Hidden Markov Models

Decoding GIVEN x = x 1 x 2 ……x N We want to find  =  1, ……,  N, such that P[ x,  ] is maximized  * = argmax  P[ x,  ] We can use dynamic programming! Let V k (i) = max {  1,…,i-1} P[x 1 …x i-1,  1, …,  i-1, x i,  i = k] = Probability of most likely sequence of states ending at state  i = k 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xKxK 2 1 K 2

The Viterbi Algorithm Similar to “aligning” a set of states to a sequence Time: O(K 2 N) Space: O(KN) x 1 x 2 x 3 ………………………………………..x N State 1 2 K V j (i)

Evaluation We demonstrated algorithms that allow us to compute: P(x)Probability of x given the model P(x i …x j )Probability of a substring of x given the model P(  I = k | x)Probability that the i th state is k, given x A more refined measure of which states x may be in

Viterbi, Forward, Backward VITERBI Initialization: V 0 (0) = 1 V k (0) = 0, for all k > 0 Iteration: V j (i) = e j (x i ) max k V k (i-1) a kj Termination: P(x,  *) = max k V k (N) FORWARD Initialization: f 0 (0) = 1 f k (0) = 0, for all k > 0 Iteration: f l (i) = e l (x i )  k f k (i-1) a kl Termination: P(x) =  k f k (N) a k0 BACKWARD Initialization: b k (N) = a k0, for all k Iteration: b k (i) =  l e l (x i +1) a kl b l (i+1) Termination: P(x) =  l a 0l e l (x 1 ) b l (1)

Posterior Decoding We can now calculate f k (i) b k (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 = argmax k P(  i = k | x)

Implementation Techniques Viterbi: Sum-of-logs V l (i) = log e k (x i ) + max k [ V k (i-1) + log a kl ] Forward/Backward: Scaling by c(i) One way to perform scaling: f l (i) = c(i)  [e l (x i )  k f k (i-1) a kl ]where c(i) = 1/(  k f k (i)) b l (i): use the same factors c(i) Details in Rabiner’s Tutorial on HMMs, 1989

A+C+G+T+ A-C-G-T- A modeling Example CpG islands in DNA sequences

Example: CpG Islands CpG nucleotides in the genome are frequently methylated (Write CpG not to confuse with CG base pair) C  methyl-C  T Methylation often suppressed around genes, promoters  CpG islands

Example: CpG Islands In CpG islands,  CG is more frequent  Other pairs (AA, AG, AT…) have different frequencies Question: Detect CpG islands computationally

A model of CpG Islands – (1) Architecture A+C+G+T+ A-C-G-T- CpG Island Not CpG Island

A model of CpG Islands – (2) Transitions How do we estimate the parameters of the model? Emission probabilities: 1/0 1.Transition probabilities within CpG islands Established from many known (experimentally verified) CpG islands (Training Set) 2.Transition probabilities within other regions Established from many known non-CpG islands + ACGT A C G T ACGT A C G T

Parenthesis – log likelihoods ACGT A C G T Another way to see effects of transitions: Log likelihoods L(u, v) = log[ P(uv | + ) / P(uv | -) ] Given a region x = x 1 …x N A quick-&-dirty way to decide whether entire x is CpG P(x is CpG) > P(x is not CpG)   i L(x i, x i+1 ) > 0

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 XY 1-p 1-q pq Length distribution of region X: P[l X = 1] = 1-p P[l X = 2] = p(1-p) … P[l X = k] = p k (1-p) E[l X ] = 1/(1-p) Geometric distribution, with mean 1/(1-p)

A model of CpG Islands – (2) Transitions No reason to favor exiting/entering (+) and (-) regions at a particular nucleotide To determine transition probabilities between (+) and (-) states 1.Estimate average length of a CpG island: l CPG = 1/(1-p)  p = 1 – 1/l CPG 2.For each pair of (+) states k, l, let a kl  p × a kl 3.For each (+) state k, (-) state l, let a kl = (1-p)/4 (better: take frequency of l in the (-) regions into account) 4.Do the same for (-) states A problem with this model: CpG islands don’t have exponential length distribution This is a defect of HMMs – compensated with ease of analysis & computation

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

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

Problem 3: Learning Re-estimate the parameters of the model based on training data

Two learning scenarios 1.Estimation when the “right answer” is known Examples: GIVEN:a genomic region x = x 1 …x 1,000,000 where we have good (experimental) annotations of the CpG islands GIVEN:the casino player allows us to observe him one evening, as he changes dice and produces 10,000 rolls 2.Estimation when the “right answer” is unknown Examples: GIVEN:the porcupine genome; we don’t know how frequent are the CpG islands there, neither do we know their composition GIVEN: 10,000 rolls of the casino player, but we don’t see when he changes dice QUESTION:Update the parameters  of the model to maximize P(x|  )

