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《统计学习方法》第10章习题

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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{argmax} 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


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