Machine Learning for natural language processing

Transcription

1 Machine Learning for natural language processing Hidden Markov Models Laura Kallmeyer Heinrich-Heine-Universität Düsseldorf Summer / 33

2 Introduction So far, we have classified texts/observations independent from the class of the previous text/observation Today: sequence classifier, assigning a sequence of classes to a sequence of observations Jurafsky & Martin (2015), chapter 8 2 / 33

3 Table of contents 1 Motivation 2 Hidden Markov Models 3 Likelihood computation 4 Decoding 5 Training 3 / 33

4 Motivation Hidden Markov Models (HMMs) are weighted finite state automata with states emitting outputs. Transitions and emissions are weighted. 4 / 33

5 Motivation HMM start q a 0.1 q b end in q a, a is emitted with probability 0.9; in q a, b is emitted with probability 0.1; in q b, b is emitted with probability 0.9; in q b, a is emitted with probability 0.1; Given a sequence bb, what is the most probable path traversed by the automaton, resulting in this output? The states in the automaton correspond to classes, i.e., the search for the most probable path amounts to the search for the most probable class sequence. Example POS Tagging: the classes are POS tags, the emissions are words 5 / 33

6 Hidden Markov Models A Markov chain is a weighted automaton in which weights are probabilities, i.e., all weights are between 0 and 1 and the sum of the weights of all outgoing edges of a state is 1, and the input sequence uniquely determines the states the automaton goes through. A Markov chain is actually a bigram language model. Markov Chain for LM example slide 9 start Sam am end I / 33

7 Hidden Markov Models Markov chains are useful when we want to compute the probability for a sequence of events that we can observe. Hidden Markov Models compute probabilities for a sequence of hidden events, given a sequence of observed events. Example from Jurafsky & Martin (2015), after Eisner (2002) HMM for inferring weather (hot = H, cold = C) from numbers of ice creams eaten by Jason. start end 0.8 H 0.4 C 0.1 P(1 H) = 0.2 P(2 H) = 0.4 P(3 H) = 0.4 P(1 C) = 0.5 P(2 C) = 0.4 P(3 C) = / 33

8 Hidden Markov Models HMM A HMM is a tuple Q, A, O, B, q 0, q F, with Q = {q 1,..., q N } a finite set of states; A a Q Q matrix, the transition probability matrix, with 0 a ij 1 for all 1 i Q, 1 j Q. O a finite set of observations; a sequence B of Q functions b i O R, the emission probabilities with o O b i (o) = 1 for all 1 i Q q 0, q F Q are special start and end states, associated with transition probabilities a 0i and a if (0 a 0i, a if 1) for all 1 i Q. For all i, 0 i Q, it holds that Q j=1 a ij + a if = 1. 8 / 33

9 Hidden Markov Models First order HMMs make the following simplifying assumptions: 1 Markov assumption: In a sequence of states q i1... q in, we assume P(q in q i1... q in 1 ) = P(q in q in 1 )(= a in 1in ) 2 Output independence: In a sequence of states q i1... q in with the associated output sequence o i1... o in, we assume P(o ij q i1... q ij... q in 1, o i1... o ij... o in 1 ) = P(o ij q ij )(= b ij(o ij ) 9 / 33

10 Hidden Markov Models The following three fundamental problems have to be solved: 1 Likelihood: Given an HMM and an observation sequence, determine the likelihood of the observation sequence. 2 Decoding: Given an HMM and an observation sequence, discover the best hidden state sequence leading to these observations. 3 Learning: Given the set of states of an HMM and an observation sequence, learn the HMM parameters A and B. 10 / 33

11 Likelihood computation Given an HMM and an observation o = o 1... o n, determine the likelihood of the observation sequence o. For a specific state sequence q = q 1... q n Q n, we have P(o, q) = P(o q)p(q) and with our independence assumptions P(o, q) = n i=1 P(o i q i ) n i=1 P(q i q i 1 ) For the probability of o, we have to sum over all possible state sequences: P(o) = n q Q n i=1 P(o i q i ) n i=1 P(q i q i 1 )P(q F q n ) 11 / 33

