Is it possible for an event A to be independent from event B, but not the other way around?
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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: $$P(A|B)=P(A)label{1}tag{1}$$ 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
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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: $$P(A|B)=P(A)label{1}tag{1}$$ 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
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.
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add a comment |
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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: $$P(A|B)=P(A)label{1}tag{1}$$ 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
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.
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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: $$P(A|B)=P(A)label{1}tag{1}$$ 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
probability-theory independence
asked 2 hours ago
MashpaMashpa
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asked 2 hours ago
MashpaMashpa
273
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asked 2 hours ago
MashpaMashpa
273
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asked 2 hours ago
MashpaMashpa
273
asked 2 hours ago
MashpaMashpa
273
273
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3 Answers
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$P(A mid B) = P(A)$ 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 $P(B) > 0$.
answered 2 hours ago
bitesizebobitesizebo
1,78828
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$P(A|B)=P(A)$ 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 $A$
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
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$P(Bmid A)$ 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.
answered 2 hours ago
amdamd
32k21053
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$P(A mid B) = P(A)$ 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 $P(B) > 0$.
answered 2 hours ago
bitesizebobitesizebo
1,78828
$endgroup$
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 $P(B) > 0$.
answered 2 hours ago
bitesizebobitesizebo
1,78828
$endgroup$
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 $P(B) > 0$.
answered 2 hours ago
bitesizebobitesizebo
1,78828
$endgroup$
$P(A mid B) = P(A)$ 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 $P(B) > 0$.
answered 2 hours ago
bitesizebobitesizebo
1,78828
answered 2 hours ago
bitesizebobitesizebo
1,78828
answered 2 hours ago
bitesizebobitesizebo
1,78828
answered 2 hours ago
bitesizebobitesizebo
1,78828
1,78828
add a comment |
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 $A$
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 $A$
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 $A$
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$P(A|B)=P(A)$ 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 $A$
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
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
76.4k53370
add a comment |
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.
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.
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.
answered 2 hours ago
amdamd
32k21053
$endgroup$
$P(Bmid A)$ 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.
answered 2 hours ago
amdamd
32k21053
answered 2 hours ago
amdamd
32k21053
answered 2 hours ago
amdamd
32k21053
answered 2 hours ago
amdamd
32k21053
32k21053
add a comment |
add a comment |
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