob’s inequality (see e.g. Theorem 11.2 (ii) in [7]) shows „ „ «« p # " Pn−1 σ (YlN )· WtN −WtN l=0 z·˜ l+1 l E sup e n∈{0,1,...,N } „ «« p # p " „PN −1 σ (YlN )· WtN −WtN p l=0 z·˜ l+1 l E e ≤ p−1 „ „ «« " p PN −2 pz ˜ (YlN )· WtN −WtN p l=0 σ l+1 l (84) E e = p−1 „ «« " „ ## N pz σ ˜ (YN −1 )· WtN −WtN FtN N N −1 ·E e N −1 =
p p−1
p
" E e
„ pz
PN −2 l=0
„ «« σ ˜ (YlN )· WtN −WtN l+1
29
l
N e (pz·˜σ(YN −1 )) 1 2
2 T N
#
due to (78) and hence, using |˜ σ (x)| ≤ L for every x ∈ R, „ „ ««
Pn−1
σ (YlN )· WtN −WtN
l=0 z·˜ l+1 l sup e
p
n∈{0,1,...,N } L „ „ «« # ( ) p1 " p PN −2 pz ˜ (YlN )· WtN −WtN 1 2 2 T p l=0 σ l+1 l ≤ E e e2p L N p−1 p p1 2 1 2 2 1 p p p L T e2 = ≤ ... ≤ e 2 pL T p−1 p−1 for every p ∈ (1, ∞), z ∈ {−1, 1} and every N ∈ N. This implies
P
z ( n−1 αN ) l l=0 sup e
n∈{0,1,...,N }
p L „ „ ««
P
n−1 nT L z ˜ (YlN )· WtN −WtN
N + l=0 σ l+1 l sup e =
n∈{0,1,...,N }
p L „ „ ««
Pn−1
N z·˜ σ (Y )· W −W l
l=0 tN tN l+1 l ≤ eT L sup e
n∈{0,1,...,N }
p L p 2 2 1 p pL T (L +L)T TL ≤ 2e 2 e2 ≤e p−1
(85)
(86)
for every p ∈ [2, ∞), z ∈ {−1, 1} and every N ∈ N. Moreover, the BurkholderDavis-Gundy inequality (50) and again Doob’s inequality yield
n−1
X P N
−( k−1 α N N ) l l=0 sgn(Yk ) e βk sup
n∈{0,1,...,N } p k=0
L N −1
Pk−1 N p
X
≤ sgn(YkN ) e−( l=0 αl ) βkN
(p − 1) k=0 Lp (87) ! 21 N −1
2 Pk−1 N X pKp
≤
sgn(YkN ) e−( l=0 αl ) βkN p (p − 1) L k=0 ! 12 N −1
X
N 2 N 2
−(Pk−1 α )
βk 2p l l=0 ≤ 2Kp
e
L2p
k=0
L
for every p ∈ [2, ∞) and every N ∈ N. The last step is H¨older’s inequality. Now
30
using Lemma 5.8 and inequality (86), we see that
n−1
P N X
−( k−1 α N N ) l l=0 sgn(Yk ) e sup βk
n∈{0,1,...,N } k=0 Lp r !2 12 N −1 X 2 2 T |µ(0)| T ≤ 2Kp 2ep(L +L)T + |σ(0)| 2p N N k=0 r ! 2 T |µ(0)| T √ = 4Kp ep(L +L)T N + |σ(0)| 2p N N holds for every p ∈ [2, ∞) and every N ∈ N. Hence, we have n−1
X
P N
−( k−1 α N N ) l l=0 sgn(Yk ) e sup βk
n∈{0,1,...,N } k=0 Lp √ p(L2 +L)T ≤ 8Kp e T |µ(0)| + p T |σ(0)| √ 2 ≤ 8LKp T + p T ep(L +L)T 2
≤ 8LpKp (T + 1) ep(L ≤ Kp ep(T +1)(L
2
+5L)
+L)T
≤ Kp e4Lp(T +1) ep(L
≤ Kp e5p(L
2
(88)
(89) 2
+L)T
+1)(T +1)
√ for every p ∈ [2, ∞) and every N ∈ N since x ≤ x2 + 1 and since 2x ≤ ex for every x ∈ [0, ∞). With these estimates at hand, we now bound the restricted p-th moment !# " N,1 p Dv,w (90) E 1˜ sup ΩN,n
v,w∈{0,1,...,n}
of the dominating process for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. By definition (42) and by the triangle inequality, we have
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L
!
Pw−1 N
≤ 1Ω˜ N,n sup e( l=v αl ) (T |µ(0)| + |σ(0)| + |ξ| + 1)
p v,w∈{0,1,...,n} L
w−1 !
X
Pw−1 N
+ 1Ω˜ N,n sup sgn(YkN ) e( l=k+1 αl ) βkN
v,w∈{0,1,...,n} k=v
Lp
31
and, using H¨ older’s inequality,
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
Pw−1 N
α ( )
l l=v sup e ≤ 1Ω˜ N,n
0≤v≤w≤n
+ 1Ω˜ N,n
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
L2p
w−1 ! X Pw−1 N sup sgn(YkN ) e( l=k+1 αl ) βkN 0≤v≤w≤n k=v
Lp
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Therefore, we obtain
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
Pw−1 N
α ( ) sup e l=v l (T |µ(0)| + |σ(0)| + kξkL2p + 1) ≤
0≤v≤w≤N
L2p
w−1 !
X Pw−1 N Pk N
α − α N N e( l=0 l ) 1Ω˜ N,n + sup sgn(Yk ) e ( l=0 l ) βk
0≤v≤w≤n k=v
Lp
and
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n}
L
Pv−1 N Pw−1 N
sup e−( l=0 αl ) sup e( l=0 αl ) ≤
4p
w∈{0,1,...,N }
4p v∈{0,1,...,N } L L
Pw−1 N
α ( ) · (T |µ(0)| + |σ(0)| + kξkL2p + 1) + sup e l=0 l
w∈{0,1,...,N }
2p L w−1
! X
Pk N
1 ˜ sgn(YkN ) e−( l=0 αl ) βkN
· sup
0≤v≤w≤n ΩN,n k=v
(91)
L2p
and, using (86),
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L 2 2p(L2 +L)T ≤ 2e (T |µ(0)| + |σ(0)| + kξkL2p + 1)
!
w−1 P N X α −( k p(L2 +L)T N N ) l=0 l + 2e 1Ω˜ N,n sgn(Yk ) e βk
sup
0≤v≤w≤n k=v
2
≤ 4e4p(L
(92)
L2p
+L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
!
w−1 Pk X N 2
α − + 4ep(L +L)T sup 1Ω˜ N,n sgn(YkN ) e ( l=0 l ) βkN
w∈{0,1,...,n} k=0
32
L2p
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and since every N ∈ N. Moreover, N (LT +L) N (ω) α (ω) ≤ T L + L, we have e−αN k α (ω) for every k ∈ − 1 ≤ e k
k
˜ N,n and every n ∈ {0, 1, . . . , N }, N ∈ N. Therefore, we {0, 1, . . . , n − 1}, ω ∈ Ω obtain w−1 X P N −( k α (ω) N N ) l=0 l sgn(Yk (ω)) e sup βk (ω) w∈{0,1,...,n} k=0 w−1 X P N N −( k−1 α (ω) N −α (ω) N ) l k l=0 sgn(Yk (ω)) e ≤ sup e − 1 βk (ω) w∈{0,1,...,n} k=0 w−1 Pk−1 N X sgn(YkN (ω)) e−( l=0 αl (ω)) βkN (ω) + sup w∈{0,1,...,n} k=0 ! n−1 Pk−1 N X N ≤ e−( l=0 αl (ω)) · e−αk (ω) − 1 · βkN (ω) k=0
w−1 P N X −( k−1 α (ω) N N ) l l=0 + sup sgn(Yk (ω)) e βk (ω) w∈{0,1,...,n} k=0
and w−1 X P N −( k α (ω) N N ) l=0 l sgn(Yk (ω)) e sup βk (ω) w∈{0,1,...,n} k=0 ! n−1 P X N −( k−1 αN (ω)) N (LT +L) l l=0 ≤e e αk (ω) · βk (ω) k=0
w−1 X Pk−1 N sgn(YkN (ω)) e−( l=0 αl (ω)) βkN (ω) + sup (93) w∈{0,1,...,n} k=0 ! ! n−1 P X N N −( w−1 α (ω) (LT +L) N ) α (ω) · β (ω) l l=0 ≤ sup e e k
w∈{0,1,...,n}
k
k=0
w−1 X P N −( k−1 α (ω) N N ) l l=0 + sup sgn(Yk (ω)) e βk (ω) w∈{0,1,...,n} k=0
˜ N,n , n ∈ {0, 1, . . . , N } and every N ∈ N. Inserting (93) into (92) for every ω ∈ Ω
33
yields
1Ω˜ N,n
≤ 4e
! N,1
Dv,w sup
v,w∈{0,1,...,n}
Lp
4p(L2 +L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
! ! n−1
P X −( w−1 αN p(L2 +L)T N N (LT +L) ) l l=0 sup e + 4e αk · βk e
w∈{0,1,...,n} k=0 L2p
w−1
P X k−1 N 2
sgn(YkN ) e−( l=0 αl ) βkN sup + 4ep(L +L)T
w∈{0,1,...,n} k=0
L2p
and, using H¨ older’s inequality and (89),
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
≤ 4e4p(L
2
+L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
Pw−1 N 2
+ 4ep(L +L)T e(LT +L) sup e−( l=0 αl )
w∈{0,1,...,n}
L
+ 4e
p(L2 +L)T
K2p e
n−1
X N N α k · βk
4p k=0
L4p
10p(L2 +1)(T +1)
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Therefore, (86) implies
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
≤ 4e
4p(L2 +L)T
+ 4e
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
p(L2 +L)T (LT +L)
e
2e
2p(L2 +L)T
n−1 X
!
