Download - Hidden Markov Models

www.bioalgorithms.infoAn Introduction to Bioinformatics Algorithms

Hidden Markov Models

1

An Introduction to Bioinformatics Algorithms www.bioalgorithms.info

Outline• CG-islandsCG-islands• The “Fair Bet Casino”The “Fair Bet Casino”• Hidden Markov ModelHidden Markov Model• Decoding AlgorithmDecoding Algorithm• Forward-Backward AlgorithmForward-Backward Algorithm• Profile HMMsProfile HMMs• HMM Parameter EstimationHMM Parameter Estimation• Viterbi training• Baum-Welch algorithm

2


CG-Islands• Given 4 nucleotides: probability of Given 4 nucleotides: probability of

occurrence is ~ 1/4. Thus, probability of occurrence is ~ 1/4. Thus, probability of occurrence of a dinucleotide is ~ 1/16.occurrence of a dinucleotide is ~ 1/16.

• However, the frequencies of dinucleotides in However, the frequencies of dinucleotides in DNA sequences vary widely.DNA sequences vary widely.

• In particular, In particular, CG CG is typically underepresented is typically underepresented (frequency of (frequency of CG CG is typically < 1/16)is typically < 1/16)

4


Why CG-Islands?• CGCG is the least frequent dinucleotide because is the least frequent dinucleotide because

CC in in CG CG is easily is easily methylated and methylated and has the has the tendency to mutate into T afterwardstendency to mutate into T afterwards

• However, the methylation is suppressed However, the methylation is suppressed around genes in a genome. So, around genes in a genome. So, CGCG appears appears at relatively high frequency within these at relatively high frequency within these CGCG islandsislands

• So, finding the So, finding the CGCG islands in a genome is an islands in a genome is an important problemimportant problem

5


CG Islands and the “Fair Bet Casino”

• The The CGCG islands problem can be modeled after islands problem can be modeled after a problem named a problem named “The Fair Bet Casino”“The Fair Bet Casino”

• The game is to flip coins, which results in only The game is to flip coins, which results in only two possible outcomes: two possible outcomes: HHead or ead or TTail.ail.

• The The FFair coin will give air coin will give HHeads and eads and TTails with ails with same probability ½.same probability ½.

• The The BBiased coin will give iased coin will give HHeads with prob. ¾.eads with prob. ¾.

6


The “Fair Bet Casino” (cont’d)

• Thus, we define the probabilities:Thus, we define the probabilities:• P(H|F) = P(T|F) = ½P(H|F) = P(T|F) = ½• P(H|B) = ¾, P(T|B) = ¼P(H|B) = ¾, P(T|B) = ¼• The crooked dealer chages between Fair The crooked dealer chages between Fair

and Biased coins with probability 10%and Biased coins with probability 10%

7


The Fair Bet Casino Problem• Input:Input: A sequence A sequence x = xx = x11xx22xx33…x…xnn of coin of coin

tosses made by two possible coins (tosses made by two possible coins (FF or or BB).). • Output:Output: A sequence A sequence π = ππ = π11 π π22 π π33… π… πnn, with , with

each each ππii being either being either F F or or BB indicating that indicating that xxii is the result of tossing the Fair or Biased is the result of tossing the Fair or Biased coin respectively.coin respectively.

8


Problem…Fair Bet Casino Fair Bet Casino ProblemProblemAny observed Any observed outcome of coin outcome of coin tosses could tosses could have been have been generated by any generated by any sequence of sequence of states!states!

Need to incorporate a way to grade different sequences differently.

Decoding ProblemDecoding Problem

9


P(x|fair coin) vs. P(x|biased coin)• Suppose first that dealer never changes Suppose first that dealer never changes

coins. Some definitions:coins. Some definitions:• P(P(xx|fair coin):|fair coin): prob. of the dealer using prob. of the dealer using the the F F coin and generating the outcome coin and generating the outcome x.x.• P(P(xx|biased coin):|biased coin): prob. of the dealer using prob. of the dealer using

the the BB coin and generating outcome coin and generating outcome x.x.

10


P(x|fair coin) vs. P(x|biased coin)

• P(P(xx|fair coin)=P(|fair coin)=P(xx11…x…xnn|fair coin)|fair coin)

ΠΠi=1,n i=1,n p (xp (xii||fair coinfair coin)= )= (1/2)(1/2)nn

• P(P(xx|biased coin)= P(|biased coin)= P(xx11…x…xnn|biased coin)=|biased coin)=

ΠΠi=1,n i=1,n p (xp (xii||biased coinbiased coin)=(3/4))=(3/4)kk(1/4)(1/4)n-kn-k== 33kk/4/4nn

• kk - - the number of the number of HHeads in eads in x.x. 11


P(x|fair coin) vs. P(x|biased coin)• P(x|fair coin) = P(x|biased coin)• 1/2n = 3k/4n • 2n = 3k

• n = k log23

• when k = n / log23 (k ~ 0.67n)

12


Log-odds Ratio• We define We define log-odds ratiolog-odds ratio as follows: as follows:

loglog22(P((P(xx|fair coin) / P(|fair coin) / P(xx|biased coin)) |biased coin)) = = ΣΣkk

i=1i=1 log log22((pp++((xxii) / ) / pp--((xxii))))

= = n – kn – k log log2233

13


Computing Log-odds Ratio in Sliding Windows

x1x2x3x4x5x6x7x8…xn

Consider a sliding window of the outcome sequence. Find the log-odds for this short window.

Log-odds value

0

Fair coin most likely used

Biased coin most likely used

Disadvantages:- the length of CG-island is not known in advance- different windows may classify the same position differently

14


Hidden Markov Model (HMM)• Can be viewed as an abstract machine with Can be viewed as an abstract machine with k hidden k hidden

states that emits symbols from an alphabet states that emits symbols from an alphabet ΣΣ..• Each state has its own probability distribution, and the Each state has its own probability distribution, and the

machine switches between states according to this machine switches between states according to this probability distribution.probability distribution.

• While in a certain state, the machine makes 2 While in a certain state, the machine makes 2 decisions:decisions:• What state should I move to next?What state should I move to next?• What symbol - from the alphabet What symbol - from the alphabet ΣΣ - should I emit? - should I emit?

15


Why “Hidden”?• Observers can see the emitted symbols of an Observers can see the emitted symbols of an

HMM but have HMM but have no ability to know which state no ability to know which state the HMM is currently inthe HMM is currently in..

• Thus, the goal is to infer the most likely Thus, the goal is to infer the most likely hidden states of an HMM based on the given hidden states of an HMM based on the given sequence of emitted symbols.sequence of emitted symbols.

16


HMM ParametersΣΣ: set of emission characters.: set of emission characters.

Ex.: Σ = {H, T} for coin tossingEx.: Σ = {H, T} for coin tossing Σ = {1, 2, 3, 4, 5, 6} for dice tossingΣ = {1, 2, 3, 4, 5, 6} for dice tossing

QQ: set of hidden states, each emitting symbols : set of hidden states, each emitting symbols from Σ.from Σ.

Q={F,B} for coin tossingQ={F,B} for coin tossing

17


HMM Parameters (cont’d)

A = (aA = (aklkl):): a |Q| x |Q| matrix of probability of a |Q| x |Q| matrix of probability of changing from state changing from state k k to state to state l.l.

aaFFFF = 0.9 a = 0.9 aFBFB = 0.1 = 0.1

aaBFBF = 0.1 a = 0.1 aBBBB = 0.9 = 0.9

E = (eE = (ekk((bb)):)): a |Q| x | a |Q| x |Σ| matrix of probability of Σ| matrix of probability of emitting symbol emitting symbol bb while being in state while being in state k.k.

eeFF(0) = ½ e(0) = ½ eFF(1) = ½ (1) = ½

eeBB(0) = ¼ e(0) = ¼ eBB(1) = ¾ (1) = ¾

18


HMM for Fair Bet Casino• The The Fair Bet CasinoFair Bet Casino in in HMM HMM terms:terms:

Σ = {0, 1} (Σ = {0, 1} (00 for for TTails and ails and 11 HHeads)eads)Q = {Q = {F,BF,B} – } – FF for Fair & for Fair & BB for Biased coin. for Biased coin.

