Cap. 6
HMMs
HMMs

HMM Basics
– Architecture
– Observable Markov Model
– Hidden Markov Model

HMM in Speech Recognition
transitions
Arquitecture
states
Observable Markov Model
The observations is enough to know which state the system is.
In other words, every state has a defined output.
Raining
Cloudy
Transition Matrix
Sun
Playing with coins
Transitions from one state
to other means that
the observation changed.
This state represents
the Tails
This state
represents
the Heads
P(H) represents the
probability of
throwing heads.
This is the
state sequence.
Hidden Markov Model (HMM)
Concept
 HMM Generator
 Types of HMMs
 Variatios of HMMs Structures
 Left-right HMM Architecture

Concept
The observations you DON´T exactly know which state the system is.
Every state output has a set of possibilities, which can be modeled with a pdf.
Two coin modeling.
Transitions from one state
to other means we change
the coin.
This state represents
one coin.
This state represents
one coin
Any coin can be
either Tail of Head
with this distribution.
This is a possible
state sequence.
Tree coin modelling.
This state represents
one coin
Transitions from one state
to another means
we change the coin.
This state represents
one coin.
This state represents
one coin.
Any coin can be
either Tail of Head
with this distribution.
This is a possible
state sequence.
We ask a person to
take balls of any URN.
All URNs have all colors.
HMM Modelling
Every URN as a different
distribution of colors.
HMM Generator
1. Choose an initial state.
2. Set t=1.
3. Choose ot=vk according to the probability
distribution (bjk).
4. Transition to a new state j according to the
state state transition probability for state.
5. Set t=t+1
6. If (t<T) return to step 3
7. end.
Types of HMMs
Full conected
Left-rigth
Parallel path
Left-rigth
Variations on HMM structures
Null transitions
Allows to generate short
sequences and still going
from state 1 to state N.
Allows alternative
word pronunciations.
Every arch
is a HMM itself.
Allows arbitrary
sequence length.
Left-right HMM Architecture
Finite State Networks, which consist of
Transition Probabilities
Initial
State
1
bk O(t)
2
Internal
States
3
b2O(t)
4
N-1
b3O(t)
b4O(t)
bk O(t) k
Internal states has the
probability of emitting a
vector.
 
O(t )  F1
k1 k 2 k 3 k 4 k 5 k 6 k 7
 
O(t )  F1  k
Final
state
N
b( N 1)1O(t)
Other example
Transition Probabilities
Initial
State
1
2
3
b2O(t)
4
b3O(t)
Final
state
N
N-1
b4O(t)
b( N 1)1O(t)
Internal states has the
probability of emitting a
vector.
Probability Calculation
a22
1
a12
a23
2
v1
P=
a33
v2
3
v3
v4
a44
a34
a45
55
4
v5
v6
a12b2v1a22b2v2a22b2v a23b3v a34b4v a44b4v a45
3
4
5
6
Known as:
Observed
sequence
/E//i//a/
a22
1
a12
 F 1
v1   
 F 2
a33
a23
2
3
a44
a34
 F 1
v 3    v   F1
 F1
 F 2 4  F 2
v2   
 F 2
a45
4
 F 1
v5   
 F 2
55
 F1
v6   
 F 2
HMMs in Speech Recognition
HMMs in Speech Recognition


Introduction
Isolated Word Recognition
– DD-HMM


Recognition
Training
– CD-HMM


Recognition
Training
– Practical Problems and Solutions
– Implementation Advise

Connected Word Recognition
DD-HMM Recognition Process
Input Data
O(t)  [o(1)o(2)o(n)]
It needs a model
for unit of recognition.
Codebook
VQ
M1
O(t)  [ k1 , k 2 , k 3 )]
M2
Probability
of each model
.
P( M q / O)
q=12
, ,,Q
MQ
Higher Probability
P( M i / O)  P( M k / O)
k  i
P( O / M i ) P( M i ) P(O / M k ) P( M k )

P(O)
P(O)
Recognised word
P(O / Mi ) P( Mi )  P(O / M k ) P( M k )
k  i
P(O / M i )  P(O / M k )
k  i
k  i; P(M k )  P(M k ) q=1,2,M
Vector Quantisation (VQ)

Convert an infinite number of parametric vectors to a
finite set of vectors.
codebook

 v 11
 2
 v1
 
 K
 v 1
v 12
v 22

v 2K
 v 1p 

 v 2p 
  

 v Kp 
codeword
Each speech vector is assigned a new vector
(codeword). Choosen from a codebook, using:


k  arg min d v, v i 
vi

In fact, every vector can be represented by a value k,
which is the index of the codeword closer to the
vector.
Constructing the codebook

Goal:
 mind v
P
p 1
p ,vk

k  1,, K;
K  2n
Table vectors
Training vectors

Clustering algorithms are used to solve
this problem.
Metodologies to calculate P(O / Mi )

Direct calculation

Total Verisimilitude (Forward Procedure)
– Takes into account all possible state’s sequences.

