1 hidden markov models hsin-min wang institute of information science, academia sinica...

Hidden Markov Models

Hsin-Min WangInstitute of Information Science, Academia Sinica

whm@iis.sinica.edu.tw

References: 1. L. R. Rabiner and B. H. Juang, (1993) Fundamentals of Speech Recognition, Chapter 6

2. X. Huang et. al., (2001) Spoken Language Processing, Chapter 8

3. L. R. Rabiner, (1989) “A Tutorial on Hidden Markov Models and Selected Applications in Speech Recognition,” Proceedings of the IEEE, vol. 77, No. 2, February 1989

Hidden Markov Model (HMM)

History– Published in Baum’s papers in late 1960s and early 1970s– Introduced to speech processing by Baker (CMU) and Jelinek

(IBM) in the 1970s– Introduced to computational biology in late1980s

• Lander and Green (1987) used HMMs in the construction of genetic linkage maps

• Churchill (1989) employed HMMs to distinguish coding from noncoding regions in DNA

Assumption– Speech signal (DNA sequence) can be characterized as a

parametric random process– Parameters can be estimated in a precise, well-defined manner

Three fundamental problems– Evaluation of probability (likelihood) of a sequence of

observations given a specific HMM– Determination of a best sequence of model states– Adjustment of model parameters so as to best account for

observed signal/sequence

{X:1/3, Y:1/3, Z:1/3}

{X:1/3, Y:1/3, Z:1/3} {X:1/3, Y:1/3, Z:1/3}

1/31/3

1/ 3 1/ 3 1/ 3

1/ 3, 1/ 3, 1/ 3

1/ 3 1/ 3 1/ 3

b X b Y b Z

Given an initial model as follows:

We can train HMMs for the following two classes using their training data respectively.

Training set for class 1:1. XYYZXYZXXYZ 2. XYZXYZ 3. XYZXXYZ 4. YYXYZXY 5. YZXXYZZXY 6. ZXZZXYZX 7. ZXYZXYZX 8. ZXYZX 9. ZXXYZX

Training set for class 2:1. YYYZZYZ 2. ZZYXYY 3. XXZZYYY 4. YYXYYXZ 5. ZZXXYYXY 6. YYYZZYXX 7. XYYYYYXYX 8. ZZZZZ 9. YYXXX

We can then decide which class the following testing sequences belong to. XYZXYZZXY XXYXYZZZZXXY

Brief Review of Probability Theorem

Consider the simple scenario of rolling two dice, labeled as die 1 and die 2. Define the following three events:

X: Die 1 lands on 3. Y: Die 2 lands on 1. Z: The dice sum to 8.

Prior probability: P(X)=P(Y)=1/6, P(Z)=5/36.

Joint probability: P(X,Y) (or P(X∩Y)) =1/36, two events X and Y are statistically independent if and only if P(X,Y) = P(X)xP(Y).

P(Y,Z)=0, two events Y and Z are mutually exclusive if and only if Y∩Z=Φ, i.e., P(Y∩Z)=0.

Conditional probability: , P(Y|X)=P(Y), P(Z|Y)=0

( , ) 1/36 1( | )

( ) 1/ 6 6

P Z XP Z X

{(2,6), (3,5), (4,4), (5,3), (6,2)}

X∩Y ={(3,1)}

Y∩Z=Φ

General Pattern Recognition Problem

ˆ arg max ( | )P

( | ) ( )arg max

arg max ( | ) ( )P P

Posterior probability

Bayes’ rule

( | ) ( )

maximum likelihood criterion

ˆ arg max ( | )P

maximum a posteriori (MAP) criterion

P(λ) is usually assumed uniformly distributed since we don’t know which class is more likely to happen.

