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

1

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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: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ doesn't make sense, so then $B$ wouldn't be independent from $A$?

Thank you

probability-theory independence

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asked 2 hours ago

MashpaMashpa

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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: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ 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.
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$endgroup$

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: $$P(A|B)=P(A)label1tag1$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label2tag2$$ but in $ref1$ if $P(A)=0$ then $ref2$ 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

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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
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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 $P(B) > 0$.

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answered 2 hours ago

bitesizebobitesizebo

1,77828

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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.

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answered 2 hours ago

amdamd

32k21053

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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