Higher Verisimilitude (Viterbi algorithm)
– Dynamic programming
– Also obtains the optimal sequence assignment.

Backward Procedure
Direct Calculation
State sequence
q=[q1,q2,...,qT]
P(O/Mk)= S P(O/q, Mk)P(q/ Mk)
Probability of the observation
given the state sequence and the
Model.
bq1,o(1) bq2,o(2) bq3,o(3) ...bqT,o(T)
The problem with this calculation
needs (2T-1) NT multiplications and
NT-1 additions.
Probability of the sequence
given the Model.
a1q1 aq1,q2 aq2,q3 ...aqT,qT
For N=5 and T=100, it needs around
1072computations!!
Total Verisimilitude
1
2
3
o(1) o( 2 )
i
o( 3)
O( t )  o(1), o( 2), , o( t )
i (t )  P(O( t ), s(t )  i / M k
1

 i (t )  0
0

i = 1; t = 0
o( t )
1 < i < N; t = 0
P(O / M k )    j (T )a jN
j 2
Addition over
all possible
predecesors.
N
Transition
Probability
form j to i
 N 1

)     j (t  1) a
b
ji  io(t)

 j 1
i = 1;0 < t  T
N 1
N-1
1 i  N
0t T
Once being in state i
Probability of being
emit h the vector o(t)
in time t 1
in a predecessor state j .
Algorithm

Initialization
1

 i (t )  0
0


i = 1; t = 0
i = 1;0 < t  T
1 < i < N; t = 0
Recursion:
 N 1

i (t )     j (t  1)a ji  biO ( t )  k
 j 1


Termination:
N 1
P(O / M k )    j (T )a jN
j 2
1 i  N
0t T
Example
Observed Signal
290 720 330 285 
O(t )  

820 1160 975 2298
Using the codebook
we obtain the
Templates(references):


O(t )  k  1 2 1 3
Codebook
index
1
2
3
F1
300
725
290
F2
850
1100
2325
1

 i (t )  0
0


i = 1;0 < t  N
1 < i < N; t = 0


i (t )     j (t  1)a ji  biO ( t )  k
 j 1

N 1
1.0
1
0.3
0.2
0.7
0.5
2
3
0.3
0.5
0.5
5
4
b2 (1)  0.8 b2 (1)  01
.
b2 (1)  01
.
b2 ( 2)  01
. b2 (2)  0.7 b2 (2)  01
.
b2 (3)  01
. b2 (3)  0.2 b2 (3)  0.8
N 1
s

O(t )  k  1 2 1 3
i = 1; t = 0
P(O / M k )    j (T )a jN
j 2
0.00477
0.0
0.0
0.024
0.01008
0.00954
0.0
0.0
0.28
0.0204
0.00311
0.0
0.8
0.16
0.00256
0.00005
1.0
0.0
0.0
0.0
0.0
t
Higher Verisimilitude
1
2
3
o(1) o( 2 )
i
o( 3)
O( t )  o(1), o(2), , o(t)
i (t )  P(O( t ), s(t )  i / M k )
1

i (t )  0
0

N-1
o( t )
Transition
Probability
form j to i
 max  j (t  1) a ji
1i  N
N

bio(t)
1 i  N
0t T
i = 1; t = 0
i = 1;0 < t  T
Once being in state i
Probability of being
emit h the vector o(t)
1 < i < N; t = 0
in time t 1
Over all possible
in a predecessor state j .
P(O / M k )  max  j (T )a jN predecesors.
1 j  N


Algorithm


Inicialisation
Recursion:
1

 i (t )  0
0

i = 1; t = 0
i = 1;0 < t  T
1 < i < N; t = 0


i (t )  max  j (t  1)a ji bio( t )  k
1 j  N
 i (t )  arg max  j (t  1)a ji bio(t )k 
1 j  N


