Markov chains (cont d)

Transcription

1 6.867 Machine learning, lecture 19 (Jaaola) 1 Lecture topics: Marov chains (cont d) Hidden Marov Models Marov chains (cont d) In the contet of spectral clustering (last lecture) we discussed a random wal over the nodes induced by a weighted graph. Let W ij 0 be symmetric weights associated with the edges in the graph; W ij = 0 whenever edge doesn t eist. We also assumed that W ii = 0 for all i. The graph defines a random wal where the probability of transitioning from node (state) i to node j is given by P (X(t + 1) = j X(t) = i) = P ij = W ij W ij j (1) Note that self-transitions (going from i to i) are disallowed because W ii = 0 for all i. We can understand the random wor as a homogeneous Marov chain: the probability of transitioning from i to j only depends on i, not the path that too the process to i. In other words, the current state summarizes the past as far as the future transitions are concerned. This is a Marov (conditional independence) property: P (X(t + 1) = j X(t) = i, X(t 1) = i t 1,..., X(1) = i 1 ) = P (X(t + 1) = j X(t) = i) (2) The term homogeneous specifies that the transition probabilities are independent of time (the same probabilities are used whenever the random wal returns to i). We also defined ergodicity as follows: Marov chain is ergodic if there eist a finite m such that P (X(t + m) = j X(t) = i) > 0 for all i and j (3) Simple weighted graphs need not define ergodic chains. Consider, for eample, a weighted graph between two nodes 1 2 where W 12 > 0. The resulting random wal is necessarily periodic, i.e., A Marov chain is ergodic only when all the states are communicating and the chain is aperiodic which is clearly not the case here. Similarly, even a graph with positive weights on the edges would not define an ergodic Marov chain. Every other state would necessarily be 2, thus the chain is periodic. The reason here is

2 6.867 Machine learning, lecture 19 (Jaaola) 2 that W ii = 0. By adding a positive self-transition, we can remove periodicity (random wal would stay in the same state a variable number of steps). Any connected weighted graph with positive weights and positive self-transitions gives rise to an ergodic Marov chain. Our definition of the random wal so far is a bit incomplete. We did not specify how the process started, i.e., we didn t specify the initial state distribution. Let q(i) be the probability that the random wal is started from state i. We will use q as a vector of probabilities across states (reserving n for the number training eamples as usual). There are two ways of describing Marov chains: through state transition diagrams or as simple graphical models. The descriptions are complementary. A transition diagram is a directed graph over the possible states where the arcs between states specify all allowed transitions (those occuring with non-zero probability). See Figure 1 for eamples. We could also add the initial state distribution as transitions from a dedicated initial (null) state (not shown in the figure) Figure 1: Eamples of transition diagrams defining non-ergodic Marov chains. In graphical models, on the other hand, we focus on eplicating variables and their dependencies. At each time point the random wal is in a particular state X(t). This is a random variable. It s value is only affected by the random variable X(t 1) specifying the state of the random wal at the previous time point. Graphically, we can therefore write a sequence of random variables where arcs specify how the values of the variables are influenced by others (dependent on others). More precisely, X(t 1) X(t) means that the value of X(t) depends on X(t 1). Put another way, in simulating the random wal, we would have to now the value of X(t 1) in order to sample a value for X(t). The graphical model is shown in Figure 2. State prediction We will cast the problem of calculating the predictive probabilities over states in a form

3 6.867 Machine learning, lecture 19 (Jaaola) 3... X(t 1) X(t) X(t + 1)... Figure 2: Marov chain as a graphical model. that will be useful for Hidden Marov Models later on. Since we can also write for any n P (X(t + m) = j X(t) = i) = [P m ] ij (4) P (X(n) = j) = q(i)p (X(n) = j X(1) = i) = q(i)[p ] ij (5) In a vector form q T P is a row vector whose j th component is P (X(n) = j). Note that the matri products involve summing over all the intermediate states until X(n) = j. More eplicitly, let s evaluate the sum over all the states 1,..., n in the matri form as n 1 times {}}{ P (X(1) = 1 ) P (X(t + 1) = t+1 X(t) = t ) = q T P P P P 1 = 1 (6) 1,..., n t=1 This is a sum over n possible state configurations (settings of 1,..., n ) but can be easily performed in terms of matri products. We can understand this in terms of recursive evaluation of t step probabilities α t (i) = P (X(t) = i). We will write α t for the corresponding column vector so that Clearly, Estimation t 1 times {}}{ q T T P P P P = α t α t 1 (i)p ij = α t (j) (10) (7) q T T = α 1 (8) T T α t 1 P = α t, t > 1 (9)

