收藏 分享(赏)

华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf

上传人:嘭** 文档编号:75559 上传时间:2023-02-15 格式:PDF 页数:40 大小:780.76KB
下载 相关 举报
华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf_第1页
第1页 / 共40页
华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf_第2页
第2页 / 共40页
华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf_第3页
第3页 / 共40页
华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf_第4页
第4页 / 共40页
华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf_第5页
第5页 / 共40页
华东师范大学《概率论与数理统计》课件-第二章上(许忠好版).pdf_第6页
第6页 / 共40页
亲,该文档总共40页,到这儿已超出免费预览范围,如果喜欢就下载吧!
资源描述

1、1 C?1?C1 C?Z(Statistics,ECNU)1?C1/10?二?上1 C?I?fy?:X1,2,3,4,5,6I n?Y0,1,2,nI,|US5?Z0,1,2,I,.?T0,)Z(Statistics,ECNU)1?C2/101 C?I?fy?:X1,2,3,4,5,6I n?Y0,1,2,nI,|US5?Z0,1,2,I,.?T0,)Z(Statistics,ECNU)1?C2/101 C?I?fy?:X1,2,3,4,5,6I n?Y0,1,2,nI,|US5?Z0,1,2,I,.?T0,)Z(Statistics,ECNU)1?C2/101 C?I?fy?:X1,2,3,

2、4,5,6I n?Y0,1,2,nI,|US5?Z0,1,2,I,.?T0,)Z(Statistics,ECNU)1?C2/101 C?C?Definition 1?(1,F1)(2,F2)m,X1?2?N?,e?B F2,X1(B)F1,KXd(1,F1)?(2,F2)?NNN?,PX F1/F2.Z(Statistics,ECNU)1?C3/101 C?Definition 2eXdm(,F)?(R,B(R)?N?,XCCC(?r.v.).Theorem 3?(,F),m,X3?,KXC?=?x,:X()x F.d4?-“?.Z(Statistics,ECNU)1?C4/101 C?Def

3、inition 2eXdm(,F)?(R,B(R)?N?,XCCC(?r.v.).Theorem 3?(,F),m,X3?,KXC?=?x,:X()x F.d4?-“?.Z(Statistics,ECNU)1?C4/101 C?Definition 2eXdm(,F)?(R,B(R)?N?,XCCC(?r.v.).Theorem 3?(,F),m,X3?,KXC?=?x,:X()x F.d4?-“?.Z(Statistics,ECNU)1?C4/101 C?Example 4?(,F),m,c?X()=c,KXC.ukx R,ex cKx :X()x=F;ex cKx :X()x=F,Z(St

4、atistics,ECNU)1?C5/101 C?Example 4?(,F),m,c?X()=c,KXC.ukx R,ex cKx :X()x=F;ex cKx :X()x=F,Z(Statistics,ECNU)1?C5/101 C?BernoulliCExample 5?q!?M1,=H,T,F=,H,T.X()=1,e=H;0,e=T.KXC.d :X()x=,x 0,T,0 x 1,x 1 F XC.kU?CBernoulliCCC.Z(Statistics,ECNU)1?C6/101 C?BernoulliCExample 5?q!?M1,=H,T,F=,H,T.X()=1,e=H

5、;0,e=T.KXC.d :X()x=,x 0,T,0 x 1,x 1 F XC.kU?CBernoulliCCC.Z(Statistics,ECNU)1?C6/101 C?BernoulliCExample 5?q!?M1,=H,T,F=,H,T.X()=1,e=H;0,e=T.KXC.d :X()x=,x 0,T,0 x 1,x 1 F XC.kU?CBernoulliCCC.Z(Statistics,ECNU)1?C6/101 C?Example 6?(,F,P)?Vm,A F.1A()=1,e A;0,e A.K1ABernoulliC(1AA?5).AO5?A=:1A()=1.:X(

6、)x=,x 0,A,0 x 1,x 1 F.Z(Statistics,ECNU)1?C7/101 C?Example 6?(,F,P)?Vm,A F.1A()=1,e A;0,e A.K1ABernoulliC(1AA?5).AO5?A=:1A()=1.:X()x=,x 0,A,0 x 1,x 1 F.Z(Statistics,ECNU)1?C7/101 C?5:(1)C,(,).(2)X=ka X b.(3)X=k=X k X k,a a,(4)?m?C.Z(Statistics,ECNU)1?C8/101 C?5:(1)C,(,).(2)X=ka X b.(3)X=k=X k X k,a

7、a,(4)?m?C.Z(Statistics,ECNU)1?C8/101 C?5:(1)C,(,).(2)X=ka X b.(3)X=k=X k X k,a a,(4)?m?C.Z(Statistics,ECNU)1?C8/101 C?5:(1)C,(,).(2)X=ka X b.(3)X=k=X k X k,a a,(4)?m?C.Z(Statistics,ECNU)1?C8/101 C?Example 7M1=H,T,F=,H,T,KY()=1T()C.5?X.Z(Statistics,ECNU)1?C9/101 C?Example 7M1=H,T,F=,H,T,KY()=1T()C.5?

