Markov Models

1
Markov Models

• Like the Bayesian network, a Markov model is a
  graph composed of
• states that represent the state of a process
• edges that indicate how to move from one state to
  another where edge is annotated with a
  probability indicating the likelihood of taking
  that transition
• Unlike the Bayesian network, the Markov models
  nodes are meant to convey temporal states so a
  transition from state1 to state2 means that in
  time 1 you are in state 1 and in time 2 you have
  moved on to state 2
• An ordinary Markov model contains states that are
  observable so that the transition probabilities
  are the only mechanism that determines the state
  transitions
• We will find a more useful version of the Markov
  model to be the hidden Markov model, covered in a
  few of slides

2
A Markov Model

• In the Markov model, we move from state to state
  based on simple probabilities
• going from S3 to S2 has a likelihood of a32
• going from S3 to S3 has a likelihood of a33
• going from S3 to S4 has a likelihood of a34
• likelihoods are usually computed stochastically
  (statistically)

We will use our Markov model to compute the
likelihoods of a number of state transitions that
might be of interest For instance, if we start
in S1, what is the probability of going from S1
to S2 to S3 to S4 to S5 and back to S1? What is
the probability of going from S1 to S1 to S1 to
S1 to S2? Etc.
3
Example Weather Forecasting

• On any day, it will either be
• rainy/snowy, cloudy or sunny
• we have the following probability matrix to
  denote given any particular day, what the weather
  will be like tomorrow
• so the probability, given today is sunny that
  tomorrow will be sunny is 0.8
• the probability, given today is rainy/snowy that
  tomorrow is cloudy is 0.2
• to compute a sequence, we multiply them together,
  so if today is sunny then the probability that
  the next two days will be sunny is 0.8 0.8, and
  the probability that the next three days will be
  cloudy is 0.1 0.1 0.1

R/S Cloudy Sunny
R/S .4 .3 .3
Cloudy .2 .6 .2
Sunny .1 .1 .8
4
Continued

• Lets assume today is cloudy and find the most
  likely sequence of three days
• There are 8 such sequences
• cloudy, cloudy, cloudy .6 .6 .36
• cloudy, cloudy, rainy .6 .2 .12
• cloudy, cloudy, sunny .6 .2 .12
• cloudy, rainy, cloudy .2 .3 .06
• cloudy, rainy, rainy .2 .4 .08
• cloudy, rainy, sunny .2 .3 .06
• cloudy, sunny, cloudy .2 .1 .02
• cloudy, sunny, rainy .2 .1 .02
• cloudy, sunny, sunny .2 .8 .16
• for simplicity, assume rainy really means rainy
  or snowy
• So the most likely sequence is three cloudy days
  in a row because today is cloudy
• But what if we didnt know what today would be?

5
Enhanced Example

• Lets assume that the probability of the first day
  being cloudy .5, rainy .2 and sunny .3
• These are our prior probabilities
• Since we do not know the first day is cloudy, we
  now have 27 possible combinations
• CCC, CCR, CCS, CRC, CRR, CRS, CSC, CSR, CSS, RCC,
  RCR, RCS, RRC, RRR, RRS, RSC, RSR, RSS, SCC, SCR,
  SCS, SRC, SRR, SRS, SSC, SSR, SSS
• The most likely sequence now is SSS .3 .8
  .8 .192 even though cloudy is the most likely
  first day (the probably for CCC .5 .6 .6
  .18)
• So, as with a Bayesian network, we have prior
  probabilities and multiply them by our
  conditional probabilities, which here are known
  as transitional probabilities

6
HMM

• Most interesting AI problems cannot be solved by
  a Markov model because there are unknown states
  in our real world problems
• in speech recognition, we can build a Markov
  model to predict the next word in an utterance by
  using the probabilities of how often any given
  word follows another
• how often does lamb follow little?
• A hidden Markov model (HMM) is a Markov model
  where the probabilities are actually
  probabilistic functions that are based in part on
  the current state, which is hidden (unknown or
  unobservable)
• determining which transition to take will require
  additional knowledge than merely the state
  transition probabilities

