Distance between the fractional Brownian motion and the space of adapted Gaussian martingales

Abstract. We consider the distance between the fractional Brownian motion defined on the interval [0, 1] and the space of Gaussian martingales adapted to the same filtration. As the distance between stochastic processes, we take the maximum over [0, 1] of mean-square deviances between the values of the processes. The aim is to calculate the function a in the Gaussian martingale representation ∫ t 0 a(s) dWs that minimizes this distance. So, we have the minimax problem that is solved by the methods of convex analysis. Since the minimizing function a cannot be either presented analytically or calculated explicitly, we perform discretization of the problem and evaluate the discretized version of the function a numerically.

It is well known that fBm is not a semimartingale and certainly is not a martingale unless H = 1/2. Therefore, a natural question of the approximation of fBm by martingales and semimartingales arises. On the one hand, as to semimartingales, the situation is more simple: fBm can be approximated by the Gaussian semimartingales in sup L 2 -norm (see, e.g., [9]). On the other hand, it is obvious that we cannot approximate the fBm by martingales with an arbitrary accuracy in any reasonable norm, see, e.g., [3]. Therefore, it is natural to put the question differently: what is the distance between the fBm and the martingale space in a reasonable metric, and on which element from the martingale space does this distance reach? This is a question of projection of fBm on the space of Gaussian martingales generated by the Wiener process. Solving this problem, we can, in the future, move forward a few steps, namely, we can consider the projection of the fBm on the space of repeated Wiener integrals, that is, to calculate the projections on the finite-dimensional spaces of Wiener chaos, so, to clarify the structure of fractional Brownian motion in comparison with the spaces of Wiener chaos. Being very simple in appearance in his statement, the question of the distance between the fBm and the space of adapted Gaussian martingales, nevertheless turned out to be very difficult. The existence and uniqueness of the minimizing function in the representation of Gaussian martingale, as well as a number of its properties were established in the paper [17] by the methods of convex analysis. However, it is still impossible to get the analytical representation of the minimizing function. Therefore, we decided to find this function approximately, for which we used the concept of the Chebyshev center and its properties. Now, let us state the problem and consider the methods of its solution in more detail.
For Gaussian stochastic process of the form X t = t 0 a(s) dW s , we define the square distance between X and B H as follows: The maximum is attained because the random processes X and B H are mean-square continuous and, as a consequence, E(X t − B H t ) 2 is a continuous function in t, which attains its minimum and maximum on [0, 1].
We are searching for the function a ∈ L 2 [0, 1] for which the square distance in minimal. The choice of the representation of the process X is justified as follows. Any square-integrable martingale with respect to F W with right-continuous trajectories admits a representation X t = x 0 + t 0 a(s) dW s , where a is a stochastic process adapted to the same filtration as W and satisfying E 1 0 a(s) 2 ds < ∞ [11,Thm. 5.5]. In [17, p. 541] it is proved that the minimum of sup t∈[0,1] E(X t −B H t ) 2 is attained for x 0 = 0 and for a being a nonrandom function. A problem of approximation of the fBm by a Gaussian martingale from some smaller classes of Gaussian martingales, e.g., from a finite-dimensional subsets, is much simpler than a general problem, and was solved in [1,2,8,13]. The reciprocal problem of approximation of the Wiener process by an integral with respect to the fractional Brownian motion, also from some specific class of integrands, was solved in [4]. It is proved in [17] that the minimum of F (a) from (3) for any H ∈ (0, 1) is attained at unique point, and recall again that even though we know some properties of a, we do not have for it an analytical expression.
In order to get function a at least numerically, in the present paper, we approximate a discrete-time slice {B H t , t = 1/N, 2/N, . . . , 1} of the fBm instead of performing continuous-time approximation.
We find a "discrete analogue" for the function a numerically. In order to do this, we use the so-called alternating minimization [5], and search for the Chebyshev center of the finite set, see, e.g., [6]. The problem of locating the Chebyshev center is a linear minimization problem with quadratic constraints, while the dual problem is the quadratic minimization with linear constraints, see [7,18]. Our goal is evaluation of min a∈L2[0,1] F (a). For that reason, we prove the following result: where F ( a, K (N ) ), N 1, is the sequence of the convex functions of the vector a ∈ R N . Vector a is related to function a, and matrices K (N ) are related to the kernel z in the representation (2). These functions are discrete analogue of the function F (a). Twosided relation (4) is the main mathematical contribution of the present paper. It allows to calculate the minimizing function a numerically.
