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This question is in reference to the equation 3.2.26 that is given as

$$ P(B,A,G,W) = P(A|B)\,P(B)\,P(W|A)\,P(G|A) $$

I would have thought that is should be

$$ P(B,A,G,W) = P(A|B)\,P(B)\,P(W|A)\,P(G|A)\,P(A) $$

Why isn't there a $P(A)$ there? Is my assumption that you always multiply by the chance of the evidence occurring wrong? Or is it the assumption that P(A) = 1, because Dr Watson said so?

BN = Belief network

Tharuka Devendra
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    I slightly edited your question. Please note some of our users may be visually impaired so they wouldn't be able to read the formulas pasted as a picture. Also notice that this site enables users to use $\TeX$ formatting for formulas. – Tim Sep 11 '17 at 17:43
  • Hi Tim, thanks. I'm not really sure how to use TEX on here even though I know the language, is there a quick guide for implementation? – Tharuka Devendra Sep 11 '17 at 17:44
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    Just put your latex stuff within dollar signs (single for in line, double for display) and you're good to go. Click edit to see how Tim pulled it off for some examples. – Matthew Drury Sep 11 '17 at 17:45
  • You can always ask on http://tex.stackexchange.com and you can find multiple tutorials online. For formula to display put it between $ for inlien formula, e.g. $E(x)$ is $E(X)$, or between $$, e.g. $$E(X)$$ would be a formula in it's own line. – Tim Sep 11 '17 at 17:48
  • Btw, drawing a graph would help, e.g. https://stats.stackexchange.com/questions/215034/bayesian-errors-in-variables-model-definition-in-jags-and-symbolically – Tim Sep 11 '17 at 18:18

1 Answers1

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The $P(A)$ you're looking for is already in the formula, in the form of $P(A|B)P(B)$. The Bayesian network has A dependent on B, and W and G both dependent on A. We can get the overall probability of A using the chain rule to compute $P(A|B)P(B)$. Then, the chain rule is applied again to get $P(W)$ and $P(G)$ from $P(W|A)P(G|A)P(A)$

Nuclear Hoagie
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