1、第6章不確定性推理局部參考答案6.8設(shè)有如下一組推理規(guī)那么ri:IFEiTHENE2 (0.6)r2： IF E2 AND E3 THEN E4 (0.7)IFE4THENH (0.8)r4：IFE5THENH (0.9)且 CF(Ei)=0.5, CF(E3)=0.6, CF(E5)=0.7.求 CF(H)=?解：(1)先由ri求CF(E2)CF(E2)=0.6 x max0,CF(Ei) =0.6 x max0,0.5=0.3(2)再由 r2 求 CF(E4)CF(E4)=0.7 x max0, minCF(E 2 ), CF(E3 ) =0.7 x max0, min0.3, 0.6=0

2、.2i再由3求CFi(H)CFi(H)= 0.8 x max0,CF(E4) =0.8 x max0, 0.2i)=0.i68 (4)再由求 CF2(H)CF2(H)= 0.9 x max0,CF(E5) =0.9 x max0, 0.7)=0.63(5)最后對(duì)CFi(H ) CF2(H)進(jìn)行合成,求出CF(H)CF(H)=CFi(H)+CF2(H)+ CFi(H) X CF2(H)=0.6926.10 設(shè)有如下推理規(guī)那么ri：IF Ei THEN(2, 0.0000i)HiIF E2 THEN(i00, 0.000i)Hir3：IF E3 THEN(200, 0.00i)H2r4：IF Hi

3、 THEN(50, 0.i) H2且 P(Ei)= P(E2)= P(H 3)=0.6, P(Hi)=0.09i,P(H2)=0.0i,又由用戶告知：P(Ei| Si)=0.84, P(E20)=0.68, P(E3|S3)=0.36請(qǐng)用主觀Bayes方法求P(H2|Si, &, &)=?解：(i)由門(mén)計(jì)算O(Hi| S)先把Hi的先驗(yàn)概率更新為在 Ei下的后驗(yàn)概率P(Hi| Ei)P(Hi| Ei)=(LS i X P(Hi) / (LSi-i) X P(Hi)+i)=(2 X 0.09i) / (2 -i) X 0.09i +i)=0.i6682由于P(Ei|Si)=0.84 P(Ei),

4、使用P(H | S)公式的后半局部,得到在當(dāng)前觀察Si下的后驗(yàn)概率P(Hi| Si)和后驗(yàn)幾率O(Hi| Si)P(Hi| Si) = P(Hi) + (P(Hi| Ei) P(Hi) / (1 - P(Ei) X (P(Ei| Si) P(Ei) =0.09I + (0.I6682 -0.09I) / (i -0.6) X (0.84 -0.6) =0.09I + 0.I8955 X 0.24 = 0.I36492O(Hi| Si) = P(Hi| Si) / (i - P(Hi| Si) =0.I5807(2)由 r2 計(jì)算 O(Hi | S2)先把Hi的先驗(yàn)概率更新為在 E2下的后驗(yàn)概率

5、P(Hi| E2)P(Hi| E2)=(LS2 X P(Hi) / (LS2-I) X P(Hi)+I)=(i00 X 0.09I) / (I00 -i) X 0.09I +i) =0.909i8由于P(E2|S2)=0.68 P(E2),使用P(H | S)公式的后半局部,得到在當(dāng)前觀察S2下的后驗(yàn)概率P(Hi| S2)和后驗(yàn)幾率O(Hi| S2)P(Hi| S2) = P(Hi) + (P(Hi| E2) - P(Hi) / (i - P(E2) X (P(E2| S2) - P(E2) =0.09I + (0.909I8 -0.09I) / (i -0.6) X (0.68 -0.6)

