Mixing
Independence and conditional independence between random variables
$begingroup$
For a family of random variables, I was wondering if independence and conditional independence under any condition among them imply each other?
If not, can these two concepts imply one another under some special cases?
probability-theory
edited Jan 18 at 20:35
nbro
2,41663174
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
$endgroup$
add a comment |
$begingroup$
For a family of random variables, I was wondering if independence and conditional independence under any condition among them imply each other?
If not, can these two concepts imply one another under some special cases?
probability-theory
edited Jan 18 at 20:35
nbro
2,41663174
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
$endgroup$
$begingroup$
The question as you have stated it is vague. Is there some problem or result that is motivating this question?
$endgroup$
– svenkatr
Feb 16 '11 at 22:31
$begingroup$
@svenkatr: Is it correct that independent random variables may not be independent conditional on some conditions? Independence conditional on some conditions may not imply independence. So I was wondering: (1) if Independence conditional on any condition and independence may imply each other; (2) what are some useful cases where one can imply the other?
$endgroup$
– Tim
Feb 16 '11 at 23:25
$begingroup$
What is independence under any condition? If you mean independence under every possible conditioning, the condition is rarely fulfilled... For example, random variables $x$ and $y$ are not independent conditionally on $[x<y]$ if you exclude some very degenerate cases.
$endgroup$
– Did
Feb 17 '11 at 11:34
add a comment |
$begingroup$
For a family of random variables, I was wondering if independence and conditional independence under any condition among them imply each other?
If not, can these two concepts imply one another under some special cases?
probability-theory
edited Jan 18 at 20:35
nbro
2,41663174
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
$endgroup$
For a family of random variables, I was wondering if independence and conditional independence under any condition among them imply each other?
If not, can these two concepts imply one another under some special cases?
probability-theory
probability-theory
edited Jan 18 at 20:35
nbro
2,41663174
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
edited Jan 18 at 20:35
nbro
2,41663174
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
edited Jan 18 at 20:35
nbro
2,41663174
edited Jan 18 at 20:35
nbro
2,41663174
edited Jan 18 at 20:35
nbro
2,41663174
2,41663174
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
asked Feb 16 '11 at 20:57
TimTim
16.5k20122328
16.5k20122328
$begingroup$
The question as you have stated it is vague. Is there some problem or result that is motivating this question?
$endgroup$
– svenkatr
Feb 16 '11 at 22:31
$begingroup$
@svenkatr: Is it correct that independent random variables may not be independent conditional on some conditions? Independence conditional on some conditions may not imply independence. So I was wondering: (1) if Independence conditional on any condition and independence may imply each other; (2) what are some useful cases where one can imply the other?
$endgroup$
– Tim
Feb 16 '11 at 23:25
$begingroup$
What is independence under any condition? If you mean independence under every possible conditioning, the condition is rarely fulfilled... For example, random variables $x$ and $y$ are not independent conditionally on $[x<y]$ if you exclude some very degenerate cases.
$endgroup$
– Did
Feb 17 '11 at 11:34
add a comment |
$begingroup$
The question as you have stated it is vague. Is there some problem or result that is motivating this question?
$endgroup$
– svenkatr
Feb 16 '11 at 22:31
$begingroup$
@svenkatr: Is it correct that independent random variables may not be independent conditional on some conditions? Independence conditional on some conditions may not imply independence. So I was wondering: (1) if Independence conditional on any condition and independence may imply each other; (2) what are some useful cases where one can imply the other?
$endgroup$
– Tim
Feb 16 '11 at 23:25
$begingroup$
What is independence under any condition? If you mean independence under every possible conditioning, the condition is rarely fulfilled... For example, random variables $x$ and $y$ are not independent conditionally on $[x<y]$ if you exclude some very degenerate cases.
$endgroup$
– Did
Feb 17 '11 at 11:34
$begingroup$
The question as you have stated it is vague. Is there some problem or result that is motivating this question?