Markov Chain

),...,,(),...,,|(),...,,|( 2212211121 nnnnn XXXPXXXXPXXXXP

),,,...,,( 1221 nnn XXXXXP

( , ) ( | ) ( )P X Y P Y X P X

),...,,|()( 1212

iXXXXPXP

),,...,,(),,...,,|( 12211221 nnnnn XXXXPXXXXXP

),(),|(...),...,,|(

),...,,|(),...,,|(

212133212

2211121

XXPXXXPXXXXP

XXXXPXXXXP

)()|(),|(...),...,,|(

),...,,|(),...,,|(

1122133212

2211121

XPXXPXXXPXXXXP

XXXXPXXXXP

)|()(),...,,( 12

121 ii

in XXPXPXXXP First-order Markov chain

N-gram Language Model

P(this is a book) = P(this) × P(is | this) × P(a | this is) × P(book | this is a)

P(this) = # “this” / # words

P(is | this) = # “this is” / # “this”

P(a | this is) = # “this is a” / # “this is”

P(book | this is a) = # “this is a book” / # “this is a”

Become more difficult to estimate

P(this is a book) ≒ P(this) × P(is | this) × P(a | is) × P(book | a)

≒ P(a | is) = # “is a” / # “is”

≒ P(book | a) = # “a book” / # “a”

bigram

The parameters of a Markov chain, with N states labeled by {1,…,N} and the state at time t in the Markov chain denoted as qt, can be described as

aij=P(qt= j|qt-1=i) 1≤i,j≤N

i =P(q1=i) 1≤i≤N

The output of the process is the

set of states at each time instant t, where each state corresponds to an observable event Xi

There is a one-to-one

correspondence between the observable sequence and the Markov chain state sequence

Observable Markov Model

Nj ij ia1 ) 1(

)1( 1 Ni i

(Rabiner 1989)

Markov Chain Model – Ex 1

A 3-state Markov Chain – State 1 generates symbol X only,

State 2 generates symbol Y only, State 3 generates symbol Z only

– Given a sequence of observed symbols O={ZXYYZXYZ}, the only one corresponding state sequence is Q={S3S1S2S2S3S1S2S3}, and the corresponding probability is

P(O|)=P(ZXYYZXYZ|)=P(Q| )=P(S3S1S2S2S3S1S2S3 |)

=π(S3)P(S1|S3)P(S2|S1)P(S2|S2)P(S3|S2)P(S1|S3)P(S2|S1)P(S3|S2)=0.10.30.30.70.20.30.30.2=0.00002268

1.05.04.0

5.02.03.0

2.07.01.0

1.03.06.0

AS2 S3

0.30.1

0.20.1

Markov Chain Model – Ex 2

A three-state Markov chain for the Dow Jones Industrial average

The probability of 5 consecutive up days

0.06480.60.5

days econsecutiv 5

111111111

,S,S,S,SSP

(Huang et al., 2001)

Extension to Hidden Markov Model

HMM: an extended version of Observable Markov Model– The observation is a probabilistic function (discrete or

continuous) of a state instead of an one-to-one correspondence of a state

– The model is a doubly embedded stochastic process with an underlying stochastic process that is not directly observable (hidden)

• What is hidden? The State Sequence!According to the observation sequence, we are not sure which state sequence generates it!

Hidden Markov Model – Ex 1

A 3-state discrete HMM

– Given an observation sequence O={XYZ}, there are 27 possible state sequences, therefore P(O|) is computed by

{X:.3, Y:.2, Z:.5}

{X:.7, Y:.1, Z:.2} {X:.3, Y:.6, Z:.1}

0.30.1

0.20.1

0.6 0.3 0.1

0.1 0.7 0.2

0.3 0.2 0.5

0.3, 0.2, 0.5

0.7, 0.1, 0.2

0.3, 0.6, 0.1

0.4 0.5 0.1

b X b Y b Z

2 2 3 2 2 3

2 2 3 2 2 2 3 2

, , , :

e.g. when ,

, 0.7 0.1 0.1 0.007

0.5 0.7 0.2 0.07

i i i ii i

P λ P λ P λ P λ state sequence

P λ P XYZ S S S P X S P Y S P Z S

P λ P S S S S P S S P S S

O O Q O Q Q Q

Initial model

Hidden Markov Model – Ex 2

Given a three-state Hidden Markov Model for the Dow Jones Industrial averageas follows:

How to find the probability P(up, up, up, up, up|)?How to find the optimal state sequence of the model which generates the observation sequence “up, up, up, up, up”?

cf. Markov chain model

(35 state sequences can generate “up, up, up, up, up”.)