12 Likelihood computation Computing each element of the sum independently from the others would lead to an exponential time complexity. Instead, we adopt the forward algorithm that implements some dynamic programming. Idea: We fill a n Q matrix α such that α ij = P(o 1... o i, q i = q j ) Forward algorithm 1 α 1j = a 0j b j (o 1 ) for 1 j Q 2 α ij = Q k=1 α i 1ka kj b j (o i ) for 1 i n and 1 j Q 3 P(o) = Q k=1 α nka kf 12 / 33

13 Likelihood computation Ice cream weather example continued P(1 H) = 0.2 P(2 H) = 0.4 P(3 H) = 0.4 P(313): start H H C P(313) = C 0.1 end P(1 C) = 0.5 P(2 C) = 0.4 P(3 C) = 0.1 α 1j = a 0j b j (o 1 ) α ij = Q k=1 α i 1ka kj b j (o i ) P(o) = Q k=1 α nka kf 13 / 33

14 Likelihood computation Alternatively, we can also compute the backward probability β. For a given observation o i in our observation sequence, the forward probability considers the part from o 1 to o i while the backward probability considers the part from o i+1 to o n. Idea: We fill a n Q matrix β such that β ij = P(o i+1... o n, q i = q j ) Backward algorithm 1 β nj = a jf for 1 j Q 2 β ij = Q k=1 β i+1ka jk b k (o i+1 ) for 1 i < n and 1 j Q 3 P(o) = Q k=1 a 0kb k (o 1 )β 1k 14 / 33

15 Likelihood computation Ice cream weather example continued start end P(1 H) = 0.2 P(2 H) = 0.4 P(3 H) = 0.4 H 0.4 C P(1 C) = 0.5 P(2 C) = 0.4 P(3 C) = 0.1 observed sequence 313: H C β nj = a jf β ij = Q k=1 β i+1ka jk b k (o i+1 ) P(o) = Q k=1 a 0kb k (o 1 )β 1k P(313) = / 33

16 Decoding Given an HMM and an observation o = o 1... o n, determine the most probable state sequence q = q 1... q n Q n for o. arg max q Q n P(o, q) = arg max q Q n and with our independence assumptions arg max q Q n P(o, q) = arg max q Q n n i=1 P(o q)p(q) P(o i q i ) n i=1 P(q i q i 1 ) 16 / 33

17 Decoding For the computation, we use the viterbi algorithm, which is very similar to the forward algorithm except that, instead of computing sums, we compute maxs. We fill a n Q matrix v such that v ij = max q Q i 1 P(o 1... o i, q 1... q i 1, q i = q j ) Viterbi algorithm 1 v 1j = a 0j b j (o 1 ) for 1 j Q 2 v ij = max 1 k Q v i 1k a kj b j (o i ) for 1 i n and 1 j Q 3 max q Q n P(o, q) = max 1 k Q v nk a kf In addition, in order to retreive the best state sequence, each entry v ij is equipped with a backpointer to the state that has lead to the maximum. 17 / 33

18 Decoding Ice cream weather example continued P(1 H) = 0.2 P(2 H) = 0.4 P(3 H) = 0.4 start 0.8 output sequence 313: H 0.4 C 0.1 end P(1 C) = 0.5 P(2 C) = 0.4 P(3 C) = 0.1 H , start , H , H C , start , H , C v 1j = a 0j b j (o 1 ) for 1 j Q v ij = max 1 k Q v i 1k a kj b j (o i ) for 1 i n and 1 j Q max q Q n P(o, q) = max 1 k Q v nk a kf most probable weather sequence is HHH with probability / 33

