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CS5263 Bioinformatics
Lecture 11: Markov Chain and Hidden Markov Models
In the next few lectures
• Hidden Markov Models– The theory– Probabilistic treatment of sequence
alignments using HMMs– Application of HMMs to biological sequence
modeling and discovery of features such as genes
Markov models
• A sequence of random variables is a k-th order Markov chain if, for all i, ith value is independent of all but the previous k values:
• First order:• Second order:
• 0th order: (independence)
Probability to generate a sequence with a 1st Markov model
CpG islands
• For biological reasons, CpG (C followed by G) is very rare in mammal genomes
• But relatively more common in promoter regions
• Detecting CpG islands help predict genes– Model: consider all 16 conditional probabilities
• P(G | C), P(T | C), P(A | C), P(C | C), P(G | T) …
A 1st order Markov model for CpG islands
• Essentially a finite state automaton• Transitions are probabilistic
• 4 states: A, C, G, T• 16 transition: ast = P(xi = t | xi-1 = s)• Begin/End states (for convenience)
Probability of a sequence
• What’s the probability of ACGGCTA?– For 0-th order Markov model, simply P(A)P(C)…– For 1st order Markov model like the one above
P(A) * P(C|A) * P(G|C) … P(A|T)
= aBA aAC aCG …aTA
Probability of a sequence
• What’s the probability of ACGGCTA?P(A) * P(C|A) * P(G|C) … P(A|T)
= aBA aAC aCG …aTA
• Equivalent: follow the path in the automaton, and multiply the probability of each arc on the path
• Given a sequence, there is only one path you can walk.
Training
• Estimate the parameters of the model– Training sequences: sequences with labels (CpG
islands or not CpG islands)– Test sequences: sequences for test (labels removed)
• Two models– CpG model learned from known CpG islands– Background model learned from known non-CpG
islands
• What to learn?– Transition frequencies– ast = #(s→t) / #(s → )
Training
• Parameters learned from known CpG islands and non-CpG islands
Discrimination / Classification
• Given a sequence, is it CpG island or not?• Log likelihood ratio• X = ACGGCTA
• Compute log (P(X | CpG+) / P(X | CpG-)) • or log(P(CpG+ | X) / P(CpG- | X)) given priors• Q: how to get a+
BA and a-BA? (B: begin state)
P(X|CpG+) = a+BA a+
AC a+CG …a+
TA
P(X|CpG-) = a-BA a-
AC a-CG …a-
TA
CpG island scores
Figure 3.2 (Durbin book) The histogram of length-normalized scores for all the sequences. CpG islands are shown with dark grey and non-CpG with light grey.
Questions
• Q1: given a short sequence, is it more likely from feature model or background model?
• Q2: Given a long sequence, where are the features in it (if any)?– Approach 1: score 100 bp (e.g.) windows
• Pro: simple• Con: arbitrary, fixed length, inflexible
– Approach 2: combine +/- models.
Combined model
Now given a sequence, we cannot direct tell which state a symbol is in. We have two states for each symbol. (hidden Markov model => hidden states)
?
Decoding
• Give you a sequence, ACGTACGTATA…
• What’s the most probable path?