Elements of an HMM

An HMM is characterized by the following:1. N, the number of states in the model

2. M, the number of distinct observation symbols per state

3. The state transition probability distribution A={aij}, where aij=P[qt+1=j|qt=i], 1≤i,j≤N

4. The observation symbol probability distribution in state j, B={bj(vk)} , where bj(vk)=P[ot=vk|qt=j], 1≤j≤N, 1≤k≤M

5. The initial state distribution ={i}, where i=P[q1=i], 1≤i≤N

For convenience, we usually use a compact notation =(A,B,) to indicate the complete parameter set of an HMM1. Requires specification of two model parameters (N and M)

Two Major Assumptions for HMM

First-order Markov assumptionFirst-order Markov assumption– The state transition depends only on the origin and destination

– The state transition probability is time invariant

Output-independent assumptionOutput-independent assumption– The observation is dependent on the state that generates it, not

dependent on its neighbor observations

aij=P(qt+1=j|qt=i), 1≤i, j≤N

tttTt qqPqPqqqPP

2111 ,,...,,..., Q

tttTtTt obqoPqqqoooPP

11 ,,,...,,...,,...,,...,, QO

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 ?

Evaluation Problem

– Problem 2:

How to choose an optimal state sequence Q=(q1,q2,……, qT)

which best explains the observations?

Decoding Problem– Problem 3:

How to adjust the model parameters =(A,B,) to maximize P(O|)? Learning/Training Problem

* arg max ( , | )P Q

)|(maxarg*iP

OP(up, up, up, up, up|)?

Solution to Problem 1

Solution to Problem 1 - Direct Evaluation

Given O and , find P(O|)= Pr{observing O given } Evaluating all possible state sequences of length T that

generate the observation sequence O

: The probability of the path Q– By the first-order Markov assumption

: The joint output probability along the path Q – By the output-independent assumption

QQOQOO

,,allall

TT qqqqqqq

ttt aaaqqPqPP

132211...,

ttt obqoPP

Solution to Problem 1 - Direct Evaluation (cont’d)

o2 o3 oT

1 2 3 T-1 T Time

Sj means bj(ot) has been computed

aij means aij has been computed

)( 133 ob )( 2232 oba )( 3323 oba … )(121 Toba

Solution to Problem 1 - Direct Evaluation (cont’d)

– A Huge Computation Requirement: O(NT) (NT state sequences)• Exponential computational complexity

A more efficient algorithm can be used to evaluate – The Forward Procedure/Algorithm

Tqqqqqq

,..,q,qqqq

allTqqqqqqqqqq

obaobaob

obobobaaa

122121

21132211

..........

ADD 1 2MUL12 : -N, TN NT- TTT Complexity

Solution to Problem 1 - The Forward Procedure

Base on the HMM assumptions, the calculation of and involves only qt-1, qt , and ot , so it is possible to compute the likelihood

with recursion on t

Forward variable : – The probability of the joint event that o1,o2,…,ot are observed and

the state at time t is i, given the model λ

,1tt qqP ,tt qoP

λiqoooPiα ttt ,,...,, 21

λjqooooPjα tttt 11211 ,,,...,,

)()( 11

iijt obaiα

, ,..., ,

α i P o o o q i λ

P α i

)(),|(,,...,,

)(),,,...,,|(,,...,,

)(,,,...,,

)(|,,...,,

),|(|,,...,,

)|(),|(,|,...,,

)|(,|,,...,,

|,,,...,,

111121

ittttt

obλiqjqPλiqoooP

obλiqooojqPλiqoooP

obλjqiqoooP

objqoooP

jqoPjqoooP

jqPjqoPjqoooP

jqPjqooooP

jqooooPj

Solution to Problem 1 - The Forward Procedure (cont’d)

( , )all Y

P X P X Y

( , , ) ( , , ) ( , )( , | ) ( | , ) ( | )

( ) ( , ) ( )

