《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

α 1 ( 1 ) = π 1 b 1 ( o 1 ) = 0.2 × 0.5 = 0.1 \alpha_1(1) = \pi_1 b_1(o_1) = 0.2 \times 0.5 = 0.1 α1(1)=π1b1(o1)=0.2×0.5=0.1

α 1 ( 2 ) = π 2 b 2 ( o 1 ) = 0.3 × 0.4 = 0.12 \alpha_1(2) = \pi_2 b_2(o_1) = 0.3 \times 0.4 = 0.12 α1(2)=π2b2(o1)=0.3×0.4=0.12

α 1 ( 3 ) = π 3 b 3 ( o 1 ) = 0.5 × 0.7 = 0.35 \alpha_1(3) = \pi_3 b_3(o_1) = 0.5 \times 0.7 = 0.35 α1(3)=π3b3(o1)=0.5×0.7=0.35

α 2 ( 1 ) = [ ∑ j = 1 3 α 1 ( j ) a j 1 ] b 1 ( o 2 ) = ( 0.1 × 0.5 + 0.12 × 0.3 + 0.35 × 0.2 ) × 0.5 = 0.078 \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 α2(1)=[j=13α1(j)aj1]b1(o2)=(0.1×0.5+0.12×0.3+0.35×0.2)×0.5=0.078

α 2 ( 2 ) = [ ∑ j = 1 3 α 1 ( j ) a j 2 ] b 2 ( o 2 ) = ( 0.1 × 0.1 + 0.12 × 0.5 + 0.35 × 0.2 ) × 0.6 = 0.084 \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 α2(2)=[j=13α1(j)aj2]b2(o2)=(0.1×0.1+0.12×0.5+0.35×0.2)×0.6=0.084

α 2 ( 3 ) = [ ∑ j = 1 3 α 1 ( j ) a j 3 ] b 3 ( o 2 ) = ( 0.1 × 0.4 + 0.12 × 0.2 + 0.35 × 0.6 ) × 0.3 = 0.0822 \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 α2(3)=[j=13α1(j)aj3]b3(o2)=(0.1×0.4+0.12×0.2+0.35×0.6)×0.3=0.0822

α 3 ( 1 ) = [ ∑ j = 1 3 α 1 ( j ) a j 1 ] b 1 ( o 3 ) = ( 0.078 × 0.5 + 0.084 × 0.3 + 0.0822 × 0.2 ) × 0.5 = 0.04032 \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 α3(1)=[j=13α1(j)aj1]b1(o3)=(0.078×0.5+0.084×0.3+0.0822×0.2)×0.5=0.04032

α 3 ( 2 ) = [ ∑ j = 1 3 α 1 ( j ) a j 2 ] b 2 ( o 3 ) = ( 0.078 × 0.1 + 0.084 × 0.5 + 0.0822 × 0.2 ) × 0.4 = 0.02650 \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 α3(2)=[j=13α1(j)aj2]b2(o3)=(0.078×0.1+0.084×0.5+0.0822×0.2)×0.4=0.02650

α 3 ( 3 ) = [ ∑ j = 1 3 α 1 ( j ) a j 3 ] b 3 ( o 3 ) = ( 0.078 × 0.4 + 0.084 × 0.2 + 0.0822 × 0.6 ) × 0.7 = 0.06812 \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 α3(3)=[j=13α1(j)aj3]b3(o3)=(0.078×0.4+0.084×0.2+0.0822×0.6)×0.7=0.06812

α 4 ( 1 ) = [ ∑ j = 1 3 α 1 ( j ) a j 1 ] b 1 ( o 4 ) = ( 0.04032 × 0.5 + 0.02650 × 0.3 + 0.06812 × 0.2 ) × 0.5 = 0.02087 \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 α4(1)=[j=13α1(j)aj1]b1(o4)=(0.04032×0.5+0.02650×0.3+0.06812×0.2)×0.5=0.02087

α 4 ( 2 ) = [ ∑ j = 1 3 α 1 ( j ) a j 2 ] b 2 ( o 4 ) = ( 0.04032 × 0.1 + 0.02650 × 0.5 + 0.06812 × 0.2 ) × 0.4 = 0.01236 \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 α4(2)=[j=13α1(j)aj2]b2(o4)=(0.04032×0.1+0.02650×0.5+0.06812×0.2)×0.4=0.01236