The paper is organized as follows. In Section 2 we construct the sequence of the functions F ( a, K (N ) ) and prove relation (4). In Section 3, we propose and justify the algorithm for numerical evaluation of F ( a, K (N ) ). In Section 4, the minimizing function is plotted. Appendix contains some auxiliary results.

Two-sided bounds and convergence result
In what follows, we denote matrices (but not matrix entries) with boldface letters. For vectors, we use notation like a. The elements of the vector are denoted with the same letter as the vector itself; that is if a ∈ R N , then a = (a 1 , . . . , a N ) .
Since z(t, s) 0, all entries k (N ) mn are nonnegative. Moreover, due to the relations z(t, s) > 0 for 0 < s < t and z(t, s) = 0 for s > t, the entries k (N ) mn form a lowertriangular matrix, which we denote by K (N ) . So, k Denote by z (N ) (t, s) the piecewise-constant approximation of z(t, s) with step 1/N : and z (N ) (t, 1) = 0. With this notation, The discrete-time distance in L 2 [0, 1] between the function z presented by its discrete counterpart z(m/N, · ) and approximation z (N ) is denoted by d mN , so that In this section, we compare minimal values of two functionals. The first functional is defined in (3). The second one has the following form: where a n are the elements of the vector a. The next result gives both nonasymptotic and asymptotic bounds for (min a∈L2[0,1] F (a)) 1/2 . Theorem 1.
(i) The following two-sided inequality holds: (ii) Additionally, the following equality holds: Proof. (i) Lower bound. Prove the first inequality in (9). In order to do this, define a linear operator of the form Let us mention several properties of the operator P (N ) . First, the operator P (N ) is the orthogonal projector in the Hilbert space L 2 [0, 1] onto the subspace of functions that are constant on the intervals [(n−1)/N, n/N ). Hence, P (N ) = 1, where · is the operator norm. Second, for the entity we have an equality According to [17], the function F (a) attains its minimum at the unique point. Let this With this notation, For every integer m, 1 m N , we have that Maximizing over m, we obtain Now we are ready to obtain the first inequality in (9): So, the first inequality in (9) is proved.
http://www.journals.vu.lt/nonlinear-analysis (i) Upper bound. Prove the second inequality in (9). Let N be fixed. Let a * = (a * 1 , . . . , a * N ) be the point where the functional F (N ) attains its minimum. Without loss of generality, we can assume that 0 min It is a consequence of the following fact: if some element a n of the vector a lies outside interval [min n m N k mn ], then, moving a n into the nearest endpoint of the interval, we do not increase the value of F ( a, K (N ) ). If the function F (·, K (N ) ) attains its minimum at the unique point, this point of minimum must satisfy inequality (10). Define the function a(s) as follows: and consider a Gaussian process {X t , t ∈ [0, 1]} of the form X t = t 0 a(s) dW s . According to isometry property for stochastic integrals, we have that for every integer 1 m N , Furthermore, Taking maximum over 1 m N , we obtain Now apply Lemma A.5. Because of (10), Now we recall that the minimum of the functional (3) is attained, a * is the minimum point of F (·, K (N ) ), hence F ( a * , K (N ) ) = min a∈R N F ( a, K (N ) ). This completes the proof of the first assertion.
and apply squeeze theorem.
Thus, instead of minimizing (3), we are searching for the minimum of the functional F ( a, K (N ) ), defined in (8), over a ∈ R N . In the next section, we propose the numerical algorithm of finding min a∈R N F ( a, K (N ) ).
3 Iterative minimization of the squared distance using alternating minimization method In this section, the dimension N ∈ N is fixed. Let K be a fixed lower-triangle N × N matrix, whose elements are denoted as k mn and rows are denoted as k m• . The function F ( a, K), a ∈ R N , is defined by the following relation: Reduce the problem of finding the minimum of the convex function F (·, K) to the problem of finding the minimum of biconvex functional of two vectors. Denote Here, by convention, Definition 1. Let Z be a nonempty bounded set in R N . Chebyshev center of the set Z is the point a ∈ R N where the minimum of sup z∈Z a − z is attained: Note that the minimum in (12)  is strongly convex, For fixed b, min F ( a, b, K) is attained for a being the Chebyshev center of the N -point Due to (11), the problem of minimization of F ( a, K) is equivalent to the problem of minimization of F ( a, b, K): If the minimum of F ( a, K) is attained for a = a * , then the minimum of F ( a, b, K) is attained for a = b = a * . If the minimum of F ( a, b, K) is attained for a = a * , b = b * , then the minimum of F ( a, K) is attained for a = a * . In what follows, min F will mean a common minimum of functions F ( a, K) and F ( a, b, K).
