不確定性推理部分參考答案

1、第6章不確定性推理部分參考答案6.8設(shè)有如下一組推理規(guī)則ri:IFEiTHENE2(0.6)r2：IFE2ANDE3THENE4(0.7)IFE4THENH(0.8)r4：IFE5THENH(0.9)且已知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)=0.6xmax0,CF(Ei)=0.6xmax0,0.5=0.3(2)再由r2求CF(E4)CF(E4)=0.7xmax0,minCF(E2),CF(E3)=0.7xmax0,min0.3,0.6=0.2i再由3求CFi(H)CFi(H)=0.8xmax0,CF(E

2、4)=0.8xmax0,0.2i)=0.i68(4)再由"求CF2(H)CF2(H)=0.9xmax0,CF(E5)=0.9xmax0,0.7)=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)XCF2(H)=0.6926.10設(shè)有如下推理規(guī)則ri：IFEiTHEN(2,0.0000i)HiIFE2THEN(i00,0.000i)Hir3：IFE3THEN(200,0.00i)H2r4：IFHiTHEN(50,0.i)H2且已知P(Ei)=P(E2)=P(H3)=0.6,P(Hi)=0.09i,P(H2)=0.0

3、i,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)=(LSiXP(Hi)/(LSi-i)XP(Hi)+i)=(2X0.09i)/(2-i)X0.09i+i)=0.i6682由于P(Ei|Si)=0.84>P(Ei),使用P(H|S)公式的后半部分，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(H

4、i)/(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.I8955X0.24=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P(Hi|E2)P(Hi|E2)=(LS2XP(Hi)/(LS2-I)XP(Hi)+I)=(i00X0.09I)/(I00-i)X0.09I+i)=0.909i8由于P(E2|S2)=0.68>P(E2),使用P(H|S)公式的后半部分，得到在當(dāng)前觀察S2

5、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)=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(Hi)X(O(

6、Hi|S2)/O(Hi)XO(Hi)=(0.I5807/0.I00II)X(0.34I63)/0.I00II)X0.I00II=0.53942再將該后驗(yàn)幾率轉(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)=(LS3xP(H2)/(LS3-I)xP(H2)+I)=(200X0.0i)/(200-i)X0.0i+i)=0.09569由于P(E3|S3)=0.36<P(E3),使用P(H|S)公式的前半部分，得到在當(dāng)前觀察S3下的后

7、驗(yàn)概率P(H2|S3)和后驗(yàn)幾率O(H2|S3)P(H2|S3)=P(H2|?E3)+(P(H2)P(H2|?E3)/P(E3)XP(E3|S3)由當(dāng)E3肯定不存在時(shí)有P(H2|?E3)=LN3XP(H2)/(LN3-I)XP(H2)+i)=0.00IX0.0i/(0.00I-i)X0.0i+i)=0.0000I因此有P(H2|S3)=P(H2|?E3)+(P(H2)-P(H2|?E3)/P(E3)XP(日|S3)=0.00001+(0.01-0.00001)/0.6)X0.36=0.00600O(H2|S3)=P(H2|S3)/(1-P(H2|S3)=0.00604(5)由"計(jì)算O

8、(H2|H1)先把H2的先驗(yàn)概率更新為在Hi下的后驗(yàn)概率P(H2|Hi)P(H2|Hi)=(LS4XP(H2)/(LS4-1)XP(H2)+1)=(50X0.01)/(50-1)X0.01+1)=0.33557由于P(Hi|Si,S2)=0.35040>P(Hi),使用P(H|S)公式的后半部分，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,S2)=P(H2)+(P(H2|Hi)P(H2)/(1-P(H1)X(P(Hi|Si,S2)P(Hi)=0.01+(0.33557-0.01)/(1-0.091)X(0.35040-0.

