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Having the following Bayesian Network:

A -> B, A -> C, B -> D, B -> F, C -> F, C -> G

$$\begin{array}{l} A&\to&B&\to& D\\\downarrow &&\downarrow \\ C&\to&F\\\downarrow\\G \end{array}$$

With the following probabilities:

$$P(+a)=...$$ $$P(+a|+b)=..., P(+a|¬b)=...$$ $$P(+b|+a)=..., P(+b|¬a)=...$$ $$P(+d|+b)=..., P(+d|¬b)=...$$ $$P(+f|+b,+c)=..., P(+f|¬b,+c)=..., P(+f|¬b,¬c)=...$$ $$P(+g|+c)=..., P(+g|¬c)=...$$

I have to compute this $P(+d, +f, \neg g)$, that I think is:

$$P(+d, +f, \neg g) = P(+a, +d, +f, \neg g) + P(\neg a, +d, +f, \neg g).$$

My question is: how can I compute each addend?

I think is: $$P(+a, +d, +f, \neg g) = P(+a)·P(+d|+b)·P(+f|+b,+c)·P(\neg g,+c)$$

But I'm using $b$ and $c$ that there aren't in $P(+a, +d, +f, \neg g)$.

NOTE: This question is related to this one: Calculate probability using brute-force method.

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1 Answer 1

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You need to marginalize all the variables from the joint distribution: $$p(a,d,f,g) = \sum_{b,c}p(a,b,c,d,f,g) = \sum_{b,c} p(a)p(b|a)p(d|b)p(f|b,c)p(g|c).$$

You plug in $+d$, $+f$, and $\neg g$ and you sum for all combinations of $\pm b$ and $\pm c$. When you plug in $+ a$ you get $p(+a,+d,+f,\neg g)$. When you plug in $\neg a$, you get $p(\neg a, +d,+f,\neg g)$. Finally, you sum these two together and you get the required marginal: $$p(+d,+f,\neg g) =\sum_{a,b,c}p(a)p(b|a)p(c|a)p(+d|b)p(+f|b,c)p(\neg g|c).$$ The sum now runs over all combinations of $\pm a$, $\pm b$, and $\pm c$. Hope this helps!

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    $\begingroup$ You missed a term $$\mathsf P({+}d,{+}f,{\lnot}g)=\sum_{a,b,c} \mathsf P(a)~\mathsf P(b\mid a)~\mathsf P(c\mid a)~\mathsf P({+}d\mid b)~\mathsf P({+}f\mid b, c)~\mathsf P({\lnot} g\mid c)$$ $\endgroup$
    – Graham Kemp
    May 6, 2019 at 6:27
  • $\begingroup$ Yes indeed, corrected! $\endgroup$
    – Riccardo Sven Risuleo
    May 6, 2019 at 6:56
  • $\begingroup$ Thanks, but if I'm trying to compute $P(a| +d, +f, \neg g) = \frac{P(a,+d,+f,\neg g)}{P(+d,+f,\neg g)}$, and both terms, the numerator and denominator will be the same term. Numerator: $P(+a, +d, +f, \neg g) = \sum_{b,c} P(+a, b, c +d, +f, \neg g)$. Denominator: $P(+a, +d, +f, \neg g) = \sum_{a,b,c} P(a, b, c +d, +f, \neg g),$ I will get that $P(a| +d, +f, \neg g) = 1$, isn't it? $\endgroup$
    – VansFannel
    May 6, 2019 at 8:32
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    $\begingroup$ No, at the numerator the sum only runs over $b$ and $c$; at the denominator it runs over $a$, $b$, and $c$. $\endgroup$
    – Riccardo Sven Risuleo
    May 6, 2019 at 8:35
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    $\begingroup$ Basically $\mathsf P(+A\mid \text{theRest})=\dfrac{\mathsf P(+A,\text{theRest})}{\mathsf P(+A,\text{theRest})+\mathsf P(\lnot A,\text{theRest})}$ $\endgroup$
    – Graham Kemp
    May 6, 2019 at 8:43

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