1 lin 6932 spring 2007 lin6932: topics in computational linguistics hana filip lecture 4: part of...

1 LIN 6932 Spring 2007

LIN6932: Topics in Computational Linguistics

Hana FilipLecture 4:

Part of Speech Tagging (II) - Introduction to Probability

February 1, 2007


Outline

Part of speech taggingParts of speechWhat’s POS tagging good for anyhow?Tag sets

2 main types of tagging algorithmsRule-basedStatistical

Important Ideas– Training sets and test sets– Unknown words– Error analysis

Examples of taggers– Rule-based tagging (Karlsson et al 1995) EngCG– Transformation-Based tagging (Brill 1995)– HMM tagging - Stochastic (Probabilitistic) taggers


3 methods for POS tagging (recap from last lecture)

1. Rule-based taggingExample: Karlsson (1995) EngCG tagger based on the Constraint Grammar architecture and ENGTWOL lexicon– Basic Idea:

Assign all possible tags to words (morphological analyzer used) Remove wrong tags according to set of constraint rules (typically more

than 1000 hand-written constraint rules, but may be machine-learned)2. Transformation-based tagging

Example: Brill (1995) tagger - combination of rule-based and stochastic (probabilistic) tagging methodologies– Basic Idea:

Start with a tagged corpus + dictionary (with most frequent tags) Set the most probable tag for each word as a start value Change tags according to rules of type “if word-1 is a determiner and

word is a verb then change the tag to noun” in a specific order (like rule-based taggers)

machine learning is used—the rules are automatically induced from a previously tagged training corpus (like stochastic approach)

3. Stochastic (=Probabilistic) taggingExample: HMM (Hidden Markov Model) tagging - a training corpus used to compute the probability (frequency) of a given word having a given POS tag in a given context


Today

ProbabilityConditional ProbabilityIndependenceBayes RuleHMM taggingMarkov ChainsHidden Markov Models


6. Introduction to Probability

Experiment (trial)Repeatable procedure with well-defined possible outcomes

Sample Space (S)– the set of all possible outcomes – finite or infinite

Example– coin toss experiment– possible outcomes: S = {heads, tails}

Example– die toss experiment– possible outcomes: S = {1,2,3,4,5,6}

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.




Introduction to Probability

Definition of sample space depends on what we are asking

Sample Space (S): the set of all possible outcomesExample– die toss experiment for whether the number is even or odd

– possible outcomes: {even,odd} – not {1,2,3,4,5,6}




More definitions

Eventsan event is any subset of outcomes from the sample space

Exampledie toss experiment let A represent the event such that the outcome of the die toss experiment is divisible by 3A = {3,6} A is a subset of the sample space S= {1,2,3,4,5,6}



Some definitionsEvents– an event is a subset of sample space– simple and compound events

Example– deck of cards draw experiment – suppose sample space S = {heart,spade,club,diamond} (four

suits)– let A represent the event of drawing a heart– let B represent the event of drawing a red card– A = {heart} (simple event)– B = {heart} u {diamond} = {heart,diamond} (compound event)

a compound event can be expressed as a set union of simple events

Example– alternative sample space S = set of 52 cards– A and B would both be compound events

QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.



Some definitionsCounting– suppose an operation oi can be performed in ni ways,

– a set of k operations o1o2...ok can be performed in n1 n2 ... nk ways

Example– dice toss experiment, 6 possible outcomes– two dice are thrown at the same time– number of sample points in sample space = 6 6 = 36



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Definition of Probability

The probability law assigns to an event a nonnegative numberCalled P(A)Also called the probability AThat encodes our knowledge or belief about the collective likelihood of all the elements of AProbability law must satisfy certain properties

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Probability Axioms

NonnegativityP(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.

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An example

An experiment involving a single coin tossThere are two possible outcomes, H and TSample space S is {H,T}If coin is fair, should assign equal probabilities to 2 outcomesSince they have to sum to 1P({H}) = 0.5P({T}) = 0.5P({H,T}) = P({H})+P({T}) = 1.0

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Another example

Experiment involving 3 coin tossesOutcome is a 3-long string of H or TS ={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}Assume each outcome is equiprobable

“Uniform distribution”What is probability of the event that exactly 2 heads occur?A = {HHT,HTH,THH} 3 events/outcomesP(A) = P({HHT})+P({HTH})+P({THH}) additivity - union of the probability of the individual events

= 1/8 + 1/8 + 1/8 total 8 events/outcomes

= 3/8

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Probability definitions

In summary:

Probability of drawing a spade from 52 well-shuffled playing cards:

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Moving toward language

What’s the probability of drawing a 2 from a deck of 52 cards with four 2s?