7
Example Speech Recognition

• We have observations, the acoustic signal
• But hidden from us is intention that created the
  signal
• For instance, at time t1, we know what the signal
  looks like in terms of data, but we dont know
  what the intended sound was (the phoneme or
  letter or word)
• The goal in speech recognition is to identify the
  actual utterance (in terms of phonetic units or
  words)
• but the phonemes/words are hidden to us
• We add to our model hidden (unobservable) states
  and appropriate probabilities for transitions
• the observables are not states in our network,
  but transition links
• the hidden states are the elements of the
  utterance (e.g., phonemes), which is what we are
  trying to identify
• we must search the HMM to determine what hidden
  state sequence best represents the input
  utterance

8
Example HMM

• Here, X1, X2 and X3 are the hidden states
• y1, y2, y3, y4 are observations
• Aij are the transition probabilities of moving
  from state i to state j
• bij make up the output probabilities from hidden
  node i to observation j that is, what is the
  probability of seeing output yj given that we are
  in state xi?

• Three problems associated with HMMs
• Given HMM, compute the probability of a given
  output sequence
• Given HMM and output sequence, compute most
  likely state transitions
• 3. Given HMM and output sequence, compute the
  transition probabilities

9
Formal Definition of an HMM

• The HMM is a graph, G V, E
• V is the set of vertices (nodes, states)
• E is the set of directed edges, or the
  transitions between pairs of nodes
• The HMM must have three sets of probabilities
• Each node in V that can be the first state of a
  sequence has a prior probability (we can denote
  nodes that cannot be the first state as having
  prior probability 0)
• For each state transition (edge in E), we need a
  transition probability
• For each node that has an associated observation,
  we need an output probability
• Commonly an HMM will represent some k distinct
  time periods where the states at time i are
  completely connected to the states at time i-1
  and time i1 although not always
• So, if there are n states and o possible
  observations at any time, there would be n prior
  probabilities, n(k-1) transition probabilities,
  and no output probabilities

10
Some Sample HMMs
11
HMM Problem 1

• As stated previously, there are three problems
  that we can solve with our HMM
• Problem 1 given an HMM and an output sequence,
  compute the probability of generating that
  particular output sequence (e.g., what is the
  likelihood of seeing this particular sequence of
  observations?)
• We have an observation sequence O O1 O2 O3 Ok
  and states
• Recall that we have 3 types of probabilities,
  prior probabilities, transition probabilities and
  output probabilities
• We generate every possible sequence of hidden
  states through the HMM from 1 to k and compute
• ps1 bs1(O1) as1s2 bs2(O2) as2s3 bs3(O3)
  ask-1sk bsk(Ok)
• Where p is the prior probability, a is the
  transition probability and b is the output
  probability
• Since there are a number of sequences through the
  HMM, we compute the above probability for each
  sequence and sum them up

12
Brief Example
We have 3 time units, t1, t2, t3 and each has 2
states, s1, s2 p(s1 at t1) .8, p(s2 at t1) .2
and there are 3 possible outputs , A, B, C Our
transition probabilities a are p(s1, s1) .7,
p(s1, s2) .3 and p(s2, s2) .6,
p(s2, s1) .4 Our output probabilities are p(A,
s1) .5, p(B, s1) .4, p(C, s1) .1
p(A, s2) .7, p(B, s2) .3, p(B, s2)
0 What is the probability of generating A, B,
C? Possible sequences are s1 s1 s1 .8 .5
.7 .4 .3 .1 0.00336 s1 s1 s2 .8
.5 .7 .4 .3 0 0.0 s1 s2 s1 .8
.5 .3 .3 .4 .1 0.00144 s1 s2 s2
.8 .5 .3 .3 .6 0 0.0 s2 s1 s1
.2 .7 .4 .4 .7 .1 0.001568 s2 s1
s2 .2 .7 .4 .4 .3 0 0.0 s2 s2
s1 .2 .7 .6 .3 .4 .1 0.001008 s2
s2 s2 .2 .7 .6 .3 .6 0 0.0
Likelihood of the sequence A, B, C is 0.00336
0.00144 0.001568 0.001008
0.007376
13
More Efficient Solution

• You might notice that there is a lot of
  repetition in our computation from the last slide
• In fact, the number of sequences is O(k nk)
• When we compute s2 s2 s2, we had already
  computed s1 s2 s2, so the last half of the
  computation was already done
• By using dynamic programming, we can reduce the
  number of computations
• this is particularly relevant when the sequence
  is far longer than 3 states and has far more
  states per time unit than 2
• We use a dynamic programming algorithm called the
  Forward algorithm (see the next slide)
• Even though we have a reasonably efficient means
  of solving problem 1, there is little need to
  solve this problem!

