.

Correctness proof of EMVariants of HMM

Sequence Alignment via HMM

Lecture #10

This class has been edited from Nir Friedman’s lecture. Changes made by Dan Geiger, then Shlomo Moran.

Background Readings: chapters 11.6, 3.4, 3.5, 4, in the Durbin et al., 2001, Chapter 3.4 Setubal et al., 1997

2

In each iteration the EM algorithm does the following. (E step): Calculate

Qθ () = ∑y p(y|x,θ)log p(x,y|)

(M step): Find * which maximizes Qθ ()

(Next iteration sets * and repeats).

The EM algorithm

Comment: At the M-step we only need that Qθ(*)>Qθ(θ). This change yields the so called Generalized EM algorithm. It is important when it is hard to find the optimal *.

3

Correctness proof of EM

Theorem:

If λ* maximizes Q (λ) = ∑y p(y|x,θ)log p(y| λ), then P(x| λ*) P(x| θ) .

Comment: The proof remains valid if we assume only that

Q(λ*) Q(θ).

4

By the definition of conditional probability, for each y we have,

p(x|) p(y|x,) = p(y,x|), and hence:

log p(x|) = log p(y,x|) – log p(y|x,)Hence

log p(x| λ) = ∑y p(y|x,θ) [log p(y|λ) – log p(y|x,λ)]

log p(x|λ)

Proof (cont.)

=1

(Next..)

5

Proof (end)

log p(x|λ) = ∑y p(y|x, θ) log p(y|λ)

- ∑y p(y|x,θ) log [p(y|x,λ)]

Qθ(λ)

Substituting λ=λ* and λ=θ, and then subtracting, we get

log p(x|λ*) - log p(x|θ) =

Q(λ*) – Q(θ) + D(p(y|x,θ) || p(y|x,λ*))

≥ 0 [since λ* maximizes Q(λ)]. Relative entropy 0 ≤

≥ 0 QED

6

EM in Practice

Initial parameters: Random parameters setting “Best” guess from other source

Stopping criteria: Small change in likelihood of data Small change in parameter values

Avoiding bad local maxima: Multiple restarts Early “pruning” of unpromising ones

7

HMM model structure:1. Duration Modeling

Markov chains are rather limited in describing sequences of symbols with non-random structures. For instance, Markov chain forces the distribution of segments in which some state is repeated for k additional times to be (1-p)pk, for some p.

Several ways enable modeling of other distributions. One is assigning k >1 states to represent the same “real” state. This may guarantee k repetitions with any desired probability.

A1 A2 A3 A4

8

HMM model structure:2. Silent states

States which do not emit symbols (as we saw in the abo locus). Can be used to model duration distributions. Also used to allow arbitrary jumps (may be used to model

deletions) Need to generalize the Forward and Backward algorithms for

arbitrary acyclic digraphs to count for the silent states (see next):

Silent states:

Regular states:

9

Eg, the forwards algorithm should look:

Directed cycles of silence (or other) states complicate things, and should be avoided.

Ezu kklkl

Evu kklkll

aufzfz

aufxevf

xv

),(

),(

)()(

)()()(

: statessilent for and

: symbolemit which statesregular For

x

v

zSilent states

Regular states

symbols

10

HMM model structure:3. High Order Markov Chains

Markov chains in which the transition probabilities depends on the last k states:

P(xi|xi-1,...,x1) = P(xi|xi-1,...,xi-k)

Can be represented by a standard Markov chain with more states. eg for k=2:

AA

BBBA

AB

11

HMM model structure:4. Inhomogenous Markov Chains

An important task in analyzing DNA sequences is recognizing the genes which code for proteins.

A triplet of 3 nucleotides – called codon - codes for amino acids (see next slide).

It is known that in parts of DNA which code for genes, the three codons positions has different statistics.

Thus a Markov chain model for DNA should represent not only the Nucleotide (A, C, G or T), but also its position – the same nucleotide in different position will have different transition probabilities. Used in GENEMARK gene finding program (93).

12

Genetic Code

There are 20 amino acids from which proteins are build.

13

Sequence Comparison using HMM

HMM for sequence alignment, which incorporates affine gap scores.

“Hidden” States Match. Insertion in x. insertion in y.

Symbols emitted Match: {(a,b)| a,b in ∑ }. Insertion in x: {(a,-)| a in ∑ }. Insertion in y: {(-,a)| a in ∑ }.

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The Transition Probabilities

ε1- ε

ε1- ε

δδ1-2δ

M X

X

M

Y

Y

Emission Probabilities Match: (a,b) with pab – only from M states

Insertion in x: (a,-) with qa – only from X state

Insertion in y: (-,a).with qa - only from Y state.

(Note that the hidden states can be reconstructed from the alignment.)

