September BASIC TECHNIQUES IN STATISTICAL NLP Word prediction n-grams smoothing

September Statistical Methods in NLE Two characteristics of NL make it desirable to endow programs with the ability to LEARN from examples of past use: – VARIETY (no programmer can really take into account all possibilities) – AMBIGUITY (need to have ways of choosing between alternatives) In a number of NLE applications, statistical methods are very common The simplest application: WORD PREDICTION

September We are good at word prediction Stocks plunged this morning, despite a cut in interestStocks plunged this morning, despite a cut in interest rates by the Federal Reserve, as Wall Stocks plunged this morning, despite a cut in interest rates by the Federal Reserve, as Wall Street began ….

September Real Spelling Errors They are leaving in about fifteen minuets to go to her house The study was conducted mainly be John Black. The design an construction of the system will take more than one year. Hopefully, all with continue smoothly in my absence. Can they lave him my messages? I need to notified the bank of this problem. He is trying to fine out.

September Handwriting recognition From Woody Allen’s Take the Money and Run (1969) – Allen (a bank robber), walks up to the teller and hands her a note that reads. "I have a gun. Give me all your cash." The teller, however, is puzzled, because he reads "I have a gub." "No, it's gun", Allen says. "Looks like 'gub' to me," the teller says, then asks another teller to help him read the note, then another, and finally everyone is arguing over what the note means.

September Applications of word prediction Spelling checkers Mobile phone texting Speech recognition Handwriting recognition Disabled users

September Statistics and word prediction The basic idea underlying the statistical approach to word prediction is to use the probabilities of SEQUENCES OF WORDS to choose the most likely next word / correction of spelling error I.e., to compute For all words w, and predict as next word the one for which this (conditional) probability is highest. P(w | W 1 …. W N-1 )

September Using corpora to estimate probabilities But where do we get these probabilities? Idea: estimate them by RELATIVE FREQUENCY. The simplest method: Maximum Likelihood Estimate (MLE). Count the number of words in a corpus, then count how many times a given sequence is encountered. ‘Maximum’ because doesn’t waste any probability on events not in the corpus

September Maximum Likelihood Estimation for conditional probabilities In order to estimate P(w|W1 … WN), we can use instead: Cfr.: – P(A|B) = P(A&B) / P(B)

September Aside: counting words in corpora Keep in mind that it’s not always so obvious what ‘a word’ is (cfr. yesterday) In text: – He stepped out into the hall, was delighted to encounter a brother. (From the Brown corpus.) In speech: – I do uh main- mainly business data processing LEMMAS: cats vs cat TYPES vs. TOKENS

September The problem: sparse data In principle, we would like the n of our models to be fairly large, to model ‘long distance’ dependencies such as: – Sue SWALLOWED the large green … However, in practice, most events of encountering sequences of words of length greater than 3 hardly ever occur in our corpora! (See below) (Part of the) Solution: we APPROXIMATE the probability of a word given all previous words

September The Markov Assumption The probability of being in a certain state only depends on the previous state: P(Xn = Sk| X1 … Xn-1) = P(Xn = Sk|Xn-1) This is equivalent to the assumption that the next state only depends on the previous m inputs, for m finite (N-gram models / Markov models can be seen as probabilistic finite state automata)

September The Markov assumption for language: n-grams models Making the Markov assumption for word prediction means assuming that the probability of a word only depends on the previous n words (N-GRAM model)

September Bigrams and trigrams Typical values of n are 2 or 3 (BIGRAM or TRIGRAM models): P(W n |W 1 ….. W n-1 ) ~ P(W n |W n-2,W n-1 ) P(W 1,…W n ) ~ П P(W i | W i-2,W i-1 ) What bigram model means in practice: – Instead of P(rabbit|Just the other day I saw a) – We use P(rabbit|a) Unigram: P(dog) Bigram: P(dog|big) Trigram: P(dog|the,big)

September The chain rule So how can we compute the probability of sequences of words longer than 2 or 3? We use the CHAIN RULE: E.g., – P(the big dog) = P(the) P(big|the) P(dog|the big) Then we use the Markov assumption to reduce this to manageable proportions:

September Example: the Berkeley Restaurant Project (BERP) corpus BERP is a speech-based restaurant consultant The corpus contains user queries; examples include – I’m looking for Cantonese food – I’d like to eat dinner someplace nearby – Tell me about Chez Panisse – I’m looking for a good place to eat breakfast

September Computing the probability of a sentence Given a corpus like BERP, we can compute the probability of a sentence like “I want to eat Chinese food” Making the bigram assumption and using the chain rule, the probability can be approximated as follows: – P(I want to eat Chinese food) ~ P(I|”sentence start”) P(want|I) P(to|want)P(eat|to) P(Chinese|eat)P(food|Chinese)