What’s the probability of a random word (from a random dictionary page) being a verb?

€

P(drawing a two) =4

52=

1

13= .077

€

P(drawing a verb) =# of ways to get a verb

all words

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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: # 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

€

P(drawing a verb) =# of ways to get a verb

all words

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Conditional Probability

A way to reason about the outcome of an experiment based on partial information

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”?How likely is it that a person has a disease given that a medical test was negative?A spot shows up on a radar screen. How likely is it that it corresponds to an aircraft?

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More precisely

Given an experiment, a corresponding sample space S, and a probability lawSuppose we know that the outcome is some event BWe want to quantify the likelihood that the outcome also belongs to some other event AWe need a new probability law that gives us the conditional probability of A given BP(A|B)

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An intuition

• Let’s say A is “it’s raining”.• Let’s say P(A) in dry Florida 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.

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S

Conditional Probability

let A and B be events in the sample space P(A|B) = the conditional probability of event A occurring given some fixed event B occurringdefinition: P(A|B) = P(A B) / P(B)

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Conditional probability

P(A|B) = P(A B)/P(B)Or

)(

),()|(

BP

BAPBAP =

A BA,B

Note: P(A,B)=P(A|B) · P(B)Also: P(A,B) = P(B,A)

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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

Note: P(A|B)=P(A) iff A,B independentAlso: P(B|A)=P(B) iff A,B independent

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Bayes Theorem

)(

)()|()|(

AP

BPBAPABP =

•Idea: The probability of an A conditional on another event B is generally different from the probability of B conditional on A. There is a definite relationship between the two.

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Deriving Bayes Rule

€

P(A | B) =P(A ∩ B)

P(B)

€

P(A | B) =P(A ∩ B)

P(B)

The probability of event A given event B is

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Deriving Bayes Rule

€

P(B | A) =P(A ∩ B)

P(A)

€

P(B | A) =P(A ∩ B)

P(A)

The probability of event B given event A is

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Deriving Bayes Rule

€

P(A | B) =P(B | A)P(A)

P(B)

€

P(A | B) =P(B | A)P(A)

P(B)

the theorem may be paraphrased as

conditional/posterior probability = (LIKELIHOOD multiplied by PRIOR) divided by NORMALIZING CONSTANT

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Hidden Markov Model (HMM) Tagging

Using an HMM to do POS taggingHMM is a special case of Bayesian inference

Foundational work in computational linguisticsn-tuple features used for OCR (Optical Character Recognition)W.W. Bledsoe and I. Browning, "Pattern Recognition and Reading by Machine," Proc. Eastern Joint Computer Conf., no. 16, pp. 225-233, Dec. 1959.F. Mosteller and D. Wallace, “Inference and disputed Authorship: The Federalist,” 1964.statistical methods applied to determine the authorship of the Federalist Papers (function words, Alexander Hamilton, James Madison)

It is also related to the “noisy channel” model in ASR (Automatic Speech Recognition)

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POS tagging as a sequence classification task

We are given a sentence (an “observation” or “sequence of observations”)

Secretariat is expected to race tomorrowsequence of n words w1…wn.

What is the best sequence of tags which corresponds to this sequence of observations?Probabilistic/Bayesian view:

Consider all possible sequences of tagsOut of this universe of sequences, choose the tag sequence which is most probable given the observation sequence of n words w1…wn.

