CMSC 723 / LING 645: Intro to Computational Linguistics February 25, 2004 Lecture 5 (Dorr): Intro to Probabilistic NLP and N-grams (chap ) Prof. Bonnie J. Dorr Dr. Nizar Habash TA: Nitin Madnani, Nate Waisbrot

Why (not) Statistics for NLP?  Pro –Disambiguation –Error Tolerant –Learnable  Con –Not always appropriate –Difficult to debug

Weighted Automata/Transducers  Speech recognition: storing a pronunciation lexicon  Augmentation of FSA: Each arc is associated with a probability

Pronunciation network for “about”

Noisy Channel

Probability Definitions  Experiment (trial) –Repeatable procedure with well-defined possible outcomes  Sample space –Complete set of outcomes  Event –Any subset of outcomes from sample space  Random Variable –Uncertain outcome in a trial

More Definitions  Probability –How likely is it to get a particular outcome? –Rate of getting that outcome in all trials Probability of drawing a spade from 52 well-shuffled playing cards:  Distribution: Probabilities associated with each outcome a random variable can take –Each outcome has probability between 0 and 1 –The sum of all outcome probabilities is 1.

Conditional Probability  What is P(A|B)?  First, what is P(A)? –P(“It is raining”) =.06  Now what about P(A|B)? –P(“It is raining” | “It was clear 10 minutes ago”) =.004 ABA,B Note: P(A,B)=P(A|B) · P(B) Also: P(A,B) = P(B,A)

Independence  What is P(A,B) if A and B are independent?  P(A,B)=P(A) · P(B) iff A,B independent. –P(heads,tails) = P(heads) · P(tails) =.5 ·.5 =.25 –P(doctor,blue-eyes) = P(doctor) · P(blue-eyes) =.01 ·.2 =.002  What if A,B independent? –P(A|B)=P(A) iff A,B independent –Also: P(B|A)=P(B) iff A,B independent

Bayes Theorem Swap the order of dependence Sometimes easier to estimate one kind of dependence than the other

What does this have to do with the Noisy Channel Model? P(H) (H) (O) P(O|H)Best H argmax H = Best H = argmax P(H|O) H likelihoodprior

Noisy Channel Applied to Word Recognition  argmax w P(w|O) = argmax w P(O|w) P(w)  Simplifying assumptions –pronunciation string correct –word boundaries known  Problem: –Given [n iy], what is correct dictionary word?  What do we need? [ni]: knee, neat, need, new

What is the most likely word given [ni]?  Now compute likelihood P([ni]|w), then multiply WordP(O|w)P(w)P(O|w)P(w) new neat need knee Wordfreq(w)P(w) new neat need knee  Compute prior P(w)

Why N-grams? WordP(O|w)P(w)P(O|w)P(w) new neat need knee P([ni]|new)P(new) P([ni]|neat)P(neat) P([ni]|need)P(need) P([ni]|knee)P(knee)  Unigram approach: ignores context  Need to factor in context (n-gram) -Use P(need|I) instead of just P(need) -Note: P(new|I) < P(need|I)  Compute likelihood P([ni]|w), then multiply

Next Word Prediction [borrowed from J. Hirschberg] From a NY Times story... –Stocks plunged this …. –Stocks plunged this morning, despite a cut in interest rates –Stocks 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

–Stocks plunged this morning, despite a cut in interest rates by the Federal Reserve, as Wall Street began trading for the first time since last … –Stocks plunged this morning, despite a cut in interest rates by the Federal Reserve, as Wall Street began trading for the first time since last Tuesday's terrorist attacks. Next Word Prediction (cont)

Human Word Prediction  Domain knowledge  Syntactic knowledge  Lexical knowledge

Claim  A useful part of the knowledge needed to allow Word Prediction can be captured using simple statistical techniques.  Compute: –probability of a sequence –likelihood of words co-occurring

Why would we want to do this?  Rank the likelihood of sequences containing various alternative alternative hypotheses  Assess the likelihood of a hypothesis

Why is this useful?  Speech recognition  Handwriting recognition  Spelling correction  Machine translation systems  Optical character recognizers

Handwriting Recognition  Assume a note is given to a bank teller, which the teller reads as I have a gub. (cf. Woody Allen)  NLP to the rescue …. –gub is not a word –gun, gum, Gus, and gull are words, but gun has a higher probability in the context of a bank

Real Word 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 a year.  Hopefully, all with continue smoothly in my absence.  Can they lave him my messages?  I need to notified the bank of….  He is trying to fine out.

