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According to 1 2, the third Kolmogorov axiom is

for disjoint sets $(A_n)_{n \in \mathbb{N}}$

$P(\cup_n A_n) = \sum_n P(A_n)$

Is that really disjoint rather than pairwise disjoint?

If we have events $A, B, C$ s.t.

$A \cap B = \emptyset$

$A \cap C = \emptyset$

$B \cap C \neq \emptyset$

$P(B \cap C) > 0$,

then A, B and C are disjoint but not pairwise disjoint...I think? (*)

I don't think it follows that $P(A \cup B \cup C) = P(A) + P(B) + P(C)$.

I think $P(A \cup B \cup C) = P(A) + P(B) + P(C \setminus B)$ ?

(*) From what I remember in advanced probability class:

$\{A_n\}_n$'s are disjoint if $\cap_n A_n = \emptyset$

$\{A_n\}_n$'s are pairwise disjoint if $A_i \cap A_j = \emptyset$ for distinct indices i,j

From Larsen and Marx (book used in my elementary probability class):

enter image description here

I find this strange. If 'disjoint' and 'pairwise disjoint' are equivalent (ie disjoint does not mean what I said above), why even say that $A_i \cap A_j = \emptyset$ for distinct indices i,j? Why not just say disjoint?

On the other hand, disjointness is used to justify the $P(\cup_n A_n) = \sum_n P(A_n)$ statements later on. Seems kind of inconsistent.

BCLC
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2 Answers2

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If the sets $A_i$ are pairwise disjoint then any intersection incorporating at least two different $A_i$ is empty, and conversely: If any intersection incorporating at least two different $A_i$ is empty then the $A_i$ are in particular pairwise disjoint. Therefore it is sufficient to call them "disjoint".

It's another thing with "independent" in probability theory: If the events $A_i$ are pairwise independent then they need not be "mutually independent", but "mutually independent" events are of course also pairwise independent.

Christian Blatter
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  • Independent implies pairwise independent. Converse is not true. Pairwise disjoint implies disjoint. Converse is not true. Right? – BCLC Sep 21 '15 at 11:04
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    BCLC It is beginning to be obvious that you do not read the answers posted to answer your questions. Hint: first sentence of Christian's answer. – Did Sep 21 '15 at 11:25
  • @Did ? Christian Blatter, {1, 2}, {2, 3}, {1, 3} are disjoint but not pairwise disjoint? I mean, are you saying that A, B and C in my example are 'disjoint' even though $A \cap B \cap C = \emptyset$ ? I thought $A \cap B \cap C = \emptyset$ was equivalent to them being disjoint. – BCLC Sep 21 '15 at 12:22
  • Christian Blatter, p. 4 here seems to agree with you. This is strange. I somewhat remember my professor making a distinction between disjoint and pairwise disjoint. So disjoint implies empty intersection but not vice versa? – BCLC Sep 21 '15 at 12:28
  • @Did and Christian Blatter, I edited the question. – BCLC Sep 21 '15 at 13:06
  • 'any intersection incorporating at least two different $A_i$ is empty' is that your definition of disjoint? – BCLC Nov 26 '15 at 01:17
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While some texts use 'disjoint' to mean 'mutually disjoint', these texts seem to use 'disjoint' as meant to be 'pairwise disjoint'.