(94)
N
αk 8p βkN 8p L L
k=0 2 2 + 4K2p e{p(L +L)T +10p(L +1)(T +1)}
and, using σ ˜ (x) ≤ L for all x ∈ R,
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L
2 2 ≤ 4e4p(L +L)T (T |µ(0)| + |σ(0)| + kξkL2p + 1) + 8e{3p(L +L)T +(LT +L)} · ! n−1
T |µ(0)|
X TL
+ L WtN − W tN + |σ(0)| WtN − WtN
8p
8p k+1 k k+1 k N N L L
k=0
+ 4K2p e{2p(L
2
+1)T +10p(L2 +1)(T +1)}
34
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Now we apply Lemma 5.8 to WtN − WtN for k ∈ {0, 1, . . . , N − 1}, N ∈ N to obtain k+1 k
1Ω˜ N,n
≤ 4e
! N,1 Dv,w sup
v,w∈{0,1,...,n}
Lp
4p(L2 +L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1) " r #" r #! n−1 X TL 2 T |µ(0)| T T + 8e{6p(L +1)T +L(T +1)} + L8p + |σ(0)|8p N N N N k=0
+ 4K2p e
12p(L2 +1)(T +1)
and
1Ω˜ N,n
! N,1 Dv,w sup
v,w∈{0,1,...,n}
Lp
2
≤ 4e8p(L
+1)(T +1)
(T |µ(0)| + |σ(0)| + kξkL2p + 1) √ √ 6p(L2 +1)T +L(T +1)} { + 8e T L + 8Lp T T |µ(0)| + |σ(0)|8p T
+ 4K2p e12p(L
2
+1)(T +1)
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Finally this implies
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L 2 (95) ≤ 4e12p(L +1)(T +1) (T |µ(0)| + |σ(0)| + kξkL2p + 1 + K2p ) √ √ 2 + 512Le{6p(L +1)T +L(T +1)} T + p T T |µ(0)| + |σ(0)|p T for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N, which yields the assertion in the case p ∈ [2, ∞). The case p ∈ [1, 2) then follows from Jensen’s inequality.
5.6
Proof of Lemma 5.5
Proof of Lemma 5.5. First of all, we have " # c − 41 ˜ P ΩN,N =P sup − W tN WtN >N n n+1 n∈{0,1,...,N −1}
N −1 − 41 = P ∪n=0 ω ∈ Ω WtN (ω) − WtN (ω) > N n n+1 ≤
N −1 X
h i − 41 N > N P WtN − W t n n+1
n=0
35
and P
˜ N,N Ω
c
i h − 41 N > N − W ≤ N · P WtN t0 1 h
= 2N · P W T > N
− 41
i
N
and Z c ˜ P ΩN,N ≤ 2N
"r = 2N · P
1 N W T > N−4 T N
r
N T
#
√ Z ∞ 1 − x2 1 x T − x2 2 2 dx dx ≤ 2N e 1 √ 1 √ 1 e 4 4 N N 4 2π 2π N √ √ T T Z ∞ 3√ x2 3√ 1 1 h − x2 ix=∞ − 1 = 2N 4 T 1 √ x e 2 dx = 2N 4 T √ −e 2 4 4 √ N x= N 2π 2π √ T T (96) r √ √ √ N N 3√ 3√ 1 2 3√ − N = 2N 4 T √ e− 2T = N 4 T e 2T ≤ N 4 T e− 2T π 2π ∞
for every N ∈ N. Additionally, note that ) ( N,1 ˜ N,N Dv,w (ω) ≤ rN ∩ Ω sup ΩN = ω ∈ Ω v,w∈{0,1,...,N } ! ) ( N,1 ˜ N,N Dv,w (ω) ≤ rN ∩ Ω sup = ω ∈ Ω 1Ω˜ N,N (ω)
(97)
v,w∈{0,1,...,N }
holds for every N ∈ N. Hence, we have ( c sup (ΩN ) = ω ∈ Ω 1Ω˜ N,N (ω)
v,w∈{0,1,...,N }
! N,1 Dv,w (ω)
) > rN
c ˜ N,N ∪ Ω
for every N ∈ N. Therefore, inequality (96) implies " ! # c N,1 c ˜ N,N Dv,w > rN + P Ω P (ΩN ) ≤ P 1 ˜ sup ΩN,N
" ≤P
1 + 1Ω˜ N,N
v,w∈{0,1,...,N }
sup v,w∈{0,1,...,N }
!!p # √ N,1 N 3√ p Dv,w > (1 + rN ) + N 4 T e− 2T
for every N ∈ N and every p ∈ [1, ∞). Now we apply Markov’s inequality to obtain h N,1 p i E 1 + 1˜ √ sup v,w∈{0,1,...,N } Dv,w ΩN,N N 3√ c − 2T 4 T e P (ΩN ) ≤ + N (1 + rN )p
! p !
√ M,1 N 3√ 1
Dv,w ≤ sup 1+ 1Ω˜ M,M sup + N 4 T e− 2T
p
(1 + r ) N M ∈N v,w∈{0,1,...,M } p L
36
and ! p ! M,1 Dv,w sup
p v,w∈{0,1,...,M } L √ 1 N 3√ p −p·min( 41 , (δ−1) − 2T ) 4 ·c N Te +N
! !#p "
M,1
Dv,w ≤ c 1 + sup 1Ω˜ M,M sup
p M ∈N v,w∈{0,1,...,M } L √ √ 1 1 N 3 −p·min( 4 , (δ−1) ) − ·N + N 4 T e 2T
c P (ΩN ) ≤
sup 1 + 1Ω˜ M,M M ∈N
(98)
for every N ∈ N and every p ∈ [1, ∞), where we used 1 1 1 ≤ c · N − min( 4 , (δ−1) ) (1 + rN )
for every N ∈ N with c ∈ (0, ∞) given by N min( 4 , (δ−1) ) (1 + rN ) 1
1
c := sup N ∈N
! < ∞.
Moreover, (98) with p := 4 max(4, δ − 1) yields c P (ΩN )
c 1 + sup 1Ω˜ M,M M ∈N
" ≤N
−4
√
−
+ N Te "
!#4 max(4,δ−1)
M,1 Dv,w sup
v,w∈{0,1,...,M }
L4 max(4,δ−1)
√ N 2T
!#4 max(4,δ−1)
M,1
Dv,w ≤N c 1 + sup 1Ω˜ M,M sup
4 max(4,δ−1) M ∈N v,w∈{0,1,...,M } L " !# √ √ x + N −4 T sup x5 e− 2T −4
x∈[0,∞)
for every N ∈ N. The right-hand side is finite by Lemma 5.3. This proves 4 c c˜ := sup N · P (ΩN ) ∈ [0, ∞). (99) N ∈N
˜ has probability 1. We have Next we show that the event Ω c 2 2 ∞ M ∞ M \ \ \ \ = 1 − P P Ωm Ωm M M M =N m=1
M =N m=1
= 1 − P
2
∞ M [ [
M =N m=1
37
c (Ωm M)
(100)
and hence 2 ∞ M ∞ X M2 \ \ X m m c P ΩM ≥ 1 − P (ΩM ) M =N m=1
M =N m=1
∞ X c 2 =1− M · P (ΩM ) ≥1− M =N
∞ X
! c˜M
−2
M =N
for every N ∈ N due to (99). This implies 2 s=∞ Z ∞ ∞ M \ \ 1 1 ds = 1 − c ˜ − P Ωm ≥ 1 − c ˜ M 2 s s=N −1 N −1 s m=1 M =N
=1−
c˜ (N − 1)
for every N ∈ {2, 3, . . .}, which shows 2 2 ∞ M ∞ M [ \ \ \ \ = lim P = 1. P Ωm Ωm M M N ∈N M =N m=1
N →∞
M =N m=1
Moreover, we have P 2 p N · f (YNN,m ) − E 1ΩN f (YNN ) m=1 1Ωm N E sup N (1+ε) N ∈N P 2 p N N,m N ∞ m · f (Y 1 ) − E 1 f (Y ) X Ω Ω N N N m=1 N ≤ E N (1+ε)p N =1
P 2
N
p ∞ · f (YNN,m ) − E 1ΩN f (YNN ) p X m=1 1Ωm N L = (1+ε)p N N =1 21 p PN 2 N N 2
K 1 f (Y ) − E 1 f (Y ) ∞ p ΩN ΩN N N m=1 X Lp ≤ N (1+ε)p N =1
38
(101)
due to Lemma 5.1 and therefore P 2 p N · f (YNN,m ) − E 1ΩN f (YNN ) m=1 1Ωm N E sup N (1+ε) N ∈N
p ∞ X Kp · N · 1ΩN f (YNN ) − E 1ΩN f (YNN ) Lp ≤ N (1+ε)p N =1 p ! p ∞ X (Kp ) · 1ΩN f (YNN ) − E 1ΩN f (YNN ) Lp = N εp N =1 ! ∞ h X p i p ≤ (Kp ) N −εp sup E 1ΩN f (YNN ) − E 1ΩN f (YNN )
(102)
N ∈N
N =1
for every p ∈ [2, ∞) and every ε ∈ (0, ∞). Hence, we obtain P 2 p N · f (YNN,m ) − E 1ΩN f (YNN ) m=1 1Ωm N E sup N (1+ε) N ∈N ! ∞ h X p i p p −εp ≤ (2 (Kp ) ) N sup E 1Ω f (YNN ) p
p
≤ (2 (Kp ) ) ≤ (2LKp )
p
N =1 ∞ X N =1 ∞ X
N
N
N =1
N
N ∈N
−εp
! sup E
h
N ∈N −εp
! sup E
N ∈N
h
1ΩN L
p
δ p i 1 + YNN δ p i
1ΩN 1 + YNN
and P 2 p N N,m N · f (Y ) − E 1 f (Y ) m=1 1Ωm Ω N N N N E sup (1+ε) N N ∈N ! ∞ h X pδ i p −εp ≤ (4LKp ) N 1 + sup E 1Ω YNN
Convergence of the stochastic Euler scheme for locally Lipschitz coefficients Martin Hutzenthaler and Arnulf Jentzen University of Munich (LMU) and Bielefeld University December 14, 2009
Abstract Stochastic differential equations are often simulated with the Monte Carlo Euler method. Convergence of this method is well understood in the case of globally Lipschitz continuous coefficients of the stochastic differential equation. The important case of superlinearly growing coefficients, however, remained an open question for a long time now. The main difficulty is that numerically weak convergence fails to hold in many cases of superlinearly growing coefficients. In this paper we overcome this difficulty and establish convergence of the Monte Carlo Euler method for a large class of one-dimensional stochastic differential equations whose drift functions have at most polynomial growth.
