Mixing
Proof that when $n to infty$, then $P(A_{n}cap B_{n})to p$
$begingroup$
Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.
I know there is a theorem that says:
If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.
I am almost sure that I have to use this Theorem but I am not able to do this.
Any help?
probability-theory elementary-set-theory convergence
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
asked Jan 8 at 19:46
LauraLaura
2288
$endgroup$
add a comment |
$begingroup$
Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.
I know there is a theorem that says:
If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.
I am almost sure that I have to use this Theorem but I am not able to do this.
Any help?
probability-theory elementary-set-theory convergence
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
asked Jan 8 at 19:46
LauraLaura
2288
$endgroup$
add a comment |
$begingroup$
Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.
I know there is a theorem that says:
If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.
I am almost sure that I have to use this Theorem but I am not able to do this.
Any help?
probability-theory elementary-set-theory convergence
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
asked Jan 8 at 19:46
LauraLaura
2288
$endgroup$
Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.
I know there is a theorem that says:
If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.
I am almost sure that I have to use this Theorem but I am not able to do this.
Any help?
probability-theory elementary-set-theory convergence
probability-theory elementary-set-theory convergence
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
asked Jan 8 at 19:46
LauraLaura
2288
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
asked Jan 8 at 19:46
LauraLaura
2288
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
edited Jan 8 at 21:09
Davide Giraudo
126k16150261
126k16150261
asked Jan 8 at 19:46
LauraLaura
2288
asked Jan 8 at 19:46
LauraLaura
2288
asked Jan 8 at 19:46
LauraLaura
2288
2288
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
$$
mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
$$
hence
$$
mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
$$
The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
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$begingroup$
Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
$$
mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
$$
hence
$$
mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
$$
The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
$$
mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
$$
hence
$$
mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
$$
The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
$$
mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
$$
hence
$$
mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
$$
The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
$$
mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
$$
hence
$$
mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
$$
The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 8 at 19:57
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
add a comment |
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