1 / 9

Bucket Elimination: A unifying framework for Probabilistic inference Rina Dechter

Bucket Elimination: A unifying framework for Probabilistic inference Rina Dechter. presented by Anton Bezuglov, Hrishikesh Goradia CSCE 582 Fall02 Instructor: Dr. Marco Valtorta. Contributions. For a Bayesian network, the paper presents algorithms for Belief Assessment

elewa
Télécharger la présentation

Bucket Elimination: A unifying framework for Probabilistic inference Rina Dechter

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Bucket Elimination: A unifying framework for Probabilistic inferenceRina Dechter presented by Anton Bezuglov, Hrishikesh Goradia CSCE 582 Fall02 Instructor: Dr. Marco Valtorta

  2. Contributions • For a Bayesian network, the paper presents algorithms for • Belief Assessment • Most Probable Explanation (MPE) • Maximum Aposteriori Hypothesis (MAP) • All of the above are bucket elimination algorithms.

  3. Belief Assessment • Definition • The belief assessment task of Xk = xk is to find • In the Visit to Asia example, the belief assessment problem answers questions like • What is the probability that a person has tuberculosis, given that he/she has dyspnea and has visited Asia recently ? • where k – normalizing constant

  4. Most Probable Explanation (MPE) • Definition • The MPE task is to find an assignment xo = (xo1, …, xon) such that • In the Visit to Asia example, the MPE problem answers questions like • What are the most probable values for allvariables such that a person doesn’t catch dyspnea ?

  5. Maximum Aposteriori Hypothesis (MAP) • Definition • Given a set of hypothesized variables A = {A1, …, Ak}, , the MAP taskis to find an assignment ao = (ao1, …, aok) such that • Inthe Visit to Asia example, the MAP problem answers questions like • What are the most probable values for a person having both lung cancerand bronchitis, given that he/she has dyspnea and that his/her X-ray is positive?

  6.        Ordering the Variables Method 1 (Minimum deficiency) Begin elimination with the node which adds the fewest number of edges 1. , ,  (nothing added) 2.  (nothing added) 3. , , ,  (one edge added) Method 2 (Minimum degree) Begin elimination with the nodewhich has the lowest degree 1. ,  (degree = 1) 2. , ,  (degree = 2) 3. , ,  (degree = 2)

  7. Elimination Algorithm for Belief Assessment P(| =“yes”, =“yes”) = X\ {} (P(|)* P(|)* P(|,)* P(|,)* P()*P(|)*P(|)*P()) Bucket : P(|)*P(), =“yes” Hn(u)=xnПji=1Ci(xn,usi) Bucket : P(|) Bucket : P(|,), =“yes” Bucket : P(|,) H(,) H() Bucket : P(|) H(,,) Bucket : P(|)*P() H(,,) Bucket : H(,) k-normalizing constant Bucket : H() H() *k P(| =“yes”, =“yes”)

  8. Elimination Algorithm for Most Probable Explanation Finding MPE = max ,,,,,,, P(,,,,,,,) MPE= MAX{,,,,,,,} (P(|)* P(|)* P(|,)* P(|,)* P()*P(|)*P(|)*P()) Bucket : P(|)*P() Hn(u)=maxxn (ПxnFnC(xn|xpa)) Bucket : P(|) Bucket : P(|,), =“no” Bucket : P(|,) H(,) H() Bucket : P(|) H(,,) Bucket : P(|)*P() H(,,) Bucket : H(,) Bucket : H() H() MPE probability

  9. Elimination Algorithm for Most Probable Explanation Forward part ’ = arg maxP(’|)*P() Bucket : P(|)*P() Bucket : P(|) ’ = arg maxP(|’) Bucket : P(|,), =“no” ’ = “no” Bucket : P(|,) H(,) H() ’ = arg maxP(|’,’)*H(,’)*H() Bucket : P(|) H(,,) ’ = arg maxP(|’)*H(’,,’) Bucket : P(|)*P() H(,,) ’ = arg maxP(’|)*P()* H(’,’,) Bucket : H(,) ’ = arg maxH(’,) Bucket : H() H() ’ = arg maxH()* H() Return: (’, ’, ’, ’, ’, ’, ’, ’)

More Related