8、X.Z(Statistics,ECNU)1?C9/101 C?Example 8?=a,b,c?m,F=,b,a,c.X()=1b()+2 1c(),KXC.X 0=a F.Z(Statistics,ECNU)1?C10/101 C?Example 8?=a,b,c?m,F=,b,a,c.X()=1b()+2 1c(),KXC.X 0=a F.Z(Statistics,ECNU)1?C10/101?1?C1?Z?Z(Statistics,ECNU)1?C1/141?Definition 1?XC,?x,F(x)=P(X x)X?.Z(Statistics,ECNU)1?C2/141?Examp

9、le 2?q!?M1,=H,T,F=,H,T,VPvP(H)=1/2.CX()=1H()?F(x)=P(X x)=0,x 0;1/2,0 x 1;1,1 x.Z(Statistics,ECNU)1?C3/141?Example 3?q!?M1,=H,T,F=,H,T,VPvP(H)=1/2.CY()=1,e=T;0,e=H.?F(y)=P(Y y)=0,y 0;1/2,0 y 1;1,1 y.2?Ck?.Z(Statistics,ECNU)1?C4/141?Example 3?q!?M1,=H,T,F=,H,T,VPvP(H)=1/2.CY()=1,e=T;0,e=H.?F(y)=P(Y y)

10、=0,y 0;1/2,0 y 1;1,1 y.2?Ck?.Z(Statistics,ECNU)1?C4/141?5(1)N;(2)k.:0 F(x)1,F(+)=1,F()=0.(3)mY:F(x+0)=F(x).Z(Statistics,ECNU)1?C5/141?5(1)?yex y,KX x X y.dV?N5,P(X x)P(X y),=F(x)F(y).Z(Statistics,ECNU)1?C6/141?5(2)?y?x,F(x)X xu)?V,?0 F(x)1.dF(x)?N5,F(+)F()3,AO/,F(+)=limnF(n),F()=limnF(n).PAk=:k 1 X(

11、)k,k=0,1,2,.,u=X =Xk=Ak.dV?,1=P()=PXk=Ak=Xk=P(Ak)=limnnXk=n+1P(Ak)=limn(F(n)F(n)=F(+)F().q0 F(x)1,?F(+),F()0,1,l?7kF(+)=1,F()=0.Z(Statistics,ECNU)1?C7/141?5(3)?y?x,dF?N5,F(x+0)3F(x+0)=limnF x+1n!.PBk=(x+1k+1 X x+1k),k=1,2,.ux X x+1=Xk=1Bk,(x+1n+1 X x+1)=nXk=1Bk,n=1,2,.Z(Statistics,ECNU)1?C8/141?5(3)

12、?yYdV?5nk5,F(x+1)F(x)=P(x X x+1)=PXk=1Bk=Xk=1P(Bk)=limnnXk=1P(Bk)=limnPnXk=1Bk=limnP x+1n+1 0;0,x 0.AB.dF(0)=0F(x)30:?mY5,A+B=0;qdF(+)=1A=1,?B=1.Z(Statistics,ECNU)1?C10/141?Example 4CX?F(x)=A+Bex22,x 0;0,x 0.AB.dF(0)=0F(x)30:?mY5,A+B=0;qdF(+)=1A=1,?B=1.Z(Statistics,ECNU)1?C10/141?Theorem 5?FCX?,K(1)

13、P(X x)=1 F(x)(2)P(X x)=F(x 0),limyxF(y)(3)P(X=x)=F(x)F(x 0)(4)P(X x)=1 F(x 0)(5)P(a X b)=F(b)F(a)(6)P(a X b)=F(b 0)F(a)(7)P(a X b)=F(b 0)F(a 0)(8)P(a X b)=F(b)F(a 0)Z(Statistics,ECNU)1?C11/141?n5(2)?ydF?N5,?x R,F(x 0)3.u,F(x 0)=limnF x 1n!.?x R,PA1=X x 1,An=(x 1n 1 X x 1n),n=2,3,.Ny,An,n 1pN?,X x=Xn=1An,P(A1)=F(x 1),P(An)=F x 1n!F x 1n 1!,n=2,3,.Z(Statistics,ECNU)1?C12/141?n5(2)?yYudV?5,P(X x)=PXn=1An=Xn=1P(An)=limnnXk=1P(Ak)=limnF(x 1)+nXk=2F x 1k!F x 1k 1!#=limnF x 1n!=F(x 0).Z(Statistics,ECNU)1?C13/141?C?XI 3?Vm(,F,P),XJCXk?,KkXA?F(x).I,e3(,)?Fvn5,y3CX,?F X?.Z(Statistics,ECNU)1?C14/14

展开阅读全文
相关资源
猜你喜欢
相关搜索

当前位置:首页 > 教育教学 > 教案课件

copyright@ 2008-2023 wnwk.com网站版权所有

经营许可证编号:浙ICP备2024059924号-2