1 hmm - part 2 review of the last lecture the em algorithm continuous density hmm
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1
HMM - Part 2
Review of the last lecture The EM algorithm Continuous density HMM
2
Three Basic Problems for HMMs
Given an observation sequence O=(o1,o2,…,oT), and an HMM =(A,B,)
– Problem 1:
How to compute P(O|) efficiently ?
The forward algorithm
– Problem 2:
How to choose an optimal state sequence Q=(q1,q2,……, qT)
which best explains the observations?
The Viterbi algorithm– Problem 3:
How to adjust the model parameters =(A,B,) to maximize P(O|)? The Baum-Welch (forward-backward) algorithm
cf. The segmental K-means algorithm maximizes P(O, Q* |)
* arg max ( , | )P Q
Q O Q
)|(maxarg*iP
i
OP(up, up, up, up, up|)?
3
The Forward Algorithm
The forward variable:– Probability of o1,o2,…,ot being observed and the state at time t
being i, given model λ
The forward algorithm
1 1 1 1
1 11
1
1. Initialization ( , | ) 1
2. Induction 1 1 1
3.Termination O
i i
N
t t ij j ti
N
Ti
α i P o q i π b o , i N
α j α i a b o , t T - , j N
P λ α i
λiqoooPi ttt ,...21
4
The Viterbi Algorithm
1. Initialization
2. Induction
3. Termination
4. Backtracking),...,,(
1,...,2.1),(**
2*1
*
*11
T
tt*t
qqq
TTtqq
Q
Ni, i
Ni, obπi ii
10)(
1
1
11
Nj,T-t, aij
Nj,T-t, obaij
ijtNi
tjijtNi
t
1 11][maxarg)(
1 11][max
11t
11
1
iq
iλP
TNi
*T
TNi
1
1
*
maxarg
max,QO
is the best state sequence
11
1 cf.
tj
N
iijtt obaiαjα
N
iT iαλP
1O
5
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 the 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
Number of " " in state ˆ Number of times in state j
k jb k
j
1Number of times ˆ
Number of training sequencesi
q i
Number of transitions from state to state ˆ
Number of transitions from state ij
i ja
i
1ˆ1
N
ii
1ˆ1
N
jija
1)(ˆ1
M
kj kb
6
Segmental K-means vs. Baum-Welch
Number of " " in state ˆ Number of times in state j
k jb k
j
1Number of times ˆ
Number of training sequencesi
q i
Number of transitions from state to state ˆ
Number of transitions from state ij
i ja
i
1 number of times ˆ
Number of training sequencesi
q i
Expected
number of " " in state ˆ number of times in state j
k jb k
j
Expected
Expected
number of transitions from state to state ˆ
number of transitions from state ij
i ja
i
Expected
Expected
7
The Backward Algorithm
The backward variable:– Probability of ot+1,ot+2,…,oT being observed, given the state at
time t being i and model
The backward algorithm
1 11
1 11
1. Initialization 1, 1
2. Induction , 1 1 1
3. Termination ( ) ( )
T
N
t ij j t tj
N
i ii
β i i N
i a b o j t T - , j N
P λ i b o
O
λiqoooPi tTttt ,,...,, 21
N
iT iαλP
1Ocf.
8
ot
The Forward-Backward Algorithm
Relation between the forward and backward variables
)(][
,...
11
21
ti
N
jjitt
ttt
obaji
iqoooPi
λ
N
jttjijt
tTttt
jobai
iqoooPi
111
21
)(
,...
