What is the difference between mutually independent and pairwise independent events in probability theory?

Question

Let us assume that a number is selected at random from $1, 2, 3$. We define

$$A = \{1, 2\},\quad B = \{2, 3\},\quad C = \{1, 3\}$$

Then are $A$, $B$ and $C$ mutually independent or pairwise independent or both?

I am confused between mutually vs pairwise independent.

Your question has omitted the one alternative that turns out to be correct—namely that the events $\ A$, $\ B,$ and $\ C\ $ are neither mutually independent nor pairwise independent, since \begin{align} P(A\cap B)&=P({2})\&=\frac{1}{3}\ ,\ \text{but}\ P(A)P(B)&=\Big(\frac{2}{3}\Big)\Big(\frac{2}{3}\Big)\&=\frac{4}{9}\&\ne P(A\cap B)\ .\end{align} — lonza leggiera, Nov 08 '21 at 10:36

JDF · Accepted Answer · 2016-09-09T14:52:54.960

41

Mutual independence: Every event is independent of any intersection of the other events.

Pairwise independence: Any two events are independent.

$A, B, C$ are mutually independent if $$P(A\cap B\cap C)=P(A)P(B)P(C)$$ $$P(A\cap B)=P(A)P(B)$$ $$P(A\cap C)=P(A)P(C)$$ $$P(B\cap C)=P(B)P(C)$$

On the other hand, $A, B, C$ are pairwise independent if $$P(A\cap B)=P(A)P(B)$$ $$P(A\cap C)=P(A)P(C)$$ $$P(B\cap C)=P(B)P(C)$$

I'm sure you can solve your problem now.

edited Sep 09 '16 at 14:52

answered Sep 09 '16 at 14:47

JDF

918

If I understood correctly, I think these are pairwise independent but not mutually independent? – exAres Sep 09 '16 at 15:03
What is your $P(A\cap B)$ and what are your $P(A)$ and $P(B)$? – JDF Sep 09 '16 at 15:04
Also, what exactly is the event $A$? – JDF Sep 09 '16 at 15:05
Question says 'Define A={1, 2}'. I assume it means A is an event that either 1 or 2 is selected. – exAres Sep 09 '16 at 15:07
If that is so, calculate the probabilities and verify. By the way, $A\cap B$ means the event that $A$ happens and $B$ happens, just to be clear. – JDF Sep 09 '16 at 15:10
Ohh Okay. So then P(A)*P(B) is 4/9 and P(A-intersect-B) is 1/3 – exAres Sep 09 '16 at 15:12
Let us continue this discussion in chat. – exAres Sep 09 '16 at 15:14
1

Jus to add, mutual independence as seen above is a harder claim and implies pairwise indolence but it is not so the other way round. – jia chen Feb 02 '20 at 11:25

Soudipta Dutta · Answer 2 · 2022-02-18T09:04:00.403

Pairwise Independent Conditions:

P(AB) = P(A)P(B)
P(BC) = P(B)P(C)
P(AC) = P(A)P(C)

Mutual Independent Conditions:

P(AB) = P(A)P(B)
P(BC) = P(B)P(C)
P(AC) = P(A)P(C)
P(ABC) = P(A)P(B)P(C)

Let us take an example to understand the situation clearly.

A lot contains 50 defective and 50 non-defective pens. Two pens are drawn at random, one at a time, with replacement. The events A, B, C are defined as :

A = ( the first pen is defective)
B = (the second pen is non-defective)
C = (the two pens are both defective or both non-defective).

Determine whether:

(i) A, B, C are pairwise independent.
(ii) A, B, C are independent. [IIT 1992]

Solution :

D := Defective , N := Not Defective
P(A) = P{the first pen is defective} = P($[D_{1}D_{2},D_{1}N_{2}] ) = (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{2}$
P(B) = P{the second pen is non-defective} = P($D_{1}N_{2},N_{1}N_{2}$) = $\frac{1}{2}$
P(C) = P{the two pens are both defective OR not defective} = P($D_{1}D_{2},N_{1}N_{2}$) = $\frac{1}{2}$

Then P($A\bigcap B$) = P(AB) = P($D_{1}N_{2}$) = $\frac{1}{2} \cdot \frac{1}{2}$=$\frac{1}{4}$, P(AC)=P($D_{1}D_{2}$) = $\frac{1}{4}$ , P(BC)= P($N_{1}N_{2}$) = $\frac{1}{4}$.

But, P($A\bigcap B\bigcap C) = P(ABC) = \varnothing = null$ = no common elements = $0$.

Therefore, $P(AB) = P(BC) = P(AC) = \frac{1}{4}$. But $P(ABC) = 0 \neq P(A) \cdot P(B) \cdot P(C)$.

Hence, A, B, C are pairwise independent BUT Not Mutually Independent.

What is the difference between mutually independent and pairwise independent events in probability theory?

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