Handling Uncertainty

1
Handling Uncertainty
2
Uncertain knowledge

• Typical example Diagnosis. Consider
• ?x Symptom(x, Toothache) ? Disease(x, Cavity).
• The problem is that this rule is wrong. Not all
  patients with toothache have cavities some of
  them have gum disease, an abscess,
• ?x Symptom(x, Toothache) ? Disease(x, Cavity) ?
  Disease(x, GumDisease) ? Disease(x, Abscess) ?
  ...
• Unfortunately, in order to make the rule true, we
  have to add an almost unlimited list of possible
  causes.
• We could try turning the rule into a causal rule
• ?x Disease(x, Cavity) ? Symptom(x, Toothache).
• But this rule isnt right either not all
  cavities cause pain. We need to make it logically
  exhaustive to augment the left side with all the
  qualifications required for a cavity to cause
  toothache.

3
So, using FOL fails

• In a domain like medical diagnosis, using FOL
  fails because of
• Laziness
• Too much work to list the complete set of
  antecedents or consequents to ensure an
  exceptionless rule.
• Too hard to use such rules.
• Theoretical Ignorance
• Medical science hasnt complete theory for the
  domain.
• Practical Ignorance
• Even if we know all the rules, we might be
  uncertain about a particular patient because not
  all necessary tests have been done

4
Belief and Probability

• The connection between toothaches and cavities is
  not a logical consequence in either direction.
• However, we can provide a degree of belief on the
  sentences. Our main tool for this is probability
  theory.
• E.g. We might not know for sure what afflicts a
  particular patient, but we believe that there is,
  say, an 80 chance that is probability 0.8
  that the patient has cavity if he has a
  toothache.
• We usually get this belief from statistical data.
• Assigning probability 0 to a sentence correspond
  to an unequivocal belief that the sentence is
  false.
• Assigning probability 1 to a sentence correspond
  to an unequivocal belief that the sentence is
  true.

5
Syntax

• Basic element random variable
• Similar to propositional logic possible worlds
  defined by assignment of values to random
  variables.
• Boolean random variables
• e.g., Cavity (do I have a cavity?)
• Discrete random variables
• e.g., Weather is one of ltsunny,rainy,cloudy,snowgt
• Domain values must be exhaustive and mutually
  exclusive
• Elementary proposition constructed by assignment
  of a value to a random variable e.g., Weather
  sunny, Cavity false
  (abbreviated as ?cavity)
• Complex propositions formed from elementary
  propositions and standard logical connectives
  e.g., Weather sunny ? Cavity false

6
Atomic events

• Atomic event A complete specification of the
  state of the world about which the agent is
  uncertain
• E.g., if the world consists of only two Boolean
  variables Cavity and Toothache, then there are 4
  distinct atomic events
• Cavity false ? Toothache false
• Cavity false ? Toothache true
• Cavity true ? Toothache false
• Cavity true ? Toothache true
• Atomic events are mutually exclusive and
  exhaustive

7
Axioms of probability

• For any propositions A, B
• 0 P(A) 1
• Necessarily true (i.e. valid) propositions have
  probability 1, and necessarily false (i.e.
  unsatisfiable) propositions have probability 0.
• P(true) 1 and P(false) 0
• P(A ? B) P(A) P(B) - P(A ? B)

8
Using the axioms of probability

• We can derive a variety of useful facts from
  basic axioms. E.g.
• P(a ? ?a) P(a) P(?a) P(a ? ?a) (by axiom
  3)
• P(true) P(a) P(?a) P(a ? ?a) (by logical
  equivalence)
• 1 P(a) P(?a) (by axiom 2)
• P(?a) 1 - P(a) (by algebra)
• Also, we can prove that for discrete variable D
  with domain ltd1,,dngt we have
• ?ni1P(Ddi) 1.

9
Prior probability and distribution

• Prior or unconditional probability associated
  with a proposition is the degree of belief
  accorded to it in the absence of any other
  information.
• e.g., P(Cavity true) 0.1 (or P(cavity)
  0.1)
• P(Weather sunny) 0.7 (or P(sunny)
  0.7)
  
• Probability distribution gives values for all
  possible assignments
• P(Weather sunny) 0.7
• P(Weather rain) 0.2
• P(Weather cloudy) 0.08
• P(Weather snow) 0.02
• As a shorthand we can use a vector notation as
• P(Weather) lt0.7, 0.2, 0.08, 0.02gt (they sum
  up to 1)

10
Joint probability

• Joint probability distribution for a set of
  random variables gives the probability of every
  atomic event on those random variables.
• E.g. for two random variables Weather and Cavity
  we have we have
  
• P(Weather,Cavity), which is a 4 2 matrix of
  values
  
• Weather sunny rainy cloudy snow
• Cavity true 0.144 0.02 0.016 0.02
• Cavity false 0.576 0.08 0.064 0.08
• We can consider the joint probability
  distribution of all the variables we use to
  describe the world. Such joint probability
  distribution is called full joint probability
  distribution.
• A full joint distribution specifies the
  probability of every atomic event.
• Any probabilistic question about a domain can be
  answered by the full joint distribution.
  

