New version page

Rutgers University CS 440 - Independencies in Bayesian Networks

Documents in this Course
Load more

This preview shows page 1-2-3 out of 9 pages.

View Full Document
View Full Document

End of preview. Want to read all 9 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document
Unformatted text preview:

Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)P(C|S,P) = P(C|S,P,E) = P(C|S,P,B) = P(C|S,P,E,B) = P(C|S,P,E,B,A) = P(C|S,P,A) = P(C|S,P,E,A) etc.Property 1: Each variable is conditionally independent of its non-descendants, given all its immediate parents.F(fatigue)B(low blood pressure)Because S and P are parents of C, and E and B and A are not descendants of CA(Asthma)S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)P(C|S,P) ≠ P(C|S,P,O) ≠ P(C|S,P,F) ≠ P(C|S,P,O,F)Property 1: Each variable is conditionally independent of its non-descendants, given all its immediate parents.F(fatigue)B(low blood pressure)HoweverBecause O and F are descendants of CA(Asthma)S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)P(C|S,P) ≠ P(C|S,P,O) ≠ P(C|S,P,F) ≠ P(C|S,P,O,F)Property 1: Each variable is conditionally independent of its non-descendants, given all its immediate parents.F(fatigue)B(low blood pressure)HoweverBecause O and F are descendants of CA(Asthma)Example: Suppose P(o|c,e) = 0P(o|c,¬e) = 0.7P(o|¬c, e) = 0P(o|¬c, ¬e) = 0Then what is P(c|s,p,o)?S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)P(C|S,P) ≠ P(C|S,P,O) ≠ P(C|S,P,F) ≠ P(C|S,P,O,F)Property 1: Each variable is conditionally independent of its non-descendants, given all its immediate parents.F(fatigue)B(low blood pressure)HoweverBecause O and F are descendants of CA(Asthma)Example: Suppose P(o|c,e) = 0P(o|c,¬e) = 0.7P(o|¬c, e) = 0P(o|¬c, ¬e) = 0P(c|s,p,o) = 1, because o has a zero chance of happening if C=false (according to the four probabilities above). Since O=true is given as evidence, then C must be true.S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)P(C|S,P) ≠ P(C|S,P,O) ≠ P(C|S,P,F) ≠ P(C|S,P,O,F)Property 1: Each variable is conditionally independent of its non-descendants, given all its immediate parents.F(fatigue)B(low blood pressure)HoweverBecause O and F are descendants of CA(Asthma)Example: Suppose P(o|c,e) = 0P(o|c,¬e) = 0.7P(o|¬c, e) = 0P(o|¬c, ¬e) = 0We can then see that P(C|S,P,O) can be different from P(C|S,P) (which can be anything), despite the fact that all parents of C are given in the evidence. This is because O is a descendent (consequence) of C, its value can change our belief about C.S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)P(C|S,P) ≠ P(C|S,P,O) ≠ P(C|S,P,F) ≠ P(C|S,P,O,F)Property 1: Each variable is conditionally independent of its non-descendants, given all its immediate parents.F(fatigue)B(low blood pressure)HoweverBecause O and F are descendants of CA(Asthma)Example: Suppose P(o|c,e) = 0P(o|c,¬e) = 0.7P(o|¬c, e) = 0P(o|¬c, ¬e) = 0A similar example can be made for P(C|S,P,F) and P(C|S,P) by just making F copy the values of O (whichever true or false value O takes, F takes the same value with probability 1).S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)Property 2: Each variable is conditionally independent of all other variables in the network, given all its immediate parents, children, and children's parents (Markov Blanket)F(fatigue)B(low blood pressure)A(Asthma)P(C|S,P,O,E) = P(C|S,P,O,E,F) = P(C|S,P,O,E,B) = P(C|S,P,O,E,F,B) = P(C|S,P,O,E,Anything in the world)S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)Property 2: Each variable is conditionally independent of all other variables in the network, given all its immediate parents, children, and children's parents (Markov Blanket)F(fatigue)B(low blood pressure)A(Asthma)Why do we have to include the other parents of the children?Why isn’t P(C|S,P,O) = P(C|S,P,O,E)?Example: Suppose P(o|c,e) = 0.7P(o|c,¬e) = 0.9P(o|¬c, e) = 0P(o|¬c, ¬e) = 0.7We can clearly see that P(c|o,e) = 1,because O=true couldn’t occur when E=true unless C=true (exercise eliminates apnoea unless there is a cancer).S(Smoking)P(Pollution)C(Cancer)O (obstructive sleep apnoea)E (Exercise)Property 2: Each variable is conditionally independent of all other variables in the network, given all its immediate parents, children, and children's parents (Markov Blanket)F(fatigue)B(low blood pressure)A(Asthma)Why do we have to include the other parents of the children?Why isn’t P(C|S,P,O) = P(C|S,P,O,E)?Example: Suppose P(o|c,e) = 0.7P(o|c,¬e) = 0.9P(o|¬c, e) = 0P(o|¬c, ¬e) = 0.7That is why all the parents of a variable’s children also need to be included in the evidence to completely isolate the variable from the rest of the network. Independence properties 1 and 2 can save us a lot of time by eliminating irrelevant variables from the network, given a query variable and a set of evidence variables (that can be either the variable’s entire Markov blanket, in which case everything else can be removed from the net, or just all its parents, in which case all non-descendent evidence variables can be


View Full Document
Loading Unlocking...
Login

Join to view Independencies in Bayesian Networks and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Independencies in Bayesian Networks and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?