6、=0.25464O(Hi| S2) = P(Hi| S2) / (i - P(Hi| S2) =0.34i63 計(jì)算 O(Hi| Si,S2)和 P(Hi| Si,S2) 先將Hi的先驗(yàn)概率轉(zhuǎn)換為先驗(yàn)幾率O(Hi) = P(Hi) / (i - P(Hi) = 0.09i/(i-0.09i)=0.i00ii再根據(jù)合成公式計(jì)算Hi的后驗(yàn)幾率O(Hi| Si,S2)= (O(Hi| Si) / O(H i) X (O(Hi| S2) / O(Hi) X O(Hi)=(0.I5807 / 0.I00II) X (0.34I63) / 0.I00II) X 0.I00II =0.53942再將該后驗(yàn)幾

7、率轉(zhuǎn)換為后驗(yàn)概率P(Hi| Si,S2) = O(Hi| Si,S2) / (i+ O(Hi| Si,S2) =0.35040(4)由 r3 計(jì)算 O(H2| S3)先把H2的先驗(yàn)概率更新為在 E3下的后驗(yàn)概率P(H2| E3)P(H2| E3)=(LS3 x P(H2)/ (LS3-I) x P(H2)+I) =(200 X 0.0i) / (200 -i) X 0.0i +i) =0.09569由于P(E3|S3)=0.36 P(Hi),使用P(H | S)公式的后半局部,得到在當(dāng)前觀察Si,S2下H2的后驗(yàn)概率P(H2| S1,S2)和后驗(yàn)幾率O(H2| S1,S2)P(H2| Si,S

8、2)= P(H2) + (P(H 2| Hi) P(H2) / (1 - P(H 1) X (P(Hi| Si,S2) P(Hi) =0.01 + (0.33557 -0.01) / (1 -0.091) X (0.35040 -0.091) =0.10291O(H2| Si,S2)= P(H2| Si, S2) / (1 - P(H2| Si, S2) =0.10291/ (1 - 0.10291) = 0.11472 (6)計(jì)算 O(H2| S1,S2,S3)和 P(H2| S1,S2,S3) 先將H2的先驗(yàn)概率轉(zhuǎn)換為先驗(yàn)幾率O(H2) = P(H2) / (1 - P(H2) )= 0.

9、01 / (1-0.01)=0.01010再根據(jù)合成公式計(jì)算H1的后驗(yàn)幾率O(H2| S1,S2,S3)= (O(H 2| S1,S2) / O(H2) X (O(H2| S3) / O(H2) X O(H2) =(0.11472 / 0.01010) X (0.00604) / 0.01010) X 0.01010 =0.06832再將該后驗(yàn)幾率轉(zhuǎn)換為后驗(yàn)概率P(H2| Si,S2,S3)= O(Hi| Si,S2,S3)/ (1+ O(Hi| Si,S2,S3) =0.06832 / (1+ 0.06832) = 0.06395可見(jiàn),H2原來(lái)的概率是0.01,經(jīng)過(guò)上述推理后得到的后驗(yàn)概率是

10、0.06395,它相當(dāng)于先驗(yàn)概率的6倍多.6.11 設(shè)有如下推理規(guī)那么r1:IFEiTHEN(100, 0.1)HiIFE2THEN(50, 0.5)H2IFE3THEN(5, 0.05)H3且P(Hi)=0.02, P(H2)=0.2, P(H3)=0.4,請(qǐng)計(jì)算當(dāng)證據(jù)Ei, E2, E3存在或不存在時(shí)P(Hi | Ei)或 P(Hi |Ei)的值各是多少(i=1,2, 3)?解：(1)當(dāng)E1、E2、E3肯定存在時(shí),根據(jù) 小2、r3有P(Hi | E1) = (LS1 X P(H1) / (LS1-1) X P(H1)+1)=(100 X 0.02) / (100 -1) X 0.02 +1

11、)=0.671P(H2 | E2)= (LS2X P(H2) / (LS2-1) X P(H2)+1) =(50 X 0.2) / (50 -1) X 0.2 +1) =0.9921P(H3 | E3) = (LS3X P(H3)/ (LS3-1) X P(H3)+1) =(5 X 0.4) / (5 -1) X 0.4 +1) =0.769 當(dāng)Ei、E2、E3肯7E存在時(shí),根據(jù)1、2、3有 P(Hi | ?E1) = (LN 1 X P(H1) / (LN 1-1) X P(H1)+1)=(0.1 X 0.02) / (0.1 -1) X 0.02 +1) =0.002P(H2 | ?E2)