λ
λiqPii ttt ,)( O
(Huang et al., 2001)
Ni tt iiλP 1 )(O
9
The Baum-Welch Algorithm (1/3)
Define two new variables:
t(i)= P(qt = i | O, ) – Probability of being in state i at time t, given O and
t( i, j )=P(qt = i, qt+1 = j | O, ) – Probability of being in state i at time t and state j at time t+1, given O
and
N
m
N
nttnmnt
ttjijtttt
nobam
jobai
λP
λjqiqPji
1 111
111 ,,,
O
O
Ni tt
tttttt
ii
ii
λP
ii
λP
iqPi
1
)|,(
OO
O
N
jtt jii
1
,
10
The Baum-Welch Algorithm (2/3)
t(i)= P(qt = i | O, ) – Probability of being in state i at time t, given O and
t( i, j )=P(qt = i, qt+1 = j | O, ) – Probability of being in state i at time t and state j at time t+1, given O
and
iiL
l
T
t
lt
l
state from ns transitioofnumber expected)( 1
1
1
iqiL
l
l
11
1 timesofnumber expected)(
jijiL
l
T
t
lt
l
state to state from ns transitioofnumber expected),( 1
1
1
11
The Baum-Welch Algorithm (3/3)
Re-estimation formulae for , A, and B are
How do you know ? ˆ( | ) ( | )P P O O
L
iiq
L
l
l
i
1
11
)(
sequences trainingofNumber
timesofnumber Expectedˆ
L
l
T
t
lt
L
l
T
t
lt
ijl
l
i
ji
i
jia
1
1
1
1
1
1
)(
),(
state from ns transitioofnumber Expected
state to state from ns transitioofnumber Expectedˆ
L
l
T
t
lt
L
l
T
vo
t
lt
jl
l
kt
j
j
j
jkkb
1 1
1 s.t.
1
)(
)(
statein timesofnumber Expected
statein "" ofnumber Expected)(ˆ
12
Maximum Likelihood Estimation for HMM
QQO
O
O
)|,(log maxarg
))|(log)(( )(maxarg
))|()(( )(maxarg
P
Pll
PLLML
,....,...,, 10 tHowever, we cannot find the solution directly.
An alternative way is to find a sequence:
....)(...)()( 10 tlll s.t.
13
])|,(
)|,([log
])|,(
)|,([log
)|,(
)|,(),|(log
)|,(
)|,(
)|(
)|,(log
)|,(
)|,(
)|(
)|,(log
)|(
)|,(log
)|(log)|,(log
)|(log)|(log)()(
),|(
),|(
tP
tP
tt
tt
t
t
t
t
t
t
tt
P
PE
P
PE
P
PP
P
P
P
P
P
P
P
P
P
P
PP
PPll
t
t
QO
QO
QO
QO
QO
QOOQ
QO
QO
O
QO
QO
QO
O
QO
O
QO
OQO
OO
OQ
OQ
Q
Q
Q
Q
Q
Jensen’s inequality
)|,(logmaxarg
)|,(log),|(maxarg
)|,(
)|,(log),|(maxarg
])|,(
)|,([logmaxarg
),|(
),|(
)1(
QO
QOOQ
QO
QOOQ
QO
QO
OQ
Q
Q
OQ
PE
PP
P
PP
P
PE
t
t
P
t
tt
tP
t
Q function
Solvable and can be proved that 1( ) ( )t tl l
1( | ) ( | )t tP P O OIf f is a concave function, and X is a r.v., thenE[f(X)]≤ f(E[X])
14
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: compute the expectation of the likelihood by including the
latent variables as if they were observed
• M step: compute the maximum likelihood estimates of the parameters by maximizing the expected likelihood found in the E step
Q QOOQ )|,(log),|( PλP
15
Three Steps for EM
Step 1. Draw a lower bound– Use the Jensen’s 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]
16
)(F
objective function
current guess
Form an Initial Guess of =(A,B,)
Given the current guess , the goal is to find a new guess such that
NEW
)()( NEWFF
)|(maxarg*
OP
