Fisher矩阵和Hessian矩阵的关系:证明Fisher为负对数似然函数的Hessian的期望
证明Fisher等于Expectation of Hessian of Negative log likelihood.
符号约定
-
f θ ( ⋅ ) f_{\theta}(\cdot) fθ(⋅): 概率密度
-
p ( x ∣ θ ) = p θ ( x ) = ∏ i N f θ ( x i ) p(x|\theta) = p_{\theta}(x) = \prod\limits_i^N f_{\theta}(x_i) p(x∣θ)=pθ(x)=i∏Nfθ(xi): 似然函数
-
s ( θ ) = ∇ θ p θ ( x ) s(\theta) = \nabla_{\theta} \ p_{\theta}(x) s(θ)=∇θ pθ(x): score function,即似然函数的梯度。
-
I = E p θ ( x ) [ ( ∇ θ l o g p θ ( x ) ) ( ∇ θ l o g p θ ( x ) ) T ] I = E_{p_{\theta}(x)}[(\nabla_{\theta} log p_{\theta}(x))(\nabla_{\theta} log p_{\theta}(x))^T] I=Epθ(x)[(∇θlogpθ(x))(∇θlogpθ(x))T]: Fisher矩阵。
-
I i , j ( θ ) = E p θ ( x ) [ ( D i l o g p θ ( x ) ) ( D j l o g p θ ( x ) ) ] I_{i,j}(\theta) = E_{p_{\theta}(x)}[(D_i log p_{\theta}(x))(D_j log p_{\theta}(x))] Ii,j(θ)=Epθ(x)[(Dilogpθ(x))(Djlogpθ(x))]: 为Fisher的第i行第j列元素。其中 D i = ∂ ∂ θ i ; D i , j = ∂ ∂ θ i ∂ θ j D_i = \frac{\partial}{\partial{\theta_i}}; \ D_{i,j} = \frac{\partial}{\partial{\theta_i} \partial{\theta_j}} Di=∂θi∂; Di,j=∂θi∂θj∂。
-
H i , j = D i , j l o g P θ ( x ) H_{i,j} = D_{i,j} log P_{\theta}(x) Hi,j=Di,jlogPθ(x): Hessian矩阵的第i行第j列元素。
证明
证明目标:
I
i
,
j
(
θ
)
=
−
E
p
θ
(
x
)
[
H
i
,
j
]
I_{i,j}(\theta) = -E_{p_{\theta}(x)}[ H_{i,j} ]
Ii,j(θ)=−Epθ(x)[Hi,j]
从
H
i
,
j
H_{i,j}
Hi,j入手。
H
i
,
j
=
D
i
,
j
l
o
g
P
θ
(
x
)
=
D
i
(
D
j
p
θ
(
x
)
p
θ
(
x
)
)
=
(
D
i
,
j
p
θ
(
x
)
)
⋅
p
θ
(
x
)
−
D
i
p
θ
(
x
)
D
j
p
θ
(
x
)
p
θ
2
(
x
)
=
D
i
,
j
p
θ
(
x
)
p
θ
(
x
)
−
D
i
p
θ
(
x
)
p
θ
(
x
)
D
j
p
θ
(
x
)
p
θ
(
x
)
\begin{align*} H_{i,j} & = D_{i,j} log P_{\theta}(x) \\ & = D_i(\frac{ D_j p_{\theta}(x) }{ p_{\theta}(x) }) \\ & = \frac{(D_{i,j}p_{\theta}(x)) \cdot p_{\theta}(x) - D_i p_{\theta}(x) D_j p_{\theta}(x)} {p_{\theta}^2(x)} \\ & = \frac{D_{i,j}p_{\theta}(x)}{p_{\theta}(x)} - \frac{D_{i}p_{\theta}(x)}{p_{\theta}(x)}\frac{D_{j}p_{\theta}(x)}{p_{\theta}(x)} \end{align*}
Hi,j=Di,jlogPθ(x)=Di(pθ(x)Djpθ(x))=pθ2(x)(Di,jpθ(x))⋅pθ(x)−Dipθ(x)Djpθ(x)=pθ(x)Di,jpθ(x)−pθ(x)Dipθ(x)pθ(x)Djpθ(x)
故右式:
− E p θ ( x ) ( H i , j ) = − E p θ ( x ) [ D i , j p θ ( x ) p θ ( x ) ] + E p θ ( x ) [ ( D i p θ ( x ) p θ ( x ) ) ⋅ ( D j p θ ( x ) p θ ( x ) ) ] \begin{align*} -E_{p_{\theta}(x)}( H_{i,j} ) & = -E_{p_{\theta}(x)}[ \frac{D_{i,j}p_{\theta}(x)}{p_{\theta}(x)}] + E_{p_{\theta}(x)}[(\frac{D_i p_{\theta}(x)}{p_{\theta}(x)}) \cdot (\frac{D_j p_{\theta}(x)}{p_{\theta}(x)})] \end{align*} −Epθ(x)(Hi,j)=−Epθ(x)[pθ(x)Di,jpθ(x)]+Epθ(x)[(pθ(x)Dipθ(x))⋅(pθ(x)Djpθ(x))]
其中:
E
p
θ
(
x
)
(
D
i
,
j
p
θ
(
x
)
p
θ
(
x
)
)
=
∫
D
i
,
j
p
θ
(
x
)
p
θ
(
x
)
⋅
p
θ
(
x
)
⋅
d
x
=
D
i
,
j
∫
p
θ
(
x
)
⋅
d
x
(
积分求导换序
)
=
D
i
,
j
1
(
对常数求导,为0
)
=
0
\begin{align*} E_{p_{\theta}(x)}( \frac{D_{i,j}p_{\theta}(x)}{p_{\theta}(x)}) & = \int \frac{D_{i,j}p_{\theta}(x)}{p_{\theta}(x)} \cdot p_{{\theta}(x)} \cdot dx \\ & = D_{i,j} \int {p_{\theta}(x) \cdot dx} \qquad & (\text{积分求导换序}) \\ & = D_{i,j} 1 \qquad & (\text{对常数求导,为0}) \\ & = 0 \end{align*}
Epθ(x)(pθ(x)Di,jpθ(x))=∫pθ(x)Di,jpθ(x)⋅pθ(x)⋅dx=Di,j∫pθ(x)⋅dx=Di,j1=0(积分求导换序)(对常数求导,为0)
且根据复合函数求导可知:
D
i
p
θ
(
x
)
p
θ
(
x
)
=
D
i
l
o
g
p
θ
(
x
)
\frac{D_i p_{\theta}(x)}{p_{\theta}(x)} = D_i log p_{\theta}(x)
pθ(x)Dipθ(x)=Dilogpθ(x)
故右式为:
E
p
θ
(
x
)
[
(
D
i
p
θ
(
x
)
p
θ
(
x
)
)
⋅
(
D
j
p
θ
(
x
)
p
θ
(
x
)
)
]
=
E
p
θ
(
x
)
[
(
D
i
l
o
g
p
θ
(
x
)
)
(
D
j
l
o
g
p
θ
(
x
)
)
]
=
I
i
,
j
(
θ
)
\begin{align*} & E_{p_{\theta}(x)}[(\frac{D_i p_{\theta}(x)}{p_{\theta}(x)}) \cdot (\frac{D_j p_{\theta}(x)}{p_{\theta}(x)})] = E_{p_{\theta}(x)}[(D_i log p_{\theta}(x))(D_j log p_{\theta}(x))] \\ & = I_{i,j}(\theta) \end{align*}
Epθ(x)[(pθ(x)Dipθ(x))⋅(pθ(x)Djpθ(x))]=Epθ(x)[(Dilogpθ(x))(Djlogpθ(x))]=Ii,j(θ)
得证
实际应用中,计算 H H H非常复杂,但是计算 I I I并将其作为 H H H的近似值是比较容易的,一些剪枝方法中就利用了这一点,如NAP [Network Automatic Pruning Start NAP and Take a Nap](基于OBS,OBD)
参考链接:
https://zhuanlan.zhihu.com/p/546885304?utm_psn=1840735001693523969
https://zhuanlan.zhihu.com/p/546885304?utm_psn=1840431492376969216
https://jaketae.github.io/study/fisher/
https://mark.reid.name/blog/fisher-information-and-log-likelihood.html
https://bobondemon.github.io/2022/01/07/Score-Function-and-Fisher-Information-Matrix/