1
Introduction
Many applications require the numerical approximation of moments or expectations of other functionals of the solution of a stochastic differential equation (SDE) whose coefficients are superlinearly growing, see Section 3 for some examples. Moments are often approximated by discretizing time using the stochastic Euler scheme (see e.g. [9, 12] and [8]) (a.k.a. Euler-Maruyama scheme) and by approximating expectations with the Monte Carlo method. This Monte Carlo Euler method has been shown to converge in the case of globally Lipschitz continuous coefficients of the SDE (see e.g. Section 14.1 in [9] and Section 12 in [12]). The important case of superlinearly growing coefficients, however, has been an open problem for a long time now. The main difficulty is that numerically weak convergence fails to hold in many cases of superlinearly growing AMS subject classifications Primary 65C05; Secondary 65C30. Euler scheme, stochastic differential equations, Monte Carlo Euler method, local Lipschitz condition.
1
coefficients, see [6]. In this paper we overcome this difficulty and establish convergence of the Monte Carlo Euler method for a large class of SDEs with at most polynomial growing drift functions and with globally Lipschitz continuous diffusion functions, see Section 2 for the exact statement. For clarity of exposition, we concentrate in this section on the following prominent example. Let T ∈ (0, ∞) be fixed and let (Xt )t∈[0,T ] be the unique strong solution of the one-dimensional SDE dXt = −Xt3 dt + σ ¯ dWt ,
X0 = x0
(1)
for all t ∈ [0, T ] where (Wt )t∈[0,T ] is a one-dimensional standard Brownian motion and where σ ¯ ∈ [0, ∞) and x0 ∈ R are given constants. Our goal is then to solve the weak approximation problem of the SDE (1). More formally, we want to compute moments and, more generally, the deterministic real number h i E f XT (2) for a given smooth function f : R → R whose derivatives have at most polynomial growth. A frequently used scheme for solving this problem is the Monte Carlo Euler method. In this method, time is discretized through the stochastic Euler scheme and expectations are approximated by the Monte Carlo method. More formally, the Euler approximation (YnN )n∈{0,1,...,N } of the solution (Xt )t∈[0,T ] of the SDE (1) is defined recursively through Y0N = x0 and N Yn+1 = YnN −
3 T YnN + σ ¯ · W (n+1)T − W nT N N N
(3)
for every n ∈ {0, 1, . . . , N − 1} and every N ∈ N := {1, 2, . . .}. Moreover, let YnN,m , n ∈ {0, 1, . . . , N }, N ∈ N, for m ∈ N be independent copies of the Euler approximation defined in (3). The Monte Carlo Euler approximation of (2) with N ∈ N time steps and M ∈ N Monte Carlo runs is then the random real number ! M X 1 N,m f YN . (4) M m=1 In order to balance the error due to the Euler method and the error due to Monte Carlo method, it turns out to be optimal to have M increasing at order of N 2 , see [2]. We say that the Monte Carlo Euler method converges h N2 i X 1 N,m lim E f XT − 2 f YN almost surely =0 N →∞ N m=1
the the if
(5)
holds for every smooth function f : R → R whose derivatives have at most polynomial growth.
2
In the literature, convergence of the Monte Carlo Euler method is usually established by estimating the time discretization error and by estimating the statistical error. More formally, the triangle inequality yields N2 h i 1 X f YNN,m E f XT − 2 N m=1 N2 h h i i h i 1 X N N − 2 ≤ E f XT − E f YN + E f YN f YNN,m N m=1
(6)
for all N ∈ N. The first summand on the right-hand side of (6) is the time discretization error due to approximating the exact solution with Euler’s method. The second summand on the right-hand side of (6) is the statistical error which is due to approximating an expectation with the arithmetic average over independent copies. The time discretization error is usually the more difficult part to estimate. This is why the concept of numerically weak convergence, which concentrates on that error part, has been studied intensively in the literature (see for instance [10, 8, 1, 13, 11, 5, 12, 15] or Part VI in [9]). To give a definition, we say that the stochastic Euler scheme converges in the numerically weak sense (not to be confused with stochastic weak convergence) if the time discretization error converges to zero, i.e., h h i i lim E f XT − E f YNN = 0 (7) N →∞
holds for every smooth function f : R → R whose derivatives have at most polynomial growth. If the coefficients of the SDE are globally Lipschitz continuous, then numerically weak convergence of Euler’s method and convergence of the Monte Carlo Euler method is well-established, see e.g. Theorem 14.1.5 in [9] and Section 12 in [12]. The case of superlinearly growing coefficients is more subtle. The main difficulty in that case is that numerically weak convergence usually fails to hold, see [6] for a large class of examples. More precisely, the sequence E (YNN )2 , N ∈ N, of second moments of the Euler approximations (3) diverges to infinity although the second moment E (XT )2 of the exact solution of the SDE (1) is finite and, hence, we have h i h i lim E (XT )2 − E (YNN )2 = ∞ (8) N →∞
instead of (7). So we should not split up the error of the Monte Carlo Euler method as in (6). To motivate our approach, we first need to understand why Euler’s method does not converge in the sense of numerically weak convergence. In the deterministic case, that is, (1) and (3) with σ ¯ = 0, the Euler approximation diverges if the starting point is sufficiently large. This divergence has been estimated in [6] and turns out to be at least double-exponentially fast. Now in the presence of noise (¯ σ > 0), the Brownian motion has an exponentially small chance to push the Euler approximation outside of [−N, N ]. On this 3
event, the Euler approximation grows at least double-exponentially fast due to the deterministic dynamics. Consequently, as being double-exponentially large over-compensates that the event has an exponentially small probability, the L2 norm of the Euler approximation diverges to infinity and, hence, numerically weak convergence fails to hold. Now we indicate for example (1) with x0 = 0 and σ ¯ = 1 why the Monte Carlo Euler method converges although the stochastic Euler scheme fails to converge in the sense of numerically weak convergence. Consider the event p ΩN := {sup0≤t≤T |Wt | ≤ N/(2T )} and note that the probability of (ΩN )c is exponentially small in N ∈ N. The key step in our proof is to show that the Euler approximation does not diverge on ΩN as N ∈ N goes to infinity. More precisely, one can show that the Euler approximations (3) are uniformly dominated on ΩN by twice the supremum of the Brownian motion, i.e., N sup 1ΩN YN ≤ 2 sup |Wt | (9) N ∈N
0≤t≤T
holds. Consequently, the restricted absolute moments are uniformly bounded p N p p sup E 1ΩN YN ≤2 ·E sup |Wt | 0, we write r− for the convergence order if the convergence order is better than r − ε for every arbitrarily small ε ∈ (0, r). We therefore choose M = N 2 in order to balance the error arising from Euler’s approximation and the error arising from the Monte Carlo approximation. Both error terms are then bounded by a random multiple of N1 . Since O (M · N ) = O N 3 function evaluations, arithmetical operations and random variables are needed to compute the Monte Carlo Euler approximation (4), the Monte Carlo Euler method converges with order 1 3 − with respect to the computational effort in the case of global Lipschitz coefficients of the SDE (see [2]). Theorem 2.1 shows that 13 − is also the convergence order in the case of superlinearly growing coefficients of the SDE. Simulations support this result, see Section 4.
2
Main result
We establish convergence of the Monte Carlo Euler method for more general one-dimensional diffusions than our introductory example (1). More precisely, we pose the following assumptions on the coefficients. The drift function is assumed to be globally one-sided Lipschitz continuous and the diffusion function is assumed to be globally Lipschitz continuous. Additionally, both the drift function and the diffusion function are assumed to have a continuous fourth derivative with at most polynomial growth. We introduce further notation for the formulation of our main result. Fix T ∈ (0, ∞) and let (Ω, F, P) be a probability space with a normal filtration (Ft )t∈[0,T ] . Let ξ (m) : Ω → R, m ∈ N, be a sequence of independent, identih p i cally distributed F0 /B(R)-measurable mappings with E ξ (1) < ∞ for every p ∈ [1, ∞) and let W (m) : [0, T ] × Ω → R, m ∈ N, be a sequence of independent scalar standard (Ft )t∈[0,T ] -Brownian motions with continuous sample paths. Furthermore, let µ, σ : R → R be four times continuously differentiable functions. Generalizing (1), let (Xt )t∈[0,T ] be a one-dimensional diffusion with drift function µ and diffusion function σ. More precisely, let X : [0, T ] × Ω → R be an (up to indistinguishability unique) adapted stochastic process with continuous sample paths which satisfies Z t Z t (1) (1) P Xt = ξ + µ(Xs ) ds + σ(Xs ) dWs =1 (13) 0
0
for every t ∈ [0, T ]. The functions µ and σ 2 are the infinitesimal mean and the infinitesimal variance respectively. Next we introduce independent versions of the Euler approximation. Define F/B(R)-measurable mappings YnN,m : Ω → R, n ∈ {0, 1, . . . , N }, N ∈ N, m ∈ N, 5
by Y0N,m (ω) := ξ (m) (ω) and by T · µ YnN,m (ω) N (m) (m) + σ YnN,m (ω) · W (n+1)T (ω) − W nT (ω)
N,m Yn+1 (ω) := YnN,m (ω) +
N
(14)
N
for every n ∈ {0, 1, . . . , N − 1}, N ∈ N and every m ∈ N. Now we formulate the main result of this article. Theorem 2.1. Suppose that f, µ, σ : R → R are four times continuously differentiable functions with (n) (n) (n) ∀x ∈ R (15) f (x) + µ (x) + σ (x) ≤ L 1 + |x|δ for every n ∈ {0, 1, . . . , 4}, where L ∈ (0, ∞) and δ ∈ (1, ∞) are fixed constants. Moreover, assume that the drift coefficient is globally one-sided Lipschitz continuous 2 (x − y) · (µ(x) − µ(y)) ≤ L (x − y) ∀ x, y ∈ R (16) and that the diffusion coefficient is globally Lipschitz continuous |σ(x) − σ(y)| ≤ L |x − y|
∀ x, y ∈ R.
