Hidden Markov Models – Concepts 1 2 K … 1 2 K … 1 2 K … … … … 1 2 K … x1x1 x2x2 x3x3 xKxK 2 1 K 2

Probability review

Markov chains Markov property: P(X 0, X 1 … X t ) = P(X 0 )P(X 1 |X 0 )…P(X t |X t-1 ) Formally: – State space = list of possible states – Transition matrix = probability of moving from one X to another – Initial distribution = initial value of X

Example: The dishonest casino 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 back-&-forth between fair and loaded die 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

Question # 1 – Evaluation GIVEN A sequence of rolls by the casino player QUESTION How likely is this sequence, given our model of how the casino works? This is the EVALUATION problem in HMMs

Question # 2 – Decoding GIVEN A sequence of rolls by the casino player 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

Question # 3 – Learning GIVEN A sequence of rolls by the casino player 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

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

The dishonest casino model FAIRLOADED 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

HMM properties: 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

HMM properties: 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

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 |  ]

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 |  ] and P[ x ] are the same, when the architecture, and the entire model, respectively, are implied Similarly, P[ x,  | M ] and P[ x,  ] are the same In the LEARNING problem we always write P[ x |  ] to emphasize that we are seeking the  that maximizes P[ x |  ]

Problem 1: Evaluation Find the likelihood of a parse

Likelihood of a parse Given a sequence x = x 1 ……x N and a parse  =  1, ……,  N, To find how likely is the parse: (given our HMM) P(x,  ) = P(x 1, …, x N,  1, ……,  N ) = P(x N,  N |  N-1 ) P(x N-1,  N-1 |  N-2 )……P(x 2,  2 |  1 ) P(x 1,  1 ) = 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

Example: the dishonest casino Let the sequence of rolls be: x = 1, 2, 1, 5, 6, 2, 1, 6, 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 = = 0.5  10 -9

Example: the dishonest casino So, the likelihood the die is fair in all this run is just  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) 8  (1/2) 2 (0.95) 9 = = 7.9  Therefore, it is after all 6.59 times more likely that the die is fair all the way, than that it is loaded all the way.

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 = = 0.5  So, it is 100 times more likely the die is loaded

Problem 2: Decoding Find the best parse (states) of a sequence

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

Decoding – main idea 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 k (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] = e l (x i+1 ) max k a kl V k (i)

Dynamic programming approach 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)

The 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)

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 ]

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

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)

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 parameters  : Given x = x 1 …x N for which the true  =  1 …  N is unknown, We can reach a more likely parameter set  if we repeat many times. 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