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当前位置:首页 > 商业/管理/HR > 信息化管理 > 测度与概率(严士健-刘秀芳)第六-七章答案
x6.11,F¾-,£¢:=f(!;!):!2g62F£F,8!i2;i=1;2;¢!1:=f!2:(!1;!2)2¢g2F;¢!2:=f!1:(!1;!2)2¢g2F;?:¢62F£F.,3.19(i)G:=fF2:FFc,g¾-.G,¾-,F=G=fF2:FFc,g.A:=fA2£:A(1Si=1(Ai£yi))S(1Sj=1(xj£Bj))(1Si=1fAi£yig)S(1Sj=1fxj£Bjg)S(1Sk=1(Ck£Dk));Ai;Bj;2F;(Ck)c;(Dk)c;xj;yi2;i;j;k¸1;xj;yig.F£F½A.C:=fF£H:F;H2Fg,F£F=¾(C).FA,C½A,A¾-.,;;2A.,A.,A.A=(1Si=1(Ai£yi))S(1Sj=1(xj£Bj))S(1Sk=1(Ck£Dk));Ai;Bj;2F;(Ck)c;(Dk)c;xj;yi2;i;j;k¸1;xj;yi:Ac=(1[i=1(Ai£yi))c\(1[j=1(xj£Bj))c\(1[k=1(Ck£Dk))c=((£fyi:i¸1gc)[1[i=1(Aci£yi))\((fxj:j¸1gc£)[1[j=1(xj£Bcj))\1\k=1((Cck£)[(£Dck))1Tk=1((Cck£)[(£Dck)),((Cc1£)[(£Dc1))\((Cc2£)[(£Dc2))=((Cc1\Cc2)£)[(Cc1£Dc2)[(Cc2£Dc1)[(£(Dc1\Dc2))Cc1\Cc2;Cc1,Cc1\Cc2=fwi:i¸1g;Cc1=fvj:j¸1g,(Cc1\Cc2£)=1Si=1(wi£),(Cc1£Dc2)=1Sj=1(vj£Dc2),((Cc1\Cc2)£)[(Cc1£Dc2)1Sj=1(xj£Bj),(Cc2£Dc1)[(£(Dc1\Dc2))11Si=1(Ai£yi),2Tk=1((Cck£)[(£Dck))(1Si=1(Ai£yi))S(1Sj=1(xj£Bj)),(Cck£)[(£Dck);k=3;4;¢¢¢,1Tk=1((Cck£)[(£Dck))(1Si=1fAi£yig)S(1Sj=1fxj£Bjg).((1Si=1fAi£yig)S(1Sj=1fxj£Bjg)(1Si=1fAi£yig)S(1Sj=1fxj£Bjg).)(£fyi:i¸1gc)S1Si=1(Aci£yi))(1Si=1fAi£yig)S(1Sk=1fCk£Dkg),(fxj:j¸1gc£)S1Sj=1(xj£Bcj)1Sj=1fxj£BjgS(1Sk=1(Ck£Dk)),Ac(1Si=1(Ai£yi))S(1Sj=1(xj£Bj)).A=(1[i=1(Ai£yi))[(1[j=1(xj£Bj));Ai;Bj;2F;xj;yi2;i;j;k¸1;xj;yi:Ac(1Si=1(Ai£yi))S(1Sj=1(xj£Bj))S(1Sk=1(Ck£Dk)).,Ac2A.A¾-,F£F½A,¢A,¢62F£F.!12;¢!1=f!1g2F,!22,¢!2=f!2g2F.\,:,.2F1£F2½F1£F2,F1£F2½F1£F2.:1=2=R;F1;F2RBorel,¹1;¹2RLebesgue.F1;F2RLebesgue.A½R;ALebesgue,!222,A£!2½1£!2,¹1£¹2(1£!2)=0,A£!2¹1£¹2-,A£!22F1£F2,(A£!2)!2=A62F1,A£!262F1£F2.:F1£F2½F1£F2.C:=fA£B:A2F1;B2F2g,F1£F2=¾(C).A£B2C,C2F1;D2F2,¹1-N1,¹2-N2,s.t.A=C[N1;B=D[N2:A£B=(C[N1)£(D[N2)=(C£D)[(C£N2)[(N1£D)[(N1£N2),C£N2;N1£D