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Unformatted text preview: Hidden Markov Models
CMSC 423 Based on Chapter 11 of Jones & Pevzner, An Introduction to
Bioinformatics Algorithms Checking a Casino Fair coin:
Pr(Heads) = 0.5 Heads/Tails: ? Biased coin:
Pr(Heads) = 0.75
Suppose either a fair or biased
coin was used to generate a
sequence of heads & tails. But
we don’t know which type of
coin was actual used. ↑↑↓↓↓↓↑↑↑↑↓↑↓↑↓↑↓↑ Checking a Casino Fair coin:
Pr(Heads) = 0.5 Heads/Tails: ? Biased coin:
Pr(Heads) = 0.75
Suppose either a fair or biased
coin was used to generate a
sequence of heads & tails. But
we don’t know which type of
coin was actual used. ↑↑↓↓↓↓↑↑↑↑↓↑↓↑↓↑↓↑ Checking a Casino Fair coin:
Pr(Heads) = 0.5 Heads/Tails: ? Biased coin:
Pr(Heads) = 0.75
Suppose either a fair or biased
coin was used to generate a
sequence of heads & tails. But
we don’t know which type of
coin was actual used. ↑↑↓↓↓↓↑↑↑↑↓↑↓↑↓↑↓↑
How could we guess which coin was more likely? Compute the Probability of the
Observed Sequence
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
X= Pr(x  Fair) = ↑ ↑ ↓ ↓ ↓ ↓ ↑ 0.5 0.5 0.5 0.5 0.5 0.5 0.5 Pr(x  Biased) = 0.75 0.75 0.25 0.25 0.25 0.25 0.75 Compute the Probability of the
Observed Sequence
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
X= Pr(x  Fair) = ↑ ↑ ↓ ↓ ↓ ↓ ↑ 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5 Pr(x  Biased) = 0.75 ×0.75× 0.25× 0.25 × 0.25× 0.25 × 0.75 = 0.57 = 0.0078125 = 0.001647949 Compute the Probability of the
Observed Sequence
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
X= Pr(x  Fair) = ↑ ↑ ↓ ↓ ↓ ↓ ↑ 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5 Pr(x  Biased) = 0.75 ×0.75× 0.25× 0.25 × 0.25× 0.25 × 0.75 = 0.57 = 0.0078125 = 0.001647949 The logodds score: 0.0078
Pr(x  Fair)
log2
= log2
= 2.245
Pr(x  Biased)
0.0016 > 0. Hence “Fair” is a
better guess. What if the casino switches coins?
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
Probability of switching coins = 0.1
0.1 Fair coin:
Pr(Heads) = 0.5 0.1 Biased coin:
Pr(Heads) = 0.75 What if the casino switches coins?
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
Probability of switching coins = 0.1
0.1 Fair coin:
Pr(Heads) = 0.5 0.1 Biased coin:
Pr(Heads) = 0.75 How could we guess which coin was more likely at each position? What if the casino switches coins?
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
Probability of switching coins = 0.1
0.1 Fair coin:
Pr(Heads) = 0.5 0.1 Biased coin:
Pr(Heads) = 0.75 How can we compute the probability of the entire sequence?
How could we guess which coin was more likely at each position? What does this have to do with biology? atg gat ggg agc aga tca gat cag atc agg gac gat aga cga tag tga What does this have to do with biology? Now:
How likely is it that this is a gene?
Which parts are the start, middle and end? atg gat ggg agc aga tca gat cag atc agg gac gat aga cga tag tga What does this have to do with biology? Now:
How likely is it that this is a gene?
Which parts are the start, middle and end? atg gat ggg agc aga tca gat cag atc agg gac gat aga cga tag tga Start
Generator Middle of
Gene Generator End
Generator What does this have to do with biology?
Before:
How likely is it that this sequence was generated by a fair coin?
Which parts were generated by a biased coin?
Now:
How likely is it that this is a gene?
Which parts are the start, middle and end? atg gat ggg agc aga tca gat cag atc agg gac gat aga cga tag tga Start
Generator Middle of
Gene Generator End
Generator Hidden Markov Model (HMM)
Fair coin: Pr(Heads) = 0.5
Biased coin: Pr(Heads) = 0.75
Probability of switching coins = 0.1 0.9 0.9
0.1
Fair Biased
0.1 Fair 0.5 0.5
H T Biased 0.75 0.25
H T Formal Deﬁnition of a HMM 7 7
Probability of
going from state
5 to state 7 6
5 A= 6
5 E= 4 Probability of
emitting T when
in state 4. 4 3 3 2 2 1 1
1 2 3 4 5 6 7 A C G T Formal Deﬁnition of a HMM
∑ = alphabet of symbols.
Q = set of states.
A = an Q x Q matrix where entry (k,l) is the probability of moving from
state k to state l.