Termination:
1 j  N


P(O / M k )  max  j (T )a jN

P (O / M k )  arg max  j (T )a jN 
1 j  N

Backtracking:
q(t)=Fi(q(t+1))
1 i  N
0t T
For sequence
tracking.
1

 i (t )  0
0


i = 1;0 < t  T
1 < i < N; t = 0

1.0

i (t )  max  j (t  1)a ji bio( t )  k
1 j  N

P(O / M k )  max  j (T )a jN
1 j  N
s

O(t )  k  1 2 1 3
i = 1; t = 0
1
0.3
0.2
0.7
0.5
2
3
0.3
0.5
0.5
5
4
b2 (1)  0.8 b2 (1)  01
.
b2 (1)  01
.
b2 ( 2)  01
. b2 (2)  0.7 b2 (2)  01
.
b2 (3)  01
. b2 (3)  0.2 b2 (3)  0.8

0.00235
0.0
0.0
0.024
0.0084
0.0047
0.0
0.0
0.28
0.0196
0.00274
0.0
0.8
0.16
0.00256
0.00005
1.0
0.0
0.0
0.0
0.0
t
Backward Procedure

Is an alternative way to calculate
P(O/Mk).

We will see later how the forward and
Backward Procedures are used to solve
the training problem.
1
N-1
i
j2
j1
o( t  1)
O' ( t )  o( t  1), o( t  2 ), , o( T )
i (t )  P(O' ( t ), s(t )  i / M k
o( t  2 ) o( t  3)
i (T )  aiN
P(O / M k )  1 (0)
N
o( T )
Transition
Probability
form i to j
 N 1

)    a b jo(t 1)  (t  1)

j
ij
 j 2

Addition over
all possible
successors.
N-1
Once being in state j
emit h the vectoro( t  1)
1 j  N
0t T
Probability of being
in time t 1
in a successor state j
and have emitted O'. (t  1)


O(t )  k  1 2 1 3
i (T )  aiN
N 1
a b
j 2
ij
jo(t 1)
1.0
1
 jo( t )
0.5
2
3
0.005
0.01
0.2
0.5
0.07356
0.0144
0.12
0.0
0.05628
0.0144
0.12
0.0
0.0
0.0
0.0
0.7
b2 (1)  0.8 b3 (1)  01
.
b2 ( 2)  01
. b3 (2)  0.7
b2 (3)  01
. b3 (3)  0.2
s
0.0045024
0.3
0.2
0.0
0.3
0.5
0.5
5
4
b4 (1)  01
.
b4 (2)  01
.
b4 (3)  0.8
t
Training
Training

Introduction

Reestimation equations

Iterating reestimation equations
Introduction
In last section is was assume that we
knew the HMM parameters. However,
we did not know it.
 HMM can learn from training data.
 Before proceed it is convenient to
define a backward probability

Input Data
O(t)  [o(1)o(2)o(n)]
Codebook
VQ
N , p, k
O(t)  [ k1 , k 2 , k 3 )]
Training
M1
M2
Probability
of each model
.
P( M q / O)
q=12
, ,,Q
MQ
1
2
i
o(1) o( 2 )
o( 3)
j2
j1
o( t )
o( t  1)
N
N-1
o( t  2 ) o( t  3)
o( T )
O  o(1), o( 2), , o( T )
P(O, s(t )  i / M k )  i (t ) i (t )
This represents the probability of emitting
O
and be at state
From this equation, we can obtain the probability of been at state
i
i
at time
in time
t
given
t.
O.
P(O, s(t )  i / M k )  P(O / M k ) P( s(t )  i / O, M k )
P(O, s(t )  i / M k )
P( s(t )  i / O, M k ) 
P(O / M k )
P( s(t )  i / O, M k ) 
1
 (t ) i (t )
p i
p  P( O / M k )
Reestimation Equations

One Observation:
– Transition Probability reestimation equation
– State Emission Discrete Probability
reestimation equation

Multiple Observations
Transition Probability

The re-estimation formula for aij is given
by:
T 1
1
i (t )aij b jo ( t )  j (t  1)

t 1 p
aij 
T 1
1
 i (t )  j (t )

t 1 p

Prove:
Expected number of transition
from state i
Transition Probability Reestimation Prove

To obtain the re-estimation formula for aij , first lets
define:
P(O, s(t )  i, s(t  1)  j,/ M k )  i (t )aij b jo(t )  j (t  1)
P(O, s(t )  i, s(t  1)  j, / M k )  P(O / M k )P(s(t )  i, s(t  1)  j / O, M k )
P(O, s(t )  i , s(t  1)  j / M k )
P( s(t )  i , s(t  1)  j / O, M k ) 
P( O / M k )
1
P( s(t )  i , s(t  1)  j ,/ O, M k )  i (t )aij b jo( t )  j (t  1)
p

p  P( O / M k )
Hence, the expected number of transition i->j is given
by:
T 1
1
i (t )aij b jo( t )  j (t  1)

t 1 p

Therefore, the re-estimation formula for aij
is given by:
T 1
1
 i (t )aij b jo(t )  j (t  1)

t 1 p
aˆ ij 
T 1
1
 i (t )  j (t )

t 1 p
Expected number of transition
from state i
State Emision Probabilty
reestimation