(17)
Then there are F/B([0, ∞))-measurable mappings Cε : Ω → [0, ∞), ε ∈ (0, 1), ˜ ∈ F with P[Ω] ˜ = 1 such that and a set Ω 2 N X 1 N,m E f (XT ) − 1 ≤ Cε (ω) · f (Y (ω)) (18) N 2 (1−ε) N N m=1 ˜ N ∈ N and every ε ∈ (0, 1). holds for every ω ∈ Ω, The proof is deferred to Section 5. Moreover, note that the assumptions of Theorem 2.1 ensure the existence of an adapted stochastic process X : [0, T ] × Ω → R with continuous sample paths which satisfies (13) and which satisfies p E sup |Xt | < ∞ (19) 0≤t≤T
for all p ∈ [1, ∞) (see Theorem 2.6.4 in [3]). Therefore, the expression E f (XT ) in (18) in Theorem 2.1 is well-defined. Since O N 3 function evaluations, arithmetical and random variP operations N2 N,m 1 (ω)) in (18) for ables are needed to compute the expression N 2 m=1 f (YN N ∈ N, Theorem 2.1 shows that the Monte Carlo Euler method converges under the above assumptions with order 13 − with respect to the computational effort. This is the standard convergence order as in the global Lipschitz case (see e.g. [2]). 6
3
Examples
In order to illustrate Theorem 2.1, we give four examples in this section. First we provide a simple condition for global one-sided Lipschitz continuity of the drift function µ. Let µ : R → R be continuously differentiable and let its derivative be globally bounded above by L ∈ (0, ∞). Then the fundamental theorem of calculus implies Z x (x − y) µ(x) − µ(y) = (x − y) µ0 (s) ds y (20) Z x 2 L ds = L(x − y) ≤ (x − y) y
for all x, y ∈ R. So every continuously differentiable function whose derivative is globally bounded above is globally one-sided Lipschitz continuous. The introductory example We begin with the introductory example (1) of Section 1. More precisely, we consider the SDE dXt = −Xt3 dt + σ ¯ dWt ,
X0 = x0
(21)
for all t ∈ [0, T ] where σ ¯ ∈ [0, ∞) and x0 ∈ R are given constants. Suppose we want to compute the second moment of XT . If we verify the assumptions of Theorem 2.1, then Theorem 2.1 reassures us that the Monte Carlo Euler method converges to the exact value of the second moment of XT . The drift function 0 µ : R → R is here given by µ(x) = −x3 for all x ∈ R. Its derivative µ (x) = −3x2 for all x ∈ R is globally bounded above by zero. So the drift function is globally one-sided Lipschitz continuous. The diffusion function σ : R → R is given by σ(x) = σ ¯ for all x ∈ R. Clearly the diffusion function is globally Lipschitz continuous. The test function at hand is f : R → R given by f (x) = x2 for all x ∈ R. All three functions µ, σ and f are polynomials and are therefore infinitely differentiable with all derivatives being again polynomials. One can check that (15), (16) and (17) hold with δ = 3 and L = 8 + σ ¯ . Hence, the assumptions of Theorem 2.1 are fulfilled in this example. Stochastic Ginzburg-Landau equation The Ginzburg-Landau equation is from the theory of superconductivity. It has been introduced by Ginzburg and Landau (1950) [4] to describe a phase transition. A stochastic version with multiplicative noise is given by the solution (Xt )t∈[0,T ] of 1 2 dXt = (η + σ ¯ )Xt − λXt3 dt + σ ¯ Xt dWt , X0 = x0 (22) 2 for all t ∈ [0, T ] where η ∈ [0, ∞) and λ, σ ¯ , x0 ∈ (0, ∞) are given constants. Its solution is known explicitly (see e.g. Section 4.4 in [9]) and is given by x0 exp (ηt + σ ¯ Wt ) Xt = q R t 1 + 2x20 λ 0 exp (2ηs + 2¯ σ Ws ) ds 7
(23)
for all t ∈ [0, T ]. As in the previous example, we want to approximate the second moment of the exact solution of the SDE (22) at time T . The assumptions of Theorem 2.1 are again easily verified. The drift function is given by µ(x) = ¯ 2 )x − λx3 for all x ∈ R. As its derivative µ0 (x) = (η + 12 σ ¯ 2 ) − 3λx2 , (η + 21 σ 1 2 ¯ , we see that the drift function x ∈ R, is globally bounded above by η + 2 σ is globally one-sided Lipschitz continuous. The diffusion function σ(x) = σ ¯ · x, x ∈ R, is globally Lipschitz continuous. All three functions µ, σ and f : R → R given by f (x) = x2 for all x ∈ R are polynomials and are infinitely differentiable with all derivatives being again polynomials. One can check that (15), (16) and ¯ 2 + 6λ + σ ¯ . Hence, the assumptions of (17) hold with δ = 3 and L = 2 + η + 12 σ Theorem 2.1 are also fulfilled in this example. Stochastic Verhulst equation The Verhulst equation is an ordinary differential equation and is a simple model for a population with competition between individuals. A stochastic version with multiplicative noise is given by the solution (Xt )t∈[0,T ] of 1 2 dXt = (η + σ ¯ )Xt − λXt2 dt + σ ¯ Xt dWt , X0 = x0 (24) 2 for all t ∈ [0, T ] where η ∈ [0, ∞) and λ, σ ¯ , x0 ∈ (0, ∞) are given constants. Its solution is also known explicitly (see e.g. Section 4.4 in [9]) and is given by Xt =
x0 exp (ηt + σ ¯ Wt ) Rt 1 + x0 λ 0 exp (ηs + σ ¯ Ws ) ds
(25)
for all t ∈ [0, T ]. Note that the unbounded interval (0, ∞) is the state space in this example. Both the drift function µ(x) = (η + 12 σ ¯ 2 )x − λx2 and the diffusion function σ(x) = σ ¯ · x are defined for x ∈ (0, ∞). In order to apply Theorem 2.1, we need to extend µ and σ to the whole real line such that µ and σ satisfy the assumptions of Theorem 2.1. One can check that this indeed is possible. Of course, the exact solution (Xt )t∈[0,T ] is not changed by such an extension of µ and σ. An SDE with non-polynomial coefficients All previous examples have polynomials as coefficients. Now we give an example with irrational drift function and rational diffusion function. Let (Xt )t∈[0,T ] be the solution of 1 + Xt dXt = sin (Xt ) − Xt5 dt + dWt , 1 + Xt2
X0 = x0
(26)
for all t ∈ [0, T ] where x0 ∈ R is a given constant. The drift function and 1+x the diffusion function are here given by µ(x) = sin(x) − x5 and σ(x) = 1+x 2 respectively for all x ∈ R. Both functions are infinitely differentiable and one can check that all derivatives have polynomial growth. The derivative of the drift function is given by µ0 (x) = cos(x) − 5x4 for all x ∈ R and is globally bounded above by 1. The derivative of the diffusion function is given by σ 0 (x) = 8
N = 20 1.1379 N = 25 0.4641
N = 21 0.9118 N = 26 0.4663
N = 22 0.4258 N = 27 0.4859
N = 23 0.2942 N = 28 0.4904
N = 24 0.4386 N = 29 0.4935
Table 1: Monte Carlo Euler approximations (29) of E (X1 )2 of the SDE (27). ˆ
1+x2 −(1+x)·2x (1+x2 )2
˜
2
= 1−2x−x (1+x2 )2 for all x ∈ R and is globally bounded. Hence, the drift function is globally one-sided Lipschitz continuous and the diffusion function is globally Lipschitz continuous. Suppose we want to compute the fourth moment of the exact solution of the SDE (26) at time T . Then the test function is the monomial f : R → R given by f (x) = x4 for all x ∈ R. Hence, the assumptions of Theorem 2.1 are fulfilled in this example.
4
Simulations
In this section, we simulate the second moment of two stochastic differential equations. First we simulate the stochastic Ginzburg-Landau equation, which we choose as there exists an explicit solution for this SDE. Let (Xt )t∈[0,T ] be the solution of (22) with the parameters being set to T = 1, η = 0, σ ¯ = 1, λ = 1 and x0 = 1. The SDE (22) then reads as 1 3 Xt − Xt dt + Xt dWt , X0 = 1 (27) dXt = 2 for all t ∈ [0, 1]. Moreover, the exact solution at time 1 is given by X1 = q
x0 exp (W1 ) . R1 1 + 2 0 exp (2Ws ) ds
(28)
The exact value of the second moment E (XT )2 = E (X1 )2 is not known. Instead we use the exact solution (28) at time 1 to approximate the second moment. For this, we approximate the Lebesgue integral in the denominator of (28) with a Riemann sum with 3 · 103 summands. Moreover, we approximate the second moment at time 1 by a Monte Carlo simulation with 107 independent approximations of X1 . This results in the approximate value E (X1 )2 ≈ 0.4945. Next we approximate the second moment at time 1 with the Monte Carlo Euler method. We will sample one random ω ∈ Ω and calculate the Monte Carlo Euler approximations for this ω ∈ Ω for different discretization step sizes. More precisely, Table 1 shows the Monte Carlo Euler approximation N2 1 X N,m 2 (ω) Y N 2 m=1 N
9
(29)
of the second moment at time 1 of the SDE (27) for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. In Figure 1, the approximation error of these Monte
Figure ˆ ˜1:
Approximation error (30) of the Monte Carlo Euler approximations (29) of E (X1 )2 of the SDE (27).
Carlo Euler approximations, i.e., the quantity N2 2 X 1 0.4945 − Y N,m (ω) , N 2 m=1 N
(30)
is plotted against N 3 for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. Note that N 3 is the computational effort up to a constant. The three order lines in Figure 1 correspond to the convergence orders 61 , 13 and 21 . Hence, Figure 1 indicates that the Monte Carlo Euler method converges in the case of the stochastic Ginzburg-Landau equation (27) with its theoretically predicted order 31 −. Next we simulate our introductory example. Let (Xt )t∈[0,T ] be the solution of the SDE (21) with T = 1, σ ¯ = 1 and x0 = 0. The SDE (21) thus reads as dXt = −Xt3 dt + dWt ,
X0 = 0
(31)
for all t ∈ [0, 1]. Here there exists no explicit expression for the or its solution second moments. As an approximation of the exact value E (X1 )2 , we now take a Monte Carlo Euler approximation with a larger N . We choose N = 212 10
N = 20 1.4516 N = 25 0.4452
N = 21 0.5166 N = 26 0.4602
N = 22 0.4329 N = 27 0.4517
N = 23 0.5308 N = 28 0.4548
N = 24 0.4285 N = 29 0.4537
Table 2: Monte Carlo Euler approximations (32) of E (X1 )2 of the SDE (31). ˆ
˜
and obtain as approximation the value E (X1 )2 ≈ 0.4529. Table 2 shows the value of the Monte Carlo Euler approximation N2 1 X N,m 2 Y (ω) N 2 m=1 N
(32)
of the second moment at time 1 of the SDE (31) for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. In Figure 2, the approximation error of these Monte
Figure ˆ ˜2:
Approximation error (33) of the Monte Carlo Euler approximations (32) of E (X1 )2 of the SDE (31).
