Proof that $P(U[X,X+1])Rightarrow N(0,1)$
$begingroup$
Let A, B and $A_{1},A_{2},...,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.
Proof that $P(A_{n} cap B)rightarrow P(B)$
This is how I am starting the answer:
Let $B = cup (A_{n}cap B)$
Is it the right way?
Any help. I am really stuck in here.
measure-theory
asked Jan 9 at 1:10
LauraLaura
1196
$endgroup$
This question has an open bounty worth +50
reputation from Laura ending tomorrow.
The current answers do not contain enough detail.
add a comment |
$begingroup$
Let A, B and $A_{1},A_{2},...,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.
Proof that $P(A_{n} cap B)rightarrow P(B)$
This is how I am starting the answer:
Let $B = cup (A_{n}cap B)$
Is it the right way?
Any help. I am really stuck in here.
measure-theory
asked Jan 9 at 1:10
LauraLaura
1196
$endgroup$
This question has an open bounty worth +50
reputation from Laura ending tomorrow.
The current answers do not contain enough detail.
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
add a comment |
$begingroup$
Let A, B and $A_{1},A_{2},...,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.
Proof that $P(A_{n} cap B)rightarrow P(B)$
This is how I am starting the answer:
Let $B = cup (A_{n}cap B)$
Is it the right way?
Any help. I am really stuck in here.
measure-theory
asked Jan 9 at 1:10
LauraLaura
1196
$endgroup$
Let A, B and $A_{1},A_{2},...,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.
Proof that $P(A_{n} cap B)rightarrow P(B)$
This is how I am starting the answer:
Let $B = cup (A_{n}cap B)$
Is it the right way?
Any help. I am really stuck in here.
measure-theory
measure-theory
asked Jan 9 at 1:10
LauraLaura
1196
asked Jan 9 at 1:10
LauraLaura
1196
edited Jan 12 at 0:53
Laura
asked Jan 9 at 1:10
LauraLaura
1196
asked Jan 9 at 1:10
LauraLaura
1196
asked Jan 9 at 1:10
LauraLaura
1196
1196
This question has an open bounty worth +50
reputation from Laura ending tomorrow.
The current answers do not contain enough detail.
This question has an open bounty worth +50
reputation from Laura ending tomorrow.
The current answers do not contain enough detail.
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
add a comment |
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
add a comment |
2 Answers
2
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$begingroup$
Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
$endgroup$
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
add a comment |
$begingroup$
Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).
Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}
answered 2 days ago
OldGodzillaOldGodzilla
1614
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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$begingroup$
Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
$endgroup$
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
add a comment |
$begingroup$
Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
$endgroup$
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
add a comment |
$begingroup$
Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
$endgroup$
Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
answered Jan 9 at 1:17
angryavianangryavian
40.1k23280
40.1k23280
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
add a comment |
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
$begingroup$
Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
$endgroup$
– Laura
Jan 9 at 1:28
add a comment |
$begingroup$
Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).
Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}
answered 2 days ago
OldGodzillaOldGodzilla
1614
$endgroup$
add a comment |
$begingroup$
Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).
Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}
answered 2 days ago
OldGodzillaOldGodzilla
1614
$endgroup$
add a comment |
$begingroup$
Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).
Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}
answered 2 days ago
OldGodzillaOldGodzilla
1614
$endgroup$
Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).
Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}
answered 2 days ago
OldGodzillaOldGodzilla
1614
answered 2 days ago
OldGodzillaOldGodzilla
1614
answered 2 days ago
OldGodzillaOldGodzilla
1614
answered 2 days ago
OldGodzillaOldGodzilla
1614
1614
add a comment |
add a comment |
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$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54