P X Y P X Y P YP X Y P X Y P Y

P P Y P

Output-independent assumption

( | , ) ( | ) ( , | )P X Y P Y P X Y

)(, 111 tjtt objqoP

( , | ) ( | ) ( | , )P X Y P X P Y X

First-order Markov assumption

3(2)=P(o1,o2,o3,q3=2|)

=[2(1)×a12+ 2(2)×a22 +2(3)×a32]×b2(o3)

o2 o3 oT

1 2 3 T-1 T Time

Sj means bj(ot) has been computed

aij means aij has been computed

a32b2(o3)

Time index

State index

Algorithm

– Complexity: O(N2T)

Based on the lattice (trellis) structure– Computed in a left-to-right time-synchronous manner, where each cell

for time t is completely computed before proceeding to time t+1

All state sequences, regardless how long previously, merge to N nodes (states) at each time instance t

1 1 1 1

1. Initialization ( , | ) 1

2. Induction 1 1 1

3.Termination O

t t ij j 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

TNN))N(T-(N-

T N)+N )(T-N(N+2

111: ADD

11 : MUL

λiqoooPi ttt ,...21

cf. O(NT) for direct evaluation

A three-state Hidden Markov Model for the Dow Jones Industrial average

b1(up)=0.7

b2(up)= 0.1

b3(up)=0.3

a11=0.6

a21=0.5

a31=0.4

b1(up)=0.7

b2(up)= 0.1

b3(up)=0.3

π1=0.5

π2=0.2

π3=0.3

α1(1)=0.5*0.7

α1(2)= 0.2*0.1

α1(3)= 0.3*0.3

α2(1)= (0.35*0.6+0.02*0.5+0.09*0.4)*0.7

α2(2)=(0.35*0.2+0.02*0.3+0.09*0.1)*0.1

α2(3)=(0.35*0.2+0.02*0.2+0.09*0.5)*0.3

P(up, up|) = α2(1)+α2(2)+α2(3)

a12=0.2

a22=0.3

a32=0.1a13=0.2a23=0.2

a33=0.5

P(up, up|) ?

Solution to Problem 2 - The Viterbi Algorithm

The Viterbi algorithm can be regarded as a dynamic programming algorithm applied to the HMM or as a modified forward algorithm– Instead of summing probabilities from different paths coming to

the same destination state, the Viterbi algorithm picks and remembers the best path

• Find a single optimal state sequence Q*

– The Viterbi algorithm also can be illustrated in a trellis framework similar to the one for the forward algorithm

* arg max ( , | )P Q

Solution to Problem 2 - The Viterbi Algorithm (cont’d)

o2 o3 oT

1 2 3 T-1 T Time

1. Initialization

2. Induction

3. Termination

4. Backtracking),...,,(

1,...,2.1),(**

Ni, obπi ii

Nj,T-t, aij

Nj,T-t, obaij

tjijtNi

1 11][maxarg)(

1 11][max

maxarg

max,QO

Complexity: O(N2T)

is the best state sequence

iijtt obaiαjα

iT iαλP

b1(up)=0.7

b2(up)= 0.1

b3(up)=0.3

a11=0.6

a21=0.5

a31=0.4

b1(up)=0.7

b2(up)= 0.1

b3(up)=0.3

π1=0.5

π2=0.2

π3=0.3

A three-state Hidden Markov Model for the Dow Jones Industrial average

δ1(1)=0.5*0.7

δ1(2)= 0.2*0.1

δ1(3)= 0.3*0.3

δ2(1)=max (0.35*0.6, 0.02*0.5, 0.09*0.4)*0.7

δ2(1)= 0.35*0.6*0.7=0.147Ψ2(1)=1

a12=0.2a22=0.3

a32=0.1a13=0.2

a23=0.2

a33=0.5

δ2(2)=max (0.35*0.2, 0.02*0.3, 0.09*0.1)*0.1

δ2(2)= 0.35*0.2*0.1=0.007Ψ2(2)=1

δ2(3)=max (0.35*0.2, 0.02*0.2, 0.09*0.5)*0.3

δ2(3)= 0.35*0.2*0.3=0.021Ψ2(3)=1

The most likely state sequence that generates “up up”: 1 1

1)1()( 2*221 qq*

1maxarg 231

Some Examples

Isolated Digit Recognition

o1 o2 o3 oT

1 2 3 T-1 T Time

)3()|( 0 TP O

)3()|( 1 TP O

Continuous Digit Recognition

o1 o2 o3 oT

1 2 3 T-1 T Time

Continuous Digit Recognition (cont’d)

