Gsyfuy

How do living politicians protect their readily obtainable signatures from misuse?

License to disallow distribution in closed source software, but allow exceptions made by owner?

Co-worker has annoying ringtone

Why datecode is SO IMPORTANT to chip manufacturers?

Why not send Voyager 3 and 4 following up the paths taken by Voyager 1 and 2 to re-transmit signals of later as they fly away from Earth?

Mounting TV on a weird wall that has some material between the drywall and stud

How much damage would a cupful of neutron star matter do to the Earth?

Can you force honesty by using the Speak with Dead and Zone of Truth spells together?

Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted?

Does the Mueller report show a conspiracy between Russia and the Trump Campaign?

Tannaka duality for semisimple groups

A `coordinate` command ignored

How can I save and copy a screenhot at the same time?

Trying to understand entropy as a novice in thermodynamics

How to write capital alpha?

Special flights

Tips to organize LaTeX presentations for a semester

White walkers, cemeteries and wights

Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?

In musical terms, what properties are varied by the human voice to produce different words / syllables?

Flight departed from the gate 5 min before scheduled departure time. Refund options

How can a team of shapeshifters communicate?

What does Turing mean by this statement?

Is there public access to the Meteor Crater in Arizona?

Is it possible for an event A to be independent from event B, but not the other way around?

1

$begingroup$

I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?

Thank you

probability-theory independence

share|cite|improve this question

asked 2 hours ago

MashpaMashpa

273

New contributor

Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?

Thank you

probability-theory independence

share|cite|improve this question

asked 2 hours ago

MashpaMashpa

273

New contributor

Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?

Thank you

probability-theory independence

share|cite|improve this question

asked 2 hours ago

MashpaMashpa

273

New contributor

Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 2 hours ago

MashpaMashpa

273

New contributor

Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 2 hours ago

MashpaMashpa

273

New contributor

Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

MashpaMashpa

273

New contributor

Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

MashpaMashpa

273

3 Answers
3

active

oldest

votes

2

$begingroup$

$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.

share|cite|improve this answer

answered 2 hours ago

bitesizebobitesizebo

1,77828

$endgroup$

add a comment | 

2

$begingroup$

$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$

share|cite|improve this answer

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

2

$begingroup$

$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.

share|cite|improve this answer

answered 2 hours ago

amdamd

32k21053

$endgroup$

add a comment | 

Your Answer

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
});


}
});

Mashpa is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3195328%2fis-it-possible-for-an-event-a-to-be-independent-from-event-b-but-not-the-other%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.

share|cite|improve this answer

answered 2 hours ago

bitesizebobitesizebo

1,77828

$endgroup$

add a comment | 

2

$begingroup$

$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.

share|cite|improve this answer

answered 2 hours ago

bitesizebobitesizebo

1,77828

$endgroup$

add a comment | 

$begingroup$

$P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.

share|cite|improve this answer

answered 2 hours ago

bitesizebobitesizebo

1,77828

$endgroup$

share|cite|improve this answer

answered 2 hours ago

bitesizebobitesizebo

1,77828

answered 2 hours ago

bitesizebobitesizebo

1,77828

answered 2 hours ago

bitesizebobitesizebo

1,77828

2

$begingroup$

$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$

share|cite|improve this answer

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

2

$begingroup$

$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$

share|cite|improve this answer

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

$begingroup$

$P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$

share|cite|improve this answer

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

2

$begingroup$

$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.

share|cite|improve this answer

answered 2 hours ago

amdamd

32k21053

$endgroup$

add a comment | 

2

$begingroup$

$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.

share|cite|improve this answer

answered 2 hours ago

amdamd

32k21053

$endgroup$

add a comment | 

$begingroup$

$P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.

share|cite|improve this answer

answered 2 hours ago

amdamd

32k21053

$endgroup$

share|cite|improve this answer

answered 2 hours ago

amdamd

32k21053

answered 2 hours ago

amdamd

32k21053

answered 2 hours ago

amdamd

32k21053