Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 2 / 11F / iv

Solution

The statement is false. Take two independent fair bits

U, V

on the four equally probable outcomes

00, 01, 10, 11

, and put

A_{1} = {U = 0}

A_{2} = {V = 0}

, and

B = {U = V}

. Each of

A_{1}, A_{2}, B

1/2

, and

P (B \cap A_{1}) = P (B \cap A_{2}) = \frac{1}{4} = P (B) P (A_{i}) .

Thus

B

is independent of each event separately. But

A_{1} \cup A_{2}

contains

00, 01, 10

, whereas its intersection with

B

contains only

00

. Therefore

P (B \cap (A_{1} \cup A_{2})) = \frac{1}{4} \neq = \frac{1}{2} \cdot \frac{3}{4} = P (B) P (A_{1} \cup A_{2}) .

This is a concrete failure of extending separate independence to the generated union. A counterexample at the permitted value

n = 2

disproves the universal claim; for any larger

n

one can repeat

A_{1}

as the additional events.

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