Joint and Conditional Probability • p(A,B) = p(A B) • p(A|B) = p(A,B) / p(B) – Estimating form counts: • p(A|B) = p(A,B) / p(B) = (c(A B) / T) / (c(B) / T) = = c(A B) / c(B) A B A B 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 21
Bayes Rule • p(A,B) = p(B,A) since p(A p(B – therefore: p(A|B) p(B) = p(B|A) p(A), and therefore p(A|B) = p(B|A) p(A) / p(B) ! A B A B 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 22
Independence • Can we compute p(A,B) from p(A) and p(B)? • Recall from previous foil: p(A|B) = p(B|A) p(A) / p(B) p(A|B) p(B) = p(B|A) p(A) p(A,B) = p(B|A) p(A) ... we’re almost there: how p(B|A) relates to p(B)? – p(B|A) = P(B) iff A and B are independent • Example: two coin tosses, weather today and weather on March 4th 1789; • Any two events for which p(B|A) = P(B)! 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 23
Chain Rule p(A 1 , A 2 , A 3 , A 4 , ..., A n ) = ! p(A 1 |A 2 ,A 3 ,A 4 ,...,A n ) p(A 2 |A 3 ,A 4 ,...,A n ) p(A 3 |A 4 ,...,A n ) ... p(A n-1 |A n ) p(A n ) • this is a direct consequence of the Bayes rule. 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 24
The Golden Rule (of Classic Statistical NLP) • Interested in an event A given B (when it is not easy or practical or desirable to estimate p(A|B)): • take Bayes rule, max over all As: • argmax A p(A|B) = argmax A p(B|A) . p(A) / p(B) = argmax A p(B|A) p(A) ! • ... as p(B) is constant when changing As 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 25
Random Variable • is a function X: Q – in general: Q = R n , typically R – easier to handle real numbers than real-world events • random variable is discrete if Q is countable (i.e. also if finite) • Example: die : natural “numbering” [1,6], coin : {0,1} • Probability distribution: – p X (x) = p(X=x) = df p(A x ) where A x = {a : X(a) = x} – often just p(x) if it is clear from context what X is 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 26
Expectation Joint and Conditional Distributions • is a mean of a random variable (weighted average) – E(X) = x X( x . p X (x) • Example: one six-sided die: 3.5, two dice (sum) 7 • Joint and Conditional distribution rules: – analogous to probability of events • Bayes: p X|Y (x,y) = notation p XY (x|y) = even simpler notation p(x|y) = p(y|x) . p(x) / p(y) • Chain rule: p(w,x,y,z) = p(z).p(y|z).p(x|y,z).p(w|x,y,z) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 27
Standard distributions • Binomial (discrete) – outcome: 0 or 1 (thus: bi nomial) – make n trials – interested in the (probability of) number of successes r • Must be careful: it’s not uniform! n • p b (r|n) = ( ) / 2 n (for equally likely outcome) r n • ( ) counts how many possibilities there are for r choosing r objects out of n; = n! / ((n-r)! r!) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 28
Continuous Distributions • The normal distribution (“Gaussian”) • p norm (x| ) = e -(x- ) 2 / (2 2 ) / • where: – is the mean (x-coordinate of the peak) (0) – is the standard deviation (1) x • other: hyperbolic, t 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 29
Essential Information Theory
The Notion of Entropy • Entropy ~ “chaos”, fuzziness, opposite of order, ... – you know it: • it is much easier to create “mess” than to tidy things up... • Comes from physics: – Entropy does not go down unless energy is applied • Measure of uncertainty: – if low... low uncertainty; the higher the entropy, the higher uncertainty, but the higher “surprise” (information) we can get out of an experiment 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 31
The Formula • Let p X (x) be a distribution of random variable X • Basic outcomes (alphabet) H(X) = - x p(x) log 2 p(x) ! • Unit: bits (log 10 : nats) • Notation: H(X) = H p (X) = H(p) = H X (p) = H(p X ) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 32
Using the Formula: Example • Toss a fair coin: = {head,tail} – p(head) = .5, p(tail) = .5 – H(p) = - 0.5 log 2 (0.5) + (- 0.5 log 2 (0.5)) = 2 ( (-0.5) (-1) ) = 2 0.5 = 1 • Take fair, 32-sided die: p(x) = 1 / 32 for every side x – H(p) = - i = 1..32 p(x i ) log 2 p(x i ) = - 32 (p(x 1 ) log 2 p(x 1 ) (since for all i p(x i ) = p(x 1 ) = 1/32) = -32 ((1/32) (-5)) = 5 (now you see why it’s called bits ?) • Unfair coin: – p(head) = .2 ... H(p) = .722 ; p(head) = .01 ... H(p) = .081 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 33
