1 statistical methods allen ’ s chapter 7 j&m ’ s chapters 8 and 12

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Statistical Methods

Allen’s Chapter 7

J&M’s Chapters 8 and 12

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Statistical Methods

• Large data sets (Corpora) of natural languages allow using statistical methods that were not possible before

• Brown Corpus includes about 1000000 words with POS

• Penn Treebank contains full syntactic annotations

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Basic Probability Theory

• Random Variable ranges over a predefined set of values

TOSS = {h or t}

• If E is a random variable with possible values of {e1 … en} then

1. P(ei) 0, for all i

2. P(ei ) 1, for all i

3. i=1,n P(ei ) =1

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An Example

• R is a random variable with values {Win, Lose}• Harry is horse had 100 race with 20 Win• P(Win) = 0.2 and P(Lose) = 0.8• In 30 races it was raining, and Harry won 15 of

those races; so in rain P(Win) = 0.5• This is captured by Conditional Probability • P(A | B) = P(A & B) / P(B)• P(Win | Rain) = 0.15 / 0.3 = 0.5

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Independent Events

• P(A | B) = P(A)• P(A & B) = P(A) * P(B)

• Assume– L is a random variable with values {F , E}– P(F) = 0.6 – P( Win & F) = 0.12

Then P(Win | F) = 0.12 / 0.6 = 0.2 = P(Win)So Win and F are independentBut Win and Rain are not independent

• P(Win & Rain) = 0.15 P(Win) * P(Rain) = 0.06

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Part of Speech Tagging

• Determining the most likely category of each word in a sentence with ambiguous words

• Example: finding POS of words that can be either nouns and/or verbs

• Need two random variables: 1. C that ranges over POS {N, V}2. W that ranges over all possible words

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Part of Speech Tagging (Cont.)

• We don’t have true probabilities• We can estimate using large data sets• Suppose:

– There is a Corpus with 1273000 words– There is 1000 uses of flies:

400 with an noun sense, and 600 with a verb sense

P(flies) = 1000 / 1273000 = 0.0008P(flies & N) = 400 / 1273000 = 0.0003P(flies & V) = 600 / 1273000 = 0.0005P(V | flies) = P(V & flies) / P(flies) = 0.0005 / 0.0008 = 0.625

So in %60 occasions flies is a verb

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Estimating Probabilities

• We want to use probability to predict the future events

• Using the information P(V | flies) = 0.625 to predict that the next “flies” is more likely to be a verb

• This is called Maximum Likelihood estimation (MLE)

• Generally the larger the data set we use, the more accuracy we get

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Estimating Probabilities (Cont.)

• Estimating the outcome probability of tossing a coin (i.e., 0.5)

• Acceptable margin of error : (0.25 - 0.75)• The more tests performed, the more accurate

estimation– 2 trials: %50 chance of reaching acceptable result– 3 trials: %75 chance– 4 trials: %87.5 chance– 8 trials: %93 chance– 12trials: %95 chance …

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Estimating tossing a coin outcome

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• So the larger data set the better, but

• The problem of sparse data

• Brown Corpus contains about a million words – but there is only 49000 different words, – so one expect each word occurs about 20 times,– But over 40000 words occur less than 5 times.

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• For a random variable X with a set of values Vi, computed from counting number times X = xi

• P(X = xi) Vi / i Vi

• Maximum Likelihood Estimation (MLE) uses Vi = |xi|

• Expected likelihood Estimation (ELE) Uses Vi = |xi| + 0.5

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MLE vs ELE

• Suppose a word w doesn’t occur in the Corpus• We want to estimate w occurring in one of 40 classes L1

… L40• We have a random variable X, • X = xi only if w appears in word category Li• By MLE, P(Li | w) is undefined because the divisor is

zero• ELE , P(Li | w) 0.5 / 20 = 0.025• Suppose w occurs 5 times (4 times as an noun and once as

a verb) • By MLE, P(N |w) = 4/5 = 0.8, • By ELE, P(N | w) =4.5/25 = 0.18

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Evaluation

• Data set is divided into:– Training set (%80-%90 of the data)– Test set (%10-%20)

• Cross-Validation:– Repeatedly removing different parts of corpus

as the test set,– Training on the reminder of the corpus,– Then evaluating the new test set.

