Symmetric matrix notes

E zt e z0 let t as before be t t x t

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: λM t) + ] 4 4 2 2 22 2 σ λx λσ Uxx + b Ux − U+ U 2 4 4M Ux = U [− sup 0 ≤σ ≤M |b|≤M λ2 x2 λM λ2 x2 tanh2 λM t − + (1 − tanh λM t) + 8 4 4 λ2 x2 λ2 x2 λM ≤U tanh2 λM t − + (1 − tanh λM t) + 4 4 4 = Ut ≤U Therefore t λ2 Zt = U (T − t , ξ (t)) exp[− 4 λ ξ (s)ds + 4M t 2 0 σ 2 (s)ds] 0 is a super martingale. E [ZT ] ≤ E [Z0 ] Let ξ (t) as before be t ξ ( t) = x + t < e(s).dβ (s) > + 0 b(s)ds 0 3 λ|x| tanh λM t 2 1 2 and denote by t η ( t) = x + < e(s).dβ (s) > ds 0 t B ( t) = b(s)ds 0 so that ξ ( t) = x + η ( t) + B ( t) Assume that σ (s) = e(s) ≤ M a.e. and |b(s)| ≤ M a.e. Lemma 2. We have T T 2 E exp λ |η (s)|ds − λ T 0 σ 2 (s)ds] ≤ C. 2 0 Proof. Apply Doob’s inequality to the non negative martingale t λ2 exp λ η (t) − 2 σ 2 (s)ds 0 to get P exp sup [λ η (t) − 0≤t≤T t λ2 2 σ 2 (s)ds] ≥ ℓ ≤ 0 1 ℓ and for λ ≥ 0, replacing λ by 2 λ, T P exp 2λ sup η (t) − 2 λ2 0≤t≤T σ 2 (s)ds] ≥ ℓ ≤ 0 This leads to 1 ℓ T σ 2 (s)ds] E exp λ sup |η (t)| − λ 0≤t≤T and E exp 0 ≤C 0 T T λ T ≤C σ 2 (s)ds] 2 |η (s)|ds − λ2 0 and replacing λ by λ T , T T E exp λ 0 |η (s)|ds − λ2 T 2 4 σ 2 (s)ds] 0 ≤C Lemma 3. For any λ ≥ 0, T E exp − 4M λ2 T 2 ξ 2 (s)ds − 0 λ 2 T λ2 T 2 2 |ξ (s)|ds + 0 T σ 2 (s)ds + 0 λ 2 T |B (s)|ds 0 ≤ C (T ) Proof: We have by the earlier lemma, for any µ > 0 T µ2 4 E exp − ξ 2 (s)ds + 0 T µ 4M σ 2 (s)ds ≤ C (T ) 0 By Schwarz’...
View Full Document

Ask a homework question - tutors are online