$endgroup$
– svenkatr
Feb 16 '11 at 22:31
$begingroup$
The question as you have stated it is vague. Is there some problem or result that is motivating this question?
$endgroup$
– svenkatr
Feb 16 '11 at 22:31
$begingroup$
@svenkatr: Is it correct that independent random variables may not be independent conditional on some conditions? Independence conditional on some conditions may not imply independence. So I was wondering: (1) if Independence conditional on any condition and independence may imply each other; (2) what are some useful cases where one can imply the other?
$endgroup$
– Tim
Feb 16 '11 at 23:25
$begingroup$
@svenkatr: Is it correct that independent random variables may not be independent conditional on some conditions? Independence conditional on some conditions may not imply independence. So I was wondering: (1) if Independence conditional on any condition and independence may imply each other; (2) what are some useful cases where one can imply the other?
$endgroup$
– Tim
Feb 16 '11 at 23:25
$begingroup$
What is independence under any condition? If you mean independence under every possible conditioning, the condition is rarely fulfilled... For example, random variables $x$ and $y$ are not independent conditionally on $[x<y]$ if you exclude some very degenerate cases.
$endgroup$
– Did
Feb 17 '11 at 11:34
$begingroup$
What is independence under any condition? If you mean independence under every possible conditioning, the condition is rarely fulfilled... For example, random variables $x$ and $y$ are not independent conditionally on $[x<y]$ if you exclude some very degenerate cases.
$endgroup$
– Did
Feb 17 '11 at 11:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Independence does not imply conditional independence: for instance, independent random variables are rarely independent conditionally on their sum or on their maximum.
Conditional independence does not imply independence: for instance, conditionally independent random variables uniform on $(0,u)$ where $u$ is uniform on $(0,1)$ are not independent.
answered Feb 17 '11 at 11:13
DidDid
248k23224463
$endgroup$
add a comment |
$begingroup$
I like to interpret these two concepts as follows:
Events $A,B$ are independent if knowing that $A$ happened would not tell you anything about whether $B$ happened (or vice versa). For instance, suppose you were considering betting some money on event $B$. Some insider comes along and offers to pass you information (for a fee) about whether or not $A$ happened. Saying $A,B$ are independent is to say that this inside information would be utterly irrelevant, and you wouldn't pay any amount of money for it.
Events $A,B$ are conditionally independent given a third event $C$ means the following: Suppose you already know that $C$ has happened. Then knowing whether $A$ happened would not convey any further information about whether $B$ happened - any relevant information that might be conveyed by $A$ is already known to you, because you know that $C$ happened.
To see independence does not imply conditional independence, one of my favorite simple counterexamples works. Flip two fair coins. Let $A$ be the event that the first coin is heads, $B$ the event that the second coin is heads, $C$ the event that the two coins are the same (both heads or both tails). Clearly $A$ and $B$ are independent, but they are not conditionally independent given $C$ - if you know that $C$ has happened, then knowing $A$ tells you a lot about $B$ (indeed, it would tell you that $B$ is guaranteed). If you want an example with random variables, consider the indicators $1_A, 1_B, 1_C$.
Of interest here is that $A,B,C$ are pairwise independent but not mutually independent (since any two determine the third).
A nice counterexample in the other direction is the following. We have a bag containing two identical-looking coins. One of them (coin #1) is biased so that it comes up heads 99% of the time, and coin #2 comes up tails 99% of the time. We will draw a coin from the bag at random, and then flip it twice. Let $A$ be the event that the first flip is heads, $B$ the event that the second flip is heads. These are clearly not independent: you can do a calculation if you like, but the idea is that if the first flip was heads, it is strong evidence that you drew coin #1, and therefore the second flip is far more likely to be heads.
But let $C$ be the event that coin #1 was drawn (where $P(C)=1/2$). Now $A$ and $B$ are conditionally independent given $C$: if you know that $C$ happened, then you are just doing an experiment where you take a 99%-heads coin and flip it twice. Whether the first flip is heads or tails, the coin has no memory so the probability of the second flip being heads is still 99%. So if you already know which coin you have, then knowing how the first flip came out is of no further help in predicting the second flip.
