続分かりやすいパターン認識 4章後半(4.7以降)

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続・わかりやすい パターン認 2015/2/10 続パタ 脇 宗也

Transcript of 続分かりやすいパターン認識 4章後半(4.7以降)

  • 2015/2/10

  • ltwitterIDwwacky n

    l n1 n

    l nKroneckerKhatri-RaoHadamard

  • l

    l n n[0,1] n()

    p(|x(n))=Be(r+1, n-r+1)

    n:r:

  • 00 Be(1,1)

    11 Be(2,1)

    22 Be(3,1)

    =0.82 0.8

    33 Be(4,1)

    44 Be(5,1)

    54 Be(5,2)

    65 Be(6,2)

    76 Be(7,2)

    86 Be(7,3)

  • (=0.8)

    100 88

    1000 805

    10 7

  • Be(3, 9)102

    100 88

    1000 805

    107

    Be(3, 9)

  • l n u

    n

    l n() u u u

    n

  • l

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( )

    = argmax

    p | x n( )( ){ }

    = argmax

    P x n( ) |( )P x n( )( )

    p ( )"#$

    %$

    &'$

    ($

    = argmax

    P x n( ) |( ) p ( ){ }= argmax

    P x n( ) |( ){ }

    P(x(n)|)x

    p()=

  • p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( )

    P(x(n)|)x

    l

    = argmax

    p | x n( )( ){ }

    = argmax

    P x n( ) |( )P x n( )( )

    p ( )"#$

    %$

    &'$

    ($

    = argmax

    P x n( ) |( ) p ( ){ }= argmax

    P x n( ) |( ){ }

    p()=

  • 4

  • K

    K

  • P n;( ) = n!n1!!nm!

    1n1!m

    nm

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( ) =p ( )P x n( )( )

    1n1!1

    nm

    x(n) P x n( ) |( ) =1n1!mnm x(n)

  • P n;( ) = n!n1!!nm!

    1n1!m

    nm

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( ) =p ( )P x n( )( )

    1n1!1

    nm

    x(n) P x n( ) |( ) =1n1!mnm x(n) (=)

  • P n;( ) = n!n1!!nm!

    1n1!m

    nm

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( ) =p ( )P x n( )( )

    1n1!1

    nm

    x(n) P x n( ) |( ) =1n1!mnm x(n) (=)

    1

  • P n;( ) = n!n1!!nm!

    1n1!m

    nm

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( ) =p ( )P x n( )( )

    1n1!1

    nm

    x(n) P x n( ) |( ) =1n1!mnm x(n)

    1/Z2

    0k1 P(|x(n))1

    kk=1

    m

    =1

  • P n;( ) = n!n1!!nm!

    1n1!m

    nm

    x(n) P x n( ) |( ) =1n1!mnm x(n)

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( ) =p ( )P x n( )( )

    1n1!1

    nm =1Z2

    1n1!1

    nm

    0k1 P(|x(n))1

    kk=1

    m

    =1

    1

    Z2 = 1n1!1nmdDm

  • p | x n( )( ) = 1Z21

    n1!1nm

    111!1

    m1dDm =

    1( )! m( ) 1,!,m( )

    k = nk+1

    p | x n( )( ) = n+m( ) n1 +1( )! nm +1( )1

    n1!1nm

    ak = n+mk=1

    m

    http://www.cis.nagasaki-u.ac.jp/~masada/DirDistNorm.pdf

  • p | x n( )( ) = 1Z21

    n1!1nm

    111!1

    m1dDm =

    1( )! m( ) 1,!,m( )

    k = nk+1

    p | x n( )( ) = n+m( ) n1 +1( )! nm +1( )1

    n1!1nm

    ak = n+mk=1

    m

    k = nk+1

    http://www.cis.nagasaki-u.ac.jp/~masada/DirDistNorm.pdf

    Dir 1,!,m( ) = ( )

    1( )! m( )1

    11!1m1

  • p | x n( )( ) = 1Z21

    n1!1nm

    111!1

    m1dDm =

    1( )! m( ) 1,!,m( )

    k = nk+1

    p | x n( )( ) = n+m( ) n1 +1( )! nm +1( )1

    n1!1nm

    ak = n+mk=1

    m

    Dir 1,!,m( ) = ( )

    1( )! m( )1

    11!1m1

    k = nk+1

    http://www.cis.nagasaki-u.ac.jp/~masada/DirDistNorm.pdf

    m=2

  • 3 v1, v2, v3

    1,4,5 3?

    Dir 1,!,m( ) = ( )

    1( )! m( )1

    11!1m1 k = nk+1

    nkvk

  • 9 n1=1, n2=3, n3=7

    9 n1=7, n2=1, n3=1

    0 9 n1=3, n2=3, n3=3

    x(n)

  • l

    l

    p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( )

  • p | x n( )( ) =P x n( ) |( )P x n( )( )

    p ( )

    =1n1!1

    nm

    P x n( )( )

    ( ) 1( )! m( )

    111!1

    m1

    =1

    P x n( )( )

    ( ) 1( )! m( )

    11+n11!1

    m+nm1

    =1Z3

    11+n11!1

    m+nm1

    =Dir 1 + n1,!,m + nm( )

  • ==

  • l

    E [ ] = 1,!,k

    ,!,m

    !

    "#

    $

    %&

    V [ ] = 1 1( ) 2 +1( )

    ,!,k k( ) 2 +1( )

    ,!,m m( ) 2 +1( )

    !

    "##

    $

    %&&

    M [ ] = 1 1 m ,!,

    k 1 m ,!,

    m 1 m

    !

    "#

    $

    %&

    http://d.hatena.ne.jp/a_bicky/20100402/1270139105