• Transition Probabilities Transition Probabilities A *** A *** Emission Probabilities Emission Probabilities EE

Fair Biased

Fair aaFFFF = 0.9 = 0.9 aaFBFB = 0.1 = 0.1

Biased aaBFBF = 0.1 = 0.1 aaBBBB = 0.9 = 0.9

Tails(0) Heads(1)

Fair eeFF(0) = ½ (0) = ½ eeFF(1) = ½(1) = ½

Biased eeBB(0) = ¼(0) = ¼ eeBB(1) = ¾(1) = ¾

19


HMM for Fair Bet Casino (cont’d)

HMM model for the HMM model for the Fair Bet Casino Fair Bet Casino ProblemProblem20


Hidden Paths• A A pathpath π = ππ = π11… π… πnn in the HMMin the HMM is defined as a is defined as a

sequence of states.sequence of states.• Consider path Consider path π π = = FFFBBBBBFFF and sequence FFFBBBBBFFF and sequence x x = =

0101110100101011101001

x 0 1 0 1 1 1 0 1 0 0 1

π π = F F F B B B B B F F = F F F B B B B B F F FFP(P(xxii|π|πii)) ½ ½ ½ ¾ ¾ ¾ ½ ½ ½ ¾ ¾ ¾ ¼¼ ¾ ½ ½ ½ ¾ ½ ½ ½ P(πP(πi-1 i-1 ππii)) ½ ½ 99//1010 99//10 10

11//10 10 99//10 10

99//10 10 99//10 10

99//10 10 11//10 10

99//10 10 99//1010

Transition probability from state ππi-1 i-1 to to state πstate πii

Probability that xi was emitted from state ππii

21


P(x|π) Calculation• P(P(xx||ππ):): Probability that sequence Probability that sequence xx was was

generated by the path generated by the path π:π: nn

P(P(xx||ππ) = P() = P(ππ00→ π→ π11) ) ·· Π PΠ P(x(xii| π| πii) ) · · P(P(ππi i → π→ πi+1i+1))

i=1i=1

= = aa ππ00, , ππ1 1 ·· Π Π ee ππii (x (xii) ) ·· aa ππii, , ππi+1i+1

22


P(x|π) Calculation• P(P(xx||ππ):): Probability that sequence Probability that sequence xx was generated was generated

by the path by the path π:π: nn

P(P(xx||ππ) = P() = P(ππ00→ π→ π11) ) ·· Π PΠ P(x(xii| π| πii) ) · · P(P(ππi i → π→ πi+1i+1))

i=1i=1

= = aa ππ00, , ππ11 ·· Π Π ee ππii (x (xii) ) ·· a a ππii, , ππi+1i+1

= Π = Π ee ππi+1i+1 (x (xi+1i+1)) ·· a a ππii, , ππi+1i+1 if we count from if we count from i=0i=0 instead of instead of i=1i=1

23


Decoding Problem• Goal:Goal: Find an optimal hidden path of states Find an optimal hidden path of states

given observations.given observations.

• Input:Input: Sequence of observations Sequence of observations x = xx = x11…x…xnn generated by an HMM generated by an HMM MM((ΣΣ, Q, A, E, Q, A, E))

• Output:Output: A path that maximizes A path that maximizes P(x|P(x|ππ)) over over all possible paths all possible paths π.π.

24


Building Manhattan for Building Manhattan for Decoding Problem• Andrew Viterbi used the Manhattan grid Andrew Viterbi used the Manhattan grid

model to solve the model to solve the Decoding ProblemDecoding Problem..

• Every choice of Every choice of π = ππ = π11… π… πn n corresponds to a corresponds to a path in the graph.path in the graph.

• The only valid direction in the graph is The only valid direction in the graph is eastward.eastward.

• This graph has |This graph has |QQ||22(n-1)(n-1) edges. edges.

25


Edit Graph for Decoding Problem

26


Decoding Problem vs. Alignment Problem

Valid directions in the Valid directions in the alignment problem.alignment problem.

Valid directions in the Valid directions in the decoding problem.decoding problem.

27


Decoding Problem as Finding a Longest Path in a DAG• The The Decoding ProblemDecoding Problem is reduced to finding is reduced to finding

a longest path in the a longest path in the directed acyclic graph directed acyclic graph (DAG)(DAG) above. above.

• Notes:Notes: the length of the path is defined as the length of the path is defined as the the productproduct of its edges’ weights, not the of its edges’ weights, not the sum.sum.

28


Decoding Problem (cont’d)

• Every path in the graph has the probability Every path in the graph has the probability P(x|P(x|ππ))..

• The Viterbi algorithm finds the path that The Viterbi algorithm finds the path that maximizes maximizes P(x|P(x|ππ) among all possible paths.) among all possible paths.

• The Viterbi algorithm runs in The Viterbi algorithm runs in O(n|Q|O(n|Q|22)) time.time.

29


Decoding Problem: weights of edges

w

The weight w is given by: ???

(k, i) (l, i+1)

30



w

The weight w is given by: ??

(k, i) (l, i+1)

nn

P(P(xx||ππ) =) = Π Π ee ππi+1i+1 (x (xi+1i+1)) . a. a ππii, , ππi+1i+1 i=0i=0

31



w

The weight w is given by: ?

(k, i) (l, i+1)

ii-th term -th term == ee ππi+1i+1 (x (xi+1i+1)) . a. a ππii, , ππi+1i+1

32



w

The weight w=el(xi+1). akl

(k, i) (l, i+1)

ii-th term -th term == ee ππii (x (xii)) . a. a ππii, , ππi+1i+1 = = el(xi+1). akl for ππi i =k, π=k, πi+1i+1==ll

33


Decoding Problem and Dynamic Programming

ssl,i+1l,i+1 = = maxmaxkk Є QЄ Q { {ssk,ik,i · · weight of edge betweenweight of edge between (k,i) (k,i) andand (l,i+1) (l,i+1)}=}=

maxmaxkk Є QЄ Q { {ssk,ik,i · a · aklkl · e · ell (x (xi+1i+1) ) }=}=

eell (x (xi+1i+1) · ) · maxmaxkk Є QЄ Q { {ssk,ik,i · a · aklkl}}

34


Decoding Problem (cont’d)

• Initialization:Initialization:• ssbegin,0begin,0 = 1= 1

• ssk,0k,0 = 0 for = 0 for k ≠ begink ≠ begin..• Let Let ππ** be the optimal path. Then,be the optimal path. Then,

P(P(xx||ππ**) = max) = maxkk Є QЄ Q { {ssk,nk,n . . aak,endk,end}}

35


Viterbi Algorithm• The value of the product can become The value of the product can become

extremely small, which leads to overflowing.extremely small, which leads to overflowing.• To avoid overflowing, use log value instead. To avoid overflowing, use log value instead.

ssk,i+1k,i+1= log= logel(xi+1) + max kk Є QЄ Q {{ssk,i k,i ++ log(akl)}}

36


Forward-Backward Problem

Given:Given: a sequence of coin tosses generated a sequence of coin tosses generated by an HMM by an HMM..Goal:Goal: find the probability that the dealer was find the probability that the dealer was using a biased coin at a particular time.using a biased coin at a particular time.

37


Forward Algorithm• Define Define ffk,ik,i ((forward probabilityforward probability) as the ) as the

probability of emitting the prefix probability of emitting the prefix xx11…x…xii and and reaching the state reaching the state ππ = k = k..

• The recurrence for the forward algorithm:The recurrence for the forward algorithm:

ffk,ik,i = = eekk(x(xii)) . . ΣΣ ffl,i-l,i-11 . a . alklk

l Є Ql Є Q

38


Backward Algorithm• However, However, forward probability forward probability is not the only is not the only

factor affecting factor affecting P(P(ππii = k|x = k|x))..

• The sequence of transitions and emissions The sequence of transitions and emissions that the HMM undergoes between that the HMM undergoes between ππi+1i+1 and and ππnn also affect also affect P(P(ππii = k|x = k|x))..

forward xi backward

39


Backward Algorithm (cont’d)• DefineDefine backward probability backward probability bbk,ik,i as the as the

probability of being in state probability of being in state ππii = k = k and emitting and emitting the the suffixsuffix xxi+1i+1…x…xnn..

• The recurrence for the The recurrence for the backward algorithmbackward algorithm::

bbk,ik,i = = ΣΣ eell(x(xi+1i+1)) . . bbl,i+1l,i+1 . a . akl kl

l Є Ql Є Q

40


Backward-Forward Algorithm• The probability that the dealer used a The probability that the dealer used a

biased coin at any moment biased coin at any moment ii::

P(x, P(x, ππii = k = k) f) fkk(i) . b(i) . bkk(i)(i) P(P(ππii = k|x = k|x) = ) = _______________ _______________ = = ____________________________

P(x) P(x)P(x) P(x)

P(x) is the sum of P(x, πP(x, πii = k) = k) over all k

41


Finding Distant Members of a Protein Family

• A distant cousin of functionally related sequences in A distant cousin of functionally related sequences in a protein family may have weak pairwise similarities a protein family may have weak pairwise similarities with each member of the family and thus fail with each member of the family and thus fail significance test. significance test.

• However, they may have weak similarities with However, they may have weak similarities with manymany members of the family. members of the family.

• The goal is to align a sequence to The goal is to align a sequence to allall members of members of the family at once.the family at once.

• Family of related proteins can be represented by Family of related proteins can be represented by their multiple alignment and the corresponding their multiple alignment and the corresponding profile.profile.

42


Profile Representation of Protein FamiliesAligned DNA sequences can be represented by a 4 ·n profile matrix reflecting the frequencies of nucleotides in every aligned position.

Protein family can be represented by a Protein family can be represented by a 20·n profile profile representing frequencies of amino acids.representing frequencies of amino acids.

43


Profiles and HMMs

• HMMs can also be used for aligning a HMMs can also be used for aligning a sequence against a profile representing sequence against a profile representing

protein family.protein family.• A A 20·n20·n profile profile PP corresponds to corresponds to n n

sequentially linked sequentially linked matchmatch states states MM11,…,M,…,Mnn

in the in the profile HMMprofile HMM of of P.P.

44


Multiple Alignments and Protein Family Classification

• Multiple alignment of a protein family shows variations in conservation along the length of a protein

• Example: after aligning many globin proteins, the biologists recognized that the helices region in globins are more conserved than others.

45


What are Profile HMMs ?

• A Profile HMM is a probabilistic representation of a multiple alignment.

• A given multiple alignment (of a protein family) is used to build a profile HMM.

• This model then may be used to find and score less obvious potential matches of new protein sequences.

46


Profile HMM

A profile HMMA profile HMM

47


Building a profile HMM• Multiple alignment is used to construct the HMM model.• Assign each column to a Match state in HMM. Add Insertion and

Deletion state. • Estimate the emission probabilities according to amino acid

counts in column. Different positions in the protein will have different emission probabilities.

• Estimate the transition probabilities between Match, Deletion and Insertion states

• The HMM model gets trained to derive the optimal parameters.

48


States of Profile HMM

• Match states Match states MM11……MMnn (plus (plus begin/endbegin/end states) states)

• Insertion states Insertion states II00II11……IInn

• Deletion states Deletion states DD11……DDnn

49


Transition Probabilities in Profile HMM• log(alog(aMIMI)+log(a)+log(aIMIM) = ) = gap initiation penaltygap initiation penalty

• log(alog(aIIII) = gap extension penaltygap extension penalty

50


Emission Probabilities in Profile HMM• Probabilty of emitting a symbol Probabilty of emitting a symbol a a at an at an

insertion stateinsertion state I Ijj::

eeIjIj(a) = p(a)(a) = p(a)

where where p(a)p(a) is the frequency of the is the frequency of the occurrence of the symbol occurrence of the symbol a a in all the in all the sequences.sequences.

51


Profile HMM Alignment• Define Define vvMM

jj (i)(i) as the logarithmic likelihood as the logarithmic likelihood score of the best path for matching score of the best path for matching xx11..x..xii to to profile HMM ending with profile HMM ending with xxii emitted by the emitted by the state state MMjj..

• vvIIj j (i) (i) andand v vDD

j j (i) (i) are defined similarly.are defined similarly.

52


Profile HMM Alignment: Dynamic Programming vvMM

j-1j-1(i-1) + log(a(i-1) + log(aMMj-1,j-1,MMj j ))

vvMMjj(i) = log (e(i) = log (eMMjj(x(xii)/p(x)/p(xii)) + max v)) + max vII

j-1j-1(i-1) + log(a(i-1) + log(aIIj-1j-1,,MMj j ))

vvDDj-1j-1(i-1) + log(a(i-1) + log(aDDj-1j-1,,MMj j ))

vvMM

jj(i-1) + log(a(i-1) + log(aMMjj, I, Ijj))

vvIIjj(i) = log (e(i) = log (eIIjj(x(xii)/p(x)/p(xii)) + max v)) + max vII

jj(i-1) + log(a(i-1) + log(aIIjj, I, Ijj))

vvDDjj(i-1) + log(a(i-1) + log(aDDjj, I, Ijj))

53


Paths in Edit Graph and Profile HMM

A path through an edit graph and the corresponding path through a profile HMM

54


Making a Collection of HMM for Protein Families

• Use Blast to separate a protein database into families of related proteins

• Construct a multiple alignment for each protein family.• Construct a profile HMM model and optimize the

parameters of the model (transition and emission probabilities).

• Align the target sequence against each HMM to find the best fit between a target sequence and an HMM

55


Application of Profile HMM to Modeling Globin Proteins• Globins represent a large collection of protein

sequences • 400 globin sequences were randomly selected

from all globins and used to construct a multiple alignment.

• Multiple alignment was used to assign an initial HMM

• This model then get trained repeatedly with model lengths chosen randomly between 145 to 170, to get an HMM model optimized probabilities.

56


How Good is the Globin HMM?• 625 remaining globin sequences in the database were

aligned to the constructed HMM resulting in a multiple alignment. This multiple alignment agrees extremely well with the structurally derived alignment.

• 25,044 proteins, were randomly chosen from the database and compared against the globin HMM.

• This experiment resulted in an excellent separation between globin and non-globin families.

57


PFAM• Pfam decribes protein domains • Each protein domain family in Pfam has:

- Seed alignment: manually verified multiple alignment of a representative set of sequences.

- HMM built from the seed alignment for further database searches.

- Full alignment generated automatically from the HMM• The distinction between seed and full alignments facilitates Pfam

updates.- Seed alignments are stable resources.- HMM profiles and full alignments can be updated with

newly found amino acid sequences.

58


PFAM Uses• Pfam HMMs span entire domains that include

both well-conserved motifs and less-conserved regions with insertions and deletions.

• It results in modeling complete domains that facilitates better sequence annotation and leads to a more sensitive detection.

59


HMM Parameter Estimation• So far, we have assumed that the transition So far, we have assumed that the transition

and emission probabilities are known.and emission probabilities are known.

• However, in most HMM applications, the However, in most HMM applications, the probabilities are not known. It’s very hard to probabilities are not known. It’s very hard to estimate the probabilities.estimate the probabilities.

60


HMM Parameter Estimation ProblemGiven HMM with states and alphabet (emission

characters) Independent training sequences x1, … xm Find HMM parameters Θ (that is, akl, ek(b))

that maximize P(x1, …, xm | Θ) the joint probability of the training sequences.

61


Maximize the likelihoodP(x1, …, xm | Θ) as a function of Θ is called the

likelihood of the model.The training sequences are assumed independent,

thereforeP(x1, …, xm | Θ) = Πi P(xi | Θ)

The parameter estimation problem seeks Θ that realizes

In practice the log likelihood is computed to avoid underflow errors

i

ixP )|(max

62


Two situationsKnown paths for training sequences

CpG islands marked on training sequencesOne evening the casino dealer allows us to see when he changes dice

Unknown paths CpG islands are not markedDo not see when the casino dealer changes dice

63


Known pathsAkl = # of times each k l is taken in the training

sequencesEk(b) = # of times b is emitted from state k in the

training sequencesCompute akl and ek(b) as maximum likelihood

estimators:

'

''

)'(/)()(

/

bkkk

lklklkl

bEbEbe

AAa

64


Pseudocounts Some state k may not appear in any of the training

sequences. This means Akl = 0 for every state l and akl cannot be computed with the given equation.

To avoid this overfitting use predetermined pseudocounts rkl and rk(b).