Example: Book Availability Entropy H(p) 1 bad bookstore good bookstore 0 0 0.5 1 p(Book Available) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 34
The Limits • When H(p) = 0? – if a result of an experiment is known ahead of time: – necessarily: x ; p(x) = 1 & y ; y x p(y) = 0 • Upper bound? – none in general – for | | = n: H(p) log 2 n • nothing can be more uncertain than the uniform distribution 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 35
Entropy and Expectation • Recall: – E(X) = x X p X (x) x • Then: E(log 2 (1/p X (x))) = x X p X (x) log 2 (1/p X (x)) = = - x X p X (x) log 2 p X (x) = = H(p X ) = notation H(p) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 36
Perplexity: motivation • Recall: – 2 equiprobable outcomes: H(p) = 1 bit – 32 equiprobable outcomes: H(p) = 5 bits – 4.3 billion equiprobable outcomes: H(p) ~= 32 bits • What if the outcomes are not equiprobable? – 32 outcomes, 2 equiprobable at .5, rest impossible: • H(p) = 1 bit – Any measure for comparing the entropy (i.e. uncertainty/difficulty of prediction) (also) for random variables with different number of outcomes ? 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 37
Perplexity • Perplexity: – G(p) = 2 H(p) • ... so we are back at 32 (for 32 eqp. outcomes), 2 for fair coins, etc. • it is easier to imagine: – NLP example: vocabulary size of a vocabulary with uniform distribution, which is equally hard to predict • the “wilder” (biased) distribution, the better: – lower entropy, lower perplexity 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 38
Joint Entropy and Conditional Entropy • Two random variables: X (space ),Y ( ) • Joint entropy: – no big deal: ((X,Y) considered a single event): H(X,Y) = - x y p(x,y) log 2 p(x,y) • Conditional entropy: H(Y|X) = - x y p(x,y) log 2 p(y|x) recall that H(X) = E (log 2 (1/p X (x))) (weighted “average”, and weights are not conditional) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 39
Conditional Entropy (Using the Calculus) • other definition: H(Y|X) = x p(x) H(Y|X=x) = for H(Y|X=x), we can use the single-variable definition (x ~ constant) = x p(x) ( - y p(y|x) log 2 p(y|x) ) = = - x y p(y|x) p(x) log 2 p(y|x) = = - x y p(x,y) log 2 p(y|x) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 40
Properties of Entropy I • Entropy is non-negative: – H(X) – proof: (recall: H(X) = - x p(x) log 2 p(x)) • log(p(x)) is negative or zero for x 1, • p(x) is non-negative; their product p(x)log(p(x) is thus negative; • sum of negative numbers is negative; • and - f is positive for negative f • Chain rule: – H(X,Y) = H(Y|X) + H(X), as well as – H(X,Y) = H(X|Y) + H(Y) (since H(Y,X) = H(X,Y)) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 41
Properties of Entropy II • Conditional Entropy is better (than unconditional): – H(Y|X) H(Y) (proof on Monday) • H(X,Y) H(X) + H(Y) (follows from the previous (in)equalities) • equality iff X,Y independent • [recall: X,Y independent iff p(X,Y) = p(X)p(Y)] • H(p) is concave (remember the book availability graph?) – concave function f over an interval (a,b): x,y (a,b), [0,1]: f f( x + (1- )y) f(x) + (1- )f(y) • function f is convex if -f is concave • [for proofs and generalizations, see Cover/Thomas] x y 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 42
“Coding” Interpretation of Entropy • The least (average) number of bits needed to encode a message (string, sequence, series,...) (each element having being a result of a random process with some distribution p): = H(p) • Remember various compressing algorithms? – they do well on data with repeating (= easily predictable = low entropy) patterns – their results though have high entropy compressing compressed data does nothing 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 43
Coding: Example • How many bits do we need for ISO Latin 1? – the trivial answer: 8 • Experience: some chars are more common, some (very) rare: • ...so what if we use more bits for the rare, and less bits for the frequent? [be careful: want to decode (easily)!] • suppose: p(‘a’) = 0.3, p(‘b’) = 0.3, p(‘c’) = 0.3, the rest: p(x) .0004 • code: ‘a’ ~ 00, ‘b’ ~ 01, ‘c’ ~ 10, rest: 11b 1 b 2 b 3 b 4 b 5 b 6 b 7 b 8 • code acbbécbaac: 0010010111000011111001000010 a c b b é c b a a c • number of bits used: 28 (vs. 80 using “naive” coding) • code length ~ 1 / probability; conditional prob OK! 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 44