The reestimator for bik can be calculated as
follows:
bik 
1
i (t ) i (t )

t o ( t )  k  p
T
1
i (t ) i (t )

t 1 p
bik

Prove:
k
State Emision Probability
reestimation equation Prove

The number of times that state i is busy
is given by:
T
1
i (t ) i (t )

p
t 1

and the number of times a specific
vector k is emitted is given by:
1
i (t )i (t )

p
t o ( t )  k 
State Emision Probabilty
reestimation

Therefore, a new estimator for bik can
be calculated as follows:
bik 
1
i (t ) i (t )

p
t o ( t )  k 
T
1
i (t ) i (t )

p
t 1
Multiple Observation Sequences


To have sufficient data to make estimations reliable
estimates of the model parameters, one have to use
multiple observations.
A multiple observation is defined as follows:
O  [O(1) , O( 2) ,O( R) ]

Since the reestimation formulas are based on
frequency of occurrence of various events, the
reestimation formulas are modified by adding
together the individual frequencies of occurrence for
each sequence.
Thus, the modified reestimation formulas are:
r
R
b j ,o ( t )
R
aij 
1
r
r


r  i ( t ) i ( t )
r 1 t o r ( t )  k  p

R
Tr
1
r
r


r  i ( t ) i ( t )
r 1 t 1 p
T r 1

r 1 t 1
1
r
r
r  i ( t ) a ij b j ,o r ( t 1 ) i ( t  1)
p
Tr
R

r 1 t 1
R
aiN 

r 1
R
1
r
r
r  i ( t ) i ( t )
p
1
r
r
r  i ( T ) i ( T )
p
Tr

r 1 t 1
1
r
r

(
t
)

r
i
i (t )
p
Iterating Reestimation Equations
Since, alphas and betas calculations
depends on alphas and betas, hence
reestimation formulas need an initial
model.
 If we define the current model as M k ,
and we define the reestimated model as
'
M
. k , then it has been proven that model
'
M
k in
M’kk is more likely than model M
Mk,
the sense that

P(O| M )  P(O| M k )
'
k
Training
Input Data
O(t)  [o(1)o(2)o(n)]
Mk
Codebook
VQ
O(t)  [ k1 , k 2 , k 3 )]
Re-estimation
equations
M k'
M1
M2
Probability
of each model
.
P( M q / O)
q=12
, ,,Q
MQ
M k  M k'
'
Mk
M
 Moreover, if we iterate
k in place of
and repeat the iteration procedure until
some limitation point is reached, we then
can improve the probability of ???
being observed.
Input Data
O(t)  [o(1)o(2)o(n)]
Codebook
VQ
O(t)  [ k1 , k 2 , k 3 )]
M k  M k'
Training
Mk
P(O| M k )
No
Re-estimation
equations
P(O| M k' )  P(O| M k )  
Yes
M k'
Mi
M2
Probability
of each model
P( M q / O)
q=12
, ,,Q
MQ
P(O| M k' )
Discrete HMM Disadvantages

Require VQ.
– Codebook with few vectors:

poor representation of the acoustic space.
– Codebook with many vectors:
needs high storage space and
 requires large training set.

Continuous HMM
Assume a parametric model for the speech vector variations in a state.
Initial
State
1
2
b2O(t)
3
4
b3O(t)
Final
state
N
N-1
b4O(t)
b( N 1)1O(t)
It common to assume a
multivariable Gaussian distribution.
characterised for the
means and variances.

In this case:
1
b j (o(t )) 
(2 ) p / 2 C j

1
2
e


1
o(t ) j
2

T

Cj 1 o ( t )   j

Assuming the observations vectors are
not correlated:
p
b j (o(t ))  
k 1

1
(2 )
p/2
 jk
e
1
2 2jk
o
k ( t )   jk

2

In this case, every speech component is
modeled as a one dimensional
Gaussian distribution:
 jk
Pj v k 
1
(2 ) 1/ 2  jk
 jk

and we calculate:
b  v   P  v   P v  P v  P v
j
j
j
1
j
2
j
p

Recognition using CD-HMM
Input Data
O(t)  [o(1)o(2)o(n)]
It needs a model
for unit of recognition.
M1
M2
Recogniser
Probability of each model.
P( M q / O)
q=12
, ,,Q
MQ
Higher Probability
P( M i / O)  P( M k / O)
k  i
Recognised word
P( O / M i ) P( M i ) P(O / M k ) P( M k )

