# Natural Language Processing COMPSCI 423/723

Natural Language Processing
COMPSCI 423/723
Rohit Kate
1
Probability Theory
Language Models
2
Probability Theory
3
What is Probability?
• In ordinary language, probability is the degree of
certainty of an event: It is very probable that it
will rain today.
• Probability theory gives a formal mathematical
framework to work with numerical estimates of
certainty of events, using this one can:
– Predict likelihood of combinations of events
– Predict most likely outcome
– Predict something given that something else
4
Why is Probability Theory
Important for NLP?
• To obtain estimates for various outcomes of an ambiguity,
for example:
• Predict most likely parse or an interpretation of an
ambiguous sentence given the context or background
knowledge
– Time flies like an arrow.
•
•
•
•
Time goes by fast: 0.9
A particular type of flies “time flies” like an arrow: 0.05
Measure speed of flies like you will measure speed of an arrow: 0.03
…
estimates of certainty, probability theory is preferable
because it has sound mathematics behind it, for example
rules for combining probabilities
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Definitions
• Sample Space (Ω): Space of possible outcomes
– Outcomes of throwing a dice: {1,2,3,4,5,6}
• Event: Subset of a sample space
– An even number will show up: {2,4,6}
– Number 2 will show up: {2}
• Probability function (or probability distribution):
Mapping from events to a real number in [0,1],
such that:
P(Ω) = 1
For any events α and β if α П β = ø (disjoint) then
P(α U β) = P(α) + P(β)
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An Example
• For throwing a dice
P({1,2,3,4,5,6}) = 1 (some outcome shows up)
• Suppose each basic outcome is equally likely, since
they are all disjoint and add up to 1, we will have
P({1}) = 1/6, P({2})=1/6, P({3})=1/6…
• P(an even number shows up) = P({2,4,6}) = P({2}) +
P({4}) + P({6}) = 1/6 + 1/6 + 1/6 = 1/2
• P(an even number or 3 shows up) = 1/2 + 1/6= 2/3
• P(an even number or 6 shows up) ≠ 1/2 + 1/6 why?
P({2,4,6}) = 1/2
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Interpretation of Probability
• Frequentist interpretation:
P({3})=1/6: If a dice is thrown multiple times then
1/6th of the times 3 will show up
P(It will rain tomorrow) = 1/2 ??
• Subjective interpretation: One’s degree of
belief that the event will happen
The mathematical rules should hold for both
interpretations.
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Estimating Probabilities
• For well defined sample spaces and events, they
can be analytically estimated:
P({3}) = 1/6 (assuming fair dice)
• For many other sample spaces it is not possible to
analytically estimate, for example P(A teenager
will drink and drive).
• For these cases they can be empirically estimated
from a good sample,
P(A teenager will drink and drive) = # of
Teenagers who drink and drive/# of teenagers
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Conditional Probability
• Updated probability of an event given that
some event happened
P({2}) = 1/6
P({2} given an even number showed up) = ?
Represented as: P({2}|{2,4,6}) or P(A|B)
P(A|B) = P(AПB)/P(B) (for P(B) > 0)
P({2}|{2,4,6}) = P({2})/P({2,4,6}) = 1/6/(1/2) = 1/3
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Multiplicative and Chain Rules
• Multiplicative rule:
P(AПB) = P(B)P(A|B) = P(A)P(B|A)
• Generalization of the rule, chain rule:
P(A1ПA2П…ПAn) =
P(A1)P(A2|A1)P(A3|A1ПA2)…P(An|A1П..ПAn-1)
Or in any order of As
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Independence
• Two events A and B are independent of each other if
P(AПB) = P(A)P(B) or equivalently P(A)=P(A|B) or
P(B)=P(B|A), i.e. happening of an event B does not change
the probability of A or vice versa. Otherwise the events are
dependent.
Example:
• P({1,2}) = 1/3 P(Even) = 1/2
• P({1,2} П Even) = P({2}) = 1/6 = P({1,2})*P(Even)
• P({1,2}|Even) = P({1,2}ПEven)/P(Even) = 1/6/(1/2) = 1/3
Hence , P({1,2}|Even)=P({1,2})
– Given that an even number showed up does not change the
probability of whether one or two showed up. Hence these are
independent events.