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
$endgroup$
3
$begingroup$
This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
$endgroup$
– dev_nut
Dec 16 '14 at 19:21
5
$begingroup$
@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
$endgroup$
– Nate Eldredge
Dec 16 '14 at 19:33
1
$begingroup$
Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
$endgroup$
– dev_nut
Dec 16 '14 at 20:32
$begingroup$
@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
$endgroup$
– Sнаđошƒаӽ
Oct 27 '15 at 16:53
1
$begingroup$
I've added another example.
$endgroup$
– Nate Eldredge
Mar 13 '18 at 21:19
|
show 1 more comment
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2 Answers
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$begingroup$
Independence does not imply conditional independence: for instance, independent random variables are rarely independent conditionally on their sum or on their maximum.
Conditional independence does not imply independence: for instance, conditionally independent random variables uniform on $(0,u)$ where $u$ is uniform on $(0,1)$ are not independent.
answered Feb 17 '11 at 11:13
DidDid
248k23224463
$endgroup$
add a comment |
$begingroup$
Independence does not imply conditional independence: for instance, independent random variables are rarely independent conditionally on their sum or on their maximum.
Conditional independence does not imply independence: for instance, conditionally independent random variables uniform on $(0,u)$ where $u$ is uniform on $(0,1)$ are not independent.
answered Feb 17 '11 at 11:13
DidDid
248k23224463
$endgroup$
add a comment |
$begingroup$
Independence does not imply conditional independence: for instance, independent random variables are rarely independent conditionally on their sum or on their maximum.
Conditional independence does not imply independence: for instance, conditionally independent random variables uniform on $(0,u)$ where $u$ is uniform on $(0,1)$ are not independent.
answered Feb 17 '11 at 11:13
DidDid
248k23224463
$endgroup$
Independence does not imply conditional independence: for instance, independent random variables are rarely independent conditionally on their sum or on their maximum.
Conditional independence does not imply independence: for instance, conditionally independent random variables uniform on $(0,u)$ where $u$ is uniform on $(0,1)$ are not independent.
answered Feb 17 '11 at 11:13
DidDid
248k23224463
edited Feb 17 '11 at 11:24
answered Feb 17 '11 at 11:13
DidDid
248k23224463
answered Feb 17 '11 at 11:13
DidDid
248k23224463
answered Feb 17 '11 at 11:13
DidDid
248k23224463
248k23224463
add a comment |
add a comment |
$begingroup$
I like to interpret these two concepts as follows:
Events $A,B$ are independent if knowing that $A$ happened would not tell you anything about whether $B$ happened (or vice versa). For instance, suppose you were considering betting some money on event $B$. Some insider comes along and offers to pass you information (for a fee) about whether or not $A$ happened. Saying $A,B$ are independent is to say that this inside information would be utterly irrelevant, and you wouldn't pay any amount of money for it.
Events $A,B$ are conditionally independent given a third event $C$ means the following: Suppose you already know that $C$ has happened. Then knowing whether $A$ happened would not convey any further information about whether $B$ happened - any relevant information that might be conveyed by $A$ is already known to you, because you know that $C$ happened.
To see independence does not imply conditional independence, one of my favorite simple counterexamples works. Flip two fair coins. Let $A$ be the event that the first coin is heads, $B$ the event that the second coin is heads, $C$ the event that the two coins are the same (both heads or both tails). Clearly $A$ and $B$ are independent, but they are not conditionally independent given $C$ - if you know that $C$ has happened, then knowing $A$ tells you a lot about $B$ (indeed, it would tell you that $B$ is guaranteed). If you want an example with random variables, consider the indicators $1_A, 1_B, 1_C$.
Of interest here is that $A,B,C$ are pairwise independent but not mutually independent (since any two determine the third).
