1. Hidden Markov Models 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xKxK 2 1 K 2

2. Outline for our next topic Hidden Markov models – the theory Probabilistic interpretation of alignments using HMMs Later in the course: Applications of HMMs to biological sequence modeling and discovery of features such as genes

3. Example: The Dishonest Casino Player A casino has two dice: Fair die P(1) = P(2) = P(3) = P(5) = P(6) = 1/6 Loaded die P(1) = P(2) = P(3) = P(5) = 1/10 P(6) = 1/2 Casino player switches between fair and loaded die randomly, on avg. once every 20 turns Game: 1.You bet $1 2.You roll (always with a fair die) 3.Casino player rolls (maybe with fair die, maybe with loaded die) 4.Highest number wins $2

4. Question # 1 – Evaluation GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION How likely is this sequence, given our model of how the casino works? This is the EVALUATION problem in HMMs Prob = 1.3 x 10 -35

5. Question # 2 – Decoding GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION What portion of the sequence was generated with the fair die, and what portion with the loaded die? This is the DECODING question in HMMs FAIRLOADEDFAIR

6. Question # 3 – Learning GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION How “loaded” is the loaded die? How “fair” is the fair die? How often does the casino player change from fair to loaded, and back? This is the LEARNING question in HMMs Prob(6) = 64%

7. The dishonest casino model FAIRLOADED 0.05 0.95 P(1|F) = 1/6 P(2|F) = 1/6 P(3|F) = 1/6 P(4|F) = 1/6 P(5|F) = 1/6 P(6|F) = 1/6 P(1|L) = 1/10 P(2|L) = 1/10 P(3|L) = 1/10 P(4|L) = 1/10 P(5|L) = 1/10 P(6|L) = 1/2

8. A HMM is memory-less At each time step t, the only thing that affects future states is the current state  t K 1 … 2

9. Definition of a hidden Markov model Definition: A hidden Markov model (HMM) Alphabet  = { b 1, b 2, …, b M } Set of states Q = { 1,..., K } Transition probabilities between any two states a ij = transition prob from state i to state j a i1 + … + a iK = 1, for all states i = 1…K Start probabilities a 0i a 01 + … + a 0K = 1 Emission probabilities within each state e i (b) = P( x i = b |  i = k) e i (b 1 ) + … + e i (b M ) = 1, for all states i = 1…K K 1 … 2 End Probabilities a i0 in Durbin; not needed

10. A HMM is memory-less At each time step t, the only thing that affects future states is the current state  t P(  t+1 = k | “whatever happened so far”) = P(  t+1 = k |  1,  2, …,  t, x 1, x 2, …, x t )= P(  t+1 = k |  t ) K 1 … 2

11. A HMM is memory-less At each time step t, the only thing that affects x t is the current state  t P(x t = b | “whatever happened so far”) = P(x t = b |  1,  2, …,  t, x 1, x 2, …, x t-1 )= P(x t = b |  t ) K 1 … 2

12. A parse of a sequence Given a sequence x = x 1 ……x N, A parse of x is a sequence of states  =  1, ……,  N 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xKxK 2 1 K 2

13. Generating a sequence by the model Given a HMM, we can generate a sequence of length n as follows: 1.Start at state  1 according to prob a 0  1 2.Emit letter x 1 according to prob e  1 (x 1 ) 3.Go to state  2 according to prob a  1  2 4.… until emitting x n 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xnxn 2 1 K 2 0 e 2 (x 1 ) a 02

14. Likelihood of a parse Given a sequence x = x 1 ……x N and a parse  =  1, ……,  N, To find how likely this scenario is: (given our HMM) P(x,  ) = P(x 1, …, x N,  1, ……,  N ) = P(x N |  N ) P(  N |  N-1 ) ……P(x 2 |  2 ) P(  2 |  1 ) P(x 1 |  1 ) P(  1 ) = a 0  1 a  1  2 ……a  N-1  N e  1 (x 1 )……e  N (x N ) 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xKxK 2 1 K 2 A compact way to write a 0  1 a  1  2 ……a  N-1  N e  1 (x 1 )……e  N (x N ) Enumerate all parameters a ij and e i (b); n params Example: a 0Fair :  1 ; a 0Loaded :  2 ; … e Loaded (6) =  18 Then, count in x and  the # of times each parameter j = 1, …, n occurs F(j, x,  ) = # parameter  j occurs in (x,  ) (call F(.,.,.) the feature counts) Then, P(x,  ) =  j=1…n  j F(j, x,  ) = = exp [  j=1…n log(  j )  F(j, x,  ) ]

15. Example: the dishonest casino Let the sequence of rolls be: x = 1, 2, 1, 5, 6, 2, 1, 5, 2, 4 Then, what is the likelihood of  = Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair? (say initial probs a 0Fair = ½, a oLoaded = ½) ½  P(1 | Fair) P(Fair | Fair) P(2 | Fair) P(Fair | Fair) … P(4 | Fair) = ½  (1/6) 10  (0.95) 9 =.00000000521158647211 ~= 0.5  10 -9

