Hidden Markov Model Lecture #6

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Hidden Markov ModelLecture #6

Reminder: Finite State Markov Chain

0, (ie, ( ))i i

x xa a p x

An integer time stochastic process, consisting of a domain D of m states {1,…,m} and

1. An m dimensional initial distribution vector ( p(1),.., p(m)).2. An m×m transition probabilities matrix M= (ast)

( )i i

1 2 1 1 1 1

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

L i i i ii

p x x x p X x p X x X x

• For each integer L, a Markov Chain assigns probability to sequences (x1…xL) over D (i.e, xi D) as follows:

Similarly, (X1,…, Xi ,…)is a sequence of probability distributions over D.

Ergodic Markov Chains

The Fundamental Theorem of Finite-state Markov Chains:If a Markov Chain is ergodic, then 1. It has a unique stationary distribution vector V > 0, which is an

Eigenvector of the transition matrix.2. The distributions Xi , as i∞, converges to V.

A Markov chain is ergodic if :1. All states are recurrent (ie, the

graph is strongly connected)2. It is not peridoic

Use of Markov Chains: Sequences with CpG Islands

Recall from last class: In human genomes the pair CG often transforms to (methyl-C) G which often transforms to TG.

Hence the pair CG appears less than expected from what is expected from the independent frequencies of C and G alone.

Due to biological reasons, this process is sometimes suppressed in short stretches of genomes such as in the start regions of many genes.

These areas are called CpG islands (p denotes “pair”).

Modeling sequences with CpG Island

The “+” model: Use transition matrix A+ = (a+st),

Where: a+

st = (the probability that t follows s in a CpG island)

The “-” model: Use transition matrix A- = (a-st),

Where: a-

st = (the probability that t follows s in a non CpG island)

Question 2: Finding CpG Islands

Given a long genomic str with possible CpG Islands, we define a Markov Chain over 8 states, all interconnected (hence it is ergodic):

C+ T+G+A+

C- T-G-A-

The problem is that we don’t know the sequence of states which are traversed, but just the sequence of letters.

Therefore we use here Hidden Markov Model

Hidden Markov Model

( , , ) ( | )L

L i ii

p s s p s s

A Markov chain (s1,…,sL):

and for each state s and a symbol x we have p(Xi=x|Si=s)

Application in communication: message sent is (s1,…,sm) but we receive (x1,…,xm) . Compute what is the most likely message sent ?

Application in speech recognition: word said is (s1,…,sm) but we recorded (x1,…,xm) . Compute what is the most likely word said ?

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

Hidden Markov Model

Notations:Markov Chain transition probabilities: p(Si+1= t|Si = s) = ast

Emission probabilities: p(Xi = b| Si = s) = es(b)

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

For Markov Chains we know:

What is p(s,x) = p(s1,…,sL;x1,…,xL) ?

( ) ( , , ) ( | )L

L i ii

p p s s p s ss

Hidden Markov Model

p(Xi = b| Si = s) = es(b), means that the probability of xi depends only on the probability of si.Formally, this is equivalent to the conditional independence assumption:

p(Xi=xi|x1,..,xi-1,xi+1,..,xL,s1,..,si,..,sL) = esi(xi)

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

1 1 11

( , ) ( , , ; ,..., ) ( | ) ( )i

L L i i s ii

p p s s x x p s s e xs x

Hidden Markov Model

Exercise: Using the definition of conditional probability:P(X|Y) = P(X,Y)/P(Y), prove formally that the equality

p(Xi = xi|x1,..,xi-1,xi+1,..,xL,s1,..,si,..,sL) = esi(xi)

implies that for any Y {x1,..,xi-1,xi+1,..,xL,s1,..,si,..,sL}, such

that si is in Y, it holds that:

p(Xi=xi|Y) = esi(xi)

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

Hidden Markov Model for CpG Islands

The states:

S1 S2 SL-1 SL

X1 X2 XL-1 XL

Domain(Si)={+, -} {A,C,T,G} (8 values)

In this representation P(xi| si) = 0 or 1 depending on whether xi is consistent with si . E.g. xi= G is consistent with si=(+,G) and with si=(-,G) but not with any other state of si.

The query of interest:

),,|,...,(argmax ),...,( 11),...,(s

L xxsspssL

Hidden Markov Model

Questions:

Given the “visible” sequence x=(x1,…,xL), find:1. A most probable (hidden) path.2. The probability of x.

3. For each i = 1,..,L, and for each state k, p(si=k| x)

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

1. Most Probable state pathS1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

First Question: Given an output sequence x = (x1,…,xL),

A most probable path s*= (s*1,…,s*

L) is one which maximizes p(s|x).

1( ,..., )

* *1 1 1* ( ,..., ) ( ,..., | ,..., )maxarg

L L Ls s s p s s x x

Most Probable path (cont.)

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

Since( , )

( | ) ( , )( )

s xs x s xx

we need to find s which maximizes p(s,x)

Viterbi’s algorithm for most probable path

1 1( ,..., )

( ,..., ; ,..., )L

L Ls s

p s s x xmaxargThe task: compute

vl(i) = the probability p(s1,..,si;x1,..,xi|si=l ) of a most probable path up to i, which ends in state l .

Let the states be {1,…,m}

Idea: for i=1,…,L and for each state l, compute:

Viterbi’s algorithm for most probable path

vl(i) = the probability p(s1,..,si;x1,..,xi|si=l ) of a most probable path up to i, which ends in state l .