Entropy of a Language • Imagine that we produce the next letter using p(l n+1 |l 1 ,...,l n ), where l 1 ,...,l n is the sequence of all the letters which had been uttered so far (i.e. n is really big!); let’s call l 1 ,...,l n the history h (h n+1 ), and all histories H: • Then compute its entropy: – - h l p(l,h) log 2 p(l|h) • Not very practical, isn’t it? 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 45
Kullback-Leibler Distance (Relative Entropy) • Remember: – long series of experiments... c i /T i oscillates around some number... we can only estimate it... to get a distribution q. • So we get a distribution q; (sample space , r.v. X) the true distribution is, however, p. (same , X) how big error are we making? • D(p||q) (the Kullback-Leibler distance): D(p||q) = x p(x) log 2 (p(x)/q(x)) = E p log 2 (p(x)/q(x)) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 46
Comments on Relative Entropy • Conventions: – 0 log 0 = 0 – p log (p/0) = (for p > 0) • Distance? (less “misleading”: Divergence) – not quite: • not symmetric: D(p||q) D(q||p) • does not satisfy the triangle inequality – but useful to look at it that way • H(p) + D(p||q): bits needed for encoding p if q is used 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 47
Mutual Information (MI) in terms of relative entropy • Random variables X, Y; p X Y (x,y), p X (x), p Y (y) • Mutual information (between two random variables X,Y): I(X,Y) = D(p(x,y) || p(x)p(y)) • I(X,Y) measures how much (our knowledge of) Y contributes (on average) to easing the prediction of X • or, how p(x,y) deviates from (independent) p(x)p(y) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 48
Mutual Information: the Formula • Rewrite the definition: [ recall: D(r||s) = v r(v) log 2 (r(v)/s(v)); substitute r(v) = p(x,y), s(v) = p(x)p(y); <v> ~ <x,y> ] I(X,Y) = D(p(x,y) || p(x)p(y)) = ! = x y p(x,y) log 2 (p(x,y)/p(x)p(y)) • Measured in bits (what else? :-) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 49
From Mutual Information to Entropy • by how many bits the knowledge of Y lowers the entropy H(X): I(X,Y) = x y p(x,y) log 2 (p(x,y)/p(y)p(x)) = ...use p(x,y)/p(y) = p(x|y) = x y p(x,y) log 2 (p(x|y)/p(x)) = ...use log(a/b) = log a - log b (a ~ p(x|y), b ~ p(x)), distribute sums = x y p(x,y)log 2 p(x|y) - x y p(x,y)log 2 p(x) = ...use def. of H(X|Y) (left term) , and y p(x,y) = p(x) (right term) = - H(X|Y) + (- x p(x)log 2 p(x) ) = ...use def. of H(X) ( right term ), swap terms = H(X) - H(X|Y) ...by symmetry, = H(Y) - H(Y|X) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 50
Properties of MI vs. Entropy • I(X,Y) = H(X) - H(X|Y) = number of bits the knowledge of Y lowers the entropy of X = H(Y) - H(Y|X) (prev. foil, symmetry) Recall: H(X,Y) = H(X|Y) + H(Y) -H(X|Y) = H(Y) - H(X,Y) • I(X,Y) = H(X) + H(Y) - H(X,Y) • I(X,X) = H(X) (since H(X|X) = 0) • I(X,Y) = I(Y,X) (just for completeness) • I(X,Y) 0 ... let’s prove that now (as promised). 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 51
Jensen’s Inequality f • Recall: f is convex on interval (a,b) iff x,y (a,b), [0,1]: f( x + (1- )y) f(x) + (1- )f(y) x y J.I.: for distribution p(x), r.v. X on , and convex f, • f ( x p(x) x ) x p(x) f(x) • Proof (idea): by induction on the number of basic outcomes; start with | | = 2 by: • • p(x 1 )f(x 1 ) + p(x 2 )f(x 2 ) f(p(x 1 )x 1 + p(x 2 )x 2 ) ( def. of convexity) • for the induction step (| | = k k+1), just use the induction hypothesis and def. of convexity (again). 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 52
Information Inequality D(p||q) 0 ! • Proof: 0 = - log 1 = - log x q(x) = - log x (q(x)/p(x))p(x) ...apply Jensen’s inequality here ( - log is convex)... x p(x) (-log(q(x)/p(x))) = x p(x) log(p(x)/q(x)) = = D(p||q) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 53
Other (In)Equalities and Facts • Log sum inequality: for r i , s i i=1..n (r i log(r i /s i )) ( i=1..n r i ) log( i=1..n r i / i=1..n s i )) • D(p||q) is convex [in p,q] ( log sum inequality) • H(p X ) log 2 | |, where is the sample space of p X Proof: uniform u(x), same sample space : p(x) log u(x) = -log 2 | |; log 2 | | - H(X) = - p(x) log u(x) + p(x) log p(x) = D(p||u) 0 • H(p) is concave [in p]: Proof: from H(X) = log 2 | | - D(p||u), D(p||u) convex H(x) concave 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 54
