Is it possible for an event A to be independent from event B, but not the other way around? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Zero probability and impossibilityExchangeable Random Variable but not independent?Probability of being away from mean for independent random variablesAn example of two random variables that are mean independent but not independentCalculating probability when order matters only sometimesIf X is independent to Y and Z, does it imply that X is independent to YZ ?Representing pairwise-independent but not independent occurrences with venn diagramPairwise independent events but not mutually independentExamples of situation in which two events are independent but one event can be predicted perfectly once we know if the other happened or not.Suppose $A $ and $B$ are independent events. For an event $C $ such that $P(C) > 0$ , prove that the event of $A$ given $C $
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Is it possible for an event A to be independent from event B, but not the other way around?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Zero probability and impossibilityExchangeable Random Variable but not independent?Probability of being away from mean for independent random variablesAn example of two random variables that are mean independent but not independentCalculating probability when order matters only sometimesIf X is independent to Y and Z, does it imply that X is independent to YZ ?Representing pairwise-independent but not independent occurrences with venn diagramPairwise independent events but not mutually independentExamples of situation in which two events are independent but one event can be predicted perfectly once we know if the other happened or not.Suppose $A $ and $B$ are independent events. For an event $C $ such that $P(C) > 0$ , prove that the event of $A$ given $C $
$begingroup$
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
asked 2 hours ago
MashpaMashpa
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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
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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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
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.
$endgroup$
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
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.
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
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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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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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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
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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
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
$P(A mid B) = P(A)$ 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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