16. Example: the dishonest casino So, the likelihood the die is fair in this run is just 0.521  10 -9 OK, but what is the likelihood of  = Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded? ½  P(1 | Loaded) P(Loaded, Loaded) … P(4 | Loaded) = ½  (1/10) 9  (1/2) 1 (0.95) 9 =.00000000015756235243 ~= 0.16  10 -9 Therefore, it somewhat more likely that all the rolls are done with the fair die, than that they are all done with the loaded die

17. Example: the dishonest casino Let the sequence of rolls be: x = 1, 6, 6, 5, 6, 2, 6, 6, 3, 6 Now, what is the likelihood  = F, F, …, F? ½  (1/6) 10  (0.95) 9 ~= 0.5  10 -9, same as before What is the likelihood  = L, L, …, L? ½  (1/10) 4  (1/2) 6 (0.95) 9 =.00000049238235134735 ~= 0.5  10 -7 So, it is 100 times more likely the die is loaded

18. Question # 1 – Evaluation GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION How likely is this sequence, given our model of how the casino works? This is the EVALUATION problem in HMMs Prob = 1.3 x 10 -35

19. Question # 2 – Decoding GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION What portion of the sequence was generated with the fair die, and what portion with the loaded die? This is the DECODING question in HMMs FAIRLOADEDFAIR

20. Question # 3 – Learning GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION How “loaded” is the loaded die? How “fair” is the fair die? How often does the casino player change from fair to loaded, and back? This is the LEARNING question in HMMs Prob(6) = 64%

21. The three main questions on HMMs 1.Evaluation GIVEN a HMM M, and a sequence x, FIND Prob[ x | M ] 2.Decoding GIVENa HMM M, and a sequence x, FINDthe sequence  of states that maximizes P[ x,  | M ] 3.Learning GIVENa HMM M, with unspecified transition/emission probs., and a sequence x, FINDparameters  = (e i (.), a ij ) that maximize P[ x |  ]

22. Let’s not be confused by notation P[ x | M ]: The probability that sequence x was generated by the model The model is: architecture (#states, etc) + parameters  = a ij, e i (.) So, P[x | M] is the same with P[ x |  ], and P[ x ], when the architecture, and the parameters, respectively, are implied Similarly, P[ x,  | M ], P[ x,  |  ] and P[ x,  ] are the same when the architecture, and the parameters, are implied In the LEARNING problem we always write P[ x |  ] to emphasize that we are seeking the  * that maximizes P[ x |  ]

23. Problem 1: Decoding Find the most likely parse of a sequence

24. Decoding GIVEN x = x 1 x 2 ……x N Find  =  1, ……,  N, to maximize P[ x,  ]  * = argmax  P[ x,  ] Maximizes a 0  1 e  1 (x 1 ) a  1  2 ……a  N-1  N e  N (x N ) Dynamic Programming! V k (i) = max {  1…  i-1} P[x 1 …x i-1,  1, …,  i-1, x i,  i = k] = Prob. 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 Given that we end up in state k at step i, maximize product to the left and right

25. Decoding – main idea Inductive assumption: Given that for all states k, and for a fixed position i, V k (i) = max {  1…  i-1} P[x 1 …x i-1,  1, …,  i-1, x i,  i = k] What is V l (i+1)? From definition, V l (i+1) = max {  1…  i} P[ x 1 …x i,  1, …,  i, x i+1,  i+1 = l ] = max {  1…  i} P(x i+1,  i+1 = l | x 1 …x i,  1,…,  i ) P[x 1 …x i,  1,…,  i ] = max {  1…  i} P(x i+1,  i+1 = l |  i ) P[x 1 …x i-1,  1, …,  i-1, x i,  i ] = max k [ P(x i+1,  i+1 = l |  i =k) max {  1…  i-1} P[x 1 …x i-1,  1,…,  i-1, x i,  i =k] ] = max k [ P(x i+1 |  i+1 = l ) P(  i+1 = l |  i =k) V k (i) ] = e l (x i+1 ) max k a kl V k (i)

26. The Viterbi Algorithm Input: x = x 1 ……x N Initialization: V 0 (0) = 1(0 is the imaginary first position) V k (0) = 0, for all k > 0 Iteration: V j (i) = e j (x i )  max k a kj V k (i – 1) Ptr j (i) = argmax k a kj V k (i – 1) Termination: P(x,  *) = max k V k (N) Traceback:  N * = argmax k V k (N)  i-1 * = Ptr  i (i)

27. 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)

28. Viterbi Algorithm – a practical detail Underflows are a significant problem P[ x 1,…., x i,  1, …,  i ] = a 0  1 a  1  2 ……a  i e  1 (x 1 )……e  i (x i ) These numbers become extremely small – underflow Solution: Take the logs of all values V l (i) = log e k (x i ) + max k [ V k (i-1) + log a kl ]

