29 page 4 33 of 239 hmms hmms as gms other hmm queries

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Unformatted text preview: +1 |qt )p(¯t+1 |qt+1 ) x (4.24) qt+1 β T ( qT ) = 1 Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 4 - Jan 23rd, 2013 (4.25) page 4-32 (of 239) HMMs HMMs as GMs Other HMM queries What HMMs can do MPE Summ HMMs and elimination - backward recursion, meaning β T (j ) = 1 βT −1 (i) = (4.26) p(¯T |QT = j )p(QT = j |QT −1 = i)βT (j ) x (4.27) j p(¯T , QT = j |QT −1 = i) = p(¯T |QT −1 = i) x x (4.28) j In general, βt (i) = p(xt+1:T |Qt = i) Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 4 - Jan 23rd, 2013 (4.29) page 4-33 (of 239) HMMs HMMs as GMs Other HMM queries What HMMs can do MPE Summ HMMs and elimination - p(x1:t ) We can get p(x1:t ) in many different ways: p(x1:T ) = p(x1:T , qt ) qt p(x1:t , xt+1:T , qt ) = qt p(xt+1:T |qt , x1:t )p(qt , x1:t ) = So this means that for any t, we can get p(x1:T ) by using the corresponding αt (i) and βt (i) quantities, for any t. qt p(xt+1:T |qt )p(qt , x1:t ) = qt βt (qt )αt (qt ) = qt Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 4 - Jan 23rd, 2013 page 4-34 (of 239) HMMs HMMs as GMs Other HMM queries What HMMs can do MPE Summ HMMs and elimination - p(x1:t ) We can get p(x1:t ) in many different ways: p(x1:T ) = p(x1:T , qt ) qt p(x1:t , xt+1:T , qt ) = qt p(xt+1:T |qt , x1:t )p(qt , x1:t ) = So this means that for any t, we can get p(x1:T ) by using the corresponding αt (i) and βt (i) quantities, for any t. Which one to choose? qt p(xt+1:T |qt )p(qt , x1:t ) = qt βt (qt )αt (qt ) = qt Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 4 - Jan 23rd, 2013 page 4-34 (of 239) HMMs HMMs as GMs Other HMM queries What HMMs can do MPE Summ HMMs and elimination - p(x1:t ) We can get p(x1:t ) in many different ways: p(x1:T ) = p(x1:T , qt ) qt p(x1:t , xt+1:T , qt ) = qt p(xt+1:T |qt , x1:t )p(qt , x1:t ) = qt p(xt+1:T |qt )p(qt , x1:t ) = qt βt (qt )αt (qt ) = So this means that for any t, we can get p(x1:T ) by using the corresponding αt (i) and βt (i) quantities, for any t. Which one to choose? primarily finite-precision arithmetic (i.e., numerical) reasons for choosing one or another. qt Prof. Jeff Bilmes EE596A/Winter 2013/DGMs – Lecture 4 - Jan 23rd, 2013 page 4-34 (of 239) HMMs HMMs as GMs Other HMM queries What HMMs can do MPE Summ HMMs and elimination - p(x1:t ) We can get p(x1:t ) in many different ways: p(x1:T ) = p(x1:T , qt ) qt p(x1:t , xt+1:T , qt ) = qt p(xt+1:T |qt , x1:t )p(qt , x1:t ) = qt p(xt+1:T |qt )p(qt , x1:t ) = qt βt (qt )αt (qt ) = qt Prof. Jeff Bilmes So this means that for any t, we can get p(x1:T ) by using the corresponding αt (i) and βt (i) quantities, for any t. Which one to choose? primarily finite-precision arithmetic (i.e., numerical) reasons for choosing one or another. p(Qt = i|x1:T ) = p(Qt =i,x1:T ) p(x1:T ) EE596A/Winter 2013/DGMs – Lecture 4 - Jan 23rd, 2013 page 4-34 (of 239) HMMs HMMs as GMs Other HMM queries What HMMs can do MPE Summ HMMs and elimination So we have the forward recursion as an elimination order and the backwards recursion as an elimination order. 3 1 10 Q1 X1 9 5 2 Q2 X2 8 7 7 4 Q3 X3 6 9 5 6 Q4 X4 4 10 3 8 Q5 X5 2 1 Green order is...
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This document was uploaded on 04/05/2014.

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