17
)(F
)()( Fg
Step 1. Draw a Lower Bound
)(glower bound function
objective function
18
)(F
Step 2. Find the Best Lower Bound
objective function
lower bound function)(g
),( g
auxiliary function
19
)(F
),( g
Step 3. Maximize the Auxiliary Function
NEW
)()( FF NEW
auxiliary function
objective function
20
)(F
Update the Model
NEW
objective function
21
)(F
),( g
Step 2. Find the Best Lower Bound
objective function
auxiliary function
22
)(F
NEW
Step 3. Maximize the Auxiliary Function
)()( FF NEW
objective function
23
Step 1. Draw a Lower Bound (cont’d)
Q
Q
Q
QOQ
QOO
)(
)|,()(log
)|,(log)|(log
p
Pp
PP
Apply Jensen’s Inequality
A lower bound function of
)(F
If f is a concave function, and X is a r.v., thenE[f(X)]≤ f(E[X])
Q Q
QOQ
)(
)|,(log)(
p
Pp
Objective function
p(Q): an arbitrary probability distribution
24
Step 2. Find the Best Lower Bound (cont’d)
– Find that makes
the lower bound function
touch the objective function
at the current guess
*
( )
( , | ) We want to maximize ( ) log w.r.t ( ) at
( )
( , | )The best ( ) arg max ( ) log
( )p
Pp p
p
Pp p
p
Q
O QQ Q
Q
O QQ Q
Q
)(Qp
25
Step 2. Find the Best Lower Bound (cont’d)
),|()|(
)|,(
)|,(
)|,()(
)|,(
1)|,(
)()|,(
)(
1)|,(log)(log
01)(log)|,(log
)(log)()|,(log)()(1
here multiplier Lagrange a introduce we,1)( Since
OQO
QO
QO
QOQ
QO
QOQ
QOQ
QOQ
QQO
QQQOQQ
Q
Q
Q
QQQ
Q
PP
P
P
Pp
P
ee
e
Pep
e
Pep
Pp
pP
ppPpp
p
Take the derivative w.r.t and set it to zero)(Qp
26
Step 2. Find the Best Lower Bound (cont’d)
Q function
)|(log)|(log),|(
),|(
)|(),|(log),|(
),|(
)|,(log),|(),(
OOOQ
OQ
OOQOQ
OQ
QOOQ
Q
Q
Q
PPP
P
PPP
P
PPg
We can check
),(),|(
)|,(log),|(
g
P
PP
Q OQ
QOOQDefine
OQOQQOOQ ),|(log),|()|,(log),|(),( PPPPg
27
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|)
Q QOOQ )|,(log),|( PλP ˆ
)|()ˆ|( λPP OO Expectation
28
Solution to Problem 3 - The EM Algorithm
The auxiliary function
where and can be expressed as
Q
Q
QOO
QO
QOOQ
,log,
,log,,
PλP
λP
PλPλλQ
T
ttq
T
tqqq
T
ttq
T
tqqq
obaP
obaλP
ttt
ttt
1
1
1
1
1
1
logloglog,log
,
11
11
QO
QO
λP QO, QO,log P
29
Solution to Problem 3 - The EM Algorithm (cont’d)
The auxiliary function can be rewritten as
wi yi
wj yj
wk yk
N
j
M
k votj
t
N
i
N
j
T
tij
tt
N
ii
kt
kbλP
λj,qPλQ
aλP
λjqi,qPλQ
λP
λi,qPλQ
1 1
1 1
1
1
1
1
1
log,
log,
,
log,
O
Ob
O
Oa
O
Oπ
b
a
π
i1
( )t j
),( jit
λQλQλQ
obaλP
λ,PλλQ
T
ttq
T
tqqq ttt
,,,
]loglog[log, all 1
1
111
baπ
O
QO
baπ
Q
example
30
Solution to Problem 3 - The EM Algorithm (cont’d)
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
ija kb ji
N
nn
jj
j
N
jj
N
jjjN
w
wyF
yyywyyygF
1
1121
: when valuemaximum a has
0 and ,1 where,log,,...,,
y
y
Mk j
Nj ij
Ni i jkbia 111 1)( , 1 ,1
λλ,Q
31
Solution to Problem 3 - The EM Algorithm (cont’d)
Proof: Apply Lagrange Multiplier
N
nn
jj
N
jj
N
jj
N
jj
N
jj
j
j
j