19 Training Supervised HMM Training: Given a sequence of observations o and a corresponding sequence of states q, learn the parameters A and B of an HMM. MLE via the counts of q i q j sequences and of pairs q i, o j of states and observations in our training data, eventually with some smoothing. More challenging: Unsupervised HMM Training: Given an observation sequence o and a state set Q, estimate A and B. Example ice cream weather case: we have a sequence of observations o {1, 2, 3} n and we know that the possible states are Q = {H, C}. POS tagging: we have a sequence of words o {w 1, w 2,..., w V } n and we know that the possible POS tags (= states) are Q = {NN, NNS, VBD, IN,... }. Task: estimate A and B of the corresponding HMMs. 19 / 33

20 Training We use the forward-backward or Baum-Welch algorithm (Baum, 1972), a special case of the EM algorithm (Dempster et al., 1977). Underlying ideas: We estimate parameters iteratively: we start with some parameters and use the estimated probabilities to derive better and better parameters. We get our estimates by computing forward probabilities and then dividing the probability mass among all the different state sequences that have contributed to a forward probability. Put differently, in each iteration loop, we count all possible state sequences for the given observation sequence. But, instead of counting each of them once, we use their probabilities according to the most recent parameters as counts. We perform some kind of frequency-based MLE with these weighted or fractional counts. 20 / 33

21 Training We start with some initial A and B. In each iteration, new parameter â ij and ˆb i (o) are computed. Intuition: and â ij = expected number of transitions from state i to state j expected number of transitions from state i ˆbi (o) = expected number of times in i observing symbol o expected number of times in state i 21 / 33

22 Training More precisely, instead of the expected numbers, we use the probabilities according to the last parameters A and B and according to the counts we retreive from the observation sequence o: and â ij = probability of transitions from i to j, given o probability of transitions from state i, given o ˆbi (o) = probability of emitting o in i, given o probability of passing through state i, given o 22 / 33

23 Training For â ij, we first calculate the probability of being in state i at time t (observation o t ) and in state j at time t + 1, with the observation sequence given. This is a product of the probability of being in i at step t after having emitted o 1... o t, which is the forward probability α ti ; the probability of the transition from i to j, which is a ij ; the probability of emitting o t+1 in j, which is b j (o t+1 ); and the probability of moving from j to q F while emitting o t+2... o n, which is given by the backward probability β t+1,j divided by the probability of the observation since this is taken as given, i.e., divided by P(o). 23 / 33

24 Training We define ξ t (i, j) as this probability, i.e., the probability of being in i at time t, i.e., at observation o t and in j when emitting o t+1, given the observation sequence o: ξ t (i, j) = P(q t = q i, q t+1 = q j o) According to the previous slide, this can be computed as Transition E-step ξ t (i, j) = α tia ij b j (o t+1 )β t+1,j P(o) for i, j {1,..., Q } This is the E (expectation) step for the transition probabilities. 24 / 33

25 Training From these ξ t (i, j) values, we can estimate new transition probabilities towards maximizing the observed data: Remember that we want to have â ij = probability of transitions from i to j, given o probability of transitions from state i, given o With the previously defined ξ t (i, j), we obtain Transition M-step â ij = n 1 t=1 ξt(i,j) n 1 t=1 Q k=1 ξt(i,k)+ α n,i a if P(o) for i, j {1,..., Q }. â 0i = a 0ib i(o 1)β 1,i Q k=1 a 0kb k (o 1)β 1,k and â if = α n,i a if P(o) n 1 t=1 Q k=1 ξt(i,k)+ α n,i a if P(o) for 1 i Q This is the M (maximization) step for the transition probabilities. 25 / 33

26 Training For ˆb i (o), we calculate the probability of being in state i at time t given the observation sequence o, which we call γ t (i). Emission E-step γ t (i) = P(q t = q i o) = α tiβ ti P(o) This is the E (expectation) step for the emission probabilities. 26 / 33

27 Training From these γ t (i) values, we can estimate new emission probabilities towards maximizing the observed data: Remember that we want to have ˆbi (o) = probability of emitting o in i, given o probability of passing through state i, given o Emission M-step ˆbi (o) = 1 t n,o t=o γ t (i) n t=1 γ t (i) for 1 i Q This is the M (maximization) step for the emission probabilities. 27 / 33