α 4 ( 3 ) = [ ∑ j = 1 3 α 1 ( j ) a j 3 ] b 3 ( o 4 ) = ( 0.04032 × 0.4 + 0.02650 × 0.2 + 0.06812 × 0.6 ) × 0.7 = 0.04361 \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 α4(3)=[j=13α1(j)aj3]b3(o4)=(0.04032×0.4+0.02650×0.2+0.06812×0.6)×0.7=0.04361

β 8 ( 1 ) = β 8 ( 2 ) = β 8 ( 3 ) = 1 \beta_8(1) = \beta_8(2) =\beta_8(3) = 1 β8(1)=β8(2)=β8(3)=1

β 7 ( 1 ) = ∑ j = 1 3 a 1 j b j ( o 8 ) β 8 ( j ) = 0.5 × 0.5 + 0.1 × 0.6 + 0.4 × 0.3 = 0.43 \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 β7(1)=j=13a1jbj(o8)β8(j)=0.5×0.5+0.1×0.6+0.4×0.3=0.43

β 7 ( 2 ) = ∑ j = 1 3 a 2 j b j ( o 8 ) β 8 ( j ) = 0.3 × 0.5 + 0.5 × 0.6 + 0.2 × 0.3 = 0.51 \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 β7(2)=j=13a2jbj(o8)β8(j)=0.3×0.5+0.5×0.6+0.2×0.3=0.51

β 7 ( 1 ) = ∑ j = 1 3 a 1 j b j ( o 8 ) β 8 ( j ) = 0.2 × 0.5 + 0.2 × 0.6 + 0.6 × 0.3 = 0.4 \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 β7(1)=j=13a1jbj(o8)β8(j)=0.2×0.5+0.2×0.6+0.6×0.3=0.4

β 6 ( 1 ) = ∑ j = 1 3 a 1 j b j ( o 7 ) β 7 ( j ) = 0.5 × 0.5 × 0.43 + 0.1 × 0.6 × 0.51 + 0.4 × 0.3 × 0.4 = 0.1861 \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 β6(1)=j=13a1jbj(o7)β7(j)=0.5×0.5×0.43+0.1×0.6×0.51+0.4×0.3×0.4=0.1861

β 6 ( 2 ) = ∑ j = 1 3 a 2 j b j ( o 7 ) β 7 ( j ) = 0.3 × 0.5 × 0.43 + 0.5 × 0.6 × 0.51 + 0.2 × 0.3 × 0.4 = 0.2415 \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 β6(2)=j=13a2jbj(o7)β7(j)=0.3×0.5×0.43+0.5×0.6×0.51+0.2×0.3×0.4=0.2415

β 6 ( 3 ) = ∑ j = 1 3 a 3 j b j ( o 7 ) β 7 ( j ) = 0.2 × 0.5 × 0.43 + 0.2 × 0.6 × 0.51 + 0.6 × 0.3 × 0.4 = 0.1762 \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 β6(3)=j=13a3jbj(o7)β7(j)=0.2×0.5×0.43+0.2×0.6×0.51+0.6×0.3×0.4=0.1762

β 5 ( 1 ) = ∑ j = 1 3 a 1 j b j ( o 6 ) β 6 ( j ) = 0.5 × 0.5 × 0.1861 + 0.1 × 0.4 × 0.2415 + 0.4 × 0.7 × 0.1762 = 0.105521 \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 β5(1)=j=13a1jbj(o6)β6(j)=0.5×0.5×0.1861+0.1×0.4×0.2415+0.4×0.7×0.1762=0.105521

β 5 ( 2 ) = ∑ j = 1 3 a 2 j b j ( o 6 ) β 6 ( j ) = 0.3 × 0.5 × 0.1861 + 0.5 × 0.4 × 0.2415 + 0.2 × 0.7 × 0.1762 = 0.100883 \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 β5(2)=j=13a2jbj(o6)β6(j)=0.3×0.5×0.1861+0.5×0.4×0.2415+0.2×0.7×0.1762=0.100883

β 5 ( 3 ) = ∑ j = 1 3 a 3 j b j ( o 6 ) β 6 ( j ) = 0.2 × 0.5 × 0.1861 + 0.2 × 0.4 × 0.2415 + 0.6 × 0.7 × 0.1762 = 0.111934 \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 β5(3)=j=13a3jbj(o6)β6(j)=0.2×0.5×0.1861+0.2×0.4×0.2415+0.6×0.7×0.1762=0.111934

β 4 ( 1 ) = ∑ j = 1 3 a 1 j b j ( o 5 ) β 5 ( j ) = 0.5 × 0.5 × 0.105521 + 0.1 × 0.6 × 0.100883 + 0.4 × 0.3 × 0.111934 = 0.0459 \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 β4(1)=j=13a1jbj(o5)β5(j)=0.5×0.5×0.105521+0.1×0.6×0.100883+0.4×0.3×0.111934=0.0459

