I am going through the presentation slides on this link: https://www.cs.cmu.edu/~tom/10701_sp11/recitations/Recitation_4.pdf

On page 11, above chain rule, they have written the following distribution:

P(F,A,S,H,N) = P(F) P(F|A) P(S|F,A) P(H|S,F,A) P(N|S,F,A,H)
  1. From where did they get this? Especially P(F|A) - aren't they independent?
  2. Why is it P(H|S,F,A) and not P(H|S,F,A,N) (like it is for the last term)?

Graph is here:

enter image description here

  • $\begingroup$ Second term on the right hand side should be P(A/F) and not P(F/A). $\endgroup$
    – Zahava Kor
    May 3, 2017 at 17:10
  • $\begingroup$ @ZahavaKor, the site mentions P(F|A) - and I again have no idea why! :( $\endgroup$ May 3, 2017 at 17:18
  • $\begingroup$ It's clearly a typo! It happens to the best of us... It has to be P(A/F) to be mathematically correct. $\endgroup$
    – Zahava Kor
    May 3, 2017 at 20:25

1 Answer 1


It's a general statement. Any joint distribution (likelihood or probability) can be written in term of a product of conditionals:

$$P(X_1, \ldots, X_n) = P(X_1) \cdot P(X_2 | X_1) \cdot \ldots \cdot P(X_n | X_{n-1}, \ldots, X_1)$$

Flu and allergy are not probabalistically independent because of Bayes Rule. They are causally independent though. If I know that someone has an allergy, then I can predict whether or not they have sinus (infection) and knowing this tells me something about the risk of flu since P(Flu | sinus) = p(sinus|flu) * p(flu)/p(sinus) and p(sinus) depends on allergy.

You get P(H|S,F,A) and don't condition whether or not there's a nose problem so you can obtain the correct marginal probability for headache problems. If both headache and nose condition on all 4 other variables, you will never be able to marginalize those likelihoods/probability models.

  • $\begingroup$ Could you please elaborate: Flu and allergy are not probabalistically independent because of Bayes Rule? They don't have any connection between them, so shouldn't they be independent? $\endgroup$ May 3, 2017 at 16:47
  • $\begingroup$ Also, if I understand your last paragraph correctly, then we do this just so that we can avoid a deadlock, wherein one depends on the other and vice versa, correct? $\endgroup$ May 3, 2017 at 16:50
  • $\begingroup$ @user6490375 reread the argument about Bayes. No no concept of a deadlock at all. It may help you to remove the arrowheads, remembering Bayes rule. $\endgroup$
    – AdamO
    May 3, 2017 at 16:57

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