统计学习方法HMM作业

最新推荐文章于 2021-09-01 15:14:03 发布

Little_sky_jty

最新推荐文章于 2021-09-01 15:14:03 发布

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分类专栏：笔记

本文链接：https://blog.youkuaiyun.com/weixin_40805392/article/details/105654398

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10.1

根据后向算法：

计算初值
$\beta_{4}(i)=1, i=1,2,3$
递推计算：
$\begin{aligned} &\text {对于} t=3,2,1 \beta_{i}(i)=\sum_{i=1} a_{i j} b_{j}\left(o_{i+1}\right) \beta_{i+1}(j), i=1,2,3\\ &\beta_{3}(1)=\sum_{i=1}^{3} a_{1 j} b_{j}\left(o_{4}\right) \beta_{4}(j)=(0.5 * 0.5+0.2 * 0.6+0.3 * 0.3) * 1=0.46\\ &\beta_{3}(2)=\sum_{j=1}^{3} a_{2 j} b_{j}\left(o_{4}\right) \beta_{4}(j)=(0.3 * 0.5+0.5 * 0.6+0.2 * 0.3) * 1=0.51\\ &\beta_{3}(3)=\sum_{j=1}^{3} a_{3 j} b_{j}\left(o_{4}\right) \beta_{4}(j)=(0.2 * 0.5+0.3 * 0.6+0.5 * 0.3) * 1=0.43\\ &\beta_{2}(1)=\sum_{i=1}^{3} a_{1 j} b_{j}\left(o_{3}\right) \beta_{3}(j)=(0.5 * 0.5+0.2 * 0.4+0.3 * 0.7) * 0.46=0.248\\ &\beta_{2}(2)=\sum_{j=1}^{3} a_{2 j} b_{j}\left(o_{3}\right) \beta_{3}(j)=(0.3 * 0.5+0.5 * 0.4+0.2 * 0.7) * 0.51=0.249\\ &\beta_{2}(3)=\sum_{i=1}^{3} a_{3} b_{j}\left(o_{3}\right) \beta_{3}(j)=(0.2 * 0.5+0.3 * 0.4+0.5 * 0.7) * 0.43=0.245\\ &\beta_{1}(1)=\sum_{j=1}^{3} a_{1 j} b_{j}\left(o_{2}\right) \beta_{2}(j)=(0.5 * 0.5+0.2 * 0.6+0.3 * 0.3) * 0.2484=0.114264\\ &\beta_{1}(2)=\sum_{j_{-1}}^{3} a_{2 j} b_{j}\left(o_{2}\right) \beta_{2}(j)=(0.3 * 0.5+0.5 * 0.6+0.2 * 0.3) * 0.2499=0.127449\\ &\beta_{1}(3)=\sum_{i=1}^{3} a_{3 j} b_{j}\left(o_{2}\right) \beta_{2}(j)=(0.2 * 0.5+0.3 * 0.6+0.5 * 0.3) * 0.2451=0.1053 \end{aligned}$
计算最终 $\lambda)=\sum_{i=1}^{3} \pi_{i} b_{i}\left(o_{1}\right) \beta_{1}(i)$ 的值：
$\lambda)=\sum_{i=1}^{3} \pi_{i} b_{i}\left(o_{1}\right) \beta_{1}(i)=0.2 * 0.5 * 0.114264+0.4 * 0.4 * 0.127449+0.4 * 0.7 * 0.105393=0.0613$

10.2

根据公式有：
$P\left(i_{4}=q_{3} | O, \lambda\right)=\gamma_{4}(3)=\frac{\alpha_{4}(3) \beta_{4}(3)}{\sum_{j=1}^{3} \alpha_{4}(j) \beta_{4}(j)}$
1.根据前向计算 $\alpha$ :