Most probable path
A C G T A C G T A T
A+
A-
C+
C-
A+
A-
A+
A-
C+
C-
G+
G-
G+
G-
T+
T-
T+
T-
T+
T-
Find a path with the following objective:
Maximize the product of probabilities for the arcs on the path
Maximize the sum of log probabilities for the arcs on the path
B
Probability of a path is the product of all probabilities for the arcs on the path
Most probable path
V(+, i+1) = max { V(+, i) + w(x+i, x+
i+1)
V(-, i) + w(x-i, x+
i+1) }
V(-, i+1) = max { V(+, i) + w(x+i, x-
i+1)
V(-, i) + w(x-i, x-
i+1) }
A+
A-
C+
C-
A+
A-
A+
A-
C+
C-
G+
G-
G+
G-
T+
T-
T+
T-
T+
T-
B
w(s, t) = log (ast)
x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
Simpler but more general case
• We can go to any state at any time. Doesn’t depend on the input• Each state can emit any symbol with some probabilities (possibly zero)
Fair LOADED
0.05
0.05
0.950.95
P(1|F) = 1/6P(2|F) = 1/6P(3|F) = 1/6P(4|F) = 1/6P(5|F) = 1/6P(6|F) = 1/6
P(1|L) = 1/10P(2|L) = 1/10P(3|L) = 1/10P(4|L) = 1/10P(5|L) = 1/10P(6|L) = 1/2
Probability of a path is the product of all probabilities for the arcs on the path, and the possibilities of emitting the sequence of symbols
• Two types of rewards– Mileage: for using an arc (fixed)– Bonus: for visiting a node (depending on the token you show)
2 5 2 1 3 6 2 6 1 6
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
B
P(s) = 1/6, for s in [1..6]
P(s) = 1/10, for s in [1..5]P(6) = 1/2
V(F, i+1) = Max { V(F,i) + w(F,F) + r(F, xi+1), V(L,i) + w(L,F) + r(F, xi+1)}
V(L, i+1) = Max { V(F,i) + w(F,F) + r(L, xi+1), V(L,i) + w(L,F) + r(L, xi+1)}
x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
B
P(s) = 1/6, for s in [1..6]
P(s) = 1/10, for s in [1..5]P(6) = 1/2
r(F, x) = log (eF(x))w(F, L) = log (aFL)
FSA interpretation
Fair LOADED
0.05
0.05
0.950.95
P(1|F) = 1/6P(2|F) = 1/6P(3|F) = 1/6P(4|F) = 1/6P(5|F) = 1/6P(6|F) = 1/6
P(1|L) = 1/10P(2|L) = 1/10P(3|L) = 1/10P(4|L) = 1/10P(5|L) = 1/10P(6|L) = 1/2
Fair LOADED
-0.7
-0.7
2.32.3
r(F,1) = 0.5r(F,2) = 0.5r(F,3) = 0.5r(F,4) = 0.5r(F,5) = 0.5r(F,6) = 0.5
r(L,1) = 0r(L,2) = 0r(L,3) = 0r(L,4) = 0r(L,5) = 0r(L,6) = 1.6
r = log (10 * p)
V(L, i+1) = max { V(L, i) + W(L, L) + r(L, xi+1), V(F, i) + W(F, L) + r(L, xi+1) }
V(F, i+1) = max { V(L, i) + W(L, F) + r(F, xi+1), V(F, i) + W(F, F) + r(F, xi+1) }
More general cases
x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2B
3
k
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3
k
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3
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3
k
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3
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3
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3
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3
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3
k
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V(1,i) + w(1,1) + r(1, xi+1),
V(2,i) + w(2,1) + r(1, xi+1),
V(1, i+1) = max V(3,i) + w(3,1) + r(1, xi+1),……
V(k,i) + w(k,1) + r(1, xi+1)
V(1,i) + w(1,2) + r(2, xi+1),
V(2,i) + w(2,2) + r(2, xi+1),
V(2, i+1) = max V(3,i) + w(3,2) + r(2, xi+1), ……
V(k,i) + w(k,2) + r(2, xi+1) ......A FSA with k states, fully connected
V(1,i) + w(1, j) + r(j, xi+1),
V(2,i) + w(2, j) + r(j, xi+1),
V(j, i+1) = max V(3,i) + w(3, j) + r(j, xi+1),
……
V(k,i) + w(k, j) + r(j, xi+1)
Or simply:
V(j, i+1) = Maxl {V(l,i) + w(l, j) + r(j, xi+1)}
Generalize
• Previous equation assumes completely connected structure. In practice may not be the case.