Let us summarize the properties of the functions F ( a, K) and F ( a, b, K) in the following proposition.  (iii) The minimal values of the functions F ( a, K) and F ( a, b, K) coincide. We denote them min F : We find the minimum of F ( a, b, K) by alternating minimization. Let a (0) ∈ R N be the initial approximation. The minimum of F ( a (0) , b, K) is attained for b = a (0) . Then minimize F ( a, a (0) , K) with respect to a: Then we have the descend of the criterion function: and since F ( a (i) , a (i) , K) = F ( a (i) , K), we get that The following theorem shows that the sequence {F ( a (i) , K), i 0} converges to min F . The minimum of F (·, K) on entire space R N is attained at a * , see explanation in the proof of Theorem 1 (and relation a * ∈ H is analogous to inequality (10)): Since a (i) ∈ H and a * ∈ H, we have a (i) − a * d for all i 0. For all a ∈ H and b ∈ H, the inequality F ( a, K) F ( a, b, K) d 2 holds true.
The case d = 0 is trivial: in this case the set H consists of the only one point, a (i) = a (0) and F (a (i) , K) = 0 for all i; therefore, the statement of this theorem holds true. So, in the rest of the proof assume that d > 0.
Denote pseudo-norms on R N : x 2 n 1/2 (for m = N , denote x N = x and x ⊥N = 0). With this notation, In what follows, we are going to use inequalities In what follows, we will be constructing an upper bound for F ( a (i) , K). It will be inequality (19).
Denote for fixed i Then Taking into account the relations 0 min F F ( a (i−1) , K) d 2 , we obtain the inequality 0 α i 1/2.
Next auxiliary result also will be applied to get (19). Namely, we construct the upper bound for F ((1 − α i ) a (i−1) + α i a * , a (i−1) , K). For every 1 m N , and Hence Take the maximum (over m = 1, . . . , N ) in the left-hand side of (16) and obtain Now we continue with upper bound for F ( a (i) , K). With Proposition 1, we get an inequality The sequence {F ( a (i) , K), i 0} is decreasing and bounded, more exactly, 0 min F F ( a (i) , K) F ( a (0) , K). Hence, it is convergent, From inequality (17) we obtain the desired upper bound In (19), take the limit as i → ∞: Therefore, In either case, lim i→∞ F ( a (i) , K) = min F (here we use inequality (18) in the latter case lim i→∞ F ( a (i) , K) 0). The first statement of the theorem is proved.
(ii) We apply Lemma A.1. We take the restriction of F (·, K) onto the set H for the function f . From the proof of the first part of Theorem 2, we take the following: a * ∈ H (at least one point of minimum of F (·, K) belongs to H; since the point of minimum is unique, it must belong to H), all elements of the sequence { a (i) , i 1} belong to H, and lim i→∞ F ( a (i) , K) = min F = min a∈H F ( a, K). Hence, the convergence a (i) → a * follows. Now we perform the implementation of computation of min F and the minimizing vector. We take the bottom row of the matrix K as the initial approximation, that is a (0) = k N • . Then we iteratively perform minimization in (14) using Chebyshev center algorithm: −1) ) .
For evaluating Chebyshev center, we adapted an algorithm presented in [6,10].
If we are evaluating min F , we stop iterations when U i−1 p 1 , where is the upper bound for F (a (i−1) ) − min F . If we are evaluating the minimizing vector a * , we stop iterations when both inequalities U i−1 p 1 and a (i) − a (i−1) p 2 hold true. Here p 1 > 0 and p 2 > 0 are thresholds. Finally, we take a (i) as an approximation of the point of minimum. Figure 1 displays the graph of the approximation of the optimal function a(t) and the square-distance

Computation of the minimizing function
for H = 1/3 and for H = 2/3. We also display the graph of approximate density of the random variable ξ in the representation a(s) = E z(ξ, s), ξ s , see [17,Thm. 4]. This density is obtained as follows. We have the following bounds: