Markov Chains Lecture #5 by 969J7X


									                    Markov Chains
                     Lecture #5

    Background Readings: Durbin et. al. Section 3.1,

       Dependencies along the genome
In previous classes we assumed every letter in a
   sequence is sampled randomly from some
   distribution q() over the alpha bet {A,C,T,G}.

This model could suffice for alignment scoring, but it
   is not the case in true genomes.
1. There are special subsequences in the genome, like
   TATA within the regulatory area, upstream a gene.
2. The pattern CG is less common than expected for
   random sampling.

We model such dependencies by Markov chains and
  hidden Markov model, which we define next.

                Finite Markov Chain

An integer time stochastic process, consisting of a
  domain D of m>1 states {s1,…,sm} and
1. An m dimensional initial distribution vector ( p(s1),.., p(sm)).
2. An m×m transition probabilities matrix M= (asisj)

For example, D can be the letters {A, C, T, G}, p(A) the
   probability of A to be the 1st letter in a sequence, and
   aAG the probability that G follows A in a sequence.

                Markov Chain (cont.)
  X1              X2                             Xn-1             Xn

• For each integer n, a Markov Chain assigns probability to
sequences (x1…xn) over D (i.e, xi D) as follows:
  p(( x1 , x2 ,...xn ))  p( X 1  x1 ) p( X i  xi | X i 1  xi 1 )
                                      i 2
   p( x1 ) axi1xi
           i 2
 Similarly, (X1,…, Xi ,…) is a sequence of probability
 distributions over D. There is a rich theory which studies the
 properties of these sequences. A bit of it is presented next.
              Markov Chain (cont.)
  X1             X2                             Xn-1              Xn

Similarly, each Xi is a probability distributions over D, which
is determined by the initial distribution (p1,..,pn) and the
transition matrix M.
There is a rich theory which studies the properties of such
“Markov sequences” (X1,…, Xi ,…). A bit of this theory is
presented next.

                 Matrix Representation
    A        B      C      D     The transition probabilities
    0.95     0     0.05   0      Matrix M =(ast)
                                 M is a stochastic Matrix:
B 0.2       0.5     0     0.3
                                        a  t   st   1
     0      0.2     0     0.8
C                                The initial distribution vector
                                 (u1…um) defines the distribution
     0       0      1     0      of X1 (p(X1=si)=ui) .

        Then after one move, the distribution is changed
        to X2 = X1M

               Matrix Representation
    A      B      C      D
    0.95   0     0.05    0     Example: if X1=(0, 1, 0, 0)
                               then X2=(0.2, 0.5, 0, 0.3)
B 0.2      0.5    0     0.3
                               And if X1=(0, 0, 0.5, 0.5)
                               then X2=(0, 0.1, 0.5, 0.4).
     0     0.2    0     0.8

     0     0      1      0

            The i-th distribution is Xi = X1Mi-1

  Representation of a Markov Chain as a
                                           A     B       C     D
 0.95                                     0.95   0     0.05   0
                                          0.2    0.5    0     0.3
         0.2            0.5                0     0.2    0     0.8
                                           0     0      1     0
0.05    0.2       0.3                 D


   Each directed edge AB is associated with the positive
   transition probability from A to B.

  Properties of Markov Chain states
States of Markov chains are classified by the digraph
representation (omitting the actual probability values)

A, C and D are recurrent states: they are in strongly connected
components which are sinks in the graph.

                        B is not recurrent – it is a transient state

                                   Alternative definitions:
                            A state s is recurrent if it can be
                            reached from any state reachable
                            from s; otherwise it is transient.

Another example of Recurrent and
        Transient States

            A and B are transient states, C and D
            are recurrent states.

            Once the process moves from B to
            D, it will never come back.

          Irreducible Markov Chains

A Markov Chain is irreducible if the corresponding graph is
strongly connected (and thus all its states are recurrent).


                       Periodic States
                               E      A state s has a period k if k is the
                                      GCD of the lengths of all the
                                      cycles that pass via s. (in the
                                      shown graph the period of A is 2).

   Exercise: All the states in the same strongly connected
   component have the same period

A Markov Chain is periodic if all the states in it have a period k >1.
It is aperiodic otherwise.

            Ergodic Markov Chains

                           A Markov chain is ergodic if :
                           1. the corresponding graph is
                              strongly connected.
                           2. It is not peridoic

Ergodic Markov Chains are important since they guarantee the
corresponding Markovian process converges to a unique
distribution, in which all states have strictly positive probability.

