( ).666 Information Extraction from Speech and Text
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1 ( ).666 Information Extraction from Speech and Text HMM Parameters Estimation for Gaussian Output Densities April 27, 205. Generalization of the Results of Section 9.4. It is suggested in Section that the results of Section 9.4. for 2-dimensional observations extend easily to d-dimensions. We will work through the details in this note. Specifically, we will consider an HMM with output densities attached to arcs. Let S denote the set of states; the arcs be indexed by t T ; the outputs or emissions take values in IR d for some finite d > 0; L(t) and R(t) respectively denote the origin- and destination-states of the arc t; and p t denote the probability of the arc t when the underlying Markov chain is in L(t). Clearly, t : L(t)s p t, s S. () For each non-null arc t, let the corresponding output density be a multivariate Gaussian, { N t (y) (2π) d 2 Ut exp } y 2 (y m t) T U t (y m t ), y IR d, (2) where m t is the mean vector and U t the covariance matrix of the emitted random vector. Note that y and m t are column vectors here, while they are row-vectors in the textbook, and the arc-dependence of m t and U t is denoted via a subscript here instead of writing m(t) and U(t). The free parameters of the HMM are θ {θ t, t T }, where θ t {p t, m t, U t }, the p t s satisfy the sum-to-one condition (), and the U t s are symmetric and positive-definite. y d
2 Given an n-length observation Y y, y 2,..., y n from this HMM, the EM auxiliary function may be constructed as Q(θ, θ) P θ (t Y) log P θ (t, Y) P θ (t Y) log p tl N tl (y l ), (3) where t t, t 2,..., t denotes any valid path through the HMM, and denotes its length. While it is not made precise in the textbook, it is to be understood in (3) that P θ (t Y) > 0 only for paths t of length n that contain exactly n non-null arcs and n null arcs, and hence other paths need not be considered in the sum over all t; the reference to the l-th output symbol y l is valid only after reindexing Y y, y 2,..., y n and (re)assigning y to the first non-null arc of t, y 2 to the second non-null arc of t, and so on, until y n to the last non-null arc of t, while no symbols are assigned to its null arcs; N tl (y l ) is computed via (2) for non-null arcs t l in t, but N tl ( ) for all null arcs in t. Next, given a θ, we try to maximize Q(θ, θ) as a function of θ. To this end, we form the Lagrangian Q(θ, θ) λ s p t. (4) s S t : L( t)s Updating the Transition Probabilities p t Note that for every arc t T, P θ (t Y) log P θ (t, Y) λ s p t p t p t p t s S t : L( t)s P θ (t Y) log p tl N tl (y l ) λ L(t) p t P θ (t Y) log p tl + log N tl (y l ) λ L(t) p t ] P θ (t Y) log p t + 0 λ L(t) p t P θ (t Y) λ L(t). p t For each arc t T, equating the derivative to 0 yields 0 P θ (t Y) p t p t λ L(t) 2
3 P θ (Y) λ L(t) P θ (Y) P θ (t, Y) p t λ L(t) P θ (t, Y) } {{ } ψ t The brute-force way to compute the double sum ψ t is to. exhaustively enumerate all paths t, 2. traverse each path t t,..., t, and p t, 3. every time t l t, i.e. the arc t is traversed, add P θ (t, Y) to an accumulator for ψ t. Once the ψ t are accumulated for all t T, the role of λ s, for every state s S, is to ensure that the probabilities of arcs leaving s sum to unity. Therefore t : L(t)s p t t : L(t)s ψ t λ s P θ (Y) λ s P θ (Y) t : L(t)s ψ t K s. Now, a n-stage trellis captures all paths t capable of producing Y: all paths t with P θ (t, Y) > 0. Furthermore, if t is a non-null arc, then it appears exactly n times in the trellis, once in each trellis stage, and every time a path t traverses the l-th copy of t, l,..., n, an output y l is produced. Therefore, the contribution of the l-th copy of t to ψ t is the sum of the probabilities of all the paths t that pass through t in the l-th stage of the trellis, namely P θ (y,..., y l, s L(t)) p t N t (y }{{} l ) P θ (y l+,..., y n s R(t)) }{{} α l (L(t)) β l (R(t)). Therefore the total contribution from all stages of the trellis for a non-null arc t is ψ t P θ (t, Y) n α l (L(t)) p t N t (y l ) β l(r(t)). Similarly, a null arc t may appear on a path t within in each column of a vertically aligned set of states. If the l-th copy of t in the trellis is designated as the one traversed before producing y l (i.e. between producing y l and y l ), l,..., n, then its contribution to ψ t from all paths t is P θ (y,..., y l, s L(t)) p t P }{{} θ(y l,..., y n s R(t)), l,..., n. }{{} α l (L(t)) β l (R(t)) Therefore, for a null arc t, the total contribution from all stages of the trellis is ψ t P θ (t, Y) 3 n α l (L(t)) p t β l (R(t)).
