Borel-Cantelli Theorem of a Finite Series of Independent Events
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Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.
We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.
However part of most of the proofs I have read do additional step where
$bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.
I don't understand why those two values are equal.
probability-theory statistics proof-explanation
asked Nov 15 at 16:28
kpr62
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Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.
We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.
However part of most of the proofs I have read do additional step where
$bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.
I don't understand why those two values are equal.
probability-theory statistics proof-explanation
asked Nov 15 at 16:28
kpr62
153
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.
We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.
However part of most of the proofs I have read do additional step where
$bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.
I don't understand why those two values are equal.
probability-theory statistics proof-explanation
asked Nov 15 at 16:28
kpr62
153
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.
We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.
However part of most of the proofs I have read do additional step where
$bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.
I don't understand why those two values are equal.
probability-theory statistics proof-explanation
probability-theory statistics proof-explanation
asked Nov 15 at 16:28
kpr62
153
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 16:28
kpr62
153
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 16:28
kpr62
153
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 16:28
kpr62
153
asked Nov 15 at 16:28
kpr62
153
153
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kpr62 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$
This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$
Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.
Can you figure out why yourself?
answered Nov 15 at 16:42
drhab
94.5k543125
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$
This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$
Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.
Can you figure out why yourself?
answered Nov 15 at 16:42
drhab
94.5k543125
add a comment |
up vote
0
down vote
Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$
This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$
Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.
Can you figure out why yourself?
answered Nov 15 at 16:42
drhab
94.5k543125
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$
This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$
Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.
Can you figure out why yourself?
answered Nov 15 at 16:42
drhab
94.5k543125
Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$
This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$
Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.
Can you figure out why yourself?
answered Nov 15 at 16:42
drhab
94.5k543125
answered Nov 15 at 16:42
drhab
94.5k543125
answered Nov 15 at 16:42
drhab
94.5k543125
answered Nov 15 at 16:42
drhab
94.5k543125
94.5k543125
add a comment |
add a comment |
kpr62 is a new contributor. Be nice, and check out our Code of Conduct.
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