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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 04/05/2014.

Ask a homework question - tutors are online