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I'm trying to understand how to calculate the likelihood of a combination of binary variables given the network structure and probabilities seen here: Given a bayesian network as follows:

As far as I understand, the decomposition of the network would be as follows: $$P(a,b,c,d,e) = P(A) * P(B) * P(C|A, B) * P(D|C) * P(E|C)$$

However applying this to the data set A=0, B=0, C=0, D=0, E=1 (as in the first example) gives me $$= (1-0.1) * (1-0.2) * (1-0.2) * (1-0.2) * 0.8 = 0.368.$$

Have I mis-represented the equation, or am I mis-reading the probabilities data?

EDIT:
I think I've resolved it: given the data entry was complete (except for A & C) the probability of the dataset did not depend on the values of B, D,E. Therefore the probability of the data vector is P(A) * P(C|A, B). Is this correct?

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  • $\begingroup$ Take a look here how to write math on this page. And I would suggest you to write here also the original problem since many people do not like to go outside MSE to study your problem. $\endgroup$ – iiivooo Mar 18 '16 at 16:17
  • $\begingroup$ Improved as much as I could, I don't have enough rep to post a picture on MSE. $\endgroup$ – Miguel Boland Mar 18 '16 at 18:06
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A bit late, but...

You have the network factorisation correct, and so, letting $X$ be the event that one of the four pentuples happens

$$\begin{align}\mathsf P([0,0,0,0,1])&={\small\mathsf P(A{=}0)\,\mathsf P(B{=}0)\,\mathsf P(C{=}0\mid A{=}0,B{=}0)\,\mathsf P(D{=}0\mid C{=}0)\,\mathsf P(E{=}1\mid C{=}0)}\\ &= (1-0.1)\cdot(1-0.2)\cdot(1-0.2)\cdot(1-0.2)\cdot0.1\\&=0.04608\\\mathsf P([0,0,1,0,1])&= (1-0.1)\cdot(1-0.2)\cdot 0.2\cdot(1-0.9)\cdot 0.8\\&= 0.01152\\\mathsf P([1,0,0,0,1])&=0.1\cdot(1-0.2)\cdot (1-0.6)\cdot(1-0.2)\cdot0.1\\&=0.00256\\\mathsf P([1,0,1,0,1])&=0.1\cdot(1-0.2)\cdot0.6\cdot(1-0.9)\cdot0.8\\ &= 0.00384\end{align}$$

Thus $\mathsf P([0,0,0,0,1]\mid X)=\dfrac{0.04608}{0.04608+0.01152+0.00256+0.00384}=0.7200$

And thusly is the last column of the "data" table generated...

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