Symmetric matrix notes

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

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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’...
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## This note was uploaded on 03/19/2013 for the course MATH GA 2931 taught by Professor Varadhan during the Fall '06 term at NYU.

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