       Stationary Distributions for Markov
 Let M be a Markov Chain of m states, and let V = (v1,…,vm) be a
 probability distribution over the m states

V = (v1,…,vm) is stationary distribution for M if VM=V.
(ie, if one step of the process does not change the distribution).

                  V is a stationary distribution

     V is a left (row) Eigenvector of M with Eigenvalue 1.

   Stationary Distributions for a Markov
                 Chain M

Exercise: A stochastic matrix always has a real left Eigenvector
with Eigenvalue 1 (hint: show that a stochastic matrix has a
right Eigenvector with Eigenvalue 1. Note that the left
Eigenvalues of a Matrix are the same as the right Eiganvlues).

[It can be shown that the above Eigenvector V can be non-
negative. Hence each Markov Chain has a stationary

           “Good” Markov chains
A Markov Chains is good if the distributions Xi , as i∞:

(1) converge to a unique distribution, independent of the initial

(2) In that unique distribution, each state has a positive

The Fundamental Theorem of Finite Markov Chains:
   A Markov Chain is good  the corresponding graph is

We will prove the  part, by showing that non-ergodic Markov
   Chains are not good.
           Examples of
       “Bad” Markov Chains

A Markov chains is not “good” if either :
1. It does not converge to a unique
2. It does converge to u.d., but some states
in this distribution have zero probability.

 Bad case 1: Mutual Unreachabaility

                         Consider two initial distributions:
                         a) p(X1=A)=1 (p(X1 = x)=0 if x≠A).

                         b) p(X1= C) = 1

In case a), the sequence will stay at A forever.
In case b), it will stay in {C,D} for ever.

Fact 1: If G has two states which are unreachable from each
other, then {Xi} cannot converge to a distribution which is
independent on the initial distribution.
       Bad case 2: Transient States

Once the process moves from B to D, it will never come back.

      Bad case 2: Transient States

                               Fact 2: For each initial distribution,
                               with probability 1 a transient state
                               will be visited only a finite number
                               of times.


Proof: Let A be a transient state, and let X be the set of states
from which A is unreachable. It is enough to show that,
starting from any state, with probability 1 a state in X is
reached after a finite number of steps (Exercise: complete the
Corollary: A good Markov
   Chain is irreducible

   Bad case 3: Periodic Markov Chains

Recall: A Markov Chain is periodic if all the states in it have a period
k >1. The above chain has period 2.
In the above chain, consider the initial distribution p(B)=1.
Then states {B, C} are visited (with positive probability) only in odd
steps, and states {A, D, E} are visited in only even steps.

           Bad case 3: Periodic States

Fact 3: In a periodic Markov Chain (of period k >1) there are initial
distributions under which the states are visited in a periodic manner.
Under such initial distributions Xi does not converge as i∞.

               Corollary: A good Markov
                 Chain is not periodic
 The Fundamental Theorem of Finite Markov
  We have proved that non-ergodic Markov Chains are not good.
  A proof of the other part (based on Perron-Frobenius theory) is
  beyond the scope of this course:

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. For any initial distribution, the distributions Xi , as i∞,
     converges to V.

 Use of Markov Chains in Genome search:
          Modeling CpG Islands
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”).

         Example: CpG Island (Cont.)

We consider two questions (and some variants):

Question 1: Given a short stretch of genomic data, does
it come from a CpG island ?

Question 2: Given a long piece of genomic data, does
it contain CpG islands in it, where, what length ?

We “solve” the first question by modeling strings with
and without CpG islands as Markov Chains over the
same states {A,C,G,T} but different transition
        Example: CpG Island (Cont.)

The “+” model: Use transition matrix A+ = (a+st),
a+st = (the probability that t follows s in a CpG

The “-” model: Use transition matrix A- = (a-st),
a-st = (the probability that t follows s in a non
       CpG island)

          Example: CpG Island (Cont.)

With this model, to solve Question 1 we need to decide
whether a given short sequence of letters is more likely
to come from the “+” model or from the “–” model.
This is done by using the definitions of Markov Chain,
in which the parameters are determined by known data
and the log odds-ratio test.