1 2 3 4 5 6 7 8 9 Time

S1 S1 S2 S6S3 S3 S4 S5 S5

Beststate sequence

CpG Island Recognition

Two QuestionsTwo Questions Q1: Given a short sequence, does it come from a CpG

island? Q2: Given a long sequence, how would we find the CpG

islands in it?

CpG Island Recognition – Q1 Given sequence x, probabilistic model M1 of CpG

islands, and probabilistic model M2 for non-CpG island regions– Compute p1=P(x|M1) and p2=P(x|M2)

– If p1 > p2, then x comes from a CpG island (CpG+)

– If p2 > p1, then x does not come from a CpG island (CpG-)

S1:A S2:C

S3:T S4:G

CpG+ A C G T

A 0.180 0.274 0.426 0.120

C 0.171 0.368 0.274 0.188

G 0.161 0.339 0.375 0.125

T 0.079 0.355 0.384 0.182

CpG- A C G T

A 0.300 0.205 0.285 0.210

C 0.322 0.298 0.078 0.302

G 0.248 0.246 0.298 0.208

T 0.177 0.239 0.292 0.292

Large CG transition probability

Small CG transition probability

CpG Island Recognition – Q2

A: 0.3C: 0.2G: 0.2T: 0.3

A: 0.2C: 0.3G: 0.3T: 0.2

a22=0.9999

a11=0.99999

a12=0.00001

a21=0.0001 CpG+CpG-

… A C T C G A G T A …

S1 S1 S1 S1S2 S2 S2 S2 S1

Observable

Hidden

A Toy Example: 5’ Splice Site Recognition

5’ splice site indicates the “switch” from an exon to an intron

Assumptions:– Uniform base composition on average in exons (25% each

base) – Introns are A/T rich (40% A/T, and 10% C/G)– The 5’SS consensus nucleotide is almost always a G (say, 95%

G and 5% A)

From “What is a hidden Markov Model?”, by Sean R. Eddy

A Toy Example: 5’ Splice Site Recognition

Solution to Problem 3 – Maximum Likelihood Estimation of Model Parameters

How to adjust (re-estimate) the model parameters =(A,B,) to maximize P(O|)?– The most difficult one among the three problems, because there

is no known analytical method that maximizes the joint probability of the training data in a closed form

• The data is incomplete because of the hidden state sequence

– The problem can be solved by the iterative Baum-Welch algorithm, also known as the forward-backward algorithm

• The EM (Expectation Maximization) algorithm is perfectly suitable for this problem

– Alternatively, it can be solved by the iterative segmental K-means algorithm

• The model parameters are adjusted to maximize P(O, Q* |), Q* is the state sequence given by the Viterbi algorithm

• Provide a good initialization of Baum-Welch training

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 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

1Number of times ˆ

Number of training sequencesi

Number of transitions from state to state ˆ

Number of transitions from state ij

Solution to Problem 3 – The Segmental K-means Algorithm (cont’d)

3 states and 2 codewords (observations)

π1=1, π2=π3=0 a11=3/4, a12=1/4, a13=0 a21=0, a22=2/3, a23=1/3 a31=0, a32=0, a33=1 b1(X)=3/4, b1(Y)=1/4 b2(X)=1/3, b2(Y)=2/3 b3(X)=2/3, b3(Y)=1/3

1 2 3 4 5 6 7 8 9 10O4

O5 O6 O9O8O7 O10Training data:

Re-estimatedparameters:

Solution to Problem 3 – Segmental K-means vs. Baum-Welch

Number of " " in state ˆ Number of times in state j

k jb k

1Number of times ˆ

Number of transitions from state to state ˆ

Number of transitions from state ij

1 number of times ˆ

Expected

number of " " in state ˆ number of times in state j

k jb k

Expected

number of transitions from state to state ˆ

number of transitions from state ij

Expected

Solution to Problem 3 – The Baum-Welch Algorithm

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

jtt jii

, expected number of transitions from state to state in T

i j i j

expected number of transitions from state in T

1 1expected number of in i q i O

Solution to Problem 3 – The Baum-Welch Algorithm (cont’d)

Re-estimation formulae for , A, and B are

111Expected number of times

ˆNumber of training sequences

1 1s.t.

Expected number of " " in state ˆ Expected number of times in state

o vj L T

k jb k

Expected number of transitions from state to state ˆ

Expected number of transitions from state

ij L T -

ξ i, ji j

Solution to Problem 3 – The Baum-Welch Algorithm (cont’d)

We can use the forward-backward algorithm to obtain

t(i)= P(qt = i | O, )

t( i, j )=P(qt = i, qt+1 = j | O, )

Solution to Problem 3 – The Forward Procedure

Forward variable: – Probability of the joint event that o1,o2,…,ot are observed and the

state at time t is i, given the model λ

– 3(1)=P(o1,o2,o3,q3=1|)

=[2(1)×a11 + 2(2)×a21 + 2(3)×a31] × b1(o3)

o2 o3 oT

1 2 3 T-1 T Time

b1(o3)

λiqoooPiα ttt ,,...,, 21

Solution to Problem 3 – The Backward Procedure

Backward variable :– Probability of the partial observation sequence ot+1,ot+2,…,oT, given

state i at time t and the model – 3(1)=P(o4,o5,…, oT | q3=1,)

=a11×b1(o4)×4(1)+a12× b2(o4)×4(2)+a13× b3(o4)×4(3)

o2 o3 oT

1 2 3 4 T-1 T Time

λ,,...,, 21 iqoooPi tTttt

3(1) b1(o4) 4(1)a11

Solution to Problem 3 – The Backward Procedure (cont’d)

Algorithm

TN))N(T-(N-T N) (T-N

Nj,T-tjobai

NiiβN

jttjijt

11 :ADD ; 12 : MUL Complexity

1 11 ,Induction 2.

1 ,1tion Initializa 1.

λiqoooPi tTttt ,,...,, 21

λiqPλiqoPλiqoooP

λiqPλiqooooPλiqooooPλP

,,,...,,

,,...,,,,,...,,,

iT iαλP

Solution to Problem 3 – The Forward-Backward Algorithm

Relation between the forward and backward variables

iqoooPi

jttjijt

iqoooPi

λiqPii ttt ,)( O

Ni tt iiλP 1 )(O

Solution to Problem 3 – The Forward-Backward Algorithm (cont’d)

iqoooP

iqPiqoooP

iqoooPiqPiqoooP

iqoooPiqoooP

tTttttt

tTtttt

)|,,...,,(

)|(),|,...,,(

),|,...,,()|(),|,...,,(

),|,...,,()|,,...,,(

it iiλiqPλP

11)()(, OO

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

nttnmnt

ttjijtttt

λjqiqPji

111 ,,,

tttttt

λiqPii ttt ,)( O

Ni tt iiλP 1 )(O

P(q3 = 1, O | )=3(1)*3(1)

o2 o3 oT

1 2 3 4 T-1 T Time

3(1) 3(1)

P(q3 = 1, q4 = 3, O | )=3(1)*a13*b3(o4)*4(3)

o2 o3 oT

1 2 3 4 T-1 T Time

b3(o4)

* * * * * * * *1 1 1 2 2 1

ˆ arg max log ( , | )

arg max log log ( ) log log ( ) log log ( )T T T

Tq q q q q q q ql

b o a b o a b o

log logL N

i iql i

Take the derivative of log (1 ) w.r.t. and set it to 0N N

i i i ii i

N N Ni i i

i i i Ni i i

n n nn

Lagrange multiplier

1 hidden markov models hsin-min wang institute of information science, academia sinica...

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