Handling Uncertainty

Uncertain knowledge Typical example: Diagnosis. Consider data instances about patients: Can we certainly derive the diagnostic rule: if Toothache=true then Cavity=true ? –The problem is that this rule isn’t right always. Not all patients with toothache have cavities; some of them have gum disease, an abscess, …: We could try turning the rule into a causal rule: if Cavity=true then Toothache=true –But this rule isn’t necessarily right either; not all cavities cause pain. NameToothache…Cavity Smithtrue… Miketrue… Maryfalse…true Quincytrue…false …………

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 rules. 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.

Syntax Basic element: random variable – corresponds to an “attribute” of the data. Boolean random variables e.g., Cavity (do I have a cavity?) Discrete random variables e.g., Weather is one of 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)

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 abbrev. P(cavity) = 0.1) P(Weather = sunny) = 0.7(or abbrev. 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) = (they sum up to 1)

Conditional probability E.g., P(cavity | toothache) = 0.8 i.e., probability of cavity given that toothache is all I know It can be interpreted as the probability that the rule if Toothache=true then Cavity=true holds. 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)

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) Useful for assessing diagnostic probability from causal probability as: P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect) 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.

Applying Bayes’ rule 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) P(StiffNeck=true | Meningitis=true) = 0.5 P(Meningitis=true) = 1/50000 P(StiffNeck=true) = 1/20 P(Meningitis=true | StiffNeck=true) = P(StiffNeck=true | Meningitis=true) P(Meningitis=true) / P(StiffNeck=true) = (0.5) x (1/50000) / (1/20) = That is, we expect only 1 in 5000 patients with a stiff neck to have meningitis. This is still a very small chance. Reason is a very small apriori probability. Also, observe that P(Meningitis=false | StiffNeck=true) = P(StiffNeck=false | Meningitis=false) P(Meningitis=false) / P(StiffNeck=true) 1/ P(StiffNeck=true) is the same. It is called the normalization constant (denoted with  ).

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 the Bayes’ rule?! Unfortunately, diagnostic knowledge is more fragile than causal knowledge. Imagine, there is sudden epidemic of meningitis. The prior probability, P( Meningitis=true ), will go up. –The doctor who derives the diagnostic probability P(Meningitis=true|StiffNeck=true) 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 from the other three values will see that P(Meningitis=true|StiffNeck=true) goes up proportionally with P(Meningitis=true). Clearly, P( StiffNeck=true|Meningitis=true ) is unaffected by the epidemic. It simply reflects the way meningitis works.

Bayes’ rule -- more vars Although the effect 1 might not be independent of effect 2, it might be that given the cause they are independent. E.g. effect 1 is ‘abilityInReading’ effect 2 is ‘lengthOfArms’ There is indeed a dependence of abilityInReading to lengthOfArms. People with longer arms read better than those with short arms…. However, given the cause ‘Age’ the abilityInReading is independent on lengthOfArms.

Conditional Independence – Naive Bayes Two assumptions: –Attributes are equally important –Conditionally independent (given the class value) This means that knowledge about the value of a particular attribute doesn’t tell us anything about the value of another attribute (if the class is known) Although based on assumptions that are almost never correct, this scheme works well in practice!

Weather Data Here we don’t really have effects, but rather evidence.

Naïve Bayes for classification Classification learning: what’s the probability of the class given an instance? –Evidence E = instance or E 1,E 2,…, E n –Event H = class value for instance Naïve Bayes assumption: evidence can be split into independent parts (i.e. attributes of instance!) P(H | E 1,E 2,…, E n ) = P(E 1 |H) P(E 2 |H)… P(E n |H) P(H) / P(E 1,E 2,…, E n )

The weather data example P(play=yes | E) = P(Outlook=Sunny | play=yes) * P(Temp=Cool | play=yes) * P(Humidity=High | play=yes) * P(Windy=True | play=yes) * P(play=yes) / P(E) = = (2/9) * (3/9) * (9/14) / P(E) = / P(E) Don’t worry for the 1/P(E); It’s alpha, the normalization constant.