14
The Forward Algorithm

• We solve the problem in three steps
• The initialization step sets the probabilities of
  starting at each initial state at time 1 as
• a1(i) pibi(O1) for all states i
• That is, the probability of starting at some
  state i is the prior probability for i the
  output probability of seeing observation O1 from
  state i
• The main step is recursive for all times after 1
• at1(j) S at(i)aijbj(Ot1) for all states j
  at time t1
• That is, at time t1, the probability of being at
  state j is the sum of all of the previous states
  at time t leading to state j (at(i)aij) times
  the output probability of seeing Ot1 at time t1
• The final step is to sum up the probabilities of
  ending in each of the states at time n (sum up
  an(j) for all states j)

15
HMM Problem 2

• Given a sequence of observations, compute the
  optimal sequence of state transitions that would
  cause those observations
• Alternatively, we could say that the optimal
  sequence best explains the observations
• We need to define what we mean by optimal
• The sequence that contains the most individual
  states with the highest likelihoods?
• this sequence would contain the most states that
  appear to be correct states notice that this
  solution does not take into account transitions
• The sequence that contains the most number of
  correct pairs of states in the sequence
• this would take into account transitions
• or most number of correct triples, most number of
  correct quadruples, etc
• The sequence that is the most likely (probable)
  overall

16
The Viterbi Algorithm

• We do not know which of the sequences that were
  generated from problem 1 is actually the best
  path, we didnt keep track of that
• But through recursion and dynamic programming, we
  did keep track of portions of paths
• So we will again use recursion
• The recursive step works like this
• Lets assume at some time t, we know the best
  paths to all states
• At time t1, we extend each of the best paths to
  time t by finding the best transition from time t
  to a state at t1
• that is, we have to find a state at time t1 such
  that the path to time t transition to t1 is
  best
• we not only compute the new probability, but
  remember the path to this point

17
Viterbi Formally Described

• Initialization step
• d1(i) pibi(O1) same as in the forward
  algorithm
• y1(i) 0 this array will represent the state
  that maximized our path leading to the prior
  state
• The recursive step
• dt1(j) max dt(i)aijbj(Ot1) here, we
  look at all of the previous states i at time t,
  and compute the state transition from t to t1
  that gives us the maximum value of dt(i)aij
  multiply that by the likelihood of this state
  being true given this time units observation
  (see the next slide for a visual representation)
• yt1(j) argmax dt(i)aij which i from the
  possible preceding states led to the maximum
  value? Store that

18
Continued

• Termination step
• p maxdn(i) the probability that the path
  selected is correct is the path that has the
  largest probability as found in the final time
  step from the last recursive call
• q argmax dn(i) this is the last state
  reached
• Path backtracking
• Now that we have found the best path, we
  backtrack using the array y starting at yq
  until we reach time unit 1

At time t-1, we know the best paths to reach each
of the states Now at time t, we look at each
state si, and try to extend the path from t-1 to t
19
Viterbi in Pseudocode
20
Example Rainy and Sunny Days

• Your colleague in another city either walks to
  work or drives every day and his decision is
  usually based on the weather
• Given daily emails that include whether he has
  walked or driven to work, you want to guess the
  most likely sequence of whether the days were
  rainy or sunny
• Two hidden states rainy and sunny
• Two observables walking and driving
• Assume equal likelihood of the first day being
  rainy or sunny
• Transitional probabilities
• rainy given yesterday was (rainy .7, sunny
  .3)
• sunny given yesterday was (rainy .4, sunny
  .6)
• Output (emission) probabilities
• rainy given walking .1, driving .9
• sunny given walking .8, driving .2
• Given that your colleague walked, drove, walked,
  what is the most likely sequence of days?

21
Example Continued
Day 1 is easy to compute, prior probability
output probability The initial path to get to day
1 is merely from state 0
22
Example Continued
We determine that from day 1, it is more likely
to reach sunny from rainy it is more likely to
reach rainy from rainy as well, so day 2s path
to sunny is from rainy, and day 2s path from
rainy is from rainy
23
Example Concluded
From day 2, it is more likely to reach sunny from
sunny and it is more likely to reach rainy from
sunny, but day 3s most likely state is rainy.
Since we reached the rainy state from sunny, and
we reached Day 2s sunny state from rainy, we now
have the most likely path rainy, sunny, rainy
24
Why Problem 2?