Load_1 Fair Load_2
P(1|F) = 1/6P(2|F) = 1/6P(3|F) = 1/6P(4|F) = 1/6P(5|F) = 1/6P(6|F) = 1/6
P(1|L1) = 1/10P(2|L1) = 1/10P(3|L1) = 1/10P(4|L1) = 1/10P(5|L1) = 1/10P(6|L1) = 1/2
P(1|L2) = 1/2P(2|L2) = 1/10P(3|L2) = 1/10P(4|L2) = 1/10P(5|L2) = 1/10P(6|L2) = 1/10
0.05 0.05
0.1
0.90.9
0.9
0.1
0
0
Load_1 Fair Load_2
0.05 0.05
0.1
0.90.9
0.9
0.1
Take advantage of sparse structure
V(L1, i) + w(L1, F) + r(F, xi+1),
V(F, i+1) = max V(L2, i) + w(L2, F) + r(F, xi+1),
V(F, i) + w(F, F) + r(F, xi+1)
V(L1, i) + w(L1, L1) + r(L1, xi+1),
V(L1, i+1) = max V(F, i) + w(F, L1) + r(L1, xi+1)
V(L2, i) + w(L2, L2) + r(L2, xi+1),
V(L2, i+1) = max V(F, i) + w(F, L2) + r(L2, xi+1)
The Viterbi AlgorithmInput: x = x1……xN
Initialization:P0(0) = 1 (zero in subscript is the start state.)Pk(0) = 0, for all k > 0 (0 in parenthesis is the imaginary first position)
Iteration:for each j = 1..N for each i = 1..k
Pj(i) = ej(xi) maxk akj Pk(i-1)Ptrj(i) = argmaxk akj Pk(i-1)
endend
Termination:Prob(x, *) = maxk Pk(N)
Traceback: N* = argmaxk Pk(N) i-1* = Ptri (i)
The Viterbi AlgorithmInput: x = x1……xN
Initialization:V0(0) = 0 (zero in subscript is the start state.)Vk(0) = -inf, for all k > 0(0 in parenthesis is the imaginary first position)
Iteration:for each j for each i
Vj(i) = rj(xi) + maxk (wkj + Vk(i-1)) rj(xi) = log(ej(xi)), wkj = log(akj)Ptrj(i) = argmaxk (wkj + Vk(i-1))
endend
Termination:Prob(x, *) = exp{maxk Vk(N)}
Traceback: N* = argmaxk Vk(N) i-1* = Ptri (i)
The Viterbi Algorithm
Similar to “aligning” a set of states to a sequence
Time:
O(K2N)
Space:
O(KN)
x1 x2 x3 ………………………………………..xN
State 1
2
K
Vj(i)
• So far so good
• but
Problem
• Is the most probable path necessarily the best one?– Single optimal vs multiple sub-optimal
• What if there are many sub-optimal paths with slightly lower probabilities?
– Global optimal vs local optimal• What’s best globally may not be the best for each
individual
For example
• Probability for path 1-2-5-7 is 0.4• Probability for path 1-3-6-7 is 0.3• Probability for path 1-4-6-7 is 0.3• What’s the most probable state at step 2?
– 0.4 goes through 5– 0.6 goes through 6– Viterbi may not be the only interesting answer
Another example
• The dishonest casino• Say x = 12341623162616364616234161221341• Most probable path: = FF……F• However: marked letters more likely to be L than
unmarked letters• Another way to interpret the problem
– With Viterbi, every pos is assigned a label– Confidence level for each assignment?
Posterior decoding
• Viterbi finds the path with the highest probability
– The probability is usually very tiny anyway
• We want to know k = 1
Probability of a sequence
• The probability that a sequence is generated by a model. P(X | M)
• Sometimes simply written as P(X)• Sometimes written as P(X | M, θ) or P(X | θ) to
emphasize that we are looking for θ to optimize the likelihood
• Not equal to the probability of a path P(X, )– There are many possible paths that can generate X– Each with a different probability
– P(X) = P(X, ) = P(X | ) P()– Why do we need this?– How to compute without summing over all paths (exponential)?
The forward algorithm
• Define fk(i) the probability to get to state k at step i– Sum over all possible previous paths– Remember the way we counted # alignments?
The forward algorithm
The forward algorithm
We can compute fk(i) for all k, i, using dynamic programming!
Initialization:f0(0) = 1fk(0) = 0, for all k > 0
Iteration:
fk(i) = ek(xi) j fj(i-1) ajk
Termination:
Prob(x) = k fk(N)
Relation between Forward and Viterbi
VITERBI
Initialization:
P0(0) = 1
Pk(0) = 0, for all k > 0
Iteration:
Pk(i) = ek(xi) maxj Pj(i-1) ajk
Termination:
Prob(x, *) = maxk Pk(N)
FORWARD
Initialization:
f0(0) = 1
fk(0) = 0, for all k > 0
Iteration:
fk(i) = ek(xi) j fj(i-1) ajk
Termination:
Prob(x) = k fk(N)
The backward algorithm
• Define bk(i) as the probability for paths starting from position i within state k, to the end of the sequence– Sum over all possible paths from i to N
1
This does not include the emission probability of xi
• fk(i): prob to get to pos i in state k and emit x i
• bk(i): prob from i to end, given i is in state k
• What is fk(i) bk(i)?
