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Proof. We shall use the notation from the last two theorems. Then for C
σ ( J )
( π 1
( ξ 1
m
(
x n
(
C
)) =
m
(
h n
(
R
×
...
×
R
×
C
)=
P
(
R
×
...
×
R
×
C
)) =
P
(
C
))
,
n
n
hence
E
( ξ n )=
tdP
ξ n
(
t
)=
tdm x n (
t
)=
E
(
x n )=
a ,
and
2
2
2 .
( ξ n
)= σ
(
)= σ
σ
x n
Moreover,
( ξ 1
1
ξ 1
( π 1
P
(
C 1 )
...
(
C n )) =
P
(
C 1 ×
...
×
C n )) =
n
n
( ξ 1
1
( ξ 1
=
m
(
h n
(
C 1
×
...
×
C n
)=
m
(
x 1
(
C 1
))
..... m
(
x n
(
C n
)) =
P
(
C 1
))
..... P
(
C n
))
,
n
n
σ
i
hence
ξ 1 , ...,
ξ n are independent for every n . Put g n (
u 1 , ..., u n )=
1 u i
a . By Theorem
=
4.2. we have
n
σ
n
i = 1 x i a )(( , t )) = m ( h n ( g 1
m
(
(
(
, t
))) =
m
(
y n ((
, t
)) =
n
; n
σ
n
i = 1 ξ i ( ω ) a < t } ) .
( η 1
n
=
((
))) =
( { ( ω )
P
, t
P
Therefore by the classical central limit theorem
t
1
n
i
1 x i
a
1
2
u 2
2 du
=
e
lim
n
m
(
(
, t
)) =
n
π
Let us have a look to the previous theorem from another point of view, say, categorial. We had
( η 1
lim
n
P
((
, t
)) = φ (
t
)
n
We can say that
in distribution. Of course, there are important
possibilities of convergencies, at least in measure and almost everywhere.
( η n ) n converges to
φ
( η n
)
A sequence
n of random variables (= measurable functions) converges to 0 in measure
S→ [
]
μ
:
0, 1
,if
μ ( η 1
lim
n
( ε
,
ε )) =
0
ε >
for every
0. And the sequence converges to 0 almost everywhere, if
1
p ,
1
p )) =
( p = 1 k = 1 n = k η 1
lim
n
P
((
1
n
( ω )
Certainly, if
η n
0,then
ε >
0
k
n
>
k :
ε < η ( ω ) < ε
 
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