• Unlike problem 1 which didnt seem to have any
  useful AI applications, problem 2 is has many
  different types of AI problems that it could
  solve
• This can be used to solve any number of credit
  assignment problems
• given a speech signal, what was uttered (what
  phonemes or words were uttered)?
• given a set of symptoms, what disease(s) is the
  patient suffering from?
• given a misspelled word, which word was intended?
• given a series of events, what caused them?
• What we have are a set of observations (symptoms,
  manifestations) and we want to explain them
• The HMM and Viterbi give us the ability to
  generate the best explanation where the term best
  means the most likely sequence through all of the
  states

25
How Do We Obtain our Probabilities?

• We saw one of the issues involved Bayesian
  probabilities was gathering accurate
  probabilities
• Like Bayesian probabilities, we need both prior
  probabilities and transition probabilities (the
  probability of moving from one state to another)
• But here we also need output (or emission)
  probabilities
• We can accumulate probabilities through counting
• Given N cases, how many started at state s1? s2?
  s3?
• although do we have enough cases to give us a
  good representative mix of probabilities?
• Given N cases, out of all state transitions, how
  often do we move from s1 to s2? From s2 to s3?
  Etc
• again, are there enough cases to give us a good
  distribution for transition probabilities?
• How do we obtain the output probabilities? That
  is, how do we determine the likelihood of seeing
  output Oi in state Sj?
• Thats trickier, and thats where HMM problem 3
  comes in

26
HMM Problem 3

• The final problem for HMMs is the most
  interesting and also the most challenging
• Given HMM and output sequence, update the various
  probabilities
• It turns out that there is an algorithm for
  modifying probabilities given a set of correct
  test cases
• The algorithm is called the Baum-Welch algorithm
  (also known as the Estimation-Modification or EM
  algorithm) which uses as a component, the
  forward-backward algorithm
• we already saw the forward portion of the forward
  algorithm, now we will take a look at the
  backward portion, which as you might expect, is
  very similar

27
Forward-Backward

• We compute the forward probabilities as before
• computing at(i) for each time unit t and each
  state i
• The backward portion is similar but reversed
• computing bt(i) for each time unit t and each
  state i
• Initialization step
• bt(i) 1 unlike the forward algorithm which
  used the prior probabilities, here we start at 1
  (notice that we also start at time t, not time 1)
• Recursive step
• bt(i) Saij bj(Ot1)bt1(j) the probability
  of reaching state i at time t backwards, is the
  sum of transitions from all states at time t1
  the probability of reaching state j at time t1
  the probability of being at state j given output
  Ot1
• this recursive step is almost the same as the
  step in the forward algorithm except that we use
  b instead of a

28
Baum-Welch (EM)

• Now that we have computed all the forward and
  backward path probabilities, how do we use them?
• First, we need to add a new value, the
  probability of being in state i at time t and
  transitioning to state j, which we will call
  xt(i, j)
• Fortunately, once we have run the
  forward-backward algorithm, this is easy to
  compute as
• xt(i, j) at(i)aijbj(Ot1)bt1(j) /
  denominator
• Before describing the denominator, lets
  understand the numerator
• this is the product of the probability of being
  at state i at time t multiplied by the transition
  probability of going from i to j multiplied by
  the output probability of seeing Ot1 at time t1
  multiplied by the probability of being at state j
  at time t1
• that is, it is the value derived by the forward
  algorithm for state i at time t the value
  derived by the backward algorithm for state j at
  time t1 transition output probabilities

29
Continued

• The denominator is a normalizing value so that
  all of our probabilities xt(i, j) for all states
  i and j add up to 1 for time t
• So this is merely the sum for all i and all j of
  at(i)aijbj(Ot1)bt1(j)
• Now we have some additional work
• We add gt(i) S xt(i, j) for all j at time t
• This represents the expected number of times we
  are at state i at time t
• If we sum up gt(i) for all times t, we have the
  number of expected times we are in state I
• Now recall that we may have started with improper
  probabilities (prior, transition and output)

