Hidden Markov Model ..

Hidden Markov Model .

Example: CpG Island 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:

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): The problem is that we don’t know the sequence of states which are traversed, but just the sequence of letters. A+ C+ G+ T+ A- C- G- T- Therefore we use here Hidden Markov Model

Hidden Markov Models - HMM
Hidden variables H1 H2 HL-1 HL X1 X2 XL-1 XL Hi Xi Observed data

Hidden Markov Model M T A Markov chain (s1,…,sL):
x1 x2 XL-1 xL M T 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 ?

Hidden Markov Model M T Notations:
SL-1 SL x1 x2 XL-1 xL M T Notations: Markov Chain transition probabilities: p(Si+1= t|Si = s) = ast Emission probabilities: p(Xi = b| Si = s) = es(b)

Example: The Dishonest Casino
A casino has two dice: Fair die P(1) = P(2) = P(3) = P(5) = P(6) = 1/6 Loaded die P(1) = P(2) = P(3) = P(5) = 1/10 P(6) = 1/2 Casino player switches back-&-forth between fair and loaded die once every 20 turns Game: You bet $1 You roll (always with a fair die) Casino player rolls (maybe with fair die, maybe with loaded die) Highest number wins $2

Question # 1 – Decoding GIVEN A sequence of rolls by the casino player
QUESTION What portion of the sequence was generated with the fair die, and what portion with the loaded die? This is the DECODING question in HMMs

Question # 2 – Evaluation
GIVEN A sequence of rolls by the casino player QUESTION How likely is this sequence, given our model of how the casino works? This is the EVALUATION problem in HMMs

Question # 3 – Learning GIVEN A sequence of rolls by the casino player
QUESTION How “loaded” is the loaded die? How “fair” is the fair die? How often does the casino player change from fair to loaded, and back? This is the LEARNING question in HMMs

The dishonest casino model
0.05 0.95 0.95 FAIR LOADED P(1|F) = 1/6 P(2|F) = 1/6 P(3|F) = 1/6 P(4|F) = 1/6 P(5|F) = 1/6 P(6|F) = 1/6 P(1|L) = 1/10 P(2|L) = 1/10 P(3|L) = 1/10 P(4|L) = 1/10 P(5|L) = 1/10 P(6|L) = 1/2 0.05

A parse of a sequence Given a sequence x = x1……xN,
A parse of x is a sequence of states  = 1, ……, N 1 2 K … 1 1 2 K … 1 2 K … 1 2 K … … 2 2 K x1 x2 x3 xK

Likelihood of a parse 1 2 K 1 1 2 K 1 2 K 1 2 K 2 2 K x1 x2 x3 xK
Given a sequence x = x1……xN and a parse  = 1, ……, N, To find how likely is the parse: (given our HMM) P(x, ) = P(x1, …, xN, 1, ……, N) = P(xN, N | x1…xN-1, 1, ……, N-1) P(x1…xN-1, 1, ……, N-1) = P(xN, N | N-1) P(x1…xN-1, 1, ……, N-1) = … = P(xN, N | N-1) P(xN-1, N-1 | N-2)……P(x2, 2 | 1) P(x1, 1) = P(xN | N) P(N | N-1) ……P(x2 | 2) P(2 | 1) P(x1 | 1) P(1) = a01 a12……aN-1N e1(x1)……eN(xN) 1 2 K … 1 1 2 K … 1 2 K … 1 2 K … … 2 2 K x1 x2 x3 xK

Example: the dishonest casino
Let the sequence of rolls be: x = 1, 2, 1, 5, 6, 2, 1, 6, 2, 4 Then, what is the likelihood of  = Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair? (say initial probs a0Fair = ½, a0Loaded = ½) ½  P(1 | Fair) P(Fair | Fair) P(2 | Fair) P(Fair | Fair) … P(4 | Fair) = ½  (1/6)10  (0.95)9 = = 0.5  10-9

So, the likelihood the die is fair in all this run is just  10-9 OK, but what is the likelihood of  = Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded? ½  P(1 | Loaded) P(Loaded, Loaded) … P(4 | Loaded) = ½  (1/10)8  (1/2)2 (0.95)9 = = 0.79  10-9 Therefore, it somewhat more likely that the die is fair all the way, than that it is loaded all the way

Let the sequence of rolls be: x = 1, 6, 6, 5, 6, 2, 6, 6, 3, 6 Now, what is the likelihood  = F, F, …, F? ½  (1/6)10  (0.95)9 = 0.5  10-9, same as before What is the likelihood  = L, L, …, L? ½  (1/10)4  (1/2)6 (0.95)9 = = 0.5  10-7 So, it is 100 times more likely the die is loaded