For Spell Checkers  Collect list of commonly substituted words –piece/peace, whether/weather, their/there...  Example: “On Tuesday, the whether …’’ “On Tuesday, the weather …”

Language Model  Definition: Language model is a model that enables one to compute the probability, or likelihood, of a sentence S, P(S).  Let’s look at different ways of computing P(S) in the context of Word Prediction

Word Prediction: Simple vs. Smart  Simple: Every word follows every other word w/ equal probability (0-gram) –Assume |V| is the size of the vocabulary –Likelihood of sentence S of length n is = 1/|V| × 1/|V| … × 1/|V| –If English has 100,000 words, probability of each next word is 1/ =  Smarter: Probability of each next word is related to word frequency (unigram) – Likelihood of sentence S = P(w 1 ) × P(w 2 ) × … × P(w n ) – Assumes probability of each word is independent of probabilities of other words.  Even smarter: Look at probability given previous words (N-gram) – Likelihood of sentence S = P(w 1 ) × P(w 2 |w 1 ) × … × P(w n |w n-1 ) – Assumes probability of each word is dependent on probabilities of other words. n times

Chain Rule  Conditional Probability –P(A 1,A 2 ) = P(A 1 ) · P(A 2 |A 1 )  The Chain Rule generalizes to multiple events –P(A 1, …,A n ) = P(A 1 ) P(A 2 |A 1 ) P(A 3 |A 1,A 2 )…P(A n |A 1 …A n-1 )  Examples: –P(the dog) = P(the) P(dog | the) –P(the dog bites) = P(the) P(dog | the) P(bites| the dog)

Relative Frequencies and Conditional Probabilities  Relative word frequencies are better than equal probabilities for all words –In a corpus with 10K word types, each word would have P(w) = 1/10K –Does not match our intuitions that different words are more likely to occur (e.g. the)  Conditional probability more useful than individual relative word frequencies –Dog may be relatively rare in a corpus –But if we see barking, P(dog|barking) may be very large

For a Word String  In general, the probability of a complete string of words w 1 …w n is: P(w ) = P(w 1 )P(w 2 |w 1 )P(w 3 |w 1..w 2 )…P(w n |w 1 …w n-1 ) = 1 n  But this approach to determining the probability of a word sequence is not very helpful in general….

Markov Assumption  How do we compute P(w n |w 1 n-1 )? Trick: Instead of P(rabbit|I saw a), we use P(rabbit|a). –This lets us collect statistics in practice –A bigram model: P(the barking dog) = P(the| )P(barking|the)P(dog|barking)  Markov models are the class of probabilistic models that assume that we can predict the probability of some future unit without looking too far into the past –Specifically, for N=2 (bigram): P(w 1 ) ≈ Π P(w k |w k-1 ) n n k=1  Order of a Markov model: length of prior context –bigram is first order, trigram is second order, …

Counting Words in Corpora  What is a word? –e.g., are cat and cats the same word? –September and Sept? –zero and oh? –Is seventy-two one word or two? AT&T? –Punctuation?  How many words are there in English?  Where do we find the things to count?

Corpora  Corpora are (generally online) collections of text and speech  Examples: –Brown Corpus (1M words) –Wall Street Journal and AP News corpora –ATIS, Broadcast News (speech) –TDT (text and speech) –Switchboard, Call Home (speech) –TRAINS, FM Radio (speech)

Training and Testing  Probabilities come from a training corpus, which is used to design the model. –overly narrow corpus: probabilities don't generalize –overly general corpus: probabilities don't reflect task or domain  A separate test corpus is used to evaluate the model, typically using standard metrics –held out test set –cross validation –evaluation differences should be statistically significant

Terminology  Sentence: unit of written language  Utterance: unit of spoken language  Word Form: the inflected form that appears in the corpus  Lemma: lexical forms having the same stem, part of speech, and word sense  Types (V): number of distinct words that might appear in a corpus (vocabulary size)  Tokens (N): total number of words in a corpus  Types seen so far (T): number of distinct words seen so far in corpus (smaller than V and N)

Simple N-Grams  An N-gram model uses the previous N-1 words to predict the next one: P(w n | w n-N+1 w n-N+2… w n-1 ) –unigrams: P(dog) –bigrams: P(dog | big) –trigrams: P(dog | the big) –quadrigrams: P(dog | chasing the big)

Using N-Grams  Recall that –N-gram: P(w n |w 1 ) ≈ P(w n |w n-N+1 ) –Bigram: P(w 1 ) ≈ Π P(w k |w k-1 ) n n k=1 n-1  For a bigram grammar –P(sentence) can be approximated by multiplying all the bigram probabilities in the sequence  Example: P(I want to eat Chinese food) = P(I | ) P(want | I) P(to | want) P(eat | to) P(Chinese | eat) P(food | Chinese)

A Bigram Grammar Fragment from BERP Eat on.16Eat Thai.03 Eat some.06Eat breakfast.03 Eat lunch.06Eat in.02 Eat dinner.05Eat Chinese.02 Eat at.04Eat Mexican.02 Eat a.04Eat tomorrow.01 Eat Indian.04Eat dessert.007 Eat today.03Eat British.001

Additional BERP Grammar I.25Want some.04 I’d.06Want Thai.01 Tell.04To eat.26 I’m.02To have.14 I want.32To spend.09 I would.29To be.02 I don’t.08British food.60 I have.04British restaurant.15 Want to.65British cuisine.01 Want a.05British lunch.01