30
Re-estimation

• By running the system on some test cases, we can
  accumulate probabilities of how likely a
  transition is, or how likely we start in a given
  state (prior probability) or how likely a state
  is for a given observation
• At this point of the Baum Welch algorithm, we
  have accumulated a summation (from the previous
  slide) of various states we have visited
• p(observation i state j) (expected number of
  times we saw observation i in the test case /
  number of times we achieved state j) (our
  observation probabilities)
• p(state i state j) (expected number of
  transitions from i to j / number of times we were
  in state j) (our transition probabilities)
• p(state i) a1(i)b1(i) / Sa1(i)b1(i) for all
  states i (this is the prior probability)

31
Continued

• The math may be elusive, and the amount of
  computations required is intensive but now we
  have the ability to
• Start with estimated probabilities (they dont
  even have to be very good)
• Use training examples to adjust the probabilities
• And continue until the probabilities stabilize
• that is, between iterations of Baum-Welch, they
  do not change (or their change is less than a
  given error rate)
• So HMMs can be said to learn the proper
  probabilities through training examples
• Each training example is merely the observations
  and the expected output (hidden states)
• The better the initial probabilities, the more
  likely it will be that the algorithm will
  converge to a stable state quickly, the worse the
  initial probabilities, the longer it will take

32
Example Determining the Weather

• Here, we have an HMM that attempts to determine
  for each day, whether it was hot or cold
• observations are the number of ice cream cones a
  person ate (1-3)
• the following probabilities are estimates that we
  will correct through learning

p(C) p(H) p(START)
p(1) 0.7 0.1   If today is cold (C) or hot (H), how many cones did I prob. eat? If today is cold (C) or hot (H), how many cones did I prob. eat?
p(2) 0.2 0.2 If today is cold (C) or hot (H), how many cones did I prob. eat? If today is cold (C) or hot (H), how many cones did I prob. eat?
p(3) 0.1 0.7 If today is cold (C) or hot (H), how many cones did I prob. eat? If today is cold (C) or hot (H), how many cones did I prob. eat?
p(C) 0.8 0.1 0.5 If today is cold or hot, what will tomorrow probably be? If today is cold or hot, what will tomorrow probably be?
p(H) 0.1 0.8 0.5 If today is cold or hot, what will tomorrow probably be? If today is cold or hot, what will tomorrow probably be?
p(STOP) 0.1 0.1 0 If today is cold or hot, what will tomorrow probably be? If today is cold or hot, what will tomorrow probably be?
33
Computing a Path Through the HMM

• Assume we know that the person ate in order, the
  following cones 2, 3, 3, 2, 3, 2, 3, 2, 2, 3,
  1,
• What days were hot and what days were cold?
• P(day i is hot j cones) ai(H) bi(H) /
  (ai(C) bi(C) ai(H) bi(H) )
• a(H), b(H), a(C) and b(C) were all computed
  using the forward-backward algorithm
• We started with guesses for our initial
  probabilities
• Now that we have run one iteration of
  forward-backward, we can apply re-estimation
• Sum up the values of our computations P(C 1)s
  and P(C)
• Recompute P(1 C) sum P(C 1) / P(C)
• we also do the same for P(C 2), and P(C 3) to
  compute P(2 C) and P(3 C) as well as the hot
  days for P(1 H), P(2 H), P(3 H)
• And we recompute P(C C), P(C H), etc
• Now our probabilities are more accurate (although
  not necessarily correct)

34
Continued

• We update the probabilities (see below)
• since our original probabilities will impact how
  good these estimates are, we repeat the entire
  process with another iteration of
  forward-backward followed by re-estimation
• we continue to do this until our probabilities
  converge into a stable state
• So, our initial probabilities will be important
  only in that they will impact the number of
  iterations required to reach these stable
  probabilities

p(C) p(H) p(START)
p(1) 0.6765 0.0584  
p(2) 0.2188 0.4251
p(3) 0.1047 0.5165
p(C) 0.8757 0.0925 0.1291
p(H) 0.109 0.8652 0.8709
p(STOP) 0.0153 0.0423 0
35
Convergence and Perplexity

• This system converged in 10 iterations to the
  probabilities shown in the table below
• Our original transition probabilities were part
  of our model of weather
• updating them is fine, but what would happen if
  we had started with different probabilities? say
  p(HC) .25 instead of .1?
• the perplexity of a model is essentially the
  degree to which we will be surprised by the
  results of our model because of the guesses we
  made when assigning a random probability like
  p(HC)
• We want our model to have a minimal perplexity so
  that it is most realistic

  p(C) p(H) p(START)
p(1) 0.6406 7.1E-05  
p(2) 0.1481 0.5343
p(3) 0.2113 0.4657
p(C) 0.9338 0.0719 5.1E-15
p(H) 0.0662 0.865 1.0
p(STOP) 1.0E-15 0.0632 0
36
Two Problems With HMMs

