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Hidden Markov Models Ellen Walker Bioinformatics Hiram College, 2008

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State Machine to Recognize “AUG” transition Start state Final state Each character causes a transition to the next state

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“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

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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

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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

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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

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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.

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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?

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“Fair Bet Casino” HMM Image from Jones & Pevner 2004

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

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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

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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

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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

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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

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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?

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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

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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

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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

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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

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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

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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

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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

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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

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