I am a bit confused about variable elimination in a Bayes Network. For example, in the following Bayes network, if I want to get P(D|B), can I directly eliminate E, F and G since they are irrelevant? So that I don't have to join them to get factors?
2026-03-25 20:40:11.1774471211
Bayesian Network Variable Elimination
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Yes. You do not need to factorise the decendents of D, since they will sum to unity by the law of total probability.
Let us do it anyway to show this.
$$\begin{align}\mathsf P(\mathrm D\mid\mathrm B)&=\sum_{A,C,E,F,G}\mathsf P(A)\,\mathsf P(C\mid A,\mathrm B)\,\mathsf P(\mathrm D\mid C)\,\mathsf P(E\mid\mathrm D)\,\mathsf P(F\mid\mathrm D)\,\mathsf P(G\mid F)\\&=\sum_{A}\mathsf P(A)\,\sum_C\mathsf P(C\mid A,\mathrm B)\,\mathsf P(\mathrm D\mid C)\,\sum_E\mathsf P(E\mid\mathrm D)\,\sum_F\mathsf P(F\mid\mathrm D)\,\sum_G\mathsf P(G\mid F)\\&=\sum_{A}\mathsf P(A)\,\sum_C\mathsf P(C\mid A,\mathrm B)\,\mathsf P(\mathrm D\mid C)\,\sum_E\mathsf P(E\mid\mathrm D)\,\sum_F\mathsf P(F\mid\mathrm D)\cdot 1\\&=\sum_{A}\mathsf P(A)\,\sum_C\mathsf P(C\mid A,\mathrm B)\,\mathsf P(\mathrm D\mid C)\,\sum_E\mathsf P(E\mid\mathrm D)\cdot 1\\&=\sum_{A}\mathsf P(A)\,\sum_C\mathsf P(C\mid A,\mathrm B)\,\mathsf P(\mathrm D\mid C)\cdot 1\end{align}$$