Transcript No Slide Title

Markov Chains
.
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 pairs C followed by G is less common than
expected for random sampling.
We model such dependencies by Markov chains and
hidden Markov model, which we define next.
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Finite Markov Chain
An integer time stochastic process, consisting of a
domain D of m 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.
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Simple Model - Markov Chains
• Markov Property: The state of the system at time t+1 only
depends on the state of the system at time t
P[Xt 1  xt 1 | X t  x t , X t -1  x t -1 , . . . , X1  x1 , X0  x0 ]
 P[Xt 1  xt 1 | X t  x t ]
X1
X2
X3
X4
X5
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Markov Chain (cont.)
X2
X1
Xn-1
Xn
• For each integer n, a Markov Chain assigns probability to
sequences (x1…xn) over D (i.e, xi D) as follows:
n
p(( x1 , x2 ,...xn ))  p( X 1  x1 ) p( X i  xi | X i 1  xi 1 )
n
 p( x1 ) axi1xi
i 2
i 2
Similarly, (X1,…, Xi ,…)is a sequence of probability
distributions over D.
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Matrix Representation
A
A
B
C
0.95
0
0.05
B 0.2
C
D
0.5
0
D
0
0.3
0
0.2
0
0.8
0
0
1
0
The transition probabilities
Matrix M =(ast)
M is a stochastic Matrix:
a
t
st
1
The initial distribution vector
(u1…um) defines the distribution
of X1 (p(X1=si)=ui) .
Then after one move, the distribution is changed
to X2 = X1M
After i moves the distribution is Xi = X1Mi-1
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Simple Example
Weather:
–raining today
rain tomorrow
prr = 0.4
–raining today
no rain tomorrow
prn = 0.6
–no raining today
rain tomorrow
pnr = 0.2
–no raining today
no rain tomorrow prr = 0.8
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Simple Example
Transition Matrix for Example
 0.4 0.6 

P  
 0.2 0.8 
• Note that rows sum to 1
• Such a matrix is called a Stochastic Matrix
• If the rows of a matrix and the columns of a
matrix all sum to 1, we have a Doubly Stochastic
Matrix
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Gambler’s Example
– At each play we have the following:
• Gambler wins $1 with probability p
or
• Gambler loses $1 with probability 1-p
– Game ends when gambler goes broke, or gains a fortune
of $100
– Both $0 and $100 are absorbing states
p
0
1
1-p
p
p
N-1
2
1-p
p
N
1-p
1-p
Start
(10$)
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Coke vs. Pepsi
Given that a person’s last cola purchase was Coke, there
is a 90% chance that her next cola purchase will also be
Coke.
If a person’s last cola purchase was Pepsi, there is an
80% chance that her next cola purchase will also be
Pepsi.
0.1
0.9
coke
0.8
pepsi
0.2
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Coke vs. Pepsi
Given that a person is currently a Pepsi purchaser, what
is the probability that she will purchase Coke two
purchases from now?
The transition matrix is:
0.9 0.1
P

0.2 0.8
(Corresponding to one
purchase ahead)
0.9 0.1 0.9 0.1 0.83 0.17
P 





0.2 0.8 0.2 0.8 0.34 0.66
2
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Coke vs. Pepsi
Given that a person is currently a Coke drinker, what is
the probability that she will purchase Pepsi three
purchases from now?
0.9 0.1 0.83 0.17 0.781 0.219
P 





0.2 0.8 0.34 0.66 0.438 0.562
3
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Coke vs. Pepsi
Assume each person makes one cola purchase per
week. Suppose 60% of all people now drink Coke, and
40% drink Pepsi.
What fraction of people will be drinking Coke three
weeks from now?
P00
Let (Q0,Q1)=(0.6,0.4) be the initial probabilities.
We will regard Coke as 0 and Pepsi as 1
We want to find P(X3=0)
0.9 0.1
P

0
.
2
0
.
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

1
( 3)
P( X 3  0)   Qi pi(03)  Q0 p00
 Q1 p10(3)  0.6  0.781 0.4  0.438  0.6438
i 0
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“Good” Markov chains
For certain Markov Chains, the distributions Xi , as i∞:
(1) converge to a unique distribution, independent of the initial
distribution.
(2) In that unique distribution, each state has a positive
probability.
Call these Markov Chain “good”.
We describe these “good” Markov Chains by considering Graph
representation of Stochastic matrices.
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Representation as a Digraph
0.95
A
0.95
B
0
C
0.05
0.2
0.5
0
0.3
0
0.2
0
0.8
0
0
1
0
D
0
A
0.2
0.5
B
C
0.05
0.2
0.3
0.8
1
D
Each directed edge AB is
associated with the positive
transition probability from A
to B.
We now define properties of this graph which guarantee:
1. Convergence to unique distribution:
2. In that distribution, each state has positive probability.
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Examples of
“Bad” Markov Chains
Markov chains are not “good” if either :
1. They do not converge to a unique
distribution.
2. They do converge to u.d., but some states
in this distribution have zero probability.
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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.
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Bad case 2: Transient States
Def: A state s is recurrent if it can be
reached from any state reachable
from s; otherwise it is transient.
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.
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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.
X
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Bad case 3: 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.
A Markov Chain is periodic if all the states in it have a period k >1.
It is aperiodic otherwise.
Example: 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.
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Bad case 3: Periodic States
E
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∞.
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Ergodic Markov Chains
0.95
0.2
0.05
0.2
0.5
0.3
A Markov chain is ergodic if :
1. All states are recurrent (ie, the
graph is strongly connected)
2. It is not periodic
0.8
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The Fundamental Theorem of Finite 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.
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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”).
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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
probabilities:
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Example: CpG Island (Cont.)
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)
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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.
[to solve Question 2 we need to decide which parts of a given
long sequence of letters is more likely to come from the “+”
model, and which parts are more likely to come from the “–”
model. This is done by using the Hidden Markov Model, to be
defined later.]
We start with Question 1:
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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:
Xi-1
A
Xi
C
G
T
A
C
0.18
0.27
0.43
0.12
0.17
p+(C | C)
0.274
p+(T|C)
G
T
0.16
p+(C|G)
p+(G|G)
p+(T|G)
0.08
p+(C |T)
p+(G|T)
p+(T|T)
(Recall: rows must add up to one; columns need not.)
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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:
Xi
Xi-1
A
C
G
T
A
C
G
T
0.3
0.2
0.29
0.21
0.32
p-(C|C)
0.078
p-(T|C)
0.25
p-(C|G)
p-(G|G)
p-(T|G)
0.18
p-(C|T)
p-(G|T)
p-(T|T)
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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)
RATIO 

p ( x |  model)
p
 ( xi 1
| xi )
i 0
L 1
 p (x

i 1
| xi )
i 0
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).
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Log Odds-Ratio test
Taking logarithm yields
p( x1...x L|  )
log Q  log

p( x1...x L|  )

i
p( xi|x i 1 )
log
p( xi|x i 1 )
If logQ > 0, then + is more likely (CpG island).
If logQ < 0, then - is more likely (non-CpG island).
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Where do the parameters (transitionprobabilities) 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,- .
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Maximum Likelihood Estimate (MLE) of
the parameters (using labeled data)
X2
X1
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:
p ( X 1  a ) 
Where Na,+ is the number of times letter a
appear in CpG islands in the dataset.
N a,
N
a ,
a
p  ( X i  a | X i 1  b) 
N ba , 
N
a
ba , 
Where Nba,+ is the number of times
letter b appears after letter a in CpG
islands in the dataset.
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