In Trifonov et al. (2013), which uses a bayesian framework to identify prioritized gene lesions, a composite posterior is defined as:
$$P(D|S) \quad \underline{deff} \quad \sum_{i}P(D|M_i) \cdot P(M_i|S) \tag{1}$$
I am trying to derive this posterior from the simple posteriors defined previously in the article, i.e.:
$$P(D|M)= \frac{\delta(D \in M) \cdot P^d(D)}{\sum_{G∈M}P^d(G)}\tag{2}$$
$$P(M|S)= \frac{\delta(M \in S)\cdot P^d(M)}{\sum_{j}P^d(M_j)}\tag{3}$$
My derivation is the following:
$$P(D|S) \quad \underline{deff} \quad \sum_{i}P(D|M_i)·P(M_i|S)\tag{4}$$
Inserting the actual posteriors:
$$P(D|S)= \sum_{i} \frac{\delta(D \in M_i) \cdot P^d(D)}{\sum_{G \in M_i} P^d(G)} \cdot \frac{\delta(M \in S) \cdot P^d(M_i)}{\sum_{j} P^d(M_j)}\tag{5}$$
If $D \in M_i$ then $D \in \bigcup S$ because $M_i \in S$, then:
$$P(D|S) = (\delta(D \in \bigcup S) \cdot P^d(D)) \cdot \sum_{i} \frac{ P^d(M_i)}{\sum_{G \in M_i} P^d(G) \cdot \sum_{j} P^d(M_j)}\tag{6}$$
Same as the previous step, since $M_i \in S$ the first sum at the denominator can be defined with respect to $\bigcup S$:
$$P(D|S) = \frac{\delta(D \in \bigcup S) \cdot P^d(D)}{\sum_{G \in \bigcup S} P^d(G)} \sum_{i} \frac{P^d(M_i)}{\sum_{j} P^d(M_j)}\tag{7}$$
At this point I'm stuck. The original article continues:
To obtain the global posterior $P^g(D|S)$ we assume that $P^d(M) \propto \sum_{G \in M} P^d(G)$ in which case: $$P^g(D|S) = \frac{\delta(D \in \bigcup S) \cdot P^d(D)}{\sum_{G \in \bigcup S} P^d(G)}\tag{8}$$
What is the passage that lead to the elimination of $\sum_{i} \frac{P^d(M_i)}{\sum_{j} P^d(M_j)}$?
It was actually simple. The elimination of $\sum_{i}\frac{P^d(M_i)}{\sum_{j}P^d(M_j)}$ is straightforward once you see that both the numerator and denominator are the sum of the same elements.