Кошкин Г.М. Основы страховой математики
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[C
1
, C
2
]
Z(u) = 1 −
u − C
1
C
2
− C
1
.
fs
N
F in
(x)
S(u) ∈ F in
S
.
fs
N
F in
s(z) ∈ N
0,1
(x), sup
x∈R
1+
f(x) ≤ C < ∞, a
N
↓ 0,
N → ∞
|b (fs
N
F in
(x))| = O (a
N
) .
E fs
N
F in
(x) =
∞
Z
0
S
x − y
a
N
dF (y) =
= s(x) +
x
Z
0
S
x − y
a
N
dF (y) +
∞
Z
x
S
x − y
a
N
− 1
dF (y).
dF (y) = f(y)dy
E fs
N
F in
(x) ≤ s(x) + C
x
Z
0
S
x − y
a
N
dy +
∞
Z
x
S
x − y
a
N
− 1
dy
.
u =
x − y
a
N
|b (fs
N
F in
(x))| ≤ Ca
N
x/a
N
Z
0
S (u) du +
0
Z
−∞
|S (u) − 1|du
.
N
x/a
N
Z
0
S (u) du ≤
C
2
Z
C
1
S (u) du < ∞,
0
Z
−∞
|S (u) − 1|du ≤
C
2
Z
C
1
|S (u) − 1|du < ∞ |b ( fs
N
(x))| = O (a
N
) . ♠
fs
N
(x) S(u) ∈ Suv
fs
N
F in
(x)
fs
N
N → ∞
D fs
N
(x) =
s(x) (1 − s(x))
N
+ o
1
N
.
D fs
N
(x) =
1
N
DS
x − X
1
a
N
=
=
1
N
Z
R
1+
S
2
x − y
a
N
dF (y) −
Z
R
1+
S
x − y
a
N
dF (y)
2
.
D fs
N
(x) =
1
N
s(x) − s
2
(x) + o(1)
=
s(x) (1 − s(x))
N
+ o
1
N
. ♠
S
LAP
(u) ∈ Suv,
D fs
N
LAP
(x) =
s(x) (1 − s(x))
N
+ o
1
N
.
fs
N
F in
N → ∞
D fs
N
F in
(x) =
s(x) (1 − s(x))
N
+ O
a
N
N
=
s(x) (1 − s(x))
N
+ o
1
N
.
♠
fs
N
F in
fs
N
LAP
(x).
fs
N
F in
N → ∞
u
2
(fs
N
F in
) =
s(x) (1 − s(x))
N
+ o
1
N
, a
N
= o
1/
√
N
;
O
1
N
, a
N
= O
1/
√
N
.
u
2
(fs
N
F in
(x)) = D fs
N
F in
(x) + b
2
(fs
N
F in
(x))
♠
fs
N
LAP
N → ∞
u
2
(fs
N
LAP
) =
s(x) (1 − s(x))
N
+ o
1
N
, a
N
= o
1/
√
N
;
O
1
N
, a
N
= O
1/
√
N
.
♠
fs
N
F in
(x).
{ξ
j,N
}
N
j=1
, N = 1, 2, . . .
ξ
j,N
N
a
N
= o
N
−1/2
N → ∞,
√
N [fs
N
F in
(x) − s(x)] =⇒ N
1
{0, s(x)(1 − s(x))}.
√
N [fs
N
F in
(x) − s(x)] =
=
√
N [fs
N
F in
(x) − E fs
N
F in
(x)] +
√
N b (fs
N
F in
(x)) .
N → ∞
√
Nb (fs
N
F in
(x)) =
√
N
h
o
N
−1/2
i
−→ 0.
ξ
j,N
=
1
√
N
S
x − X
j
a
N
− ES
x − X
j
a
N
.
fs
N
F in
(x) − E fs
N
F in
(x) =
1
√
N
N
X
j=1
ξ
j,N
.
Eξ
j,N
= 0
Eξ
2
j,N
=
1
N
DS
x − X
j
a
N
< ∞.
lim
N→∞
nEξ
2
1,N
= s(x) (1 − s(x)) .
sup
u∈R
1
S(u) ≤ 1 τ > 0
β
N
= nE
|ξ
1,N
|
2
, |ξ
1,N
| > τ
<
N
τ
E |ξ
1,N
|
3
≤
≤
C
√
N
"
E
S
x − X
1
a
N
3
+
ES
x − X
1
a
N
3
#
<
2C
√
N
C < ∞
β
N
= O
N
−1/2
−→ 0 N → ∞
√
N ξ
N
=
√
N
N
X
j=1
ξ
j,N
=⇒ N
1
{0, s(x)(1 − s(x))}.
♠
F(u)
Dis F(u) ∈ Dis F(u)
F(·) : R
1
→
R
1
F(−∞) = 0, F(∞) = 1.
fs
N
(x)
f
F
N
(x) =
1
N
N
X
i=1
F
x − X
i
a
N
,
F(u) ∈ Dis a
N
↓ 0.
F(u)
F(u) = 1 − S(u) S(u) ∈ Suv
S(u) ∈ F in
S
f
F
N
(x) = 1 −
1
N
N
X
i=1
S
x − X
i
a
N
= 1 − fs
N
(x).
f
F
N
(x)
f
F
N
F (z) ∈ N
0,1
(x), F(u) ∈ Dis,
a
N
↓ 0,
E
f
F
N
(x) = F (x) + o(1), D
f
F
N
(x) =
1
N
F (x) (1 − F (x)) + o
1
N
.
