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 λx
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.
 Fall '06
 Varadhan

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