Uncertainty

1
Uncertainty

• Chapter 13

2
Outline

• Uncertainty
• Probability
• Syntax and Semantics
• Inference
• Independence and Bayes' Rule

3
Sources of Uncertainty

• Information is partial
• Information is not fully reliable.
• Representation language is inherently imprecise.
• Information comes from multiple sources and it is
  conflicting.
• Information is approximate
• Non-absolute cause-effect relationships exist

4
Basic Probability

• Probability theory enables us to make rational
  decisions.
• Which mode of transportation is safer
• Car or Plane?
• What is the probability of an accident?

5
Basic Probability Theory

• An experiment has a set of potential outcomes,
  e.g., throw a dice
• The sample space of an experiment is the set of
  all possible outcomes, e.g., 1, 2, 3, 4, 5, 6
• An event is a subset of the sample space.
• 2
• 3, 6
• even 2, 4, 6
• odd 1, 3, 5

6
Language of probability

• random variables Boolean or discrete
• e.g., Cavity (do I have a cavity?)
• e.g., Weather is one of ltsunny,rainy,cloudy
  ,snowgt
• Domain values must be exhaustive and mutually
  exclusive
• Elementary propositions
• e.g., Weather sunny, Cavity false
  (or ?cavity)
• Complex propositions formed from elementary
  propositions and standard logical connectives
  e.g., Weather sunny ? Cavity false
  

7
Language of probability

• 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

8
Axioms of probability

• For any propositions A, B
• 0 P(A) 1
• P(true) 1 and P(false) 0
• P(A ? B) P(A) P(B) - P(A ? B)

9
Prior probability

• Prior or unconditional probabilities of
  propositions
  
• e.g., P(Cavity true) 0.1
• P(Weather sunny) 0.72
• belief prior to arrival of any (new) evidence
• Notation for prior probability distribution
• E.g., suppose domain of Weather is sunny,
  rain, cloudy, snow
• We may write
• P(Weather) lt0.7, 0.2, 0.08,
  0.02gt
• (note we use bold P in this
  case)
• instead of
• P(Weathersunny) 0.7
• P(Weatherrain) 0.2

10
Prior probability

• Joint probability distribution for a set of
  random variables gives the probability of every
  atomic event on random variables
  
• P(Weather,Cavity) 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
• All questions about a domain can be answered by
  the full joint distribution
  

11

• Example
• 100 attempts are made to swim a length in 30
  secs. The swimmer succeeds on 20 occasions
  therefore the probability that a swimmer can
  complete the length in 30 secs is
• 20/100 0.2
• Failure 1-.2 or 0.8

12
Conditional probability

• Conditional or posterior probabilities
• e.g., P(cavity toothache) 0.8
• i.e., given that toothache is all I know
• If we know more, e.g., cavity is also given, then
  we have
  
• P(cavity toothache,cavity) 1
• New evidence may be irrelevant, allowing
  simplification, e.g.,
  
• P(cavity toothache, sunny)
• P(cavity toothache) 0.8

13
Conditional probability

• Definition of conditional probability
• P(a b) P(a ? b) / P(b) if P(b) gt 0
• The definition suggests that conditional
  probability can be computed from unconditional
  probabilities.
• Product rule joint probability in terms of cond.
  probability
• P(a ? b) P(a b) P(b) P(b a) P(a)
  

14
Probabilistic Reasoning

• Evidence
• What we know about a situation.
• Hypothesis
• What we want to conclude.
• Compute
• P( Hypothesis Evidence )

15
Credit Card Authorization

• E is the data about the applicant's age, job,
  education, income, credit history, etc,
• H is the hypothesis that the credit card will
  provide positive return.
• The decision of whether to issue the credit card
  to the applicant is based on the probability
  P(HE).

16
Medical Diagnosis

• E is a set of symptoms, such as, coughing,
  sneezing, headache, ...
• H is a disorder, e.g., common cold, SARS, flu.
• The diagnosis problem is to find an H (disorder)
  such that P(HE) is maximum.

17
How to Compute P(AB)?
B
A
18
Business Students

• Of 100 students completing a course, 20 were
  business major. 10 students received an A in the
  course, and 3 of these were business majors.
  Suppose A is the event that a randomly selected
  student got an A in the course, B is the event
  that a randomly selected student is a business
  major. What is the probability of A? What is the
  probability of A after knowing B is true?