4 6.867 Machine learning, lecture 19 (Jaaola) 4 Marov models can be estimated easily from observed sequences of states. Given 1,..., n (e.g., ), the log-lielihood of the observed sequence is given by [ ] log P ( 1,..., n ) = log P (X(1) = 1 ) P (X(t + 1) = t+1 X(t) = t ) (11) t=1 = log q( 1 ) + log P t,t+1 (12) t=1 = log q( 1 ) + nˆ(i, j) log P ij (13) i,j where ˆn(i, j) is the number of observed transitions from i to j in the sequence 1,..., n. The resulting maimum lielihood setting of P ij is obtained as an empirical fraction nˆ(i, j) Pˆij = j nˆ(i, j ) (14) Note that q(i) can only be reliably estimated from multiple observed sequences. For eample, based on 1,..., n, we would simply set ˆq(i) = δ(i, 1 ) which is hardly accurate (sample size one). Regularization is useful here, as before. Hidden Marov Models Hidden Marov Models (HMMs) etend Marov models by assuming that the states of the Marov chain are not observed directly, i.e., the Marov chain itself remains hidden. We therefore also model how the states relate to the actual observations. This assumption of a simple Marov model underlying a sequence of observations is very useful in many practical contets and has made HMMs very popular models of sequence data, from speech recognition to bio-sequences. For eample, to a first approimation, we may view words in speech as Marov sequences of phonemes. Phonemes are not observed directly, however, but have to be related to acoustic observations. Similarly, in modeling protein sequences (sequences of amino acid residues), we may, again approimately, describe a protein molecule as a Marov sequence of structural characteristics. The structural features are typically not observable, only the actual residues. We can understand HMMs by combining miture models and Marov models. Consider the simple eample in Figure 3 over four discrete time points t = 1, 2, 3, 4. The figure summarizes multiple sequences of observations y 1,..., y 4, where each observation sequence

5 6.867 Machine learning, lecture 19 (Jaaola) 5 corresponds to a single value y t per time point. Let s begin by ignoring the time information and instead collapse the observations across the time points. The observations form two clusters are now well modeled by a two component miture: 2 P (y) = P (j)p (y j) (15) j=1 where, e.g., P (y j) could be a Gaussian N(y; µ j, σj 2 ). By collapsing the observations we are effectively modeling the data at each time point with the same miture model. If we generate data form the resulting miture model we would select a miture component at random at each time step and generate the observation from the corresponding component (cluster). There s nothing that ties the selection of miture components in time so that samples from the miture yield phantom clusters at successive time points (we select the wrong component/cluster with equal probability). By omitting the time information, we therefore place half of the probability mass in locations with no data. Figure 4 illustrates the miture model as a graphical model. y y a) t = 1 t = 2 t = 3 t = 4 time b) t = 1 t = 2 t = 3 t = 4 time Figure 3: a) Eample data over four time points, b) actual data and ranges of samples generated from a miture model (red ovals) estimated without time information. The solution is to model the selection of the miture components as a Marov model, i.e., the component at t = 2 is selected on the basis of the component used at t = 1. Put another way, each state in the Marov model now uses one of the components in the miture model to generate the corresponding observation. As a graphical model, the miture model is a combination of the two as shown in Figure 5. Probability model One advantage of representing the HMM as a graphical model is that we can easily write down the joint probability distribution over all the variables. The graph eplicates how the

