Hidden Markov Models
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1 s Ben angmead Department of Computer Science You are free to use these slides. If you do, please sign the guestbook ( or me and tell me briefly how you re using them. or original Keynote files, me.
2 Sequence models Can we use Markov chains to pick out CpG islands from the rest of the genome? Markov chain assigns a score to a string; doesn t naturally give a running score across a long sequence Probability of being in island We could use a sliding window Genome position (a) Pick window size w, (b) score every w-mer using Markov chains, (c) use a cutoff to find islands Smoothing before (c) might also be a good idea
3 Sequence models Probability of being in island Genome position Choosing w involves an assumption about how long the islands are If w is too large, we ll miss small islands If w is too small, we ll get many small islands where perhaps we should see fewer larger ones In a sense, we want to switch between Markov chains when entering or exiting a CpG island
4 Sequence models Something like this: A inside C inside A outside C outside inside G inside outside G outside All inside-outside edges
5 rellis diagram p 1 p 2 p 3 p 4 p n x 1 x 2 x 3 x 4 x n Steps 1 through n p = { p 1, p 2,, p n } is a sequence of states (AKA a path). Each p i takes a value from set Q. We do not observe p. x = { x 1, x 2,, x n } is a sequence of emissions. Each x i takes a value from set. We do observe x.
6 p 1 p 2 p 3 p 4 p n x 1 x 2 x 3 x 4 x n ike for Markov chains, edges capture conditional independence: x 2 is conditionally independent of everything else given p 2 p 4 is conditionally independent of everything else given p 3 Probability of being in a particular state at step i is known once we know what state we were in at step i-1. Probability of seeing a particular emission at step i is known once we know what state we were in at step i.
7 Example: occasionally dishonest casino Dealer repeatedly flips a coin. Sometimes the coin is fair, with P(heads) = 0.5, sometimes it s loaded, with P(heads) = 0.8. Dealer occasionally switches coins, invisibly to you. ow does this map to an MM? p 1 p 2 p 3 p 4 p n x 1 x 2 x 3 x 4 x n
8 Example: occasionally dishonest casino Dealer repeatedly flips a coin. Sometimes the coin is fair, with P(heads) = 0.5, sometimes it s loaded, with P(heads) = 0.8. Dealer occasionally switches coins, invisibly to you. ow does this map to an MM? loaded Emissions encode flip outcomes (observed), states encode loadedness (hidden)
9 States encode which coin is used = fair = loaded p 1 p 2 p 3 p 4 p n Emissions encode flip outcomes = heads = tails x 1 x 2 x 3 x 4 x n
10 Example with six coin flips: p 1 p 2 p 3 p 4 p 5 p 6 p = x 1 x 2 x 3 x 4 x 5 x 6 x =
11 p 1 p 2 p 3 p 4 p n.. x 1 x 2 x 3 x 4 x n.. P( p 1, p 2,, p n, x 1, x 2,, x n ) = n P( x k p k ) k = 1 n P( p k p k-1 ) P( p 1 ) k = 2 Q emission matrix E encodes P( x i p i )s E [p i, x i ] = P( x i p i ) Q Q transition matrix A encodes P( p i p i-1 )s A [p i-1, p i ] = P( p i p i-1 ) Q array I encodes initial probabilities of each state I [p i ] = P( p 1 )
12 Dealer repeatedly flips a coin. Coin is sometimes fair, with P(heads) = 0.5, sometimes loaded, with P(heads) = 0.8. Dealer occasionally switches coins, invisibly to you. After each flip, dealer switches coins with probability 0.4 A: E: Q emission matrix E encodes P( x i p i )s E [p i, x i ] = P( x i p i ) Q Q transition matrix A encodes P( p i p i-1 )s A [p i-1, p i ] = P( p i p i-1 )
13 A E Given A & E (right), what is the joint probability of p & x? p x P( x i p i ) P( p i p i-1 ) If P( p 1 = ) = 0.5, then joint probability = =
14 Given flips, can we say when the dealer was using the loaded coin? We want to find p*, the most likely path given the emissions. p* = argmax P( p x ) p = argmax P( p, x ) p his is decoding. Viterbi is a common decoding algorithm.
