📂Mathematical Statistics

Bernstein Distributions: Pairwise Independence Does Not Imply Mutual Independence

Definition

For $(x,y,z) \in \left\{ (1,0,0), (0,1,0), (0,0,1), (1,1,1) \right\}$, the distribution with the following probability mass function is called the Bernstein Distribution. $$p(x,y,z) = {{1} \over {4} }$$

Explanation

Although the Bernstein Distribution satisfies all the conditions for a distribution, it is hard to consider it as a distribution that actually exists in nature. It is presented as a counterexample to the proposition that ‘if pairs are independent, then they are mutually independent,’ and it has no other significance. However, as a counterexample, it is quite intuitive and greatly assists in familiarizing oneself with the fact.

Counterproof

The marginal probability density function for one random variable is as follows: $$f_{X} (0) = f_{Y} (0) = f_{Z} (0) = {{1} \over {2}} \\ f_{X} (1) = f_{Y} (1) = f_{Z} (1) = {{1} \over {2}}$$ The marginal probability density function for two random variables is as follows: $$f_{X,Y} (0,0) = f_{X,Y} (1,0) = f_{X,Y} (0,1) = f_{X,Y} (0,1) = {{1} \over {4}} \\ f_{Y,Z} (0,0) = f_{Y,Z} (1,0) = f_{Y,Z} (0,1) = f_{Y,Z} (0,1) = {{1} \over {4}} \\ f_{X,Z} (0,0) = f_{X,Z} (1,0) = f_{X,Z} (0,1) = f_{X,Z} (0,1) = {{1} \over {4}}$$ Therefore, $X,Y$, $Y,Z$, and $X,Z$ are independent, and $X,Y,Z$ is pairwise independent. However, since $${{1} \over {4}} = f_{X,Y,Z} (1,1,1) \ne f_{X} (1) f_{Y} (1) f_{Z} (1) = {{1} \over {8}}$$ $X,Y,Z$ is not mutually independent.

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