Introduction to Markov Models Estimating the probability of phrases - - PowerPoint PPT Presentation
Introduction to Markov Models Estimating the probability of phrases - - PowerPoint PPT Presentation
Introduction to Markov Models Estimating the probability of phrases of words, sentences, etc . But first: A few preliminaries CIS 391 - Intro to AI 2 What counts as a word? A tricky question. CIS 391 - Intro to AI 3 How to find
But first: A few preliminaries
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What counts as a word? A tricky question….
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How to find Sentences??
Q1: How to estimate the probability of a given sentence W?
A crucial step in speech recognition (and lots of
- ther applications)
First guess: products of unigrams Given word lattice: Unigram counts (in 1.7 * 106 words of AP text): Not quite right…
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ˆ( ) ( )
w W
P W P w
form subsidy for farm subsidies far form 183 subsidy 15 for 18185 farm 74 subsidies 55 far 570
Predicting a word sequence II
Next guess: products of bigrams
- For W=w1w2w3… wn,
Given word lattice: Bigram counts (in 1.7 * 106 words of AP text): Better (if not quite right) … (But the counts are tiny! Why?)
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1 1 1
ˆ( ) ( )
n i i i
P W P w w
form subsidy for farm subsidies far form subsidy 0 subsidy for 2 form subsidies 0 subsidy far 0 farm subsidy 0 subsidies for 6 farm subsidies 4 subsidies far 0
How can we estimate P correctly?
Problem: Naïve Bayes model for bigrams violates independence assumptions.
Let’s do this right….
Let W=w1w2w3… wn. Then, by the chain rule, We can estimate P(w2|w1) by the Maximum Likelihood Estimator and P(w3|w1w2) by and so on…
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1 1 1
1 2 3 2 1
( ) ( )* ( | )* ( | )*...* ( | ... )
n n
P W P w P w w P w w w P w w w
1 2 1
( ) ( ) Count w w Count w
1 2 3 1 2
( ) ( ) Count w w w Count w w
1 1 1
1 2 3 2 1
( ) ( )* ( | )* ( | )*...* ( | ... )
n n
P W P w P w w P w w w P w w w
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and finally, Estimating P(wn|w1w2…wn-1)
Again, we can estimate P(wn|w1w2…wn-1) with the MLE So to decide pat vs. pot in Heat up the oil in a large p?t, compute for pot
1 2 1 2 1
( ... ) ( ... )
n n
Count w w w Count w w w ("Heat up the oil in a large pot") ("Heat up the oil in a larg ") e Count Count
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Hmm..The Web Changes Things (2008 or so)
Even the web in 2008 yields low counts!
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Statistics and the Web II
So, P(“pot”|”heat up the oil in a large___”) = 8/49 0.16
But the web has grown!!!
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….
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165/891=0.185
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So….
A larger corpus won’t help much unless it’s
HUGE …. but the web is!!!
But what if we only have 100 million words for our estimates??
A BOTEC Estimate of What We Can Estimate
What parameters can we estimate with 100 million words of training data??
Assuming (for now) uniform distribution over only 5000 words So even with 108 words of data, for even trigrams we encounter the sparse data problem…..
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The Markov Assumption: Only the Immediate Past Matters
The Markov Assumption: Estimation
We estimate the probability of each wi given previous context by which can be estimated by So we’re back to counting only unigrams and bigrams!! AND we have a correct practical estimation method for P(W) given the Markov assumption!
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P(wi|w1w2…wi-1) = P(wi|wi-1)
1 1
( ) ( )
i i i
Count w w Count w
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Markov Models
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Visualizing an n-gram based language model: the Shannon/Miller/Selfridge method To generate a sequence of n words given unigram estimates:
- Fix some ordering of the vocabulary v1 v2 v3 …vk.
- For each word wi , 1 ≤ i ≤ n
—Choose a random value ri between 0 and 1 — wi = the first vj such that
1
( )
j m i m
P v r
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Visualizing an n-gram based language model: the Shannon/Miller/Selfridge method To generate a sequence of n words given a 1st
- rder Markov model (i.e. conditioned on one
previous word):
- Fix some ordering of the vocabulary v1 v2 v3 …vk.
- Use unigram method to generate an initial word w1
- For each remaining wi , 2 ≤ i ≤ n
—Choose a random value ri between 0 and 1
— wi = the first vj such that
1 1
( | )
j m i i m
P v w r
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The Shannon/Miller/Selfridge method trained on Shakespeare
(This and next two slides from Jurafsky)
Wall Street Journal just isn’t Shakespeare
Shakespeare as corpus N=884,647 tokens, V=29,066 Shakespeare produced 300,000 bigram types
- ut of V2= 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
The Sparse Data Problem Again
How likely is a 0 count? Much more likely than I let on!!!
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English word frequencies well described by Zipf’s Law
Zipf (1949) characterized the relation between word frequency and rank as: Purely Zipfian data plots as a straight line on a log- log scale
*Rank (r): The numerical position of a word in a list sorted by decreasing frequency (f ).
(f) log
- log(C)
log(r) C/f r ) constant (for C C r f
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Word frequency & rank in Brown Corpus vs Zipf
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From: Interactive mathematics http://www.intmath.com Lots of area under the tail
- f this curve!
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Zipf’s law for the Brown corpus
Smoothing
This black art is why NLP is taught in the engineering school – Jason Eisner
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Smoothing
At least one unknown word likely per sentence given Zipf!! To fix 0’s caused by this, we can smooth the data.
- Assume we know how many types never occur in the data.
- Steal probability mass from types that occur at least once.
- Distribute this probability mass over the types that never occur.
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Smoothing
….is like Robin Hood:
- it steals from the rich
- and gives to the poor
Review: Add-One Smoothing
Estimate probabilities by assuming every possible word type v V actually occurred one extra time (as if by appending an unabridged dictionary) So if there were N words in our corpus, then instead of estimating we estimate
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ˆ P
( ) ˆ( ) Count w P w N
( 1) ˆ( ) Count w P w N V
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Add-One Smoothing (again)
Pro: Very simple technique Cons:
- Probability of frequent n-grams is underestimated
- Probability of rare (or unseen) n-grams is overestimated
- Therefore, too much probability mass is shifted towards unseen n-
grams
- All unseen n-grams are smoothed in the same way