j
j
N
j
N
jjjj
N
jjj
w
wy
wwyy
jy
w
y
w
y
F
yywywF
1
1111
1 11
0Then
0 Letting
1loglog that Suppose
Multiplier Lagrange applyingBy
Constraint
xe
xxh
x
xhxh
h
xhx
h
xhx
dx
xd
he
hx
xh
xhx
h
h
h
hh
h
h
1ln
1/1lnlim
1
/1lnlim/1lnlim
/lnlim
)ln()ln(lim
ln
...71828.21lim
/
0/
/1/
0
/1
0
00
/1
0
32
Solution to Problem 3 - The EM Algorithm (cont’d)
N
ii
λP
λi,qPλQ
1
1log,
O
Oππ
wi yi
N
nn
ii
w
wy
1 i
P
iqPi 1
1,ˆ
O
O
λiqPi tt ,)( O
1
1
1
1
N
n
N
nn
λP
λn,qP
w
O
O
33
Solution to Problem 3 - The EM Algorithm (cont’d)
N
nn
jj
w
wy
1
1
1
1
11
1
1
11 ,
,
,,ˆ
T
tt
T
tt
T
tt
T
ttt
ij
i
ji
iqP
jqiqPa
O
O
N
i
N
j
T
tij
tta
λP
λjqi,qPλQ
1 1
1
1
1log
,,
O
Oaa
wj yj
34
Solution to Problem 3 - The EM Algorithm (cont’d)
N
nn
kk
w
wy
1
wk yk
1 1s.t. s.t.
1 1
,
ˆ
,
t k t k
T T
t tt to v o v
j T T
t tt t
P q j j
b kP q j j
O
O
N
j
M
k votj
t
kt
kbλP
λj,qPλQ
1 1log,
O
Obb
35
Solution to Problem 3 - The EM Algorithm (cont’d)
The new model parameter set can be expressed as:
BAπ ˆ,ˆ,ˆ=̂
1
1
1 1
11 1
1 1
1 1
1 1s.t. s.t.
1 1
,ˆ
, , ,ˆ
,
,
ˆ
,
t k t k
i
T T
t t tt t
ij T T
t tt t
T T
t tt to v o v
j T T
t tt t
P q ii
P
P q i q j i ja
P q i i
P q j j
b kP q j j
O
O
O
O
O
O
λjqiqPji
λiqPi
ttt
tt
,,,
,)(
1 O
O
36
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., V={v1, v2, v3, ……, vM}, vkRL
• In this case, the observation probability distribution in state j, B={bj(k)}, is defined as bj(k)=P(ot=vk|qt=j), 1kM, 1jNot : 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., V={v| vRL}
• In this case, the observation probability distribution in state j, B={bj(v)}, is defined as bj(v)=f(ot=v|qt=j), 1jNot : 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
37
Gaussian Distribution
A continuous random variable X is said to have a Gaussian distribution with mean μ and variance σ2(σ>0) if X has a continuous pdf in the following form:
2
2
2/12
2exp
2
1),|(
x
μxXf
38
Multivariate Gaussian Distribution
If X=(X1,X2,X3,…,Xd) is an d-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,…,Xd are independent random variables, the covariance matrix is reduced to diagonal, i.e.,
))((
oft determinan : ((
2
1exp
2
1),;()(
2
TTT
1T2/12/
jjiiij
d
xxE
E))E
E
Nf
ΣΣμμxxμxμxΣ
xμ
μxΣμxΣ
ΣμxxX
jiij ,02
d
i ii
ii
ii
xf
12
2
2/1 2exp
2
1),|(
ΣμxX
39
Multivariate Mixture Gaussian Distribution
An d-dimensional random vector X=(X1,X2,X3,…,Xd) 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
M
kjkjkjk
jkd
jkj cb1
1T2/12/ 2
1exp
2
1μvΣμv
Σv
M
kjkjk cc
1
1and0 Covariance matrix of the k-th mixture of the j-th state
Mean vectorof the k-th mixture of the j-th state
Observation vector
wNwfM
kk
M
kkkk 1 ,),;()(
11
Σμxx
40
Solution to Problem 3 – The Segmental K-means Algorithm for CDHMM
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
1Number of times ˆ
Number of training sequencesi
