Stochastic Methods

1
Stochastic Methods

• A Review

2
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

3
Problem Areas

• Diagnostic reasoning
• Natural language understanding
• Planning and scheduling
• Learning

4
Pascal Laplace

• Developed probabalistic 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 Christs crown.

5
Sets

• 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 B is a subset of set A if 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
• Intersetcion
• The intersection of sets A and B is the set of
  all elements that are elements of both sets
• Denoted

6
Rules for Sets

• Addition rule
• Cartesian Product
• Multiplication Principle

7
Permutations and Combinations

• Permutation of elements in a set is an arranged
  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!)

8
Elements of Probability Theory

• Elementary Event
• An occurrence that cannot be made up of other
  events
• Event, E
• Set of elementary events
• Sample Space, S
• The set of all possible outcomes of an event E
• Probability, p
• The ratio of the cardinality of E to that of S
• p(E) E / S and

9
Union
10
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 late or early
• Early(E) Late(L) Totals
• Red (R) 5 8 13
• Yellow (y) 3 4 7
• Totals 8 12 20

11

• If one bulb is selected at random, the
  probability that the bulb will be red is P(R)
  13/20
• Now, we want to know the probability of grabbing
  a red bulb given that it is early.
• P(RE) 5/8 R E/E
• (R E/S) / (E/S)
• P(R E)/P(E)

12
Conditional Probability

• Definition
• For any two events, with p(B) gt 0, the
  conditional probability of A given that B has
  occurred is
• P(AB) P(A B)/P(B)

13
Independence

• Definition
• Two events A and B are independent if p(AB)
  p(A)
• The second piece follows
• p(BA) p(A B)/p(A) (def. of cond.prob)
• p(AB) p(B)/p(A)
• p(B)

14
Theorem

• A and B are independent if and only if
• P(A B) p(A) p(B)
• Proof (left to right)
• Since p(A B) p(AB) p(B) and since A and
  B are independent
• P(A B) p(A) p(B)
• Proof (right to left)
• Since p(A B) p(A) p(B)
• And p(A B) p(AB) p(B)
• Then P(A) p(AB) so A and B meet the definition
  of independence

15
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 8/36
  0
• .2222

16
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 all ways to pick four cards of
  the same type
• R is the set of all remaining cards
• So
• E T W R
• E 13C1 4C4 48C1 13 1 48 624
• So, p(E) E/S .00024

17
Example 3

• Recall that 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
• So A and B are independent