# Mutual independence of three events

As I understand it, if three events are mutually independent, then $P(A\cap B\cap C) = P(A)\cdot P(B)\cdot P(C)$. I'd like to know if the converse is also true, that is, if $P(A)\cdot P(B)\cdot P(C) = P(A\cap B\cap C)$, are A, B, and C mutually independent? The question here seems to suggest that that isn't true, and the last response (see below) supports that conclusion, but I'm wondering if both the question and the response might be flawed.

Useful example. Whole space has numbers 1 through 8, each with probability = 1/8.

Let A = B = {1,2,3,4}

Let C = {1,5,6,7}

P(A)=P(B)=P(C)=1/2

P(A and B and C)=P({1})=1/8.

However A and B are obviously not independent. Also A and C are not independent.

I appreciate any light you can shed on the issue.

• The counter example appears valid. Can you be more specific about what isnt clear to you? – stephematician Sep 18 '17 at 21:33
• It seems to me that the counter example provides an example of events that are not mutually exclusive rather than ones that are not independent. – Jon Sep 18 '17 at 23:41
• An event can be both not mutually exclusive and not independent. On the other hand, I don't think an event can be both mutually exclusive and independent (although I am not sure about events that almost-surely do not occur). So being mutually exclusive does not prohibit this from being a counter-example. The (pair-wise) not-independent status of the events is easily determined as shown in the answer below. – stephematician Sep 19 '17 at 0:43
• To more precise, I'm having trouble with the phrase "A and B are obviously not independent" is that simply b/c it's clear that $P(A)\cdot P(B) \neq P(A\cap B)$? Are A = {1, 2, 3, 4} and B = {3, 4, 5, 6} independent since P(A) $\cdot$ P(B) = .25 and P(A $\cap$ B) also equals .25? – Jon Sep 19 '17 at 2:28
• I hope this may be of help: math.stackexchange.com/a/3157866/351322 – Lerner Zhang Mar 22 '19 at 8:04

The answer to your confusion is that in order for three events $$A$$, $$B$$ and $$C$$ to be mutually independent it is necessary but not sufficient that $$P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$$ (condition 1).

The other condition that must be met is that each pair of events must also be independent [so $$A$$ and $$B$$ must be independent, $$B$$ and $$C$$ must be independent and $$A$$ and $$C$$ must be independent - also known as pairwise independence (condition 2)].

As shown in your example, condition 1 can be met by events that violate condition 2. Events $$A$$ and $$B$$ are dependent ($$P(A|B) \ne P(A) \times P(B)$$ and $$P(B|A) \ne P(A) \times P(B)$$) and so events $$A$$, $$B$$ and $$C$$ are not mutually independent.

Mutual Independence of three events For any three events $$A$$, $$B$$ and $$C$$ to be mutually independent the following two conditions must be met:

1. $$P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$$

2. $$A$$ and $$B$$ must be independent, $$B$$ and $$C$$ must be independent and $$A$$ and $$C$$ must be independent

Mutual Independence of $$n$$ events Events $$A_1, A_2,...,A_n$$ are mutually independent iff for any sub-collection of $$k$$ events, $$A_{i1}, A_{i2},...,A_{ik}$$, it is true that:

$$P(A_{i1} \cap A_{i2} \cap ... \cap A_{ik}) = P(A_{i1}) \times P(A_{i2}) \times ... \times P(A_{ik})$$

This was growing too long for a comment. I'm going to construct a similar (although not equivalent) counterexample using three coins so that independence makes more 'sense'.

To determine if a pair of events are independent, the pair need only satisfy that $P(A \cap B) = P(A) P(B)$.

The state space for three coins could be written as:

$$\{(h,h,h), (h,h,t), (h,t,h), (h,t,t), (t,h,h), (t,h,t), (t,t,h), (t,t,t)\}$$

This can also be written as $\{1, 2, \ldots, 8\}$, i.e. by enumerating each outcome.

Consider the following pair of dependet events:

1. $A$ the event that the first coin is a head (i.e. state 1, 2, 3 or 4),
2. $B$ the event that the first two coins are heads, or the last two are tails (i.e. state 1, 2, 4 or 8).

The dependence is physically obvious; they both rely on the outcome of the first coin flip. We can also see that they don't obey the definition of independence, as

• $P(A \cap B) = P(\text{state 1, 2, or 4}) = 3/8$ and
• $P(A) P(B) = (1/2) \times (1/2) = 1/4$.

Let's consider a third event:

1. $C$ the event that the last two coins are not the same (state 2, 3, 6, 7).

In physical terms, it seems that this event is independent from $A$, because event $C$ doesn't depend on the flip of the first coin, while event $A$ only depends on the first coin. It is easy to show, as well, that they obey the definition of independence.

Meanwhile, event $B$ and $C$ are dependent, in physical terms because they both rely on the outcome of the last two coins; and it is also easy to show that the definition of independence is not satisfied.

Now of course $P(A \cap B \cap C) = P(\text{state 2}) = 1/8$, and $P(A) P(B) P(C) = 1/8$, too.

So the question is, still, are they 'mutually independent'? No. As per the definition of independence for multiple events, we require that each pair of events is independent - which we have already seen is false.