C-G Islands Example C-G island? A/C/G/T H1 H2 HL-1 HL X1 X2 XL-1 XL Hi
change T C A G T C C-G island? A/C/G/T H1 H2 HL-1 HL X1 X2 XL-1 XL Hi Xi

Hidden Markov Model for CpG Islands
The states: Domain(Si)={+, -} (2 values) S1 S2 SL-1 SL X1 X2 XL-1 XL 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: ) , | ,..., ( arg max 1 (s * L s x p K =

1. Most Probable state path = decoding
SL-1 SL x1 x2 XL-1 xL M T 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).

Decoding GIVEN x = x1x2……xN 1 We want to find  = 1, ……, N,
such that P[ x,  ] is maximized * = argmax P[ x,  ] We can use dynamic programming! Let Vk(i) = max{1,…,i-1} P[x1…xi-1, 1, …, i-1, xi, i = k] = Probability of most likely sequence of states ending at state i = k 1 2 K … x1 x2 x3 xK

Decoding – main idea Given that for all states k,
and for a fixed position i, Vk(i) = max{1,…,i-1} P[x1…xi-1, 1, …, i-1, xi, i = k] What is Vj(i+1)? From definition, Vj(i+1) = max{1,…,i}P[ x1…xi, 1, …, i, xi+1, i+1 = j ] = max{1,…,i}P(xi+1, i+1 = j | x1…xi,1,…, i) P[x1…xi, 1,…, i] = max{1,…,i}P(xi+1, i+1 = j | i ) P[x1…xi-1, 1, …, i-1, xi, i] = maxk [P(xi+1, i+1 = j | i = k) max{1,…,i-1}P[x1…xi-1,1,…,i-1, xi,i=k]] = ej(xi+1) maxk akj Vk(i)

The Viterbi Algorithm Input: x = x1……xN Initialization:
V0(0) = 1 (0 is the imaginary first position) Vk(0) = 0, for all k > 0 Iteration: Vj(i) = ej(xi)  maxk akj Vk(i-1) Ptrj(i) = argmaxk akj Vk(i-1) Termination: P(x, *) = maxk Vk(N) Traceback: N* = argmaxk Vk(N) i-1* = Ptri (i)

The Viterbi Algorithm x1 x2 x3 ………………………………………..xN State 1 2 Vj(i) K Similar to “aligning” a set of states to a sequence Time: O(K2N) Space: O(KN)

Viterbi Algorithm – a practical detail
Underflows are a significant problem P[ x1,…., xi, 1, …, i ] = a01 a12……ai e1(x1)……ei(xi) These numbers become extremely small – underflow Solution: Take the logs of all values Vl(i) = log ek(xi) + maxk [ Vk(i-1) + log akl ]

Example Let x be a sequence with a portion of ~ 1/6 6’s, followed by a portion of ~ ½ 6’s… x = … … Then, it is not hard to show that optimal parse is : FFF…………………...F LLL………………………...L 6 characters “123456” parsed as F, contribute .956(1/6) = 1.610-5 parsed as L, contribute .956(1/2)1(1/10)5 = 0.410-5 “162636” parsed as F, contribute .956(1/6) = 1.610-5 parsed as L, contribute .956(1/2)3(1/10)3 = 9.010-5

2. Computing p(x) = evaluation
S1 S2 SL-1 SL x1 x2 XL-1 xL M T Given an output sequence x = (x1,…,xL), Compute the probability that this sequence was generated: The summation taken over all state-paths s generating x.

Evaluation We will develop algorithms that allow us to compute:
P(x) Probability of x given the model P(xi…xj) Probability of a substring of x given the model P(i = k | x) Probability that the ith state is k, given x

The Forward Algorithm We want to calculate
P(x) = probability of x, given the HMM Sum over all possible ways of generating x: P(x) =  P(x, ) =  P(x | ) P() To avoid summing over an exponential number of paths , define fk(i) = P(x1…xi, i = k) (the forward probability)

The Forward Algorithm – derivation
Define the forward probability: fk(i) = P(x1…xi, i = k) = 1…i-1 P(x1…xi-1, 1,…, i-1, i = k) ek(xi) = j 1…i-2 P(x1…xi-1, 1,…, i-2, i-1 = j) ajk ek(xi) = ek(xi) j fj(i-1) ajk

The Forward Algorithm We can compute fk(i) for all k, i, using dynamic programming! Initialization: f0(0) = 1 fk(0) = 0, for all k > 0 Iteration: fk(i) = ek(xi) j fj(i-1) ajk Termination: P(x) = k fk(N) ak0 Where, ak0 is the probability that the terminating state is k (usually = a0k)

