1. . Correctness proof of EM Variants of HMM Sequence Alignment via HMM Lecture # 10 This class has been edited from Nir Friedman’s lecture. Changes made by Dan Geiger, then Shlomo Moran. Background Readings: chapters 11.6, 3.4, 3.5, 4, in the Durbin et al., 2001, Chapter 3.4 Setubal et al., 1997

2. 2 In each iteration the EM algorithm does the following. u (E step): Calculate Q θ ( ) = ∑ y p(y|x,θ)log p(x,y| ) u (M step): Find * which maximizes Q θ ( ) (Next iteration sets   * and repeats). The EM algorithm Comment: At the M-step we only need that Q θ ( *)>Q θ (θ). This change yields the so called Generalized EM algorithm. It is important when it is hard to find the optimal *.

3. 3 Correctness proof of EM Theorem: If λ* maximizes Q  (λ) = ∑ y p(y|x,θ)log p(y| λ), then P(x| λ*)  P(x| θ). Comment: The proof remains valid if we assume only that Q  (λ*)  Q  (θ).

4. 4 By the definition of conditional probability, for each y we have, p(x| ) p(y|x, ) = p(y,x| ), and hence: log p(x| ) = log p( y,x| ) – log p( y |x, ) Hence log p(x| λ) = ∑ y p(y|x,θ) [log p(y|λ) – log p(y|x,λ)] log p(x|λ) Proof (cont.) =1 (Next..)

5. 5 Proof (end) log p(x|λ) = ∑ y p(y|x, θ) log p(y|λ) - ∑ y p(y|x,θ) log [p(y|x,λ)] Qθ(λ)Qθ(λ) Substituting λ=λ* and λ=θ, and then subtracting, we get log p(x|λ*) - log p(x|θ) = Q(λ*) – Q(θ) + D(p(y|x,θ) || p(y|x,λ*)) ≥ 0 [since λ* maximizes Q(λ)]. Relative entropy 0 ≤ ≥ 0 QED

6. 6 EM in Practice Initial parameters: u Random parameters setting u “Best” guess from other source Stopping criteria: u Small change in likelihood of data u Small change in parameter values Avoiding bad local maxima: u Multiple restarts u Early “pruning” of unpromising ones

7. 7 HMM model structure: 1. Duration Modeling Markov chains are rather limited in describing sequences of symbols with non-random structures. For instance, Markov chain forces the distribution of segments in which some state is repeated for k additional times to be (1-p)p k, for some p. Several ways enable modeling of other distributions. One is assigning k >1 states to represent the same “real” state. This may guarantee k repetitions with any desired probability. A1A1 A2A2 A3A3 A4A4

8. 8 HMM model structure: 2. Silent states u States which do not emit symbols (as we saw in the abo locus). u Can be used to model duration distributions. u Also used to allow arbitrary jumps (may be used to model deletions) u Need to generalize the Forward and Backward algorithms for arbitrary acyclic digraphs to count for the silent states (see next): Silent states: Regular states:

9. 9 Eg, the forwards algorithm should look: Directed cycles of silence (or other) states complicate things, and should be avoided. x v z Silent states Regular states symbols

10. 10 HMM model structure: 3. High Order Markov Chains Markov chains in which the transition probabilities depends on the last k states: P(x i |x i-1,...,x 1 ) = P(x i |x i-1,...,x i-k ) Can be represented by a standard Markov chain with more states. eg for k=2: AA BB BA AB

11. 11 HMM model structure: 4. Inhomogenous Markov Chains u An important task in analyzing DNA sequences is recognizing the genes which code for proteins. u A triplet of 3 nucleotides – called codon - codes for amino acids (see next slide). u It is known that in parts of DNA which code for genes, the three codons positions has different statistics. u Thus a Markov chain model for DNA should represent not only the Nucleotide (A, C, G or T), but also its position – the same nucleotide in different position will have different transition probabilities. Used in GENEMARK gene finding program (93).

12. 12 Genetic Code There are 20 amino acids from which proteins are build.

13. 13 Sequence Comparison using HMM HMM for sequence alignment, which incorporates affine gap scores. “Hidden” States u Match. u Insertion in x. u insertion in y. Symbols emitted u Match: {(a,b)| a,b in ∑ }. u Insertion in x: {(a,-)| a in ∑ }. u Insertion in y: {(-,a)| a in ∑ }.

