Probability Theory Bayes Theorem and Naïve Bayes classification

1. Probability Theory Bayes Theorem and Naïve Bayes classification

2. Definition of Probability • Probability theory encodes our knowledge or belief about the collective likelihood of the outcome of an event. • We use probability theory to try to predict which outcome will occur for a given event. Slide adapted from Dan Jurafsky's

3. Sample Spaces • We think about the “sample space” as being set of all possible outcomes: • For tossing a coin, the possible outcomes are Heads or Tails. • For competing in the Olympics, the set of outcomes for a given contest are {gold, silver, bronze, no_award}. • For computing part-of-speech, the set of outcomes are {JJ, DT, NN, RB, … etc.} • We use probability theory to try to predict which outcome will occur for a given event. Slide adapted from Dan Jurafsky's

4. Axioms of Probability Theory • Probabilities are real numbers between 0­1 representing the a priori likelihood that a proposition is true. • Necessarily true propositions have probability 1, necessarily false have probability 0. P(true) = 1 P(false) = 0 Slides adapted from Mary Ellen Califf

5. Probability Axioms • Probability law must satisfy certain properties: • Nonnegativity • P(A) >= 0, for every event A • Additivity • If A and B are two disjoint events, then the probability of their union satisfies: • P(A U B) = P(A) + P(B) • Normalization • The probability of the entire sample space S is equal to 1, i.e., P(S) = 1. P(Cold) = 0.1 (the probability that it will be cold) P(¬Cold) = 0.9 (the probability that it will be not cold) Slide adapted from Dan Jurafsky's

6. An example • An experiment involving a single coin toss • There are two possible outcomes, Heads and Tails • Sample space S is {H,T} • If coin is fair, should assign equal probabilities to 2 outcomes • Since they have to sum to 1 • P({H}) = 0.5 • P({T}) = 0.5 • P({H,T}) = P({H})+P({T}) = 1.0 Slide adapted from Dan Jurafsky's

7. Another example • Experiment involving 3 coin tosses • Outcome is a 3-long string of H or T • S ={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} • Assume each outcome is equally likely (equiprobable) • “Uniform distribution” • What is probability of the event A that exactly 2 heads occur? • A = {HHT,HTH,THH} • P(A) = P({HHT})+P({HTH})+P({THH}) • = 1/8 + 1/8 + 1/8 • =3/8 Slide adapted from Dan Jurafsky's

8. Probability definitions • In summary: number of outcomes corresponding to event E P(E) = --------------------------------------------------------- total number of outcomes • Probability of drawing a spade from 52 well-shuffled playing cards: • 13/52 = ¼ = 0.25 Slide adapted from Dan Jurafsky's

9. How about non-uniform probabilities? An example: • A biased coin, • twice as likely to come up tails as heads, • is tossed twice • What is the probability that at least one head occurs? • Sample space = {hh, ht, th, tt} (h = heads, t = tails) • Sample points/probability for the event: • (each head has probability 1/3; each tail has prob 2/3): • ht 1/3 x 2/3 = 2/9 hh 1/3 x 1/3= 1/9 • th 2/3 x 1/3 = 2/9 tt 2/3 x 2/3 = 4/9 • Answer: 5/9 = 0.56 • (sum of weights in red since want outcome with at least one heads) • By contrast, probability of event for the unbiased coin = 0.75 Slide adapted from Dan Jurafsky's

10. Moving toward language • What’s the probability of drawing a 2 from a deck of 52 cards? • What’s the probability of a random word (from a random dictionary page) being a verb? Slide adapted from Dan Jurafsky's

11. Probability and part of speech tags • What’s the probability of a random word (from a random dictionary page) being a verb? • How to compute each of these? • All words = just count all the words in the dictionary • # of ways to get a verb: number of words which are verbs! • If a dictionary has 50,000 entries, and 10,000 are verbs…. P(V) is 10000/50000 = 1/5 = .20 Slide adapted from Dan Jurafsky's

12. Independence • Most things in life depend on one another, that is, they are dependent: • If I drive to SF, this may effect your attempt to go there. • It may also effect the environment, the economy of SF,etc. • If 2 events are independent, this means that the occurrence of one event makes it neither more nor less probable that the other occurs. • If I flip my coin, it won’t have an effect on the outcome of your coin flip. P(A,B)= P(A) ·P(B) if and only if A and B are independent. P(heads,tails) = P(heads) · P(tails) = .5 · .5 = .25 Note: P(A|B)=P(A) iff A,B independent P(B|A)=P(B) iff A,B independent Slide adapted from Dan Jurafsky's

13. Conditional Probability • A way to reason about the outcome of an experiment based on partial information • How likely is it that a person has a disease given that a medical test was negative? • In a word guessing game the first letter for the word is a “t”. What is the likelihood that the second letter is an “h”? • A spot shows up on a radar screen. How likely is it that it corresponds to an aircraft? Slide adapted from Dan Jurafsky's