β 4 ( 2 ) = ∑ j = 1 3 a 2 j b j ( o 5 ) β 5 ( j ) = 0.3 × 0.5 × 0.105521 + 0.5 × 0.6 × 0.100883 + 0.2 × 0.3 × 0.111934 = 0.0528 \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 β4(2)=j=13a2jbj(o5)β5(j)=0.3×0.5×0.105521+0.5×0.6×0.100883+0.2×0.3×0.111934=0.0528

β 4 ( 3 ) = ∑ j = 1 3 a 3 j b j ( o 5 ) β 5 ( j ) = 0.2 × 0.5 × 0.105521 + 0.2 × 0.6 × 0.100883 + 0.6 × 0.3 × 0.111934 = 0.0428 \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 β4(3)=j=13a3jbj(o5)β5(j)=0.2×0.5×0.105521+0.2×0.6×0.100883+0.6×0.3×0.111934=0.0428

P ( i 4 = q 3 ∣ O , λ ) = α 4 ( 3 ) β 4 ( 3 ) α 4 ( 1 ) β 4 ( 1 ) + α 4 ( 2 ) β 4 ( 2 ) + α 4 ( 3 ) β 4 ( 3 ) = 0.04361 × 0.0428 0.02087 × 0.0459 + 0.01236 × 0.0528 + 0.04361 × 0.0428 ≈ 0.537 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 P(i4=q3O,λ)=α4(1)β4(1)+α4(2)β4(2)+α4(3)β4(3)α4(3)β4(3)=0.02087×0.0459+0.01236×0.0528+0.04361×0.04280.04361×0.04280.537

10.3

δ 1 ( 1 ) = π 1 b 1 ( 1 ) = 0.2 × 0.5 = 0.1 , δ 1 ( 2 ) = π 2 b 2 ( 1 ) = 0.4 × 0.4 = 0.16 \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 δ1(1)=π1b1(1)=0.2×0.5=0.1,δ1(2)=π2b2(1)=0.4×0.4=0.16

δ 1 ( 3 ) = π 3 b 3 ( 1 ) = 0.4 × 0.7 = 0.28 \delta_1(3) = \pi_3 b_3(1) = 0.4 \times 0.7 = 0.28 δ1(3)=π3b3(1)=0.4×0.7=0.28

δ 2 ( 1 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 1 ( j ) a j 1 ] b 1 ( o 2 ) = max ⁡ { 0.1 × 0.5 , 0.16 × 0.3 , 0.28 × 0.2 } × 0.5 = 0.028 \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 δ2(1)=max1j3[δ1(j)aj1]b1(o2)=max{0.1×0.5,0.16×0.3,0.28×0.2}×0.5=0.028

Ψ 2 ( 1 ) = a r g m a x 1 ≤ j ≤ 3 δ 1 ( j ) a j 1 = 3 \Psi_2(1) = argmax_{1 \le j \le 3} \delta_1(j)a_{j1} = 3 Ψ2(1)=argmax1j3δ1(j)aj1=3

δ 2 ( 2 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 1 ( j ) a j 2 ] b 2 ( o 2 ) = max ⁡ { 0.1 × 0.2 , 0.16 × 0.5 , 0.28 × 0.3 } × 0.6 = 0.0504 \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 δ2(2)=max1j3[δ1(j)aj2]b2(o2)=max{0.1×0.2,0.16×0.5,0.28×0.3}×0.6=0.0504

Ψ 2 ( 2 ) = a r g m a x 1 ≤ j ≤ 3 δ 1 ( j ) a j 2 = 3 \Psi_2(2) = argmax_{1 \le j \le 3} \delta_1(j)a_{j2} = 3 Ψ2(2)=argmax1j3δ1(j)aj2=3

δ 2 ( 3 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 1 ( j ) a j 3 ] b 3 ( o 2 ) = max ⁡ { 0.1 × 0.3 , 0.16 × 0.2 , 0.28 × 0.5 } × 0.3 = 0.042 \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 δ2(3)=max1j3[δ1(j)aj3]b3(o2)=max{0.1×0.3,0.16×0.2,0.28×0.5}×0.3=0.042

Ψ 2 ( 3 ) = a r g m a x 1 ≤ j ≤ 3 δ 1 ( j ) a j 3 = 3 \Psi_2(3) = argmax_{1 \le j \le 3} \delta_1(j)a_{j3} = 3 Ψ2(3)=argmax1j3δ1(j)aj3=3