4 As noted above, for every state s, K s n ψ t α l (s)p t N t (y l )β l(r(t)) + α l (s)p tβ l (R(t)) t : L(t)s non null t : L(t)s n α l (s) null arcs t : L(t)s p t N t (y l )β l(r(t)) + non null t : L(t)s null arcs t : L(t)s n α l (s)β l (s). p tβ l (R(t)) Updating the Mean Vectors m t Next, note that for every non-null arc t T, if we let m t m t,... m t,d ] T, then m t,i 2 P θ (t Y) log P θ (t, Y) λ s m t,i m t,i s S P θ (t Y) log p tl N tl (y l ) 0 m t,i P θ (t Y) m t,i t : L( t)s log p tl + log N tl (y l ) P θ (t Y) 0 + ] log N t (y l ) m t,i P θ (t Y) { } 2 (y l m t ) T U t (y l m t ) log (2π) d 2 Ut m t,i P θ (t Y) { (yl m t ) T U t (y l m t ) + 0 }. m t,i The partial derivatives of with respect to the components of the mean vector m t may therefore be compactly written as the vector { m t, P θ (t Y) (yl m t, m t ) T U t (y l m t ) } 2. { m t,d (yl m t,d m t ) T U t (y l m t ) }. (5) Next, note that x T b d m x mb m, and thus x i x T b x i 4 d x m b m b i, m p t
5 from which it follows that for any d vectors x x... x d ] T and b b... b d ] T x x T b b. x d x T b Similarly x T Ax d m d n x ma mn x n, and therefore x i x T Ax x i d d d d x m a mn x n x m a mi + a in x n + 2a ii x i m n m, m i n, n i d a mi +a im ]x m. m Therefore, for a symmetric d d matrix A, x x T Ax A T x + Ax 2Ax. x d x T Ax Set A U t and x (y t m t ) and note that m t, (y l m t ) T U t (y l m t ). (y (y l, m t, ) l m t ) T U t (y l m t )., m t,d (y l m t ) T U t (y l m t ) (y (y l,d m t,d ) l m t ) T U t (y l m t ) where the negative sign is due to the fact that m t, (y l, m t, ).. m t,d (y l,d m t,d ) This easily provides the partial derivatives (5) of with respect to the components of m t. To obtain the update equation for m t we must set it to zero. i.e. Set m t, 0 P θ (t Y) 2U t (y l m t ) 0 m t,d P θ (t Y)(y l m t ) 0 2U t P θ (t Y)y l P θ (t, Y) 5 P θ (t Y)m t P θ (t, Y)y l m t
6 n n α l (L(t)) p t N t (y l ) β l (R(t))] y l α l (L(t)) p t N t (y l ) β l (R(t))] m t, (6) where the last step, once again, follows from the argument that the overall double sum, similar to ψ t above, may be obtained by separately computing the contribution of all paths t to the arc t in a particular (l-th) stage of the trellis, and accumulating such contributions for l,..., n. To interpret the mean update equation (6) qualitatively, recall that α l (L(t)) p t N t (y l ) β l (R(t)) P θ (Y) P θ (t l t Y) P θ (y l was emitted by arc t Y). The value of m t in (6) that maximizes may thus be seen as a sample mean, where each observation y l in the sample has a fractional count the probability that it came from arc t and the sample size is the expected number of times the arc t was traversed, or as a weighted mean, where the weight of the sample y l is the probability that it was emitted from t. Updating the Covariance Matrices U t To find the U t that maximizes, let V t U t, and v t,ij denote the ij-th element of V t. v t,ij 