12、 = (LN 2 X P(H2) / (LN 2-1) X P(H2)+1) =(0.5 X 0.2) / (0.5 -1) X 0.2 +1) =0.111P(H3 | ?E3) = (LN 3 X P(H3) / (LN 3-1) X P(H3)+1) =(0.05 X 0.4) / (0.05 -1) X 0.4 +1) =0.0326.13 設(shè)有如下一組推理規(guī)那么r1:IFE1AND E2 THEN AIFE2AND (E3 OR E4)3:IFATHEN H=h1, h2, h34:IFBTHEN H=h1, h2, h3且初始證據(jù)確實(shí)定性分別為:a (CF=0.9)THEN B=b

13、1, b2(CF=0.8, 0.7)(CF=0.6, 0.5, 0.4)(CF=0.3, 0.2, 0.1)CER(E1)=0.6, CER(E2)=0.7, CER(E3)=0.8, CER(E4)=0.9.假設(shè) | Q|=10 ,求 CER(H).解：其推理過(guò)程參考例6.9具體過(guò)程略6.15 設(shè)U=V=1 , 2, 3, 4且有如下推理規(guī)那么：IF x is 少 THEN y is 多其中,“少與“多分別是 U與V上的模糊集,設(shè)少=0.9/1+0.7/2+0.4/3多=0.3/2+0.7/3+0.9/4事實(shí)為x is 較少“較少的模糊集為較少=0.8/1+0.5/2+0.2/3請(qǐng)用模糊關(guān)系

14、Rm求出模糊結(jié)論.解：先用模糊關(guān)系Rm求出規(guī)那么IF x is 少 THEN y is 多 所包含的模糊關(guān)系RmRm (1,1)=(0.9 A0) V (1-0.9)=0.1Rm (1,2)=(0.9 A 0.3) V (1-0.9)=0.3 Rm (1,3)=(0.9 A 0.7) V (1-0.9)=0.7 Rm (1,4)=(0.9 A 0.9) V (1-0.9)=0.7 Rm (2,1)=(0.7 A0) V (1-0.7)=0.3 Rm (2,2)=(0.7 A 0.3) V (1-0.7)=0.3 Rm (2,3)=(0.7 A 0.7) V (1-0.7)=0.7 Rm (2,

15、4)=(0.7 A 0.9) V (1-0.7)=0.7 Rm (3,1)=(0.4 A0) V (1-0.4)=0.6 Rm (3,2)=(0.4 A 0.3) V (1-0.4)=0.6 Rm (3,3)=(0.4 A 0.7) V (1-0.4)=0.6 Rm (3,4)=(0.4 A 0.9) V (1-0.4)=0.6 Rm (4,1)=(0 A0) V(1-0)=1Rm (4,2)=(0 A0.3)V (1-0)=1Rm (4,3)=(0 A0.7)V (1-0)=1 Rm (3,4)=(0 A0.9)V (1-0)=1 即：0.1 0.3 0.7 0.90.3 0.3 0.7 0.7Rm =0.6 0.6 0.6 0.6J 1111 因此有；0.8,0.5,0.2,00.10.30.60.30.30.610.7 0.90.7 0.70.6 0.6110.3,030.7,0.8 )即,模糊結(jié)論為Y =0.3, 0.3, 0.7, 0.86.16 設(shè)U=V=W=1,2,3,4 且設(shè)有如下規(guī)那么：r1： IF x is F THEN y is Gr2：IFyisGTHENzisHr3：IFxisFTHENzisH其中,F、G、H的模糊集分別為：F=1/1+0.8/2+0.5/3+0.4/4G=0.1/2+0.2/3+0.4