( ) ( ) max{ ( 1) }l l i k klk

v i e x v i a

Exercise: For i = 1,…,L and for each state l:

s1 Si-1

X1 Xi-1

Viterbi’s algorithm

s1 s2 sL-1 sL

X1 X2 XL-1 XL

For i=1 to L do for each state l :

vl(i) = el(xi) MAXk {vk(i-1)akl }ptri(l)=argmaxk{vk(i-1)akl} [storing previous state for reconstructing the path]Termination:

Initialization: v0(0) = 1 , vk(0) = 0 for k > 0

We add the special initial state 0.

Result: p(s1*,…,sL

*;x1,…,xl) = { ( )}max kk

2. Computing p(x)

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

Given an output sequence x = (x1,…,xL),Compute the probability that this sequence was generated:

( ) ( ),p px x sS

The summation taken over all state-paths s generating x.

Forward algorithm for computing p(x)

( ) ( ),s

p px x s

The task: compute

Idea: for i=1,…,L and for each state l, compute:

fl(i) = p(x1,…,xi;si=l ), the probability of all the paths which emit (x1,..,xi) and end in state si=l.

( ) ( ) ( 1)l l i k klk

f i e x f i a Use the recursive formula:

Forward algorithm for computing p(x)

s1 s2 sL-1 sL

X1 X2 XL-1 XL

fl(i) = el(xi) ∑k fk(i-1)akl

Initialization: f0(0) := 1 , fk(0) := 0 for k>0

Similar to Viterbi’s algorithm:

Result: p(x1,…,xL) = ( )kkf L

3. The distribution of Si, given x

S1 S2 SL-1 SL

x1 x2 XL-1 xL

M M M M

Given an output sequence x = (x1,…,xL),

Compute for each i=1,…,l and for each state k the probability that si = k.

This helps to reply queries like: what is the probability that si is in a CpG island, etc.

Solution in two stages

1. For each i and each state k, compute p(si=k | x1,…,xL).

s1 s2 sL-1 sL

X1 X2 XL-1 XL

2. Do the same computation for every i = 1,..,L but without repeating the first task L times.

Computing for a single i:

s1 s2 sL-1 sL

X1 X2 XL-1 XL

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

( ,..., )

1 ( , ,..., )

i Li L

p s x xp s x x

i Lp s x x

Decomposing the computation

P(x1,…,xL,si) = P(x1,…,xi,si) P(xi+1,…,xL | x1,…,xi,si)

(by the equality p(A,B) = p(A)p(B|A ).

s1 s2 sL-1 sL

X1 X2 XL-1 XL

P(x1,…,xi,si)= fsi(i) ≡ F(si), so we are left with

the task to compute P(xi+1,…,xL | x1,…,xi,si) ≡ B(si)

s1 s2 Si+1 sL

X1 X2 Xi+1 XL

Exercise: Show from the definitions of Markov Chain and Hidden Markov Chain that:P(xi+1,…,xL | x1,…,xi,si) = P(xi+1,…,xL | si)

Denote P(xi+1,…,xL | si) ≡ B(si).

Summary:P(x1,…,xL,si) = P(x1,…,xi,si) P(xi+1,…,xL | x1,…,xi,si)

s1 s2 sL-1 sL

X1 X2 XL-1 XL

Equality due to independence of {xi+1,…,xL}, and {x1,…,xi} | si} – by the Exercise.

= P(x1,…,xi,si) P(xi+1,…,xL | si) ≡ F(si)·B(si)

F(si): The Forward algorithm:

s1 s2 sL-1 sL

X1 X2 XL-1 XL

F(si) = esi(xi)·∑si-1

F (si-1)asi-1si

Initialization: F (0) = 1

The algorithm computes F(si) = P(x1,…,xi,si) for i=1,…,L (namely, considering evidence up to time slot i).

B(si): The backward algorithm

The task: Compute B(si) = P(xi+1,…,xL|si) for i=L-1,…,1 (namely, considering evidence after time slot i).

SL-1 SL

XL-1 XL

Si Si+1

{first step, step L-1: Compute B(sL-1).}

{step i: compute B(si) from B(si+1)}

P(xi+1,…,xL|si) = P(si+1 | si) P(xi+1 | si+1) P(xi+2,…,xL| si+1)si+1

B(si) B(si+1)

P(xL| sL-1) = sLP(xL ,sL |sL-1) = s

L P(sL |sL-1) P(xL |sL )

The combined answer

1. To compute the probability that Si=si given that {x1,…,xL} run the forward algorithm and compute F(si) = P(x1,…,xi,si), run the backward algorithm to compute B(si) = P(xi+1,…,xL|si), the product F(si)B(si) is the answer (for every possible value si).

2. To compute these probabilities for every si simply run the forward and backward algorithms once, storing F(si) and B(si) for every i (and every value of si). Compute F(si)B(si) for every i.

s1 s2 sL-1 sL

X1 X2 XL-1 XL

Time and Space Complexity of the forward/backward algorithms

Time complexity is O(m2L) where m is the number of states. It is linear in the length of the chain, provided the number of states is a constant.

s1 s2 sL-1 sL

X1 X2 XL-1 XL

Space complexity is also O(m2L).

Hidden Markov Model Lecture #6

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