P(O)
P(O)
k  i
P(O / Mi ) P( Mi )  P(O / M k ) P( M k )
k  i
P(O / Mi )  P(O / M k )
k  i; P( M k )  P( M k ) q=12
, , M
It uses the same equations of DD-HMM,
however bi ,o( k ) is a continuous
distribution.
 Thus, we can use either:

– Forward Procedure
– Viterbi Algorithm
– Backward Procedure
Training CD-HMM
Alphas and betas are calculated as
before.
 The transition probabilities are also
calculated as before.


The re-estimation formula are:
– For the mean
 1

r
r
r



(
t
)

(
t
)
o
(
t
)


i
 pr i

r 1 t 1 

ˆ i 
R
Tr
 1

r
r



 p r  i (t )  i (t ) 

r 1 t 1 

Tr
R
– For the covariance:
Number of times
in state i was visited.
 1
T
r
r
r
r


(
t
)

(
t
)
o
(
t
)


o
(t )   j


i
i
i
r
 p
r 1 t 1 

R Tr
 1

 r  ir (t )  ir (t ) 


r 1 t 1  p

R
ˆ ij
Values of the vector
in state i.

Tr

– For the variance (diagonal elements):
R

2
ii

Tr
 1
   p
r 1 t 1

r
2

 1

r
r



(
t
)

(
t
)


i
 pr i
r 1 t 1 

R

 ir (t )  ir (t )o r (t )   ik  
T
r



Sometimes, the Gaussian Distribution is
not very good distribution model:

In this case a mixture of Gaussians can
be used:
b ( v )   c N v ,  , c 
M
j
m1
jm
jm
M
c
m1
jm
1
jm
Practical Problems and Solutions

Computational Underflow

Insufficient data training

Initial Estimates

State Duration HMM Modelling
Computational Underflow


Alfa beta’s values get very small as number of states
and times increases:
Solutions:
– Scaling alphas at every time t.
c( t ) 
1
N .1

s 1
si
(t )
i (t )  c(t )i (t )

The LogProbability can be obtain
as follows:
T
Plog S (O / Mk )   log(c(i ))
– Using logProbabilities
– Scaling betas
t 1
1

 i (t )  0
0


i = 1;0 < t  T
1 < i < N; t = 0

1.0

i (t )  max  j (t  1)a ji bio( t )  k
1 j  N

P(O / M k )  max  j (T )a jN
1 j  N
s

O(t )  k  1 2 1 3
i = 1; t = 0
1
0.3
0.2
0.5
2
3
0.3
0.5
0.5
5
4
b2 (1)  0.8 b2 (1)  01
.
b2 (1)  01
.
b2 ( 2)  01
. b2 (2)  0.7 b2 (2)  01
.
b2 (3)  01
. b2 (3)  0.2 b2 (3)  0.8

T
Plog S (O / Mk )   log(c(i ))  4.6
t 1
1.0
0.0
0.0
0.075
0.2529
0.6338
0.0
0.0
0.875
0.3877
0.2038
0.0
1.0
0.05
0.3594
0.1624
1.0
0.0
0.0
0.0
1.25
0.7
2.5
4.49
0.0
2.25
3.1556
t
1

 i (t )  0
0


i = 1;0 < t  T
1 < i < N; t = 0

1.0

i (t )  max  j (t  1)a ji bio( t )  k
1 j  N
s

O(t )  k  1 2 1 3
i = 1; t = 0
1
0.7
0.5
2
3
0.3
0.5
0.5
5
4
b2 (1)  0.8 b2 (1)  01
.
b2 (1)  01
.
b2 ( 2)  01
. b2 (2)  0.7 b2 (2)  01
.
b2 (3)  01
. b2 (3)  0.2 b2 (3)  0.8
T
Plog S (O / Mk )   log(c(i ))  4.6
t 1
Demostración
0.0
0.0
0.3*01=0.03
0.0
0.0
0.5*0.7=0.35
0.0
1.0
0.2*0.1=0.02
1.0
0.0
0.0
1.25
0.3
0.2
2.5
0.0
0.0
t
Using log Probability
1
2
3
o(1) o( 2 )
i
o( 3)
O( t )  o(1), o( 2), , o( t )
 (t )   log P(O( t ), s(t )  i / M k )
'
i
0

i' (t )  


N-1
o( t )
min 
1 j  N
N
Transition
logProbability
form j to i

 j' (t  1) log(a ji )  log(bio(t) )
1 i  N
0t T
i=1;t=0
i=1;0<t  T
logProbability of being
1<i<N;t=0
in time t 1
Over all possible
in a predecessor state j .
predecessors.

