This preview shows page 1. Sign up to view the full content.
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 432 (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 433 (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 434 (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 434 (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 ﬁniteprecision
arithmetic (i.e.,
numerical) reasons for
choosing one or another. qt Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 4  Jan 23rd, 2013 page 434 (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 ﬁniteprecision
arithmetic (i.e.,
numerical) reasons for
choosing one or another.
p(Qt = ix1:T ) =
p(Qt =i,x1:T )
p(x1:T ) EE596A/Winter 2013/DGMs – Lecture 4  Jan 23rd, 2013 page 434 (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.
 Winter '14

Click to edit the document details