Cross-Entropy • Typical case: we’ve got series of observations T = {t 1 , t 2 , t 3 , t 4 , ..., t n }(numbers, words, ...; t i ); estimate (simple): y (y) = c(y) / |T|, def. c(y) = |{t ; t = y}| p • ...but the true p is unknown; every sample is too small! • Natural question: how well do we do using [instead of p] ? p • Idea: simulate actual p by using a different T’ (or rather: by using different observation we simulate the insufficiency of T vs. some other data (“random” difference)) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 55
Cross Entropy: The Formula • H p’ ( ) = H(p’) + D(p’|| ) p p ( x ) ! H p’ ( ) = - x p’(x) log 2 p p • p’ is certainly not the true p, but we can consider it the “real world” distribution against which we test p • note on notation (confusing...): p/p’ , also H T’ (p) p • (Cross)Perplexity: G p’ (p) = G T’ (p)= 2 H p ’( ) p 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 56
Conditional Cross Entropy • So far: “unconditional” distribution(s) p(x), p’(x)... • In practice: virtually always conditioning on context • Interested in: sample space , r.v. Y, y ; context: sample space , r.v. X, x : “our” distribution p(y|x), test against p’(y,x), which is taken from some independent data: H p’ (p) = - y x p’(y,x) log 2 p(y|x) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 57
Sample Space vs. Data • In practice, it is often inconvenient to sum over the sample space(s) , (especially for cross entropy!) • Use the following formula: H p’ (p) = - y x p’(y,x) log 2 p(y|x) = ! - 1/|T’| i = 1..|T’| log 2 p(y i |x i ) • This is in fact the normalized log probability of the “test” data: H p’ (p) = - 1/|T’| log 2 i = 1..|T’| p(y i |x i ) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 58
Computation Example = {a,b,..,z}, prob. distribution (assumed/estimated from data) : • p(a) = .25, p(b) = .5, p( ) = 1/64 for {c..r}, = 0 for the rest: s,t,u,v,w,x,y,z • Data (test): barb p’(a) = p’(r) = .25, p’(b) = .5 • Sum over : a b c d e f g ... p q r s t ... z -p’( )log 2 p( ) .5+.5+0+0+0+0+0+0+0+0+0+1.5+0+0+0+0+0 = 2.5 • Sum over data: i / s i 1/b 2/a 3/r 4/b 1/|T’| = 10 (1/4) 10 = 2.5 -log 2 p(s i ) 1 + 2 + 6 + 1 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 59
Cross Entropy: Some Observations • H(p) ?? > ?? H p’ (p): ALL! • Previous example: [p(a) = .25, p(b) = .5, p( ) = 1/64 for {c..r}, = 0 for the rest: s,t,u,v,w,x,y,z] H(p) = 2.5 bits = H(p’) (barb) • Other data: probable: (1/8)(6+6+6+1+2+1+6+6)= 4.25 H(p) < 4.25 bits = H(p’) (probable) • And finally: abba: (1/4)(2+1+1+2)= 1.5 H(p) > 1.5 bits = H(p’) (abba) • But what about: baby -p’(‘y’)log 2 p(‘y’) = -.25 log 2 0 = (??) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 60
Cross Entropy: Usage • Comparing data?? – NO! (we believe that we test on real data!) • Rather: comparing distributions ( vs. real data) • Have (got) 2 distributions: p and q (on some , X) – which is better? – better: has lower cross-entropy (perplexity) on real data S • “Real” data: S • H S (p) = - 1/|S| i = 1..|S| log 2 p(y i |x i ) ?? H S (q) = - 1/|S| i = 1..|S| log 2 q(y i |x i ) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 61
Comparing Distributions Test data S: probable • p(.) from prev. example: H S (p) = 4.25 p(a) = .25, p(b) = .5, p( ) = 1/64 for {c..r}, = 0 for the rest: s,t,u,v,w,x,y,z • q(.|.) (conditional; defined by a table): q(.|.) a b e l o p r other a 0 .5 0 0 0 .125 0 0 ex.: q(o|r) = 1 b 1 0 0 0 1 .125 0 0 e 0 0 0 1 0 .125 0 0 q(r|p) = .125 l 0 .5 0 0 0 .125 0 0 o 0 0 0 0 0 .125 1 0 p 0 0 0 0 0 .125 0 1 r 0 0 0 0 0 .125 0 0 other 0 0 1 0 0 .125 0 0 (1/8) (log(p|oth.)+log(r|p)+log(o|r)+log(b|o)+log(a|b)+log(b|a)+log(l|b)+log(e|l)) (1/8) ( 0 + 3 + 0 + 0 + 1 + 0 + 1 + 0 ) H S (q) = .625 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 62
Language Modeling (and the Noisy Channel)
The Noisy Channel • Prototypical case: Input Output (noisy) The channel 0,1,1,1,0,1,0,1,... (adds noise) 0,1,1,0,0,1,1,0,... • Model: probability of error (noise): • Example: p(0|1) = .3 p(1|1) = .7 p(1|0) = .4 p(0|0) = .6 • The Task: known: the noisy output; want to know: the input ( decoding ) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 64
Noisy Channel Applications • OCR – straightforward: text print (adds noise), scan image • Handwriting recognition – text neurons, muscles (“noise”), scan/digitize image • Speech recognition (dictation, commands, etc.) – text conversion to acoustic signal (“noise”) acoustic waves • Machine Translation – text in target language translation (“noise”) source language • Also: Part of Speech Tagging – sequence of tags selection of word forms text 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 65