Carlo Euler approximations, i.e., the quantity N2 X 1 N,m 2 0.4529 − (YN (ω)) , 2 N m=1 11
(33)
is plotted against N 3 for every N ∈ 20 , 21 , 22 , . . . , 29 and one random ω ∈ Ω. Note that N 3 is the computational effort up to a constant. The three order lines in Figure 2 correspond to the convergence orders 16 , 31 and 12 . Therefore, Figure 2 suggests that the Monte Carlo Euler method converges in the case of the SDE (31) with its theoretically predicted order 31 −.
5
Proof of Theorem 2.1
First we introduce more notation. Recall the Brownian motion W (1) : [0, T ] × Ω → R and the Euler approximations YnN,1 : Ω → R, n ∈ {0, 1, . . . , N }, N ∈ N, from Section 2. Throughout this section, we use the stochastic process W : [0, T ] × Ω → R and the F/B(R)-measurable mappings YnN : Ω → R, n ∈ {0, 1, . . . , N }, N ∈ N, given by (1)
Wt (ω) := Wt (ω)
and
YnN (ω) := YnN,1 (ω)
(34)
for every t ∈ [0, T ], n ∈ {0, 1, . . . , N }, ω ∈ Ω and every N ∈ N.
5.1
Outline
For general drift and diffusion functions, our proof of Theorem 2.1 somewhat buries the main new ideas. To explain these ideas, we give now an outline. The main step will be to establish uniform boundedness of the restricted absolute moments of the Euler approximations h i sup E YNN 1ΩN < ∞ (35) N ∈N
where (ΩN )N ∈N is a sequence of events whose probabilities converge to 1 sufficiently fast. From here, one can then adapt the arguments of the global Lipschitz case to derive the modified numerically weak convergence (12) and to obtain Theorem 2.1. The idea behind (35) is now explained on the example of negative cubic drift and multiplicative noise. Formally, we consider µ(x) = −x3 , σ(x) = x for all x ∈ R and ξ (1) (ω) = 1 for all ω ∈ Ω. The Euler approximation (34) is then given by Y0N = 1 and N Yn+1 = YnN −
3 T YnN + YnN · W (n+1)T − W nT N N N
(36)
for all n ∈ {0, 1, . . . , N − 1} and all N ∈ N. Fix N ∈ N and assume that the Euler approximation (YkN )k∈{0,1,...,N } does not change sign until and including the n-th approximation step for some n ∈ {0, 1, . . . , N } which we fix for now, that is, YkN ≥ 0 for all k ∈ {0, 1, . . . , n}. Then, using 1 + x ≤ exp(x) for all
12
x ∈ R, we have 3 T N N N YkN = Yk−1 − Yk−1 + Yk−1 · W kT − W (k−1)T N N N N ≤ Yk−1 1 + W kT − W (k−1)T N N N ≤ Yk−1 exp W kT − W (k−1)T N
(37)
N
and iterating this inequality shows N YkN ≤ Yk−2 exp W (k−1)T − W (k−2)T exp W kT − W (k−1)T N N N N N = Yk−2 exp W kT − W (k−2)T N N N ˜ kN ≤ . . . ≤ Y0 exp W kT =: D
(38)
N
for all k ∈ {0, 1, . . . , n}. Thus the Euler approximation (YkN )k∈{0,1,...,n} is ˜ N )k∈{0,1,...,n} . This dominatbounded above by the dominating process (D k ing process has uniformly bounded absolute moments. So the absolute moments of the Euler approximation can only be unbounded if the Euler approximation changes its sign. Now if Yk happens to be very large for one k ∈ {0, 1, . . . , N − 1}, then Yk+1 is negative with absolute value being very large because of the negative cubic drift. Through a sequence of changes in sign, it could happen that the absolute value of the Euler approximation increases more and more. To avoid this, we restrict the Euler approximation to an event ΩN on which the drift alone cannot change the sign of the Euler approximation. On ΩN , the Euler approximation changes sign only due to the diffusive part. As the diffusion function is at most linearly growing, these changes of sign can be controlled. In between consecutive changes of sign, the Euler approximation is again bounded by a dominating process as above. Through this pathwise comparison with a dominating process, we will establish the uniform boundedness (35) of the restricted absolute moments. For the details, we refer to Lemma 5.2, which is the key result in our proof of Theorem 2.1.
5.2
Notation and auxiliary lemmas
In order to show Theorem 2.1, the following objects are needed. First of all, nT define tN n := N for every n ∈ {0, 1, . . . , N } and every N ∈ N and let the B(R)/B(R)-measurable mapping σ ˜ : R → R be given by ( (σ(x)−σ(0)) : x 6= 0 x (39) σ ˜ (x) := 0 :x=0 for all x ∈ R. Moreover, let the F/B(R)-measurable mappings αnN,m : Ω → R and βnN,m : Ω → R be given by TL (m) (m) αnN,m (ω) := +σ ˜ (YnN,m (ω)) · WtN (ω) − WtN (ω) (40) n n+1 N 13
and
T µ(0) (m) (m) + σ(0) · WtN (ω) − WtN (ω) (41) n n+1 N for every ω ∈ Ω, n ∈ {0, 1, . . . , N − 1} and every N, m ∈ N. For simplicity we also use αnN , βnN : Ω → R given by αnN (ω) := αnN,1 (ω) and βnN (ω) := βnN,1 (ω) for every ω ∈ Ω, n ∈ {0, 1, . . . , N − 1} and every N ∈ N. Using these ingredients, we now define the dominating process. Let the F/B(R)-measurable mapping N,m Dv,w : Ω → R be given by Pw−1 N,m N,m Dv,w (ω) := e( l=v αl (ω)) T |µ(0)| + |σ(0)| + ξ (m) (ω) + 1 w−1 Pw−1 X N,m + sgn(YkN,m (ω)) e( l=k+1 αl (ω)) βkN,m (ω) (42) βnN,m (ω) :=
k=v
for every ω ∈ Ω, v, w ∈ {0, 1, . . . , N } and every N, m ∈ N, where sgn : R → {−1, 1} is given by sgn(x) := 1 for every x ∈ [0, ∞) and sgn(x) := −1 for N,m : Ω → R only depends on the Brownian every x ∈ (−∞, 0). Note that Dv,w (m) motion W : [0, T ] × Ω → R and the initial random variable ξ (m) : Ω → R N,m , m ∈ N, for every v, w ∈ {0, 1, . . . , N } and every N, m ∈ N. Therefore, Dv,w is a sequence of independent random variables for every v, w ∈ {0, 1, . . . , N } and every N ∈ N. We will show that the Euler approximation is dominated by the dominating process since the last change of sign. More formally, let the F/P ({0, 1, . . . , n})-measurable mapping τnN : Ω → {0, 1, . . . , n} be given by N N N τn (ω) := max {0} ∪ k ∈ {1, 2, . . . , n} sgn(Yk−1 (ω)) 6= sgn(Yk (ω)) (43) for every ω ∈ Ω, n ∈ {0, 1, . . . , N } and every N ∈ N. The random time τnN : Ω → {0, 1, . . . , n} is the last time of a change of sign of YkN , k ∈ {0, 1, . . . , n}, for every n ∈ {0, 1, . . . , N } and every N ∈ N. Lemma 5.2 below shows that |YnN | is on a certain event ΩN,n for every n ∈ {0, 1, . . . , N } and every bounded by DτN,1 N n ,n N ∈ N. Next we define these events ΩN,n , n ∈ {0, 1, . . . , N }, N ∈ N, such that the drift alone cannot cause a change of sign and such that the increment of the Brownian motion is not extraordinary large. Let the real number rN ∈ [0, ∞) be given by 1 (δ−1) 1 N N 4 − 1 , max 0, rN := min (44) L 0 T sup |µ (s)| + 3L s∈[−1,1]
˜ N,n , ΩN,n , Ωm ∈ F and sets ΩN ∈ F by for every N ∈ N. Now define sets Ω N − 41 ˜ N,n := ω ∈ Ω N N Ω sup , (45) (ω) − W (ω) ≤ N Wtk+1 tk k∈{0,1,...,n−1}
by ( ΩN,n :=
ω ∈ Ω
) N,1 Dv,w (ω) ≤ rN
sup v,w∈{0,1,...,n}
14
˜ N,n , ∩Ω
(46)
by
Ωm N
:=
ω ∈ Ω
sup v,w∈{0,1,...,N }
N,m Dv,w (ω) ≤ rN ,
sup n∈{0,1,...,N −1}
(m) (m) − 41 WtN (ω) − WtN (ω) ≤ N n+1
(47)
n
and by ΩN := ΩN,N = Ω1N
(48)
˜ ∈ F by for every n ∈ {0, 1, . . . , N } and every N, m ∈ N. Finally, define Ω ˜ := Ω
2
[
∞ M \ \
Ωm M
N ∈N M =N m=1
P 2 N (ω)·f (YNN,m (ω)) − E 1ΩN f (YNN ) m=1 1Ωm \ N < ∞ . ω ∈ Ω sup N (1+ε) N ∈N
ε>0
(49) p
1
˜ is indeed in F. Moreover, we write kZk p := (E [|Z| ]) p ∈ [0, ∞] Note that Ω L for all p ∈ [1, ∞) and all F/B(R)-measurable mappings Z : Ω → R. Our proof of Theorem 2.1 uses the following lemmas. Lemma 5.1 (Burkholder-Davis-Gundy inequality). Let N ∈ N and let Z1 , . . . , ZN : Ω → R be F/B(R)-measurable mappings with E|Zn |2 < ∞ for all n ∈ {1, . . . , N } and with E [Zn+1 |Z1 , . . . , Zn ] = 0 for all n ∈ {1, . . . , N − 1}. Then we obtain 12 2 2 kZ1 + . . . + ZN kLp ≤ Kp · kZ1 kLp + . . . + kZN kLp
(50)
for every p ∈ [2, ∞), where Kp , p ∈ [2, ∞), are universal constants. The following lemma is the key result in our proof of Theorem 2.1. N,1 Lemma 5.2 (Dominator Lemma). Let YnN : Ω → R, Dn,m : Ω → R, τnN : Ω → R and ΩN,n ∈ F for n, m ∈ {0, 1, . . . , N } and N ∈ N be given by (34), (42), (43) and (46). Then we have N Yn (ω) ≤ DN,1 (ω) (51) τ N (ω),n n
for every ω ∈ ΩN,n , n ∈ {0, 1, . . . , N } and every N ∈ N. The domination (51) might not look helpful at first view since DτN,1 depends N n ,n N on the Euler approximation and since τn is in general not a stopping time for all n ∈ {1, . . . , N } and all N ∈ N. However, the dependence of the dominating 15
process on the Euler approximation can be controlled as σ ˜ is bounded and the N dependence of DτN,1 on τ for all n ∈ {0, 1, . . . , N } and all N ∈ N is no problem N n n ,n N,1 as Dv,w can be controlled uniformly in v, w ∈ {0, 1, . . . , N } and N ∈ N. This is subject of the following lemma. Lemma 5.3 (Bounded moments of the dominator). N,1 ˜ N,n ∈ F and Dn,m Let Ω : Ω → [0, ∞) for n, m ∈ {0, 1, . . . , N } and N ∈ N be given by (45) and (42). Then we have " !# N,1 p Dv,w sup sup E 1Ω˜ N,n sup Yn+1 (ω) ≥ YnN (ω) · 1 + (ω) +σ ˜ YnN (ω) · WtN (ω) − W tn n+1 N T µ(0) + (ω) − WtN (ω) + σ(0) · WtN n n+1 N „
TL σ N +˜
„
(YnN (ω))·
«« WtN
(ω)−WtN (ω)
n n+1 ≥ YnN (ω) · e T µ(0) N + + σ(0) · WtN (ω) − W (ω) tn n+1 N N
= YnN (ω) · eαn (ω) + βnN (ω) N = − YnN (ω) · eαn (ω) + sgn(YnN (ω)) · βnN (ω) .