BCLC
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    Any example of texts using 'disjoint' to mean 'mutually disjoint'? – Did Nov 15 '15 at 11:34
  • @Did My probability professor deducted points from those wrote disjoint instead of pairwise disjoint. :| I think some students were confused because our elem prob book uses disjoint. I forgot if Williams made a distinction between disjoint and PW disjoint, but since you're very familiar w/ the book, I guess not. Maybe Rosenthal. My prof referred to that book as well – BCLC Nov 15 '15 at 13:25
  • Accepting instantaneously an answer posted by yourself and unsourced (to stay polite) when another user made the effort to provide a (better) answer... Why do I find this typical? – Did Nov 15 '15 at 13:56
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    @Did Any book that uses the term 'pairwise disjoint' hence uses 'disjoint' to mean 'mutually disjoint'. Why do we even have the term 'pairwise disjoint' if it means the same thing as 'disjoint' ? Why doesn't the Larsen and Marx book say 'disjoint' instead of $A_i \cap A_j = \emptyset \ \forall i \ne j$? – BCLC Nov 26 '15 at 01:18
  • Not even sure whom you are talking to--but do you have any canonical source for "pairwise disjoint"? – Did Nov 26 '15 at 06:27
  • @Did What do you mean by canonical? Since pairwise disjoint is an easy find on Google, I'm guessing you mean a real analysis, probability, set or measure theory book that uses 'pairwise disjoint' ? Why is that needed? The fact that the term 'pairwise disjoint' exists means that it is either unnecessary or 'disjoint' collection means a collection w/ empty intersection. Personally, I think most probability people will find it unnecessary as such a situation that needs to distinguish rarely arises. I don't think I've ever encountered such a situation in my entire post-highschool life. But my – BCLC Nov 26 '15 at 07:00
  • @Did my adv prob professor gave deductions for those people who said disjoint instead of pairwise disjoint despite our elem prob textbook using 'disjoint' to refer to what e calls 'pairwise' disjoint. Assuming I'm telling the truth, how is that not good enough for you? It's not like e's making this stuff up to piss people off or anything – BCLC Nov 26 '15 at 07:02
  • The trouble is that, as often, you make circles, rehashing ad infinitum the same trivia already explained clearly to you. Sooo... once again: 1. When one mentions disjoint sets $(A_i)$, one refers to the situation where $A_i\cap A_j=\varnothing$ for every $i\ne j$. 2. To state that $\bigcap\limits_i A_i=\varnothing$, one says that the collection $(A_i)$ has an empty intersection. Thus, "pairwise disjoint" is unnecessary and might be mostly used in hastily written internet sources. All this is crystal clear since at least @Christian's answer, but you do not listen. // Re your professor, ... – Did Nov 26 '15 at 07:34
  • ... if really they deducted points for using "disjoint" instead of "pairwise disjoint" to mean that $A_i\cap A_j=\varnothing$ for every $i\ne j$ and not for another reason you would have missed, and if they did not specify in their lectures that they wished to use "pairwise disjoint" (not in accordance with the literature) to mean "disjoint", then they are liable to some explanations for this departure from maths standards. Note the two conditionals. // Re your answer, it might be the most confused part of the whole page since 'disjoint', 'mutually disjoint' (your invention) and ... – Did Nov 26 '15 at 07:35
  • ... 'pairwise disjoint' obviously all mean the same thing (but I seem to understand you now backtracked from the assertion that some sources were using any of them for anything else than the fact that $A_i\cap A_j=\varnothing$ for every $i\ne j$, hence it might be time to leave this fascinating subject). – Did Nov 26 '15 at 07:36
  • @Did Christian say 'empty intersection' ? – BCLC Dec 06 '15 at 17:42
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    @BCLC Can you explain what that last comment refers to? As it stands, it makes no sense. – Daniel Fischer Dec 06 '15 at 19:17
  • @DanielFischer Did said 'empty intersection' and 'All this is crystal clear since at least Christian's answer'. The part about the 'empty intersection', to me, was not crystal clear or not in Christian's answer – BCLC Dec 07 '15 at 11:22
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    I swear I have the exact same thought process as you and an internal war right now. Why do we need the label "pairwise disjoint" if it means the same thing as disjoint? It's for this reason I've always believed disjoint means if you intersect all of the sets in the collection, it's empty, while pairwise disjoint means any pair have empty intersection. We are both thinking about this the same way, and it looks like we're wrong. However weird it might feel, I guess people accept that "disjoint" really is another word for the concept of pairwise disjointness. – layman Dec 14 '16 at 13:02
  • @user46944 you think L&M are inconsistent? – BCLC Dec 14 '16 at 14:00
  • @BCLC By L&M do you mean Larson and Marx? If so, I haven't checked out their book. – layman Dec 14 '16 at 19:39
  • @user46944 screenshots? – BCLC Dec 15 '16 at 22:51
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    @BCLC It doesn't seem like they're being inconsistent, but I haven't seen where they define "disjoint". From the screenshot, I only see them using the word "disjoint" once, to apply countable additivity of the measure, which means they are referring to "pairwise disjoint" as just "disjoint". – layman Dec 15 '16 at 23:22
  • @user46944 I'm a dumbass. I just looked it up. They don't define disjoint. They define however mutually exclusive but only for two sets. It looks like there's an implicit definition for mutually exclusive for more than two sets by which they mean pairwise disjoint rather than empty intersection (Of course I avoided the word disjoint because we don't know if it means pairwise disjoint or empty intersection) – BCLC Dec 21 '16 at 03:15
  • @BCLC But you had said your professor insisted "disjoint" means empty intersection and is different from "pairwise disjoint," which just confuses matters. – layman Dec 21 '16 at 09:25