非负矩阵分解证明

j
]2},
(1.4)
σi j is the weight of its corresponding observation,then:
L(B,C )
=
∑ [Xi
j
− (BC )i
j ]2.
ij
(1.5)
1
The maximum likelihood solution is to minimize the following function:
{B,C } = arg max p(X |B,C )
B ,C
= arg min[− log p(X |B,C )].
B ,C
(1.3)
Assuming that noise obeys Gauss distribution:
p(Xi
j
|B,C )
=
1 exp{− [
2
Xi
j
− (BC )i σi j
L(B,C ) = 1 ∑ [Xi j − (BC )i j ]2 2 + ∑ log(
2 ij
σi j
ij
2 ∗ πσi j ).
We assume that σi j =1 and ignore the effect of the constant,then:
LED (B,C )
=
ห้องสมุดไป่ตู้
∑ [Xi
j
− (BC )i
YOUR SCHOOL NAME
Proof of NMF
Author
November 20, 2013
1 INTRODUCTION TO ENSEMBLE RDC
A NMF model XN×M (x) is constructed as follows:
Xn×m = Bm×r Cr ×m + En×m .
En×m is the matrix of noise. Furthermore, the equation can be written as :
(1.1)
Xi j = (BC )i j + Ei j
(1.2)
In order to get the matrix of B,C,consider the following the maximum likelihood function solution:
(1.6) (1.7)
(1.8) (1.9) (1.10) (1.11) (1.12)
2
= −2[(B T X )k j
− (B T BC )k j ],
Then we can get the additive iteration as follows:
Bi k ←−, Bik + ϕik [(X C T )ik − (BCC T )i k ], Ck j ←−,Ck j + φk j [(B T X )k j − (B T BC )k j ].
j
]2.
ij
By using the matrix derivation principle:
d(X T Y )
dXT
dY T
=Y ∗
+X ∗
,
dB
dB
dB
Then ,we can get the result:
∂L E D ∂Bi k
= −2[(X C T )ik
− (BCC T )ik ],
∂L E D ∂Ck j