Akl = # of transitions kl + rkl

Ek(b) = # of emissions of b from k + rk(b)The pseudocounts reflect our prior biases about the

probability values.65


Unknown paths: Viterbi trainingIdea: use Viterbi decoding to compute the most

probable path for training sequence xStart with some guess for initial parameters and compute π* the most probable path for x using initial parameters.Iterate until no change in π* :

1. Determine Akl and Ek(b) as before2. Compute new parameters akl and ek(b) using the

same formulas as before3. Compute new π* for x and the current parameters

66


Viterbi training analysis The algorithm converges precisely

There are finitely many possible paths.New parameters are uniquely determined by the current π*.There may be several paths for x with the same probability, hence must compare the new π* with all previous paths having highest probability.

Does not maximize the likelihood Πx P(x | Θ) but the contribution to the likelihood of the most probable path Πx P(x | Θ, π*)

In general performs less well than Baum-Welch

67


Unknown paths: Baum-Welch Idea:1. Guess initial values for parameters.

art and experience, not science2. Estimate new (better) values for parameters.

how ?3. Repeat until stopping criteria is met.

what criteria ?

68


Better values for parametersWould need the Akl and Ek(b) values but cannot

count (the path is unknown) and do not want to use a most probable path.

For all states k,l, symbol b and training sequence x

Compute Akl and Ek(b) as expected values, given the current parameters

69


NotationFor any sequence of characters x emitted

along some unknown path π, denote by πi = k the assumption that the state at position i (in which xi is emitted) is k.

70


Probabilistic setting for Ak,lGiven x1, … ,xm consider a discrete probability space

with elementary eventsεk,l, = “k l is taken in x1, …, xm ”

For each x in {x1,…,xm} and each position i in x let Yx,i be a random variable defined by

Define Y = Σx Σi Yx,i random var that counts # of times the event εk,l happens in x1,…,xm.

otherwiselandkif

Y iilkix ,0

,1)( 1

,,

71


The meaning of Akl

Let Akl be the expectation of Y

E(Y) = Σx Σi E(Yx,i) = Σx Σi P(Yx,i = 1) = ΣxΣi P({εk,l | πi = k and πi+1 = l}) = ΣxΣi P(πi = k, πi+1 = l | x)

Need to compute P(πi = k, πi+1 = l | x)

72


Probabilistic setting for Ek(b)Given x1, … ,xm consider a discrete probability

space with elementary events εk,b = “b is emitted in state k in x1, … ,xm ”

For each x in {x1,…,xm} and each position i in x let Yx,i be a random variable defined by

Define Y = Σx Σi Yx,i random var that counts # of times the event εk,b happens in x1,…,xm.

otherwise

kandbxifY ii

bkix ,0,1

)( ,,

73


The meaning of Ek(b)

Let Ek(b) be the expectation of Y

E(Y) = Σx Σi E(Yx,i) = Σx Σi P(Yx,i = 1) =ΣxΣi P({εk,b | xi = b and πi = k})

Need to compute P(πi = k | x)

x bxi

ix bxi

iibkii

xkPkbxP}|{}|{

, )|(}),|({

74


Computing new parametersConsider x = x1…xn training sequence

Concentrate on positions i and i+1

Use the forward-backward values: fki = P(x1 … xi , πi = k)bki = P(xi+1 … xn | πi = k)

75


Compute Akl (1)Prob k l is taken at position i of x

P(πi = k, πi+1 = l | x1…xn) = P(x, πi = k, πi+1 = l) / P(x)

Compute P(x) using either forward or backward valuesWe’ll show that P(x, πi = k, πi+1 = l) = bli+1 ·el(xi+1) ·akl ·fki

Expected # times k l is used in training sequences

Akl = Σx Σi (bli+1 ·el(xi+1) ·akl ·fki) / P(x)

76


Compute Akl (2)P(x, πi = k, πi+1 = l) = P(x1…xi, πi = k, πi+1 = l, xi+1…xn) =P(πi+1 = l, xi+1…xn | x1…xi, πi = k)·P(x1…xi,πi =k)=P(πi+1 = l, xi+1…xn | πi = k)·fki =P(xi+1…xn | πi = k, πi+1 = l)·P(πi+1 = l | πi = k)·fki =P(xi+1…xn | πi+1 = l)·akl ·fki =P(xi+2…xn | xi+1, πi+1 = l) · P(xi+1 | πi+1 = l) ·akl ·fki =P(xi+2…xn | πi+1 = l) ·el(xi+1) ·akl ·fki =bli+1 ·el(xi+1) ·akl ·fki