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Independence
• P({2}|Even) = P({2}ПEven)/P(Even) =
P({2})/P{2,4,6} = 1/6/(1/2) = 1/3
P({2}) ≠P({2}|Even)
– Given that even number showed up increases the
probability that the number was 2, hence these
are not independent events
• Outcome of second throw of dice is supposed
to be independent of the first throw of dice
P(consecutive {2}) = P({2})*P({2}) = 1/6*1/6 = 1/36
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Independence
• If A1, A2..An are independent then,
P(A1ПA2П…ПAn) = P(A1)P(A2|A1)P(A3|A1ПA2)…P(An|A1П..ПAn-1)
(chain rule)
= P(A1)P(A2)P(A3)…P(An)
Independence assumption is often used in NLP to simplify
computations of complicated probabilities.
S
For example, probability of a
parse tree is often simplified as the
product of probabilities of
generating individual productions.
NP
VP
Article NN
The
girl
Verb
ate
NP
Article
NN
the
cake
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Conditional Independence
• Two events A and B are conditionally
independent given C if
P(AПB|C) = P(A|C)P(B|C)
Conditional independence is encountered more
often than unconditional independence.
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Maximum Likelihood Estimate
• P(A teenager will drink and drive) = # of
Teenagers who drink and drive/# of teenagers
• Suppose out of a sample of 5 teenagers one
drinks and drives
P(A teenager will drink and drive) = 1/5
• Relative frequency estimates can be proven to be
maximum likelihood estimates (MLE) because
they maximize the probability that it will generate
the sampled data
• Any other probability value will explain the data
with less probability
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Maximum Likelihood Estimates
• If P(TDD) = 1/5 then P(not TDD)=4/5
– Probability of the sampled data making assumption
that each teenager is independent
• P(Data)=1/5*(4/5)4=0.08192
– It will be less for any other value of P(TDD), for
P(TDD)=1/6,
• P(Data) = 1/6*(5/6)4 = 0.0803.
– For P(TDD)=1/4
• P(Data) = 1/4*(3/4)4 = 0.07091
• Whenever possible to compute, simple frequency
counts are not only intuitive but theoretically also
the best probability estimates
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Bayes’ Theorem
• Lets us calculate P(B|A) in terms of P(A|B)
• For example, using Bayes’ theorem we can
calculate P(Hypothesis|Evidence) in terms of
P(Evidence|Hypothesis) which is usually easier
to estimate.
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Bayes Theorem
P( E | H ) P( H )
P( H | E ) 
P( E )
Simple proof from definition of conditional probability:
P( H  E )
P( H | E ) 
P( E )
(Def. cond. prob.)
P( H  E )
(Def. cond. prob.)
P( E | H ) 
P( H )
P( H  E)  P( E | H ) P( H )
QED: P( H | E ) 
P( E | H ) P( H )
P( E )
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Random Variable
• Represents a measurable value associated with
events, for example, the number that showed up
on the dice, sum of the numbers of consecutive
throws of a dice
• Let X represent the number that showed up
P(X=2) is the probability that 2 showed up
• Let Z represent sum of the numbers that showed
up on throwing the dice twice
P(Z > 5) is the probability that the sum was greater than
5
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Probability Distribution
• Probability distribution: Specification of probabilities
for all the values of a random variable
Example: X represents the number that shows up when a
dice is thrown, a probability distribution (should add to 1)
X
P(X)
1
1/6
2
1/6
3
2/6
4
1/6
5
1/12
6
1/12
Given a probability distribution, probability of any event over
the random variable can be computed, P(X>5), P(X=2 or 3)
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Joint Probability Distribution
• The joint probability distribution for a set of
random variables, X1,…,Xn gives the probability of
every combination of values: P(X1,…,Xn)
• Given a joint probability distribution, probability
of any event over the random variables can be
computed
• Example:
S: shape (circle, square)
C: color (red, blue)
L: label (positive, negative)
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Joint Probability Distribution
• Joint distribution of P(S,C,L):
S,C,L
P(S,C,L)
Circle, Red, Positive
0.2
Circle, Red, Negative
0.05
Circle, Blue, Positive
0.02
Circle, Blue, Negative
0.2
Square, Red, Positive
0.02
Square, Red, Negative
0.3
Square, Blue, Positive
0.01
Square, Blue, Negative
0.2
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Marginal Probability
Distributions
• The probability of all possible conjunctions
(assignments of values to some subset of
variables) can be calculated by summing the
appropriate subset of values from the joint
distribution
P(red  circle)  0.20  0.05  0.25
P(red )  0.20  0.02  0.05  0.3  0.57
• In general, distribution of subset of random
variables, for e.g. P(C) or P(C^S), can be
computed from joint distribution, these are called
marginal probability distributions
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Conditional Probability
Distributions
• Once marginal probabilities are computed,
conditional probabilities can also be
calculated
P( positive| red  circle) 
P( positive red  circle) 0.20

 0.80
P(red  circle)
0.25
• In general, distributions of subsets of random
variables with conditions, for e.g. P(L|C^S),
can also be computed, these are called
conditional probability distributions
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Language Models
Most of these slides have been adapted from
Raymond Mooney’s slides from his NLP course at UT
Austin.
