If A is dependent upon B, is B necessarily dependent upon A?
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I read that independence of two events has a symmetric relationship. Does this relationship hold for dependent events as well?
If so, what would be the intuition?
Thanks
probability probability-theory
asked Nov 22 at 13:52
Kohler Fryer
1031
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up vote
0
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favorite
I read that independence of two events has a symmetric relationship. Does this relationship hold for dependent events as well?
If so, what would be the intuition?
Thanks
probability probability-theory
asked Nov 22 at 13:52
Kohler Fryer
1031
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I read that independence of two events has a symmetric relationship. Does this relationship hold for dependent events as well?
If so, what would be the intuition?
Thanks
probability probability-theory
asked Nov 22 at 13:52
Kohler Fryer
1031
I read that independence of two events has a symmetric relationship. Does this relationship hold for dependent events as well?
If so, what would be the intuition?
Thanks
probability probability-theory
probability probability-theory
asked Nov 22 at 13:52
Kohler Fryer
1031
asked Nov 22 at 13:52
Kohler Fryer
1031
asked Nov 22 at 13:52
Kohler Fryer
1031
asked Nov 22 at 13:52
Kohler Fryer
1031
asked Nov 22 at 13:52
Kohler Fryer
1031
1031
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Yes, two events are dependent if and only if they are not independent.
Since, if $mu$ is a probability measure, then events A and B are independent if and only if
begin{equation}
mu(Acap B)=mu(A)cdotmu(B)
end{equation}
and dependent otherwise. Since multiplication is commutative you easily get exactly what you're looking for.
An easy example is a double coin toss: intuitively the result of tossing the first coin should have no effect on the second and vice-versa.
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
add a comment |
up vote
0
down vote
The language $A$ dependent on $B$ is a bit vague, since it may involve something wherein there is a causal dependency that $A$ happens because $B$ does. Stochastic independence avoids that: Events $A,B$ are independent iff $P(A land B)=P(A)P(B).$ So $A,B$ are dependent iff $P(A land B)neq P(A)P(B),$ and the latter is symmetric in $A,B.$ I don't know what rules one uses in case "A depends on B" means something other than stochastic dependence.
answered Nov 22 at 14:02
coffeemath
1,9801413
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, two events are dependent if and only if they are not independent.
Since, if $mu$ is a probability measure, then events A and B are independent if and only if
begin{equation}
mu(Acap B)=mu(A)cdotmu(B)
end{equation}
and dependent otherwise. Since multiplication is commutative you easily get exactly what you're looking for.
An easy example is a double coin toss: intuitively the result of tossing the first coin should have no effect on the second and vice-versa.
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
add a comment |
up vote
1
down vote
accepted
Yes, two events are dependent if and only if they are not independent.
Since, if $mu$ is a probability measure, then events A and B are independent if and only if
begin{equation}
mu(Acap B)=mu(A)cdotmu(B)
end{equation}
and dependent otherwise. Since multiplication is commutative you easily get exactly what you're looking for.
An easy example is a double coin toss: intuitively the result of tossing the first coin should have no effect on the second and vice-versa.
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, two events are dependent if and only if they are not independent.
Since, if $mu$ is a probability measure, then events A and B are independent if and only if
begin{equation}
mu(Acap B)=mu(A)cdotmu(B)
end{equation}
and dependent otherwise. Since multiplication is commutative you easily get exactly what you're looking for.
An easy example is a double coin toss: intuitively the result of tossing the first coin should have no effect on the second and vice-versa.
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
Yes, two events are dependent if and only if they are not independent.
Since, if $mu$ is a probability measure, then events A and B are independent if and only if
begin{equation}
mu(Acap B)=mu(A)cdotmu(B)
end{equation}
and dependent otherwise. Since multiplication is commutative you easily get exactly what you're looking for.
An easy example is a double coin toss: intuitively the result of tossing the first coin should have no effect on the second and vice-versa.
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
answered Nov 22 at 14:04
Jean-Pierre de Villiers
435
435
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
add a comment |
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
Thanks, any recommendation on becoming more familiar with this stuff? I'm currently taking a class at a University that is covering this content a bit too fast for my liking.
– Kohler Fryer
Nov 22 at 14:46
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
I'm a big fan of the MIT open courseware and would thus recommend taking a peek at that and the books they used. This also depends on the flavour of probility you are doing. For a more statistical/applied approach (or to find the intuiton/motivation behind theory) you can check out link. For a more mathematical (measure-theoretic) approach have a look at say link.
– Jean-Pierre de Villiers
Nov 22 at 15:05
add a comment |
up vote
0
down vote
The language $A$ dependent on $B$ is a bit vague, since it may involve something wherein there is a causal dependency that $A$ happens because $B$ does. Stochastic independence avoids that: Events $A,B$ are independent iff $P(A land B)=P(A)P(B).$ So $A,B$ are dependent iff $P(A land B)neq P(A)P(B),$ and the latter is symmetric in $A,B.$ I don't know what rules one uses in case "A depends on B" means something other than stochastic dependence.
answered Nov 22 at 14:02
coffeemath
1,9801413
add a comment |
up vote
0
down vote
The language $A$ dependent on $B$ is a bit vague, since it may involve something wherein there is a causal dependency that $A$ happens because $B$ does. Stochastic independence avoids that: Events $A,B$ are independent iff $P(A land B)=P(A)P(B).$ So $A,B$ are dependent iff $P(A land B)neq P(A)P(B),$ and the latter is symmetric in $A,B.$ I don't know what rules one uses in case "A depends on B" means something other than stochastic dependence.
answered Nov 22 at 14:02
coffeemath
1,9801413
add a comment |
up vote
0
down vote
up vote
0
down vote
The language $A$ dependent on $B$ is a bit vague, since it may involve something wherein there is a causal dependency that $A$ happens because $B$ does. Stochastic independence avoids that: Events $A,B$ are independent iff $P(A land B)=P(A)P(B).$ So $A,B$ are dependent iff $P(A land B)neq P(A)P(B),$ and the latter is symmetric in $A,B.$ I don't know what rules one uses in case "A depends on B" means something other than stochastic dependence.
answered Nov 22 at 14:02
coffeemath
1,9801413
The language $A$ dependent on $B$ is a bit vague, since it may involve something wherein there is a causal dependency that $A$ happens because $B$ does. Stochastic independence avoids that: Events $A,B$ are independent iff $P(A land B)=P(A)P(B).$ So $A,B$ are dependent iff $P(A land B)neq P(A)P(B),$ and the latter is symmetric in $A,B.$ I don't know what rules one uses in case "A depends on B" means something other than stochastic dependence.
answered Nov 22 at 14:02
coffeemath
1,9801413
edited Nov 22 at 14:07
answered Nov 22 at 14:02
coffeemath
1,9801413
answered Nov 22 at 14:02
coffeemath
1,9801413
answered Nov 22 at 14:02
coffeemath
1,9801413
1,9801413
add a comment |
add a comment |
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