Noisy Channel: The Golden Rule of ... OCR, ASR, HR, MT, ... • Recall: p(A|B) = p(B|A) p(A) / p(B) (Bayes formula) A best = argmax A p(B|A) p(A) (The Golden Rule) • p(B|A): the acoustic/image/translation/lexical model – application-specific name – will explore later • p(A): the language model 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 66
The Perfect Language Model • Sequence of word forms [forget about tagging for the moment] • Notation: A ~ W = (w 1 ,w 2 ,w 3 ,...,w d ) • The big (modeling) question: p(W) = ? • Well, we know (Bayes/chain rule ): p(W) = p(w 1 ,w 2 ,w 3 ,...,w d ) = = p(w 1 ) p(w 2 |w 1 ) p(w 3 |w 1 ,w 2 ) p(w d |w 1 ,w 2 ,...,w d-1 ) • Not practical (even short W too many parameters) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 67
Markov Chain • Unlimited memory (cf. previous foil): – for w i , we know all its predecessors w 1 ,w 2 ,w 3 ,...,w i-1 • Limited memory: – we disregard “too old” predecessors – remember only k previous words: w i-k ,w i-k+1 ,...,w i-1 – called “k th order Markov approximation” • + stationary character (no change over time): p(W) i=1..d p(w i |w i-k ,w i-k+1 ,...,w i-1 ), d = |W| 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 68
n-gram Language Models • (n-1) th order Markov approximation n-gram LM: prediction history p(W) df i=1..d p(w i |w i-n+1 ,w i-n+2 ,...,w i-1 ) ! • In particular (assume vocabulary |V| = 60k): • 0-gram LM: uniform model, p(w) = 1/|V|, 1 parameter 6 10 4 parameters • 1-gram LM: unigram model, p(w), p(w i |w i-1 ) 3.6 10 9 parameters • 2-gram LM: bigram model, p(w i |w i-2 ,w i-1 ) 2.16 10 14 parameters • 3-gram LM: trigram model, 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 69
LM: Observations • How large n ? – nothing is enough (theoretically) – but anyway: as much as possible ( close to “perfect” model) – empirically: 3 • parameter estimation? (reliability, data availability, storage space, ...) • 4 is too much: |V|=60k 1.296 10 19 parameters • but: 6-7 would be (almost) ideal (having enough data): in fact, one can recover the original text ssequence from 7-grams! • Reliability ~ (1 / Detail) ( need compromise) • For now, keep word forms (no “linguistic” processing) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 70
The Length Issue n; w n p(w) = 1 n=1..∞ w n p(w) >> 1 ( ∞) • • We want to model all sequences of words – for “fixed” length tasks: no problem - n fixed, sum is 1 • tagging, OCR/handwriting (if words identified ahead of time) – for “variable” length tasks: have to account for • discount shorter sentences • General model: for each sequence of words of length n, define p’(w) = n p(w) such that n=1..∞ n = 1 n=1.. ∞ w n p’(w)=1 e.g., estimate n from data; or use normal or other distribution 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 71
Parameter Estimation • Parameter: numerical value needed to compute p(w|h) • From data (how else?) • Data preparation: • get rid of formatting etc. (“text cleaning”) • define words (separate but include punctuation, call it “word”) • define sentence boundaries (insert “words” <s> and </s>) • letter case: keep, discard, or be smart: – name recognition – number type identification [these are huge problems per se!] • numbers: keep, replace by <num>, or be smart (form ~ pronunciation) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 72
Maximum Likelihood Estimate • MLE: Relative Frequency... – ...best predicts the data at hand (the “training data”) • Trigrams from Training Data T: – count sequences of three words in T: c 3 (w i-2 ,w i-1 ,w i ) [NB: notation: just saying that the three words follow each other] – count sequences of two words in T: c 2 (w i-1 ,w i ): • either use c 2 (y,z) = w c 3 (y,z,w) • or count differently at the beginning (& end) of data! p(w i |w i-2 ,w i-1 ) = est. c 3 (w i-2 ,w i-1 ,w i ) / c 2 (w i-2 ,w i-1 ) ! 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 73
Character Language Model • Use individual characters instead of words: p(W) df i=1..d p(c i |c i-n+1 ,c i-n+2 ,...,c i-1 ) • Same formulas etc. • Might consider 4-grams, 5-grams or even more • Good only for language comparison • Transform cross-entropy between letter- and word-based models: H S (p c ) = H S (p w ) / avg. # of characters/word in S 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 74