     Question 1: Using two Markov chains
                      A+ (For CpG islands):

We need to specify p+(xi | xi-1) where + stands for CpG
Island. From Durbin et al we have:
                            A        C           G         T
                  A        0.18     0.27       0.43       0.12
          Xi-1    C        0.17   p+(C | C)    0.274    p+(T|C)
                  G        0.16   p+(C|G)     p+(G|G)   p+(T|G)
                  T        0.08   p+(C |T)    p+(G|T)   p+(T|T)

       (Recall: rows must add up to one; columns need not.)
Question 1: Using two Markov chains

             A- (For non-CpG islands):
…and for p-(xi | xi-1) (where “-” stands for Non CpG
island) we have:
                      A        C          G        T
            A        0.3       0.2       0.29     0.21
   Xi-1     C        0.32    p-(C|C)   0.078     p-(T|C)
            G        0.25    p-(C|G)   p-(G|G)   p-(T|G)
            T        0.18    p-(C|T)   p-(G|T)   p-(T|T)

  Discriminating between the two models

      X1           X2                          XL-1            XL

Given a string x=(x1….xL), now compute the ratio
                                    L 1

              p( x |  model)
                                    p (x
                                    i 0
                                              i 1   | xi )
      RATIO                       L 1
              p( x |  model)
                                    p (x
                                    i 0
                                              i 1   | xi )

    If RATIO>1, CpG island is more likely.
    Actually – the log of this ratio is computed:

    Note: p+(x1|x0) is defined for convenience as p+(x1).
    p-(x1|x0) is defined for convenience as p-(x1).
                  Log Odds-Ratio test
  Taking logarithm yields

              p(x1...xL|  )
                                            p(xi|xi 1 )
  log Q  log                          log
              p(x1...xL|  )        i
                                            p(xi|xi 1 )

If logQ > 0, then + is more likely (CpG island).
If logQ < 0, then - is more likely (non-CpG island).

      Where do the parameters (transition-
          probabilities) come from ?

Learning from complete data, namely, when the label is
  given and every xi is measured:

Source: A collection of sequences from CpG islands, and a
collection of sequences from non-CpG islands.
Input: Tuples of the form (x1, …, xL, h), where h is + or -

Output: Maximum Likelihood parameters (MLE)

Count all pairs (Xi=a, Xi-1=b) with label +, and
with label -, say the numbers are Nba,+ and Nba,- .

      Maximum Likelihood Estimate (MLE) of
       the parameters (using labeled data)
          X1                     X2                            XL-1         XL

The needed parameters are:
P+(x1), p+ (xi | xi-1), p-(x1), p-(xi | xi-1)
The ML estimates are given by:
                    N a,
p ( X 1  a )                       Where Nx,+ is the number of times letter x
                                      appear in CpG islands in the dataset.

                                 N ba ,        Where Nbx,+ is the number of times
 p ( X i  a | X i 1  b ) 
                                 N    bx , 
                                                letter x appears after letter b in CpG
                                 x              islands in the dataset.
Using MLE is justified when we have a large sample. The numbers appearing in
the text book are based on 60,000 nucleotides. When only small samples are
available, Bayesian learning is an attractive alternative.                               34
      Question 2: Finding CpG Islands

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

                           The problem is that we don’t know
 A+     C+    G+     T+    the sequence of states which are
                           traversed, but just the sequence of
 A-     C-    G-     T-    letters.

Therefore we use here Hidden Markov Model
                Hidden Markov Model
            M              M             M                       M
      S1            S2                                 SL-1               SL

     T             T                                   T                  T
      x1            x2                                 XL-1               xL

      A Markov chain (s1,…,sL):      p( s1 , , sL )   p( si | si 1 )
                                                       i 1
    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 ?

             Hidden Markov Model
         M                M            M                      M
  S1              S2                              SL-1             SL

 T               T                                T               T
  x1              x2                              XL-1             xL

Markov Chain transition probabilities: p(Si+1= t|Si = s) = ast
Emission probabilities: p(Xi = b| Si = s) = es(b)
For Markov Chains we know: p( s)  p( s1 , , sL )   p( si | si 1 )
                                                         i 1

What is p(s,x) = p(s1,…,sL;x1,…,xL) ?

              Hidden Markov Model
        M                     M                  M                        M
  S1               S2                                          SL-1               SL

 T                T                                            T                  T
  x1                x2                                         XL-1               xL

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)

Thus    p( s, x)  p( s1 , , sL ; x1 ,..., xL )   p( si | si 1 ) esi ( xi )
                                                     i 1


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