19
Contd

• If you look at the picture on the last slide,
• you see clearly
• P(AB) 3/200.15
• More formally, you can also calculate it by
• P(AB) P(A,B)/P(B) 0.03/0.20.15

20
Inference by enumeration

• Start with the joint probability distribution
• For any proposition f, sum the atomic events
  where it is true P(f) S??f P(?)
  
• E.g.
• P(toothache) 0.1080.0120.0160.0640.2

21
Inference by enumeration

• Start with the joint probability distribution
• For any proposition f, sum the atomic events
  where it is true P(f) S??f P(?)
  
• E.g.
• P(toothache) 0.108 0.012 0.016
  0.064 0.2
  

22
Inference by enumeration

• Start with the joint probability distribution
• Can also compute conditional probabilities
• P(?cavity toothache) P(?cavity ? toothache)
• P(toothache)
• 0.0160.064
• 0.2
• 0.4

23
Normalization

• Denote 1/P(toothache) by a, which can be viewed
  as a normalization constant a for the
  distribution P(Cavity toothache), ensuring it
  adds up to 1.
• We thus write
• P(Cavity toothache) a
  P(Cavity,toothache)
• a P(Cavity,toothache,catch)
  P(Cavity,toothache,? catch)
• a lt0.108,0.016gt lt0.012,0.064gt
• a lt0.12,0.08gt lt0.6,0.4gt
• General idea compute distribution on query
  variable by fixing evidence variables and summing
  over hidden variables

24
Inference by enumeration

• 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
  

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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 distribution form
• P(YX) P(XY) P(Y) / P(X) aP(XY) P(Y)
• Useful for assessing diagnostic probability from
  causal probability
  
• P(CauseEffect) P(EffectCause) P(Cause) /
  P(Effect)
  
• E.g., let M be meningitis, S be stiff neck
• P(ms) P(sm) P(m) / P(s) 0.8 0.0001 / 0.1
  0.0008
  
• Note posterior probability of meningitis still
  very small!
  

26
Exercise
A patient takes a lab test and the result comes
back positive. The test has a false negative rate
of 2 and false positive rate of 3. Furthermore,
0.01 of the entire population have this
disease. What is the probability of disease if
we know the test result is positive? Some info
(below, d for disease, t for test pos. and -t for
negative) P(td) 0.98
P(t-d)0.03 P(d) 0.0001
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Rough calculation If 10000 people take the
test, we expect 1 to have disease and likely to
test positive. While the rest do not have the
disease, 300 of them will test positive anyway.
So, the chance of a positive test having disease
is 1/300, a very small number.
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More precisely, with P(td) 0.98,
P(t-d)0.03, P(d) 0.0001 we get P(dt)
P(td)P(d)/P(t)
by Bayes rule P(td)P(d)/P(t,d)P(t,-d)
summing out
P(td)P(d)/P(td)P(d) P(t-d)P(-d)
product rule 0.980.0001/0.980.00010.030.9999
0.00325
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Independence

• A and B are independent iff
• P(AB) P(A) or P(BA) P(B) or P(A, B)
  P(A) P(B)
  
• P(Toothache, Catch, Cavity, Weather)
• P(Toothache, Catch, Cavity) P(Weather)
• 32 entries reduced to 12
• Absolute independence is powerful but rare
• Dentistry is a large field with hundreds of
  variables, none of which are independent. What to
  do?
  

30
Conditional independence

• P(Toothache, Cavity, Catch) has 23 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)
  

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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)
  
• I.e., 2 2 1 5 independent numbers
• 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.
  

32
Naïve Bayes model

• This is an example of a naïve Bayes model
• P(Cause,Effect1, ,Effectn) P(Cause)piP(Effecti
  Cause)
  
• This is correct if Effects are all conditionally
  independent given Cause.
  
• Total number of parameters is linear in n

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Summary

• Probability is a rigorous formalism for uncertain
  knowledge
  
• Joint probability distribution specifies
  probability of every atomic event
• Queries can be answered by summing over atomic
  events
  
• For nontrivial domains, we must find a way to
  reduce the joint size
  
• Independence and conditional independence provide
  the tools