Solution

Fix

A_{1} \in A_{1}

and let

D_{A_{1}} = {B \in F : P (A_{1} \cap B) = P (A_{1}) P (B)} .

This is a Dynkin system. It contains

Ω

. If

B \in D_{A_{1}}

, then

P (A_{1} \cap B^{c}) = P (A_{1}) - P (A_{1} \cap B) = P (A_{1}) (1 - P (B)) = P (A_{1}) P (B^{c}),

so it is closed under complements. Countable additivity gives closure under countable disjoint unions.

The hypothesis says

A_{2} \subseteq D_{A_{1}}

. Since

A_{2}

σ (A_{2}) \subseteq D_{A_{1}} .

Thus the factorization holds for every

A_{1} \in A_{1}

and every

B \in σ (A_{2})

Now fix such a

B

and define

E_{B} = {A \in F : P (A \cap B) = P (A) P (B)} .

The identical calculation makes

E_{B}

a Dynkin system. The first step gives

A_{1} \subseteq E_{B}

, so a second application of Dynkin lemma gives

σ (A_{1}) \subseteq E_{B} .

Therefore the probability factorization holds for every

A \in σ (A_{1}), B \in σ (A_{2}),

Solved by gpt-5.6-sol high.

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