《统计学习方法》第11章习题答案
11.1
图11.3中最大团是{ Y 1 , Y 2 , Y 3 Y_1,Y_2,Y_3 Y1,Y2,Y3},{ Y 2 , Y 3 , Y 4 Y_2,Y_3,Y_4 Y2,Y3,Y4}
根据Hammersley-Clifford定理可知概率图模型可分解为
P ( Y ) = 1 Z ∏ C Ψ C ( Y C ) = 1 Z Ψ C 1 ( Y 1 , Y 2 , Y 3 ) Ψ C 2 ( Y 2 , Y 3 , Y 4 ) P(Y) = \frac{1}{Z}\prod_C \Psi_C(Y_C) = \frac{1}{Z} \Psi_{C_1}(Y_1,Y_2,Y_3)\Psi_{C_2}(Y_2,Y_3,Y_4) P(Y)=Z1∏CΨC(YC)=Z1ΨC1(Y1,Y2,Y3)ΨC2(Y2,Y3,Y4)
Z = ∑ Y Ψ C 1 ( Y 1 , Y 2 , Y 3 ) Ψ C 2 ( Y 2 , Y 3 , Y 4 ) Z= \sum_Y \Psi_{C_1}(Y_1,Y_2,Y_3)\Psi_{C_2}(Y_2,Y_3,Y_4) Z=∑YΨC1(Y1,Y2,Y3)ΨC2(Y2,Y3,Y4)
11.2
α n T 1 = α n − 1 T M n 1 = α n − 2 T M n − 1 M n 1 = α 0 T M 1 … M n − 1 M n 1 = Z ( x ) \alpha_n^T \pmb{1} = \alpha_{n-1}^T M_n\pmb{1} = \alpha_{n-2}^T M_{n-1} M_n\pmb{1}=\alpha_{0}^T M_{1} \ldots M_{n-1} M_n\pmb{1} = Z(x) αnT111=αn−1TMn111=αn−2TMn−1Mn111=α0TM1…Mn−1Mn111=Z(x)
1 T β 0 = 1 T M 1 β 1 = … = 1 T M 1 … M n + 1 β n + 1 = Z ( x ) \pmb{1}^T \beta_0 = \pmb{1}^TM_1 \beta_1 = \ldots = \pmb{1}^T M_1 \ldots M_{n+1} \beta_{n+1} =Z(x) 111Tβ0=111TM1β1=…=111TM1…Mn+1βn+1=Z(x)
11.4
| 路径 | 非规范化概率 | 数值 |
|---|---|---|
| (1,1,1) | a 21 b 11 c 11 a_{21}b_{11}c_{11} a21b11c11 | = 0.5 × 0.3 × 0.5 = 0.075 =0.5 \times 0.3 \times 0.5=0.075 =0.5×0.3×0.5=0.075 |
| (1,1,2) | a 21 b 11 c 12 a_{21}b_{11}c_{12} a21b11c12 | = 0.5 × 0.3 × 0.5 = 0.075 =0.5 \times 0.3 \times 0.5=0.075 =0.5×0.3×0.5=0.075 |
| (1,2,1) | a 21 b 12 c 21 a_{21}b_{12}c_{21} a21b12c21 | = 0.5 × 0.7 × 0.6 = 0.21 =0.5 \times 0.7 \times 0.6=0.21 =0.5×0.7×0.6=0.21 |
| (1,2,2) | a 21 b 12 c 22 a_{21}b_{12}c_{22} a21b12c22 | = 0.5 × 0.7 × 0.4 = 0.14 =0.5 \times 0.7 \times 0.4=0.14 =0.5×0.7×0.4=0.14 |
| (2,1,1) | a 22 b 21 c 11 a_{22}b_{21}c_{11} a22b21c11 | = 0.5 × 0.7 × 0.5 = 0.175 =0.5 \times 0.7 \times 0.5=0.175 =0.5×0.7×0.5=0.175 |
| (2,1,2) | a 22 b 21 c 12 a_{22}b_{21}c_{12} a22b21c12 | = 0.5 × 0.7 × 0.5 = 0.175 =0.5 \times 0.7 \times 0.5=0.175 =0.5×0.7×0.5=0.175 |
| (2,2,1) | a 22 b 22 c 21 a_{22}b_{22}c_{21} a22b22c21 | = 0.5 × 0.3 × 0.6 = 0.09 =0.5 \times 0.3 \times 0.6=0.09 =0.5×0.3×0.6=0.09 |
| (2,2,2) | a 22 b 22 c 22 a_{22}b_{22}c_{22} a22b22c22 | = 0.5 × 0.3 × 0.4 = 0.06 =0.5 \times 0.3 \times 0.4=0.06 =0.5×0.3×0.4=0.06 |
概率最大的状态序列是(1,2,1)
Z ( x ) = ∑ y P ( y ) = 0.075 + 0.075 + 0.21 + 0.14 + 0.175 + 0.175 + 0.09 + 0.06 = 1 Z(x)=\sum_y P(y) = 0.075 + 0.075 + 0.21 + 0.14+0.175+0.175+0.09+0.06=1 Z(x)=∑yP(y)=0.075+0.075+0.21+0.14+0.175+0.175+0.09+0.06=1
改编参考：统计学习方法 第十一章习题答案
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