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Noisy Channel

noisy channel X Y

real language X

noisy language Y

p(X)

p(Y | X)

p(X|Y)

*

want to recover xX from yYchoose x that maximizes p(x | y)

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Part of speech tagging

Simplest Algorithm: choose the interpretation that occurs most frequently

“flies” in the sample corpus was %60 a verbThis algorithm success rate is %90Over %50 of words appearing in most corpora are

unambiguous To improve the success rate,

Use the tags before or after the word under examinationIf “flies” is preceded by the word “the” it is definitely a

noun

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Part of speech tagging (Cont.)

• General form of the POS Problem:• There is a sequence of words w1 … wt, and • We want to find a sequence of lexical categories C1 … Ct,

such that1. P(C1 … Ct | w1 … wt) is maximized

• Using the Bayes rule:2. P(C1 … Ct) * P(w1 … wt | C1 … Ct) / P(w1 … wt)

• The problem is reduced to finding C1 … Ct, such that3. P(C1 … Ct) * P(w1 … wt | C1 … Ct) is maximized

• But no effective method for calculating the probability of these long sequences accurately exists, as it would require too much data

• The probabilities can be estimated by some independence assumptions

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• Using the information about– The previous word category: bigram– Or two previous word categories: trigram– Or n-1 previous word categories: n-gram

• Using the bigram model• P(C1 … Ct) i=1,t P(Ci | Ci-1)

• P(Art N V N) = P(Art, ) * P(N | ART) * P( V | N) * P(N | V)

• P(w1 … wt | C1 … Ct) i=1,t P(wi | Ci)• Therefore we are looking for a sequence C1 … Ct such

that i=1,t P(Ci | Ci-1) * P(wi | Ci) is maximized

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• The information needed by this new formula can be extracted from the corpus

• P(Ci = V | Ci-1 = N) = Count( N at position i-1 & V at position i) / Count (N at

position i-1) (Fig. 7-4)

• P( the | Art) = Count(# times the is an Art) /Count(# times an Art occurs) (Fig. 7-6)

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Using an Artificial corpus

• An artificial corpus generated with 300 sentences of categories Art, N, V, P

• 1998 words, 833 nouns, 300 verbs, 558 article, and 307 propositions,

• To deal with the problem of the problem of the sparse data, a minimum probability of 0.0001 is assumed

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Bigram probabilities from the generated corpus

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Word counts in the generated corpus

N V ART P TOTAL

flies 21 23 0 0 44

fruit 49 5 1 0 55

like 10 30 0 21 61

a 1 0 201 0 202

the 1 0 300 2 303

flower 53 15 0 0 68

flowers 42 16 0 0 58

birds 64 1 0 0 65

others 592 210 56 284 1142

TOTAL 833 300 558 307 1998

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• How to find the sequence C1 … Ct that maximizes i=1,t P(Ci | Ci-1) * P(wi | Ci)

• Brute Force search:Finding all possible sequencesWith N categories and T words, there are NT sequences

• Using the independence assumption and bigram probabilities, the probability wi to be in category Ci depends only on Ci-1

• The process can be modeled by a special form of probabilistic finite state machine (Fig. 7-7)

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Markov Chain

• Probability of a sequence of 4 words being in cats: ART N V N0.71 * 1 * 0.43 * 0.35 = 0.107

• The representation is accurate only if the probability of a category occurring depends only the one category before it.

• This called the Markov assumption • The network is called Markov chain

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Hidden Markov Model (HMM)

• Markov network can be extended to include the lexical-generation probabilities, too.

