Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 25J / Solution

A pi-system is a collection closed under finite intersections. The uniqueness theorem says that two probability measures agreeing on a generating pi-system agree on its generated sigma-field. For finite measures one also fixes the total mass; for a sigma-finite version one requires a covering sequence from the pi-system with finite common masses. Adding the whole space to a pi-system of events causes no change to its generated sigma-field and preserves the stated independence condition.

Fix $H\in \mathcal{H}$. On $\sigma (\mathcal{G})$ the finite measures $A\mapsto P(A\cap H)$ and $A\mapsto P(A)P(H)$ agree on $\mathcal{G}$ and on the whole space. Uniqueness makes them agree for every $A\in \sigma (\mathcal{G})$. Now fix such an $A$. On $\sigma (\mathcal{H})$ the measures $B\mapsto P(A\cap B)$ and $B\mapsto P(A)P(B)$ agree on $\mathcal{H}$ and the whole space; a second application of uniqueness provesThus the two generated sigma-fields are independent. The extension is done one argument at a time, not by assuming the required independence for arbitrary unions.

For the independent random variables, let $\mathcal{G}$ be the rectangles ${\bigcap }_{i=1}^m\{Y_i\in B_i\}$ and let $\mathcal{H}$ be the analogous rectangles in the $Z_j$. Both are pi-systems. Joint independence of the full list factors each combined rectangular probability into the product of the individual probabilities, hence into $P(G)P(H)$. Their generated sigma-fields are exactly $\sigma (Y_1,\dots ,Y_m)$ and $\sigma (Z_1,\dots ,Z_n)$. The proved result therefore gives independence of the two vector-generated sigma-fields.