Independent, Pairwise Independent and Mutually Independent events
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I want to understand the difference between Independent, Pairwise Independent and Mutually Independent events. I have read multiple answers related to this like in here, here and here. Most of them talk with 3 events and explain the difference between pairwise independent and mutually independent. I understand that. But what happens when there are $n$ events?
Suppose, $A_1$, $A_2$,....$A_n$ are n events, if
$$P(A_1 cap A_2 cap .... A_n) = P(A_1)P(A_2)....P(A_n)$$
but they are neither pairwise independent nor mutually independent. Only the above statement holds. Now, are these events still called Independent Events? Or is there any separate nomenclature for that?
probability independence
asked Aug 22 at 5:37
Nagabhushan S N
399
add a comment |Â
up vote
1
down vote
favorite
I want to understand the difference between Independent, Pairwise Independent and Mutually Independent events. I have read multiple answers related to this like in here, here and here. Most of them talk with 3 events and explain the difference between pairwise independent and mutually independent. I understand that. But what happens when there are $n$ events?
Suppose, $A_1$, $A_2$,....$A_n$ are n events, if
$$P(A_1 cap A_2 cap .... A_n) = P(A_1)P(A_2)....P(A_n)$$
but they are neither pairwise independent nor mutually independent. Only the above statement holds. Now, are these events still called Independent Events? Or is there any separate nomenclature for that?
probability independence
asked Aug 22 at 5:37
Nagabhushan S N
399
1
For a silly example, consider what happens if $A_1=emptyset$ in which case trivially $P(A_1cap A_2capdots A_n)=P(A_1)P(A_2)dots P(A_n)=0$, regardless of the nature of the relationships between $A_2,dots,A_n$
– JMoravitz
Aug 22 at 5:40
Sorry, I didn't get how it answers my question? Or, are you just adding an example? How does it help?
– Nagabhushan S N
Aug 22 at 5:41
4
I'm saying if you allow one of the events to be impossible, then any and every choice of $A_2,dots,A_n$, in particular those which are not mutually nor pairwise independent, will satisfy the condition you give. They are neither pairwise nor mutually independent, and we do not have much if any information as to anything else related to how they act, so it seems unnecessary to give such a situation a name.
– JMoravitz
Aug 22 at 5:52
1
Oh! Okay. Understood. But you're giving one situation where it doesn't make sense. That doesn't mean there aren't any situations where it makes sense. The equation might hold true for some situation. Right?
– Nagabhushan S N
Aug 22 at 6:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to understand the difference between Independent, Pairwise Independent and Mutually Independent events. I have read multiple answers related to this like in here, here and here. Most of them talk with 3 events and explain the difference between pairwise independent and mutually independent. I understand that. But what happens when there are $n$ events?
Suppose, $A_1$, $A_2$,....$A_n$ are n events, if
$$P(A_1 cap A_2 cap .... A_n) = P(A_1)P(A_2)....P(A_n)$$
but they are neither pairwise independent nor mutually independent. Only the above statement holds. Now, are these events still called Independent Events? Or is there any separate nomenclature for that?
probability independence
asked Aug 22 at 5:37
Nagabhushan S N
399
I want to understand the difference between Independent, Pairwise Independent and Mutually Independent events. I have read multiple answers related to this like in here, here and here. Most of them talk with 3 events and explain the difference between pairwise independent and mutually independent. I understand that. But what happens when there are $n$ events?
Suppose, $A_1$, $A_2$,....$A_n$ are n events, if
$$P(A_1 cap A_2 cap .... A_n) = P(A_1)P(A_2)....P(A_n)$$
but they are neither pairwise independent nor mutually independent. Only the above statement holds. Now, are these events still called Independent Events? Or is there any separate nomenclature for that?
probability independence
asked Aug 22 at 5:37
Nagabhushan S N
399
asked Aug 22 at 5:37
Nagabhushan S N
399
asked Aug 22 at 5:37
Nagabhushan S N
399
asked Aug 22 at 5:37
Nagabhushan S N
399
399
1
For a silly example, consider what happens if $A_1=emptyset$ in which case trivially $P(A_1cap A_2capdots A_n)=P(A_1)P(A_2)dots P(A_n)=0$, regardless of the nature of the relationships between $A_2,dots,A_n$
– JMoravitz
Aug 22 at 5:40
Sorry, I didn't get how it answers my question? Or, are you just adding an example? How does it help?