14. Conditional Probability • Conditional probability specifies the probability given that the values of some other random variables are known. P(Sneeze | Cold) = 0.8 P(Cold | Sneeze) = 0.6 • The probability of a sneeze given a cold is 80%. • The probability of a cold given a sneeze is 60%. Slides adapted from Mary Ellen Califf

15. More precisely • Given an experiment, a corresponding sample space S, and the probability law • Suppose we know that the outcome is within some given event B • The first letter was ‘t’ • We want to quantify the likelihood that the outcome also belongs to some other given event A. • The second letter will be ‘h’ • We need a new probability law that gives us the conditional probability of A given B • P(A|B) “the probability of A given B” Slide adapted from Dan Jurafsky's

16. Joint Probability Distribution • The joint probability distribution for a set of random variables X1…Xn gives the probability of every combination of values P(X1,...,Xn) Sneeze ¬Sneeze Cold 0.08 0.01 ¬Cold 0.01 0.9 • The probability of all possible cases can be calculated by summing the appropriate subset of values from the joint distribution. • All conditional probabilities can therefore also be calculated • P(Cold |¬Sneeze) • BUT it’s often very hard to obtain all the probabilities for a joint distribution Slides adapted from Mary Ellen Califf

17. Conditional Probability Example • Let’s say A is “it’s raining”. • Let’s say P(A) in dry California is .01 • Let’s say B is “it was sunny ten minutes ago” • P(A|B) means • “what is the probability of it raining now if it was sunny 10 minutes ago” • P(A|B) is probably way less than P(A) • Perhaps P(A|B) is .0001 • Intuition: The knowledge about B should change our estimate of the probability of A. Slide adapted from Dan Jurafsky's

18. Conditional Probability • Let A and B be events • P(A,B) and P(A  B) both means “the probability that BOTH A and B occur” • p(B|A) = the probability of event B occurring given event A occurs • definition: p(A|B) = p(A  B) / p(B) • P(A, B) = P(A|B) * P(B) (simple arithmetic) • P(A, B) = P(B, A) (AND is symmetric) Slide adapted from Dan Jurafsky's

19. Bayes Theorem • We start with conditional probability definition: • So say we know how to compute P(A|B). What if we want to figure out P(B|A)? We can re-arrange the formula using Bayes Theorem:

20. Deriving Bayes Rule Slide adapted from Dan Jurafsky's

21. How to compute probabilities? • We don’t have the probabilities for most NLP problems • We can try to estimate them from data • (that’s the learning part) • Usually we can’t actually estimate the probability that something belongs to a given class given the information about it • BUT we can estimate the probability that something in a given class has particular values. Slides adapted from Mary Ellen Califf

22. Simple Bayesian Reasoning • If we assume there are n possible disjoint rhetorical zones, z1 … zn P(zi | w) = P(w | zi) P(zi) P(w) • Want to know the probability of the zone given the word. • Can count how often we see the word in a sentence from a given zone in the training set, and how often the zone itself occurs. • P(w| zi ) = number of times we see this zone with this word divided by how often we see the zone • This is the learning part. • Since P(w) is always the same, we can ignore it. • So now only need to compute P(w | zi) P(zi) Slides adapted from Mary Ellen Califf

23. Bayes Independence Example • Imagine there are diagnoses ALLERGY, COLD, and WELL and symptoms SNEEZE, COUGH, and FEVER Prob Well Cold Allergy P(d) 0.9 0.05 0.05 P(sneeze|d) 0.1 0.9 0.9 P(cough | d) 0.1 0.8 0.7 P(fever | d) 0.01 0.7 0.4 Slides adapted from Mary Ellen Califf

24. Bayes Independence Example • By assuming independence, we ignore the possible interactions. • If symptoms are: sneeze & cough & no fever: P(well | e) = (0.9)(0.1)(0.1)(0.99)/P(e) = 0.0089/P(e) P(cold | e) = (.05)(0.9)(0.8)(0.3)/P(e) = 0.01/P(e) P(allergy | e) = (.05)(0.9)(0.7)(0.6)/P(e) = 0.019/P(e) P(e) = .0089 + .01 + .019 = .0379 P(well | e) = .23 P(cold | e) = .26 P(allergy | e) = .50 • Diagnosis: allergy Slides adapted from Mary Ellen Califf

25. Kupiec et al. Feature Representation • Fixed-phrase feature • Certain phrases indicate summary, e.g. “in summary” • Paragraph feature • Paragraph initial/final more likely to be important. • Thematic word feature • Repetition is an indicator of importance • Uppercase word feature • Uppercase often indicates named entities. (Taylor) • Sentence length cut-off • Summary sentence should be > 5 words.

26. Training • Hand-label sentences in training set (good/bad summary sentences) • Train classifier to distinguish good/bad summary sentences • Model used: Naïve Bayes • Can rank sentences according to score and show top n to user.