LM: an Example • Training data: <s> <s> He can buy the can of soda. – Unigram: p 1 (He) = p 1 (buy) = p 1 (the) = p 1 (of) = p 1 (soda) = p 1 (.) = .125 p 1 ( can ) = .25 – Bigram: p 2 ( He|<s> ) = 1 , p 2 ( can|He ) = 1 , p 2 ( buy|can ) = .5 , p 2 ( of|can ) = .5 , p 2 ( the|buy ) = 1 ,... – Trigram: p 3 ( He|<s>,<s> ) = 1 , p 3 ( can|<s>,He ) = 1 , p 3 ( buy|He,can ) = 1 , p 3 ( of|the,can ) = 1 , ..., p 3 ( .|of,soda ) = 1 . – Entropy: H(p 1 ) = 2.75, H(p 2 ) = .25, H(p 3 ) = 0 Great?! 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 75
LM: an Example (The Problem) • Cross-entropy: • S = <s> <s> It was the greatest buy of all. • Even H S (p 1 ) fails (= H S (p 2 ) = H S (p 3 ) = ), because: – all unigrams but p 1 (the), p 1 (buy), p 1 (of) and p 1 (.) are 0. – all bigram probabilities are 0. – all trigram probabilities are 0. • We want: to make all (theoretically possible * ) probabilities non-zero. * in fact, all: remember our graph from day 1? 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 76
LM Smoothing (And the EM Algorithm)
The Zero Problem • “Raw” n-gram language model estimate: – necessarily, some zeros • !many: trigram model 2.16 10 14 parameters, data ~ 10 9 words – which are true 0? • optimal situation: even the least frequent trigram would be seen several times, in order to distinguish it’s probability vs. other trigrams • optimal situation cannot happen, unfortunately (open question: how many data would we need?) – we don’t know – we must eliminate the zeros • Two kinds of zeros: p(w|h) = 0, or even p(h) = 0! 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 78
Why do we need Nonzero Probs? • To avoid infinite Cross Entropy: – happens when an event is found in test data which has not been seen in training data H(p) = prevents comparing data with 0 “errors” • To make the system more robust – low count estimates: • they typically happen for “detailed” but relatively rare appearances – high count estimates: reliable but less “detailed” 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 79
Eliminating the Zero Probabilities: Smoothing • Get new p’(w) (same ): almost p(w) but no zeros • Discount w for (some) p(w) > 0: new p’(w) < p(w) w discounted (p(w) - p’(w)) = D • Distribute D to all w; p(w) = 0: new p’(w) > p(w) – possibly also to other w with low p(w) • For some w (possibly): p’(w) = p(w) • Make sure w p’(w) = 1 • There are many ways of smoothing 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 80
Smoothing by Adding 1 • Simplest but not really usable: – Predicting words w from a vocabulary V, training data T: p’(w|h) = (c(h,w) + 1) / (c(h) + |V|) • for non-conditional distributions: p’(w) = (c(w) + 1) / (|T| + |V|) – Problem if |V| > c(h) (as is often the case; even >> c(h)!) • Example: Training data: <s> what is it what is small ? |T| = 8 • V = { what, is, it, small, ?, <s>, flying, birds, are, a, bird, . }, |V| = 12 • p(it)=.125, p(what)=.25, p(.)=0 p(what is it?) = .25 2 .125 2 .001 p(it is flying.) = .125 .25 0 2 = 0 • p’(it) =.1, p’(what) =.15, p’(.)=.05 p’(what is it?) = .15 2 .1 2 .0002 p’(it is flying.) = .1 .15 .05 2 .00004 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 81
Adding less than 1 • Equally simple: – Predicting words w from a vocabulary V, training data T: p’(w|h) = (c(h,w) + ) / (c(h) + |V|), • for non-conditional distributions: p’(w) = (c(w) + ) / (|T| + |V|) • Example: Training data: <s> what is it what is small ? |T| = 8 • V = { what, is, it, small, ?, <s>, flying, birds, are, a, bird, . }, |V| = 12 • p(it)=.125, p(what)=.25, p(.)=0 p(what is it?) = .25 2 .125 2 .001 p(it is flying.) = .125 .25 0 2 = 0 • Use = .1: • p’(it) .12, p’(what) .23, p’(.) .01 p’(what is it?) = .23 2 .12 2 .0007 p’(it is flying.) = .12 .23 .01 2 .000003 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 82
Good - Turing • Suitable for estimation from large data – similar idea: discount/boost the relative frequency estimate: p r (w) = (c(w) + 1) N(c(w) + 1) / (|T| N(c(w))) , where N(c) is the count of words with count c (count-of- counts) specifically, for c(w) = 0 (unseen words), p r (w) = N(1) / (|T| N(0)) – good for small counts (< 5-10, where N(c) is high) – variants ( see MS ) – normalization! (so that we have w p’(w) = 1) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 83