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What is a Language Model (LM)?
• Given a sentence how likely is it a sentence of the
language?
The dog bit the man. => very likely or 0.75
Dog man the the bit. => very unlikely or 0.002
The dog bit man. => likely or 0.15
• A probabilistic model is better than a formal
grammar model which will only give a binary
decision
• To specify a correct probability distribution,
the probability of all sentences in a language
must sum to 1
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What are the Uses of an LM?
• Speech recognition
– “I ate a cherry” is a more likely sentence than “Eye eight
uh Jerry”
• OCR & Handwriting recognition
– More probable sentences are more likely correct readings
• Machine translation
– More likely sentences are probably better translations
• Generation
– More likely sentences are probably better NL generations
• Context sensitive spelling correction
– “Their are problems wit this sentence.”
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What are the Uses of an LM?
• A language model also supports predicting
the completion of a sentence.
– Your program does not ______
• Predictive text input systems can guess what
you are typing and give choices on how to
complete it.
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What is the probability of a
sentence?
P(A1ПA2П…ПAn) = P(A1)P(A2|A1)P(A3|A1ПA2)…P(An|A1П..ПAn-1)
(chain rule)
ff,your,cell)
Estimate the above probabilities from a large corpus
–
–
Too many probabilities (parameters) to estimate
They become sparse, cannot be estimated well.
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What is the probability of a
sentence?
Approximate the probability by making independence
assumptions
Bigram approximation:
Trigram approximation:
your)P(phone|your,cell)
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N-Gram Model Formulas
• Word sequences
w1n  w1...wn
• Chain rule of probability
n
P( w )  P( w1 ) P( w2 | w1 ) P(w3 | w )...P( wn | w )   P( wk | w1k 1 )
n
1
2
1
• Bigram approximation
n 1
1
k 1
n
P( w )   P( wk | wk 1 )
n
1
k 1
• N-gram approximation
n
P( w )   P( wk | wkk1N 1 )
n
1
k 1
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Estimating Probabilities
• N-gram conditional probabilities can be estimated
from raw text based on the relative frequency of
word sequences, they are the maximum
likelihood estimates
Bigram:
N-gram:
C ( wn 1wn )
P( wn | wn 1 ) 
C ( wn 1 )
n 1
n  N 1
P(wn | w
C (wnn1N 1wn )
)
C (wnn1N 1 )
• To have a consistent probabilistic model, append
a unique start (<s>) and end (</s>) symbol to
every sentence and treat these as additional
words
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Generative Model of a Language
• An N-gram model can be seen as a
probabilistic automata for generating
sentences.
Until </s> is generated do:
Stochastically pick the next word based on the conditional
probability of each word given the previous N 1 words.
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Training and Testing a Language Model
• A language model must be trained on a
large corpus of text to estimate good
parameter (probability) values
• Model can be evaluated based on its ability
to predict a high probability for a disjoint
(held-out) test corpus
• Ideally, the training (and test) corpus
should be representative of the actual
application data
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Unknown Words
• How to handle words in the test corpus
that did not occur in the training data, i.e.
out of vocabulary (OOV) words?
• Train a model that includes an explicit
symbol for an unknown word (<UNK>).
– Choose a vocabulary in advance and replace
other words in the training corpus with <UNK>.
– Replace the first occurrence of each word in
the training data with <UNK>.
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Evaluation of an LM
• Ideally, evaluate use of model in end application
(extrinsic)
– Realistic
– Expensive
• Evaluate on ability to model test corpus (intrinsic)
– Less realistic
– Cheaper
37
Perplexity
• Measure of how well a model “fits” the test data
• Uses the probability that the model assigns to the
test corpus
• Normalizes for the number of words in the test
corpus and takes the inverse
PP (W )  N
1
P ( w1w2 ...wN )
• Measures the weighted average branching factor
in predicting the next word (lower is better)
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Sample Perplexity Evaluation
• Models trained on 38 million words from
the Wall Street Journal (WSJ) using a
19,979 word vocabulary.
