概
本文讨论identifiability的问题, 即
p θ ( x ) = p θ ~ ( x ) ⇒ θ = θ ~ p_{\theta}(x) = p_{\tilde{\theta}}(x) \Rightarrow \theta = \tilde{\theta}pθ(x)=pθ~(x)⇒θ=θ~
在何种情况下能够成立, 或者近似成立.
主要内容
假设观测数据x xx和隐变量z zz满足联合分布:
p θ ∗ ( x , z ) = p θ ∗ ( x ∣ z ) p θ ∗ ( z ) , p_{\theta^*}(x, z) = p_{\theta^*}(x|z) p_{\theta^*}(z),pθ∗(x,z)=pθ∗(x∣z)pθ∗(z),
因为隐变量是未知的, 所以我们接触到的实际上只有边际分布
p θ ∗ ( x ) = ∫ z p θ ∗ ( x , z ) d z . p_{\theta^*}(x) = \int_z p_{\theta^*}(x, z)\mathrm{d}z.pθ∗(x)=∫zpθ∗(x,z)dz.
在实际估计参数θ \thetaθ的时候, 很有可能发生:
p θ ( x ) = p θ ~ ( x ) ≈ p θ ∗ ( x ) , θ ≠ θ ~ . p_{\theta}(x) = p_{\tilde{\theta}}(x) \approx p_{\theta^*}(x), \: \theta \not = \tilde{\theta}.pθ(x)=pθ~(x)≈pθ∗(x),θ=θ~.
即两个不同的联合分布p θ ( x , z ) , p θ ~ ( x , z ) p_{\theta}(x, z), p_{\tilde{\theta}}(x, z)pθ(x,z),pθ~(x,z)但是却对应着同一个边际分布, 这就identifiability的问题.
在经典的VAE框架中, 已经有工作指出, 无监督下, 即仅凭观测数据x xx, 是无法保证identifiability的.
本文的模型
本文需要用到一些额外的信息u uu, 考虑如下分布:
p θ ( x , z ∣ u ) = p f ( x ∣ z ) p T , λ ( z ∣ u ) , θ = ( f , T , λ ) . p_{\theta}(x, z|u) = p_f(x|z) p_{T,\lambda}(z|u), \: \theta = (f, T, \lambda).pθ(x,z∣u)=pf(x∣z)pT,λ(z∣u),θ=(f,T,λ).
注: x ∈ R d , z ∈ R n , u ∈ R m x \in \mathbb{R}^d, z \in \mathbb{R}^n, u \in \mathbb{R}^mx∈Rd,z∈Rn,u∈Rm.
其中,
x = f ( z ) + ϵ ⇒ p f ( x ∣ z ) = p ϵ ( x − f ( z ) ) . x = f(z) + \epsilon \Rightarrow p_{f}(x|z) = p_{\epsilon}(x - f(z)).x=f(z)+ϵ⇒pf(x∣z)=pϵ(x−f(z)).
p T , λ ( z ∣ u ) = ∏ i Q i ( z i ) Z i ( u ) exp [ ∑ j = 1 k T i , j ( z i ) λ i , j ( u ) ] , p_{T, \lambda}(z|u) = \prod_{i}\frac{Q_i(z_i)}{Z_i(u)}\exp [\sum_{j=1}^k T_{i,j}(z_i) \lambda_{i,j}(u)],pT,λ(z∣u)=i∏Zi(u)Qi(zi)exp[j=1∑kTi,j(zi)λi,j(u)],
即假设先验z ∣ μ z|\muz∣μ满足的是指数族的分布.
套用VAE的框架:
- encoder:
T ^ , λ ^ = h ( x , u ; ϕ ) , z ∼ p T ^ , λ ^ ( z ∣ u ) . \hat{T}, \hat{\lambda} = h(x, u;\phi), z \sim p_{\hat{T}, \hat{\lambda}}(z|u).T^,λ^=h(x,u;ϕ),z∼pT^,λ^(z∣u). - decoder:
x ^ = f ( z ) + ϵ . \hat{x} = f(z) + \epsilon.x^=f(z)+ϵ.
既估计的后验分布为q ϕ ( z ∣ x , μ ) q_{\phi}(z|x,\mu)qϕ(z∣x,μ), 则ELBO:
E q D ( x , u ) [ E q ϕ ( z ∣ x , u ) [ log p θ ( x , z ∣ u ) − log q ϕ ( z ∣ x , u ) ] ] . \mathbb{E}_{q_{\mathcal{D}(x,u)}}[\mathbb{E}_{q_{\phi}(z|x,u)}[\log p_{\theta}(x, z|u) - \log q_{\phi}(z|x,u)]].EqD(x,u)[Eqϕ(z∣x,u)[logpθ(x,z∣u)−logqϕ(z∣x,u)]].
Identifiability
∼ \sim∼定义: 定义∼ \sim∼等价关系如下:
( f , T , λ ) ∼ ( f ~ , T ~ , λ ~ ) ⇔ ∃ A , c , s . t . T ( f − 1 ( x ) ) = A T ~ ( f ~ − 1 ( x ) ) + c , ∀ x ∈ X , (f, T, \lambda) \sim (\tilde{f}, \tilde{T}, \tilde{\lambda}) \Leftrightarrow \\ \exist A, c, \: \mathrm{s.t.} \: T(f^{-1}(x)) = A\tilde{T}(\tilde{f}^{-1}(x)) + c, \forall x \in \mathcal{X},(f,T,λ)∼(f~,T~,λ~)⇔∃A,c,s.t.T(f−1(x))=AT~(f~−1(x))+c,∀x∈X,
其中A ∈ R n k × n k A \in \mathbb{R}^{nk \times nk}A∈Rnk×nk. 若A AA还是个可逆矩阵, 则
( f , T , λ ) ∼ A ( f ~ , T ~ , λ ~ ) . (f, T, \lambda) \sim_{A} (\tilde{f}, \tilde{T}, \tilde{\lambda}).(f,T,λ)∼A(f~,T~,λ~).
