Hidden Markov Models

Ellen WalkerBioinformatics

Hiram College, 2008

State Machine to Recognize “AUG”

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Start state

Final state

Each character causes a transition to the next state

“AUG” anywhere in a string

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“AUG” in frame

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Deterministic Finite Automaton (DFA)

• States– One start state– One or more accept states

• Transitions– For every state, for every character

• Outputs– Optional: states can emit outputs, e.g.

“Stop” at accept state

Why DFAs?

• Every regular expression has an associated state machine that recognizes it (and vice versa)

• State machines are easy to implement in very low level code (or hardware)

• Sometimes the state machine is easier to describe than the regular expression

Hidden Markov Models

• Also a form of state machine• Transitions based on probabilities, not

inputs• Every state has (probabilistic) output (or

emission)• “Hidden” because only emissions are

visible, not states or transitions

HMM vs. DFA

• DFA is deterministic– Each decision (which state next? What to

output?) is fully determined by the input string

• HMM is probabilistic– HMM makes both decisions based on

probability distributions

HMM vs. DFA (2)

• DFA model is explicit and used directly like a program.

• HMM model must be inferred from data. Only emissions (outputs) can be observed. States and transitions, as well as the probability distributions for transitions and outputs are hidden.

HMM Example: Fair Bet Casino

• The casino has two coins, a Fair coin (F) and a Biased coin (B)– Fair coin has 50% H, 50% T– Biased coin has 75% H, 25% T

• Before each flip, with probability 10%, the dealer will switch coins.

• Can you tell, based only on a sequence of H and T which coin is used when?

“Fair Bet Casino” HMM

Image from Jones & Pevner 2004

The Decoding Problem

• Given an HMM and a sequence of outputs, what is the most likely path through the HMM that generated the outputs?

Viterbi Algorithm

• Uses dynamic programming• Starting point:

– When the output string is “”, the most likely state is the start state (and there is no path)

• Taking a step:– Likelihood of this state is maximum of all ways to

get here, measured as:• Likelihood of previous state *

Likelihood of transition to this state * Likelihood of output from this state

Example: “HHT”

• Initial -> F – Prev= 1, Trans = 0.5, Out=0.5, total = 0.25

• Initial -> B– Prev =1, Trans = 0.5, Out=0.75, total =

0.375• Result: F = 0.25, B=0.375

Example: “HHT”• F -> F

– Prev=0.25, Trans = 0.9, Out=0.5, total = 0.1125• B -> F

– Prev=0.375, Trans = 0.1, Out=0.5, total = 0.01875• F -> B

– Prev =.25, Trans = 0.1, Out=0.75, total = 0.01875• B -> B

– Prev =.375, Trans = 0.9, Out=0.75, total = 0.253125

• Result: F = 0.1125, B=0.253125

Example: HHT

• F -> F – Prev=.1125, Trans = 0.9, Out=0.5, total = 0.0506

• B -> F – Prev=.253125, Trans = 0.1, Out=0.5, total = 0.0127

• F -> B– Prev =.1125, Trans = 0.1, Out=0.25, total = 0.00281

• B -> B– Prev=.253125, Trans = 0.9, Out=0.25, total = 0.0570

• Result: F = 0.0506, B=0.0570

Tracing Back

• Pick the highest result from the last step, follow the highest transition from each previous step (just like Smith-Waterman)

• Result: initial->B->B->B• Biased coin always used• What if the next flip is T?

Log Probabilities

• Probabilities are increasingly small, as you multiply numbers less than one

• Computers have limits to precision• Therefore, it’s better to use a log

probability format• 1/10*1/10 = 1/100 (10-1 *10-1 = 10-2)• -1 + -1 = -2

GC Rich Islands

• A GC Rich Island is an area of a genome where GC content is significantly greater than the genome as a whole

• GC Rich Islands are like Biased Coins• Can recognize them using the same HMM

– GC content is p(H) for fair coin– Larger number is p(H) for biased coin– Estimate probability of entering vs. leaving GC

Rich island for “changing coin” probability

Probability of State Sequence, Given Output Sequence

• Given HMM and output string, what is probability that HMM is in state S at time t?– Forward: similar formulation as decoding

problem, except take sum of all paths, instead of max of all paths (times from 0 to t-1)

– Backward: similar, but work from end of string (times from t+1 to end of sequence

Parameter Estimation

• Given many strings, what are the parameters of the HMM that generated them?– Assume we know the states and

transitions, but not the probabilities of transitions or outputs

– This is an optimization problem

Characteristics of an Optimization Problem

• Each potential solution has a “goodness” value (in this case, probability)

• We want the best solution• Perfect answer: try all possibilities (not

usually possible)• Good, but not perfect answer: use a

heuristic

Hill Climbing (an Optimization Heuristic)

• Start with a solution (could be random)• Consider one or more “steps”, or

perturbations to the solution• Choose the “step” that most improves

the score• Repeat until the score is good enough,

or no better score can be reached

Hill Climbing for HMM

• Guess a state sequence• Using the string(s), estimate transition

and emission probabilities• Using the probabilities, generate a new

state sequence using the decoding algorithm

• Repeat until the sequence stabilizes

HMM for Sequence Profiles

• Three kinds of states:– Insertion– Deletion– Match

• Probability estimations indicate how often each occurs

• Logos are direct representations of HMMs in this format