考虑三维正态分布p(x∣ω)∼N(μ∣Σ)p({\bf x}\mid\omega)\sim N({\bf \mu}\mid\Sigma)p(x∣ω)∼N(μ∣Σ),其中μ=(122){\bf \mu}=\left( \begin{matrix}1\\2\\2\end{matrix} \right)μ=⎝⎛122⎠⎞及Σ=(100052025)\Sigma=\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 5 & 2 \\ 0 & 2 &5 \end{matrix} \right)Σ=⎝⎛100052025⎠⎞
(a) 求点x0=(0.5,0,1)Tx_0=(0.5,0,1)^Tx0=(0.5,0,1)T处的概率密度;
解:根据正态分布密度公式:
N(μ∣Σ)=1(2π)D/21∣Σ∣1/2exp{ −12(x−μ)TΣ−1(x−μ)}N({\bf \mu}\mid\Sigma)= \frac {1}{(2\pi)^{D/2}}\frac {1}{|\Sigma|^{1/2}}exp\left \{ \frac{-1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu) \right \}N(μ∣Σ)=(2π)D/21∣Σ∣1/21exp{ 2−1(x−μ)TΣ−1(x−μ)}
其中,由题目条件可知 x0−μ=(−0.5−2−1),∣Σ∣−1=(1000521−2210−221521),∣Σ∣=21x_0-\mu=\left( \begin{matrix}-0.5\\-2\\-1\end{matrix} \right),|\Sigma|^{-1}=\left( \begin{matrix} 1&0&0 \\ 0&\frac{5}{21}&-\frac{2}{21} \\ 0&-\frac{2}{21}&\frac{5}{21} \end{matrix} \right),|\Sigma|=21x0−μ=⎝⎛−0.5−2−1⎠⎞,∣Σ∣−1=⎝⎛1000215−2120−212215⎠⎞,∣Σ∣=21
因此,
p(x0∣μ)=1(2π)3/21∣Σ∣1/2exp{
−12(x0−μ)TΣ−1(x0−μ)}=1168π3exp{
−89168}\begin{aligned} p(x_0\mid\mu)&=\frac {1}{(2\pi)^{3/2}}\frac {1}{|\Sigma|^{1/2}}exp\left \{ \frac{-1}{2}(x_0-\mu)^T\Sigma^{-1}(x_0-\mu) \right \} \\ &=\frac{1}{\sqrt{168\pi^3}}exp \left \{ -\frac{89}{168} \right \} \end{aligned}p(x0∣μ)=(2π)3/21∣Σ∣1/21exp{
2−1(x0−μ)TΣ−1(x0−μ)}=168π31exp{
−16889}
(b) 构造白化变换矩阵Aω(Aω=ΦΛ−1/2)A_\omega(A_\omega=\Phi\Lambda^{-1/2})Aω(Aω=ΦΛ