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Handling Uncertainty
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 isn’t 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.
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 hasn’t 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…
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.
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 <sunny,rainy,cloudy,snow>
• 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
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
Cavity = false
Cavity = true
Cavity = true




Toothache = false
Toothache = true
Toothache = false
Toothache = true
• Atomic events are mutually exclusive and exhaustive
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)
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
<d1,…,dn> we have
ni=1P(D=di) = 1.
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
P(Weather = sunny) = 0.7
(or P(cavity) = 0.1)
(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:
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 =
Cavity = true
Cavity = false
sunny
0.144
0.576
rainy
0.02
0.08
cloudy snow
0.016 0.02
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.
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), P(cavity | toothache)
•
P(Cavity
| Toothache)
is a 2-element
vector| 
oftoothache
2-element vectors
P(cavity
| toothache
), P(cavity
)
Conditional probability
• Definition of conditional probability:
•
P(a | b) = P(a  b) / P(b)
if P(b) > 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
Chain rule
• Chain rule is derived by successive application of product
rule:
Inference by enumeration
• Start with the joint probability distribution:
•
• For any proposition φ, sum up the probabilities of the atomic events
where it is true:
Inference by enumeration
Inference by enumeration
Inference by enumeration
• Can also compute conditional probabilities:
•
Normalization
• Denominator can be viewed as a normalization constant α and
we can write in vector notation:
Inference by enumeration, cont’d
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) = αP(Y,E = e) = αΣhP(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
•
Independence
• Let’s 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? Let’s use the product rule:
P(toothache, catch, cavity, cloudy) =
P(cloudy | toothache, catch, cavity) P(toothache, catch, cavity)
Of course, one’s dental problems don’t influence the weather, so
P(cloudy | toothache, catch, cavity) = P(cloudy)
Independence (cont’d)
• 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(A|B) = P(A)
or P(B|A) = P(B)
or P(A, B) = P(A) P(B)
• Absolute independence powerful but rare
•
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(Y|X) = P(X|Y) P(Y) / P(X) = α P(X|Y) P(Y)
• Useful for assessing diagnostic probability from causal probability:
•
– P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect)
–
Applying Bayes’ rule
• Bayes’s 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.
Bayes’ rule (cont’d)
Let s be the proposition that the patient has a stiff neck
m be the proposition that the patient has meningitis,
P(s|m) = 0.5
P(m) = 1/50000
P(s) = 1/20
P(m|s) = P(s|m) 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.
Bayes’ rule (cont’d)
• 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(m|s) 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(m|s) from the
other three values will see that P(m|s) goes up proportionally with
P(m).
• Clearly, P(s|m) is unaffected by the epidemic. It simply reflects the way
meningitis works.
Difficulty with more than two vars
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:
•
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.
In general…
Bayes' Rule and conditional
independence
P(Cavity | toothache  catch)
= αP(toothache  catch | Cavity) P(Cavity)
= αP(toothache | Cavity) P(catch | Cavity) P(Cavity)
• This is an example of a naïve Bayes model:
•
P(Cause,Effect1, … ,Effectn) = P(Cause)
π P(Effect |Cause)
i
i
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
• What’s 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.
Athens Example (cont’d)
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 wasn’t blue, given that it
looked blue. Let’s 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) = < *0.075, *0.225>. 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)
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(Category=c) 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.
Text Categorization (cont’d)
• Now we can use naïve Bayes for each c:
P(Category = c | Word1 = true, …, Wordn = true) =
*P(Category = c)ni=1 P(Wordi = true | Category = c)
P(Category = c | Word1 = true, …, Wordn = true) =
*P(Category = c)ni=1 P(Wordi = true | Category = c)
where  is the normalization constant.