$ P\left(E_e\right)=\sum _{\eta =1}^{\mathbb{H}} P\left(H_{\eta }\right) P\left(E_e|H_{\eta }\right) $ How to prove?

Question

Is it that true?

If yes, how to prove this?

$$ P\left(E_e\right)=\sum _{\eta =1}^{\mathbb{H}} P\left(H_{\eta }\right) P\left(E_e|H_{\eta }\right), $$

where $E_e$ is an generic evidence, $H_\eta$ it's the $\eta$-esim hyphotesis and $\mathbb{H}$ is the cardinality of the hypothesis ($H$) set.

Answer 1

The formula you quote seems to be just the law of total probability. Assume that the set of events $\{ H_\eta \}_{1 \leq \eta \leq \mathbb H}$ forms a partition of the sample space $\Omega$; i.e., the $H_\eta$'s are pairwise disjoint, and $\bigcup \limits_{\eta = 1}^{\mathbb H} H_\eta = \Omega$.

Now for any event $E_e$, the set of events $\{ E_e \cap H_\eta \}$ forms a partition of $E_e$. Therefore, by additivity, we have $$ P(E_e) = \sum_{\eta = 1}^{\mathbb H} P(E_e \cap H_\eta). \tag{1} $$ Now, by the definition of conditional probability, we have $P(E_e \cap H_\eta) = P(H_\eta) \cdot P(E_e \mid H_\eta)$. Plugging this in $(1)$ we get the claim.*

The following sentence taken from the wikipedia article explains what this theorem means intuitively (notation changed to match ours):

The summation can be interpreted as a weighted average, and consequently the marginal probability, $P(E_e)$, is sometimes called "average probability"; "overall probability" is sometimes used in less formal writings.

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*The formula is true even for $\{ H_\eta \}_{\eta \geq 1}$ forms a countably infinite partition of $\Omega$. The proof has to be modified only slightly for this.