Let ((X, mathscr{A}, mu)) be a finite measure space, (G_{1}, ldots, G_{n} in mathscr{A}) such that (bigcup_{i=1}^{n}

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Let \((X, \mathscr{A}, \mu)\) be a finite measure space, \(G_{1}, \ldots, G_{n} \in \mathscr{A}\) such that \(\bigcup_{i=1}^{n} G_{i}=X\) and \(\mu\left(G_{i}ight)>0\) for all \(i=1,2, \ldots, n\). Then

\[\mathbb{E}^{\mathscr{G}} u=\sum_{i=1}^{n}\left[\int_{G_{i}} u(x) \frac{\mu(d x)}{\mu\left(G_{i}ight)}ight] \mathbb{1}_{G_{i}}\]

Remark. The measure \(\mathbb{1}_{G_{i}} \mu / \mu\left(G_{i}ight)=\mu\left(\cdot \cap G_{i}ight) / \mu\left(G_{i}ight)\) is often called the conditional probability given \(G_{i}\).

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