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.

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    $\begingroup$ The corresponding Wikipedia page formulates "Any countable sequence of disjoint (synonymous with mutually exclusive) ..." (my emphasis) $\endgroup$ – Hagen von Eitzen Sep 21 '15 at 10:52
  • $\begingroup$ @HagenvonEitzen So? $\endgroup$ – BCLC Sep 21 '15 at 10:59
  • $\begingroup$ So the wikipedia article clearly and unambiguously states what "disjoint" means there. $\endgroup$ – Daniel Fischer Dec 6 '15 at 19:56

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.

  • $\begingroup$ Independent implies pairwise independent. Converse is not true. Pairwise disjoint implies disjoint. Converse is not true. Right? $\endgroup$ – BCLC Sep 21 '15 at 11:04
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    $\begingroup$ 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. $\endgroup$ – Did Sep 21 '15 at 11:25
  • $\begingroup$ @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. $\endgroup$ – BCLC Sep 21 '15 at 12:22
  • $\begingroup$ 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? $\endgroup$ – BCLC Sep 21 '15 at 12:28
  • $\begingroup$ @Did and Christian Blatter, I edited the question. $\endgroup$ – BCLC Sep 21 '15 at 13:06

While some texts use 'disjoint' to mean 'mutually disjoint', these texts seem to use 'disjoint' as meant to be 'pairwise disjoint'.

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    $\begingroup$ Any example of texts using 'disjoint' to mean 'mutually disjoint'? $\endgroup$ – Did Nov 15 '15 at 11:34
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    $\begingroup$ @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 $A_i \cap A_j = \emptyset \ \forall i \ne j$? $\endgroup$ – BCLC Nov 26 '15 at 1:18
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    $\begingroup$ @BCLC Can you explain what that last comment refers to? As it stands, it makes no sense. $\endgroup$ – Daniel Fischer Dec 6 '15 at 19:17
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    $\begingroup$ 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. $\endgroup$ – layman Dec 14 '16 at 13:02
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    $\begingroup$ @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". $\endgroup$ – layman Dec 15 '16 at 23:22

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