模式识别作业 11.1

1.

$$\begin{aligned}\because p(x|w_1)p(w_1)&=0.18 \\ p(x|w_2)p(w_2)&=0.04 \\ \therefore p(x|w_1)p(w_1) &> p(x|w_2)p(w_2) \\ \therefore x \in w_1 \end{aligned}$$

2.

$$\begin{aligned} r_1(x)=L_{11}p(x|w_1)p(w_1) & + L_{21}p(x|w_2)p(w_2)=0.24 \\ r_2(x)=L_{12}p(x|w_1)p(w_1) & + L_{22}p(x|w_2)p(w_2)=0.18 \\ r_1(x) &> r_2(x) \\ \therefore x \in w_2 \end{aligned}$$

3.

$$\begin{aligned} p(x|\theta)=&\frac{1}{\sqrt{(2\pi^m)|\Sigma|}}e^{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)} \\ H(\theta)=&\sum_{k=1}^{N}ln(p(x|\theta))\\ =&\sum_{k=1}^{N}(ln(2\pi|\Sigma|)-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)) \\ set\ the\ derivative\ to\ 0\\ \frac{\partial H(\theta)}{\partial \mu}&=-\sigma^{-1}\sum_{k=1}^{N}(x_k-\mu) =0 \\ \therefore \hat{\mu}&=\frac{1}{N}\sum_{k=1}^{N} x_k \\ \frac{\partial H(\theta)}{\partial \Sigma}&=N\Sigma^{-1}-\Sigma^{-1}\sum_{k=1}^{N}(x_k-\mu)(x_k-\mu)^T\Sigma^{-1}=0\\ \therefore \hat{\Sigma}&=\frac{1}{N}\sum_{k=1}^N(x_k-\hat{\mu})(x_k-\hat{\mu})^T \end{aligned}$$