q i Number of transitions from state to state
ˆ Number of transitions from state ij
i ja
i
By partitioning the observation vectors within each state into clusters
number of vectors classified into cluster of state ˆ
number of vectors in state
ˆ sample mean of vectors classified
jm
jm
j M
m jc
j
μ into cluster of state
ˆ sample covariance matrix of vectors classified into cluster of state jm
m j
m jΣ
41
Solution to Problem 3 – The Segmental K-means Algorithm for CDHMM
(cont’d) 3 states and 4 Gaussian mixtures per state
O1
State
O2
1 2 tOt
s2
s3
s1
s2
s3
s1
s2
s3
s1
s2
s3
s1
s2
s3
s1
s2
s3
s1
s2
s3
s1
s2
s3
s1
s2
s3
s1
Global mean Cluster 1 mean
Cluster 2mean
K-means {11,11,c11}{12,12,c12}
{13,13,c13} {14,14,c14}
42
Solution to Problem 3 – The Baum-Welch Algorithm for 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, given O and
M
mjmjmtjm
jkjktjk
N
stt
tt
tt
tttt
t
tttttt
tttttt
Nc
Nc
ss
jj
λjqP
λjqkmPj
λjqP
λjqkmPjλjqkmPj
λjqkmPλjqPλkmjqPkj
11,;
,;
,
,,
,
,,,,
,,,,,,
Σμo
Σμo
o
o
O
OO
OOO
Observation-independent assumption
λjqP
λjqkmP
tTttt
tTtttt
,,...,,,,...,
,,...,,,,...,,
111
111
ooooo
ooooo
λjqP
λkmjqPλjqkmP
tt
ttttt
,
,,,
o
o
43
Solution to Problem 3 – The Baum-Welch Algorithm for CDHMM (cont’d)
Re-estimation formulae for are
1
1
,ˆ Weighted average (mean) of observations in state and mixture
,
T
t tt
jk T
tt
j kj k
j k
oμ
1 1
1 1 1
Expected number of times in state and mixture ˆ
Expected number of times in state
T T
t tt t
jk T M T
t tt m t
j,k j,kj k
cj j,m j
T
1
1
ˆ Weighted covariance of observations in state and mixture
ˆ ˆ,
,
jk
T
t t jk t jkt
T
tt
j k
j k
j k
Σ
o μ o μ
, , jk jk jkc μ Σ
44
A Simple Example
o1
State
o2 o3
1 2 3 Time
S1
S2
S1
S2
S1
S2
1 1 11
2 2 11 2 2 22
2 2 33
1 1 22 1 1 33
The Forward/Backward Procedure
N
jtt
tt
N
jt
t
tt
jj
ii
λjqP
λiqP
λP
λiqPi
1
1
,
,
,
O
O
O
O
N
jt1tjijt
N
i
t1tjijt
N
jtt
N
i
tt
ttt
jobai
jobai
λjqiqP
λjqiqP
λP
λjqiqPji
11
1
1
11
1
1
1
)(
)(
,,
,,
,,,
O
O
O
O
45
A Simple Example (cont’d) 1
2
1
2
1
2
4v 7v 4v
start1
2
11a
12a
22a
21a
4,1117,1114,11 babab 1 4,1117,1114,11 loglogloglogloglog babab
4,2127,1114,11 babab 2 4,2127,1114,11 loglogloglogloglog babab
4,1217,2124,11 babab 3 4,1217,2124,11 loglogloglogloglog babab
4,2227,2124,11 babab 4 4,2227,2124,11 loglogloglogloglog babab
4,1117,1214,22 babab 5 4,1117,1214,22 loglogloglogloglog babab
4,2127,1214,22 babab 6 4,2127,1214,22 loglogloglogloglog babab
4,1217,2224,22 babab 7 4,1217,2224,22 loglogloglogloglog babab
4,2227,2224,22 babab 8 4,2227,2224,22 loglogloglogloglog babab
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Total 8 paths
q: 1 1 1
q: 1 1 2
46
A Simple Example (cont’d)
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back
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