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Getting to HMM

Let T = t1,t2,…,tn

Let W = w1,w2,…,wn

Goal: Out of all sequences of tags t1…tn, get the the most probable sequence of POS tags T underlying the observed sequence of words w1,w2,…,wn

Hat ^ means “our estimate of the best = the most probable tag sequence”

Argmaxx f(x) means “the x such that f(x) is maximized” it maximazes our estimate of the best tag sequence

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Getting to HMM

This equation is guaranteed to give us the best tag sequence

But how do we make it operational? How do we compute this value?Intuition of Bayesian classification:

Use Bayes rule to transform it into a set of other probabilities that are easier to computeThomas Bayes: British mathematician (1702-1761)

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Bayes Rule

Breaks down any conditional probability P(x|y) into three other probabilities

P(x|y): The conditional probability of an event x assuming that y has occurred

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Bayes Rule

We can drop the denominator: it does not change for each tag sequence; we are looking for the best tag sequence for the same observation, for the same fixed set of words

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Bayes Rule

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Likelihood and prior

n

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Likelihood and prior Further Simplifications

n

1. the probability of a word appearing depends only on its own POS tag, i.e, independent of other words around it

2. BIGRAM assumption: the probability of a tag appearing depends only on the previous tag

3. The most probable tag sequence estimated by the bigram tagger

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n

1. the probability of a word appearing depends only on its own POS tag, i.e, independent of other words around it

thekoalaputthekeysonthetable

WORDSTAGS

NVPDET

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2. BIGRAM assumption: the probability of a tag appearing depends only on the previous tag

Bigrams are groups of two written letters, two syllables, or two words; they are a special case of N-gram.

Bigrams are used as the basis for simple statistical analysis of text

The bigram assumption is related to the first-order Markov assumption

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3. The most probable tag sequence estimated by the bigram tagger

n

biagram assumption

---------------------------------------------------------------------------------------------------------------

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Two kinds of probabilities (1)

Tag transition probabilities p(ti|ti-1)Determiners likely to precede adjs and nouns– That/DT flight/NN– The/DT yellow/JJ hat/NN– So we expect P(NN|DT) and P(JJ|DT) to be high– But P(DT|JJ) to be:?

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Tag transition probabilities p(ti|ti-1)Compute P(NN|DT) by counting in a labeled corpus:

# of times DT is followed by NN

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Word likelihood probabilities p(wi|ti)P(is|VBZ) = probability of VBZ (3sg Pres verb) being “is”

Compute P(is|VBZ) by counting in a labeled corpus:

If we were expecting a third person singular verb, how likely is it that

this verb would be is?

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An Example: the verb “race”

Secretariat/NNP is/VBZ expected/VBN to/TO race/VB tomorrow/NRPeople/NNS continue/VB to/TO inquire/VB the/DT reason/NN for/IN the/DT race/NN for/IN outer/JJ space/NNHow do we pick the right tag?

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Disambiguating “race”

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Disambiguating “race”

P(NN|TO) = .00047P(VB|TO) = .83

The tag transition probabilities P(NN|TO) and P(VB|TO) answer the question: ‘How likely are we to expect verb/noun given the previous tag TO?’

P(race|NN) = .00057P(race|VB) = .00012

Lexical likelihoods from the Brown corpus for ‘race’ given a POS tag NN or VB.

P(NR|VB) = .0027P(NR|NN) = .0012

tag sequence probability for the likelihood of an adverb occurring given the previous tag verb or noun

P(VB|TO)P(NR|VB)P(race|VB) = .00000027P(NN|TO)P(NR|NN)P(race|NN)=.00000000032

Multiply the lexical likelihoods with the tag sequence probabiliies: the verb wins

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Hidden Markov Models

What we’ve described with these two kinds of probabilities is a Hidden Markov Model (HMM)Let’s just spend a bit of time tying this into the modelIn order to define HMM, we will first introduce the Markov Chain, or observable Markov Model.

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Definitions

A weighted finite-state automaton adds probabilities to the arcs

The sum of the probabilities leaving any arc must sum to one

A Markov chain is a special case of a WFST in which the input sequence uniquely determines which states the automaton will go throughMarkov chains can’t represent inherently ambiguous problems

Useful for assigning probabilities to unambiguous sequences

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Markov chain = “First-order observed Markov Model”

a set of states Q = q1, q2…qN; the state at time t is qt

a set of transition probabilities: a set of probabilities A = a01a02…an1…ann.

