Stochastic MethodsA 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:

BA

A

BA

BA

Rules for Sets

Addition rule– This can be generalized for N sets

Multiplication Principle– If we have two sets, A and B, the number of

unique ways the elements can be combined is |A| x |B|

Cartesian Product of two set, A and B:

CACACA

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

nPr = n!/(n-r)!– The permutation of n objects taken r at a time,

duplicates allowed nr

Combination of a set of n elements is any subset that can be formed– The combination of n elements taken r at a

time nCr = 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

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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) TotalsRed (R) 5 8 13Yellow (y) 3 4 7Totals 8 12 20

Let S designate the sample space.Then |S| = 20If one bulb is selected at random, the probability that the

bulb will be red is: P(R) = 13/20Now, we want to know the probability of grabbing a red bulb

given that it blooms in April.

P(R|A) = 5/8 = =

=

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 ifP(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,

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). 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| = 52C5 = 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| = 13C1 * 4C4 * 48C1 = 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| = 24 = 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

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