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## PowerPoint Slideshow about 'CS498-EA Reasoning in AI Lecture #5' - carney

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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?)
- Give me your answer; NO IMPACT on your final score in this class

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:OR 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 treesNoisy 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 AssumptionAncestor

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

Radio

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

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)

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