Relation between Forward and Viterbi
Initialization: V0(0) = 1 Vk(0) = 0, for all k > 0 Iteration: Vj(i) = ej(xi) maxk Vk(i-1) akj Termination: P(x, *) = maxk Vk(N) FORWARD Initialization: f0(0) = 1 fk(0) = 0, for all k > 0 Iteration: fj(i) = ej(xi) k fk(i-1) akj Termination: P(x) = k fk(N) ak0

Motivation for the Backward Algorithm
We want to compute P(i = k | x), the probability distribution on the ith position, given x We start by computing P(i = k, x) = P(x1…xi, i = k, xi+1…xN) = P(x1…xi, i = k) P(xi+1…xN | x1…xi, i = k) = P(x1…xi, i = k) P(xi+1…xN | i = k) Then, P(i = k | x) = P(i = k, x) / P(x) Forward, fk(i) Backward, bk(i)

The Backward Algorithm – derivation
Define the backward probability: bk(i) = P(xi+1…xN | i = k) = i+1…N P(xi+1,xi+2, …, xN, i+1, …, N | i = k) = j i+1…N P(xi+1,xi+2, …, xN, i+1 = j, i+2, …, N | i = k) = j ej(xi+1) akj i+2…N P(xi+2, …, xN, i+2, …, N | i+1 = j) = j ej(xi+1) akj bj(i+1)

The Backward Algorithm
We can compute bk(i) for all k, i, using dynamic programming Initialization: bk(N) = ak0, for all k Iteration: bk(i) = j ej(xi+1) akj bj(i+1) Termination: P(x) = j a0j ej(x1) bj(1)

Computational Complexity
What is the running time, and space required, for Forward, and Backward? Time: O(K2N) Space: O(KN) Useful implementation technique to avoid underflows Viterbi: sum of logs Forward/Backward: rescaling at each position by multiplying by a constant

Viterbi, Forward, Backward
Initialization: V0(0) = 1 Vk(0) = 0, for all k > 0 Iteration: Vj(i) = ej(xi) maxk Vk(i-1) akj Termination: P(x, *) = maxk Vk(N) FORWARD Initialization: f0(0) = 1 fk(0) = 0, for all k > 0 Iteration: fj(i) = ej(xi) k fk(i-1) akj Termination: P(x) = k fk(N) ak0 BACKWARD Initialization: bk(N) = ak0, for all k Iteration: bj(i) = k ej(xi+1) akj bk(i+1) Termination: P(x) = k a0k ek(x1) bk(1)

Posterior Decoding We can now calculate fk(i) bk(i)
P(i = k | x) = ––––––– P(x) Then, we can ask What is the most likely state at position i of sequence x: Define ^ by Posterior Decoding: ^i = argmaxk P(i = k | x)

Posterior Decoding For each state,
Posterior Decoding gives us a curve of likelihood of state for each position That is sometimes more informative than Viterbi path * Posterior Decoding may give an invalid sequence of states Why?

Posterior Decoding =  {:[i] = k} P( | x)
x1 x2 x3 …………………………………………… xN State 1 l P(i=l|x) k P(i = k | x) =  P( | x) 1(i = k) =  {:[i] = k} P( | x)

Posterior Decoding x1 x2 x3 …………………………………………… xN State 1 l k
P(i=l|x) P(j=l’|x) k Example: How do we compute P(i = l, ji = l’ | x)? fl(i) bl(j) P(i = l, iI = l’ | x) = ––––––– P(x)

Applications of the model
Given a DNA region x, The Viterbi algorithm predicts locations of CpG islands Given a nucleotide xi, (say xi = A) The Viterbi parse tells whether xi is in a CpG island in the most likely general scenario The Forward/Backward algorithms can calculate P(xi is in CpG island) = P(i = A+ | x) Posterior Decoding can assign locally optimal predictions of CpG islands ^i = argmaxk P(i = k | x)

A model of CpG Islands – (2) Transitions
What about transitions between (+) and (-) states? They affect Avg. length of CpG island Avg. separation between two CpG islands 1-p Length distribution of region X: P[lX = 1] = 1-p P[lX = 2] = p(1-p) … P[lX= k] = pk(1-p) E[lX] = 1/(1-p) Geometric distribution, with mean 1/(1-p) X Y p q 1-q

Hidden Markov Model ..

Similar presentations

Presentation on theme: "Hidden Markov Model .."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Hidden Markov Model ..

Similar presentations

Presentation on theme: "Hidden Markov Model .."— Presentation transcript:

Similar presentations

About project

Feedback