《统计学习方法》第10章习题答案

参考：李航统计学习方法第二版第十章习题解答

10.2

${\alpha }_1(1)={\pi }_1{b}_1(o_1)=0.2\times 0.5=0.1$

${\alpha }_1(2)={\pi }_2{b}_2(o_1)=0.3\times 0.4=0.12$

${\alpha }_1(3)={\pi }_3{b}_3(o_1)=0.5\times 0.7=0.35$

${\alpha }_2(1)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j1}]b_1(o_2)=(0.1\times 0.5+0.12\times 0.3+0.35\times 0.2)\times 0.5=0.078$

${\alpha }_2(2)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j2}]b_2(o_2)=(0.1\times 0.1+0.12\times 0.5+0.35\times 0.2)\times 0.6=0.084$

${\alpha }_2(3)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j3}]b_3(o_2)=(0.1\times 0.4+0.12\times 0.2+0.35\times 0.6)\times 0.3=0.0822$

${\alpha }_3(1)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j1}]b_1(o_3)=(0.078\times 0.5+0.084\times 0.3+0.0822\times 0.2)\times 0.5=0.04032$

${\alpha }_3(2)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j2}]b_2(o_3)=(0.078\times 0.1+0.084\times 0.5+0.0822\times 0.2)\times 0.4=0.02650$

${\alpha }_3(3)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j3}]b_3(o_3)=(0.078\times 0.4+0.084\times 0.2+0.0822\times 0.6)\times 0.7=0.06812$

${\alpha }_4(1)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j1}]b_1(o_4)=(0.04032\times 0.5+0.02650\times 0.3+0.06812\times 0.2)\times 0.5=0.02087$

${\alpha }_4(2)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j2}]b_2(o_4)=(0.04032\times 0.1+0.02650\times 0.5+0.06812\times 0.2)\times 0.4=0.01236$

${\alpha }_4(3)=[{\sum }_{j=1}^3{\alpha }_1(j)a_{j3}]b_3(o_4)=(0.04032\times 0.4+0.02650\times 0.2+0.06812\times 0.6)\times 0.7=0.04361$

${\beta }_8(1)={\beta }_8(2)={\beta }_8(3)=1$

${\beta }_7(1)={\sum }_{j=1}^3{a}_{1j}{b}_j(o_8){\beta }_8(j)=0.5\times 0.5+0.1\times 0.6+0.4\times 0.3=0.43$

${\beta }_7(2)={\sum }_{j=1}^3{a}_{2j}{b}_j(o_8){\beta }_8(j)=0.3\times 0.5+0.5\times 0.6+0.2\times 0.3=0.51$

${\beta }_7(1)={\sum }_{j=1}^3{a}_{1j}{b}_j(o_8){\beta }_8(j)=0.2\times 0.5+0.2\times 0.6+0.6\times 0.3=0.4$

${\beta }_6(1)={\sum }_{j=1}^3{a}_{1j}{b}_j(o_7){\beta }_7(j)=0.5\times 0.5\times 0.43+0.1\times 0.6\times 0.51+0.4\times 0.3\times 0.4=0.1861$

${\beta }_6(2)={\sum }_{j=1}^3{a}_{2j}{b}_j(o_7){\beta }_7(j)=0.3\times 0.5\times 0.43+0.5\times 0.6\times 0.51+0.2\times 0.3\times 0.4=0.2415$

${\beta }_6(3)={\sum }_{j=1}^3{a}_{3j}{b}_j(o_7){\beta }_7(j)=0.2\times 0.5\times 0.43+0.2\times 0.6\times 0.51+0.6\times 0.3\times 0.4=0.1762$

${\beta }_5(1)={\sum }_{j=1}^3{a}_{1j}{b}_j(o_6){\beta }_6(j)=0.5\times 0.5\times 0.1861+0.1\times 0.4\times 0.2415+0.4\times 0.7\times 0.1762=0.105521$

${\beta }_5(2)={\sum }_{j=1}^3{a}_{2j}{b}_j(o_6){\beta }_6(j)=0.3\times 0.5\times 0.1861+0.5\times 0.4\times 0.2415+0.2\times 0.7\times 0.1762=0.100883$

${\beta }_5(3)={\sum }_{j=1}^3{a}_{3j}{b}_j(o_6){\beta }_6(j)=0.2\times 0.5\times 0.1861+0.2\times 0.4\times 0.2415+0.6\times 0.7\times 0.1762=0.111934$

${\beta }_4(1)={\sum }_{j=1}^3{a}_{1j}{b}_j(o_5){\beta }_5(j)=0.5\times 0.5\times 0.105521+0.1\times 0.6\times 0.100883+0.4\times 0.3\times 0.111934=0.0459$

${\beta }_4(2)={\sum }_{j=1}^3{a}_{2j}{b}_j(o_5){\beta }_5(j)=0.3\times 0.5\times 0.105521+0.5\times 0.6\times 0.100883+0.2\times 0.3\times 0.111934=0.0528$

${\beta }_4(3)={\sum }_{j=1}^3{a}_{3j}{b}_j(o_5){\beta }_5(j)=0.2\times 0.5\times 0.105521+0.2\times 0.6\times 0.100883+0.6\times 0.3\times 0.111934=0.0428$

$P(i_4=q_3|O,\lambda )=\frac{{\alpha }_4(3){\beta }_4(3)}{{\alpha }_4(1){\beta }_4(1)+{\alpha }_4(2){\beta }_4(2)+{\alpha }_4(3){\beta }_4(3)}=\frac{0.04361\times 0.0428}{0.02087\times 0.0459+0.01236\times 0.0528+0.04361\times 0.0428}\approx 0.537$