Is your guess correct?
• What we need is
• But:
• We have P(x) already in the forward algorithm
• Can verify: k = 1
The forward-backward algorithm
• Compute fk(i), for each state k and pos i
• Compute bk(i), for each state k and pos I
• Compute P(x) = kfk(N)
• Compute P(i=k | x) = fk(i) * bk(i) / P(x)
The prob of x, with the constraint that it has to pass state k on step i
stateSequence
+
P(i=k | x)Space: O(KN)
Time: O(K2N)
/ P(X)Forward probabilities Backward probabilities
Relation to another F-B algorithm
• We’ve learned a forward-backward algorithm in linear-space sequence alignment– Hirscheberg’s algorithm– Also known as forward-backward alignment algorithm
xy
What’s P(i=k | x) good for?
• For each position, you can assign a probability (in [0, 1]) to the states that the system might be in at that point – confidence level
• Assign each symbol to the most-likely state according to this probability rather than the state on the most-probable path – posterior decoding
^i = argmaxk P(i = k | x)
For example
If P(fair) > 0.5, the roll is more likely to be generated by a fair die than a loaded die
CpG islands again
• Data: 41 human sequences, totaling 60kbp, including 48 CpG islands of about 1kbp each
• Viterbi: Post-process:– Found 46 of 48 46/48– plus 121 “false positives” 67 false pos
• Posterior Decoding:– same 2 false negatives 46/48– plus 236 false positives 83 false pos
Post-process: merge within 500; discard < 500
What if a new genome comes?We just sequenced the porcupine genome
We know CpG islands play the same role in this genome
However, we have no known CpG islands for porcupines
We suspect the frequency and characteristics of CpG islands are quite different in porcupines
How do we adjust the parameters in our model?
- LEARNING
Learning
• When the states are known– We’ve already done that– Estimate parameters from labeled data
(known CpG or non-CpG)– “supervised” learning
• When the states are unknown– Estimate parameters without labeled data– “unsupervised” learning
Basic idea
1. We estimate our “best guess” on the model parameters θ
2. We use θ to predict the unknown labels
3. We re-estimate a new set of θ
4. Repeat 2 & 3
Two ways
Viterbi Traininggiven θ estimate π; then re-estimate θ
1. Make initial estimates of parameters θ2. Find Viterbi path π for each training sequence3. Count transitions/emissions on those paths,
getting new θ4. Repeat 2&3
• Not rigorously optimizing desired likelihood, but still useful & commonly used.
• (Arguably good if you’re doing Viterbi decoding.)
Baum-Welch Traininggiven θ, estimate π ensemble; then re-estimate θ
• Instead of estimating the new θ from the most probable path
• We can re-estimate θ from all possible paths• For example, according to Viterbi, pos i is in
state k and pos (i+1) is in state l– This contributes 1 count towards the frequency that
transition kl is used– In Baum-Welch, this transition is counted only
partially, according the probability that this arc is taken by some path
Estimated # of kl transition
Why is this working?
• Proof is complicated (ch 11 in Durbin book)• But basically,
– The backward-forward algorithm computes the likelihood of the data given the HMM
– When we re-estimate θ, we maximize P(sequence | θ).
– Effect: in each iteration, the likelihood of the sequence will be improved
– Therefore, guaranteed to converge (not necessarily to a global optimal)
expectation-maximization (EM)
• Baum-Welch algorithm is a special case of the expectation-maximization (EM) algorithm, a widely used technique in statistics for learning parameters from unlabeled data
• E-step: compute the expectation (e.g. prob for each pos to be in a certain state)
• M-step: maximum-likelihood parameter estimation
• We’ll see EM and similar techniques again in motif finding
• k-means clustering has some similar flavor
HMM summary
• Viterbi – best single path
• Forward – sum over all paths
• Backward – similar
• Baum-Welch – training via EM and forward-backward
• Viterbi – another “EM”, but Viterbi-based
