# Pairwise Independence does not imply Independence

## Theorem

Just because all the events in a family of events in a probability space are pairwise independent, it does not mean that the family is independent.

## Proof

Consider throwing a fair four-sided die.

This gives us an event space $\Omega = \set {1, 2, 3, 4}$, with each $\omega \in \Omega$ equally likely to occur:

$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 4$

Consider the set of events:

$\SS = \set {A, B, C}$

where:

$A = \set {1, 2}, B = \set {1, 3}, C = \set {1, 4}$

We have that:

$\map \Pr A = \map \Pr B = \map \Pr C = \dfrac 1 2$

We also have that:

$\map \Pr {A \cap B} = \map \Pr {A \cap C} = \map \Pr {B \cap C} = \map \Pr {\set 1} = \dfrac 1 4$

Thus:

$\map \Pr A \map \Pr B = \map \Pr {A \cap B}$
$\map \Pr A \map \Pr C = \map \Pr {A \cap C}$
$\map \Pr B \map \Pr C = \map \Pr {B \cap C}$

Thus the events $A, B, C$ are pairwise independent.

Now, consider:

$\map \Pr {A \cap B \cap C} = \map \Pr {\set 1} = \dfrac 1 4$

But:

$\map \Pr A \map \Pr B \map \Pr C = \dfrac 1 8 \ne \map \Pr {A \cap B \cap C}$

So, although $\SS$ is pairwise independent, it is not independent.

$\blacksquare$