Symmetric matrix notes

# Sds 4 0 0 0 t 2m 2 t 2 0 sds 0 b sds 0

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Unformatted text preview: s inequality, in combination with Lemma 2, with the choice of µ = 4M λ2 T 2 T 2 E exp − 4M λ T T λ2 T 2 ξ (s)ds + 2 0 T λ σ (s)ds + 2 2 2 2 0 |η (s)|ds 0 ≤ C (T ) This in turn implies T 2 E exp − 4M λ T λ ξ (s)ds − 2 T λ2 T 2 |ξ (s)|ds + 2 2 2 0 0 T λ σ (s)ds + 2 T 2 0 |B (s)|ds 0 ≤ C (T ) If X and Y are two random variables such that E [exp[−aX + bY ]] ≤ C then √ 1 a b b a E exp[− X ] = E exp[− X + Y − Y ] ≤ C [E [exp[−bY ]]] 2 2 2 2 2 Therefore T 2 E exp −2M λ T 0 T ≤ C (T )E exp − T 2 2M λ T λ ξ (s)ds + 4 0 0 0 T ≤ 2M λ2 T 2 0 |ξ (s)|ds] 0 |B (s)|ds 0 |ξ (s)|ds √ λT ξ 2 (s)ds + 4 1 2 T ξ 2 (s)ds 0 T ≤ 2M λ2 T 2 1 2 T λ λ2 T 2 2 σ (s)ds − 2 2 T 2 2 T λ ξ (s)ds − 4 2 2 T ξ 2 (s)ds + 2λ2 T 0 ξ 2 (s)ds + 0 T ≤ 2λ2 T (1 + M T ) 5 ξ 2 (s)ds + 0 1 32 1 32 Therefore T E exp −2λ2 T (1 + M T ) ξ 2 (s)ds 0 T λ2 T 2 ≤ C (T )E exp − 2 λ σ (s)ds − 2 0 |B (s)|ds 0 1 2 T λ2 T 2 ≤ C (T )E exp − 2 1 2 T 2 σ 2 (s)ds 0 and T ξ 2 (s)ds 2 E exp −2λ T (1 + M T ) 0 ∞ 2 e−Ax xk dx = 0 1 2 ∞ e−Ay y k −1 2 (u(x)) xk dx ≤ 0 dy = Γ( k+1 ) − k+1 2 A2 2 ∞ (u(x)) xk dx + (u(x)) xk dx 0 1 1 2 1 2k u(x) x dx ≤ |B (s)|ds 0 0 1 ∞ 1 2 T λ ≤ C (T )E exp − 2 ∞ + 0 1 ∞ u(x) x2k dx = 1 2 dx x2 1 2 u( x) x 2k+2 dx 1 ∞ u(x) x2k+2 dx + 1 2 ∞ 1 2 0 0 Therefore T ξ 2 (s)ds E − k+1 2 σ 2 (s)ds ≤ C (T , k )E 1 + 0 +3 − 2k2 T 0 1 2 and T ξ 2 (s)ds E − k+1 2 ≤ C (T , k )E 1 + 0 The ﬁnal step is to estimate −2k T |B (s)|ds E 0 in terms of −k T E 0 |b(s)|2 ds 6 −(2k+3) T 0 |B (s)|ds 1 2 where t b(s)ds B ( t) = x + 0 Should depend on the simple estimate (Sobolev) b L2 [0,T ] ≤ CT ( B First if we consider T T Θ2 = L1 [0,T ] ) a (b | b ( t) − b ( s ) | 2 7 0 p Hα [0,T ] ) |t − s| 4 0 dtds then Θ= b 3 with p = 2 and α = 8 . p Hα [0,T ] E [Θk ] ≤ C (T , k ) and one can get b L2 [0,T ] ≤ CT Θ1−a ( B for some a > 0. 7 L1 [0,T ] ) a 1−a...
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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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