P(O / M k )  min  j' (T )  log(a jN )
1 j  N

Once being in state i
emit the vector o(t)
Algorithm


Inicialisation
Recursion:
0

i' (t )  


i=1;t=0
i=1;0<t  T
1<i<N;t=0


i' (t )  min  j' '(t  1)  log a ji  log bio( t )
1 j  N

0t T

 i' (t )  arg min  j' (t  1)  log a ji  log bio(t )
1 j  N

Termination:

P(O / M k )  min  j' (t  1)  log a jN
1 j  N


P(O / M k )  arg min  j' (t  1)  log a jN
1 j  N

Backtracking:
q(t )   i' (q(t  1))
For sequence
tracking.

0

 i (t )  



i = 1;0 < t  T
1 < i < N; t = 0


1.0
 (t )  min  '(t  1)  log a ji  log bio( t )
'
i

O(t )  k  1 2 1 3
i = 1; t = 0
1 j  N
'
j


1
s

P(O / M k )  min  j' (t  1)  log a jN
1 j  N


P(O / M k )  arg min  j' (t  1)  log a jN
1 j  N
0.7
0.5
2
3
b2 (1)  0.8 b3 (1)  01
.
b2 ( 2)  01
. b3 (2)  0.7
b2 (3)  01
. b3 (3)  0.2
 i' (t )  arg min  j' (t  1)  log a ji  log bio( t )
1 j  N
0.3
0.2

0.3
0.5
0.5
5
4
b4 (1)  01
.
b4 (2)  01
.
b4 (3)  0.8
6.046
Inf
Inf
3.72
4.77
5.3531
Inf
Inf
1.273
3.93
5.897
Inf
0.223
4.1331
5.976
09.886
0.0
Inf
Inf
Inf
Inf
t
Scaling Betas


The betas also suffer the underflow problem.
Hence, we have to scale them or use
logProbability.
The scaling factor for alpha and beta at time t
should be the same. Hence, the
recommended procedure is
– for time t


calculate the alphas values
calculate the scaling factor and store it.
– for time t


calculate the betas;
scale the betas with the scaled factor used for the alphas
at time t.
Insufficient data training

Since there are low-probability events, hence
poor parameter estimation.
In example, if the training set is to small
some symbols could not be in the training set.
bik 

1
i (t ) i (t )

t o ( t )  k  p
This is zero
if not symbol is counted.
T
1
i (t ) i (t )

t 1 p
Solutions:
– set parameter thresholds.
– Train with more data (if possible).
Initial Estimates

Reestimation equations give parameter
values which converge to a local
maximum.
– Experience shown that
aij parameters converge to the global parameter
without problem.
 state distribution function (bi,o(t) ) or parameters
(mean and variance) need good initial estimates.

– Segmental k-means Segmentation into States
Segmental k-means
Segmentation into states
State Duration HMM Modelling


For most physical signals, the exponential
distribution is inappropriate.
In order to improve modeling, we have to
incorporate state duration information in a
HMM.
– Incorporate state duration information into the
mechanics of HMMs.
– Heuristic method.
Heuristic for incorporating state
duration into HMM

At training:
– Segmental k-means algorithm is used.
– Calculate the state duration probability pj(d).

At recognition:
– Viterbi algorithm to obtain


The logProbability and
the best segmentation via backtracking.
– The duration of each state is measure from the
state segmentation.
– A post processor increases the logProbability as
follows:
N
log Pˆ q,O/Mk   log Pq,O/Mk   α d  log [p j (d j )]
j 1
Implementation Advise