F in
S
F in
F
F(u)
F in
F
F(u) ∈ F in
F
F(u) =
(
0, −∞ < u < C
1
,
Y (u), C
1
≤ u ≤ C
2
,
1, C
2
< u < ∞,
Y (u)
Y (C
1
) = 0, Y (C
2
) = 1, C
1
< C
2
F(u) ∈ F in
F
F(u) ∈ F in
F
F ∈ Dis
f
F
N
F in
(x)
F(u) ∈ F in
F
.
f
F
N
F in
F (z) ∈ N
0,1
(x), sup
t∈R
1+
f(x) < ∞, a
N
↓ 0, N → ∞
b
f
F
N
F in
(x)
= O (a
N
) .
f
F
N
N → ∞
u
2
f
F
N
F in
=
F (x) (1 − F (x))
N
+ o
1
N
, a
N
= o
1/
√
N
,
O
1
N
, a
N
= O
1/
√
N
.
a
N
= o
N
−1/2
,
√
N
h
f
F
N
(x) − F (x)
i
=⇒ N
1
{0, F (x)(1 − F (x))}.
F(u)
f(x) = F
0
(x)
f
N
(x) =
d
dx
f
F
N
(x) =
1
Na
N
N
X
i=1
K
x − X
i
a
N
,
K(u) = F
0
(u)
a
N
↓ 0
f
N
(x) =
1
Nh
N
N
X
i=1
K
x − X
i
h
N
,
h
N
↓ 0 K(u)
h
N
,
K(·)
f(x)
K(u)
A,
sup
u∈R
1
|K(u)| < ∞,
∞
Z
−∞
|K(u)|du < ∞,
∞
Z
−∞
K(u)du = 1; K(u) ∈ A
ν
,
K(u) ∈ A K(u)
∞
Z
−∞
|u
ν
K(u)|du < ∞, T
j
=
∞
Z
−∞
u
j
K(u)du = 0, j = 1, . . . , ν − 1.
f(x),
X
f
N
f(x) ∈ N
0,1
(R), f(x) R
1
=
(−∞, ∞) sup
x∈R
1+
f(x) ≤ C
1
< ∞,
∞
Z
−∞
|K(u)|du < ∞,
∞
Z
−∞
K(u)du = 1,
h
N
↓ 0,
lim
N→∞
Ef
N
(x) = f(x).
Ef
N
(x) =
1
h
N
∞
Z
0
K
x − y
h
N
f(y)dy.
u =
x − y
h
N
Ef
N
(x) =
1
h
N
−∞
Z
x/h
N
(−h
N
)K (u) f(x − uh
N
)du =
=
∞
Z
−∞
K (u) f(x − uh
N
)du −
∞
Z
x/h
N
K (u) f(x − uh
N
)du.
lim
N→∞
∞
Z
−∞
K (u) f(x − uh
N
)du = f(x)
∞
Z
−∞
K (u) du = f (x),
lim
N→∞
−
∞
Z
x/h
N
K (u) f(x − uh
N
)du ≤ C
1
lim
N→∞
∞
Z
x/h
N
K (u) du = 0.
♠
h
N
> 0 lim
N→∞
h
N
= 0, lim
N→∞
h
N
+
1
Nh
N
= 0,
H
1
, H
2
.
f
N
∞
Z
−∞
K
2
(u)du < ∞ h
N
∈ H
2
, N → ∞
Df
N
(x) =
f(x)
Nh
N
∞
Z
−∞
K
2
(u)du + o
1
Nh
N
.
Df
N
(x) =
1
Nh
2
N
DK
x − X
1
h
N
=
1
Nh
2
N
∞
Z
0
K
2
x − y
h
N
f(y)dy−
−
∞
Z
0
K
x − y
h
N
f(y)dy
2
.
u =
x − y
h
N
Df
N
(x) =
=
1
Nh
N
∞
Z
−∞
K
2
(u)f(x − uh
N
)du − h
N
∞
Z
−∞
K(u)f(x − uh
N
)du−
−
∞
Z
x/h
N
K(u)f(x − uh
N
)du
2
−
∞
Z
x/h
N
K
2
(u)f(x − uh
N
)du
.
Df
N
(x) =
1
Nh
N
f(x)
∞
Z
−∞
K
2
(u)du + h
N
(f(x) + o(1))
2
+ o(1)
=
=
f(x)
Nh
N
∞
Z
−∞
K
2
(u)du + o
1
Nh
N
. ♠
f
N
(x)
b (f
N
(x))
w
ν
(x) =
f
(ν)
(x)
ν!
T
ν
,
f
(ν)
(x) =
∂
ν
f(x)
∂x
ν
, f
(0)
(x) = f(x).
b(f
N
)
f(z) ∈ N
ν,1
(R) ;
sup
x∈R
1
f
(m)
(x)
< ∞, m = 0, ν;
K(u) ∈ A
ν
sup
u∈R
1
|K(u)| < ∞;
1 − K(x) = o
x
−ν
x → ∞, K(x) =
x
Z
−∞
K(u)du;
h
N
↓ 0.
N → ∞ b(f
N
)
|b(f
N
(x)) − w
ν
(x)h
ν
N
| = o (h
ν
N
) .
Ef
N
(x) =
∞
Z
−∞
f(x − uh
N
)K(u)du −
∞
Z
x/h
N
f(x − uh
N
)K(u)du.
f(x) N → ∞
∞
Z
x/h
N
f(x − uh
N
)K(u)du ≤ C
∞
Z
x/h
N
K(u)du =
= C
1 − K
x
h
N
= o(h
ν
N
).
f(x −uh
N
)
Ef
N
(x) = f(x) +
ν−1
X
i=1
(−1)
i
f
(i)
(x)
h
i
N
i!
T
i
+
+(−1)
ν
f
(ν)
(x)
h
ν
N
ν!
T
ν
+ γ
N
−
∞
Z
x/h
N
f(x − uh
N
)K(u)du,