Transitions probabilities(note the forbidden ones).δ = probability for 1st gapε = probability for tailing gap.

15

Scoring alignments

Each aligned pair is generated by the above HMM with certain probability. For each pair of sequences x (of length m) and y (of length n), there are many alignments of x and y, each corresponds to a different state-path (the length of the paths are between max{m,n} and m+n).Our task is to score alignments by using this model. The score should reflect the probability of the alignment.

16

Most probable alignment

Let vM(i,j) be the probability of the most probable alignment of x(1..i) and y(1..j), which ends with a match. Similarly, vX(i,j) and vY(i,j), the probabilities of the most probable alignment of x(1..i) and y(1..j), which ends with an insertion to x or y.

Then using a recursive argument, we get:

(1 2 ) ( 1, 1)

[ , ] max (1 ) ( 1, 1)

(1 ) ( 1, 1)i j

M

M Xx y

Y

v i j

v i j p v i j

v i j

17

Most probable alignment

Similar argument for vX(i,j) and vY(i,j), the probabilities of the most probable alignment of x(1..i) and y(1..j), which ends with an insertion to x or y, are:

( 1, )[ , ] max

( 1, )i

MX

x X

v i jv i j q

v i j

( , 1)[ , ] max

( , 1)j

MY

y Y

v i jv i j q

v i j

18

Adding termination probabilities

For this, an END state is added, with transition probability τ from any other state to END. This assumes expected sequence length of 1/ τ.

M X Y END

M 1-2δ -τ δ δ τ

X 1-ε -τ ε τ

Y1-ε -τ

ε τ

END 1

We may want a model which defines a probability distribution over all possible sequences.

The last transition in each alignment is to the END state, with probability τ

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The log-odds scoring function

We wish to know if the alignment score is above or below the score of random alignment.

For gapless alignments we used the log-odds ratio:

s(a,b) = log (pab / qaqb). log (pab/qaqb)>0 iff the probability that a and b are related by our model is larger than the probability that they are picked at random.

To adapt this for the HMM model, we need to model random sequence by HMM, with end state. This model assigns probability to each pair of sequences x and y of arbitrary lengths m and n.

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scoring function for random model

X Y END

X 1- η η

Y 1- η η

END 1

The transition probabilities for the random model, with termination probability η:

(x is the start state)

2

1 1

( , | ) (1 )i j

n mn m

x yi j

p x y Random q q

The emission probability for a is qa.

Thus the probability of x (of length n) and y (of length m) being random is:

And the corresponding log-odds score is:

m

iy

n

ix ii

qqmnRandomyxp11

12 loglog)log()(log)|,(log

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Markov Chains for “Random” and “Model”

X Y END

X 1- η η

Y 1- η η

END 1

M X Y END

M 1-2δ -τ δ δ τ

X 1-ε -τ ε τ

Y1-ε -τ

ε τ

END 1

“Model”

“Random”

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Combining models in the log-odds scoring function

In order to compare the M score to the R score of sequences x and y, we can find an optimal M score, and then subtract from it the R score.

This is insufficient when we look for local alignments, where the optimal substrings in the alignment are not known in advance. A better way:

1. Define a log-odds scoring function which keeps track of the difference Match-Random scores of the partial strings during the alignment.

2. At the end add to the score (logτ – 2logη).

We get the following:

23

The log-odds scoring function

)log(

],[)log(

],[)log(

],[)log(

maxlog],[

12

111

111

1121

bY

X

M

yx

yxM

jiV

jiV

jiV

qq

pjiV

ji

ji

And at the end add to the score (logτ – 2logη).

)log(],[log

],[logmax],[

11

1

jiV

jiVjiV

Y

MY

)log(],[log

],[logmax],[

1

1

1

jiV

jiVjiV

X

MX

(assuming that letters at insertions/deletions are selected by the random model)

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The log-odds scoring function

Another way (Durbin et. al., Chapter 4.1):Define scoring function s with penalties d and e for a first gap and a

tailing gap, resp.

Then modify the algorithm to correct for extra prepayment, as follows:

-1

states) Yor X tong when movi"prepayment(" 211

1

state)from M move(assume 1

212

log

))((

)(log

)(

)(loglog),(

-e

-d

qq

pbas

ba

ab

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Log-odds alignment algorithm

( 1, 1)

[ , ] ( , ) max ( 1, 1)

( 1, 1)

M

M Xi j

Y

V i j

V i j s x y V i j

V i j

( 1, )[ , ] max

( 1, )

MX

X

V i j dV i j

V i j e

( , 1)[ , ] max

( , 1)

MY

Y

V i j dV i j

V i j e

Initialization: VM(0,0)=logτ - 2logη.

Termination:V = max{ VM(m,n), VX(m,n)+c, VY(m,n)+c}Where c= log (1-2δ-τ) – log(1-ε-τ)