29. Example Let x be a long sequence with a portion of ~ 1/6 6’s, followed by a portion of ~ ½ 6’s… x = 123456123456…12345 6626364656…1626364656 Then, it is not hard to show that optimal parse is (exercise): FFF…………………...F LLL………………………...L 6 characters “123456” parsed as F, contribute.95 6  (1/6) 6 = 1.6  10 -5 parsed as L, contribute.95 6  (1/2) 1  (1/10) 5 = 0.4  10 -5 “162636” parsed as F, contribute.95 6  (1/6) 6 = 1.6  10 -5 parsed as L, contribute.95 6  (1/2) 3  (1/10) 3 = 9.0  10 -5

30. Problem 2: Evaluation Find the likelihood a sequence is generated by the model

31. Generating a sequence by the model Given a HMM, we can generate a sequence of length n as follows: 1.Start at state  1 according to prob a 0  1 2.Emit letter x 1 according to prob e  1 (x 1 ) 3.Go to state  2 according to prob a  1  2 4.… until emitting x n 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xnxn 2 1 K 2 0 e 2 (x 1 ) a 02

32. A couple of questions Given a sequence x, What is the probability that x was generated by the model? Given a position i, what is the most likely state that emitted x i ? Example: the dishonest casino Say x = 12341…23162616364616234112…21341 Most likely path:  = FF……F (too “unlikely” to transition F  L  F) However: marked letters more likely to be L than unmarked letters P(box: FFFFFFFFFFF) = (1/6) 11 * 0.95 12 = 2.76 -9 * 0.54 = 1.49 -9 P(box: LLLLLLLLLLL) = [ (1/2) 6 * (1/10) 5 ] * 0.95 10 * 0.05 2 = 1.56*10 -7 * 1.5 -3 = 0.23 -9 FF

33. Evaluation We will develop 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)“Posterior” probability that the i th state is k, given x A more refined measure of which states x may be in

34. The Forward Algorithm We want to calculate P(x) = probability of x, given the HMM Sum over all possible ways of generating x: P(x) =   P(x,  ) =   P(x |  ) P(  ) To avoid summing over an exponential number of paths , define f k (i) = P(x 1 …x i,  i = k) (the forward probability) “generate i first characters of x and end up in state k”

35. The Forward Algorithm – derivation Define the forward probability: f k (i) = P(x 1 …x i,  i = k) =   1…  i-1 P(x 1 …x i-1,  1,…,  i-1,  i = k) e k (x i ) =  l   1…  i-2 P(x 1 …x i-1,  1,…,  i-2,  i-1 = l) a lk e k (x i ) =  l P(x 1 …x i-1,  i-1 = l) a lk e k (x i ) = e k (x i )  l f l (i – 1 ) a lk

36. The Forward Algorithm We can compute f k (i) for all k, i, using dynamic programming! Initialization: f 0 (0) = 1 f k (0) = 0, for all k > 0 Iteration: f k (i) = e k (x i )  l f l (i – 1) a lk Termination: P(x) =  k f k (N)

37. Relation between Forward and Viterbi 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)

38. Motivation for the Backward Algorithm We want to compute P(  i = k | x), the probability distribution on the i th position, given x We start by computing P(  i = k, x) = P(x 1 …x i,  i = k, x i+1 …x N ) = P(x 1 …x i,  i = k) P(x i+1 …x N | x 1 …x i,  i = k) = P(x 1 …x i,  i = k) P(x i+1 …x N |  i = k) Then, P(  i = k | x) = P(  i = k, x) / P(x) Forward, f k (i)Backward, b k (i)

39. The Backward Algorithm – derivation Define the backward probability: b k (i) = P(x i+1 …x N |  i = k) “starting from i th state = k, generate rest of x” =   i+1…  N P(x i+1,x i+2, …, x N,  i+1, …,  N |  i = k) =  l   i+1…  N P(x i+1,x i+2, …, x N,  i+1 = l,  i+2, …,  N |  i = k) =  l e l (x i+1 ) a kl   i+1…  N P(x i+2, …, x N,  i+2, …,  N |  i+1 = l) =  l e l (x i+1 ) a kl b l (i+1)

40. The Backward Algorithm We can compute b k (i) for all k, i, using dynamic programming Initialization: b k (N) = 1, 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)

41. Computational Complexity What is the running time, and space required, for Forward, and Backward? Time: O(K 2 N) 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

42. 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) P(  i = k | x) = P(  i = k, x)/P(x) = P(x 1, …, x i,  i = k, x i+1, … x n ) / P(x) = P(x 1, …, x i,  i = k) P(x i+1, … x n |  i = k) / P(x) = f k (i) b k (i) / P(x)

43. 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?

44. Posterior Decoding P(  i = k | x) =   P(  | x) 1(  i = k) =  {  :  [i] = k} P(  | x) x 1 x 2 x 3 …………………………………………… x N State 1 l P(  i =l|x) k 1(  ) = 1, if  is true 0, otherwise

45. Viterbi, Forward, Backward VITERBI Initialization: V 0 (0) = 1 V k (0) = 0, for all k > 0 Iteration: V l (i) = e l (x i ) max k V k (i-1) a kl 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) BACKWARD Initialization: b k (N) = 1, for all k Iteration: b l (i) =  k e l (x i +1) a kl b k (i+1) Termination: P(x) =  k a 0k e k (x 1 ) b k (1)