2 2 P θ (t Y) log P θ (t, Y) v t,ij λ s v t,ij s S P θ (t Y) log p tl N tl (y l ) 0 v t,ij P θ (t Y) P θ (t Y) v t,ij P θ (t Y) t : L( t)s log p tl + log N tl (y l ) 0 + v t,ij ] log N t (y l ) v t,ij { 2 (y l m t ) T U t (y l m t ) log (2π) d 2 Ut P θ (t Y) { (yl m t ) T U t (y l m t ) + log U t } v t,ij P θ (t Y) { (yl m t ) T V t (y l m t ) log V t }. (7) v t,ij V t, the inverse of the covariance matrix, is sometimes called the precision matrix, and is often of interest in multivariate statistics and factor analysis. p t } 6
7 Next, for any d vector x ] T x... x d and symmetric positive-definite d d matrix A, the partial derivatives of the scalar x T Ax with respect to the components of the matrix A are a x T Ax a d x T Ax a ij x T Ax. xx T, a d x T Ax a dd x T Ax and the partial derivatives of the scalar log A with respect to the components of A are a log A a d log A a ij log A. A. a d log A a dd log A The derivatives v ij are obtained by setting x (y l m t ) and A U t in the formulae above: v t, v t,d v t,ij. P θ (t Y) { } (yl m t )(y l m t ) T Vt, 2 v t,d v t,dd and the choice of V t (equivalently U t ) that makes 2 P θ (t Y) P θ (t, Y) v t,ij 0 for all i and j is P θ (t Y) { } (yl m t )(y l m t ) T V 0 P θ (t Y) (y l m t )(y l m t ) T V t P θ (t, Y) (y l m t )(y l m t ) T U t, (8) where, once again, the parameters p t and N t correspond to θ, and α ( ) and β ( ) are the forward- and backward-probabilities computed using θ on a n-stage trellis, with the null arcs going between vertically aligned states and only non-null arcs traversing left-to-right. Finally, the covariance update equation (9) once again follows by observing that, similar to ψ t, the double sum may be obtained by first accumulating the contribution of all paths t in the trellis to the l-th copy of an arc t, and then summing these contributions for l,..., n. t U t n α l (L(t)) p t N t (y l ) β l (R(t))] (y l m t )(y l m t ) T. (9) α l (L(t)) p t N t (y l ) β l (R(t))] n It remains to verify that the updated matrix U t of (9) is symmetric and positive-(semi)definite, thereby justifying why the Lagrangian of (4) did not impose any constraints on the components of the parameter set θ corresponding to the U t s. 7
8 It is straightforward to see that the updated U t of (9) satisfies U T t n P θ (t ] l t, Y) (y l m t )(y l m t ) T T n P θ (t l t, Y) n P θ (t l t, Y) (y l m t )(y l m t ) T] T n P θ (t l t, Y) and that for any vector x, U t, x T U t x n P θ (t l t, Y) x T (y l m t )(y l m t ) T x n P θ (t l t, Y) n P θ (t l t, Y) ( x T (y l m t ) ) 2 n P θ (t l t, Y) 0. This guarantees that U t will always be a bona fide covariance matrix. 8
p(d θ ) l(θ ) 1.2 x x x
p(d θ ).2 x 0-7 0.8 x 0-7 0.4 x 0-7 l(θ ) -20-40 -60-80 -00 2 3 4 5 6 7 θ ˆ 2 3 4 5 6 7 θ ˆ 2 3 4 5 6 7 θ θ x FIGURE 3.. The top graph shows several training points in one dimension, known or assumed to
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