Download Stochastic Calculus for Fractional Brownian Motion and by Yuliya Mishura PDF
By Yuliya Mishura
The conception of fractional Brownian movement and different long-memory procedures are addressed during this quantity. attention-grabbing issues for PhD scholars and experts in chance concept, stochastic research and fiscal arithmetic reveal the fashionable point of this box. between those are effects approximately Levy characterization of fractional Brownian movement, maximal second inequalities for Wiener integrals together with the values 0<H<1/2 of Hurst index, the stipulations of lifestyles and area of expertise of suggestions to SDE concerning additive Wiener integrals, and of ideas of the combined Brownian—fractional Brownian SDE. the writer develops optimum filtering of combined types together with linear case, and reviews monetary purposes and statistical inference with hypotheses checking out and parameter estimation. She proves that the industry with inventory guided by means of the combined version is arbitrage-free with none limit at the dependence of the elements and deduces varied types of the Black-Scholes equation for fractional market.
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Additional info for Stochastic Calculus for Fractional Brownian Motion and Related Processes
Example text
21) s−α (u − s)−α |v − s|2α−1 ds 0 × (1{v
Also, let g be a function with g(v) ≥ 0, g ∈ L1 [0, T ], such that for any 0 ≤ s < t ≤ T , it holds t that ϕ(ρ(s, t)) ≤ s g(v)dv. Then T 0 g(v)dv . ϕ(2u) N ([0, T ], u) ≤ 1 + Proof. Consider 0 = s0 < s1 < . . < sM < T , where |sk+1 −sk | = 2u, 0 ≤ k ≤ M − 1, |T − SM | ≤ 2u. Such a partition exists, because our condition ensures s the continuity of ρ(s, t). Evidently, ϕ(2u) ≤ skk+1 g(v)dv, 0 ≤ k ≤ M − 1, and N ([0, T ], u) ≤ M + 1. e. M ≤ T 0 sM g(v)dv = T g(u)du ≤ g(v)dv, 0 a g(v)dv · (ϕ(2u))−1 .
4. Now, let H ∈ (1/2, 1). 1 we can take here only 1 ≤ p < α−1 since −α H LH 2 (R) = {f ∈ D(I− ) : M− f ∈ L2 (R)}. Note, that α L2 (R) = ∪1≤p<α−1 I− (Lp (R)). 5) α I− (Lp (R)) Indeed, it was proved in (SKM93) that all spaces coincide for 1 < α (Lp (R)) does not coincide with any space Lr (R), 1 ≤ r ≤ ∞. 5) follows from these theorems. 5. The space LH 2 is incomplete for H ∈ (1/2, 1). H H : LH Proof. The operator M− 2 (R) → L2 (R) is isometric. So, L2 (R) can be identified with its image in L2 (R).