14. 14 The Transition Probabilities ε1- ε ε δδ 1-2 δ MX X M Y Y Emission Probabilities u Match: (a,b) with p ab – only from M states u Insertion in x: (a,-) with q a – only from X state u Insertion in y: (-,a).with q a - only from Y state. (Note that the hidden states can be reconstructed from the alignment.) Transitions probabilities (note the forbidden ones).  δ = probability for 1 st gap  ε = probability for tailing gap.

15. 15 Scoring alignments Each aligned pair is generated by the above HMM with certain probability. For each pair of sequences x (of length m) and y (of length n), there are many alignments of x and y, each corresponds to a different state-path (the length of the paths are between max{m,n} and m+n). Our task is to score alignments by using this model. The score should reflect the probability of the alignment.

16. 16 Most probable alignment Let v M (i,j) be the probability of the most probable alignment of x(1..i) and y(1..j), which ends with a match. Similarly, v X (i,j) and v Y (i,j), the probabilities of the most probable alignment of x(1..i) and y(1..j), which ends with an insertion to x or y. Then using a recursive argument, we get:

17. 17 Most probable alignment Similar argument for v X (i,j) and v Y (i,j), the probabilities of the most probable alignment of x(1..i) and y(1..j), which ends with an insertion to x or y, are:

18. 18 Adding termination probabilities For this, an END state is added, with transition probability τ from any other state to END. This assumes expected sequence length of 1/ τ. MXY END M 1-2δ -τ δδτ X 1-ε -τ ε τ Y ετ END 1 We may want a model which defines a probability distribution over all possible sequences. The last transition in each alignment is to the END state, with probability τ

19. 19 The log-odds scoring function u We wish to know if the alignment score is above or below the score of random alignment. u For gapless alignments we used the log-odds ratio: s(a,b) = log (p ab / q a q b ). log (p ab /q a q b )>0 iff the probability that a and b are related by our model is larger than the probability that they are picked at random. u To adapt this for the HMM model, we need to model random sequence by HMM, with end state. This model assigns probability to each pair of sequences x and y of arbitrary lengths m and n.

20. 20 scoring function for random model XYEND X1- ηη Y η END1 The transition probabilities for the random model, with termination probability η: (x is the start state) The emission probability for a is q a. Thus the probability of x (of length n) and y (of length m) being random is: And the corresponding log-odds score is:

21. 21 Markov Chains for “Random” and “Model” XYEND X1- ηη Y η END 1 MXY M1-2δ -τ δδτ X1-ε -τ ε τ Y ετ END 1 “Model” “Random”

22. 22 Combining models in the log-odds scoring function In order to compare the M score to the R score of sequences x and y, we can find an optimal M score, and then subtract from it the R score. This is insufficient when we look for local alignments, where the optimal substrings in the alignment are not known in advance. A better way: 1. Define a log-odds scoring function which keeps track of the difference Match-Random scores of the partial strings during the alignment. 2. At the end add to the score (logτ – 2logη). We get the following:

23. 23 The log-odds scoring function And at the end add to the score (logτ – 2logη). (assuming that letters at insertions/deletions are selected by the random model)

24. 24 The log-odds scoring function Another way (Durbin et. al., Chapter 4.1): Define scoring function s with penalties d and e for a first gap and a tailing gap, resp. Then modify the algorithm to correct for extra prepayment, as follows:

25. 25 Log-odds alignment algorithm Initialization: V M (0,0)=logτ - 2logη. Termination: V = max{ V M (m,n), V X (m,n)+c, V Y (m,n)+c} Where c= log (1-2δ-τ) – log(1-ε-τ)

26. 26 Total probability of x and y Rather then computing the probability of the most probable alignment, we look for the total probability that x and y are related by our model. Let f M (i,j) be the sum of the probabilities of all alignments of x(1..i) and y(1..j), which end with a match. Similarly, f X (i,j) and f Y (i,j) are the sum of these alignments which end with insertion to x (y resp.). A “forward” type algorithm for computing these functions. Initialization: f M (0,0)=1, f X (0,0)= f Y (0,0)=0 (we start from M, but we could select other initial state).

27. 27 Total probability of x and y (cont.) The total probability of all alignments is: P(x,y|model)= f M [m,n] + f X [m,n] + f Y [m,n]