24
Hence, the induction hypothesis yields N Yn+1 (ω) ≤ DN,1 N =e
N
τn (ω),n “P ” n N α (ω) l=τ N (ω) l
(T |µ(0)| + |σ(0)| + |ξ(ω)| + 1)
n
n−1 X
+
N (ω) k=τn “P n
Pn
sgn(YkN (ω)) e(
N (ω) l=τn
=e
n X
+
(ω) · eαn (ω) + sgn(YnN (ω)) · βnN (ω)
” αN l (ω)
l=k+1
αN l (ω)) β N (ω) k
+ sgn(YnN (ω)) · βnN (ω)
(T |µ(0)| + |σ(0)| + |ξ(ω)| + 1) Pn
sgn(YkN (ω)) e(
l=k+1
αN l (ω)) β N (ω) k
N (ω) k=τn
N,1 = DτN,1 N (ω),n+1 (ω) = Dτ N
n+1 (ω),n+1
n
(ω),
which shows that (51) also holds in this case for ω and n + 1. N (ω) < 0 holds. Then 3.) In the third case assume that YnN (ω) ≥ 0 and that Yn+1 N we obtain τn+1 (ω) = n + 1. Additionally, note that N N,1 Yn (ω) ≤ sup (69) Dk,l (ω) ≤ rN k,l∈{0,1,...,n}
holds due to the induction hypothesis and since ω ∈ ΩN,n+1 ⊂ ΩN,n . Hence, (64) yields T µ(0) T ≥0 (70) YnN (ω) + µ YnN (ω) − N N and therefore N 0 ≥ Yn+1 (ω)
T N (ω) (ω) − W = YnN (ω) + µ YnN (ω) + σ YnN (ω) · WtN tn n+1 N T T µ(0) = YnN (ω) + µ YnN (ω) − N N T µ(0) + (ω) − WtN (ω) + σ YnN (ω) · WtN n n+1 N T µ(0) N (ω) + σ YnN (ω) · WtN (ω) − W ≥ t n n+1 N and T µ(0) − σ YnN (ω) · WtN (ω) − WtN (ω) 0≥ ≥ − n n+1 N N N ≥ −T |µ(0)| − L · Yn (ω) + |σ(0)| · WtN (ω) − W (ω) . tn n+1 N Yn+1 (ω)
25
(71)
This implies N N N (ω) − W (ω) 0 ≥ Yn+1 (ω) ≥ −T |µ(0)|− L · DτN,1 (ω) + |σ(0)| · W N tn tn+1 n (ω),n (ω) − WtN (ω) ≥ −T |µ(0)| − (L · rN + |σ(0)|) · WtN n n+1 ! 1 N4 N (ω) (ω) − W + |σ(0)| · WtN ≥ −T |µ(0)| − L · t n n+1 L and 1 1 N 0 ≥ Yn+1 (ω) ≥ −T |µ(0)| − |σ(0)| + N 4 · 1 N4 |σ(0)| = −T |µ(0)| − − 1 ≥ −T |µ(0)| − |σ(0)| − 1 1 N4 and finally N Yn+1 (ω) ≤ T |µ(0)| + |σ(0)| + 1 + |ξ(ω)| Pn
= e( +
l=n+1
n X
αN l (ω))
sgn(YkN (ω)) e(
k=n+1 „ Pn
=e +
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|)
l=τ N (ω) n+1
n X
αN l (ω)
Pn
l=k+1
αN l (ω)) β N (ω) k
«
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|) Pn
sgn(YkN (ω)) e(
l=k+1
αN l (ω)) β N (ω) k
N (ω) k=τn+1
= DτN,1 N
n+1 (ω),n+1
(ω),
which shows that (51) also holds in this case for ω and n + 1. N 4.) In the last case assume that YnN (ω) < 0 and that Yn+1 (ω) ≥ 0 holds. Then N we also obtain τn+1 (ω) = n + 1. Note that N N,1 Yn (ω) ≤ sup (72) Dk,l (ω) ≤ rN k,l∈{0,1,...,n}
holds due to the induction hypothesis and since ω ∈ ΩN,n+1 ⊂ ΩN,n . Therefore, (65) implies T µ(0) T YnN (ω) + µ YnN (ω) − ≤0 (73) N N
26
and hence N 0 ≤ Yn+1 (ω)
T N (ω) = YnN (ω) + µ YnN (ω) + σ YnN (ω) · WtN (ω) − W t n n+1 N T µ(0) T = YnN (ω) + µ YnN (ω) − N N T µ(0) N (ω) + + σ YnN (ω) · WtN (ω) − W t n n+1 N T µ(0) N (ω) + σ YnN (ω) · WtN (ω) − W ≤ t n n+1 N
(74)
and T µ(0) N + σ YnN (ω) · WtN (ω) − WtN (ω) 0 ≤ Yn+1 (ω) ≤ n n+1 N N N (ω) ≤ T |µ(0)| + L · Yn (ω) + |σ(0)| · WtN (ω) − W t n n+1 N N (ω) + |σ(0)| · (ω) − W (ω) ≤ T |µ(0)| + L · DτN,1 W . N (ω),n tn tn+1
(75)
n
This shows N Yn+1 (ω) ≤ T |µ(0)| + (L · rN + |σ(0)|) · WtN (ω) − WtN (ω) n n+1 ! 1 N4 ≤ T |µ(0)| + L · + |σ(0)| · WtN (ω) − WtN (ω) n n+1 L 1 1 |σ(0)| ≤ T |µ(0)| + |σ(0)| + N 4 · 1 = T |µ(0)| + +1 1 4 N N4
(76)
and finally N Yn+1 (ω) ≤ T |µ(0)| + |σ(0)| + 1 + |ξ(ω)| Pn
= e( +
l=n+1
n X
αN l (ω))
sgn(YkN (ω)) e(
k=n+1 „ Pn
=e +
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|)
l=τ N (ω) n+1
n X
αN l (ω)
Pn
l=k+1
αN l (ω)) β N (ω) k
«
(T |µ(0)| + |σ(0)| + 1 + |ξ(ω)|) Pn
sgn(YkN (ω)) e(
l=k+1
N (ω) k=τn+1
αN l (ω)) β N (ω) k
= DτN,1 N
n+1 (ω),n+1
(ω),
which shows that (51) holds in this case for ω and n + 1 and which finally yields (51) for every ω ∈ ΩN,n , n ∈ {0, 1, 2, . . . , N } and every N ∈ N by induction.