A nice counterexample in the other direction is the following. We have a bag containing two identical-looking coins. One of them (coin #1) is biased so that it comes up heads 99% of the time, and coin #2 comes up tails 99% of the time. We will draw a coin from the bag at random, and then flip it twice. Let $A$ be the event that the first flip is heads, $B$ the event that the second flip is heads. These are clearly not independent: you can do a calculation if you like, but the idea is that if the first flip was heads, it is strong evidence that you drew coin #1, and therefore the second flip is far more likely to be heads.
But let $C$ be the event that coin #1 was drawn (where $P(C)=1/2$). Now $A$ and $B$ are conditionally independent given $C$: if you know that $C$ happened, then you are just doing an experiment where you take a 99%-heads coin and flip it twice. Whether the first flip is heads or tails, the coin has no memory so the probability of the second flip being heads is still 99%. So if you already know which coin you have, then knowing how the first flip came out is of no further help in predicting the second flip.
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
$endgroup$
3
$begingroup$
This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
$endgroup$
– dev_nut
Dec 16 '14 at 19:21
5
$begingroup$
@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
$endgroup$
– Nate Eldredge
Dec 16 '14 at 19:33
1
$begingroup$
Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
$endgroup$
– dev_nut
Dec 16 '14 at 20:32
$begingroup$
@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
$endgroup$
– Sнаđошƒаӽ
Oct 27 '15 at 16:53
1
$begingroup$
I've added another example.
$endgroup$
– Nate Eldredge
Mar 13 '18 at 21:19
|
show 1 more comment
$begingroup$
I like to interpret these two concepts as follows:
Events $A,B$ are independent if knowing that $A$ happened would not tell you anything about whether $B$ happened (or vice versa). For instance, suppose you were considering betting some money on event $B$. Some insider comes along and offers to pass you information (for a fee) about whether or not $A$ happened. Saying $A,B$ are independent is to say that this inside information would be utterly irrelevant, and you wouldn't pay any amount of money for it.
Events $A,B$ are conditionally independent given a third event $C$ means the following: Suppose you already know that $C$ has happened. Then knowing whether $A$ happened would not convey any further information about whether $B$ happened - any relevant information that might be conveyed by $A$ is already known to you, because you know that $C$ happened.
To see independence does not imply conditional independence, one of my favorite simple counterexamples works. Flip two fair coins. Let $A$ be the event that the first coin is heads, $B$ the event that the second coin is heads, $C$ the event that the two coins are the same (both heads or both tails). Clearly $A$ and $B$ are independent, but they are not conditionally independent given $C$ - if you know that $C$ has happened, then knowing $A$ tells you a lot about $B$ (indeed, it would tell you that $B$ is guaranteed). If you want an example with random variables, consider the indicators $1_A, 1_B, 1_C$.
Of interest here is that $A,B,C$ are pairwise independent but not mutually independent (since any two determine the third).
A nice counterexample in the other direction is the following. We have a bag containing two identical-looking coins. One of them (coin #1) is biased so that it comes up heads 99% of the time, and coin #2 comes up tails 99% of the time. We will draw a coin from the bag at random, and then flip it twice. Let $A$ be the event that the first flip is heads, $B$ the event that the second flip is heads. These are clearly not independent: you can do a calculation if you like, but the idea is that if the first flip was heads, it is strong evidence that you drew coin #1, and therefore the second flip is far more likely to be heads.
But let $C$ be the event that coin #1 was drawn (where $P(C)=1/2$). Now $A$ and $B$ are conditionally independent given $C$: if you know that $C$ happened, then you are just doing an experiment where you take a 99%-heads coin and flip it twice. Whether the first flip is heads or tails, the coin has no memory so the probability of the second flip being heads is still 99%. So if you already know which coin you have, then knowing how the first flip came out is of no further help in predicting the second flip.
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
$endgroup$
3
$begingroup$
This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
$endgroup$
– dev_nut
Dec 16 '14 at 19:21
5
$begingroup$
@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
$endgroup$
– Nate Eldredge
Dec 16 '14 at 19:33
1
$begingroup$
Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
$endgroup$
– dev_nut
Dec 16 '14 at 20:32
$begingroup$
@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
$endgroup$
– Sнаđошƒаӽ
Oct 27 '15 at 16:53
1
$begingroup$
I've added another example.
