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Il-Lemma ta' Borel-Cantelli hu riżultat fit-teorija tal-probabbiltà u t-teorija tal-miżura fundamentali għall-prova tal-liġi qawwija tan-numri kbar .
Il-lemma hi msemmija għal Émile Borel u Francesco Paolo Cantelli .
Lemma ta' Borel-Cantelli :
Prova
Bil-monotonija ta'
μ
{\displaystyle {\displaystyle \mu }}
, għandna
μ
(
lim sup
n
→
∞
S
n
)
=
μ
(
⋂
n
=
1
∞
⋃
i
=
n
∞
S
i
)
=
lim
n
→
∞
μ
(
⋃
i
>
n
S
i
)
.
{\displaystyle \mu \left(\limsup _{n\to \infty }S_{n}\right)=\mu \left({\bigcap _{n=1}^{\infty }}{\bigcup _{i=n}^{\infty }}S_{i}\right)=\lim _{n\to \infty }\mu \left(\bigcup _{i>n}S_{i}\right).}
Issa bis-subadditività:
lim
n
→
∞
μ
(
⋃
i
>
n
S
i
)
≤
lim
n
→
∞
∑
i
=
n
∞
μ
(
S
i
)
.
{\displaystyle \lim _{n\to \infty }\mu \left(\bigcup _{i>n}S_{i}\right)\leq \lim _{n\to \infty }\sum _{i=n}^{\infty }\mu (S_{i}).}
Għalhekk billi ta' l-aħħar hu l-limitu tal-bqija ta' serje konverġenti, għandna
μ
(
lim sup
n
→
∞
S
n
)
≤
lim
n
→
∞
∑
i
=
n
∞
μ
(
S
i
)
=
0.
{\displaystyle \mu \left(\limsup _{n\to \infty }S_{n}\right)\leq \lim _{n\to \infty }\sum _{i=n}^{\infty }\mu (S_{i})=0.}
In partikulari, fi spazju tal-probabbiltà
(
Ω
,
A
,
P
)
{\displaystyle (\Omega ,{\mathcal {A}},\operatorname {P} )}
, għal suċċessjoni ta' ġrajjiet
{
E
n
}
n
∈
N
{\displaystyle \{E_{n}\}_{n\in \mathbb {N} }}
, għandna:
(
1
)
∑
n
=
0
∞
P
(
E
n
)
∈
R
⇒
P
(
lim sup
n
→
∞
E
n
)
=
0.
{\displaystyle (1)\sum _{n=0}^{\infty }\operatorname {P} (E_{n})\in \mathbb {R} \Rightarrow \operatorname {P} \left(\limsup _{n\to \infty }E_{n}\right)=0.}
Fil-każ ta' spazji tal-probabbiltà, hi veru wkoll il-propożizzjoni li ġejja (spiss imsejjħa "it-tieni lemma ta' Borel-Cantelli"):
(
2
)
∑
n
=
0
∞
P
(
E
n
)
=
∞
{\displaystyle (2)\,\sum _{n=0}^{\infty }\operatorname {P} (E_{n})=\infty }
u
E
n
{\displaystyle {\displaystyle E_{n}}}
huma indipendenti
⇒
P
(
lim sup
n
→
∞
S
n
)
=
1.
{\displaystyle \Rightarrow \operatorname {P} \left(\limsup _{n\to \infty }S_{n}\right)=1.}
Prova ta' l-asserzjoni 2
P
(
lim sup
n
→
∞
E
n
)
=
lim
n
→
∞
P
(
⋃
i
≥
n
E
i
)
;
{\displaystyle \operatorname {P} \left(\limsup _{n\to \infty }E_{n}\right)=\lim _{n\to \infty }\operatorname {P} \left(\bigcup _{i\geq n}E_{i}\right);}
P
(
⋃
i
≥
n
E
i
)
=
1
−
P
(
⋂
i
≥
n
E
i
¯
)
=
1
−
lim
k
→
∞
P
(
⋂
i
=
n
k
E
i
¯
)
.
{\displaystyle \operatorname {P} \left(\bigcup _{i\geq n}E_{i}\right)=1-\operatorname {P} \left(\bigcap _{i\geq n}{\overline {E_{i}}}\right)=1-\lim _{k\to \infty }\operatorname {P} \left(\bigcap _{i=n}^{k}{\overline {E_{i}}}\right).}
Issa bl-indipendenza u d-diżugwaljanza
e
−
x
≥
1
−
x
{\displaystyle e^{-x}\geq 1-x}
;
1
−
lim
k
→
∞
P
(
⋂
i
=
n
k
E
i
¯
)
=
1
−
lim
k
→
∞
(
∏
i
=
n
k
P
(
E
i
¯
)
)
=
1
−
lim
k
→
∞
(
∏
i
=
n
k
(
1
−
P
(
E
i
)
)
)
≥
1
−
lim
k
→
∞
∏
i
=
n
k
e
−
P
(
E
i
)
{\displaystyle 1-\lim _{k\to \infty }\operatorname {P} \left(\bigcap _{i=n}^{k}{\overline {E_{i}}}\right)=1-\lim _{k\to \infty }\left(\prod _{i=n}^{k}\operatorname {P} ({\overline {E_{i}}})\right)=1-\lim _{k\to \infty }\left(\prod _{i=n}^{k}\left(1-\operatorname {P} (E_{i})\right)\right)\geq 1-\lim _{k\to \infty }\prod _{i=n}^{k}e^{-\operatorname {P} (E_{i})}}
=
1
−
lim
k
→
∞
e
−
∑
i
=
n
k
P
(
E
i
)
=
1
,
{\displaystyle =1-\lim _{k\to \infty }e^{-\sum _{i=n}^{k}\operatorname {P} (E_{i})}=1,}
billi s-somma tiddiverġi u hekk l-esponenzjali tersaq lejn 0. Għalhekk:
P
(
lim sup
n
→
∞
E
n
)
≥
lim
n
→
∞
1
=
1
⇒
P
(
lim sup
n
→
∞
E
n
)
=
1
{\displaystyle \operatorname {P} \left(\limsup _{n\to \infty }E_{n}\right)\geq \lim _{n\to \infty }1=1\Rightarrow \operatorname {P} \left(\limsup _{n\to \infty }E_{n}\right)=1}
Fi kliem ieħor jekk suċċessjoni ta' ġrajjiet għandha probabbiltà sommabbli, kważi żgurament, in-numru ta' drabi li jiġru hu finit. Jekk minflok il-probabbiltà mhijiex sommabili u l-ġrajjiet huma indipendenti kważi żgurament in-numru ta' drabi li jiġru hu infinit. In partikulari f'numru infinit ta' provi indipendenti kull ġrajja b'probabbiltà pożittiva tiġri numru infinit ta' drabi.