If two events co-occur, then is their probability equal?
$begingroup$
If two events co-occur, then is their probability equal?
I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.
In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.
But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?
$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .
Why aren’t these two probabilities always equal?
real-analysis probability probability-theory
asked Jan 9 at 6:50
jajajaja
8711
$endgroup$
add a comment |
$begingroup$
If two events co-occur, then is their probability equal?
I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.
In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.
But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?
$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .
Why aren’t these two probabilities always equal?
real-analysis probability probability-theory
asked Jan 9 at 6:50
jajajaja
8711
$endgroup$
1
$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08
1
$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20
1
$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21
$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25
add a comment |
$begingroup$
If two events co-occur, then is their probability equal?
I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.
In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.
But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?
$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .
Why aren’t these two probabilities always equal?
real-analysis probability probability-theory
asked Jan 9 at 6:50
jajajaja
8711
$endgroup$
If two events co-occur, then is their probability equal?
I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.
In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.
But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?
$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .
Why aren’t these two probabilities always equal?
real-analysis probability probability-theory
real-analysis probability probability-theory
asked Jan 9 at 6:50
jajajaja
8711
asked Jan 9 at 6:50
jajajaja
8711
edited Jan 9 at 7:23
jaja
asked Jan 9 at 6:50
jajajaja
8711
asked Jan 9 at 6:50
jajajaja
8711
asked Jan 9 at 6:50
jajajaja
8711
8711
1
$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08
1
$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20
1
$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21
$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25
add a comment |
1
$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08
1
$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20
1
$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21
$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25
1
1
$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08
$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08
1
1
$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20
$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20
1
1
$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21
$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21
$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25
$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.
Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.
That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.
answered Jan 9 at 7:06
ShashiShashi
7,3621628
$endgroup$
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
2
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
1
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
|
show 3 more comments
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%2f3067164%2fif-two-events-co-occur-then-is-their-probability-equal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.
Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.
That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.
answered Jan 9 at 7:06
ShashiShashi
7,3621628
$endgroup$
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
2
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
1
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
|
show 3 more comments
$begingroup$
What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.
Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.
That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.
answered Jan 9 at 7:06
ShashiShashi
7,3621628
$endgroup$
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
2
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
1
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
|
show 3 more comments
$begingroup$
What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.
Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.
That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.
answered Jan 9 at 7:06
ShashiShashi
7,3621628
$endgroup$
What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.
Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.
That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.
answered Jan 9 at 7:06
ShashiShashi
7,3621628
edited Jan 9 at 7:45
answered Jan 9 at 7:06
ShashiShashi
7,3621628
answered Jan 9 at 7:06
ShashiShashi
7,3621628
answered Jan 9 at 7:06
ShashiShashi
7,3621628
7,3621628
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
2
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
1
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
|
show 3 more comments
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
2
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
1
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31
2
2
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03
1
1
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00
|
show 3 more comments
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%2f3067164%2fif-two-events-co-occur-then-is-their-probability-equal%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
$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08
1
$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20
1
$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21
$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25