11
Conditional probability

• Conditional or posterior probabilities
• e.g., P(cavity toothache) 0.8
• i.e., given that toothache is all I know
• Notation for conditional distributions
• P(Cavity Toothache) is a 2-element vector of
  2-element vectors

12
Conditional probability

• Definition of conditional probability
• P(a b) P(a ? b) / P(b) if P(b) gt 0
• Product rule gives an alternative formulation
• P(a ? b) P(a b) P(b) P(b a) P(a)
• A general version holds for whole distributions,
  e.g.,
  
• P(Weather,Cavity) P(Weather Cavity) P(Cavity)
• i.e. shorthand for
• P(sunny ? cavity) P(sunny cavity) P(cavity)
• P(rainy ? cavity) P(rainy cavity) P(cavity)
• (View it as a set of 4 2 equations, not matrix
  multiplication)
  

13
Chain rule

• Chain rule is derived by successive application
  of product rule

14
Inference by enumeration

• Start with the joint probability distribution
• For any proposition f, sum up the probabilities
  of the atomic events where it is true
• P(f) S??f P(?)

15
Inference by enumeration
16
Inference by enumeration
17
Inference by enumeration

• Can also compute conditional probabilities

18
Normalization

• Denominator can be viewed as a normalization
  constant a and we can write in vector notation

19
Inference by enumeration, contd

• Typically, we are interested in
• the posterior joint distribution of the query
  variables Y
• given specific values e for the evidence
  variables E
  
• Let the hidden variables be H X - Y - E
• Then the required summation of joint entries is
  done by summing out the hidden variables
  
• P(Y E e) aP(Y,E e) aShP(Y,E e, H h)
• The terms in the summation are joint entries
  because Y, E and H together exhaust the set of
  random variables
  
• Obvious problems
• Worst-case time complexity O(dn) where d is the
  largest arity
  
• Space complexity O(dn) to store the joint
  distribution
  
• How to find the numbers for O(dn) entries?

20
Independence

• Lets add a fourth variable, Weather.
• The full joint probability distribution
• P(Toothache, Catch, Cavity, Weather)
• has 32 entries (because Weather has 4 values)!!
• It contains 4 editions of the previous table, one
  for each kind of weather.
• Naturally, we ask what relationship these
  editions have to each other and to the original
  table?
• E.g. how are
• P(toothache, catch, cavity, cloudy) and
• P(toothache, catch, cavity) related? Lets use
  the product rule
• P(toothache, catch, cavity, cloudy)
  P(cloudy toothache, catch, cavity)
  P(toothache, catch, cavity)
• Of course, ones dental problems dont influence
  the weather, so
• P(cloudy toothache, catch, cavity) P(cloudy)

21
Independence (contd)

• So, we can write
• P(toothache, catch, cavity, cloudy)
  P(cloudy toothache, catch, cavity)
  P(toothache, catch, cavity)
• P(cloudy) P(toothache, catch, cavity)
• Thus, the 32 element table for four variables can
  be constructed from one 8-element table and one
  4-element table!!
• This property is called independece.
• A and B are independent iff
• P(AB) P(A) or P(BA) P(B) or P(A, B)
  P(A) P(B)
  
• Absolute independence powerful but rare
• Dentistry is a large field with hundreds of
  variables, none of which are independent.

22
Bayes' Rule

• Product rule
• P(a?b) P(a b) P(b) P(b a) P(a)
• Bayes' rule P(a b) P(b a) P(a) / P(b)
• or in vector form
• P(YX) P(XY) P(Y) / P(X) a P(XY) P(Y)
• Useful for assessing diagnostic probability from
  causal probability
  
• P(CauseEffect) P(EffectCause) P(Cause) /
  P(Effect)
  

23
Applying Bayes rule

• Bayess rule is useful in practice because there
  are many cases where we do have good probability
  estimates for these three numbers and need to
  compute the fourth.
• For example,
• A doctor knows that the meningitis causes the
  patient to have a stiff neck 50 of the time.
• The doctor also knows some unconditional facts
• the prior probability that a patient has
  meningitis is 1/50,000, and
• the prior probability that any patient has a
  stiff neck is 1/20.

24
Bayes rule (contd)

• Let  s  be the proposition that the patient has a
  stiff neck
•            m  be the proposition that the patient
  has meningitis,  
•             P(sm) 0.5
•             P(m) 1/50000
•             P(s) 1/20
•             P(ms) P(sm) P(m) / P(s) (0.5) x
  (1/50000) / (1/20) 0.0002
•                                               
• That is, we expect only 1 in 5000 patients with a
  stiff neck to have meningitis.

