Некоторые главы анализа и приложение к финансовой математике
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φξn (t) := 01 exp(itξn(x)) dx |
t R1 |
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n → ∞ pn → p (0, 1) |
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, . . . , xn) := (1(xk ≤ pn) − 1(xk > pn)) |
Sn = S |
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k=1
n
≡ 2 (1(xk ≤ pn) − n,
k=1


x1, . . . , xn [0, 1] 





f
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f( |
Sn − n(2pn − 1) |
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. . . dx |
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2 npn(1 − pn) |
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∞ |
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/2 dz → 0, n → ∞. |
− −∞ f(z) √2π e−z |
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ˆ |
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Sn = 2Sn − n |
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f( |
Sn − n(2pn − 1) |
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Sn − npn |
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npn(1 |
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pn) |
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Sn |
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Sn |
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pn → p (0, 1) |
n → ∞ |
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n |
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Sn = S(x1, . . . , xn) = |
1(xk ≤ pn), |
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k=0 |
x1, . . . , xn [0, 1] |
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ε > 0 |
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c > 0 |
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1(| n − pn| ≥ ε)dx1 . . . dxn ≤ 2 exp(−cn), n → ∞. |
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A = |
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Sn |
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A ≤ exp(−λεn) |
exp(λ|Sn − npn|)dx1 . . . dxn |
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≤ exp(−λεn) exp(λ(Sn − npn))dx1 . . . dxn
+ exp(−λεn) exp(λ(−Sn + npn))dx1 . . . dxn
≡ exp(−λεn)ϕn(λ) + exp(−λεn)ψn(λ).
ϕn(λ) = exp(λ(1(xk ≤ pn) − pn))dxk = (ϕ1(λ))n,
22
ψn(λ) = exp(λ(−1(xk ≤ pn) + pn))dxk = (ψ1(λ))n, ϕ1(0) = ψ1(0) = 1,
ϕ1(0) = |
(1(x1 ≤ pn) − pn))dx1 = |
0pn |
1 dx1 − pn = 0, |
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ψ1(0) = |
(−1(x1 ≤ pn) + pn))dx1 = − 0pn |
1 dx1 + pn = 0. |
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λ > 0
exp(−λε)ϕ1(λ) < 1, exp(−λε)ψ1(λ) < 1.
λ > 0
exp(−c) = max(exp(−λε)ϕ1(λ), exp(−λε)ψ1(λ)),
c = − ln (max(exp(−λε)ϕ1(λ), exp(−λε)ψ1(λ))) .





λ R1
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npn(1 pn) |
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C(λ) := sup |
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exp λ |
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Sn − npn |
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dx |
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. . . dx |
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npn(1 |
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pn) |
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exp(λ |
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Sn |
− npn |
) dx |
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. . . dx |
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n |
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= 1 exp(λ |
1(x1 ≤ pn) − pn |
) dx1 |
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0npn(1 − pn)
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np(1 pn) |
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np(1 pn) |
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= p |
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exp(λ |
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− pn |
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exp(λ |
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−pn |
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= 1 + |
cλ |
+ o(1/n) . |
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n→∞ |
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= exp cλ < ∞, |
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lim |
1 + |
cλ |
+ o(1/n) |
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λ > 0 |
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n |
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npn(1 pn) ≥ |
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sup |
1( |
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Sn |
− npn |
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N) dx |
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. . . dx |
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n |
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npn(1 pn) |
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exp( |
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λN) sup |
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exp(λ |
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Sn |
− npn |
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. . . dx |
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≤ C(λ) exp(−λN). |
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λ > 0 |
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1( |
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Sn − npn |
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N) = 1(λ |
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Sn − npn |
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λN) |
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npn(1 − pn) ≥ |
npn(1 − pn) ≥ |
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≤ |
exp( |
− |
λN) exp(λ |
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Sn − npn |
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npn(1 − pn) |
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24
0 









δ > 0
S0




exp(rδ) 










δ 




















r > 0





















r = 0











S+ 


















S−
25
S− < erδS0 < S+ 



















S− ≤ erδS0 ≤ S+ 






















nδ


























































K 




















K
S− ≤ K ≤ S+.
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S− 





























S+ 






















































K 




























S+ 










K 





















































r > 0

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S
+
S
0
S
−
δ






















δ 









S0 














S+
S− 






























r ≥ 0



















r < 0 















r = 0 


































11
9
8
δ
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2 |
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16 |
16 |
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P0 = |
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S0 |
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t = δ
S1 = 11







P0 |N1 |
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2 |
× |
11 |
− |
16 |
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22 − 16 |
= 2. |
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S1 = 8





P0 |N2 |
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2 |
× |
8 |
− |
16 |
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16 − 16 |
= 0. |
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