δ 3 ( 1 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 2 ( j ) a j 1 ] b 1 ( o 3 ) = max ⁡ { 0.028 × 0.5 , 0.0504 × 0.3 , 0.042 × 0.2 } × 0.5 = 0.00756 \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 δ3(1)=max1j3[δ2(j)aj1]b1(o3)=max{0.028×0.5,0.0504×0.3,0.042×0.2}×0.5=0.00756

Ψ 3 ( 1 ) = a r g m a x 1 ≤ j ≤ 3 δ 2 ( j ) a j 1 = 2 \Psi_3(1) = argmax_{1 \le j \le 3} \delta_2(j)a_{j1} = 2 Ψ3(1)=argmax1j3δ2(j)aj1=2

δ 3 ( 2 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 2 ( j ) a j 2 ] b 2 ( o 3 ) = max ⁡ { 0.028 × 0.2 , 0.0504 × 0.5 , 0.042 × 0.3 } × 0.4 = 0.01008 \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 δ3(2)=max1j3[δ2(j)aj2]b2(o3)=max{0.028×0.2,0.0504×0.5,0.042×0.3}×0.4=0.01008

Ψ 3 ( 2 ) = a r g m a x 1 ≤ j ≤ 3 δ 2 ( j ) a j 2 = 2 \Psi_3(2) = argmax_{1 \le j \le 3} \delta_2(j)a_{j2} = 2 Ψ3(2)=argmax1j3δ2(j)aj2=2

δ 3 ( 3 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 2 ( j ) a j 3 ] b 3 ( o 3 ) = max ⁡ { 0.028 × 0.3 , 0.0504 × 0.2 , 0.042 × 0.5 } × 0.7 = 0.0147 \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 δ3(3)=max1j3[δ2(j)aj3]b3(o3)=max{0.028×0.3,0.0504×0.2,0.042×0.5}×0.7=0.0147

Ψ 3 ( 3 ) = a r g m a x 1 ≤ j ≤ 3 δ 2 ( j ) a j 3 = 3 \Psi_3(3) = argmax_{1 \le j \le 3} \delta_2(j)a_{j3} = 3 Ψ3(3)=argmax1j3δ2(j)aj3=3

δ 4 ( 1 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 3 ( j ) a j 1 ] b 1 ( o 4 ) = max ⁡ { 0.00756 × 0.5 , 0.01008 × 0.3 , 0.0147 × 0.2 } × 0.5 = 0.00189 \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 δ4(1)=max1j3[δ3(j)aj1]b1(o4)=max{0.00756×0.5,0.01008×0.3,0.0147×0.2}×0.5=0.00189

Ψ 4 ( 1 ) = a r g m a x 1 ≤ j ≤ 3 δ 3 ( j ) a j 1 = 1 \Psi_4(1) = argmax_{1 \le j \le 3} \delta_3(j)a_{j1} = 1 Ψ4(1)=argmax1j3δ3(j)aj1=1

δ 4 ( 2 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 3 ( j ) a j 2 ] b 2 ( o 4 ) = max ⁡ { 0.00756 × 0.2 , 0.01008 × 0.5 , 0.0147 × 0.3 } × 0.6 = 0.003024 \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 δ4(2)=max1j3[δ3(j)aj2]b2(o4)=max{0.00756×0.2,0.01008×0.5,0.0147×0.3}×0.6=0.003024

Ψ 4 ( 2 ) = a r g m a x 1 ≤ j ≤ 3 δ 3 ( j ) a j 2 = 2 \Psi_4(2) = argmax_{1 \le j \le 3} \delta_3(j)a_{j2} = 2 Ψ4(2)=argmax1j3δ3(j)aj2=2

δ 4 ( 3 ) = max ⁡ 1 ≤ j ≤ 3 [ δ 3 ( j ) a j 3 ] b 3 ( o 4 ) = max ⁡ { 0.00756 × 0.3 , 0.01008 × 0.2 , 0.0147 × 0.5 } × 0.3 = 0.002205 \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 δ4(3)=max1j3[δ3(j)aj3]b3(o4)=max{0.00756×0.3,0.01008×0.2,0.0147×0.5}×0.3=0.002205

Ψ 4 ( 1 ) = a r g m a x 1 ≤ j ≤ 3 δ 3 ( j ) a j 1 = 3 \Psi_4(1) = argmax_{1 \le j \le 3} \delta_3(j)a_{j1} = 3 Ψ4(1)=argmax1j3δ3(j)aj1=3

