3、S3)+d2(X2),B的取值可能为0,123,4,5,6,7,8,f3(S3)+d2(X2)?2(S2)♦x2以S2)012345678000814447226915266340311940405S244545414460604550505566647070365155608086747386275256658510096777410018535766901051109978751100第三步:f](Sj=fXS2)+fXS3)+d
4、(X]),A的取值可能为0,1,2,3,4,5,6,7,8,fl(Sl)込(S
5、2)+$(S3)+di(xi)fiO)X
6、*012345678Si1101051011101401301211021001404求得最大值为140,此吋的x「,x2*,x3*,分别是4,4,0,即A项戸投资4千万元,B项PI投资4千万元,C项F1部投资。马尔科夫的抛硬币问题解:1、若O(HTHT)当t=l时,即d](l)=l/3*0.5=l/65(2)=1/3*0.75=0.255(3)=1/3*0.25=1/12当t=2吋?a2(l)=(l/6*0.9+0.25*0.45+l/12*0.45)*0.5=0.15
7、a2(2)=(l/6*0.05+0.25*0.1+1/12*0.45)*0.25=17/960a2(3)=(l/6*0.05+0.25*0.45+l/12*0.1)*0.75=31/320当t=3时,a3(l)=(0.15*0.9+l7/960*0.45+31/320*0.45)*0.5=0.02180252a3(2)=(0.15*0.05+17/960*0.1+31/320*0.45)*0.75=0.03269718a3(2)=(0.15*0.05+17/960*0.45+31/320*0.1)*0.25=0.
8、00242250当1=4吋,O4(l)=(0.02180252*0.9+0.03269718*0.45+0.00242250*0.45)*0.5=0.005842O4(2)=(0.02180252*0.05+0.03269718*0.1+0.00242250*0.45)*0.25=0.001175a5(3)=(0.02180252*0.05+0.03269718*0.45+0.00242250*0.1)*0.75=0.001322所以P(0
9、X)=0.005842+0.001175+0.001322=0.0083
10、392、利用以下方法计算最可能的硬币选择序列当t=l时,8](1)=1/3*0.5=1/6,屮1=051(2)=1/3*0.75=0.25,屮]=0§1(3)=1/3*0.25=1/12,屮1=0当t=2时,52(1)=max{l/6*0.9,0.25*0.45,l/12*0.45}*0.5=0.075屮2(l)=argmax{l/6*0.9?0.25*0.45,1/12*0.45)=182(2)=max{l/6*0.05,0.25*0.1?l/12*0.45}*0.25=3/320T2(2)=argmax{l/
11、6*0.05?0.25*0.1,1/12*0.45)=382(3)=max{1/6*0.05,0.25*0.45,1/12*0.1}*0.75=27/320K?2(3)=argmax{1/6*0.05,0.25*0.45,1/12*0.1}=2当t=3时,§3(l)=max{0.075*0.9,3/320*0.45,27/320*0.45}*0.5=0.03375屮3(1