If the number of models, in the decoding
process, is very large, then try to save
memory. Observe that for a given time t, it
only need a limited quantity of information:
s
0.00477
0.0
0.0
0.024
0.01008
0.00954
0.0
0.0
0.28
0.0204
0.00311
0.0
0.8
0.16
0.00256
0.00005
0.0
0.0
0.0
0.0
t
s
0.00477
0.0
0.0
0.024
0.01008
0.00954
0.0
0.0
0.28
0.0204
0.00311
0.0
0.8
0.16
0.00256
0.00005
0.0
0.0
0.0
0.0
s
0.00932
0.0
0.0
0.037
0.02008
0.0154
0.0
0.0
0.61
0.074
0.00561
0.0
0.8
0.32
0.0046
0.00025
0.0
0.0
0.0
0.0
s s
0.00932
0.0
0.0
0.037
0.02008
0.0154
0.0
0.0
0.61
0.074
0.00561
0.01972
s
0.0
s
0.0 0.0
0.0 0.8
0.32
0.037
0.0046
0.02008
0.00025
0.0154
0.0
0.0 0.0
0.610.0
0.0
0.074
0.0
0.00561
0.0
0.8
0.0046
0.02008
0.00025
0.0154
0.0
0.0
0.0
0.0
0.8
0.32
0.037
0.0
0.61
0.0
0.0
0.074
0.0
0.0
0.32
0.037
0.0
0.61
0.0
0.8
0.32
0.0046
0.0
0.0
0.0
0.0
0.0
0.0
0.0133
0.0
0.00561
0.00025 0.0035
0.0154
0.0046
0.02008
0.074
0.00561
0.00025
0.0
0.0
0.0
Connected Word Recognition

Problems with isolated word recognition:
– we don´t know the limits of the words.
– Increases variability



coarticulation of words
Speech velocity
Solutions:
–
–
–
–
Unconstrained End Point Matching
DP in two dimensions
one-pass DP algorithm.
Viterbi
Viterbi Algorithm
0.3
1.0
0.2
1.0
0.1
1.0
0.2
0.5
0.5
0.3
0.3
0.6
0.7
0.3
0.3
0.7
0.8
0.2
0.5
0.5
0.4
0.6
0.9
0.1
1. Inicialisation:
Each state as a token t
2. Calculate every state of every model using:
t.probj
t.probj+log(aij)
t.startj
3. Update every internal state of every model:
max(t.probj+log(aij))
frame at which the token at
state j enter the model
0.3
0.2
4. Update state N of every model:
1.0
5. Find mode with higher logProbability
0.7
0.5
0.5
0.5
0.3
6. Update limits table
7. Copy token in
state 1 of each
model.
0.0
0
Inf
0
-Inf
4.5
4.5
0
-Inf
0
1.2
0
3.2
0
2.1
0
0
2.2
0
2.7
0
5.2
0
0
2.5
0
5.3
0
5.3
0.2
0.1
1.0
w.model
w.start
2 1
0 0
0.3
0.7
0.2
0.1
0.7
0.0
0
Inf
0
-Inf
0
-Inf
0
4.5
0
2.3
0
3.5
0
4.3
0
4.5
0
7.3
0
4.5
0
1.3
0
4.5
4.5
0
2.7
0
0
0
For N words
0.3
1.0
0.2
1.0
0.1
1.0
0.2
0.5
0.5
0.3
0.3
0.6
0.7
0.3
0.3
0.7
0.8
0.2
0.5
0.5
0.4
0.6
0.9
0.1
Recovering the uttered words.



Array w, is the same length as the number of
observations.
This array gives information of the limits of
the words.
At the end of the utterance, value w.model[M]
stores the last HMM recognised sequence,
and the predecesor models are obtain by
“tracking back” trough the array.
Finite State Syntax (FSS)

With token passing the FSS are
straightforward implemented.
Monterrey
Monterrey
to
Frankfort
Frankfort
from
Houston
Houston
Cap. 8 Large Vocabulary
Continuous Speech Recognition

Problems with isolated word recognition:
– Do not easily account for variations in word
pronunciation accross different dialects, etc.

Solution:
– Use subword speech units.

Problems with large vocabulary continuous
Speech Recognition:
– we don´t know the limits of the subword units.
– Increases variability


coarticulation of subword units
Speech velocity

La probabilidad MAP (Maximum A
Posteriory) de la cadena de palabras W
dadas la observaciones:
P(W / O)  max P(O / W )
W

Utilizando la regla de Bayes:
P(O / W ) P(W )
P(W / O) 
P(O)

ya que P(O) es independiente de W
P( M i / O)  arg max P(O / M i ) P( M i )
Mi
Modelo Acústico:
•Sub-palabras.
•PhoneLike Units(PLU):50
•Syllable-like Units (SLU): 10,000
•Dyad: 2000
•Acústicas (Clustering): 256-512
Modelo de Lenguaje:
•Restricciones
•Sintácticas:
•“parser” o
•n-gram (n=2,3,4)
•Par de palabras (word pair)
•semánticas del Lenguaje.
Issues in the choice of subword
units sets

Depende de la:
– Sensibilidad del contexto
– De lo fácil de entrenar
– Ejemplo:
PLU exageradamente sensibles al contexto
(varian mucho al variar el contexto), pero por
ser tan pocas son sencillas de entrenar.
 SLU muy poco sensibles al contexto, pero son
muchos y muy difícil de entrenar (mejor
entrenamos todas las palabra).

