Open Study Answer #133

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WORK IN PROGRESS

Scope product

\[\begin{eqnarray*} \pi \times \rho & := & \{ A \cap B : A \in \pi, B \in \rho \} \\ \pi \times B & := & \{ A \cap B : A \in \pi \} \\ \end{eqnarray*}\]

Consider any two scopes \(\theta\) and \(\phi\) sharing probability space \(\Omega\). Define \[\begin{eqnarray*} \theta \times \phi & := & \bigcup_{i=0}^\infty \bigcup_{\substack{\pi \in \operatorname{dom}_{ i}{ \theta} \\ \rho \in \operatorname{dom}_{ i}{ \phi} \\ ({\cup{ \pi}}) \cap ({\cup{ \rho}}) \not= \emptyset }} \mathcal{C}(\pi, \rho) \end{eqnarray*}\] where \[\begin{eqnarray*} \mathcal{C}(\pi, \rho) & = & \begin{cases} \{ (\pi \times \rho) \mapsto \theta(\pi) \} & \text{ when } \theta(\pi) = \phi(\rho) \\ \left\{ \pi \mapsto \theta(\pi) \right\} \cup \bigcup_{A \in \pi} \left\{ \rho \times A \mapsto \phi(\rho) \right\} & \text{ when } \theta(\pi) > \phi(\rho) \\ \left\{ \rho \mapsto \phi(\rho) \right\} \cup \bigcup_{A \in \rho} \left\{ \pi \times A \mapsto \theta(\pi) \right\} & \text{ when } \theta(\pi) < \phi(\rho) \\ \end{cases} \\ \end{eqnarray*}\]

Theorem (Scope product is a scope of relevance)

For any two scopes of relevance \(\theta\) and \(\phi\) sharing probability space \(\Omega\), \(\theta \times \phi\) is also a scope of relevance.

Proof

TBD

References