September Bigram counts

September How the bigram probabilities are computed Example of P(I,I): – C(“I”,”I”): 8 – C(“I”): …. = 3437 – P(“I”|”I”) = 8 / 3437 =.0023

September Bigram probabilities

September The probability of the example sentence P(I want to eat Chinese food)  P(I|”sentence start”) * P(want|I) * P(to|want) * P(eat|to) * P(Chinese|eat) * P(food|Chinese) =.25 *.32 *.65 *.26 *.002 *.60 =

September Examples of actual bigram probabilities computed using BERP

September Visualizing an n-gram based language model: the Shannon/Miller/Selfridge method For unigrams: – Choose a random value r between 0 and 1 – Print out w such that P(w) = r For bigrams: – Choose a random bigram P(w| ) – Then pick up bigrams to follow as before

September The Shannon/Miller/Selfridge method trained on Shakespeare

September Approximating Shakespeare, cont’d

September A more formal evaluation mechanism Entropy Cross-entropy

September The downside The entire Shakespeare oeuvre consists of – 884,647 tokens (N) – 29,066 types (V) – 300,000 bigrams All of Jane Austen’s novels (on Manning and Schuetze’s website): – N = 617,091 tokens – V = 14,585 types

September Comparing Austen n-grams: unigrams In person shewasinferiorto 1-gramP(.) 1the.034the.034the.034the.034 2to.032to.032to.032to.032 3and.030and.030and.030 … 8was.015was.015 … 13she.011 … 1701inferior.00005

September Comparing Austen n-grams: bigrams In person shewasinferiorto 2-gramP(.|person)P(.|she)P(.|was)P(.inferior) 1and.099had.0141not.065to.212 2who.099was.122a.052 … 23she.009 … inferior0

September Comparing Austen n-grams: trigrams In person shewasinferiorto 3-gramP(.|In,person)P(.|person, she) P(.|she, was) P(.was, inferior) 1UNSEENdid.05not.057UNSEEN 2was.05very.038 … inferior0

September Maybe with a larger corpus? Words such as ‘ergativity’ unlikely to be found outside a corpus of linguistic articles More in general: Zipf’s law

September Zipf’s law for the Brown corpus

September Addressing the zeroes SMOOTHING is re-evaluating some of the zero- probability and low-probability n-grams, assigning them non-zero probabilities – Add-one – Witten-Bell – Good-Turing BACK-OFF is using the probabilities of lower order n- grams when higher order ones are not available – Backoff – Linear interpolation

September Add-one (‘Laplace’s Law’)

September Effect on BERP bigram counts

September Add-one bigram probabilities

September The problem

September The problem Add-one has a huge effect on probabilities: e.g., P(to|want) went from.65 to.28! Too much probability gets ‘removed’ from n- grams actually encountered – (more precisely: the ‘discount factor’

September Witten-Bell Discounting How can we get a better estimate of the probabilities of things we haven’t seen? The Witten-Bell algorithm is based on the idea that a zero-frequency N-gram is just an event that hasn’t happened yet How often these events happen? We model this by the probability of seeing an N-gram for the first time (we just count the number of times we first encountered a type)

September Witten-Bell: the equations Total probability mass assigned to zero-frequency N- grams: (NB: T is OBSERVED types, not V) So each zero N-gram gets the probability:

September Witten-Bell: why ‘discounting’ Now of course we have to take away something (‘discount’) from the probability of the events seen more than once:

September Witten-Bell for bigrams We `relativize’ the types to the previous word:

September Add-one vs. Witten-Bell discounts for unigrams in the BERP corpus WordAdd-OneWitten-Bell “I’” “want” “to” “eat” “Chinese” “food” “lunch”.22.91

September One last discounting method …. The best-known discounting method is GOOD- TURING (Good, 1953) Basic insight: re-estimate the probability of N- grams with zero counts by looking at the number of bigrams that occurred once For example, the revised count for bigrams that never occurred is estimated by dividing N 1, the number of bigrams that occurred once, by N 0, the number of bigrams that never occurred

September Combining estimators A method often used (generally in combination with discounting methods) is to use lower-order estimates to ‘help’ with higher-order ones Backoff (Katz, 1987) Linear interpolation (Jelinek and Mercer, 1980)

September Backoff: the basic idea

September Backoff with discounting

September Readings Jurafsky and Martin, chapter 6 The Statistics GlossaryStatistics Glossary Word prediction: – For mobile phones For mobile phones – For disabled users For disabled users Further reading: Manning and Schuetze, chapters 6 (Good-Turing)

September Acknowledgments Some of the material in these slides was taken from lecture notes by Diane Litman & James Martin