基于联合贝叶斯的人脸验证

特征表示 人脸由两个高斯变量的和表征：$x=\mu +\epsilon$ 这里$x$代表人脸，$\mu$代表固有身份，$\epsilon$代表脸部变化（光照，姿态，表情等）。隐变量$\mu$代表固有身份，$\epsilon$服从高斯分布$N(0,\Sigma \epsilon )$,$N\left(0,S_{\epsilon }\right)$ 以照片$x_i$和$x_j$为例,$x_i={\mu }_i+{ϵ}_i,x_j={\mu }_j+{ϵ}_j,i,j\in \{1,2\}$ $\begin{array}{rl}\mathrm{cov}\left(x_i,x_j\right)& =\mathrm{cov}\left({\mu }_i,{\mu }_j\right)+\mathrm{cov}\left({\mu }_i,{ϵ}_j\right)+\mathrm{cov}\left({ϵ}_i,{\mu }_j\right)+\mathrm{cov}\left({ϵ}_i,{ϵ}_j\right)\\ & =\mathrm{cov}\left({\mu }_i,{\mu }_j\right)+0+0+\mathrm{cov}\left({ϵ}_i,{ϵ}_j\right)(\text{ Because }ϵ\text{ and }\mu \text{ are independent })\\ & =\mathrm{cov}\left({\mu }_i,{\mu }_j\right)+\mathrm{cov}\left({ϵ}_i,{ϵ}_j\right)\end{array}$ 判断依据推导 $p\left(x_1,x_2\mid H_I\right)\sim \mathcal{N}\left(0,{\Sigma }_I\right),p\left(x_1,x_2\mid H_E\right)\sim \mathcal{N}\left(0,{\Sigma }_E\right)$ 在HI假设下，即两张脸属于同一个人，本征变量μ1,μ2是相同的，而且ϵ1,ϵ2是相互独立的 $${\Sigma }_I=\left[\begin{array}{cc}{S}_{\mu }+S_{ϵ}& S_{\mu }\\ S_{\mu }& S_{\mu }+S_{ϵ}\end{array}\right]$$ 在HE假设下，μ,ϵ都是独立的 $${\Sigma }_E=\left[\begin{array}{cc}{S}_{\mu }+S_{ϵ}& 0\\ 0& S_{\mu }+S_{ϵ}\end{array}\right]$$ $${\Sigma }_I^{-1}(仅仅作为标记)=\left(\begin{array}{cc}F+G& G\\ G& F+G\end{array}\right)$$ $${\Sigma }_E^{-1}=\left(\begin{array}{cc}{\left(S_{\mu }+S_{ϵ}\right)}^{-1}& 0\\ 0& {\left(S_{\mu }+S_{ϵ}\right)}^{-1}\end{array}\right)$$ ${\Sigma }_E$，${\Sigma }_I$带入高斯公式可得： $$\begin{array}{l}p\left(x_1,x_2\mid H_I\right)=\frac{1}{(2\pi {)}^{m/2}{|{\Sigma }_I|}^{\frac{1}{2}}}\exp \left(-\frac{1}{2}\left(x_1{x}_2\right){\Sigma }_I^{-1}\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\right)\\ p\left(x_1,x_2\mid H_E\right)=\frac{1}{(2\pi {)}^{m/2}{|{\Sigma }_E|}^{\frac{1}{2}}}\exp \left(-\frac{1}{2}\left(x_1{x}_2\right){\Sigma }_E^{-1}\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\right)\end{array}$$ 可以得到： $$r\left(x_1,x_2\right)=\log \frac{p\left(x_1,x_2\mid H_I\right)}{p\left(x_1,x_2\mid H_E\right)}$$ $$=\log \frac{{\left|{\Sigma }_I\right|}^{-\frac{1}{2}}\exp \left(-\frac{1}{2}\left(x_1{x}_2\right){\Sigma }_I^{-1}\left(\begin{array}{l}{x}_1\\ x_2\end{array}\right)\right)}{{\left|{\Sigma }_E\right|}^{-\frac{1}{2}}\exp \left(-\frac{1}{2}\left(x_1{x}_2\right){\Sigma }_E^{-1}\left(\begin{array}{l}{x}_1\\ x_2\end{array}\right)\right)}$$ $$\begin{array}{rl}& =\log [\exp \left(-\frac{1}{2}\left(x_1{x}_2\right)\left(\begin{array}{c}{\Sigma }_E^{-1}-{\Sigma }_I^{-1})\\ x_2\end{array}\right)\right)\cdot \frac{{\left|{\Sigma }_I\right|}^{-\frac{1}{2}}}{{\left|{\Sigma }_E\right|}^{-\frac{1}{2}}}\\ & =-\frac{1}{2}\left(\begin{array}{ll}{x}_1& x_2\end{array}\right)\left(\begin{array}{l}{\Sigma }_E^{-1}-{\Sigma }_I^{-1})\\ x_2\end{array}\right)+C_1\left(\text{ here }{C}_1=\log \frac{{\left|{\Sigma }_I\right|}^{-\frac{1}{2}}}{{\left|{\Sigma }_E\right|}^{-\frac{1}{2}}}\right)\\ & =\frac{1}{2}\left(\begin{array}{ll}{x}_1& x_2\end{array}\right)\left(\left(\begin{array}{cc}{\left(S_{\mu }+S_{ϵ}\right)}^{-1}& 0\\ 0& {\left(S_{\mu }+S_{ϵ}\right)}^{-1}\end{array}\right)-\left(\begin{array}{cc}F+G& G\\ G& F+G\end{array}\right)\right)\left(\begin{array}{l}{x}_1\\ x_2\end{array}\right)+C_1\\ & =\frac{1}{2}{\left(\begin{array}{ll}{x}_1& x_2\end{array}\right)}^T\left(\begin{array}{cc}A& -G\\ -G& A\end{array}\right)\left(\begin{array}{ll}{x}_1& x_2\end{array}\right)+C_1\\ & =\frac{1}{2}\left(x_1^{T}A{x}_1-2x_1^{T}G{x}_2+x_2^T{x}_2\right)+C_1\end{array}$$ 可得$$A={\left(S_{\mu }+S_{ϵ}\right)}^{-1}-(F+G)$$ 推导${\Sigma }_x$ $$x_1=\mu +{ϵ}_1,x_2=\mu +{ϵ}_2,\dots ,x_m=\mu +{ϵ}_m$$ $$x={\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_m\end{array}\right)}_{m\ast 1},P={\left(\begin{array}{ccccc}1& 1& 0& \dots & 0\\ 1& 0& 1& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& 0& 0& \dots & 1\end{array}\right)}_{(m)\ast (m+1)},h={\left(\begin{array}{c}{\mu }_1\\ {ϵ}_2\\ ⋮\\ {ϵ}_m\end{array}\right)}_{(m+1)\ast 1}$$ $$x=Ph$$ $$h\sim \mathcal{N}\left(0,{\Sigma }_h\right),{\Sigma }_h=\left(\begin{array}{cccc}{S}_{\mu }& 0& \cdots & 0\\ 0& S_{ϵ}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & S_{ϵ}\end{array}\right)$$ 依据$\mathrm{Cov}[\mathbf{A}\mathbf{x},\mathbf{B}\mathbf{y}]=\mathbf{A}\mathrm{Cov}[\mathbf{x},\mathbf{y}]{\mathbf{B}}^T$ $$x\sim \mathcal{N}\left(0,{\Sigma }_x\right),{\Sigma }_x=P{\Sigma }_h{P}^T=\left(\begin{array}{cccc}{S}_{\mu }+S_{ϵ}& S_{\mu }& \dots & S_{\mu }\\ S_{\mu }& S_{\mu }+S_{ϵ}& \dots & S_{\mu }\\ ⋮& ⋮& \ddots & ⋮\\ S_{\mu }& S_{\mu }& \dots & S_{\mu }+S_{ϵ}\end{array}\right)$$ 关于推导