Good-Turing: An Example • Example: remember: p r (w) = (c(w) + 1) N(c(w) + 1) / (|T| N(c(w))) Training data: <s> what is it what is small ? |T| = 8 • V = { what, is, it, small, ?, <s>, flying, birds, are, a, bird, . }, |V| = 12 p(it)=.125, p(what)=.25, p(.)=0 p(what is it?) = .25 2 .125 2 .001 p(it is flying.) = .125 .25 0 2 = 0 • Raw reestimation (N(0) = 6, N(1) = 4, N(2) = 2, N(i) = 0 for i > 2): p r (it) = (1+1) N(1+1)/(8 N(1)) = 2 2/(8 4) = .125 p r (what) = (2+1) N(2+1)/(8 N(2)) = 3 0/(8 2) = 0: keep orig. p(what) p r (.) = (0+1) N(0+1)/(8 N(0)) = 1 4/(8 6) .083 • Normalize (divide by 1.5 = w |V| p r (w)) and compute: p’(it) .08, p’(what) .17, p’(.) .06 p’(what is it?) = .17 2 .08 2 .0002 p’(it is flying.) = .08 .17 .06 2 .00004 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 84
Smoothing by Combination: Linear Interpolation • Combine what? • distributions of various level of detail vs. reliability • n-gram models: • use (n-1)gram, (n-2)gram, ..., uniform reliability detail • Simplest possible combination: – sum of probabilities, normalize: • p(0|0) = .8, p(1|0) = .2, p(0|1) = 1, p(1|1) = 0, p(0) = .4, p(1) = .6: • p’(0|0) = .6, p’(1|0) = .4, p’(0|1) = .7, p’(1|1) = .3 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 85
Typical n-gram LM Smoothing • Weight in less detailed distributions using =( 0 , , , ): p’ (w i | w i-2 ,w i-1 ) = p 3 (w i | w i-2 ,w i-1 ) + p 2 (w i | w i-1 ) + p 1 (w i ) + 0 /|V| • Normalize: i > 0, i=0..n i = 1 is sufficient ( 0 = 1 - i=1..n i ) (n=3) • Estimation using MLE: – fix the p 3 , p 2 , p 1 and |V| parameters as estimated from the training data – then find such { i } which minimizes the cross entropy (maximizes probability of data): -(1/|D|) i=1..|D| log 2 (p’ (w i |h i )) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 86
Held-out Data • What data to use? – try the training data T: but we will always get = 1 • why? (let p iT be an i-gram distribution estimated using r.f. from T) • minimizing H T (p’ ) over a vector , p’ = p 3T + p 2T + p 1T + /|V| – remember: H T (p’ ) = H(p 3T )+D(p 3T ||p’ ); • (p 3T fixed H(p 3T ) fixed, best) – which p’ minimizes H T (p’ )? ... a p’ for which D(p 3T || p’ )=0 – ...and that’s p 3T (because D(p||p) = 0, as we know). – ...and certainly p’ = p 3T if = 1 (maybe in some other cases, too). (p’ = 1 p 3T + 0 p 2T + 0 p 1T + 0/|V|) – – thus: do not use the training data for estimation of • must hold out part of the training data ( heldout data, H): • ...call the remaining data the (true/raw) training data, T • the test data S (e.g., for comparison purposes): still different data! 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 87
The Formulas • Repeat: minimizing -(1/|H|) i=1..|H| log 2 (p’ (w i |h i )) over p’ (w i | h i ) = p’ (w i | w i-2 ,w i-1 ) = p 3 (w i | w i-2 ,w i-1 ) + ! p 2 (w i | w i-1 ) + p 1 (w i ) + 0 /|V| • “Expected Counts (of lambdas)”: j = 0..3 ! c( j ) = i=1..|H| ( j p j (w i |h i ) / p’ (w i |h i )) • “Next ”: j = 0..3 ! j,next = c( j ) / k=0..3 (c( k )) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 88
The (Smoothing) EM Algorithm 1. Start with some , such that j > 0 for all j 0..3. 2. Compute “Expected Counts” for each j . 3. Compute new set of j , using the “Next ” formula. 4. Start over at step 2, unless a termination condition is met. • Termination condition: convergence of . – Simply set an , and finish if | j - j,next | < for each j (step 3). • Guaranteed to converge: follows from Jensen’s inequality, plus a technical proof. 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 89
Remark on Linear Interpolation Smoothing • “Bucketed” smoothing: – use several vectors of instead of one, based on (the frequency of) history: (h) • e.g. for h = (micrograms,per) we will have ( h ) = (.999,.0009,.00009,.00001) (because “cubic” is the only word to follow...) – actually: not a separate set for each history, but rather a set for “similar” histories (“bucket”): (b(h)), where b: V 2 N (in the case of trigrams) b classifies histories according to their reliability (~ frequency) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 90