• Evaluate on a disjoint set of 1.5 million WSJ
words.
Unigram
Perplexity
962
Bigram
170
Trigram
109
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Smoothing
• Since there are a combinatorial number of
possible word sequences, many rare (but not
impossible) combinations never occur in training,
so MLE incorrectly assigns zero to many
parameters (also know as sparse data problem).
• If a new combination occurs during testing, it is
given a probability of zero and the entire
sequence gets a probability of zero
• In practice, parameters are smoothed (also know
as regularized) to reassign some probability mass
to unseen events.
– Adding probability mass to unseen events requires
removing it from seen ones (discounting) in order to
maintain a joint distribution that sums to 1.
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• “Hallucinate” additional training data in which each
word occurs exactly once in every possible (N1)gram context and adjust estimates accordingly.
C ( wn1wn )  1
C ( wn1 )  V
Bigram:
P( wn | wn1 ) 
N-gram:
n 1
C
(
w
n  N 1wn )  1
P(wn | wnn1N 1 ) 
C (wnn1N 1 )  V
where V is the total number of possible words (i.e.
the vocabulary size)
• Tends to reassign too much mass to unseen events,
• More advanced smoothing techniques have also
41
been developed
Model Combination
• As N increases, the power (expressiveness)
of an N-gram model increases, but the
ability to estimate accurate parameters
from sparse data decreases (i.e. the
smoothing problem gets worse).
• A general approach is to combine the
results of multiple N-gram models of
increasing complexity (i.e. increasing N).
42
Interpolation
• Linearly combine estimates of N-gram
models of increasing order
Interpolated Trigram Model:
Pˆ (wn | wn2, wn1 )  1P(wn | wn2, wn1 )  2 P(wn | wn1 )  3 P(wn )
Where:

i
1
i
• Learn proper values for i by training to
(approximately) maximize the likelihood of
an independent development (also known as
tuning) corpus
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Backoff
• Only use lower-order model when data for higher-order
model is unavailable (i.e. count is zero).
• Recursively back-off to weaker models until data is
available.
n 1
n 1

P
*
(
w
|
w
)
if
C
(
w
n 1
n
n  N 1
n  N 1 )  1
Pkatz ( wn | wn N 1 )  
n 1
n 1

(
w
)
P
(
w
|
w
otherwise
n  N 1
katz
n
n N 2 )

Where P* is a discounted probability estimate to reserve
mass for unseen events and ’s are back-off weights
44
A Problem for N-Grams LMs:
Long Distance Dependencies
• Many times local context does not provide the
most useful predictive clues, which instead are
provided by long-distance dependencies
– Syntactic dependencies
• “The man next to the large oak tree near the grocery store on
the corner is tall.”
• “The men next to the large oak tree near the grocery store on
the corner are tall.”
– Semantic dependencies
• “The bird next to the large oak tree near the grocery store on
the corner flies rapidly.”
• “The man next to the large oak tree near the grocery store on
the corner talks rapidly.”
• More complex models of language that use syntax
and semantics are needed to handle such
dependencies
45
Domain-specific LMs
• Using a domain-specific corpus one can
build a domain-specific language model
• For example, train a language model for
each difficulty level (grades 1 to 12)
• Then automatically predict difficulty level
of a new document and recommend for the
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A Large LM
trillion words in 2006
• Available through Linguistic Data Consortium
(LDC) (not free)
http://www.ldc.upenn.edu/Catalog/CatalogEntry.jsp?catalogId=LDC2006T13
• Data Sizes
•
•
•
•
•
•
•
File sizes: approx. 24 GB compressed (gzip'ed) text files
Number of tokens: 1,024,908,267,229
Number of sentences: 95,119,665,584
Number of unigrams:
13,588,391
Number of bigrams:
314,843,401
Number of trigrams:
977,069,902
Number of fourgrams: 1,313,818,354
•
Number of fivegrams:
1,176,470,663
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Homework 1, Due next week in class
Find marginal distributions for P(C^S) and
P(C) from the joint distribution shown in
slide #23. Compute the conditional
distribution P(S|C) from this.
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