依据： $$\begin{array}{l}\text{ Assume }\mathbf{x}\sim {\mathcal{N}}_{\mathbf{x}}(\mu ,\mathbf{\Sigma })\text{ where }\\ \mathbf{x}=\left[\begin{array}{c}{\mathbf{x}}_a\\ {\mathbf{x}}_b\end{array}\right]\mathbf{\mu }=\left[\begin{array}{c}{\mathbf{\mu }}_a\\ {\mathbf{\mu }}_b\end{array}\right]\mathbf{\Sigma }=\left[\begin{array}{cc}{\mathbf{\Sigma }}_a& {\mathbf{\Sigma }}_c\\ {\mathbf{\Sigma }}_c^T& {\mathbf{\Sigma }}_b\end{array}\right]\end{array}$$ $$p\left({\mathbf{x}}_a\mid {\mathbf{x}}_b\right)={\mathcal{N}}_{{\mathbf{x}}_a}\left({\hat{\mu }}_a,{\hat{\mathbf{\Sigma }}}_a\right)\{\begin{array}{l}{\hat{\mathbf{\mu }}}_a={\mathbf{\mu }}_a+{\mathbf{\Sigma }}_c{\mathbf{\Sigma }}_b^{-1}\left({\mathbf{x}}_b-{\mathbf{\mu }}_b\right)\\ {\hat{\mathbf{\Sigma }}}_a={\mathbf{\Sigma }}_a-{\mathbf{\Sigma }}_c{\mathbf{\Sigma }}_b^{-1}{\mathbf{\Sigma }}_c^T\end{array}$$ 因此 求解F,G 根据相乘玩对角线为一，非对角线为零可得

流程图： 1.初始化 2，EMlike算法 E步： M步：更新参数{Sμ,Sϵ}

3.联合贝叶斯判据