# Bayesian Network - Inference

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

EDITED:

Here are my solutions to questions a and b:

a)

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


b)

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


c)

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.

• What alpha constant? I don't see any alpha in the question or in your attempted answer.
– D.W.
Aug 19 '13 at 0:27
• @D.W. the alpha constant is donoted "a". Aug 20 '13 at 9:04
• OK, got it. Thanks. So where did you get those equations from? You are trying to calculate $P(e|a,c,\neg b)$, and then some equations just appear from nowhere. I suggest editing the question to show us your reasoning and why you got to those equations. For instance, why did you introduce the constant $a$ into those equations? (what you called alpha) Do you know how to compute the answer to parts (a) and (b)? That would be an excellent warmup; why don't you edit the question to show us how you solved parts (a) and (b)?
– D.W.
Aug 20 '13 at 12:43
• @D.W. Thanks for the feedback, the question is now edited :) Aug 20 '13 at 19:51

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.

• Thank you so much, I really appreciate your effort. Can you take a look at my answer below to see if it's correct ? Thanks once again. Aug 20 '13 at 20:51

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

$$\Pr[e,a,c,\neg b] = \sum_{d} Pr[a,\neg b,c, D = d, e] = Pr[a, \neg b, c, d, e] + Pr[a, \neg b, c, \neg d, e] =$$

$$= P(a) * P(\neg b) * P(c | a,\neg b) * P(d | \neg b) * P(e|c) +$$

$$P(a) * P(\neg b) * P(c | a,\neg b) * P(\neg d | \neg b) * P(e|c) =$$

$$..... = 0.0148$$

$$\Pr[a, \neg b, c] = P[a, \neg b, c, D, E] = \sum_{d} \sum_{e} P[a, \neg b, c, D= d, E = e] =$$

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

$$= / ........ / = C$$

Now we have that:

$$\Pr[e|a,c,\neg b] = {\Pr[e,a,c,\neg b] \over \Pr[a,c,\neg b]} = 0.0148 / C$$

• You'd better keep going, to compute $C$ and make sure you get to a final answer. Your final answer should be a number (not an expression involving variables like $C$).
– D.W.
Aug 20 '13 at 22:11
• But: Yes, it looks like you've got the right approach now. You've got the idea down. Well done!
– D.W.
Aug 20 '13 at 22:12
• The only reason that i didn't computed the final expression was of laziness, that's why I introduced the constant C. Thanks again ! Aug 21 '13 at 8:46