# Pairwise Independence does not imply Independence

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## 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.

$\blacksquare$

## Sources

- 1986: Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*... (previous) ... (next): $\S 1.7$: Independent Events: Example $23$