10.3

${\delta }_1(1)={\pi }_1{b}_1(1)=0.2\times 0.5=0.1,{\delta }_1(2)={\pi }_2{b}_2(1)=0.4\times 0.4=0.16$

${\delta }_1(3)={\pi }_3{b}_3(1)=0.4\times 0.7=0.28$

${\delta }_2(1)={\max }_{1\le j\le 3}[{\delta }_1(j)a_{j1}]b_1(o_2)=\max \{0.1\times 0.5,0.16\times 0.3,0.28\times 0.2\}\times 0.5=0.028$

${\Psi }_2(1)=argma{x}_{1\le j\le 3}{\delta }_1(j)a_{j1}=3$

${\delta }_2(2)={\max }_{1\le j\le 3}[{\delta }_1(j)a_{j2}]b_2(o_2)=\max \{0.1\times 0.2,0.16\times 0.5,0.28\times 0.3\}\times 0.6=0.0504$

${\Psi }_2(2)=argma{x}_{1\le j\le 3}{\delta }_1(j)a_{j2}=3$

${\delta }_2(3)={\max }_{1\le j\le 3}[{\delta }_1(j)a_{j3}]b_3(o_2)=\max \{0.1\times 0.3,0.16\times 0.2,0.28\times 0.5\}\times 0.3=0.042$

${\Psi }_2(3)=argma{x}_{1\le j\le 3}{\delta }_1(j)a_{j3}=3$

${\delta }_3(1)={\max }_{1\le j\le 3}[{\delta }_2(j)a_{j1}]b_1(o_3)=\max \{0.028\times 0.5,0.0504\times 0.3,0.042\times 0.2\}\times 0.5=0.00756$

${\Psi }_3(1)=argma{x}_{1\le j\le 3}{\delta }_2(j)a_{j1}=2$

${\delta }_3(2)={\max }_{1\le j\le 3}[{\delta }_2(j)a_{j2}]b_2(o_3)=\max \{0.028\times 0.2,0.0504\times 0.5,0.042\times 0.3\}\times 0.4=0.01008$

${\Psi }_3(2)=argma{x}_{1\le j\le 3}{\delta }_2(j)a_{j2}=2$

${\delta }_3(3)={\max }_{1\le j\le 3}[{\delta }_2(j)a_{j3}]b_3(o_3)=\max \{0.028\times 0.3,0.0504\times 0.2,0.042\times 0.5\}\times 0.7=0.0147$

${\Psi }_3(3)=argma{x}_{1\le j\le 3}{\delta }_2(j)a_{j3}=3$

${\delta }_4(1)={\max }_{1\le j\le 3}[{\delta }_3(j)a_{j1}]b_1(o_4)=\max \{0.00756\times 0.5,0.01008\times 0.3,0.0147\times 0.2\}\times 0.5=0.00189$

${\Psi }_4(1)=argma{x}_{1\le j\le 3}{\delta }_3(j)a_{j1}=1$

${\delta }_4(2)={\max }_{1\le j\le 3}[{\delta }_3(j)a_{j2}]b_2(o_4)=\max \{0.00756\times 0.2,0.01008\times 0.5,0.0147\times 0.3\}\times 0.6=0.003024$

${\Psi }_4(2)=argma{x}_{1\le j\le 3}{\delta }_3(j)a_{j2}=2$

${\delta }_4(3)={\max }_{1\le j\le 3}[{\delta }_3(j)a_{j3}]b_3(o_4)=\max \{0.00756\times 0.3,0.01008\times 0.2,0.0147\times 0.5\}\times 0.3=0.002205$

${\Psi }_4(1)=argma{x}_{1\le j\le 3}{\delta }_3(j)a_{j1}=3$

$P^{\ast }={\max }_{1\le l\le 3}{\delta }_4(i)=0.003024$

最优路径的终点是$i_4^{\ast }=argma{x}_{1\le i\le 3}[{\delta }_4(i)]=2$

$i_2^{\ast }={\Psi }_4(i_4^{\ast })=2$

$i_2^{\ast }={\Psi }_3(i_3^{\ast })=2$

$i_1^{\ast }={\Psi }_2(i_2^{\ast })=3$

$I^{\ast }=(i_1^{\ast },i_2^{\ast },i_3^{\ast },i_4^{\ast })=(3,2,2,2)$

10.4

${\sum }_{i=1}^N{\sum }_{j=1}^N{\alpha }_t(i)a_{ij}{b}_j(o_{t+1}){\beta }_{t+1}(j)$

$={\sum }_{i=1}^N{\sum }_{j=1}^{N}P(o_1,o_2,\dots ,o_t,I_t=q_i|\lambda )P(I_{t+1}=q_j|I_t=q_i,\lambda )P(o_{t+1}|I_{t+1}=q_j,\lambda )P(o_{t+2},\dots ,o_T|I_{t+1}=q_j,\lambda )$

$={\sum }_{i=1}^N{\sum }_{j=1}^{N}P(o_1,o_2,\dots ,o_t,I_t=q_i|\lambda )P(I_{t+1}=q_j|I_t=q_i,\lambda )P(o_{t+1}|I_{t+1}=q_j,\lambda )P(o_{t+2},\dots ,o_T|o_{t+1},I_{t+1}=q_j,\lambda )$

$={\sum }_{i=1}^N{\sum }_{j=1}^{N}P(o_1,o_2,\dots ,o_t,I_t=q_i|\lambda )P(I_{t+1}=q_j|I_t=q_i,\lambda )P(o_{t+1},o_{t+2},\dots ,o_T|,I_{t+1}=q_j,\lambda )$

$={\sum }_{i=1}^{N}P(o_1,o_2,\dots ,o_t,I_t=q_i|\lambda )P(o_{t+1},o_{t+2},\dots ,o_T|I_t=q_i,\lambda )$

$=P(O|\lambda )$