Subword unit Models based on
HMMs

Normalmente es de tres estados.
0.1
0.3
1.0
0.6
0.8
0.9
0.1
0.2
Subword Training
1. Given a label training set of speech
sentences, where every sentece
consists of
speech waveforms and
 Its transcripts into words

SW  W1W2W3W4 ... WI
– Assuming that the waveform segmentation
into words is NOT available.

A word lexicon thats provide a transcription of
every word in the training set in terms of the
set of subword units being trained is
available.
Table 8.2 Rabiner & Juang).
Therefore, the sentences are represented in terms
of its subwords, as follow:
S U  U 1 W1 U 2 W1 U L W1  W1   U1 W2 U 2 W2 U L W2  W2  
U 1 W3 U 2 W3 U L W3  W3     U1 WI U 2 WI U L WI  WI 
2. Create an extended (composite) HMM
for each sentence as follow:
– Replace each subword unit by its HMM
W2
W1
– Incorporate Silences between words and at
the begining and end of each sentence.
Sil
W1
Sil
W2
Sil
3. Estimate the subword unit model
parameters which maximize the
likelihood of the models for all the given
training data using:
– Forward-backward procedure
– Segmental K-means training procedure
W1
Sil
Sil
Sil
Segmental K-means training
procedure
Initialization
 Clustering
 Estimation
 Segmentation
 Iteration

Fig. 8.5
Fig. 8.6
Language Models for Large
Vocabulary Speech Recognitin

Goal:
– Provide an estimate of the probability of a
word sequence
PW   Pw1 w2 w3 wQ 
for the given recognition task.

This can be solved as follows:
PW   Pw1 w2 w3  wQ   Pw1 Pw2 | w1 Pw3 | w1 w2 
PwQ | w1 w2  wQ 1 

Since, it is impossible to reliable estimate the
conditional probabilities,
Pw j | w1 w2 w j 1 

hence in practice it is used an N-gram word
model:
PwQ | w1w2 wQ1   PwQ | w j  N 1w j  N 2 w j 1 

En practice, realiable estimators are obtained
for N=2 (bigram) or possible N=3 (trigram).
Smoothing

Since many trigrams are rarerly found,
even in large amount of text. Smoothing
is used as follows:
Pw3 | w1 w2   3 Pw3 | w1 w2   2 Pw3 | w2   1 Pw3 
3  2  1  1
Optimal Linear Smoothing
Recognition
0.3
1.0
0.2
1.0
0.1
1.0
0.2
0.5
0.5
0.3
0.3
0.6
0.7
0.3
0.3
0.7
0.8
0.2
0.5
0.5
0.4
0.6
0.9
0.1
1. Inicialisation:
Each state as a token t
2. Calculate every state of every model using:
t.probj
t.probj+log(aij)
t.startj
3. Update every internal state of every model:
max(t.probj+log(aij))
frame at which the token at
state j enter the model
0.3
0.2
4. Update state N of every model:
1.0
5. Find mode with higher logProbability
0.7
0.5
0.5
0.5
0.3
6. Update limits table
7. Copy token in
state 1 of each
model.
0.0
0
Inf
0
-Inf
4.5
4.5
0
-Inf
0
1.2
0
3.2
0
2.1
0
0
2.2
0
2.7
0
5.2
0
0
2.5
0
5.3
0
5.3
0.2
0.1
1.0
w.model
w.start
2 1
0 0
0.3
0.7
0.2
0.1
0.7
0.0
0
Inf
0
-Inf
0
-Inf
0
4.5
0
2.3
0
3.5
0
4.3
0
4.5
0
7.3
0
4.5
0
1.3
0
4.5
4.5
0
2.7
0
0
0
For N subunits
0.3
1.0
0.2
1.0
0.1
1.0
0.2
0.5
0.5
0.3
0.3
0.6
0.7
0.3
0.3
0.7
0.8
0.2
0.5
0.5
0.4
0.6
0.9
0.1
Recovering the uttered words.



Array w, is the same length as the number of
observations.
This array gives information of the limits of
the words.
At the end of the utterance, value w.model[M]
stores the last HMM recognised sequence,
and the predecesor models are obtain by
“tracking back” trough the array.
Finite State Syntax (FSS)

With token passing the FSS are
straightforward implemented.
Monterrey
Monterrey
to
Frankfort
Frankfort
from
Houston
Houston