Proof that $P(A_{n} cap B)rightarrow P(B)$ given $A_{n} nearrow A$ and $P(A)=1$
$begingroup$
Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ 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.
probability measure-theory
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
asked Jan 9 at 1:10
LauraLaura
2408
$endgroup$
add a comment |
$begingroup$
Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ 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.
probability measure-theory
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
asked Jan 9 at 1:10
LauraLaura
2408
$endgroup$
1
$begingroup$
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 $A_{1},A_{2},ldots,A_{n}$ 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.
probability measure-theory
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
asked Jan 9 at 1:10
LauraLaura
2408
$endgroup$
Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ 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.
probability measure-theory
probability measure-theory
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
asked Jan 9 at 1:10
LauraLaura
2408
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
asked Jan 9 at 1:10
LauraLaura
2408
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
edited Jan 18 at 10:42
Lee David Chung Lin
4,26031141
4,26031141
asked Jan 9 at 1:10
LauraLaura
2408
asked Jan 9 at 1:10
LauraLaura
2408
asked Jan 9 at 1:10
LauraLaura
2408
2408
1
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
add a comment |
1
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54
1
1
$begingroup$
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
active
oldest
votes
$begingroup$
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 Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
$endgroup$
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
40.9k23380
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066945%2fproof-that-pa-n-cap-b-rightarrow-pb-given-a-n-nearrow-a-and-pa%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
$endgroup$
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 Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
$endgroup$
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 Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
$endgroup$
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 Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
answered Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
answered Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
answered Jan 15 at 17:26
OldGodzillaOldGodzilla
58227
58227
add a comment |
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
40.9k23380
$endgroup$
$begingroup$
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
40.9k23380
$endgroup$
$begingroup$
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
40.9k23380
$endgroup$
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
40.9k23380
answered Jan 9 at 1:17
angryavianangryavian
40.9k23380
answered Jan 9 at 1:17
angryavianangryavian
40.9k23380
answered Jan 9 at 1:17
angryavianangryavian
40.9k23380
40.9k23380
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066945%2fproof-that-pa-n-cap-b-rightarrow-pb-given-a-n-nearrow-a-and-pa%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54