习题10.1
由题， \(T=4, N=3,M=2\)
根据算法10.3
第一步，计算终期 \(\beta\) ：
\(\beta_4(1) = 1, \beta_4(2) = 1, \beta_4(3) = 1\)
第二步，计算中间每期 \(\beta\) ：
\(\beta_3(1) = a_{11}b_1(o_4)\beta_4(1) + a_{12}b_2(o_4)\beta_4(2) + a_{13}b_3(o_4)\beta_4(3) = 0.46\)
\(\beta_3(2) = a_{21}b_1(o_4)\beta_4(1) + a_{22}b_2(o_4)\beta_4(2) + a_{23}b_3(o_4)\beta_4(3) = 0.51\)
\(\beta_3(3) = a_{31}b_1(o_4)\beta_4(1) + a_{32}b_2(o_4)\beta_4(2) + a_{33}b_3(o_4)\beta_4(3) = 0.43\)
\(\beta_2(1) = a_{11}b_1(o_3)\beta_3(1) + a_{12}b_2(o_3)\beta_3(2) + a_{13}b_3(o_3)\beta_3(3) = 0.2461\)
\(\beta_2(2) = a_{21}b_1(o_3)\beta_3(1) + a_{22}b_2(o_3)\beta_3(2) + a_{23}b_3(o_3)\beta_3(3) = 0.2312\)
\(\beta_2(3) = a_{31}b_1(o_3)\beta_3(1) + a_{32}b_2(o_3)\beta_3(2) + a_{33}b_3(o_3)\beta_3(3) = 0.2577\)
\(\beta_1(1) = a_{11}b_1(o_2)\beta_2(1) + a_{12}b_2(o_2)\beta_2(2) + a_{13}b_3(o_2)\beta_2(3) = 0.112462\)
\(\beta_1(2) = a_{21}b_1(o_2)\beta_2(1) + a_{22}b_2(o_2)\beta_2(2) + a_{23}b_3(o_2)\beta_2(3) = 0.121737\)
\(\beta_1(3) = a_{31}b_1(o_2)\beta_2(1) + a_{32}b_2(o_2)\beta_2(2) + a_{33}b_3(o_2)\beta_2(3) = 0.104881\)
第三步，计算 \(P(O|\lambda)\)
\(P(O|\lambda) = \pi_1b_1(o_1)\beta_1(1) + \pi_2b_2(o_1)\beta_1(2) + \pi_3b_3(o_1)\beta_1(3) = 0.0601088\)
习题10.2
根据定义， \(P(i_4 = q_3|O,\lambda) = \gamma_4(3)\)
根据公式，可得 \(\gamma_4(3) = \frac{\alpha_4(3) \beta_4(3)}{P(O|\lambda)} = \frac{\alpha_4(3) \beta_4(3)}{\sum \alpha_4(j) \beta_4(j)}\)
通过程序计算，可得 \(P(i_4 = q_3|O,\lambda) = \gamma_4(3) = 0.536952\)
习题10.3
根据算法10.5
第一步，初始化
\(\delta_1(1) = \pi_1 b_1(o_1) = 0.2*0.5=0.1\) ， \(\psi_1(1) = 0\)
\(\delta_1(2) = \pi_2 b_2(o_1) = 0.4*0.4=0.16\) ， \(\psi_1(2) = 0\)
\(\delta_1(3) = \pi_3 b_3(o_1) = 0.4*0.7=0.28\) ， \(\psi_1(3) = 0\)
第二步，递推
\(\delta_2(1) = \mathop{max} \limits_j [\delta_1(j)a_{j1}] b_1(o_2) = max\{0.1*0.5, 0.16*0.3, 0.28*0.2\}*0.5=0.028\) ， \(\psi_2(1) = 3\)
\(\delta_2(2) = \mathop{max} \limits_j [\delta_1(j)a_{j2}] b_2(o_2) = max\{0.1*0.2, 0.16*0.5, 0.28*0.3\}*0.6=0.0504\) ， \(\psi_2(2) = 3\)
\(\delta_2(3) = \mathop{max} \limits_j [\delta_1(j)a_{j3}] b_3(o_2) = max\{0.1*0.3, 0.16*0.2, 0.28*0.5\}*0.3=0.042\) ， \(\psi_2(3) = 3\)
