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Apologies in advance if this is considered an easy topic. I am absolutely mired and feel so defeated.

I am working with the following Bayesian Network: enter image description here

I am being asked to compute the following: P(H, ~B, L, ~F, ~C) and P(F|L)

I do not know much of where to start. I have reviewed the following resources:

My attempts are below:

Calculating P(F|L)

I know that this is considered a Top Down approach, and thus need to perform the following:

  1. Rewrite the goal conditional probability of query variable Q in terms of Q and all of its parents (that are not evidence) given the evidence
  2. Re-express each joint probability back to the probability of Q given all of its parents
  3. Lookup values in the Bayesian Network

Therefore:

P(F|L) =

= P(F,L)/P(L)

= P(F,L,B)/P(L)+P(F,L,~B)/P(L) (Total Probability)

= P(F,B|L) + P(F, ~B|L)

= P(F|B,L)P(B|L) + P(F|~B, L)P(~B|L) (Condtionalized Chain Rule)

= P(F|B,L)P(B) + P(F|~B, L)P(~B) (Independence)

But I cannot see how that relates back to the graph. I also have no clue where to start for P(H, ~B, L, ~F, ~C)

Thank you in advance.

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1 Answer 1

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$H$ has no parents. $H$ is the parent of $B$. $H$ is the parent of $L$. $B$ and $L$ are the parents of $F$. $L$ is the parent of $C$.

$\def\P{\operatorname{\sf P}}$Thus the factorisation for $\P(H,\neg B, L, \neg F, \neg C)$ is $$\begin{align}\P(H,\neg B, L, \neg F, \neg C)~&=~\mathsf P(H)\P(\neg B\mid H)\P(L\mid H)\P(\neg F\mid \neg B, L)\P(\neg C\mid L)\end{align}$$

Likewise the factorisation for $\P(F\mid L)$ is (by total probability)

$$\begin{align}\P(F\mid L)&=\P(F,B\mid L)+\P(F,\neg B\mid L)\\&=\P(F\mid B,L)\P(B\mid L)+\P(F\mid \neg B, L)\P(\neg B\mid L)\end{align}$$

Which you have correct until here. However, $B$ and $L$ are not independent. They are conditionally independent given $H$.

$\begin{align}\P(B\mid L)&= \dfrac{\P(B\mid H)\P(L\mid H)\P(H)+\P(B\mid\neg H)\P(L\mid\neg H)\P(\neg H)}{\P(L\mid H)\P(H)+\P(L\mid\neg H)\P(\neg H)}\end{align}$

and likewise

$\begin{align}\P(\neg B\mid L)&= \dfrac{\P(\neg B\mid H)\P(L\mid H)\P(H)+\P(\neg B\mid\neg H)\P(L\mid\neg H)\P(\neg H)}{\P(L\mid H)\P(H)+\P(L\mid\neg H)\P(\neg H)}\end{align}$

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  • $\begingroup$ Isn't P(F,L) different than P(F|L)? Additionally -- I have a lurking suspicion I am unable to truly compute these probabilities with the network given, and that I can only factor it down to a certain point, as you have done, and that I will not end up with a true value. Is that suspicion correct? (I believe the professor has provided an incorrect graphic). $\endgroup$
    – user615962
    Mar 26, 2020 at 2:30
  • $\begingroup$ Yes, they are different. By definition of conditional probability, and the DAG factorisation. $$\begin{align}\mathsf P(F\mid L)&=\dfrac{\mathsf P(F,L)}{\mathsf P(L)}\\&=\dfrac{\sum_B\mathsf P(F,B,L)}{\mathsf P(L)}\\&=\dfrac{\sum_B \mathsf P(F\mid B,L)~\mathsf P(B\mid L)\mathsf P(L)}{\mathsf P(L)}\\&=\sum_B \mathsf P(F\mid B,L)~\mathsf P(B\mid L)\end{align}$$ $\endgroup$
    – Graham Kemp
    Mar 26, 2020 at 2:57
  • $\begingroup$ The graph is correct, but on it, $H$ is $h_1$ and $\neg H$ is $h_2$, and so forth. Everything you need is there. (Remember complementation $P(b_2\mid h_1)=1-P(b_1\mid h_1)$) $\endgroup$
    – Graham Kemp
    Mar 26, 2020 at 3:01

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