Lets say we have the events $A,B,C$. We know that if they are independants, then the following occurs: $$P(A \cap B \cap C)=P(A)P(B)P(C)$$ But does it work the opposite way? If some $A,B,C$ events satisfies the equation above, does it mean they are necessarily independent?

Another question: when we say that $A_1,A_2,...,A_n$ are independants, does it mean they are independants in pairs or in any $2 \le k \le n$ groups?

  • $\begingroup$ What if $A$ is a probability zero event? What can you say about $B$ and $C$? $\endgroup$ Jan 11 at 10:38
  • $\begingroup$ hmm nothing. $A \cap B = \emptyset$. same for $A \cap C$. It's doesn't tell nothing about the relation between $B$ and $C$ $\endgroup$
    – ryden
    Jan 11 at 10:41
  • $\begingroup$ @BurakKaraosmanoğlu Links to brilliant.org are not very helpful because one has to login $\endgroup$
    – miracle173
    Jan 11 at 10:47
  • $\begingroup$ @ryden what means independent? Pairwise independent? Mutually independent? en.wikipedia.org/wiki/… $\endgroup$
    – miracle173
    Jan 11 at 10:53
  • $\begingroup$ @miracle173 My mistake! I actually wanted to link another site. I deleted my first comment with your warning. Thanks! New comment in response to the question: You need to impose three extra mutual independence conditions for the independence of three events, independence of A and B, independence of A and C, and independence of B and C, as stated here. You may also refer this post. $\endgroup$ Jan 11 at 10:56

1 Answer 1


$A_1,...,A_n$ are independent when they are $k$-wise independent for all $2\le k \le n$. That is, $A_1,\dots,A_n$ are independent if and only if for any subset of $k$ events with indices $i(1),\dots,i(k)$, you have $$ P(A_{i(1)}\cap \dots \cap A_{i(k)})=P(A_{i(1)})\cdots P(A_{i(k)}) $$

Your question then boils down to whether $P(A\cap B\cap C)=P(A)P(B)P(C)$ implies $P(A\cap B)=P(A)P(B)$, $P(A\cap C)=P(A)P(C)$ and $P(B\cap C)=P(B)P(C)$.

A counterexample is $P(A\cap B \cap C)=1/8$, $P(A\cap B)=1/8$, $P(A\cap C)=3/8$, $P(B\cap C)=1/4$, $P(A)=P(B)=P(C)=1/2$.

Additional: To answer miracle173's question, consider the set $\{0,...,7\}$ and the subsets $A=\{0,...,3\}$, $B=\{0,1,4,5\}$, and $C=\{0,2,4,6\}$. We then take $P(i)=1/8,0,1/4,1/8,1/8,1/4,0,1/8$ for $i=0,...,7$. Since these numbers sum to one, we get a valid probability measure.

  • $\begingroup$ so, "independent" only means they are pairwise independent? $\endgroup$
    – ryden
    Jan 11 at 11:08
  • $\begingroup$ You can define other kinds of 'independence' as you like, but as far as I know pairwise independence is the only useful one. Wikipedia does define something other than pairwise independence: en.wikipedia.org/wiki/… $\endgroup$ Jan 11 at 11:11
  • $\begingroup$ alright, now i get it. thank you very much! $\endgroup$
    – ryden
    Jan 11 at 11:12
  • $\begingroup$ A counterexample consists of the sets A,B,C and not of some numbers for the probabilities. Maybe there do not exist sets that result in such probabilities. $\endgroup$
    – miracle173
    Jan 11 at 11:53
  • $\begingroup$ Well, a counterexample is both the probability space (the sets) and the probability measure (the numbers). These numbers do in fact come from a measure on a valid sigma algebra/probability space. $\endgroup$ Jan 11 at 11:56

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