;N1£N2¹1£¹2-,(C£N2)[(N1£D)[(N1£N2)¹1£¹2-,C£D2F1£F2,A£B2F1£F2,C½F1£F2,F1£F2¾-,F1£F2½F1£F2.Lebesgue:Lebesgue,Lebesgue.32(1):P(X1+X22B)=ZRnP(X12B¡y;X22dy)=ZRnP(X12B¡y)P(X22dy)=ZRnP1(B¡y)P2(dy)X1;X2.X1+X2P1¤P2.P(X1+X2·x)=RRnP1((¡1;x¡y])P2(dy)=RRnF1(x¡y)dF2(y),8x2Rn,X1+X2F1¤F2.(2)X1;X2p1;p2,(1),P(X1+X2·x)=ZRnF1(x¡y)dF2(y)=ZRnZ(¡1;x¡y]p1(z)dzp2(y)dy=Z(¡1;x]ZRnp1(z0+y)p2(y)dydz0z=z0y,.X1+X2p1¤p2.(3),:.(i)P1;P2;P3,(P1¤P2)¤P3=P1¤(P2¤P3):(P1¤P2)¤P3(B)=ZRnP1¤P2(B¡y)P3(dy)=ZRnZRnP1(B¡y¡z)P2(dz)P3(dy)=ZRnZRnP1(B¡z0)P2(dz0¡y)P3(dy)=ZRnP1(B¡z0)ZRnP2(dz0¡y)P3(dy)=ZRnP1(B¡z0)P2¤P3(dz0)=P1¤(P2¤P3)(B).(ii).P1¤P2;P2¤P1.(1),P1¤P2X1+X2,P2¤P1X2+X1,P1¤P2=P2¤P1..43:A:=f(t;!):f(t;!)=1g,At:=f!:f(t;!)=1g;8t2R,A!:=ft:f(t;!)=1g;8!2.,P(At)=0;8t2R,(¸£P)(A)=RRP(At)d¸=0:R¸(A!)dP=(¸£P)(A)=0.¸(A!)¸0;8!2;¸(A!)=0;a:e:!(P);¸(ft:f(t;!)=1g)=0;a:e:!(P):5:F1£F2£¢¢¢£FnA(!i1;!i2;¢¢¢;!ik)Fj1£Fj2£¢¢¢£Fjn¡k.¤=fA2F1£F2£¢¢¢£Fn:A(!i1;!i2;¢¢¢;!ik)Fj1£Fj2£¢¢¢£Fjn¡kg:A=A1£A2£¢¢¢£An;Ai2Fi;i=1;¢¢¢;n,A(!i1;!i2;¢¢¢;!ik)=Aj1£Aj2¢¢¢£Ajn¡k2Fj1£Fj2£¢¢¢£Fjn¡k,F1£F2£¢¢¢£Fn¤.¤¾-.(i);;2¤.(ii)A2¤,A(!i1;!i2;¢¢¢;!ik)2Fj1£Fj2£¢¢¢£Fjn¡k,Ac(!i1;!i2;¢¢¢;!ik)=(A(!i1;!i2;¢¢¢;!ik))c2Fj1£Fj2£¢¢¢£Fjn¡k,Ac2¤.(iii)An2¤;n¸1;(An)(!i1;!i2;¢¢¢;!ik)2Fj1£Fj2£¢¢¢£Fjn¡k;8n¸1,(Sn¸1An)(!i1;!i2;¢¢¢;!ik)=Sn¸1(An)(!i1;!i2;¢¢¢;!ik)2Fj1£Fj2£¢¢¢£Fjn¡k.¤=F1£F2£¢¢¢£Fn:fF1£F2£¢¢¢£Fn,B(Borel),f(!j1;!j2;¢¢¢;!jn¡k):f(!i1;!i2;¢¢¢;!ik)(!j1;!j2;¢¢¢;!jn¡k)2Bg=(f(!1;!2;¢¢¢;!n):f(!1;!2;¢¢¢;!n)2Bg)(!i1;!i2;¢¢¢;!ik)2Fj1£Fj2£¢¢¢£Fjn¡k,.f(!i1;!i2;¢¢¢;!ik)Fj1£Fj2£¢¢¢£Fjn¡k.6:(1)A=A1£A2,g(w1):=Z21A(w1;w2)¸(w1;dw2)(6:1)F1-,A12F1A22F2.¸¾,i=1;2,FifBi;ngn2Ni=[n2NBi;nm;n2Nsupw12B2;m¸(w1;B2;n)1:g(w1)=Xn2NZB2;n1A(w1;w2)¸(w1;dw2)=Xn2N¸(w1;B2;n\A2):4¸(w1;B2;n\A2)F1,gF1.C.(2)A2F1£F2,(6.1)gF1-.