显然, 如果
p θ ( x ∣ u ) = p θ ~ ( x ∣ u ) ⇒ θ ∼ A θ ~ , p_{\theta}(x|u) = p_{\tilde{\theta}}(x|u) \Rightarrow \theta \sim_A \tilde{\theta},pθ(x∣u)=pθ~(x∣u)⇒θ∼Aθ~,
那么可以说是在线性变换允许范围内是identifiable的.
接下来给出的定理说明了什么时候θ , θ ~ \theta, \tilde{\theta}θ,θ~是∼ A \sim_A∼A-identifiable的.
定理: 在前述定义的模型下, 对于θ = ( f , T , λ ) \theta = (f, T, \lambda)θ=(f,T,λ), 以及任意θ ~ = ( f ~ , T ~ , λ ~ ) \tilde{\theta} =(\tilde{f}, \tilde{T}, \tilde{\lambda})θ~=(f~,T~,λ~)满足
p θ ( x ∣ u ) = p θ ~ ( x ∣ u ) , a . e . , p_{\theta}(x|u)= p_{\tilde{\theta}}(x|u), \: a.e.,pθ(x∣u)=pθ~(x∣u),a.e.,
若一下条件成立, 则θ ∼ A θ ~ \theta \sim_A \tilde{\theta}θ∼Aθ~:
若φ ϵ \varphi_{\epsilon}φϵ为p ϵ p_{\epsilon}pϵ的特征函数(这里即为对于的傅里叶变换), 且φ ϵ ≠ 0 , a . e . \varphi_{\epsilon} \not = 0, \: a.e.φϵ=0,a.e..
f ff是一个单射.
T i , j T_{i, j}Ti,j几乎处处可微, 且( T i j ) j ( x ) (T_{ij})_j(x)(Tij)j(x)线性独立, 即
∑ j k α i j T i , j ( x ) = c i , ∀ x , ⇒ c i = 0 , α i j = 0 , ∀ j , \sum_j^k \alpha_{ij} T_{i, j}(x) = c_{i}, \forall x, \Rightarrow c_i = 0, \alpha_{ij} = 0, \forall j,j∑kαijTi,j(x)=ci,∀x,⇒ci=0,αij=0,∀j,
对于i = 1 , … , n i=1,\ldots, ni=1,…,n均成立.存在不同的点u 0 , ⋯ , u n k u^0, \cdots, u^{nk}u0,⋯,unk, 使得
L = ( λ ( u 1 ) − λ ( u 0 ) , ⋯ , λ ( u n k ) − λ ( u 0 ) ) ∈ R n k × n k . L = (\lambda(u_1) - \lambda(u_0), \cdots, \lambda(u_{nk}) - \lambda(u_0)) \in \mathbb{R}^{nk \times nk}.L=(λ(u1)−λ(u0),⋯,λ(unk)−λ(u0))∈Rnk×nk.
可逆.
证明流程:
利用条件1, 2证明
p T , λ ( f − 1 ( x ) ∣ u ) v o l J f − 1 ( x ) = p T ~ , λ ~ ( f − 1 ( x ) ∣ u ) v o l J f ~ − 1 ( x ) . p_{T,\lambda}(f^{-1}(x)|u) \mathrm{vol} J_{f^{-1}}(x) =p_{\tilde{T},\tilde{\lambda}}(f^{-1}(x)|u) \mathrm{vol} J_{\tilde{f}^{-1}}(x).pT,λ(f−1(x)∣u)volJf−1(x)=pT~,λ~(f−1(x)∣u)volJf~−1(x).
利用条件4证明
T ( f − 1 ( x ) ) = A T ~ ( f ~ − 1 ( x ) ) + c , A = L − T L ~ T . T(f^{-1}(x)) = A\tilde{T}(\tilde{f}^{-1}(x)) + c, \: A = L^{-T}\tilde{L}^T.T(f−1(x))=AT~(f~−1(x))+c,A=L−TL~T.
利用条件3证明A AA可逆.
注: 显然条件四一定程度熵说明了为什么无监督不行(因为其相当于λ ( u ) \lambda(u)λ(u)为常数).
注: 关于引理2的证明我有疑问, 我认为应当这般证明:
令X i = { x ∈ R , T i ′ ( x ) = 0 } \mathcal{X}_i = \{x \in \mathbb{R}, T_i'(x) = 0\}Xi={x∈R,Ti′(x)=0}, 取θ i ≠ 0 , θ j = 0 , j ≠ i \theta_i\not=0, \theta_j = 0, j\not=iθi=0,θj=0,j=i, 则
⟨ T ′ ( x ) , θ ⟩ = 0 , ∀ x ∈ X i ⇒ ⟨ T ( x ) , θ ⟩ = c o n s t , \langle T'(x), \theta \rangle = 0, \forall x \in \mathcal{X_i} \Rightarrow \langle T(x), \theta \rangle = \mathrm{const},⟨T′(x),θ⟩=0,∀x∈Xi⇒⟨T(x),θ⟩=const,
由定义知X i \mathcal{X}_iXi的测度为0.
注: 本文还有一些别的identifiability的讨论, 这里不多赘述.