The weather data example P(play=no | E) = P(Outlook=Sunny | play=no) * P(Temp=Cool | play=no) * P(Humidity=High | play=no) * P(Windy=True | play=no) * P(play=no) / P(E) = = (3/5) * (1/5) * (4/5) * (3/5) * (5/14) / P(E) = / P(E)

Normalization constant P(play=yes, E) + P(play=no, E) = P(E) i.e. P(play=yes, E)/P(E) + P(play=no, E)/P(E) = 1 i.e. P(play=yes | E) + P(play=no | E) = 1 i.e / P(E) / P(E) = 1 i.e. P(E) = So, P(play=yes | E) = / ( ) = 20.5% P(play=no | E) = / ( ) = 79.5% E play=yes play=no 20.5% 79.5%

The “zero-frequency problem” What if an attribute value doesn’t occur with every class value (e.g. “Humidity = High” for class “Play=Yes”)? –Probability P(Humidity=High|play=yes) will be zero! A posteriori probability will also be zero! –No matter how likely the other values are! Remedy: –Add 1 to the count for every attribute value-class combination (Laplace estimator); –Add k (no of possible attribute values) to the denumerator. (see example on the right). P(play=yes | E) = P(Outlook=Sunny | play=yes) * P(Temp=Cool | play=yes) * P(Humidity=High | play=yes) * P(Windy=True | play=yes) * P(play=yes) / P(E) = = (2/9) * (3/9) * (3/9) * (3/9) *(9/14) / P(E) = / P(E) It will be instead: = ((2+1)/(9+3)) * ((3+1)/(9+3)) * ((3+1)/(9+2)) * ((3+1)/(9+2)) *(9/14) / P(E) = / P(E) Number of possible values for ‘Outlook’ Number of possible values for ‘Windy’

Missing values Training: instance is not included in frequency count for attribute value-class combination Classification: attribute will be omitted from calculation Example: P(play=yes | E) = P(Temp=Cool | play=yes) * P(Humidity=High | play=yes) * P(Windy=True | play=yes) * P(play=yes) / P(E) = = (3/9) * (3/9) * (3/9) *(9/14) / P(E) = / P(E) P(play=no | E) = P(Temp=Cool | play=no) * P(Humidity=High | play=no) * P(Windy=True | play=no) * P(play=no) / P(E) = = (1/5) * (4/5) * (3/5) *(5/14) / P(E) = / P(E) After normalization: P(play=yes | E) = 41%, P(play=no | E) = 59%

Dealing with numeric attributes Usual assumption: attributes have a normal or Gaussian probability distribution (given the class). Probability density function for the normal distribution is: We approximate  by the sample mean: We approximate    by the sample variance:

Weather Data f(temperature=66 | yes) =e(- ((66-m)^2 / 2*var) ) / sqrt(2*3.14*var) m = ( )/ 9 = 73 var = ( (83-73)^2 + (70-73)^2 + (68-73)^2 + (64-73)^2 + (69- 73)^2 + (75-73)^2 + (75-73)^2 + (72-73)^2 + (81-73)^2 )/ (9-1) = 38 f(temperature=66 | yes) =e(- ((66-73)^2 / (2*38) ) ) / sqrt(2*3.14*38) =.034

Statistics for the weather data

Classifying a new day A new day E: P(play=yes | E) = P(Outlook=Sunny | play=yes) * P(Temp=66 | play=yes) * P(Humidity=90 | play=yes) * P(Windy=True | play=yes) * P(play=yes) / P(E) = = (2/9) * (0.0340) * (0.0221) * (3/9) *(9/14) / P(E) = / P(E) P(play=no | E) = P(Outlook=Sunny | play=no) * P(Temp=66 | play=no) * P(Humidity=90 | play=no) * P(Windy=True | play=no) * P(play=no) / P(E) = = (3/5) * (0.0291) * (0.0380) * (3/5) *(5/14) / P(E) = / P(E) After normalization: P(play=yes | E) = 20.9%, P(play=no | E) = 79.1%

Probability densities Relationship between probability and density: But: this doesn’t change calculation of a posteriori probabilities because  cancels out Exact relationship:

Discussion of Naïve Bayes Naïve Bayes works surprisingly well (even if independence assumption is clearly violated) Why? Because classification doesn’t require accurate probability estimates as long as maximum probability is assigned to correct class However: adding too many redundant attributes will cause problems (e.g. identical attributes) Note also: numeric attributes might not be normally distributed

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(Word i | Category). For each category c, P(Category=c) is estimated as the fraction of all the “training” documents that are of that category. Similarly, P(Word i = true | Category = c) is estimated as the fraction of documents of category that contain word. Also, P(Word i = 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 classifying a new document: P(Category = c | Word 1 = true, …, Word n = true) =  *P(Category = c)  n i=1 P(Word i = true | Category = c) P(Category =  c | Word 1 = true, …, Word n = true) =  *P(Category =  c)  n i=1 P(Word i = true | Category =  c) Word 1, …, Word n are the words occurring in the new document  is the normalization constant. Observe that similarly with the “missing values” the new documents doesn’t contain every word for which we computed the probabilities.