27. Naïve Bayes Classifier • The simpler version of Bayes was: • P(B|A) = P(A|B)P(B)/P(A) • P(Sentence | feature) = P(feature | S) P(S)/P(feature) • Using Naïve Bayes, we expand the number of feaures by defining a joint probability distribution: • P(Sentence, f1, f2, … fn) = P(Sentence)P(fi| Sentence)/ P(fi ) • We learn P(Sentence) and P(fi| Sentence) in training • Test: we need to state P(Sentence | f1, f2, … fn) • P(Sentence| f1, f2, … fn) = P(Sentence, f1, f2, … fn) / P(f1, f2, … fn)

28. Probability of feature-value pair occurring in a source sentence which is also in the summary compression rate Probability that sentence s is included in summary S, given that sentence’s feature value pairs Probability of feature-value pair occurring in a source sentence Details: Bayesian Classifier • Assuming statistical independence:

29. Bayesian Classifier (Kupiec at el 95) • Each Probability is calculated empirically from a corpus • See how often each feature is seen with a sentence selected for a summary, vs how often that feature is seen in any sentence. • Higher probability sentences are chosen to be in the summary • Performance: • For 25% summaries, 84% precision From lecture notes by Nachum Dershowitz & Dan Cohen

30. How to compute this? • For training, for each feature f: • For each sentence s: • Is the sentence in the target summary S? Increment T • Increment F no matter which sentence it is in. • P(f|S) = T/N • P(f) = F/N • For testing, for each document: • For each sentence • Multiply the probabilities of all of the features occurring in the sentence times the probability of any sentence being in the summary (a constant). Divide by the probability of the features occurring in any sentence.

31. Learning Sentence Extraction Rules (Kupiec et al. 95) • Results • About 87% (498) of all summary sentences (568) could be directly matched to sentences in the source (79% direct matches, 3% direct joins, 5% incomplete joins) • Location was best feature at 163/498 = 33% • Para+fixed-phrase+sentence length cut-off gave best sentence recall performance … 217/498=44% • At compression rate = 25% (20 sentences), performance peaked at 84% sentence recall J. Kupiec, J. Pedersen, and F. Chen. "A Trainable Document Summarizer", Proceedings of the 18th ACM-SIGIR Conference, pages 68--73, 1995.

32. Language Identification

33. Language identification • Tutti gli esseri umani nascono liberi ed eguali in dignità e diritti. Essi sono dotati di ragione e di coscienza e devono agire gli uni verso gli altri in spirito di fratellanza. • Alle Menschen sind frei und gleich an Würde und Rechten geboren. Sie sind mit Vernunft und Gewissen begabt und sollen einander im Geist der Brüderlichkeit begegnen. Universal Declaration of Human Rights, UN, in 363 languages http://www.unhchr.ch/udhr/navigate/alpha.htm

34. Language identification • égaux • eguali • iguales • edistämään • Ü • ¿ • How to do determine, for a stretch of text, which language it is from?

35. Language Identification • Turns out to be really simple • Just a few character bigrams can do it (Sibun & Reynar 96) • Based on language models for sample languages

36. Language model basics • Unigram and bigram models • Evaluating N-gram language models • Perplexity • The intuition of perpexity is that given two probabilistic models, the better model is the one that better predicts new data (not used to train the model) • We can measure better prediction by comparing the probability the model assigns to the test data • The better probability will assign a higher probability

37. Perplexity minimazing perplexity is equivalent to maximazing the test set probability according to the language model

38. Entropy • X---a random variable with probability distribution p(x) Measures how predictive a given N-gram model is about what the next word could be

39. KL divergence (relative entropy) Basis of comparing two probability distributions

40. Language Identification • Turns out to be really simple • Just a few character bigrams can do it (Sibun & Reynar 96) • Used Kullback Leibler distance (relative entropy) • Compare probability distribution of the test set to those for the languages trained on • Smallest distance determines the language • Using special character sets helps a bit, but barely

41. Language Identification (Sibun & Reynar 96)

42. Confusion Matrix • A table that shows, for each class, which ones your algorithm got right and which wrong Gold standard Algorithm’s guess

44. Author Identification(Stylometry)

45. Author Identification • Also called Stylometry in the humanities • An example of a Classification Problem • Classifiers: • Decide which of N buckets to put an item in • (Some classifiers allow for multiple buckets)

46. The Disputed Federalist Papers • In 1787-1788, Jay, Madison, and Hamilton wrote a series of anonymous essays to convince the voters of New York to ratify the new U. S. Constitution. • Scholars have consensus that: • 5 authored by Jay • 51 authored by Hamilton • 14 authored by Madison • 3 jointly by Hamilton and Madison • 12 remain in dispute … Hamilton or Madison?

47. Author identification • Federalist papers • In 1963 Mosteller and Wallace solved the problem • They identified function words as good candidates for authorships analysis • Using statistical inference they concluded the author was Madison • Since then, other statistical techniques have supported this conclusion.

48. Function vs. Content Words High rates for “by” favor M, low favor H High rates for “from” favor M, low says little High rates for “to” favor H, low favor M

49. Function vs. Content Words No consistent pattern for “war”

50. Federalist Papers Problem Fung, The Disputed Federalist Papers: SVM Feature Selection Via Concave Minimization, ACM TAPIA’03