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[Solved]: Bayesian Network - Inference

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Problem Detail: 

I have the following Bayesian Network and need help with answering the following query.

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Here are my solutions to questions a and b:


P(A,B,C,D,E) = P(A) * P(B) * P(C | A, B) * P(D | E) * P(E | C) 


P(a, ¬b, c ¬d, e) = P(a) * P(¬b) * P(c | a, b) * P(¬d | ¬b) * P(e | c)  = 0.02 * 0.99 * 0.5 * 0.99 * 0.88 = 0.0086 


P(e | a, c, ¬b)

This is my attempt:

a ×  ∑ P(a, ¬b, c, D = d, e) =      d   a × ∑  { P(a) * P(¬b) * P(c | a, b) * P(d) * P(e | c) + P(a) * P(¬b) * P(c | a,b)    *P(¬d)      d                                                                  + P(e | c) } 

Note that a is the alpha constant and that a = 1/ P(a,¬b, c)

The problem I have is that I don't know how to compute the constant a that the sum is multiplied by. I would appreciate help because I'm preparing for an exam and have no solutions available to this old exam question.

Asked By : mrjasmin

Answered By : D.W.

You're on the right path. Here's my suggestion. First, apply the definition of conditional probability:

$$ \Pr[e|a,c,\neg b] = {\Pr[e,a,c,\neg b] \over \Pr[a,c,\neg b]}. $$

So, your job is to compute both $\Pr[e,a,c,\neg b]$ and $\Pr[a,c,\neg b]$. I suggest that you do each of them separately.

To compute $\Pr[a,\neg b,c,e]$, it is helpful to notice that

$$ \Pr[a,\neg b,c,e] = \Pr[a,\neg b,c,d,e] + \Pr[a,\neg b,c,\neg d,e]. $$

So, if you can compute terms on the right-hand side, then just add them up and you've got $\Pr[a,\neg b,c,e]$. You've already computed $\Pr[a,\neg b,c,\neg d,e]$ in part (b). So, just use the same method to compute $\Pr[a,\neg b,c,d,e]$, and you're golden.

Another way to express the last relation above is to write

$$ \Pr[a,\neg b,c,e] = \sum_d \Pr[a,\neg b,c,D=d,e]. $$

If you think about it, that's exactly the same equation as what I wrote, just using $\sum$ instead of $+$. You can think about whichever one is easier for you to think about.

Anyway, now you've got $\Pr[e,a,c,\neg b]$. All that remains is to compute $\Pr[a,c,\neg b]$. You can do that using exactly the same methods. I'll let you fill in the details: it is a good exercise. Finally, plug into the first equation at the top of my answer, and you're done.

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