P ∗ = max ⁡ 1 ≤ l ≤ 3 δ 4 ( i ) = 0.003024 P^* = \max_{1 \le l \le 3} \delta_4(i) = 0.003024 P=max1l3δ4(i)=0.003024

最优路径的终点是 i 4 ∗ = a r g m a x 1 ≤ i ≤ 3 [ δ 4 ( i ) ] = 2 i_4^* = argmax_{1 \le i \le 3} [\delta_4(i)] = 2 i4=argmax1i3[δ4(i)]=2

i 2 ∗ = Ψ 4 ( i 4 ∗ ) = 2 i_2^* = \Psi_4(i_4^*) = 2 i2=Ψ4(i4)=2

i 2 ∗ = Ψ 3 ( i 3 ∗ ) = 2 i_2^* = \Psi_3(i_3^*) = 2 i2=Ψ3(i3)=2

i 1 ∗ = Ψ 2 ( i 2 ∗ ) = 3 i_1^* = \Psi_2(i_2^*) = 3 i1=Ψ2(i2)=3

I ∗ = ( i 1 ∗ , i 2 ∗ , i 3 ∗ , i 4 ∗ ) = ( 3 , 2 , 2 , 2 ) I^* = (i_1^*, i_2^*, i_3^*, i_4^*) = (3,2,2,2) I=(i1,i2,i3,i4)=(3,2,2,2)

10.4

∑ i = 1 N ∑ j = 1 N α t ( i ) a i j b j ( o t + 1 ) β t + 1 ( j ) \sum_{i=1}^N \sum_{j=1}^N \alpha_t(i)a_{ij}b_j(o_{t+1})\beta_{t+1}(j) i=1Nj=1Nαt(i)aijbj(ot+1)βt+1(j)

= ∑ i = 1 N ∑ j = 1 N P ( o 1 , o 2 , … , o t , I t = q i ∣ λ ) P ( I t + 1 = q j ∣ I t = q i , λ ) P ( o t + 1 ∣ I t + 1 = q j , λ ) P ( o t + 2 , … , o T ∣ I t + 1 = q j , λ ) = \sum_{i=1}^N \sum_{j=1}^N P(o_1,o_2,\ldots,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},\ldots,o_T|I_{t+1}=q_j,\lambda) =i=1Nj=1NP(o1,o2,,ot,It=qiλ)P(It+1=qjIt=qi,λ)P(ot+1It+1=qj,λ)P(ot+2,,oTIt+1=qj,λ)

= ∑ i = 1 N ∑ j = 1 N P ( o 1 , o 2 , … , o t , I t = q i ∣ λ ) P ( I t + 1 = q j ∣ I t = q i , λ ) P ( o t + 1 ∣ I t + 1 = q j , λ ) P ( o t + 2 , … , o T ∣ o t + 1 , I t + 1 = q j , λ ) = \sum_{i=1}^N \sum_{j=1}^N P(o_1,o_2,\ldots,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},\ldots,o_T|o_{t+1},I_{t+1}=q_j, \lambda) =i=1Nj=1NP(o1,o2,,ot,It=qiλ)P(It+1=qjIt=qi,λ)P(ot+1It+1=qj,λ)P(ot+2,,oTot+1,It+1=qj,λ)

= ∑ i = 1 N ∑ j = 1 N P ( o 1 , o 2 , … , o t , I t = q i ∣ λ ) P ( I t + 1 = q j ∣ I t = q i , λ ) P ( o t + 1 , o t + 2 , … , o T ∣ , I t + 1 = q j , λ ) =\sum_{i=1}^N \sum_{j=1}^N P(o_1,o_2,\ldots,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},\ldots,o_T|,I_{t+1}=q_j, \lambda) =i=1Nj=1NP(o1,o2,,ot,It=qiλ)P(It+1=qjIt=qi,λ)P(ot+1,ot+2,,oT,It+1=qj,λ)

= ∑ i = 1 N P ( o 1 , o 2 , … , o t , I t = q i ∣ λ ) P ( o t + 1 , o t + 2 , … , o T ∣ I t = q i , λ ) =\sum_{i=1}^N P(o_1,o_2,\ldots,o_t,I_t=q_i|\lambda)P(o_{t+1},o_{t+2},\ldots,o_T|I_t=q_i,\lambda) =i=1NP(o1,o2,,ot,It=qiλ)P(ot+1,ot+2,,oTIt=qi,λ)

= P ( O ∣ λ ) =P(O|\lambda) =P(Oλ)

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