27
5.5
Proof of Lemma 5.3
In order to bound the restricted moments of the dominating process, we need to know the exponential moments of a standard normal distributed random variable and we also need to estimate the p-th moment of a normal distributed random variable for every p ∈ [1, ∞). Lemma 5.7. Let Y : Ω → R be a standard normal distributed F/B(R)-measurable mapping. Then we obtain that c2 (77) E ecY = e 2 holds for every c ∈ R. Proof. By completing the square, we obtain Z Z 2 1 x2 1 1 √ e− 2 (x −2cx) dx E ecY = ecx √ e− 2 dx = 2π 2π R ZR Z 2 c2 1 1 − 12 (x2 −2cx+c2 ) c2 1 √ e √ e− 2 (x−c) dx = e 2 dx = e 2 2π 2π R R Z c2 c2 1 − x2 √ e 2 dx = e 2 =e2 2π R
(78)
for every c ∈ R. Lemma 5.8. Let Y : Ω → R be a normal distributed F/B(R)-measurable mapping. Then we obtain that kY kLp ≤ p kY kL2
(79)
holds for every p ∈ [1, ∞). Proof. First of all, H¨ older’s inequality implies kY kLp ≤ kY kL2 ≤ p kY kL2
(80)
for every p ∈ [1, 2), which shows (79) in the case p ∈ [1, 2). Denote now the mean r hof Y : Ω i→ R by c := E [Y ] ∈ R and the standard deviation by 2 σ := E (Y − c) ∈ [0, ∞). If Yˆ : Ω → R is a standard normal distributed F/B(R)-measurable mapping, then
Y
Lp
= σ Yˆ + c
Lp
≤ σ Yˆ + c
Ldpe
1 dpe dpe ˆ = E σ Y + c
1 dpe dpe k X dpe k (dpe−k) ≤ E |σ| Yˆ |c| k
k=0
1 dpe dpe k X dpe k (dpe−k) = |σ| E Yˆ |c| k
k=0
28
(81)
k/2 ˆ k holds for for all k ∈ {2, 3, . . . }, qevery p ∈ [2, ∞). Using E|Y | ≤ (k − 1) 2 2 2 E|Yˆ | ≤ E(Yˆ )2 = 1 and (σ + c) ≤ 2(σ + c ) yields 1 dpe dpe X dpe dpe (dpe−k) k ≤ |σ| (dpe − 1) 2 |c| k k=0 p p √ = (dpe − 1) (σ + c) ≤ p (σ + c) ≤ p (σ 2 + c2 ) = p kY kL2
kY kLp
for every p ∈ [2, ∞), which finally shows (79). Proof of Lemma 5.3. Note that the discrete time stochastic integral n−1 X
N z·σ ˜ (YlN ) · WtN − W t l+1 l
(82)
l=0
is a martingale in n ∈ {0, 1, . . . , N } for every z ∈ {−1, 1} and every N ∈ N. Since the exponential function is convex, we obtain that ! n−1 X N exp z·σ ˜ (Yl ) · WtN − WtN (83) l+1 l l=0
is a positive submartingale in n ∈ {0, 1, . . . , N } for every z ∈ {−1, 1} and every N ∈ N. Therefore, Doob’s inequality (see e.g. Theorem 11.2 (ii) in [7]) shows „ „ «« p # " Pn−1 σ (YlN )· WtN −WtN l=0 z·˜ l+1 l E sup e n∈{0,1,...,N } „ «« p # p " „PN −1 σ (YlN )· WtN −WtN p l=0 z·˜ l+1 l E e ≤ p−1 „ „ «« " p PN −2 pz ˜ (YlN )· WtN −WtN p l=0 σ l+1 l (84) E e = p−1 „ «« " „ ## N pz σ ˜ (YN −1 )· WtN −WtN FtN N N −1 ·E e N −1 =
p p−1
p
" E e
„ pz
PN −2 l=0
„ «« σ ˜ (YlN )· WtN −WtN l+1
29
l
N e (pz·˜σ(YN −1 )) 1 2
2 T N
#
due to (78) and hence, using |˜ σ (x)| ≤ L for every x ∈ R, „ „ ««
Pn−1
σ (YlN )· WtN −WtN
l=0 z·˜ l+1 l sup e
p
n∈{0,1,...,N } L „ „ «« # ( ) p1 " p PN −2 pz ˜ (YlN )· WtN −WtN 1 2 2 T p l=0 σ l+1 l ≤ E e e2p L N p−1 p p1 2 1 2 2 1 p p p L T e2 = ≤ ... ≤ e 2 pL T p−1 p−1 for every p ∈ (1, ∞), z ∈ {−1, 1} and every N ∈ N. This implies
P
z ( n−1 αN ) l l=0 sup e
n∈{0,1,...,N }
p L „ „ ««
P
n−1 nT L z ˜ (YlN )· WtN −WtN
N + l=0 σ l+1 l sup e =
n∈{0,1,...,N }
p L „ „ ««
Pn−1
N z·˜ σ (Y )· W −W l
l=0 tN tN l+1 l ≤ eT L sup e
n∈{0,1,...,N }
p L p 2 2 1 p pL T (L +L)T TL ≤ 2e 2 e2 ≤e p−1
(85)
(86)
for every p ∈ [2, ∞), z ∈ {−1, 1} and every N ∈ N. Moreover, the BurkholderDavis-Gundy inequality (50) and again Doob’s inequality yield
n−1
X P N
−( k−1 α N N ) l l=0 sgn(Yk ) e βk sup
n∈{0,1,...,N } p k=0
L N −1
Pk−1 N p
X
≤ sgn(YkN ) e−( l=0 αl ) βkN
(p − 1) k=0 Lp (87) ! 21 N −1
2 Pk−1 N X pKp
≤
sgn(YkN ) e−( l=0 αl ) βkN p (p − 1) L k=0 ! 12 N −1
X
N 2 N 2
−(Pk−1 α )
βk 2p l l=0 ≤ 2Kp
e
L2p
k=0
L
for every p ∈ [2, ∞) and every N ∈ N. The last step is H¨older’s inequality. Now
30
using Lemma 5.8 and inequality (86), we see that
n−1
P N X
−( k−1 α N N ) l l=0 sgn(Yk ) e sup βk
n∈{0,1,...,N } k=0 Lp r !2 12 N −1 X 2 2 T |µ(0)| T ≤ 2Kp 2ep(L +L)T + |σ(0)| 2p N N k=0 r ! 2 T |µ(0)| T √ = 4Kp ep(L +L)T N + |σ(0)| 2p N N holds for every p ∈ [2, ∞) and every N ∈ N. Hence, we have n−1
X
P N
−( k−1 α N N ) l l=0 sgn(Yk ) e sup βk
n∈{0,1,...,N } k=0 Lp √ p(L2 +L)T ≤ 8Kp e T |µ(0)| + p T |σ(0)| √ 2 ≤ 8LKp T + p T ep(L +L)T 2
≤ 8LpKp (T + 1) ep(L ≤ Kp ep(T +1)(L
2
+5L)
+L)T
≤ Kp e4Lp(T +1) ep(L
≤ Kp e5p(L
2
(88)
(89) 2
+L)T
+1)(T +1)
√ for every p ∈ [2, ∞) and every N ∈ N since x ≤ x2 + 1 and since 2x ≤ ex for every x ∈ [0, ∞). With these estimates at hand, we now bound the restricted p-th moment !# " N,1 p Dv,w (90) E 1˜ sup ΩN,n
v,w∈{0,1,...,n}
of the dominating process for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. By definition (42) and by the triangle inequality, we have
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L
!
Pw−1 N
≤ 1Ω˜ N,n sup e( l=v αl ) (T |µ(0)| + |σ(0)| + |ξ| + 1)
p v,w∈{0,1,...,n} L
w−1 !
X
Pw−1 N
+ 1Ω˜ N,n sup sgn(YkN ) e( l=k+1 αl ) βkN
v,w∈{0,1,...,n} k=v
Lp
31
and, using H¨ older’s inequality,
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
Pw−1 N
α ( )
l l=v sup e ≤ 1Ω˜ N,n
0≤v≤w≤n
+ 1Ω˜ N,n
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
L2p
w−1 ! X Pw−1 N sup sgn(YkN ) e( l=k+1 αl ) βkN 0≤v≤w≤n k=v
Lp
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Therefore, we obtain
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
Pw−1 N
α ( ) sup e l=v l (T |µ(0)| + |σ(0)| + kξkL2p + 1) ≤
0≤v≤w≤N
L2p
w−1 !
X Pw−1 N Pk N
α − α N N e( l=0 l ) 1Ω˜ N,n + sup sgn(Yk ) e ( l=0 l ) βk
0≤v≤w≤n k=v
Lp
and
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n}
L
Pv−1 N Pw−1 N
sup e−( l=0 αl ) sup e( l=0 αl ) ≤
4p
w∈{0,1,...,N }
4p v∈{0,1,...,N } L L
Pw−1 N
α ( ) · (T |µ(0)| + |σ(0)| + kξkL2p + 1) + sup e l=0 l
w∈{0,1,...,N }
2p L w−1
! X
Pk N
1 ˜ sgn(YkN ) e−( l=0 αl ) βkN
· sup
0≤v≤w≤n ΩN,n k=v
(91)
L2p
and, using (86),
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L 2 2p(L2 +L)T ≤ 2e (T |µ(0)| + |σ(0)| + kξkL2p + 1)
!
w−1 P N X α −( k p(L2 +L)T N N ) l=0 l + 2e 1Ω˜ N,n sgn(Yk ) e βk
sup
0≤v≤w≤n k=v
2
≤ 4e4p(L
(92)
L2p
+L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
!
w−1 Pk X N 2
α − + 4ep(L +L)T sup 1Ω˜ N,n sgn(YkN ) e ( l=0 l ) βkN
w∈{0,1,...,n} k=0
32
L2p
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and since every N ∈ N. Moreover, N (LT +L) N (ω) α (ω) ≤ T L + L, we have e−αN k α (ω) for every k ∈ − 1 ≤ e k
k
˜ N,n and every n ∈ {0, 1, . . . , N }, N ∈ N. Therefore, we {0, 1, . . . , n − 1}, ω ∈ Ω obtain w−1 X P N −( k α (ω) N N ) l=0 l sgn(Yk (ω)) e sup βk (ω) w∈{0,1,...,n} k=0 w−1 X P N N −( k−1 α (ω) N −α (ω) N ) l k l=0 sgn(Yk (ω)) e ≤ sup e − 1 βk (ω) w∈{0,1,...,n} k=0 w−1 Pk−1 N X sgn(YkN (ω)) e−( l=0 αl (ω)) βkN (ω) + sup w∈{0,1,...,n} k=0 ! n−1 Pk−1 N X N ≤ e−( l=0 αl (ω)) · e−αk (ω) − 1 · βkN (ω) k=0
w−1 P N X −( k−1 α (ω) N N ) l l=0 + sup sgn(Yk (ω)) e βk (ω) w∈{0,1,...,n} k=0
and w−1 X P N −( k α (ω) N N ) l=0 l sgn(Yk (ω)) e sup βk (ω) w∈{0,1,...,n} k=0 ! n−1 P X N −( k−1 αN (ω)) N (LT +L) l l=0 ≤e e αk (ω) · βk (ω) k=0
w−1 X Pk−1 N sgn(YkN (ω)) e−( l=0 αl (ω)) βkN (ω) + sup (93) w∈{0,1,...,n} k=0 ! ! n−1 P X N N −( w−1 α (ω) (LT +L) N ) α (ω) · β (ω) l l=0 ≤ sup e e k
w∈{0,1,...,n}
k
k=0
w−1 X P N −( k−1 α (ω) N N ) l l=0 + sup sgn(Yk (ω)) e βk (ω) w∈{0,1,...,n} k=0
˜ N,n , n ∈ {0, 1, . . . , N } and every N ∈ N. Inserting (93) into (92) for every ω ∈ Ω
33
yields
1Ω˜ N,n
≤ 4e
! N,1
Dv,w sup
v,w∈{0,1,...,n}
Lp
4p(L2 +L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
! ! n−1
P X −( w−1 αN p(L2 +L)T N N (LT +L) ) l l=0 sup e + 4e αk · βk e
w∈{0,1,...,n} k=0 L2p
w−1
P X k−1 N 2
sgn(YkN ) e−( l=0 αl ) βkN sup + 4ep(L +L)T
w∈{0,1,...,n} k=0
L2p
and, using H¨ older’s inequality and (89),
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
≤ 4e4p(L
2
+L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
Pw−1 N 2
+ 4ep(L +L)T e(LT +L) sup e−( l=0 αl )
w∈{0,1,...,n}
L
+ 4e
p(L2 +L)T
K2p e
n−1
X N N α k · βk
4p k=0
L4p
10p(L2 +1)(T +1)
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Therefore, (86) implies
!