$endgroup$
– Nate Eldredge
Mar 13 '18 at 21:19
|
show 1 more comment
$begingroup$
I like to interpret these two concepts as follows:
Events $A,B$ are independent if knowing that $A$ happened would not tell you anything about whether $B$ happened (or vice versa). For instance, suppose you were considering betting some money on event $B$. Some insider comes along and offers to pass you information (for a fee) about whether or not $A$ happened. Saying $A,B$ are independent is to say that this inside information would be utterly irrelevant, and you wouldn't pay any amount of money for it.
Events $A,B$ are conditionally independent given a third event $C$ means the following: Suppose you already know that $C$ has happened. Then knowing whether $A$ happened would not convey any further information about whether $B$ happened - any relevant information that might be conveyed by $A$ is already known to you, because you know that $C$ happened.
To see independence does not imply conditional independence, one of my favorite simple counterexamples works. Flip two fair coins. Let $A$ be the event that the first coin is heads, $B$ the event that the second coin is heads, $C$ the event that the two coins are the same (both heads or both tails). Clearly $A$ and $B$ are independent, but they are not conditionally independent given $C$ - if you know that $C$ has happened, then knowing $A$ tells you a lot about $B$ (indeed, it would tell you that $B$ is guaranteed). If you want an example with random variables, consider the indicators $1_A, 1_B, 1_C$.
Of interest here is that $A,B,C$ are pairwise independent but not mutually independent (since any two determine the third).
A nice counterexample in the other direction is the following. We have a bag containing two identical-looking coins. One of them (coin #1) is biased so that it comes up heads 99% of the time, and coin #2 comes up tails 99% of the time. We will draw a coin from the bag at random, and then flip it twice. Let $A$ be the event that the first flip is heads, $B$ the event that the second flip is heads. These are clearly not independent: you can do a calculation if you like, but the idea is that if the first flip was heads, it is strong evidence that you drew coin #1, and therefore the second flip is far more likely to be heads.
But let $C$ be the event that coin #1 was drawn (where $P(C)=1/2$). Now $A$ and $B$ are conditionally independent given $C$: if you know that $C$ happened, then you are just doing an experiment where you take a 99%-heads coin and flip it twice. Whether the first flip is heads or tails, the coin has no memory so the probability of the second flip being heads is still 99%. So if you already know which coin you have, then knowing how the first flip came out is of no further help in predicting the second flip.
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
$endgroup$
I like to interpret these two concepts as follows:
Events $A,B$ are independent if knowing that $A$ happened would not tell you anything about whether $B$ happened (or vice versa). For instance, suppose you were considering betting some money on event $B$. Some insider comes along and offers to pass you information (for a fee) about whether or not $A$ happened. Saying $A,B$ are independent is to say that this inside information would be utterly irrelevant, and you wouldn't pay any amount of money for it.
Events $A,B$ are conditionally independent given a third event $C$ means the following: Suppose you already know that $C$ has happened. Then knowing whether $A$ happened would not convey any further information about whether $B$ happened - any relevant information that might be conveyed by $A$ is already known to you, because you know that $C$ happened.
To see independence does not imply conditional independence, one of my favorite simple counterexamples works. Flip two fair coins. Let $A$ be the event that the first coin is heads, $B$ the event that the second coin is heads, $C$ the event that the two coins are the same (both heads or both tails). Clearly $A$ and $B$ are independent, but they are not conditionally independent given $C$ - if you know that $C$ has happened, then knowing $A$ tells you a lot about $B$ (indeed, it would tell you that $B$ is guaranteed). If you want an example with random variables, consider the indicators $1_A, 1_B, 1_C$.
Of interest here is that $A,B,C$ are pairwise independent but not mutually independent (since any two determine the third).