1.When the right answer is known Given x = x 1 …x N for which the true  =  1 …  N is known, Define: A kl = # times k  l transition occurs in  E k (b) = # times state k in  emits b in x We can show that the maximum likelihood parameters  are: A kl E k (b) a kl = ––––– e k (b) = –––––––  i A ki  c E k (c)

1.When the right answer is known Intuition: When we know the underlying states, Best estimate is the average frequency of transitions & emissions that occur in the training data Drawback: Given little data, there may be overfitting: P(x|  ) is maximized, but  is unreasonable 0 probabilities – VERY BAD Example: Given 10 casino rolls, we observe x = 2, 1, 5, 6, 1, 2, 3, 6, 2, 3  = F, F, F, F, F, F, F, F, F, F Then: a FF = 1; a FL = 0 e F (1) = e F (3) =.2; e F (2) =.3; e F (4) = 0; e F (5) = e F (6) =.1

Pseudocounts Solution for small training sets: Add pseudocounts A kl = # times k  l transition occurs in  + r kl E k (b) = # times state k in  emits b in x+ r k (b) r kl, r k (b) are pseudocounts representing our prior belief Larger pseudocounts  Strong priof belief Small pseudocounts (  < 1): just to avoid 0 probabilities

Pseudocounts Example: dishonest casino We will observe player for one day, 500 rolls Reasonable pseudocounts: r 0F = r 0L = r F0 = r L0 = 1; r FL = r LF = r FF = r LL = 1; r F (1) = r F (2) = … = r F (6) = 20(strong belief fair is fair) r F (1) = r F (2) = … = r F (6) = 5(wait and see for loaded) Above #s pretty arbitrary – assigning priors is an art

2.When the right answer is unknown We don’t know the true A kl, E k (b) Idea: We estimate our “best guess” on what A kl, E k (b) are We update the parameters of the model, based on our guess We repeat

2.When the right answer is unknown Starting with our best guess of a model M, parameters  : Given x = x 1 …x N for which the true  =  1 …  N is unknown, We can get to a provably more likely parameter set  Principle: EXPECTATION MAXIMIZATION 1.Estimate A kl, E k (b) in the training data 2.Update  according to A kl, E k (b) 3.Repeat 1 & 2, until convergence

Estimating new parameters To estimate A kl : At each position i of sequence x, Find probability transition k  l is used: P(  i = k,  i+1 = l | x) = [1/P(x)]  P(  i = k,  i+1 = l, x 1 …x N ) = Q/P(x) where Q = P(x 1 …x i,  i = k,  i+1 = l, x i+1 …x N ) = = P(  i+1 = l, x i+1 …x N |  i = k) P(x 1 …x i,  i = k) = = P(  i+1 = l, x i+1 x i+2 …x N |  i = k) f k (i) = = P(x i+2 …x N |  i+1 = l) P(x i+1 |  i+1 = l) P(  i+1 = l |  i = k) f k (i) = = b l (i+1) e l (x i+1 ) a kl f k (i) f k (i) a kl e l (x i+1 ) b l (i+1) So: P(  i = k,  i+1 = l | x,  ) = –––––––––––––––––– P(x |  )

Estimating new parameters So, f k (i) a kl e l (x i+1 ) b l (i+1) A kl =  i P(  i = k,  i+1 = l | x,  ) =  i ––––––––––––––––– P(x |  ) Similarly, E k (b) = [1/P(x)]  {i | xi = b} f k (i) b k (i)

Estimating new parameters If we have several training sequences, x 1, …, x M, each of length N, f k (i) a kl e l (x i+1 ) b l (i+1) A kl =  x  i P(  i = k,  i+1 = l | x,  ) =  x  i –––––––––––––––– P(x |  ) Similarly, E k (b) =  x (1/P(x))  {i | x i = b} f k (i) b k (i)

The Baum-Welch Algorithm Initialization: Pick the best-guess for model parameters (or arbitrary) Iteration: 1.Forward 2.Backward 3.Calculate A kl, E k (b) 4.Calculate new model parameters a kl, e k (b) 5.Calculate new log-likelihood P(x |  ) GUARANTEED TO BE HIGHER BY EXPECTATION-MAXIMIZATION Until P(x |  ) does not change much

The Baum-Welch Algorithm – comments Time Complexity: # iterations  O(K 2 N) Guaranteed to increase the log likelihood of the model P(  | x) = P(x,  ) / P(x) = P(x |  ) / ( P(x) P(  ) ) Not guaranteed to find globally best parameters Converges to local optimum, depending on initial conditions Too many parameters / too large model:Overtraining

Alternative: Viterbi Training Initialization:Same Iteration: 1.Perform Viterbi, to find  * 2.Calculate A kl, E k (b) according to  * + pseudocounts 3.Calculate the new parameters a kl, e k (b) Until convergence Notes:  Convergence is guaranteed – Why?  Does not maximize P(x |  )  In general, worse performance than Baum-Welch