HMM - Part 2

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HMM - Part 2

• The EM algorithm
• Continuous density HMM

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The EM Algorithm

• EM Expectation Maximization
• Why EM?
• Simple optimization algorithms for likelihood
  functions rely on the intermediate variables,
  called latent dataFor HMM, the state sequence is
  the latent data
• Direct access to the data necessary to estimate
  the parameters is impossible or difficultFor
  HMM, it is almost impossible to estimate (A, B,
  ?) without considering the state sequence
• Two Major Steps
• E step computes an expectation of the likelihood
  by including the latent variables as if they were
  observed
• M step computes the maximum likelihood estimates
  of the parameters by maximizing the expected
  likelihood found in the E step

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Three Steps for EM

• Step 1. Draw a lower bound
• Use the Jensens inequality
• Step 2. Find the best lower bound ? auxiliary
  function
• Let the lower bound touch the objective function
  at the current guess
• Step 3. Maximize the auxiliary function
• Obtain the new guess
• Go to Step 2 until converge

Minka 1998
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Form an Initial Guess of ?(A,B,?)
objective function
current guess
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Step 1. Draw a Lower Bound
objective function
lower bound function
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Step 2. Find the Best Lower Bound
objective function
lower bound function
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Step 3. Maximize the Auxiliary Function
objective function
auxiliary function
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Update the Model
objective function
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Step 2. Find the Best Lower Bound
objective function
auxiliary function
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Step 3. Maximize the Auxiliary Function
objective function
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Step 1. Draw a Lower Bound (contd)
Objective function
If f is a concave function, and X is a r.v.,
then Ef(X) f(EX)
Apply Jensens Inequality
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Step 2. Find the Best Lower Bound (contd)

• Find that makes
• the lower bound function
• touch the objective function
• at the current guess

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Step 2. Find the Best Lower Bound (contd)
Set it to zero
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Step 2. Find the Best Lower Bound (contd)
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EM for HMM Training

• Basic idea
• Assume we have ? and the probability that each Q
  occurred in the generation of O
• i.e., we have in fact observed a complete
  data pair (O,Q) with frequency proportional to
  the probability P(O,Q?)
• We then find a new that maximizes
  
  
• It can be guaranteed that
• EM can discover parameters of model ? to maximize
  the log-likelihood of the incomplete data,
  logP(O?), by iteratively maximizing the
  expectation of the log-likelihood of the complete
  data, logP(O,Q?)

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Solution to Problem 3 - The EM Algorithm

• The auxiliary function
• where and
  can be expressed as

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Solution to Problem 3 - The EM Algorithm (contd)

• The auxiliary function can be rewritten as

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Solution to Problem 3 - The EM Algorithm (contd)

• The auxiliary function is separated into three
  independent terms, each respectively corresponds
  to , , and
• Maximization procedure on can be
  done by maximizing the individual terms
  separately subject to probability constraints
• All these terms have the following form

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Solution to Problem 3 - The EM Algorithm (contd)

• Proof Apply Lagrange Multiplier

Constraint
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Solution to Problem 3 - The EM Algorithm (contd)
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Solution to Problem 3 - The EM Algorithm (contd)
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Solution to Problem 3 - The EM Algorithm (contd)
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Solution to Problem 3 - The EM Algorithm (contd)

• The new model parameter set
  can be expressed as

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Discrete vs. Continuous Density HMMs

• Two major types of HMMs according to the
  observations
• Discrete and finite observation
• The observations that all distinct states
  generate are finite in number, i.e., Vv1, v2,
  v3, , vM, vk?RL
• In this case, the observation probability
  distribution in state j, Bbj(k), is defined as
  bj(k)P(otvkqtj), 1?k?M, 1?j?Not
  observation at time t, qt state at time t
• ? bj(k) consists of only M probability values
• Continuous and infinite observation
• The observations that all distinct states
  generate are infinite and continuous, i.e., Vv
  v?RL
• In this case, the observation probability
  distribution in state j, Bbj(v), is defined as
  bj(v)f(otvqtj), 1?j?Not observation at
  time t, qt state at time t
• ? bj(v) is a continuous probability density
  function (pdf) and is often a mixture of
  Multivariate Gaussian (Normal) Distributions

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Gaussian Distribution

• A continuous random variable X is said to have a
  Gaussian distribution with mean µand variance
  s2(sgt0) if X has a continuous pdf in the
  following form

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Multivariate Gaussian Distribution

• If X(X1,X2,X3,,XL) is an L-dimensional random
  vector with a multivariate Gaussian distribution
  with mean vector ? and covariance matrix ?, then
  the pdf can be expressed as
• If X1,X2,X3,,XL are independent random
  variables, the covariance matrix is reduced to
  diagonal, i.e.,

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Multivariate Mixture Gaussian Distribution

• An L-dimensional random vector X(X1,X2,X3,,XL)
  is with a multivariate mixture Gaussian
  distribution if
• In CDHMM, bj(v) is a continuous probability
  density function (pdf) and is often a mixture of
  multivariate Gaussian distributions

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Solution to Problem 3 The Segmental K-means
Algorithm

• Assume that we have a training set of
  observations and an initial estimate of model
  parameters
• Step 1 Segment the training data
• The set of training observation sequences is
  segmented into states, based on the current
  model, by Viterbi Algorithm
• Step 2 Re-estimate the model parameters
• Step 3 Evaluate the model If the difference
  between the new and current model scores exceeds
  a threshold, go back to Step 1 otherwise, return

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Solution to Problem 3 The Segmental K-means
Algorithm (contd)

• 3 states and 4 Gaussian mixtures per state

State
s3
s3
s3
s3
s3
s3
s3
s3
s3
s2
s2
s2
s2
s2
s2
s2
s2
s2
s1
s1
s1
s1
s1
s1
s1
s1
s1
1 2 N
O1
O2
ON
?12,?12,c12
?11,?11,c11
K-means
Global mean
Cluster 1 mean
Cluster 2mean
?13,?13,c13
?14,?14,c14
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Solution to Problem 3 The Intuitive View
(CDHMM)

• Define a new variable ?t(j,k)
• probability of being in state j at time t with
  the k-th mixture component accounting for ot

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Solution to Problem 3 The Intuitive View
(CDHMM) (contd)

• Re-estimation formulae for
  are

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A Simple Example
The Forward/Backward Procedure
S1
S1
S1
State
S2
S2
S2
1 2 3 Time
o1
o2
o3
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A Simple Example (contd)









q 1 1 1
q 1 1 2
Total 8 paths
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A Simple Example (contd)
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