15 : Viterbi algorithm Bottom-up dynamic programming s air, 3 s k, i = score of the most likely path up to step i with p i = k Start at step 1, calculate successively longer s k, i s p 1 p 2 p 3 p n x 1 x 2 x 3 x n
16 : Viterbi algorithm Given transition matrix A and emission matrix E (right), what is the most probable path p for the following x? A E Initial probabilities of / are p??????????? x s air, i 0.25?????????? s oaded, i 0.1?????????? Viterbi fills in all the question marks
17 : Viterbi algorithm s air,i = P(eads air) max { s k,i-1 P(air k) } k { air, oaded } Emission prob ransition prob p i-1 s air, i-1 P(air air) max s oaded, i-1 P(air oaded) x i-1 p i x i P(eads air)
18 : Viterbi algorithm x s, i p( ) = 0.5 Step 1 Step 2 Step 3 Step 4 p() = 0.5 S, 1 = p( ) = 0.5 max: p( ) = p( ) = wins max S, 2 = p( ) = 0.5 max: p( ) = p( ) = wins max S,3 = etc s, i p( ) = 0.2 p() = 0.5 p( ) = 0.8 max: p( ) = p( ) = etc etc S, 1 = wins max S, 2 =
19 : Viterbi algorithm Pick state in step n with highest score; backtrace for most likely path Backtrace according to which state k won the max in: Step n-3 Step n-2 Step n-1 Step n s, n-3 s, n-2 s, n-1 s, n s, n-3 s, n-2 s, n-1 s, n s,n > s,n
20 : Viterbi algorithm ow much work did we do, given Q is the set of states and n is the length of the sequence? n Q # s k, i values to calculate = n Q, each involves max over Q products O(n Q 2 ) Matrix A has Q 2 elements, E has Q elements, I has Q elements
21 : Implementation def viterbi(self, x): ''' Given sequence of emissions, return the most probable path along with its joint probability. ''' x = map(self.smap.get, x) # turn emission characters into ids nrow, ncol = len(self.q), len(x) mat = numpy.zeros(shape=(nrow, ncol), dtype=float) # prob matb = numpy.zeros(shape=(nrow, ncol), dtype=int) # backtrace # ill in first column for i in xrange(0, nrow): mat[i, 0] = self.e[i, x[0]] * self.i[i] # ill in rest of prob and b tables for j in xrange(1, ncol): for i in xrange(0, nrow): ep = self.e[i, x[j]] mx, mxi = mat[0, j- 1] * self.a[0, i] * ep, 0 for i2 in xrange(1, nrow): pr = mat[i2, j- 1] * self.a[i2, i] * ep if pr > mx: mx, mxi = pr, i2 mat[i, j], matb[i, j] = mx, mxi # ind final state with maximal probability omx, omxi = mat[0, ncol- 1], 0 for i in xrange(1, nrow): if mat[i, ncol- 1] > omx: omx, omxi = mat[i, ncol- 1], i # Backtrace i, p = omxi, [omxi] for j in xrange(ncol- 1, 0, - 1): i = matb[i, j] p.append(i) p = ''.join(map(lambda x: self.q[x], p[::- 1])) return omx, p # Return probability and path Calculate s k,i s ind maximal s k,n Backtrace mat holds the s k,i s matb holds traceback info self.e holds emission probs self.a holds transition probs self.i holds initial probs
22 : Viterbi algorithm >>> hmm = MM({"":0.6, "":0.4, "":0.4, "":0.6}, {"":0.5, "":0.5, "":0.8, "":0.2}, {"":0.5, "":0.5}) >>> prob, _ = hmm.viterbi("") >>> print prob e- 06 >>> prob, _ = hmm.viterbi("" * 100) >>> print prob 0.0 Repeat string 100 times Occasionally dishonest casino setup What happened? Underflow! Python example:
23 : Viterbi algorithm >>> hmm = MM({"":0.6, "":0.4, "":0.4, "":0.6}, {"":0.5, "":0.5, "":0.8, "":0.2}, {"":0.5, "":0.5}) >>> prob, _ = hmm.viterbi("") >>> print prob e- 06 >>> prob, _ = hmm.viterbi("" * 100) >>> print prob 0.0 >>> logprob, _ = hmm.viterbi("" * 100) >>> print logprob log-space Viterbi Repeat string 100 times When multiplying many numbers in (0, 1], we quickly approach the smallest number representable in a machine word. Past that we have underflow and processor rounds down to 0. Switch to log space. Multiplies become adds. Python example:
24 : Implementation def viterbi(self, x): ''' Given sequence of emissions, return the most probable path along with log2 of its joint probability. ''' x = map(self.smap.get, x) # turn emission characters into ids nrow, ncol = len(self.q), len(x) mat = numpy.zeros(shape=(nrow, ncol), dtype=float) # prob matb = numpy.zeros(shape=(nrow, ncol), dtype=int) # ill in first column for i in xrange(0, nrow): mat[i, 0] = self.elog[i, x[0]] + self.ilog[i] * # ill in rest of log prob and b tables for j in xrange(1, ncol): for i in xrange(0, nrow): ep = self.elog[i, x[j]] mx, mxi = mat[0, j- 1] + self.alog[0, i] + ep, 0 for i2 in xrange(1, nrow): pr = mat[i2, j- 1] + self.alog[i2, i] + ep if pr > mx: mx, mxi = pr, i2 mat[i, j], matb[i, j] = mx, mxi # ind final state with maximal log probability omx, omxi = mat[0, ncol- 1], 0 for i in xrange(1, nrow): if mat[i, ncol- 1] > omx: omx, omxi = mat[i, ncol- 1], i # Backtrace i, p = omxi, [omxi] for j in xrange(ncol- 1, 0, - 1): i = matb[i, j] p.append(i) p = ''.join(map(lambda x: self.q[x], p[::- 1])) return omx, p # Return log probability and path # backtrace * * * log-space version
25 ask: design an MM for finding CpG islands? p 1 p 2 p 3 p 4 p n x 1 x 2 x 3 x 4 x n
26 Idea 1: Q = { inside, outside }, = { A, C, G, } p 1 p 2 p 3 p 4 p n I I O I O I O I O O x 1 x 2 x 3 x 4 x n A C A C A C A C A C G G G G G
27 Idea 1: Q = { inside, outside }, = { A, C, G, } I inside A I O O outside E A C G Estimate as fraction of positions where we transition from inside to outside I O ransition matrix I O Emission matrix Estimate as fraction of nucleotides inside islands that are C
28 Example 1 using MM idea 1: A I O I O E A C G I O x: AAAACGCGCGCGCGCGCGAAAAAAA p: (from Viterbi) OOOOOOOIIIIIIIIIIIIIIOOOOOOOOOOOOO Python example:
29 Example 2 using MM idea 1: A I O I O E A C G I O x: AACGCGCGCGAAACGCGCGCGAAAA p: (from Viterbi) OOOOIIIIIIIIOOOOOOIIIIIIIIOOOOOOOO Python example:
30 Example 3 using MM idea 1: A I O I O E A C G I O x: AAAACCCCCCCCCCCCCCAAAAAAA p: OOOOOOOIIIIIIIIIIIIIIOOOOOOOOOOOOO (from Viterbi) Oops - not a CpG island! Python example:
31 Idea 2: Q = { A i, C i, G i, i, A o, C o, G o, o }, = { A, C, G, } p 1 p 2 p n A i C i G i i A i C i G i i A i C i G i i A o C o G o o A o C o G o o A o C o G o o x 1 x 2 x n A C A C A C G G G
32 Idea 2: Q = { A i, C i, G i, i, A o, C o, G o, o }, = { A, C, G, } A inside C inside A outside C outside inside G inside outside G outside All inside-outside edges
33 Idea 2: Q = { A i, C i, G i, i, A o, C o, G o, o }, = { A, C, G, } A A i C i G i i A o C o G o o A i C i G i i A o C o G o o Estimate P(C i i ) as fraction of all dinucleotides where first is an inside, second is an inside C E A C G A i C i G i i A o C o G o o
34 Actual trained transition matrix A: A: [[ e e e e e e e e- 04] [ e e e e e e e e- 04] [ e e e e e e e e- 04] [ e e e e e e e e- 04] [ e e e e e e e e- 01] [ e e e e e e e e- 01] [ e e e e e e e e- 01] [ e e e e e e e e- 01]]
35 Actual trained transition matrix A: Red & orange: low probability Yellow: high probability White: probability = 0 A i C i G i i A o C o G o o A i C i G i i Once inside, we likely stay inside for a while CpG islands end with G (in fact, they end with CG) A o C o Same for outside G o o CpG islands start with C (in fact, they start with CG)
36 Viterbi result; lowercase = outside, uppercase = inside: atatatatatatatatatatatatatatatatatatatatcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcgcg CGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCG CGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGatatatatatatatatatatatatatatatatatatatatatatatatatatat
37 Viterbi result; lowercase = outside, uppercase = inside: atatatatatatatatatatatatatatatatatatatatgggggggggggggggggggggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg ggggggggggggggggggggggggggggggggggggggggggggatatatatatatatatatatatatatatatatatatatatatatatatatatat
38 Many of the Markov chains and MMs we ve discussed are first order, but we can also design models of higher orders irst-order Markov chain: Second-order Markov chain:
39 or higher-order MMs, Viterbi s k, i no longer depends on just the previous state assignment p i-2 p i-1 p i x i-1 x i-1 x i Can sidestep the issue by expanding the state space
40 Now one state encodes the last two loadedness es of the coin p i-1 p i p i-2 p i-1 p i Q = {, } Q = {, } {, } After expanding, usual Viterbi works fine.
41 We also expanded the state space here: p j p j I A i C i G i i O A o C o G o o Q = { I, O } Q = { I, O } { A, C, G, }
Hidden Markov Models
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