1. travel
2. tourist destinations
3. eastern europe

Introduction to Markov Models
But first:
A few preliminaries
Estimating the probability of phrases of words,
sentences, etc.…
Corrected 3/30
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What’s a Word? Tokenization Is Tricky…
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How to find Sentences??
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Q1: How to estimate the probability of a
given sentence?
Predicting a word sequence II
 A crucial step in speech recognition (and lots of
other applications)
 First guess: products of unigrams
Pˆ (W )   P( w)
wW
Proposed word lattice:
form
subsidy
for
farm
subsidies
far
 Next guess: products of bigrams
• For W=w1w2w3… wn, Pˆ (W ) 
i 1
form
subsidy
for
farm
subsidies
far
form subsidy 0
106
subsidy 15
for 18185
form subsidies 0
subsidy far 0
subsidies 55
far 570
farm subsidy 0
subsidies for 6
farm subsidies 4
subsidies far 0
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)
words of AP text):
farm 74
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i 1
subsidy for 2
form 183
Not quite right…
i
Proposed word lattice:
Bigram counts (in 1.7 *
Unigram counts (in 1.7 * 106 words of AP text):
n 1
 P( w w
Just right…
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But the counts are tiny! Why?
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and finally, Estimating P(wn|w1w2…wn-1)
How can we estimate P correctly?
 Problem: Naïve Bayes model for bigrams violates independence
assumptions.
Again, we can estimate P(wn|w1w2…wn-1) with the MLE
Count ( w1w2 ...wn )
Count ( w1 w2 ...wn 1 )
Let’s do this right….
 Let W=w1w2w3… wn. Then, by the chain rule,
P(W )  P( w1 ) * P( w2 | w1 ) * P( w3 | w1 w2 ) *...* P( wn | w1 ...wn 1 )
 We can estimate P(w2|w1) by the Maximum Likelihood Estimator
and P(w3|w1w2) by
So to decide pat vs. pot in Heat up the oil in a large p?t,
compute for pot
Count ( w1w2 )
Count ( w1 )
Count ("Heat up the oil in a large pot") 0

Count ("Heat up the oil in a large")
0
Count ( w1w2 w3 )
Count ( w1 w2 )
and so on…
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Hmm..The Web Changes Things (2008 or so)
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Statistics and the Web II
So, P(“pot”|”heat up the oil in a large___”) = 8/49  0.16
Even the web in 2008 yields low counts!
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….
But the web has grown!!!
165/891=0.185
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So….
A BOTEC Estimate of What We Can Estimate
 A larger corpus won’t help much unless it’s
HUGE ….
but the web is!!!
What parameters can we estimate with 100 million
words of training data??
Assuming (for now) uniform distribution over only 5000 words
What if we only have 100 million words for our
estimates??
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
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The Markov Assumption: Estimation
We estimate the probability of each wi given previous context by
P(wi|w1w2…wi-1) = P(wi|wi-1)
which can be estimated by
Count ( wi 1wi )
Count ( wi 1 )
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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Visualizing an n-gram based language model:
the Shannon/Miller/Selfridge method
Markov Models
 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
j
— wi = the first vj such that
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 P (v
m 1
m
)  ri
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Visualizing an n-gram based language model:
the Shannon/Miller/Selfridge method
The Shannon/Miller/Selfridge method trained on
Shakespeare
 To generate a sequence of n words given a 1st
order 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
j
— wi
= the first vj such that
 P (v
m 1
m
| wi 1 )  ri
(This and next two slides from Jurafsky)
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Wall Street Journal just isn’t Shakespeare
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Shakespeare as corpus
 N=884,647 tokens, V=29,066
 Shakespeare produced 300,000 bigram types
out 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
English word frequencies well described
by Zipf’s Law
 Zipf (1949) characterized the relation between word
frequency and rank as:
f  r  C (for constant C )
r  C/f
log(r)  log(C) - log (f)
 Purely Zipfian data plots as a straight line on a loglog scale
 How likely is a 0 count? Much more likely than I let on!!!
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*Rank (r): The numerical position of a word in a list sorted by
decreasing frequency (f ).
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Word frequency & rank in Brown Corpus
vs Zipf
Zipf’s law for the Brown corpus
Lots of area
under the tail
of this curve!
From: Interactive mathematics http://www.intmath.com
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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.
Smoothing
This black art is why
NLP is taught in the
engineering school –
Jason Eisner
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Smoothing
Solution: Add-One Smoothing (again)
….is like Robin Hood:
 Pro: Very simple technique
 Cons:
• it steals from the rich
• 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 ngrams
• All unseen n-grams are smoothed in the same way
• and gives to the poor
 Using a smaller added-count improves things but only some
 More advanced techniques (Kneser Ney, Witten-Bell) use
properties of component n-1 grams and the like...
(Hint for this homework  )
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