# if $P(A \cap B \cap C)=P(A)P(B)P(C)$, does it necessarily means $A,B,C$ are independants?

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?

• What if $A$ is a probability zero event? What can you say about $B$ and $C$? Jan 11 at 10:38
• hmm nothing. $A \cap B = \emptyset$. same for $A \cap C$. It's doesn't tell nothing about the relation between $B$ and $C$ Jan 11 at 10:41
• @BurakKaraosmanoğlu Links to brilliant.org are not very helpful because one has to login Jan 11 at 10:47
• @ryden what means independent? Pairwise independent? Mutually independent? en.wikipedia.org/wiki/… Jan 11 at 10:53
• @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. Jan 11 at 10:56

$$A_1,...,A_n$$ 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.

• so, "independent" only means they are pairwise independent? Jan 11 at 11:08
• 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/… Jan 11 at 11:11
• alright, now i get it. thank you very much! Jan 11 at 11:12
• 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. Jan 11 at 11:53
• 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. Jan 11 at 11:56