Некоторые главы анализа и приложение к финансовой математике
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S+ − K 
S − K
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δ
P0 = ϕS0 + ψB0.
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ϕ 
S0 
ψ 
B0 
ϕ 
ϕ ψ 
ϕS+ + erδψ 
ϕS− + erδψ 
ϕS+ + erδψ = S+ − K
ϕS− + erδψ = 0.
ϕS0 + ψ
C0 = ϕS0 + ψ.
ϕ 
ψ
ϕS+ + erδψ = S+ − K, ϕS− + erδψ = 0.
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ϕ = |
S+ − K |
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ψ = |
− |
e−rδS |
S+ − K |
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S+ − S− |
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− S+ − S− |
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C |
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= S |
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S+ − K |
− |
e−rδS |
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S+ − K |
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0 |
0 S+ − S− |
− S+ − S− |
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= (S |
0 − |
e−rδS |
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S+ − K |
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S+ − S− |
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= e−rδ |
erδS0 − S− |
(S |
+ − |
K). |
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S+ − S− |
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ϕ 
ψ 
K 
p˜ 
ϕ
ψ 
C0 = e−rδ p˜(S+ − K),
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p˜ = erδS0 − S− .
S+ − S−
e−rδ p˜(S+ − K) = e−rδ erδS0 − S− (S+ − K), S+ − S−
(S− K)+ := max(S − K, 0) 
S−K 
S−K < 0 
S+ 
p˜ = 01 1(x ≤ p˜) dx
1
C0 = e−rδ (S+ − K)+1(x ≤ p˜) dx.
0
(S+ − K)+ 
x
0 ≤ p˜ ≤ 1
1 p˜ 1
1(x ≤ p˜) dx = 1(x ≤ p˜) dx + 1(x ≤ p˜) dx
0 0 p˜
p˜ 1
= 1 dx + 0 dx = p˜.
0p˜
1
e−rδ (S+ − K)+1(x ≤ p˜) dx
0
1
= e−rδ(S+ − K)+ 1(x ≤ p˜) dx = e−rδ(S+ − K)+ p,˜
0
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r = 0 
K = 9
11
9
8
δ
2 (= 11 − 9)
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0
δ
r = 0
p˜ = |
10 |
− 8 |
= |
2 |
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11 |
− 8 |
3 |
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C0 = e0 × 23 × 1 = 2/3.
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P0 = 9ϕ + ψ 
8ϕ + ψ = 0. |
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ψ = 8ϕ |
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ψ = |
3 |
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11ϕ + ψ = 2 |
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3ϕ = 2 |
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ϕ = |
2 |
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− |
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− 3 |
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P0 = |
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S0 − |
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3 |
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t = 0 |
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P0 |N0= 9ϕ + ψ = |
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− |
16 |
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3 |
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$ 0, 5
2 − 0.5 = 1.5
P0 = C0 + αS0 − 0.5 − α × 9.
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+C0 |
0.5 + α × 9 |
α > 0 α |
0.5 + α × 9 < 0
α
−(0.5 + α × 9) 
2 + α × 11 − 0.5 − α × 9,
2 + α × 11 − 0 5 − α × 9 > 0
0 + α × 8 − 0.5 − α × 9,
0 + α × 8 − 0.5 − α × 9 > 0
2α + 1.5 > 0, |
α < −0.5, |
2α > −1.5, |
α < −0.5, |
−0.75 < α < −0.5.
α 
α = −0.7
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f(S+) f(S−) |
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ϕS+ + erδψ = f(S+), |
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ϕS− + erδψ = f(S−). |
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(ϕ, ψ) |
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ϕ = |
f(S+) − f(S−) |
, ψ = e−rδf(S |
− |
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− |
e−rδS |
− |
f(S+) − f(S−) |
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S+ − S− |
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S+ − S− |
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P0 |
= |
f(S+) − f(S−) |
S |
0 |
+ e−rδf(S |
− |
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− |
e−rδS |
− |
f(S+) − f(S−) |
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S+ − S− |
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S+ − S− |
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= e−rδ |
f(S |
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erδS0 − S− |
+ f(S |
− |
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S+ − erδS0 |
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+ |
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S+ − S− |
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S+ − S− |
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p˜ 
q˜ := S+ − erδS0 ,
S+ − S−
P0 = e−rδ(f(S+)˜p + f(S−)˜q).
p˜ + q˜ = 1, 0 ≤ p,˜ q˜ ≤ 1.
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1
P0 = e−rδ (f(S+)1(x ≤ p˜) + f(S−)1(x > p˜)) dx.
0
S3
S1
S0 S4
S2
S5
δ 2δ
Ni 
Si 
0 ≤ i ≤ 5
N4 
39
(S3 − K)+
?
? (S4 − K)+
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(S5 − K)+
δ 2δ
S1 < erδS0 < S2, S4 < erδS1 < S3, S5 < erδS2 < S4.
(x−K)+ = max(x−K, 0)
2δ
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N0 
δ
t = 0 
2δ 
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