ThmDex – An index of mathematical definitions, results, and conjectures.
Pairwise independent event collection need not be independent
Formulation 0
Let $P = (\{ 1, 2, 3, 4 \}, \mathcal{F}, \mathbb{P})$ be a D6122: Standard uniform discrete probability space.
Then
(1) $\{ 1, 2 \}, \{ 1, 3 \}, \{ 2, 3 \}$ is a D2148: Pairwise independent event collection
(2) \begin{equation} \mathbb{P}(\{ 1, 2 \} \cap \{ 1, 3 \} \cap \{ 2, 3 \}) = \mathbb{P} (\emptyset) = 0 \neq \frac{1}{2^3} = \mathbb{P} \{ 1, 2 \} \mathbb{P} \{ 1, 3 \} \mathbb{P} \{ 2, 3 \} \end{equation}
Proofs
Proof 0
Let $P = (\{ 1, 2, 3, 4 \}, \mathcal{F}, \mathbb{P})$ be a D6122: Standard uniform discrete probability space.
On (1), observe that the intersection of any two events in the collection yields a singleton, each of which is assigned probability $1/4$. Claim (2) is clear. $\square$