Hidden Markov Models 1 2 K … x1 x2 x3 xK

Generating a sequence by the model Given a HMM, we can generate a sequence of length n as follows: Start at state 1 according to prob a01 Emit letter x1 according to prob e1(x1) Go to state 2 according to prob a12 … until emitting xn 1 2 K … 1 2 K … 1 1 2 K … 1 2 K … … a02 2 2 K e2(x1) x1 x2 x3 xn

Evaluation We will develop algorithms that allow us to compute: P(x) Probability of x given the model P(xi…xj) Probability of a substring of x given the model P(i = k | x) “Posterior” probability that the ith state is k, given x A more refined measure of which states x may be in

The Forward Algorithm fk(i) = P(x1…xi, i = k) (the forward probability) Initialization: f0(0) = 1 fk(0) = 0, for all k > 0 Iteration: fk(i) = ek(xi) l fl(i – 1) alk Termination: P(x) = k fk(N)

Motivation for the Backward Algorithm We want to compute P(i = k | x), the probability distribution on the ith position, given x We start by computing P(i = k, x) = P(x1…xi, i = k, xi+1…xN) = P(x1…xi, i = k) P(xi+1…xN | x1…xi, i = k) = P(x1…xi, i = k) P(xi+1…xN | i = k) Then, P(i = k | x) = P(i = k, x) / P(x) Forward, fk(i) Backward, bk(i)

The Backward Algorithm – derivation Define the backward probability: bk(i) = P(xi+1…xN | i = k) “starting from ith state = k, generate rest of x” = i+1…N P(xi+1,xi+2, …, xN, i+1, …, N | i = k) = l i+1…N P(xi+1,xi+2, …, xN, i+1 = l, i+2, …, N | i = k) = l el(xi+1) akl i+1…N P(xi+2, …, xN, i+2, …, N | i+1 = l) = l el(xi+1) akl bl(i+1)

The Backward Algorithm We can compute bk(i) for all k, i, using dynamic programming Initialization: bk(N) = 1, for all k Iteration: bk(i) = l el(xi+1) akl bl(i+1) Termination: P(x) = l a0l el(x1) bl(1)

Computational Complexity What is the running time, and space required, for Forward, and Backward? Time: O(K2N) Space: O(KN) Useful implementation technique to avoid underflows Viterbi: sum of logs Forward/Backward: rescaling at each few positions by multiplying by a constant

Posterior Decoding We can now calculate fk(i) bk(i) P(i = k | x) = P(i = k , x)/P(x) = P(x1, …, xi, i = k, xi+1, … xn) / P(x) = P(x1, …, xi, i = k) P(xi+1, … xn | i = k) / P(x) = fk(i) bk(i) / P(x) We can now calculate fk(i) bk(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 = argmaxk P(i = k | x)

Posterior Decoding For each state, Posterior Decoding gives us a curve of likelihood of state for each position That is sometimes more informative than Viterbi path * Posterior Decoding may give an invalid sequence of states (of prob 0) Why?

Posterior Decoding =  {:[i] = k} P( | x) x1 x2 x3 …………………………………………… xN State 1 l P(i=l|x) k P(i = k | x) =  P( | x) 1(i = k) =  {:[i] = k} P( | x) 1() = 1, if  is true 0, otherwise

Viterbi, Forward, Backward Initialization: V0(0) = 1 Vk(0) = 0, for all k > 0 Iteration: Vl(i) = el(xi) maxk Vk(i-1) akl Termination: P(x, *) = maxk Vk(N) FORWARD Initialization: f0(0) = 1 fk(0) = 0, for all k > 0 Iteration: fl(i) = el(xi) k fk(i-1) akl Termination: P(x) = k fk(N) BACKWARD Initialization: bk(N) = 1, for all k Iteration: bl(i) = k el(xi+1) akl bk(i+1) Termination: P(x) = k a0k ek(x1) bk(1)

Variants of HMMs

Higher-order HMMs How do we model “memory” larger than one time point? P(i+1 = l | i = k) akl P(i+1 = l | i = k, i -1 = j) ajkl … A second order HMM with K states is equivalent to a first order HMM with K2 states aHHT state HH state HT aHT(prev = H) aHT(prev = T) aHTH state H state T aTHH aHTT aTHT state TH state TT aTH(prev = H) aTH(prev = T) aTTH

Similar Algorithms to 1st Order P(i+1 = l | i = k, i -1 = j) Vlk(i) = maxj{ Vkj(i – 1) + … } Time? Space?