¸.M:=fA2F1£F2:8n¸1;ZB2;n1A(w1;w2)¸(w1;dw2)F1-g:(1)C½M.¾(C)=F1£F2.M¸¾(C)½M,F1£F2=M.,1£22C½M.A;B2MB½A,Z21AnB(w1;w2)¸(w1;dw2)=1Xn=1ZB2;n1AnB(w1;w2)¸(w1;dw2)=1Xn=1(ZB2;n1A(w1;w2)¸(w1;dw2)¡ZB2;n1B(w1;w2)¸(w1;dw2))AnB2M.fAngn2N½M½F1,An,Z21[n2NAn(w1;w2)¸(w1;dw2)=Z2limn2N1An(w1;w2)¸(w1;dw2)=limn2NZ21An(w1;w2)¸(w1;dw2)[n2NAn2M.M¸.(3)F1£F2f(1)(2),Z2f(w1;w2)¸(w1;dw2)(6:2)F1-.F1£F2f(6.2)F1-.8:(1)B2B([0;1]),f(x;y)IB(y),f(x;y)IB(y),6.1.9,¸(¢;B)=RBf(¢;y)dy=RIB(y)f(¢;y)dyB([0;1]).(2)x2[0;1],f(x;¢)([0;1];B([0;1])),dy([0;1];B([0;1])),5.3.15,¸(x;B)=RBf(x;y)dy([0;1];B([0;1])).(1)(2),¸(x;B)[0;1]£B([0;1]).9:º(A)=Z1Z2IA(!1;!2)¸2(!1;d!2)¸1(d!1)=Z1¸2(!1;A!1)¸1(d!1)5A!1=f!2:(!1;!2)2Ag.¸2(!1;A!1)!1,5.2.3(3),R1¸2(!1;A!1)¸1(d!1)=0¸2(!1;A!1)=0;a.s.!1-¸1.º(A)=0N21,s.t.¸1(N)=0;!12Nc,¸2(!1;A!1)=0:10:(1)A=A1£A3A12F1A32F3g(w1;w3)=Z31A(w1;w3)¸(w2;dw3)(10:1)F1£F2-.g(w1;w2)=1A1(w1)Z31A3(w3)¸(w2;dw3)=1A1(w1)¸(w2;A3);F1£F2-.CF1£F3.(2)A2F1£F3,(6.1)gF1£F2-.¸.M:=fA2F1£F3:8n¸1;ZB3;n1A(w1;w3)¸(w2;dw3)F1£F2g:¸¾,i=2;3,FifBi;ngn2Ni=[n2NBi;nm;n2Nsupw22B3;m¸(w2;B3;n)1:(1)C½M.¾(C)=F1£F3.M¸¾(C)½M,F1£F3=M.,1£32C½M.A;B2MB½A,Z31AnB(w1;w3)¸(w2;dw3)=1Xn=1ZB3;n1AnB(w1;w3)¸(w2;dw3)=1Xn=1(ZB3;n1A(w1;w3)¸(w2;dw3)¡ZB3;n1B(w1;w3)¸(w2;dw3));AnB2M.fAngn2N½M½f1£F3,An,Z21[n2NAn(w1;w3)¸(w2;dw3)=Z2limn2N1An(w1;w3)¸(w2;dw3)=limn2NZ21An(w1;w3)¸(w2;dw3)[n2NAn2M.M¸.6(3)F1£F3f(1)(2),Z2f(w1;w3)¸(w2;dw3)(10:2)F1£F2-.F1£F3f(10.2)F1£F2-.14:(1)A2F,¼(x;A)¸0,º¼(A)=R¼(x;A)º(dx)¸0.º¼()=R¼(x;)º(dx)=1.(2)An2F;8n¸1,.x2,¼(x;¢)F,,¼(x;Sn¸1An)=Pn¸1¼(x;An),º¼(Sn¸1An)=R¼(x;Sn¸1An)º(dx)=RPn¸1¼(x;An)º(dx)=Pn¸1R¼(x;An)º(dx)=Pn¸1¼º(An).(1)(2)¼º(¢)F.7x6.2P1531:E»n=ZRxnF(dx)=Z10Zx0ntn¡1dtF(d
本文标题:测度与概率(严士健-刘秀芳)第六-七章答案
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