Stochastic Methods A Review (Mostly)

Relationship between Heuristic and Stochastic Methods  Heuristic and stochastic methods useful where –Problem does not have an exact solution –Full state space is too costly to search  In addition, stochastic methods are useful when –One samples an information base –Causal models are learned from the data

Problem Areas  Diagnostic reasoning  Natural language understanding  Speech recognition  Planning and scheduling  Learning

Pascal  Developed probabilistic techniques to develop a mathematical foundation for gambling  You might remember that Pascal abjured mathematics when in 1657 his niece was cured of a painful infection while being in the proximity of nuns who kissed a thorn from Christ’s crown.

Sets  Set –A set is an unordered collection of elements  Cardinality –Number of elements in a set –For a set A, its cardinality is denoted: |A|  Universe –The domain of interest –Denoted U  Complement of a set A –The set of all elements from U that are not part of A –Denoted  Subset –Set A is a subset of set B iff every element of A is also a an element of B –Denoted:  Union –The union of sets of A and B is the set of all elements of either set –Denoted:  Intersectcion –The intersection of sets A and B is the set of all elements that are elements of both sets –Denoted:

Rules for Sets

Permutations and Combinations  Permutation of elements is a unique sequence of elements in that set –The permutation of n elements taken r at a time, duplicates not allowed  n P r = n!/(n-r)! –The permutation of n objects taken r at a time, duplicates allowed  n r  Combination of a set of n elements is any subset that can be formed –The combination of n elements taken r at a time  n C r = n!/((n-r)! r!)

Elements of Probability Theory  Elementary Event –An occurrence that cannot be made up of other events  Event, E –Set of elementary events (e.g., two rolls of two fair dice)  Sample Space, S –The set of all possible outcomes of an event E –E.g. 2 … 12 for dice example  Probability, p –The ratio of the cardinality of E to that of S –p(E) = |E| / |S| and

Union Given that: Then

Intersection and Conditional Probability Suppose we have the following tulip table, telling us whether the tulip bulbs we have are red or yellow and bloom in April or May April(A)May(M)Totals Red (R)5813 Yellow (y)347 Totals81220

Conditional Probability  In general: For any two events, with p(B) > 0, the conditional probability of A given that B has occurred is: P(A|B) = P(A B)/P(B)

Independence  Two events are independent if the occurrence of one of them does not affect the probability of occurrence of the other  More formally, two events A and B are independent if: –p(A|B) = p(A) provided that p(B) > 0 –p(B|A) = p(B) provided that p(A) > 0

Theorem (intersection/independence) A and B are independent if and only if P(A B) = p(A) * p(B) Proof (if A and B are independent then …) We know: p(A B) = p(A|B) * p(B). Since A and B are independent, p(A|B) = p(A) So, So, P(A B) = p(A) * p(B) Proof (if p(A B) = p(A) * p(B) then …) We know that p(A B) = p(A|B) * p(B). We know that p(A B) = p(A|B) * p(B). So, p(A) = P(A|B) So, p(A) = P(A|B) So A is independent of B. The reverse can easily be demonstrated.

Example 1  What is the probability of rolling a 7 or an 11 using two fair dice? –Sample space is the cardinality of the cartesian product of the set of values from each die: namely, 36 –The subset of the cartesian product that can produce 7 is: A ={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} –The subset of the cartesian product that can produce 11 is: B = {(5,6),(6,5)} –|E| = |A| + |B| –p(E)=8/36 =.2222 –Another way is to use the probability of the union of sets –P(A U B) = p(A) + p(B) – p(A B) = 6/36 + 2/36 – 0 =.2222

Example 2  What is the probability of being dealt a four-of-a- kind hand in a five card poker hand?  S is the set of all five card hands. So, |S| = 52 C 5 = 2,598,960  E is the set of all four-of-a-kind hands  T is the set of card types  W is the set of ways to pick four of the same type  R is the set of all ways to pick 1 card from the remaining 48  So |E| = |T| * |W| * |R| |E| = 13 C 1 * 4 C 4 * 48 C 1 = 13 * 1 * 48 = 624  So, p(E) = |E|/|S| ≈.00024

Example 3  Recall (by the intersection/independence theorem) two events are independent if and only  Consider a situation where bit strings of length 4 are randomly generated.  Let A = the event of the bit strings containing an even number of 1s.  Let B = the event of the bit strings ending in 0.  Are A and B independent?  |S| = 2 4 = 16  A = {1111,1100,1010,1001,0110,0101,0011,0000}  |A| = 8  B = {1110,1100,1010,1000,0010,0100,0110,0000}  |B| = 8  P(A B) = |A B|/|S| = 4/16 =.25  P(A) * p(B) = 8/16 * 8/16 =.25  So A and B are independent