A nice counterexample in the other direction is the following. We have a bag containing two identical-looking coins. One of them (coin #1) is biased so that it comes up heads 99% of the time, and coin #2 comes up tails 99% of the time. We will draw a coin from the bag at random, and then flip it twice. Let $A$ be the event that the first flip is heads, $B$ the event that the second flip is heads. These are clearly not independent: you can do a calculation if you like, but the idea is that if the first flip was heads, it is strong evidence that you drew coin #1, and therefore the second flip is far more likely to be heads.
But let $C$ be the event that coin #1 was drawn (where $P(C)=1/2$). Now $A$ and $B$ are conditionally independent given $C$: if you know that $C$ happened, then you are just doing an experiment where you take a 99%-heads coin and flip it twice. Whether the first flip is heads or tails, the coin has no memory so the probability of the second flip being heads is still 99%. So if you already know which coin you have, then knowing how the first flip came out is of no further help in predicting the second flip.
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
edited Mar 13 '18 at 21:19
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
answered Feb 17 '11 at 13:44
Nate EldredgeNate Eldredge
63.9k682174
63.9k682174
3
$begingroup$
This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
$endgroup$
– dev_nut
Dec 16 '14 at 19:21
5
$begingroup$
@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
$endgroup$
– Nate Eldredge
Dec 16 '14 at 19:33
1
$begingroup$
Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
$endgroup$
– dev_nut
Dec 16 '14 at 20:32
$begingroup$
@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
$endgroup$
– Sнаđошƒаӽ
Oct 27 '15 at 16:53
1
$begingroup$
I've added another example.
$endgroup$
– Nate Eldredge
Mar 13 '18 at 21:19
|
show 1 more comment
3
$begingroup$
This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
$endgroup$
– dev_nut
Dec 16 '14 at 19:21
5
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@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
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– Nate Eldredge
Dec 16 '14 at 19:33
1
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Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
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– dev_nut
Dec 16 '14 at 20:32
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@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
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– Sнаđошƒаӽ
Oct 27 '15 at 16:53
1
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I've added another example.
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– Nate Eldredge
Mar 13 '18 at 21:19
3
3
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This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
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– dev_nut
Dec 16 '14 at 19:21
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This is a great answer. Should be the accepted one in my opinion. It would be great if there's an intuitive real world example for the inverse as well, where a conditional independence does not imply independence?
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– dev_nut
Dec 16 '14 at 19:21
5
5
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@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
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– Nate Eldredge
Dec 16 '14 at 19:33
$begingroup$
@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
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– Nate Eldredge
Dec 16 '14 at 19:33
1
1
$begingroup$
Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
$endgroup$
– dev_nut
Dec 16 '14 at 20:32
$begingroup$
Thanks. It's trivial, but real world? :). Not as intuitive as your answer.
$endgroup$
– dev_nut
Dec 16 '14 at 20:32
$begingroup$
@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
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– Sнаđошƒаӽ
Oct 27 '15 at 16:53
$begingroup$
@NateEldredge any more obvious example? This trivial one is kind of hard to fathom using intuition ;-)
$endgroup$
– Sнаđошƒаӽ
Oct 27 '15 at 16:53
1
1
$begingroup$
I've added another example.
$endgroup$
– Nate Eldredge
Mar 13 '18 at 21:19
$begingroup$
I've added another example.
$endgroup$
– Nate Eldredge
Mar 13 '18 at 21:19
|
show 1 more comment
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The question as you have stated it is vague. Is there some problem or result that is motivating this question?
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– svenkatr
Feb 16 '11 at 22:31
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@svenkatr: Is it correct that independent random variables may not be independent conditional on some conditions? Independence conditional on some conditions may not imply independence. So I was wondering: (1) if Independence conditional on any condition and independence may imply each other; (2) what are some useful cases where one can imply the other?
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– Tim
Feb 16 '11 at 23:25
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What is independence under any condition? If you mean independence under every possible conditioning, the condition is rarely fulfilled... For example, random variables $x$ and $y$ are not independent conditionally on $[x<y]$ if you exclude some very degenerate cases.
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– Did
Feb 17 '11 at 11:34