$\begin{aligned} &\text { } \alpha_{1}(1)=\pi_{1} b_{1}\left(0_{1}\right)=0.10\\ &\begin{array}{l} \alpha_{1}(1)=\pi_{2} b_{2}(q)=0.12 \\ \alpha_{1}(1)=\pi_{3} b_{3}(0,1)=0.35 \\ \alpha_{2}(1)=\left[\sum_{17}^{3} \alpha_{1}(i) a_{i 1}\right] b_{1}\left(0_{2}\right)=0.156 \times 0.5=0.078 \\ \alpha_{2}(2)=\left[\sum_{i=1}^{3} \alpha_{1}(i) a_{i 2}\right] b_{2}\left(0_{1}\right)=0.14 \times 0.6=0.084 \\ \alpha_{2}(3)=\left[\sum_{i=1}^{3} \alpha_{1}(i) a_{i 3}\right] b_{3}\left(0_{2}\right)=0.274 \times 0.3=0.882 \end{array}\\ &\alpha_{3}(i)=\left[\sum_{i=1}^{3} \alpha_{2}(i) a_{i i}\right] b_{i}\left(0_{3}\right)=0.05004 \times 0.5=0.04032\\ &\alpha_{3}(x)=\left[\sum_{i=1}^{3} \alpha_{2}(i) a_{i 2}\right] b_{2}\left(\theta_{i}\right)=0.06624 \times 04=0.02650\\ &\alpha_{3}(3)=\left[\sum_{i=1}^{3} \alpha_{2}(i) a_{i j}\right] b_{3}\left(0_{3}\right)=0.09732 \times a_{1}=0.06812\\ &\alpha_{4}(1)=\left[\sum_{i=1}^{3} \alpha_{3}(i) a_{i 1}\right] b_{q_{i}}\left(0_{4}\right)=0.04173 \times 0.5=0.02087\\ &\alpha_{4}(2)=\left[\sum_{i=1}^{3} \alpha_{3}(i) a_{i i}\right] b_{i}\left(0_{4}\right)=0.03091 \times 0.9=0.01236\\ &\alpha_{4}(3)=\left[\sum_{i=1}^{3} \alpha_{3}(i) a_{i 3}\right] b_{i j}\left(0_{4}\right)=0.0623 \times 0.7=0.04361 \end{aligned}$

根据后向计算 $\beta$ ：

$\begin{aligned} &\beta_{8}(i)=1 . \quad i=1.2 \cdot N\\ \\ &\beta_{7}(1)=\sum_{j=1}^{3} a_{1 j} b_{j}\left(0_{8}\right) \cdot \beta_{8}(j)=0.43\\ &\beta_{7}(2)=\sum_{j=1}^{3} a_{2 j} b_{j}\left(0_{8}\right) \cdot \beta_{8}(j)=0.51\\ &\beta_{7}(3)=\sum_{j=1}^{3} a_{3 j} b_{j}\left(0_{8}\right) \cdot \beta_{8}(j)=0.40\\ \\ &\beta_{6}(1)=\sum_{j=1}^{3} a_{1 j} b_{j}\left(0_{7}\right) \cdot \beta_{7}(j)=0.1861\\ &\beta_{6}(2)=\sum_{j=1}^{3} a_{2 j} b_{j}\left(0_{7}\right) \cdot \beta_{7}(j)=0.2415\\ &\beta_{6}(3)=\sum_{j=1}^{3} a_{3 j} b_{j}\left(0_{7}\right) \cdot \beta_{7}(j)=0.762\\ \\ &\beta_{5}(1)=\sum_{j=1}^{3} a_{1 j} b_{j}\left(0_{6}\right) \cdot \beta_{6}(j)=0.10552\\ &\beta_{5}(2)=\sum_{j=1}^{3} a_{2 j} b_{j}\left(0_{6}\right) \cdot \beta_{6}(j)=0.10088\\ &\beta_{5}(3)=\sum_{j=1}^{3} a_{3 j} b_{j}\left(0_{6}\right) \cdot \beta_{6}(j)=0.11193\\ \\ &\beta_{4}(1)=\sum_{j=1}^{3} a_{1 j} b_{j}\left(0_{5}\right) \cdot \beta_{5}(j)=0.04586\\ &\beta_{4}(2)=\sum_{j=1}^{3} a_{2 j} b_{j}\left(0_{5}\right) \cdot \beta_{5}(j)=0.05281\\ &\beta_{4}(3)=\sum_{j=1}^{3} a_{3 j} b_{j}(0_5) \cdot \beta_{5}(j)=0.04281 \end{aligned}$