• Each node could have an output probability for its every possible corresponding output

• node N is associated with a probability table indicating, for each word, how likely that word is to be selected if we randomly select a noun

• The output probabilities are exactly the lexical-generation probabilities shown in fig 7-6

• Markov network with output probabilities is called Hidden Markov Model (HMM)

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• The word hidden indicates that for a specific sequence of words, it is not clear what state the Markov model is in

• For instance, the word “flies” could be generated from state N with a probability of 0.25, or from state V with a probability of 0.076

• Now, it is not trivial to compute the probability of a sequence of words from the network

• But, if you are given a particular sequence, the probability that it generates a particular output is easily computed by multiplying the probabilities on the path times the probabilities of each output.

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• The probability that the sequence N V ART N generates the output Flies Like a flower is:– The probability of path N V ART N is

0.29 * 0.43 * 0.65 * 1 = 0.081– The probability of the output being Flies like a flower is

P(flies | N) * P(like | V) * P(a | ART) * P(flower | N) = 0.025 * 0.1 * 0.36 * 0.063 = 5.4 * 10-5

• The likelihood that HMM would generate the sentence is 0.000054 * 0.081 = 4.374 * 10-6

• Therefore, the probability of a sentence w1 … wt, given a sequence C1 … Ct, is

i=1,t P(Ci | Ci-1) * P(wi | Ci)

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Markov Chain

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Viterbi Algorithm

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Flies like a flower

• SEQSCORE(i, 1) = P(flies | Li) * P(Li | )

• P(flies/V) = 0.076 * 0.0001 = 7.6 * 10-6

• P(flies/N) = 0.035 * 0.29 = 0.00725

• P(likes/V) = max( P(flies/N) * P(V | N), P(flies/V) * P(V | V)) *

P(like | V)= max (0.00725 * 0.43, 7.6 * 10-6 * 0.0001) * 0.1 = 0.00031

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Flies like a flower

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Flies like a flower

• Brute force search steps are NT

• Viterbi algorithm steps are K* T * N2

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Getting Reliable Statistics (smoothing)

• Suppose we have 40 categories• To collect unigrams, at least 40 samples,

one for each category, are needed• For bigrams, 1600 samples are needed• For trigerams, 64000 samples are needed• For 4-grams, 2560000 samples are needed• P(Ci | C … Ci-1) =

1P(Ci) + 2P(Ci | Ci-1) + 3P(Ci | Ci-2Ci-1)1+ 2 + 3 = 1

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Statistical Parsing

• Corpus-based methods offer new ways to control parsers

• We could use statistical methods to identify the common structures of a Language

• We can choose the most likely interpretation when a sentence is ambiguous

• This might lead to much more efficient parsers that are almost deterministic

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Statistical Parsing

• What is the input of an statistical parser?• Input is the output of a POS tagging algorithm• If POSs are accurate, lexical ambiguity is removed• But if tagging is wrong, parser cannot find the

correct interpretation, or, may find a valid but implausible interpretation

• With %95 accuracy, the chance of correctly tagging a sentence of 8 words is 0.67, and that of 15 words is 0.46

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Obtaining Lexical probabilities

A better approach is:1. computing the probability that each word

appears in the possible lexical categories.2. combining these probabilities with some

method of assigning probabilities to rule use in the grammar

The context independent Lexical category of a word w be Lj can be estimated by:

P(Lj | w) = count (Lj & w) / i=1, N count( Li & w)

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Context-independent lexical categories

• P(Lj | w) = count (Lj & w) / i=1,N count( Li & w)

• P(Art | the) = 300 /303 =0.99

• P(N | flies) = 21 / 44 = 0.48

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Context dependent lexical probabilities

• A better estimate can be obtained by computing how likely it is that category Li occurs at position t, in all sequences of the input w1 … wt

• Instead of just finding the sequence with the maximum probability, we add up the probabilities of all sequences that end in wt/Li

• The probability that flies is a noun in the sentence The flies like flowers is calculated by adding the probability of all sequences that end with flies as a noun

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Context-dependent lexical probabilities

• Similarly, there are three nonzero sequences ending with flies as a V with a total value of 1.13 * 10 -5

