How can I compute probability P(E|D) from Bayes Net

37 Views Asked by At
1

There are 1 best solutions below

0
On

The network encodes the factorisation:$$\def\P{\operatorname{\mathsf P}}\P(A,B,C,D,E,F)=\P(A)\P(B)\P(C\mid A,B)\P(D\mid C)\P(E\mid A)\P(F\mid B)$$

And therefore by the law of total probability: $$\require{cancel}\begin{align}\P(A,B,C,D,E)&=\P(A)\P(B)\P(C\mid A,B)\P(D\mid C)\P(E\mid A)\color{blue}{\cancelto 1{\color{silver}{\sum_{\dot F}\P(\dot F\mid B)}}}\\[2ex]\P(A,B,C,D)&=\P(A)\P(B)\P(C\mid A,B)\P(D\mid C)\color{blue}{\cancelto 1{\color{silver}{\sum_{\dot E}\P(\dot E\mid A)}}}\end{align}$$

So applying the Law of Total Probability (again):

$$\def\tilda#1{\overset\sim #1}\begin{align}\P(D,E)&=\sum_{\dot A,\dot B,\dot C}\P(\dot A)\P(E\mid\dot A)\P(\dot B)\P(\dot C\mid\dot A,\dot B)\P(D\mid\dot C)\\[2ex]\P(D)&=\sum_{\dot A,\dot B,\dot C}\P(\dot A)\P(\dot B)\P(\dot C\mid\dot A,\dot B)\P(D\mid\dot C)\end{align}$$

(With the usual understanding that $\small\raise{1.5ex}{\sum\limits_{\dot X}}$ represents $\small\raise{1.5ex}{\sum\limits_{\dot X\in\{X,{\tilda{X}}\}}}$ )

And so...