根据上式计算，得到结果：

$P\left(i_{4}=q_{3} | O, \lambda\right)=0.537$

10.3

初始化为：
$\delta_{1}(i)=\pi_{i} b_{i}\left(o_{1}\right)=\pi_{i} b_{i}\text{（红）}, \quad i=1,2,3 \quad \psi_{1}(i)=0, i=1,2,3$
自然，有：
$\delta_{1}(1)=0.2^{*} 0.5=0.1, \quad \delta_{1}(2)=0.4^{*} 0.4=0.16, \quad \delta_{1}(3)=0.4^{*} 0.7=0.28$
递推公式：
对t=2,3,4：
$\delta_{t}(i)=\max _{1 \leq i \leq 3}\left[\delta_{t-1}(j) a_{j j}\right] b_{i}\left(o_{t}\right), \quad \psi_{t}(i)=\arg \max _{1 \leq j \leq 3}\left[\delta_{t-1}(j) a_{j i}\right], \quad i=1,2,3$
当t=4时：
$\begin{array}{c} \delta_{4}(1)=\max _{1 \leq j \leq 3}\left[\delta_{3}(j) a_{j 1}\right] b_{1}\left(o_{4}\right)=\max \{0.00756 * 0.5,0.01008 * 0.3,0.0147 * 0.2\} * 0.5=0.00189 \\ \psi_{4}(1)=\arg \max _{1 \leq j \leq 3}\left[\delta_{3}(j) a_{j 1}\right]=1 \\ \delta_{4}(2)=\max _{1 \leq j \leq 3}\left[\delta_{3}(j) a_{j 2}\right] b_{2}\left(o_{4}\right)=\max \{0.00756 * 0.2,0.01008 * 0.5,0.0147 * 0.3\} * 0.6=0.003024 \\ \psi_{4}(2)=\arg \max _{1 \leq j \leq 3}\left[\delta_{3}(j) a_{j 2}\right]=2 \\ \delta_{4}(3)=\max _{1 \leq \leq 3}\left[\delta_{3}(j) a_{j 3}\right] b_{3}\left(o_{4}\right)=\max \{0.00756 * 0.3,0.01008 * 0.2,0.0147 * 0.5\} * 0.3=0.002205 \\ \psi_{4}(3)=\arg \max _{1 \leq j \leq 3}\left[\delta_{3}(j) a_{j 3}\right]=3 \end{array}$
最终结果：

最优路径终点： $i_{4}^{*}=\arg \max _{1 \leq j \leq 3}\left[\delta_{a}(i)\right]=2$
最优路径概率： $P^{*}=\max _{1 \leq j \leq 3} \delta_{4}(i)=\delta_{4}(2)=0.003024$
4. 最优路径：
对于t=3,2,1有， $i_{t}^{*}=\psi_{t+1}\left(i_{t+1}^{*}\right)$
于是有，
$i_{3}^{*}=\psi_{4}\left(i_{4}^{*}\right)=\psi_{4}(2)=2, \quad i_{2}^{*}=\psi_{3}\left(i_{3}^{*}\right)=\psi_{3}(2)=2, \quad i_{1}^{*}=\psi_{2}\left(i_{2}^{*}\right)=\psi_{2}(2)=3$
所以最优的状态序列为： $I^{*}=\left(i_{1}^{*}, i_{2}^{*}, i_{3}^{*}, i_{4}^{*}\right)=(3,2,2,2)$