– Nagabhushan S N
Aug 22 at 5:41
4
I'm saying if you allow one of the events to be impossible, then any and every choice of $A_2,dots,A_n$, in particular those which are not mutually nor pairwise independent, will satisfy the condition you give. They are neither pairwise nor mutually independent, and we do not have much if any information as to anything else related to how they act, so it seems unnecessary to give such a situation a name.
– JMoravitz
Aug 22 at 5:52
1
Oh! Okay. Understood. But you're giving one situation where it doesn't make sense. That doesn't mean there aren't any situations where it makes sense. The equation might hold true for some situation. Right?
– Nagabhushan S N
Aug 22 at 6:13
add a comment |Â
1
For a silly example, consider what happens if $A_1=emptyset$ in which case trivially $P(A_1cap A_2capdots A_n)=P(A_1)P(A_2)dots P(A_n)=0$, regardless of the nature of the relationships between $A_2,dots,A_n$
– JMoravitz
Aug 22 at 5:40
Sorry, I didn't get how it answers my question? Or, are you just adding an example? How does it help?
– Nagabhushan S N
Aug 22 at 5:41
4
I'm saying if you allow one of the events to be impossible, then any and every choice of $A_2,dots,A_n$, in particular those which are not mutually nor pairwise independent, will satisfy the condition you give. They are neither pairwise nor mutually independent, and we do not have much if any information as to anything else related to how they act, so it seems unnecessary to give such a situation a name.
– JMoravitz
Aug 22 at 5:52
1
Oh! Okay. Understood. But you're giving one situation where it doesn't make sense. That doesn't mean there aren't any situations where it makes sense. The equation might hold true for some situation. Right?
– Nagabhushan S N
Aug 22 at 6:13
1
1
For a silly example, consider what happens if $A_1=emptyset$ in which case trivially $P(A_1cap A_2capdots A_n)=P(A_1)P(A_2)dots P(A_n)=0$, regardless of the nature of the relationships between $A_2,dots,A_n$
– JMoravitz
Aug 22 at 5:40
For a silly example, consider what happens if $A_1=emptyset$ in which case trivially $P(A_1cap A_2capdots A_n)=P(A_1)P(A_2)dots P(A_n)=0$, regardless of the nature of the relationships between $A_2,dots,A_n$
– JMoravitz
Aug 22 at 5:40
Sorry, I didn't get how it answers my question? Or, are you just adding an example? How does it help?
– Nagabhushan S N
Aug 22 at 5:41
Sorry, I didn't get how it answers my question? Or, are you just adding an example? How does it help?
– Nagabhushan S N
Aug 22 at 5:41
4
4
I'm saying if you allow one of the events to be impossible, then any and every choice of $A_2,dots,A_n$, in particular those which are not mutually nor pairwise independent, will satisfy the condition you give. They are neither pairwise nor mutually independent, and we do not have much if any information as to anything else related to how they act, so it seems unnecessary to give such a situation a name.
– JMoravitz
Aug 22 at 5:52
I'm saying if you allow one of the events to be impossible, then any and every choice of $A_2,dots,A_n$, in particular those which are not mutually nor pairwise independent, will satisfy the condition you give. They are neither pairwise nor mutually independent, and we do not have much if any information as to anything else related to how they act, so it seems unnecessary to give such a situation a name.
– JMoravitz
Aug 22 at 5:52
1
1
Oh! Okay. Understood. But you're giving one situation where it doesn't make sense. That doesn't mean there aren't any situations where it makes sense. The equation might hold true for some situation. Right?
– Nagabhushan S N
Aug 22 at 6:13
Oh! Okay. Understood. But you're giving one situation where it doesn't make sense. That doesn't mean there aren't any situations where it makes sense. The equation might hold true for some situation. Right?