25
Bayes rule (contd)

• Well, we might say that doctors know that a stiff
  neck implies meningitis in 1 out of 5000 cases
• That is the doctor has quantitative information
  in the diagnostic direction from symptoms
  (effects) to causes.
• Such a doctor has no need for Bayes rule?!
• Unfortunately, diagnostic knowledge is more
  fragile than causal knowledge.
• Imagine, there is sudden epidemic of meningitis.
  The prior probability, P(m), will go up.
• The doctor who derives the diagnostic probability
  P(ms) from his statistical observations of
  patients before the epidemic will have no idea
  how to update the value.
• The doctor who derives the diagnostic
  probability P(ms) from the other three values
  will see that P(ms) goes up proportionally with
  P(m).
• Clearly, P(sm) is unaffected by the epidemic. It
  simply reflects the way meningitis works.

26
Difficulty with more than two vars
27
Conditional independence

• P(Toothache, Cavity, Catch) has 23 8
  independent entries
  
• If I have a cavity, the probability that the
  probe catches in it doesn't depend on whether I
  have a toothache
  
• (1) P(catch toothache, cavity) P(catch
  cavity)
• The same independence holds if I haven't got a
  cavity
  
• (2) P(catch toothache,?cavity) P(catch
  ?cavity)
• Catch is conditionally independent of Toothache
  given Cavity
  
• P(Catch Toothache,Cavity) P(Catch Cavity)
• Equivalent statements
• P(Toothache Catch, Cavity) P(Toothache
  Cavity)
  
• P(Toothache, Catch Cavity) P(Toothache
  Cavity) P(Catch Cavity)
  

28
Conditional independence contd.

• Write out full joint distribution using chain
  rule
  
• P(Toothache, Catch, Cavity)
• P(Toothache Catch, Cavity) P(Catch, Cavity)
• P(Toothache Catch, Cavity) P(Catch Cavity)
  P(Cavity)
  
• P(Toothache Cavity) P(Catch Cavity)
  P(Cavity)
  
• In most cases, the use of conditional
  independence reduces the size of the
  representation of the joint distribution from
  exponential in n to linear in n.
• Conditional independence is our most basic and
  robust form of knowledge about uncertain
  environments.
  

29
In general
30
Bayes' Rule and conditional independence

• P(Cavity toothache ? catch)
• aP(toothache ? catch Cavity) P(Cavity)
• aP(toothache Cavity) P(catch Cavity)
  P(Cavity)
• This is an example of a naïve Bayes model
• P(Cause,Effect1, ,Effectn) P(Cause)
  piP(EffectiCause)
  
• Total number of parameters is linear in n

31
Athens Example

• Suppose you are a witness to a nighttime
  hit-and-run accident involving a taxi in Athens.
• All taxis in Athens are blue or green.
• You swear, under oath, that the taxi was blue.
• Extensive testing shows that, under the dim
  lighting conditions, discrimination between blue
  and green is 75 reliable.
• 9 out of 10 Athenian taxis are green
• Whats most likely color for the taxi?
• Hint distinguish carefully between the
  proposition that the taxi is blue and the
  proposition that the taxi appears blue.

32
Athens Example (contd)

• Two random variables.
• B taxi was blue with domain b, ?b
• LB taxi looked blue with domain lb, ?lb
• The information on the reliability of color
  identification can be written as
• P(lb b) 0.75 P(?lb ?b) 0.75
• We need to know the probability that the taxi was
  blue, given that it looked blue.
• Then, we need to know the probability that the
  taxi wasnt blue, given that it looked blue.
  Lets use the Bayes rule
• P(b lb) ?P(lb b) P(b) ? 0.75 0.1 ?
  0.075
• P(?b lb) ?P(lb ?b) P(?b)
• ?(1 - P(?lb ?b)) (1 - P(b))
• ?(1 - 0.75) (1 0.1) ? 0.25 0.9 ?
  0.225
• Hence, P(B lb) lt ?0.075, ?0.225gt. So, even
  if the witness has seen a blue it is more
  probable that the taxi was green.
• ? 1/P(lb) 1/( P(b lb) P(?b lb) )
  1/(0.075 0.225)

33
Text Categorization

• Text categorization is the task of assigning a
  given document to one of a fixed set of
  categories, on the basis of the text it contains.
  
• Naïve Bayes models are often used for this task.
• In these models, the query variable is the
  document category, and the effect variables are
  the presence or absence of each word in the
  language.
• How such a model can be constructed, given as
  training data a set of documents that have been
  assigned to categories?
• The model consists of the prior probability
  P(Category) and the conditional probabilities
  P(Wordi Category).
• For each category c, P(Categoryc) is estimated
  as as the fraction of all the training
  documents that are of that category.
• Similarly, P(Wordi true Category c) is
  estimated as the fraction of documents of
  category that contain word.
• Also, P(Wordi true Category ?c) is
  estimated as the fraction of documents not of
  category that contain word.

34
Text Categorization (contd)

• Now we can use naïve Bayes for each c
• P(Category c Word1 true, , Wordn true)
  
• ?P(Category c)?ni1 P(Wordi true
  Category c)
• P(Category ?c Word1 true, , Wordn true)
  
• ?P(Category ?c)?ni1 P(Wordi true
  Category ?c)
• where ? is the normalization constant.