Modeling the Duration of States 1-p Length distribution of region X: E[lX] = 1/(1-p) Geometric distribution, with mean 1/(1-p) This is a significant disadvantage of HMMs Several solutions exist for modeling different length distributions X Y p q 1-q

Example: exon lengths in genes Why the algorithm has to be generalized is because in a standard HMM the output in each step would be one base, leading to state durations of geometric length. If we look at empirical data, such as these plots, we see that the intron lengths seem to follow the geometric distribution fairly well, but for the exon that would be a pretty bad model. So in our state space the exons are generalized states, choosing a length from a general distribution and outputting the entire exon in one step, while the intron and intergene states still output one base at a time and follow the geometric distribution.

Solution 1: Chain several states p 1-p X X X Y q 1-q Disadvantage: Still very inflexible lX = C + geometric with mean 1/(1-p)

Solution 2: Negative binomial distribution p p p 1 – p 1 – p 1 – p X(1) X(2) X(n) Y …… Duration in X: m turns, where During first m – 1 turns, exactly n – 1 arrows to next state are followed During mth turn, an arrow to next state is followed m – 1 m – 1 P(lX = m) = n – 1 (1 – p)n-1+1p(m-1)-(n-1) = n – 1 (1 – p)npm-n

Example: genes in prokaryotes EasyGene: Prokaryotic gene-finder Larsen TS, Krogh A Negative binomial with n = 3

Solution 3: Duration modeling Upon entering a state: Choose duration d, according to probability distribution Generate d letters according to emission probs Take a transition to next state according to transition probs Disadvantage: Increase in complexity of Viterbi: Time: O(D) Space: O(1) where D = maximum duration of state F d<Df xi…xi+d-1 Pf Warning, Rabiner’s tutorial claims O(D2) & O(D) increases

Viterbi with duration modeling emissions emissions F L d<Df d<Dl Pf Pl transitions xi…xi + d – 1 xj…xj + d – 1 Precompute cumulative values Recall original iteration: Vl(i) = maxk Vk(i – 1) akl  el(xi) New iteration: Vl(i) = maxk maxd=1…Dl Vk(i – d)  Pl(d)  akl  j=i-d+1…iel(xj)

Proteins, Pair HMMs, and Alignment

A state model for alignment (+1,+1) Alignments correspond 1-to-1 with sequences of states M, I, J I (+1, 0) J (0, +1) -AGGCTATCACCTGACCTCCAGGCCGA--TGCCC--- TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

Let’s score the transitions s(xi, yj) M (+1,+1) Alignments correspond 1-to-1 with sequences of states M, I, J s(xi, yj) s(xi, yj) -d -d I (+1, 0) J (0, +1) -e -e -AGGCTATCACCTGACCTCCAGGCCGA--TGCCC--- TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

Alignment with affine gaps – state version Dynamic Programming: M(i, j): Optimal alignment of x1…xi to y1…yj ending in M I(i, j): Optimal alignment of x1…xi to y1…yj ending in I J(i, j): Optimal alignment of x1…xi to y1…yj ending in J The score is additive, therefore we can apply DP recurrence formulas

Alignment with affine gaps – state version Initialization: M(0,0) = 0; M(i, 0) = M(0, j) = -, for i, j > 0 I(i,0) = d + ie; J(0, j) = d + je Iteration: M(i – 1, j – 1) M(i, j) = s(xi, yj) + max I(i – 1, j – 1) J(i – 1, j – 1) e + I(i – 1, j) I(i, j) = max d + M(i – 1, j) e + J(i, j – 1) J(i, j) = max d + M(i, j – 1) Termination: Optimal alignment given by max { M(m, n), I(m, n), J(m, n) }