N,1
Dv,w sup
1Ω˜ N,n
p
v,w∈{0,1,...,n} L
≤ 4e
4p(L2 +L)T
+ 4e
(T |µ(0)| + |σ(0)| + kξkL2p + 1)
p(L2 +L)T (LT +L)
e
2e
2p(L2 +L)T
n−1 X
!
(94)
N
αk 8p βkN 8p L L
k=0 2 2 + 4K2p e{p(L +L)T +10p(L +1)(T +1)}
and, using σ ˜ (x) ≤ L for all x ∈ R,
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L
2 2 ≤ 4e4p(L +L)T (T |µ(0)| + |σ(0)| + kξkL2p + 1) + 8e{3p(L +L)T +(LT +L)} · ! n−1
T |µ(0)|
X TL
+ L WtN − W tN + |σ(0)| WtN − WtN
8p
8p k+1 k k+1 k N N L L
k=0
+ 4K2p e{2p(L
2
+1)T +10p(L2 +1)(T +1)}
34
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Now we apply Lemma 5.8 to WtN − WtN for k ∈ {0, 1, . . . , N − 1}, N ∈ N to obtain k+1 k
1Ω˜ N,n
≤ 4e
! N,1 Dv,w sup
v,w∈{0,1,...,n}
Lp
4p(L2 +L)T
(T |µ(0)| + |σ(0)| + kξkL2p + 1) " r #" r #! n−1 X TL 2 T |µ(0)| T T + 8e{6p(L +1)T +L(T +1)} + L8p + |σ(0)|8p N N N N k=0
+ 4K2p e
12p(L2 +1)(T +1)
and
1Ω˜ N,n
! N,1 Dv,w sup
v,w∈{0,1,...,n}
Lp
2
≤ 4e8p(L
+1)(T +1)
(T |µ(0)| + |σ(0)| + kξkL2p + 1) √ √ 6p(L2 +1)T +L(T +1)} { + 8e T L + 8Lp T T |µ(0)| + |σ(0)|8p T
+ 4K2p e12p(L
2
+1)(T +1)
for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N. Finally this implies
!
N,1
Dv,w sup
1Ω˜ N,n
p v,w∈{0,1,...,n} L 2 (95) ≤ 4e12p(L +1)(T +1) (T |µ(0)| + |σ(0)| + kξkL2p + 1 + K2p ) √ √ 2 + 512Le{6p(L +1)T +L(T +1)} T + p T T |µ(0)| + |σ(0)|p T for every n ∈ {0, 1, . . . , N }, p ∈ [2, ∞) and every N ∈ N, which yields the assertion in the case p ∈ [2, ∞). The case p ∈ [1, 2) then follows from Jensen’s inequality.
5.6
Proof of Lemma 5.5
Proof of Lemma 5.5. First of all, we have " # c − 41 ˜ P ΩN,N =P sup − W tN WtN >N n n+1 n∈{0,1,...,N −1}
N −1 − 41 = P ∪n=0 ω ∈ Ω WtN (ω) − WtN (ω) > N n n+1 ≤
N −1 X
h i − 41 N > N P WtN − W t n n+1
n=0
35
and P
˜ N,N Ω
c
i h − 41 N > N − W ≤ N · P WtN t0 1 h
= 2N · P W T > N
− 41
i
N
and Z c ˜ P ΩN,N ≤ 2N
"r = 2N · P
1 N W T > N−4 T N
r
N T
#
√ Z ∞ 1 − x2 1 x T − x2 2 2 dx dx ≤ 2N e 1 √ 1 √ 1 e 4 4 N N 4 2π 2π N √ √ T T Z ∞ 3√ x2 3√ 1 1 h − x2 ix=∞ − 1 = 2N 4 T 1 √ x e 2 dx = 2N 4 T √ −e 2 4 4 √ N x= N 2π 2π √ T T (96) r √ √ √ N N 3√ 3√ 1 2 3√ − N = 2N 4 T √ e− 2T = N 4 T e 2T ≤ N 4 T e− 2T π 2π ∞
for every N ∈ N. Additionally, note that ) ( N,1 ˜ N,N Dv,w (ω) ≤ rN ∩ Ω sup ΩN = ω ∈ Ω v,w∈{0,1,...,N } ! ) ( N,1 ˜ N,N Dv,w (ω) ≤ rN ∩ Ω sup = ω ∈ Ω 1Ω˜ N,N (ω)
(97)
v,w∈{0,1,...,N }
holds for every N ∈ N. Hence, we have ( c sup (ΩN ) = ω ∈ Ω 1Ω˜ N,N (ω)
v,w∈{0,1,...,N }
! N,1 Dv,w (ω)
) > rN
c ˜ N,N ∪ Ω
for every N ∈ N. Therefore, inequality (96) implies " ! # c N,1 c ˜ N,N Dv,w > rN + P Ω P (ΩN ) ≤ P 1 ˜ sup ΩN,N
" ≤P
1 + 1Ω˜ N,N
v,w∈{0,1,...,N }
sup v,w∈{0,1,...,N }
!!p # √ N,1 N 3√ p Dv,w > (1 + rN ) + N 4 T e− 2T
for every N ∈ N and every p ∈ [1, ∞). Now we apply Markov’s inequality to obtain h N,1 p i E 1 + 1˜ √ sup v,w∈{0,1,...,N } Dv,w ΩN,N N 3√ c − 2T 4 T e P (ΩN ) ≤ + N (1 + rN )p
! p !
√ M,1 N 3√ 1
Dv,w ≤ sup 1+ 1Ω˜ M,M sup + N 4 T e− 2T
p
(1 + r ) N M ∈N v,w∈{0,1,...,M } p L
36
and ! p ! M,1 Dv,w sup
p v,w∈{0,1,...,M } L √ 1 N 3√ p −p·min( 41 , (δ−1) − 2T ) 4 ·c N Te +N
! !#p "
M,1
Dv,w ≤ c 1 + sup 1Ω˜ M,M sup
p M ∈N v,w∈{0,1,...,M } L √ √ 1 1 N 3 −p·min( 4 , (δ−1) ) − ·N + N 4 T e 2T
c P (ΩN ) ≤
sup 1 + 1Ω˜ M,M M ∈N
(98)
for every N ∈ N and every p ∈ [1, ∞), where we used 1 1 1 ≤ c · N − min( 4 , (δ−1) ) (1 + rN )
for every N ∈ N with c ∈ (0, ∞) given by N min( 4 , (δ−1) ) (1 + rN ) 1
1
c := sup N ∈N
! < ∞.
Moreover, (98) with p := 4 max(4, δ − 1) yields c P (ΩN )
c 1 + sup 1Ω˜ M,M M ∈N
" ≤N
−4
√
−
+ N Te "
!#4 max(4,δ−1)
M,1 Dv,w sup
v,w∈{0,1,...,M }
L4 max(4,δ−1)
√ N 2T
!#4 max(4,δ−1)
M,1
Dv,w ≤N c 1 + sup 1Ω˜ M,M sup
4 max(4,δ−1) M ∈N v,w∈{0,1,...,M } L " !# √ √ x + N −4 T sup x5 e− 2T −4
x∈[0,∞)
for every N ∈ N. The right-hand side is finite by Lemma 5.3. This proves 4 c c˜ := sup N · P (ΩN ) ∈ [0, ∞). (99) N ∈N
˜ has probability 1. We have Next we show that the event Ω c 2 2 ∞ M ∞ M \ \ \ \ = 1 − P P Ωm Ωm M M M =N m=1
M =N m=1
= 1 − P
2
∞ M [ [
M =N m=1
37
c (Ωm M)
(100)
and hence 2 ∞ M ∞ X M2 \ \ X m m c P ΩM ≥ 1 − P (ΩM ) M =N m=1
M =N m=1
∞ X c 2 =1− M · P (ΩM ) ≥1− M =N
∞ X
! c˜M
−2
M =N
for every N ∈ N due to (99). This implies 2 s=∞ Z ∞ ∞ M \ \ 1 1 ds = 1 − c ˜ − P Ωm ≥ 1 − c ˜ M 2 s s=N −1 N −1 s m=1 M =N
=1−
c˜ (N − 1)
for every N ∈ {2, 3, . . .}, which shows 2 2 ∞ M ∞ M [ \ \ \ \ = lim P = 1. P Ωm Ωm M M N ∈N M =N m=1
N →∞
M =N m=1
Moreover, we have P 2 p N · f (YNN,m ) − E 1ΩN f (YNN ) m=1 1Ωm N E sup N (1+ε) N ∈N P 2 p N N,m N ∞ m · f (Y 1 ) − E 1 f (Y ) X Ω Ω N N N m=1 N ≤ E N (1+ε)p N =1
P 2
N
p ∞ · f (YNN,m ) − E 1ΩN f (YNN ) p X m=1 1Ωm N L = (1+ε)p N N =1 21 p PN 2 N N 2
K 1 f (Y ) − E 1 f (Y ) ∞ p ΩN ΩN N N m=1 X Lp ≤ N (1+ε)p N =1
38
(101)
due to Lemma 5.1 and therefore P 2 p N · f (YNN,m ) − E 1ΩN f (YNN ) m=1 1Ωm N E sup N (1+ε) N ∈N
p ∞ X Kp · N · 1ΩN f (YNN ) − E 1ΩN f (YNN ) Lp ≤ N (1+ε)p N =1 p ! p ∞ X (Kp ) · 1ΩN f (YNN ) − E 1ΩN f (YNN ) Lp = N εp N =1 ! ∞ h X p i p ≤ (Kp ) N −εp sup E 1ΩN f (YNN ) − E 1ΩN f (YNN )
(102)
N ∈N
N =1
for every p ∈ [2, ∞) and every ε ∈ (0, ∞). Hence, we obtain P 2 p N · f (YNN,m ) − E 1ΩN f (YNN ) m=1 1Ωm N E sup N (1+ε) N ∈N ! ∞ h X p i p p −εp ≤ (2 (Kp ) ) N sup E 1Ω f (YNN ) p
p
≤ (2 (Kp ) ) ≤ (2LKp )
p
N =1 ∞ X N =1 ∞ X
N
N
N =1
N
N ∈N
−εp
! sup E
h
N ∈N −εp
! sup E
N ∈N
h
1ΩN L
p
δ p i 1 + YNN δ p i
1ΩN 1 + YNN
and P 2 p N N,m N · f (Y ) − E 1 f (Y ) m=1 1Ωm Ω N N N N E sup (1+ε) N N ∈N ! ∞ h X pδ i p −εp ≤ (4LKp ) N 1 + sup E 1Ω YNN