28 Training Forward-backward algorithm (input observations o O n, state set Q): initialize A and B iterate until convergence: E-step γ t (i) = αtiβti P(o) for all 1 t n, 1 i Q ξ t (i, j) = αtiaijbj(ot+1)βt+1,j P(o) for all 1 t n 1, 1 i, j Q M-step â ij = â 0i = n 1 t=1 ξt(i,j) n 1 t=1 Q k=1 ξt(i,k)+ α n,i a if P(o) a 0ib i(o 1)β 1,i Q k=1 a 0kb k (o 1)β 1,k and â if = ˆbi (o) = 1 t n,ot =o γ t(i) n t=1 γt(i) return A, B for all 1 i, j Q α n,i a if P(o) n 1 t=1 Q k=1 ξt(i,k)+ α n,i a if P(o) for 1 i Q 28 / 33

29 Training (Simplified) ice cream weather example Initialized HMM: P(1 H) = 0.2 P(3 H) = 0.8 start H observed sequence 31: α: H C t 1 2 E-step: γ: M-step: P(1 H) = 0.2 P(3 H) = t H C start C β: ξ 1 : end H C end P(1 C) = 0.8 P(3 C) = 0.2 H C t 1 2 j =H j =C i =H i =C P(1 C) = 0.8 P(3 C) = 0.2 P(31) = 0.27 P(31) = / 33

30 Training (Simplified) ice cream weather example Second step: P(1 H) = 0.2 P(3 H) = end H C start P(1 C) = 0.8 P(3 C) = 0.2 α H C t 1 2 β H C t 1 2 E-step: γ: M-step: P(1 H) = P(3 H) = 0.98 t H C end H C start ξ j =H j =C i =H i =C P(1 C) = 0.98 P(3 C) = P(31) = / 33

31 Training (Simplified) ice cream weather example Third step: P(1 H) = P(3 H) = end H C start P(1 C) = 0.98 P(3 C) = α H C t 1 2 β H C t 1 2 E-step: γ: t H C ξ j =H j =C i =H 0 1 i =C 0 0 M-step: P(1 H) = P(3 H) = 1 start H 0 C end 1 P(1 C) = 1 P(3 C) = 0 P(31) = 1 31 / 33

32 Training Note that we can also leave out the factor initialize A and B iterate until convergence: 1 P(o), the result is the same: E-step φ t (i) = α ti β ti for all 1 t n, 1 i Q ψ t (i, j) = α ti a ij b j (o t+1 )β t+1,j for all 1 t n 1, 1 i, j Q M-step â ij = â 0i = n 1 t=1 ξt(i,j) n 1 t=1 Q k=1 ψt(i,k)+αn,ia if a 0ib i(o 1)β 1,i Q k=1 a 0kb k (o 1)β 1,k and â if = ˆbi (o) = 1 t n,ot =o φ t(i) n t=1 φt(i) return A, B for all 1 i, j Q α n,ia if n 1 t=1 Q k=1 ξt(i,k)+αn,ia if for 1 i Q 32 / 33

33 References Baum, L. E An inequality and ssociated maximization technique in statistical estimation for jprobabilistic functions of markov processes. In O. Shisha (ed.), Inequalities iii: Proceedigns of the 3rd symposiom on inequalities, 1 8. University of California, Los Angeles: Academic Press. Dempster, A. P., N. M. Laird & D. B. Rubin Maximum likelihood from incomplete data via the EM algorithm. Journal of the Royal Statistical Society 39(1) Eisner, Jason An interactive spreadsheet for teaching the forward-backward algorithm. In Proceedings of the acl workshop on effective tools and methodologies for teaching nlp and cl, Jurafsky, Daniel & James H. Martin Speech and language processing. an introduction to natural language processing, computational linguistics, and speech recognition. Draft of the 3rd edition. 33 / 33