9、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)=0.01/(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(H2|S1,S2)/O(H2)X(O(H2|S3)/O(H2)XO(H2)=(0.11472/0.01010)X(0.00604)/0.01010)X0.01010=0.06832再將該后驗(yàn)幾

10、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，H2原來的概率是0.01,經(jīng)過上述推理后得到的后驗(yàn)概率是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)則r1:IFEiTHEN(100,0.1)HiIFE2THEN(50,0.5)H2IFE3THEN(5,0.05)H3且已知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)

11、E1、E2、E3肯定存在時(shí)，根據(jù)小2、r3有P(Hi|E1)=(LS1XP(H1)/(LS1-1)XP(H1)+1)=(100X0.02)/(100-1)X0.02+1)=0.671P(H2|E2)=(LS2XP(H2)/(LS2-1)XP(H2)+1)=(50X0.2)/(50-1)X0.2+1)=0.9921P(H3|E3)=(LS3XP(H3)/(LS3-1)XP(H3)+1)=(5X0.4)/(5-1)X0.4+1)=0.769當(dāng)Ei、E2、E3肯7E存在時(shí)，根據(jù)1、2、3有P(Hi|?E1)=(LN1XP(H1)/(LN1-1)XP(H1)+1)=(0.1X0.02)/(0.1-1)

12、X0.02+1)=0.002P(H2|?E2)=(LN2XP(H2)/(LN2-1)XP(H2)+1)=(0.5X0.2)/(0.5-1)X0.2+1)=0.111r1:IFE1ANDE2THENAIFE2AND(E3ORE4)3:IFATHENH=h1,h2,h34:IFBTHENH=h1,h2,h36.13設(shè)有如下一組推理規(guī)則且已知初始證據(jù)的確定性分別為:P(H3|?E3)=(LN3XP(H3)/(LN3-1)XP(H3)+1)=(0.05X0.4)/(0.05-1)X0.4+1)=0.032a(CF=0.9)THENB=b1,b2(CF=0.8,0.7)(CF=0.6,0.5,0.4)(

13、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：其推理過程參考例6.9具體過程略6.15 設(shè)U=V=1,2,3,4且有如下推理規(guī)則：IFxis少THENyis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已知事實(shí)為xis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Rm求出模糊結(jié)論。解：先用模糊關(guān)系Rm求出規(guī)則IFxis少THENyis多所包含的模糊關(guān)系RmRm(1,1)=(

14、0.9A0)V(1-0.9)=0.1Rm(1,2)=(0.9A0.3)V(1-0.9)=0.3Rm(1,3)=(0.9A0.7)V(1-0.9)=0.7Rm(1,4)=(0.9A0.9)V(1-0.9)=0.7Rm(2,1)=(0.7A0)V(1-0.7)=0.3Rm(2,2)=(0.7A0.3)V(1-0.7)=0.3Rm(2,3)=(0.7A0.7)V(1-0.7)=0.7Rm(2,4)=(0.7A0.9)V(1-0.7)=0.7Rm(3,1)=(0.4A0)V(1-0.4)=0.6Rm(3,2)=(0.4A0.3)V(1-0.4)=0.6Rm(3,3)=(0.4A0.7)V(1-0.4)

15、=0.6Rm(3,4)=(0.4A0.9)V(1-0.4)=0.6Rm(4,1)=(0A0)V(1-0)=1Rm(4,2)=(0A0.3)V(1-0)=1Rm(4,3)=(0A0.7)V(1-0)=1Rm(3,4)=(0A0.9)V(1-0)=1即：0.10.30.70.90.30.30.70.7Rm二0.60.60.60.6J111-0.30.30.610.70.90.70.70.60.611因此有0.1Y'=o.8,O.5,O.2,oQ0.30.6"0.3,030.7,0.8)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)則：r1：IFxisFTHENyisGr2：IFyisGTHENzisHr3：IFxisFTHENzisH其中，F(xiàn)、G、H的模糊集分別為：F=1/1+0.8/2+0.5/3+0.4/