example of three events that are pairwise independent but not independent

Question

If we have $(\{1,2,3,4,5,6\}, \sigma(\{\{1\},\{2\},\{3\},\{4\},\{5\},\{6\}\}), k)$ and $k$ assigns to each singleton measure $\displaystyle\frac{1}{6}$, what would be an example of a collection of $3$ events that are pairwise independent but not independent? $\sigma$ here denotes sigma algebra.

Answer 1

It isn't possible in this case. Indeed, suppose $A,B$ and $C$ are pairwise independent events that are not independent. If $k(A)=0$, $A=\emptyset$ so $k(A\cap B\cap C)=k(\emptyset)=0=k(A)k(B)k(C)$, a contradiction. If $k(A)=1$, $A=\Omega$ so $k(A\cap B\cap C)=k(B\cap C)=k(B)k(C)=k(A)k(B)k(C)$, again a contradiction. Hence $k(A)=\frac{i}6$ for some $i\in\{1,2,3,4,5\}$. Similarly $k(B)=\frac{j}6$, $k(C)=\frac{l}6$. Hence $$\frac{N}6=k(A\cap B)=k(A)k(B)=\frac{jk}{36}$$ and so $6$ divides $jk$. Given $j,k\in\{1,2,3,4,5\}$ the only possibilities are $jk=6$ and $jk=12$, and repeating for the other pairs we have $jk,kl,jl\in\{6,12\}$. One can verify directly that this is impossible.

For an example of a probability space where you can find such events, let $\Omega=\{1,2,3,4\}$, $\mathcal{F}=\mathcal P(\Omega)$ and $P(\{i\})=\frac14$. Let $A=\{1,2\}$, $B=\{2,3\}$ and $C=\{1,3\}$. Then $$P(A)=P(B)=P(C)=\tfrac12,\\ P(A\cap B)=P(B\cap C)=P(A\cap C)=\tfrac14,\\ P(A\cap B\cap C)=0.$$