Computing Sentence Probability  P(I want to eat British food) = P(I| ) P(want|I) P(to|want) P(eat|to) P(British|eat) P(food|British) =.25×.32×.65×.26×.001×.60 =  vs. I want to eat Chinese food =  Probabilities seem to capture “syntactic” facts, “world knowledge” –eat is often followed by a NP –British food is not too popular  N-gram models can be trained by counting and normalization

BERP Bigram Counts IWantToEatChineseFoodlunch I Want To Eat Chinese Food Lunch

BERP Bigram Probabilities: Use Unigram Count  Normalization: divide bigram count by unigram count of first word. IWantToEatChineseFoodLunch  Computing the probability of I I –P(I|I) = C(I I)/C(I) = 8 / 3437 =.0023  A bigram grammar is an NxN matrix of probabilities, where N is the vocabulary size

Learning a Bigram Grammar  The formula P(w n |w n-1 ) = C(w n-1 w n )/C(w n-1 ) is used for bigram “parameter estimation”  Relative Frequency  Maximum Likelihood Estimation (MLE): Parameter set maximizes likelihood of training set T given model M — P(T|M).

What do we learn about the language?  What about... –P(I | I) =.0023 –P(I | want) =.0025 –P(I | food) =.013  What's being captured with... –P(want | I) =.32 –P(to | want) =.65 –P(eat | to) =.26 –P(food | Chinese) =.56 –P(lunch | eat) =.055

Approximating Shakespeare  As we increase the value of N, the accuracy of the n-gram model increases  Generating sentences with random unigrams... –Every enter now severally so, let –Hill he late speaks; or! a more to leg less first you enter  With bigrams... –What means, sir. I confess she? then all sorts, he is trim, captain. –Why dost stand forth thy canopy, forsooth; he is this palpable hit the King Henry.

More Shakespeare  Trigrams –Sweet prince, Falstaff shall die. –This shall forbid it should be branded, if renown made it empty.  Quadrigrams –What! I will go seek the traitor Gloucester. –Will you not tell me who I am?

Dependence of N-gram Models on Training Set  There are 884,647 tokens, with 29,066 word form types, in about a one million word Shakespeare corpus  Shakespeare produced 300,000 bigram types out of 844 million possible bigrams: so, 99.96% of the possible bigrams were never seen (have zero entries in the table).  Quadrigrams worse: What's coming out looks like Shakespeare because it is Shakespeare.  All those zeroes are causing problems.

N-Gram Training Sensitivity  If we repeated the Shakespeare experiment but trained on a Wall Street Journal corpus, there would be little overlap in the output  This has major implications for corpus selection or design unigram: Months the my and issue of year foreign new exchange’s september were recession exchange new endorsed a acquire to six executives. bigram: Last December through the way to preserve the Hudson corporation N.B.E.C. Taylor would seem to complet the major central planners one point five percent of U.S.E has laready old M.X. … trigram: They also point to ninety nine point six billion dollars from twohundred four oh six three percent …

Some Useful Empirical Observations  A small number of events occur with high frequency  A large number of events occur with low frequency  You can quickly collect statistics on the high frequency events  You might have to wait an arbitrarily long time to get valid statistics on low frequency events  Some of the zeroes in the table are really zeroes. But others are simply low frequency events you haven't seen yet. How to fix?

Smoothing Techniques  Every ngram training matrix is sparse, even for very large corpora (Zipf’s law)  Computing perplexity requires non-zero probabilities  Unsmoothed: P(w i )=c i /N  Solution: estimate the likelihood of unseen ngrams  Add-one smoothing: –Add 1 to every ngram count –Normalize by N/(N+V) –Smoothed count is: c i ′= (c i +1) · N/(N+V) –Smoothed probability: P′(w i ) = c i ′/N

Add-One Smoothed Bigrams P(w n |w n-1 ) = C(w n-1 w n )/C(w n-1 ) P ′ (w n |w n-1 ) = [C(w n-1 w n )+1]/[C(w n-1 )+V]

Add-One Smoothing Flaw c i ′=(c i +1) cici

Discounted Smoothing: Witten-Bell  Witten-Bell Discounting –A zero ngram is just an ngram you haven’t seen yet… –Model unseen bigrams by the ngrams you’ve only seen once: total number of n-gram types in the data. –Total probability of unseen bigrams estimated as: T/(N+T) –View training corpus as series of events, one for each token (N) and one for each new type (T): MLE. –We can divide the probability mass equally among unseen bigrams. Let Z = number of unseen n-grams: c i ′ = (T/Z) · N/(N+T) if c i =0; else c i ′ = c i · N/(N+T)

Witten-Bell Bigram Counts c i ′=(c i +1) cici c i ′ = T/Z · if c i =0 c i · otherwise

Other Discounting Methods  Good-Turing Discounting –Re-estimate amount of probability mass for zero (or low count) ngrams by looking at ngrams with higher counts –Estimate –Assumes: word bigrams follow a binomial distribution We know number of unseen bigrams (VxV-seen)

Readings for next time  J&M Chapter