核PCA投影平面公式推导

样本方差推导

样本方差公式$$S = \frac{1}{n-1}\sum_{i=1}n(x_i-\mu_i)2$$

扩展开来得到$$S = \frac{1}{n-1}[(X-\frac{1}{n}XTI_nI_nT)T(X-\frac{1}{n}XTI_nI_n^T)]$$

\[S = \frac{1}{n-1}X^T(I_n - \frac{1}{n}I_nI_n^T)(I_n - \frac{1}{n}I_nI_n^T)X \]

\(H = I_n - \frac{1}{n}I_nI_n^T\)得$$S = \frac{1}{n-1}X^THX$$

其中H为等幂矩阵HH=H和中心矩阵\(H_n*I_n = 0\)

核PCA推导

核函数:设X是输入空间(\(R^n\)的子集或离散子集),又F为特征空间(希尔伯特空间),如果存在一个从X到F的隐射$$\phi (X):X -> F$$使得对所有x,z\in X,函数K(x,z)满足条件$$K(x,z) = \phi (x)\bullet \phi (z)$$

下面推导F投影到的主成分定义的平面,根据F样本方差的特征值分解得(为推导方便去掉前面的(\(\frac{1}{n-1}\))$$F^THFV_i = \lambda _i V_i$$由于H为等逆矩阵,则$$F^THHFV_i = \lambda _i V_i$$

由于想得到F很难,我们换一种思路将求F转移求K上,根据AAT与ATA的关系:非零特质值相同,得到$$HFF^THU_i = \lambda _iU_i $$

两边同时乘以\(F^TH\)得到$$FTHHFFTHU_i = \lambda _iF^THU_i$$

从上式可以得到\(F^THU_i\)\(F^THHF\)的特征向量

\(F^THU_i\)进行归一化$$U_{normal} = \frac{FTHU_i}{{||U_iTHFF^THU_i||}_2}$$

由于\(HFF^TH = HKH = \lambda _i\),则$$U_{normal} = \lambda {-\frac{1}{2}}FTHU_i$$

F投影到\(U_normal\)定义的平面$$P = F_{center} U_{normal}$$

\[P= (F-\frac{1}{n}\sum_{i=1}^nF_i)(\lambda ^{-\frac{1}{2}}F^THU_i) \]

\[P= (F-\frac{1}{n}F^TI_n)(\lambda ^{-\frac{1}{2}}F^THU_i) \]

\[P= \lambda ^{-\frac{1}{2}}(K - \frac{1}{n}K(x,x_i))HU_i \]

posted @ 2019-04-22 13:45  Fate0729  阅读(597)  评论(0编辑  收藏  举报