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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.
 Fall '06
 Varadhan

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