Stochastic Methods

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Stochastic Methods

• A Review

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Random Variables

• A random variable is a function whose domain is a
  sample space and whose range is a set of
  outcomes, usually real numbers
• A boolean random variable is a function from an
  even space to true,false or 1.0, 2.0
• A bernoulli experiment
• Is performed n times
• The trials are independent
• The probability of success on each trial is a
  constant p the probability of failure is q
  1-p.
• A random variable, Y, counts the number of
  success in the n trials

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Example

• A fair die is cast six times
• Success a six is rolled
• Failure all other outcomes
• An observed sequence is a sequence of 4 1s and
  0s (0,0,1,0) means that a six has been rolled
  on the third trial.
• Call this event, X.
• Since every trial in the sequence is independent,
  the probability of this event is the product of
  the probabilities of the atomic events composing
  it.
• So, p(X) 5/6 5/6 1/6 5/6 (1/6)(5/6)3

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Binomial Probabilities

• In a sequence of Bernoulli trials, we are
  interested in the total number of successes, not
  their order
• Let the event, X, be the number of observed
  successes in n Bernoulli trials.
• If x individual successes occur, where x 0, 1,
  2, , n
• then n-x failures occur

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• Facts
• The number of ways of selecting x positions from
  n in total is just
• nXx
• Since the trials are independent with x successes
  and n-x failures, the probability of each success
  is just the probability of success times the
  probability of failure px(1-p)n-x
• The probability of the event X is the sum of the
  probabilities of all individual events
• (nCx)px(1-p)n-x
• The random variable, X, is said to have a
  binomial distribution

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What is the probability of obtaining 5 heads in 7
flips of a fair coin

• The probability of the event X, p(X), is the sum
  of the probabilities of all individual events
• (nCx)px(1-p)n-x
• The Event X is 5 successes
• n 7, x 5
• p(of a single success) ½
• p(of a single failure) ½
• P(X) (7C5)(1/2)5(1/2)2

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Expectation

• If the reward for the occurrence of an event E,
  with probabiliy p(E), is r, and the cost of the
  event not occurring, 1-p(E), is c, then the
  expectation for an even occuring, ex(E), is
• ex(E) r x p(E) c (1-p(E))

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Expectation Example

• A fair roulette wheel has integers, 0 to 36.
• Each player places 5 dollars on any slot.
• If the wheel stops on the spot, the player wins
  35, else she loses 1
• So,
• p(winning) 1/37
• P(losing) 36/37
• ex(E) 35(1/37) (-5)(36/37)
• (approx) -3.92

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Bayes Theorem For Two Events

• Recall that we defined conditional (posterior)
  probability like this
• We can also express s in terms of d
• Multiplying (2) by p(d) we get
• Substituting (3) into (1) gives Bayes theorem
  for two events

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• If d is a disease and s is a symptom, the theorem
  tells us that the probability of the disease
  given the symptom is the probability of the
  symptom given the disease times the probability
  of the disease divided by the probability of the
  symptom

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The Probability of the Intersection of Multiple
Events (that are not necessarily independent)

• Called the Multiplication Rule

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This can be extended to multiple events
Given
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Lets do one More
Called the Chain rule and can be proven by
induction
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Example Four cards are to be dealt one after
another at random and without replacement from an
fair deck. What is the probability of receiving a
spade, a heart, a diamond, and a club in that
order? A event of being dealt a spade B event
of being dealt a heart C event of being dealt a
diamond D event of being dealt a club P(A)
13/52 P(BA) 13/51 P(CA B) 13/50 P(DA B
C) 13/49 Total Probability
13/5213/5113/5013/49
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An Application

• Def A probabilistic finite state machine is a
  finite state machine where each arc is associated
  with a probability, indicating how likely that
  path is to be taken. The sum of the probabilities
  of all arcs leaving a node must sum to 1.0.
• A PFSM is an acceptor when one or more states are
  indicated as the start states and one or more
  states is indicated as the accept state.

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Phones/Phonemes

• Def A phone is a speech sound
• Def A phoneme is a collection of related phones
  (allophones) that are pronounced differently in
  different contexts
• So t is phoneme.
• The t sound in tunafish differs from the t
  sound in starfish. The first t is aspirated,
  meaning the vocal chords briefly dont vibrate,
  producing a sound like a puff a air. A t
  followed by an s is unaspirated
• FSA showing the probabilities of allophones in
  the word tomato

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More Phonemes

• This happens with a k and gboth are
  unaspirated, leading to the mishearing of the
  Jimi Hendrix song
• Scuse me, while I kiss the sky
• Scuse me, while I kiss this guy

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PFSA for the pronunciation of tomatoe
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Phoneme Recognition Problem

• Computational Linguists have collections of
  spoken and written language called corpora.
• The Brown Corpus and the Switchboard Corpus are
  two examples. Together, they contain 2.5 million
  written and spoken words that we can use as a base

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Now Suppose

• Our machine identified the phone I
• Next the machine has identified the phone ni (as
  in knee)
• Turns out that an investigation of the
  Switchboard corpus shows 7 words that can be
  pronounced ni after I
• the,neat, need, new, knee, to, you

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How can this be?