77


Compute Ek(b)

Prob xi of x is emitted in state kP(πi = k | x1…xn) = P(πi = k, x1…xn)/P(x)

P(πi = k, x1…xn) = P(x1…xi,πi = k,xi+1…xn) = P(xi+1…xn | x1…xi,πi = k) · P(x1…xi,πi = k) =P(xi+1…xn | πi = k) · fki = bki · fki

Expected # times b is emitted in state k

x bxi

kikiki

xPbfbE:

)()(78


Finally, new parameters

Can add pseudocounts as before.

'

''

)'(/)()(

/

bkkk

lklklkl

bEbEbe

AAa

79


Stopping criteriaCannot actually reach maximum (optimization of

continuous functions)Therefore need stopping criteria

Compute the log likelihood of the model for current Θ

Compare with previous log likelihoodStop if small differenceStop after a certain number of iterations

x

xP )|(log

80


The Baum-Welch algorithmInitialization:

Pick the best-guess for model parameters(or arbitrary)

Iteration:1. Forward for each x2. Backward for each x3. Calculate Akl, Ek(b)4. Calculate new akl, ek(b)5. Calculate new log-likelihood

Until log-likelihood does not change much

81


Baum-Welch analysis Log-likelihood is increased by iterationsBaum-Welch is a particular case of the EM (expectation maximization) algorithmConvergence to local maximum. Choice of initial parameters determines local maximum to which the algorithm converges

82


Speech Recognition• Create an Create an HMMHMM of the words in a language of the words in a language

• Each word is a hidden state in Each word is a hidden state in QQ..• Each of the basic sounds in the language is Each of the basic sounds in the language is

a symbol in a symbol in Σ.Σ.

• Input:Input: use speech as the input sequence. use speech as the input sequence.• Goal:Goal: find the most probable sequence of find the most probable sequence of

states.states.

91


Speech Recognition: Building the Model• Analyze some large source of English Analyze some large source of English

sentences, such as a database of newspaper sentences, such as a database of newspaper articles, to form probability matrixes.articles, to form probability matrixes.

• AA0i0i: the chance that word : the chance that word ii begins a begins a sentence.sentence.

• AAijij: the chance that word : the chance that word jj follows word follows word ii..

92


Building the Model (cont’d)

• Analyze English speakers to determine what Analyze English speakers to determine what sounds are emitted with what words.sounds are emitted with what words.

• EEkk(b):(b): the chance that sound the chance that sound bb is spoken in is spoken in word word kk. Allows for alternate pronunciation . Allows for alternate pronunciation of words.of words.

93


Speech Recognition: Using the Model• Use the same dynamic programming Use the same dynamic programming

algorithm as beforealgorithm as before• Weave the spoken sounds through the Weave the spoken sounds through the

model the same way we wove the rolls of model the same way we wove the rolls of the die through the casino model.the die through the casino model.

• ππ represents the most likely set of words.represents the most likely set of words.

94


Using the Model (cont’d)

• How well does it work?How well does it work?• Common words, such as Common words, such as ‘‘thethe’’, , ‘‘aa’’, , ‘‘ofof’’

make prediction less accurate, since there make prediction less accurate, since there are so many words that follow normally.are so many words that follow normally.

95


Improving Speech Recognition• Initially, we were using a Initially, we were using a ‘bigram,’‘bigram,’ a graph a graph

connecting every two words.connecting every two words.• Expand that to a Expand that to a ‘trigram’‘trigram’

• Each state represents two words spoken in Each state represents two words spoken in succession.succession.

• Each edge joins those two words (Each edge joins those two words (A BA B) to another ) to another state representing (state representing (B CB C))

• Requires nRequires n33 vertices and edges, where n is the vertices and edges, where n is the number of words in the language.number of words in the language.

• Much better, but still limited context.Much better, but still limited context.

96


References• Slides for CS 262 course at Stanford given by

Serafim Batzoglou

97