6 6.867 Machine learning, lecture 19 (Jaaola) 6 X(1) X(2) X(3) X(4) Y (1) Y (2) Y (3) Y (4) Figure 4: A graphical model view of the miture model over the four time points. The variables are indeed by time (different samples would be drawn at each time point) but the parameters are shared across the four time points. X(t) refers to the selection of the miture component while Y (t) refers to the observations. variables depend on each other (who influences who) and thus highlights which conditional probabilities we need to write down: P ( 1,..., n, y 1,..., y n ) = P ( 1 )P (y 1 1 )P ( 2 1 )P (y 2 2 )... (16) = P ( 1 )P (y 1 1 ) [P ( t+1 t )P (y t+1 t+1 )] (17) t=1 [ ] = q( 1 )P (y 1 1 ) P t,t+1 P (y t+1 t+1 ) (18) where we have used the same notation as before for the Marov chains. t=1 X(1) X(2) X(3) X(4) Y (1) Y (2) Y (3) Y (4) Figure 5: HMM as a graphical model. It is a Marov model where each state is associated with a distribution over observations. Alternatively, we can view it as a miture model where the miture components are selected in a time dependent manner. Three problems to solve We typically have to be able to solve the following three problems in order to use these models effectively:

7 6.867 Machine learning, lecture 19 (Jaaola) 7 1. Evaluate the probability of observed data or P (y 1,..., y n ) = P ( 1,..., n, y 1,..., y n ) (19) 1,..., n 2. Find the most liely hidden state sequence 1,..., n given observations y 1,..., y n, i.e., {1,..., n} = arg ma P ( 1,..., n, y 1,..., y n ) (20) 1,..., n 3. Estimate the parameters of the model from multiple sequences of y (l) ) 1,..., y n (ll, l = 1,..., L. Problem 1 As in the contet of Marov chains we can efficiently sum over the possible hidden state sequences. Here the summation means evaluating P (y 1,..., y n ). We will perform this in two ways depending on whether the recursion moves forward in time, computing α t (j), or bacward in time, evaluating β t (i). The only change from before is the fact that whatever state we happen to visit at time t, we will also have to generate the observation y t from that state. This additional requirement of generating the observations can be included via diagonal matrices P (y 1) 0 D y = (21) 0 P (y ) So, for eample, Similarly, q T D y1 1 = q(i)p (y 1 i) = P (y 1 ) (22) q T D y1 P D y2 1 = q(i)p (y 1 i) P ij P (y 2 j) = P (y 1, y 2 ) (23) j=1 We can therefore write the forward and bacward algorithms as methods that perform the matri multiplications in q T D y1 P D y2 P P D yn 1 = P (y 1,..., y n ) (24)

8 6.867 Machine learning, lecture 19 (Jaaola) 8 either in the forward or bacward direction. In terms of the forward pass algorithm: q T D y1 = α1 T (25) α T P D yt = αt T, or equivalently (26) ( ) t 1 α t 1 (i)p ij P (y t j) = α t (j) (27) These values hold eactly α t (j) = P (y 1,..., y t, X(t) = j) since we have generated all the observations up to and including y t and have summed over all the states ecept for the last one X(t). The bacward pass algorithm is similarly defined as: β n = 1 (28) β t = P D yt+1 β t+1, or equivalently (29) β t (i) = P ij P (y t+1 j)β t+1 (j) (30) j=1 In this case β t (i) = P (y t+1,..., y n X(t) = i) since we have summed over all the possible values of the state variables X(t + 1),..., X(n), starting from a fied X(t) = i, and the first observation we have generated in the recursion is y t+1. By combining the two recursions we can finally evaluate T P (y 1,..., y n ) = α t β t = α t (i)β t (i) (31) which holds for any t = 1,..., n. You can understand this result in two ways: either in terms of performing the remaining matri multiplication corresponding to the two parts α T t {}}{{}}{ P (y 1,..., y n ) = (q T D y1 P P D yt ) (P D yt+1 P D yn 1) (32) or as an illustration of the Marov property: α t(i) β t(i) {}}{{}}{ P (y 1,..., y n ) = P (y 1,..., y t, X(t) = i) P (y t+1,..., y n X(t) = i) (33) β t

9 6.867 Machine learning, lecture 19 (Jaaola) 9 Also, since β n (i) = 1 for all i, clearly P (y 1,..., y n ) = α n (i) = P (y 1,..., y t, X(t) = i) (34)