E = a Q x ∑ matrix, where entry (k,b) is the probability of emitting b when
in state k.
7 7
Probability of
going from state
5 to state 7 6
5 A= 6
5 E= 4 Probability of
emitting T when
in state 4. 4 3 3 2 2 1 1
1 2 3 4 5 6 7 A C G T Constraints on A and E 7 7
Probability of
going from state
5 to state 7 6 A= 5 6 E= 5 4 4 3 3 2 2 1 Probability of
emitting T when
in state 4. 1
1 2 3 4 5 6 7 A C G T Sum of the # in each row must be 1. Computing Probabilities Given Path
0.9 0.9
0.1
Fair Biased
0.1 Fair 0.5 0.5
H T Biased 0.75 0.25
H T x= ↓↑↓↑↑↑↓↑↑↓ π= FFFBBB Pr(xi  πi) = B F F F 0.5 0.5 0.5 0.75 0.75 0.75 0.25 0.5 0.5 0.5 Pr(πi → πi+1) =0.1 0.9 0.9 0.1 0.9 0.9 0.9 0.1 0.1 0.1 The Decoding Problem
Given x and π, we can compute: • Pr(x  π): product of Pr(xi  πi)
• Pr(π): product of Pr(πi → πi+1)
• Pr(x, π): product of all the Pr(xi  πi) and Pr(πi → πi+1)
n
Pr(x, π ) = Pr(π0 → π1 )
Pr(xi  πi ) Pr(πi → πi+1 )
i=1 But they are “hidden” Markov models because π is unknown.
Decoding Problem: Given a sequence x1x2x3...xn generated by an
HMM (∑, Q, A, E), ﬁnd a path π that maximizes Pr(x, π). The Viterbi Algorithm to Find Best Path
A[a, k] := the probability of the best path for x1...xk that ends at state a.
Q 2
1
1 2 3 4 5 6 7 8 9 10 ↓↑ ↓↑↑↑↓ ↑↑ ↓ A[a, k] = the path for x1...xk1 that goes to some state b times cost of a
transition from b to i, and then to output xk from state a.
b Biased Q a Fair
1 2 3 4 5 6 7 8 9 10 ↓↑ ↓↑↑↑↓ ↑↑ ↓ =k Viterbi DP Recurrence b Biased Q a Fair
1 2 3 4 5 6 7 8 9 10 = k ↓↑ ↓↑↑↑↓ ↑↑ ↓ A[a, k ] = max {A[b, k − 1] × Pr(b → a) × Pr(xk  πk = a)}
b∈Q Over all possible
previous states. Best path for
x1..xk ending
in state b Probability of
transitioning
from state b to
state a Base case: A[a, 1] = Pr(π1 = a) × Pr(x1  π1 = a)
Probability
that the ﬁrst
state is a Probability of
emitting x1
given the ﬁrst
state is a. Probability of
outputting xk
given that the
kth state is a. Which Cells Do We Depend On?
7
6
5
4
3 Q 2
1
1 x= 2 3 4 5 6 7 8 9 10 ↓↑ ↓↑↑↑↓ ↑↑ ↓ Order to Fill in the Matrix:
7
6
5
4
3 Q 2
1
1 x= 2 3 4 5 6 7 8 9 10 ↓↑ ↓↑↑↑↓ ↑↑ ↓ Where’s the answer?
7
6 max value in
these red cells 5
4
3 Q 2
1
1 x= 2 3 4 5 6 7 8 9 10 ↓↑ ↓↑↑↑↓ ↑↑ ↓ Graph View of Viterbi Q x1 x2 x3 x4 x5 x6 Running Time
• # of subproblems = O(nQ), where n is the length of the sequence. • Time to solve a subproblem = O(Q) • Total running time: O(nQ2) Using Logs
Typically, we take the log of the probabilities to avoid multiplying a lot
of terms:
log(A[a, k ]) = max {log(A[b, k − 1] × Pr(b → a) × Pr(xk  πk = a))} = max {log(A[b, k − 1]) + log(Pr(b → a)) + log(Pr(xk  πk = a))} b∈Q
b∈Q Remember: log(ab) = log(a) + log(b)
Why do we want to avoid multiplying lots of terms? Using Logs
Typically, we take the log of the probabilities to avoid multiplying a lot
of terms:
log(A[a, k ]) = max {log(A[b, k − 1] × Pr(b → a) × Pr(xk  πk = a))} = max {log(A[b, k − 1]) + log(Pr(b → a)) + log(Pr(xk  πk = a))} b∈Q
b∈Q Remember: log(ab) = log(a) + log(b)
Why do we want to avoid multiplying lots of terms?
Multiplying leads to very small numbers:
0.1 x 0.1 x 0.1 x 0.1 x 0.1 = 0.00001
This can lead to underﬂow.