\(\delta_3(1) = \mathop{max} \limits_j [\delta_2(j)a_{j1}] b_1(o_3) = max\{0.028*0.5, 0.0504*0.3, 0.042*0.2\}*0.5=0.00756\) ， \(\psi_3(1) = 2\)
\(\delta_3(2) = \mathop{max} \limits_j [\delta_2(j)a_{j2}] b_2(o_3) = max\{0.028*0.2, 0.0504*0.5, 0.042*0.3\}*0.4=0.01008\) ， \(\psi_3(2) = 2\)
\(\delta_3(3) = \mathop{max} \limits_j [\delta_2(j)a_{j3}] b_3(o_3) = max\{0.028*0.3, 0.0504*0.2, 0.042*0.5\}*0.7=0.0147\) ， \(\psi_3(3) = 3\)
\(\delta_4(1) = \mathop{max} \limits_j [\delta_3(j)a_{j1}] b_1(o_4) = max\{0.00756*0.5, 0.01008*0.3, 0.0147*0.2\}*0.5=0.00189\) ， \(\psi_4(1) = 1\)
\(\delta_4(2) = \mathop{max} \limits_j [\delta_3(j)a_{j2}] b_2(o_4) = max\{0.00756*0.2, 0.01008*0.5, 0.0147*0.3\}*0.6=0.003024\) ， \(\psi_4(2) = 2\)
\(\delta_4(3) = \mathop{max} \limits_j [\delta_3(j)a_{j3}] b_3(o_4) = max\{0.00756*0.3, 0.01008*0.2, 0.0147*0.5\}*0.3=0.002205\) ， \(\psi_4(3) = 3\)
第三步，终止
\(P^* = \mathop{max} \limits_i \delta_4(i) = 0,003024\)
\(i_4^* = \mathop{\arg\max} \limits_i [\delta_4(i)] = 2\)
第四步，最优路径回溯
\(i_3^* = \psi_4(i_4^*) = 2\)
\(i_2^* = \psi_3(i_3^*) = 2\)
\(i_1^* = \psi_2(i_2^*) = 3\)
因此最优路径 \(I^* = (i_1^*,i_2^*,i_3^*,i_4^*)=(3,2,2,2)\)
习题10.4
用前向概率和后向概率证明：\(P(O|\lambda) = \sum \limits_{i=1}^N \sum \limits_{j=1}^N \alpha_t(i)a_{ij}b_j(o_{t+1})\beta_{t+1}(j)\)
\(\begin{aligned} P(O|\lambda) &= P(o_1,o_2,...,o_T|\lambda) \\ &= \sum_{i=1}^N P(o_1,..,o_t,i_t=q_i|\lambda) P(o_{t+1},..,o_T|i_t=q_i,\lambda) \\ &= \sum_{i=1}^N \sum_{j=1}^N P(o_1,..,o_t,i_t=q_i|\lambda) P(o_{t+1},i_{t+1}=q_j|i_t=q_i,\lambda)P(o_{t+2},..,o_T|i_{t+1}=q_j,\lambda) \\ &= \sum_{i=1}^N \sum_{j=1}^N [P(o_1,..,o_t,i_t=q_i|\lambda) P(o_{t+1}|i_{t+1}=q_j,\lambda) P(i_{t+1}=q_j|i_t=q_i,\lambda) \\ & \quad \quad \quad \quad P(o_{t+2},..,o_T|i_{t+1}=q_j,\lambda)] \\ &= \sum_{i=1}^N \sum_{j=1}^N \alpha_t(i) a_{ij} b_j(o_{t+1}) \beta_{t+1}(j),{\quad}t=1,2,...,T-1 \end{aligned}\)
习题10.5
维特比算法：
初始化：\(\delta_1(i) = \pi_1b_i(o_1)\)
递推：\(\delta_{t+1}(i) = \mathop{max} \limits_j [\delta_ta_{ji}]b_i(o_{t+1})\)
前向算法：
初值：\(\alpha_1(i) = \pi_ib_i(o_1)\)
递推： \(\alpha_{t+1}(i) = [\sum \limits_j \alpha_t(j)a_{ji}]b_i(o_{t+1})\)
维特比算法需要在前一期的计算结果的基础上选择最大值
前向算法直接对上一期的结果进行数值计算