Bucketed Smoothing: The Algorithm • First, determine the bucketing function b (use heldout!): – decide in advance you want e.g. 1000 buckets – compute the total frequency of histories in 1 bucket (f max (b)) – gradually fill your buckets from the most frequent bigrams so that the sum of frequencies does not exceed f max (b) (you might end up with slightly more than 1000 buckets) • Divide your heldout data according to buckets • Apply the previous algorithm to each bucket and its data 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 91
Simple Example • Raw distribution (unigram only; smooth with uniform): p(a) = .25, p(b) = .5, p( ) = 1/64 for {c..r}, = 0 for the rest: s,t,u,v,w,x,y,z • Heldout data: baby; use one set of ( 1 : unigram, 0 : uniform) • Start with 1 = .5; p’ (b) = .5 x .5 + .5 / 26 = .27 p’ (a) = .5 x .25 + .5 / 26 = .14 p’ (y) = .5 x 0 + .5 / 26 = .02 c( 1 ) = .5x.5/.27 + .5x.25/.14 + .5x.5/.27 + .5x0/.02 = 2.72 c( 0 ) = .5x.04/.27 + .5x.04/.14 + .5x.04/.27 + .5x.04/.02 = 1.28 Normalize: 1,next = .68, 0,next = .32. Repeat from step 2 (recompute p’ first for efficient computation, then c( i ), ...) Finish when new lambdas almost equal to the old ones (say, < 0.01 difference). 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 92
Some More Technical Hints • Set V = {all words from training data}. • You may also consider V = T H, but it does not make the coding in any way simpler (in fact, harder). • But: you must never use the test data for you vocabulary! • Prepend two “words” in front of all data: • avoids beginning-of-data problems • call these index -1 and 0: then the formulas hold exactly • When c n (w,h) = 0: • Assign 0 probability to p n (w|h) where c n-1 (h) > 0, but a uniform probability (1/|V|) to those p n (w|h) where c n-1 (h) = 0 [this must be done both when working on the heldout data during EM, as well as when computing cross-entropy on the test data!] 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 93
Words and the Company They Keep
Motivation • Environment: – mostly “not a full analysis (sentence/text parsing)” • Tasks where “words & company” are important: – word sense disambiguation (MT, IR, TD, IE) – lexical entries: subdivision & definitions (lexicography) – language modeling (generalization, [kind of] smoothing) – word/phrase/term translation (MT, Multilingual IR) – NL generation (“natural” phrases) (Generation, MT) – parsing (lexically-based selectional preferences) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 95
Collocations • Collocation – Firth: “word is characterized by the company it keeps”; collocations of a given word are statements of the habitual or customary places of that word. – non-compositionality of meaning • cannot be derived directly from its parts (heavy rain) – non-substitutability in context • for parts (red light) – non-modifiability (& non-transformability) • kick the yellow bucket; take exceptions to 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 96
Association and Co-occurence; Terms • Does not fall under “collocation”, but: • Interesting just because it does often [rarely] appear together or in the same (or similar) context: • (doctors, nurses) • (hardware,software) • (gas, fuel) • (hammer, nail) • (communism, free speech) • Terms: – need not be > 1 word (notebook, washer) 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 97
Collocations of Special Interest • Idioms: really fixed phrases • kick the bucket, birds-of-a-feather, run for office • Proper names: difficult to recognize even with lists • Tuesday (person’s name), May, Winston Churchill, IBM, Inc. • Numerical expressions – containing “ordinary” words • Monday Oct 04 1999, two thousand seven hundred fifty • Phrasal verbs – Separable parts: • look up, take off 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 98
Further Notions • Synonymy: different form/word, same meaning: • notebook / laptop • Antonymy: opposite meaning: • new/old, black/white, start/stop • Homonymy: same form/word, different meaning: • “true” (random, unrelated): can (aux. verb / can of Coke) • related: polysemy; notebook, shift, grade, ... • Other: • Hyperonymy/Hyponymy: general vs. special: vehicle/car • Meronymy/Holonymy: whole vs. part: body/leg 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 99
How to Find Collocations? • Frequency – plain – filtered • Hypothesis testing – t test – test • Pointwise (“poor man’s”) Mutual Information • (Average) Mutual Information 2018/9 UFAL MFF UK NPFL067/Intro to Statistical NLP I/Jan Hajic - Pavel Pecina 100
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