• P(The flies) = 9.58 * 10 -3 + 1.13 * 10 -5 = 9.591 * 10 -3

• P(flies/N | The flies) = P(flies/N & The flies) / P(The flies)

= 9.58 * 10 -3 / 9.591 * 10 -3 = 0.9988• P(flies/V | The flies) = P(flies/V & The flies) / P(The flies)

= 1.13 * 10 -5 / 9.591 * 10 -3 = 0.0012

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Forward Probabilities

• The probability of producing the wordsw1 … wt and ending is state wt/Li is called the forward probability i(t) and is defined as:

i(t) = P(wt/Li & w1 … wt) • In the flies like flowers, 2(3) is the sum of values

computed for all sequences ending in a V (the second category) in position 3, for the input the flies like

• P(wt/Li | w1 … wt) = P(wt/Li & w1 … wt) / P(w1 … wt) i(t) / j=1, N j(t)

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Forward Probabilities

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Context dependent lexical Probabilities

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Context dependent lexical Probabilities

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Backward Probability

• Backward probability, j(t)), is the probability of producing the sequence

wt … wT beginning from the state wt/Lj

• P(wt/Li) (i(t) * i(t) ) / j=1, N (j(t) * i(t))

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Probabilistic Context-free Grammars

• CFGs can be generalized to PCFGs

• We need some statistics on rule use

• The simplest approach is to count the number of times each rule is used in a corpus with parsed sentences

• If category C has rules R1 … Rm, then

P(Rj | C) = count(# times Rj used) /

i=1,m count(# times Ri used)

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Probabilistic Context-free Grammars

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Independence assumption

• You can develop algorithm similar to the Veterbi algorithm that finds the most probable parse tree for an input

• Certain independence assumptions must be made• The probability of a constituent being derived by a

rule Rj is independent of how the constituent is used as a sub constituent

• The probabilities of NP rules are the same no matter the NP is the subject, the object of a verb, or the object of a proposition

• This assumption is not valid; a subject NP is much more likely to be a pronoun than an object NP

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Inside Probability

• The probability that a constituent C generates a sequence of words wi, wi+1, …, wj (written as wi,j) is called the inside probability and is denoted as P(wi,j | C)

• It is called inside probability because it assigns a probability to the word sequence inside the constituent

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Inside Probabilities

• How to derive inside probabilities?

• For lexical categories, these are the same as lexical-generation probabilities

• P(flower | N) is the inside probability that the constituent N is realized as the word flower (0.06 in fig. 7-6)

• Using lexical-generation probabilities, inside probabilities of Non-lexical constituents can be computed

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Probabilistic chart parsing

• In parsing, we are interested in finding the most likely parse rather than the overall probability of a given sentence.

• We can a Chart Parser for this propose• When entering an entry E of category C using rule

i with n sub constituents corresponding to entries E1 … En, then

• P(E) = P(Rule i | C) * P(E1) * … * P(En)• For lexical categories, it is better to use forward

probabilities rather than lexical-generation probs.

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A flower

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Probabilistic Parsing

• This technique identifies the correct parse %50 times• The reason is that the independence assumption is too radical• One of crucial issues is handling of lexical items• A context-free model does not consider lexical preferences• Parser prefers that PP attached to V rather than NP, and fails to find

the correct structure of those that PP should be attached to NP

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Best-First Parsing

• Exploring higher probability constituents first

• Much of the search space, containing lower-rated probabilities is not explored

• Chart parser’s Agenda is organized as a priority queue

• Arc extension algorithm need to be modified

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New arc extension for Prob. Chart Parser

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The man put a bird in the house

• Best first parser finds the correct parse after generating 65 constituents,

• Standard bottom-up parser generates 158 constituents

• Standard algorithm generates 106 constituents to find the first answer

• So, the best-first parsing is a significant improvement

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Best First Parsing

• It finds the most probable interpretation first• Probability of a constituent is always lower or

equal to the probability of any of its sub constituents

• If S2 with probability of p2 is found after S1 with the probability of p1, then p2 cannot be higher than p1, otherwise:

• Sub constituents of S2 would have higher probabilities than p1 and would be found sooner than S1 and thus S2 would be found sooner, too

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Problem of multiplication

• In practice with large grammars, probabilities would drop quickly because of multiplications

• Other functions can be used

• Score(C) = MIN (Score(C C1 … Cn) ,

• Score(C1), …, Score(Cn)

• But MIN leads to a %39 correct result

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Context-dependent probabilistic parsing

• The best-first algorithm improves the efficiency, but has no effect on accuracy

• Computing rules probability based on some context-dependent lexical information can improve accuracy

• The first word of a constituent is often its head word

• Computing the probability of rules based on the first word of constituents : P(R | C, w)

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Context-dependent probabilistic parsing

• P(R | C, w) = • Count( # times R used for cat. C starting with w) /

• Count(# times cat. C starts with w)

• Singular names rarely occur alone as a noun phrase (NP N)

• Plural nouns rarely act as a modifying name (NP N N)

• Context-dependent rules also encode verb preferences for sub categorizations

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Rule probabilities based on the first word of constituents

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Context-Dependent Parser Accuracy

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The man put the bird in the house

• P(VP V NP PP | VP, put) = • 0.93 * 0.99 * 0.76 * 0.76 = 0.54

• P(VP V NP | VP, put) = 0.0038

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The man Likes the bird in the house

• P(VP V NP PP | VP, like) = 0.1• P(VP V NP | VP, like) = 0.054

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Context-dependent rules

• The accuracy of the parser is still %66• Make the rule probabilities relative to larger fragment of

input (bigram, trigram, …)• Using other important words, such as prepositions• The more selective the lexical categories, the more

predictive the estimates can be (provided that there is enough data)

• Other closed class words such as articles, quantifiers, conjunctions can also be used (i.e., treated individually)

• But what about open class words such as verbs and nouns (cluster similar words)

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Handling Unknown Words

• An unknown word will disrupt the parse• Suppose we have a trigram model of data• If w3 in the sequence of words w1 w2 w3 is unknown, and if

w1 and w2 are of categories C1 and C2

• Pick the category C for w3 such that P(C | C1 C2) is maximized.

• For instance, if C2 is ART, then C will probably be a NOUN (or an ADJECTIVE)

• Morphology can also help• Unknown words ending with –ing are likely a VERB, and

those ending with –ly are likely an ADVERB

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Human preference in Parsing

Allen’s Chapter 6

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• Parsing techniques seen so far have depended on a Search

• But human seem to parse more deterministically

• However, they may fall in a garden-path:

The raft floated down the river sank

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• Some of principles that appears to be used by people to choose the correct interpretation are:– Minimal Attachment– Right Association– Lexical Preferences

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Minimal Attachment

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Minimal Attachment

• The M.A. Principle may cause misparsing1. We painted all the walls with cracks

PP tends to attach to VP rather to NP

2. The horse [that was] raced past the barn fellThe reduced relative clause introduce more nodes, so

“raced” is taken as the main verb, but it is rejected when “fell” is seen

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Right Association (or Late Closure)

• George said that Henry left in his car

• George said that Henry left in his car

• I thought it would rain yesterday

• I thought it would rain yesterday

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Right Association (or Late Closure)

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Lexical Preference

• M.A. and R.A. principle may have conflict

The man keeps the dog in the house

R.A. SuggestsThe man keeps the dog in the house

M.A. SuggestsThe man keeps the dog in the house

Should M.A. be given more priority?

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Lexical Preference

1. I wanted the dog in the house (R.A. is preferred)I wanted the dog in the house

2. I kept the dog in the houseI kept the dog in the house (M.A is preferred)

3. I put the dog in the houseI put the dog in the house (R.A. = wrong)

So L.P. overrides both M.A. and R.A.

1 statistical methods allen ’ s chapter 7 j&m ’ s chapters 8 and 12

Documents

possible values

jm s chapters

predefined set of values

syntactic annotations

pos penn treebank

brown corpus