– Nagabhushan S N
Aug 22 at 6:13
add a comment |Â
1 Answer
1
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oldest
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up vote
0
down vote
I got the below answer.
Mutual Independence and Pairwise Independence can be defined on a collection of events only. When it is said that a collection of events is independent, it means that all the events in the collection are mutually independent.
Suppose it is said that some events $A$, $B$, $C$, $D$ are independent, it means that
$$ P(A cap B cap C cap D) = P(A) * P(B) * P(C) * P(D)$$
and nothing else. We cannot assume that these events are pairwise or mutually independent.
answered Aug 23 at 4:58
Nagabhushan S N
399
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I got the below answer.
Mutual Independence and Pairwise Independence can be defined on a collection of events only. When it is said that a collection of events is independent, it means that all the events in the collection are mutually independent.
Suppose it is said that some events $A$, $B$, $C$, $D$ are independent, it means that
$$ P(A cap B cap C cap D) = P(A) * P(B) * P(C) * P(D)$$
and nothing else. We cannot assume that these events are pairwise or mutually independent.
answered Aug 23 at 4:58
Nagabhushan S N
399
add a comment |Â
up vote
0
down vote
I got the below answer.
Mutual Independence and Pairwise Independence can be defined on a collection of events only. When it is said that a collection of events is independent, it means that all the events in the collection are mutually independent.
Suppose it is said that some events $A$, $B$, $C$, $D$ are independent, it means that
$$ P(A cap B cap C cap D) = P(A) * P(B) * P(C) * P(D)$$
and nothing else. We cannot assume that these events are pairwise or mutually independent.
answered Aug 23 at 4:58
Nagabhushan S N
399
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I got the below answer.
Mutual Independence and Pairwise Independence can be defined on a collection of events only. When it is said that a collection of events is independent, it means that all the events in the collection are mutually independent.
Suppose it is said that some events $A$, $B$, $C$, $D$ are independent, it means that
$$ P(A cap B cap C cap D) = P(A) * P(B) * P(C) * P(D)$$
and nothing else. We cannot assume that these events are pairwise or mutually independent.
answered Aug 23 at 4:58
Nagabhushan S N
399
I got the below answer.
Mutual Independence and Pairwise Independence can be defined on a collection of events only. When it is said that a collection of events is independent, it means that all the events in the collection are mutually independent.
Suppose it is said that some events $A$, $B$, $C$, $D$ are independent, it means that
$$ P(A cap B cap C cap D) = P(A) * P(B) * P(C) * P(D)$$
and nothing else. We cannot assume that these events are pairwise or mutually independent.
answered Aug 23 at 4:58
Nagabhushan S N
399
answered Aug 23 at 4:58
Nagabhushan S N
399
answered Aug 23 at 4:58
Nagabhushan S N
399
answered Aug 23 at 4:58
Nagabhushan S N
399
399
add a comment |Â
add a comment |Â
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1
For a silly example, consider what happens if $A_1=emptyset$ in which case trivially $P(A_1cap A_2capdots A_n)=P(A_1)P(A_2)dots P(A_n)=0$, regardless of the nature of the relationships between $A_2,dots,A_n$
– JMoravitz
Aug 22 at 5:40
Sorry, I didn't get how it answers my question? Or, are you just adding an example? How does it help?
– Nagabhushan S N
Aug 22 at 5:41
4
I'm saying if you allow one of the events to be impossible, then any and every choice of $A_2,dots,A_n$, in particular those which are not mutually nor pairwise independent, will satisfy the condition you give. They are neither pairwise nor mutually independent, and we do not have much if any information as to anything else related to how they act, so it seems unnecessary to give such a situation a name.
– JMoravitz
Aug 22 at 5:52
1
Oh! Okay. Understood. But you're giving one situation where it doesn't make sense. That doesn't mean there aren't any situations where it makes sense. The equation might hold true for some situation. Right?
– Nagabhushan S N
Aug 22 at 6:13