Taking logs and adding keeps numbers bigger. Some probabilities we are interested in
What is the probability of observing a string x under the assumed HMM? Pr(x) = Pr(x, π ) π What is the probability of observing x using a path where the ith state is a? Pr(x, πi = a) = Pr(x, π ) π :πi = a What is the probability that the ith state is a? Pr(x, πi = a)
Pr(πi = ax) =
Pr(x) The Forward Algorithm
How do we compute this: Pr(x, πk = a) = Pr(x1 , . . . , xi , πi = a) Pr(xi+1 , . . . , xn  πi = a)
Recall the recurrence to compute best path for x1...xk that ends at state a: A[a, k ] = max {A[b, k − 1] × Pr(b → a) × Pr(xk  πk = a)}
b∈Q We can compute the probability of emitting x1,...,xk using some path that
ends in a: F [a, k ] = b∈Q F [b, k − 1] × Pr(b → a) × Pr(xk  πk = a) The Forward Algorithm
How do we compute this: Pr(x, πk = a) = Pr(x1 , . . . , xi , πi = a) Pr(xi+1 , . . . , xn  πi = a)
Recall the recurrence to compute best path for x1...xk that ends at state a: A[a, k ] = max {A[b, k − 1] × Pr(b → a) × Pr(xk  πk = a)}
b∈Q We can compute the probability of emitting x1,...,xk using some path that
ends in a: F [a, k ] = b∈Q F [b, k − 1] × Pr(b → a) × Pr(xk  πk = a) The Forward Algorithm
Computes the total probability
of all the paths of length k
ending in state a. Q
a x1 x2 x3 x4 F[a,4] x5 x6 The Forward Algorithm
Computes the total probability
of all the paths of length k
ending in state a. Still need to compute the
probability of paths leaving a
and going to the end. Q
a x1 x2 x3 x4 F[a,4] x5 x6 The Backward Algorithm
The same idea as the forward algorithm, we just start from the end of the
input string and work towards the beginning:
B[a,k] = “the probability of generating string xk+1,...,xn starting from state b” B [a, k ] = b∈Q B [b, k + 1] × Pr(a → b) × Pr(xk+1  πk+1 = b)
Prob for
xk+1..xn
starting in
state b Probability
going from
state a to b Probability of emitting
xk+1 given that the next
state is b. The ForwardBackward Algorithm
Pr(x, πi = k )
F [a, i] · B [a, i]
Pr(πi = a  x) =
=
Pr(x)
Pr(x) F[a,i] B[a,i]
a Estimating HMM Parameters
(x(1),π(1)) = (x(2),π(2)) = (1) (1) (1) (1) (1)
(1)
x1 x2 x3 x4 x5 . . . xn
(1) (1) (1) (1) (1)
(1)
π1 π2 π3 π4 π5 . . . πn
(2) (2) (2) (2) (2)
(2)
x1 x2 x3 x4 x5 . . . xn
(2) (2) (2) (2) (2)
(2)
π1 π2 π3 π4 π5 . . . πn # of times transition
a → b is observed. Aab
Pr(a → b) =
q ∈Q Aaq Training examples
where outputs and
paths are known. # of times x was
observed to be
output from state a. Exa
Pr(x  a) =
x∈Σ Exq Pseudocounts
# of times transition
a → b is observed. Aab
Pr(a → b) =
q ∈Q Aaq # of times x was
observed to be
output from state a. Exa
Pr(x  a) =
x∈Σ Exq What if a transition or emission is never observed in the training data?
0 probability Meaning that if we observe an example with that transition or emission in the real
world, we will give it 0 probability.
But it’s unlikely that our training set will be large enough to observe every possible
transition.
Hence: we take Aab = (#times a → b was observed) + 1
Similarly for Exa. “pseudocount” Viterbi Training
• Problem: typically, in the real would we only have examples of the
output x, and we don’t know the paths π. Viterbi Training Algorithm:
1. Choose a random set of parameters.
2. Repeat:
1. Find the best paths.
2. Use those paths to estimate new parameters.
This is an local search algorithm.
It’s also an example of a “Gibbs sampling” style algorithm.
The BaumWelch algorithm is similar, but doesn’t commit to a single
best path for each example. Recap
• Hidden Markov Model (HMM) model the generation of strings. • They are governed by a string alphabet (∑), a set of states (Q), a set of
transition probabilities A, and a set of emission probabilities for each
state (E). • Given a string and an HMM, we can compute: •
• The most probable path the HMM took to generate the string (Viterbi).
The probability that the HMM was in a particular state at a given step (forwardbackward algorithm). • Algorithms are based on dynamic programming. • Finding good parameters is a much harder problem.
The BaumWelch algorithm is an oftused heuristic algorithm. ...
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This note was uploaded on 01/13/2012 for the course CMSC 423 taught by Professor Staff during the Fall '07 term at Maryland.
 Fall '07
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