• Phoneme t is often deleted at the end of the
  word say neat little quickly
• the can be pronounced like ni after in or.
  Talk like Jersey gangster here or Bob Marley

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Strategy

• Compile the probabilities of each of the
  candidate words from the corpora
• Applies Bayes theorem for two events

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• Word Frequency Probability
• knee 61 .000024
• the 114834 .046
• neat 338 .00013
• need 1417 .00056
• new 2625 .001

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Apply Simplified Bayes
Since all of the candidtates will be divided by
pni, we can drop it off givingp(wordni)
p(niword)p(word))

• But where does p(niword) come from?
• Rules of pronunciation variation in English are
  well-known.
• Run them through the corpora and generate
  probabilities for each.
• So, for example, that word initial th becomes
  n if the preceding word ended in n is .15
• This can be done for other pronunciation rules

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Result

• Word p(niword) p(word) p(niword)p(word)
• New .36 .001 .00036
• Neat .52 .00013 .000068
• Need .11 .00056 .000062
• Knee 1.0 .000024 .000024
• The 0.0 .046 0.0
• The has a probability of 0.0 since the previous
  phone was the not n
• Notice that new seems to be the most likely
  candidate. This might be resolved at the
  syntactic level
• Another possibility is to look at the probability
  of two word combinations in the corpora
• I new is less probable than I need
• This is referred to as N-Gram analysis

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General Bayes Theorem

• Recall Bayes Theorem for two events
• P(AB) p(BA)p(A)/p(B)
• We would like to generalize this to multiple
  events

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Example

• Suppose
• Bowl A contains 2 red and 4 white chips
• Bowl B contains 1 red and 2 white chips
• Bowl C contains 5 red and 4 white chips
• We want to select the bowls and compute the p of
  drawing a red chip
• Suppose further
• P(A) 1/3
• P(B) 1/6
• P(C) ½
• Where A,B,C are the events that A,B,C are chosen

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• P(R) is dependent upon two probabilities p(which
  bowl) then the p(drawing a red chip)
• So, p(R) is the union of the probability of
  mutually exclusive events

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• Now suppose that the outcome of the experiment is
  a red chip, but we dont know which bowl it was
  drawn from.
• So we can compute the conditional probability for
  each of the bowls.
• From the definition of conditional probability
  and the result above, we know

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• We can do the same thing for the other bowls
• p(BR) 1/8
• P(CR) 5/8
• This accords with intuition. The probability
  that the red bowl was chosen increases over the
  original probability, because since it has more
  red chips, it is the more likely candidate.
• The original probabilities are called prior
  probabilities
• The conditional probabilities (e.g., p(AR)) are
  called the posterior probabilities.

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To Generalize

• Let events B1,B2,,Bm constitute a partition of
  the sample space S.
• That is

Suppose R is an event with B1 Bm its prior
probabilities, all of which 0, then R is the
union m mutually exclusive events, namely,
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• Now,
• If p(A) 0, we have from the definition of
  conditional probability that

P(BkR) is the posterior probability
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Example

• Machines A,B,C produce bolts of the same size.
• Each machine produces as follows
• Machine A 35, with 2 defective
• Machine B 25,
• with 1 defective
• Machine C 40
• with 3 defective
• Suppose we select one bolt at the end of the day.
  The probability that it is defective is

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• Now suppose the selected bolt is defective. The
  probability that it was produced by machine 3 is

Notice how the posterior probability increased,
once we concentrated on C since C produces both
more bolts and a more defective bolts.
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Evidence and Hypotheses

• We can think of these various events as evidence
  (E) and hypotheses (H).

Where p(HkE) is the probability that hypothesis
i is true given the evidence, E p(Hk) is the
probability that hypothesis I is true
overall p(EHk) is the probability of observing
evidence, E, when Hi is true m is the number of
hypotheses
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Why Bayes Works

• The probability of evidence given hypotheses is
  often easier to determine than the probability of
  hypotheses given the evidence.
• Suppose the evidence is a headache.
• The hypothesis is meningitis.
• It is easier to determine the number of patients
  who have headaches given that they have
  meningitis than it is to determine the number of
  patients who have meningitis, given that that
  they have headaches.
• Because the population of headache sufferers is

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But There Are Issues

• When we thought about bowls (hypotheses) and
  chips (evidence), the probability of a kind of
  bowl given a red chip required that we compute 3
  posterior probabilities for each of three bowls.
  If we also worked it out for white chips, we
  would have to compute 3X2 6 posterior
  probabilities.
• Now suppose our hypotheses are drawn from the set
  of m diseases and our evidence from the set of n
  symptoms, we have to compute mXn posterior
  probabilities.

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But Theres More

• Bayes assumes that the hypothesis partitions the
  set of evidence into disjoint sets.
• This is fine with bolts and machines or red chips
  and bowls, but much less fine with natural
  phenomena. Pneumonia and strep probably doesnt
  partition the set of fever sufferers (since they
  could overlap)

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That is

• We have to use a form of Bayes theorem that that
  considers any single hypothesis, hi, in the
  context of the union of multiple symptoms ei

If n is the number of symptoms and m the number
of diseases, this works out to be mxn2 n2 m
pieces of information to collect. In a expert
system that is to classify 200 diseases using
2000 symptoms, this is 800,000,000 pieces of
information to collect.
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Naïve Bayes to the Rescue

• Naive Bayes classification assumes that variables
  are independent.
• The probability that a fruit is an apple, given
  that it is red, round, and firm, can be
  calculated from the independent probabilities
  that the observed fruit is red, that it is round,
  and that it is firm.
• The probability that a person has strep, given
  that he has a fever, and a sore throat, can be
  calculated from the independent probabilities
  that a person has a fever and has a sore throat.

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• In effect, we want to calculate this

Since the intersection of sets is a set, Bayes
lets us write
Since we only want to classify and the
denominator is constant, we can ignore it giving
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Independent Events to the Rescue

• Assume that all pieces of evidence are
  independent given a particular hypothesis.
• Recall the chain rule

Since p(BA) p(B) and p(C)A B) p(C), that
is, the events are mutually exclusive, then
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Becomes (with a little hand-waving) P(hiE)
p(e1h)Xp(e2hiXXp(enhi)
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Leading to the naïve Bayes Classifier

• P(EHj)