CS498-EA Reasoning in AI Lecture #5

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# CS498-EA Reasoning in AI Lecture #5 - PowerPoint PPT Presentation

CS498-EA Reasoning in AI Lecture #5. Instructor: Eyal Amir Fall Semester 2009. Last Time. Propositional Logic Inference in different representations CNF: SAT hard; small representation sometimes DNF: SAT easy; large representation OBDDs: SAT easy; large representation sometimes

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### CS498-EAReasoning in AILecture #5

Instructor: Eyal Amir

Fall Semester 2009

Last Time
• Propositional Logic
• Inference in different representations
• CNF: SAT hard; small representation sometimes
• DNF: SAT easy; large representation
• OBDDs: SAT easy; large representation sometimes
• NNF: SAT hard; fewest large representations
• Applications:
• Circuit and program verification; computational biology
Pop Quiz (5 min)
• Prove or disprove:
• Every CNF representation with n variables of propositional formulas takes O(2n) space to represent some propositional theories on n variables (hint: how many non-equivalent theories of n variables are there?)
Today
• Probabilistic graphical models
• Treewidth methods:
• Variable elimination
• Clique tree algorithm
• Applications du jour: Sensor Networks
Probability
• A sample space Omega (O) is a set of outcomes of a random experiment
• A probability P is a function from a sigma-field (e.g., all measurable subsets) A on O (the events) to [0,1].
• A random variable X is a function X:OR such that for all B Borel set in R, X-1(B) is in A.
Independent Random Variables
• Two variables X and Y are independent if
• P(X = x|Y = y) = P(X = x) for all values x,y
• That is, learning the values of Y does not change prediction of X
• If X and Y are independent then
• P(X,Y) = P(X|Y)P(Y) = P(X)P(Y)
• In general, if X1,…,Xp are independent, then P(X1,…,Xp)= P(X1)...P(Xp)
• Requires O(n) parameters
Conditional Independence
• Unfortunately, most of random variables of interest are not independent of each other
• A more suitable notion is that of conditional independence
• Two variables X and Y are conditionally independent given Z if
• P(X = x|Y = y,Z=z) = P(X = x|Z=z) for all values x,y,z
• That is, learning the values of Y does not change prediction of X once we know the value of Z
• notation: I ( X , Y | Z )
Marge

Homer

Lisa

Maggie

Bart

Example: Family trees

Noisy stochastic process:

Example: Pedigree

• A node represents an individual’sgenotype
• Modeling assumptions:
• Ancestors can effect descendants' genotype only by passing genetic materials through intermediate generations
Y1

Y2

X

Non-descendent

Markov Assumption

Ancestor

Parent

• We now make this independence assumption more precise for directed acyclic graphs (DAGs)
• Each random variable X, is independent of its non-descendents, given its parents Pa(X)
• Formally,I (X, NonDesc(X) | Pa(X))

Non-descendent

Descendent

Burglary

Earthquake

Alarm

Call

Markov Assumption Example
• In this example:
• I ( E, B )
• I ( B, {E, R} )
• I ( R, {A, B, C} | E )
• I ( A, R | B,E )
• I ( C, {B, E, R} | A)
X

Y

X

Y

I-Maps
• A DAG G is an I-Map of a distribution P if all Markov assumptions implied by G are satisfied by P

(Assuming G and P both use the same set of random variables)

Examples:

X

Y

Factorization
• Given that G is an I-Map of P, can we simplify the representation of P?
• Example:
• Since I(X,Y), we have that P(X|Y) = P(X)
• Applying the chain ruleP(X,Y) = P(X|Y) P(Y) = P(X) P(Y)
• Thus, we have a simpler representation of P(X,Y)