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Proof. Let C 1 , C 2 , ..., C n
∈B (
R
)
. Then by Definition 3.3. and Definition 3.1
μ n + 1 (
C 1 ×
C 2 ×
...
×
C n
×
R
)=
m
(
x 1 (
C 1 ))
. x 2 (
C 2 )
..... x n
(
C n
)
. x n + 1 (
R
)) =
=
m
(
x 1 (
C 1 ))
. x 2 (
C 2 )
..... x n
(
C n
)
.
(
1, 0
)) =
=
m
(
x 1 (
C 1 ))
. x 2 (
C 2 )
..... x n (
C n )) =
= μ n
(
C 1
×
C 2
×
...
×
C n
)
,
| ( J
×
)= μ n
|J
J
hence
μ n + 1
R
n . Of course, if two measures coincide on
n then they coincide
n
σ ( J
)
on
, too.
n
Now we shall formulate a translation formula between sequences of observables in
( F
, m
)
R N ,
and corresponding random variables in
(
σ ( C )
, P
)
([67]).
Let g n : R n
Theorem 4.2.
Let the assumptions of Theorem 4.1 be satisfied.
R be Borel
be the family of all cylinders in R N ,
ξ n : R N
=
C
measurable functions n
1, 2, .... Let
R be
((
t i ) i )=
defined by the formula
ξ n
t n ,
η n : R N
η n
=
( ξ 1 ., ...,
ξ n
)
R ,
g n
,
R n
g 1
n
y n :
B (
) →F
, y n
=
h n
.
Then
( η 1
n
(
)) =
(
(
))
P
B
m
y n
B
for any B
∈B (
R
)
.
g 1
n
Proof. Put A
=
(
B
)
. By Theorem 4.1.
g 1
n
( π 1
g 1
n
m
(
y n
(
B
)) =
m
(
h n
(
(
B
))) =
P
(
(
B
))) =
n
) 1
( η 1
n
=
((
◦ π n
(
)) =
(
))
P
g n
B
P
B
.
As an easy corollary of Theorem 4.2 we obtain a variant of central limit theorem.
In the
classical case
t
1
n
n
i
1 ξ i ( ω )
a
1
2
u 2
2 du
=
e
lim
n
P
( { ω
;
<
t
} )=
n
π
Of course, we must define for observables the element
n
σ
n
i = 1 x i a )( , t )
(
It is sufficient to put
n
σ
n
i = 1 u i a
(
)=
g n
u 1 , ..., u n
Theorem 4.3.
Let
(
x n
)
n be a sequence of square integrable, equally distributed, independent
2
2
observables, E
(
x n )=
a ,
σ
(
x n )= σ
(
n
=
1, 2, ...
)
. Then
t
1
n
i
1 x i
a
1
2
u 2
2 du
=
e
lim
n
m
(
(
, t
)=
n
π
 
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