Each aij represents the probability of transitioning from state i to state jThe set of these is the transition probability matrix A

Distinguished start and end states

Special initial probability vector

i the probability that the MM will start in state i, each i

expresses the probability p(qi|START)

€

aij = P(qt = j | qt−1 = i) 1≤ i, j ≤ N

€

aij =1; 1≤ i ≤ Nj=1

N

∑

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Markov chain = “First-order observed Markov Model”Markov Chain for weather: Example 1three types of weather: sunny, rainy, foggywe want to find the following conditional probabilities:P(qn|qn-1, qn-2, …, q1)

- I.e., the probability of the unknown weather on day n, depending on the (known) weather of the preceding days- We could infer this probability from the relative frequency (the

statistics) of past observations of weather sequencesProblem: the larger n is, the more observations we must collect.

Suppose that n=6, then we have to collect statistics for 3(6-1) =

243 past histories

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Markov chain = “First-order observed Markov Model”Therefore, we make a simplifying assumption, called the (first-order) Markov assumption

for a sequence of observations q1, … qn,

current state only depends on previous state

the joint probability of certain past and current observations

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Markov chain = “First-order observable Markov Model”

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Markov chain = “First-order observed Markov Model”

Given that today the weather is sunny, what's the probability that tomorrow is sunny and the day after is rainy?

Using the Markov assumption and the probabilities in table 1, this translates into:

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The weather figure: specific example

Markov Chain for weather: Example 2

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Markov chain for weather

What is the probability of 4 consecutive rainy days?Sequence is rainy-rainy-rainy-rainyI.e., state sequence is 3-3-3-3P(3,3,3,3) =

1a11a11a11a11 = 0.2 x (0.6)3 = 0.0432

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Hidden Markov Model

For Markov chains, the output symbols are the same as the states.

See sunny weather: we’re in state sunny

But in part-of-speech tagging (and other things)

The output symbols are wordsBut the hidden states are part-of-speech tags

So we need an extension!A Hidden Markov Model is an extension of a Markov chain in which the output symbols are not the same as the states.This means we don’t know which state we are in.

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Markov chain for weather

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Markov chain for words

Observed events: words

Hidden events: tags

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States Q = q1, q2…qN; Observations O = o1, o2…oN;

Each observation is a symbol from a vocabulary V = {v1,v2,…vV}

Transition probabilities (prior)

Transition probability matrix A = {aij}

Observation likelihoods (likelihood)

Output probability matrix B={bi(ot)}a set of observation likelihoods, each expressing the probability of an observation ot being generated from a state i, emission probabilities

Special initial probability vector

i the probability that the HMM will start in state i, each i expresses the probability

p(qi|START)

Hidden Markov Models

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Assumptions

Markov assumption: the probability of a particular state depends only on the previous state

Output-independence assumption: the probability of an output observation depends only on the state that produced that observation

€

P(qi | q1...qi−1) = P(qi | qi−1)

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HMM for Ice Cream

You are a climatologist in the year 2799Studying global warmingYou can’t find any records of the weather in Boston, MA for summer of 2007But you find Jason Eisner’s diaryWhich lists how many ice-creams Jason ate every date that summerOur job: figure out how hot it was

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Noam task

GivenIce Cream Observation Sequence: 1,2,3,2,2,2,3…

(cp. with output symbols)

Produce:Weather Sequence: C,C,H,C,C,C,H …

(cp. with hidden states, causing states)

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HMM for ice cream

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Different types of HMM structure

Bakis = left-to-right Ergodic = fully-connected

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HMM Taggers

Two kinds of probabilitiesA transition probabilities (PRIOR) (slide 36)B observation likelihoods (LIKELIHOOD) (slide 36)

HMM Taggers choose the tag sequence which maximizes the product of word likelihood and tag sequence probability

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Weighted FSM corresponding to hidden states of HMM, showing A probs

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B observation likelihoods for POS HMM

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HMM Taggers

The probabilities are trained on hand-labeled training corpora (training set)Combine different N-gram levels Evaluated by comparing their output from a test set to human labels for that test set (Gold Standard)

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Next Time

Minimum Edit DistanceA “dynamic programming” algorithmA probabilistic version of this called “Viterbi” is a key part of the Hidden Markov Model!

1 lin 6932 spring 2007 lin6932: topics in computational linguistics hana filip lecture 4: part of...

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