• There are two primary problems with using HMMs
• The first is minor what if a probability
  (whether output or transition) is 0?
• Because we are multiplying probabilities
  together, this would cause a path that goes
  through the state will have a probability of 0
  and so will never be selected
• To get around this problem, we will replace any 0
  probabilities with some minimum probability (say
  .001)
• The other is the complexity of the search
• Imagine we are using an HMM for speech
  recognition where the hidden states are the
  possible phonemes (say there are 30 of them) and
  the utterance consists of some 100 phonemes
  (perhaps 20 words)
• Recall that the complexity for the forward
  algorithm is O(TNT) where N is 30 and T is 100!
  Ouch
• So we might use a beam search to reduce the
  number of possible paths searched

37
Beam Search

• A beam search is a combination of the heuristic
  search idea along with a breadth-first search
• The beam search algorithm examines all of the
  next states accessible and evaluates them
• for an HMM, the evaluation is the probability a
  or b depending on whether we are doing a forward
  or backward pass
• In order to reduce the complexity of the search,
  only some of the states at this time interval are
  retained
• we might either keep the top k where k is a
  constant (known as the beam width) or we can use
  a threshold value and prune away states that do
  not exceed the threshold value
• if we discard a state, we are actually discarding
  the entire path that led us to that state (recall
  that the path would be the path that had the
  highest probability leading to that particular
  state at that time)

38
Forms of HMMs

• One of the most common form of HMM is called an
  Ergodic model this is a fully connected model,
  that is, every state has an edge to every other
  state
• From ealier in the lecture, we saw a slide of
  examples the bull/bear market and the
  cloudy/sunny/rainy day are examples
• The weather/ice cream cone example could be
  thought of as an Ergodic model, but instead we

would prefer to envision each day as being in a
new state, so this leads us to the Forward
Trellis model
Each variant of HMM has its own training
algorithms although they are all based on
Baum-Welch
39
Bakis and Factorial HMMs

• The Bakis model is one used to denote precise
  temporal changes where states transition left to
  right across the model where each state
  represents a new time unit
• States may also loop back onto themselves
• This is often used in speech recognition for
  instance to represent portions of a phonetic unit
• see below to the left
• Factorial HMMs when the system is so complex
  that a given state cannot represent the process
  of a single state in the model
• at time i, there will be multiple states, all of
  which lead to multiple successor states and all
  of which have emission probabilities from the
  observations input (see the figure below)

40
Hierarchical HMM

• We use this model when each state is itself a
  self-contained probabilistic model including
  their own hidden nodes
• That is, a state has its own internal HMM

• The rationale for having a HHMM is that each
  state can represent a sequence of observations
  instead of a one-to-one mapping of observation
  and state
• For instance, q2 might consist of 3 or more
  observations as shown in the figure

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N-Grams

• N-Gram HMMs the transition probabilities here
  are not just from the previous time unit, but
  from n-1 prior time units
• The N-gram is primarily used in natural language
  understanding or genetics types of problems where
  we can accumulate the transition probabilities
  from some corpus of data

• The bi-gram is the most common form of n-gram
  used in natural language understanding
• To the right is some bigram data for the
  frequency of two-letter pairs in English (out of
  2000 words)
• Tri-grams are also somewhat commonly used but it
  is rare to go beyond tri-grams

TH 50 AT 25 ST 20 ER 40 EN 25 IO 18 ON 39
ES 25 LE 18 AN 38 OF 25 IS 17 RE 36 OR 25
OU 17 HE 33 NT 24 AR 16 IN 31 EA 22 AS 16
ED 30 TI 22 DE 16 ND 30 TO 22 RT 16 HA 26
IT 20 VE 16
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Applications for HMMs

• The first impressive use of HMMs in AI was for
  speech recognition (in the late 80s)
• Since then, a lot of other applications have been
  tested
• Hand written character recognition
• Natural language understanding
• word sense disambiguation
• machine translation
• word matching (for misspelled words)
• semantic tagging of words (could be useful for
  the semantic web)
• Bioinformatics (e.g., protein structure
  predictions, gene analysis and sequencing
  predictions)
• Market predictions
• Diagnosis of mechanical systems