概率论与数理统计(茆诗松)第二版课后第四章习题参考答案.pdf
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1第四章第四章大数定律与中心极限定理大数定律与中心极限定理习题习题4.11如果XXPn,且YXPn试证:
PX=Y=1证:
因|XY|=|(XnX)+(XnY)|XnX|+|XnY|,对任意的0,有+2|2|0YXPXXPYXPnn,又因XXPn,且YXPn,有02|lim=+XXPnn,02|lim=+YXPnn,则P|XY|=0,取k1=,有01|=kYXP,即11|=kYXP,故11|lim1|1=0,有+2|2|)()(|0YYPXXPYXYXPnnnn,又因XXPn,YYPn,有02|lim=+XXPnn,02|lim=+YYPnn,故0|)()(|lim=+YXYXPnnn,即YXYXPnn+;
(2)因|XnYnXY|=|(XnX)Yn+X(YnY)|XnX|Yn|+|X|YnY|,对任意的0,有+2|2|0YYXPYXXPXYYXPnnnnn,对任意的h0,存在M10,使得4|1hMXP0,使得8|2hMYP0,当nN1时,81|hYYPn,因|Yn|=|(YnY)+Y|YnY|+|Y|,有4|1|1|22hMYYYPMYPnn0,当nN2时,4)1(2|2hMXXPnmaxN1,N2时,有22441|)1(2|2|22hhhMYPMXXPYXXPnnnn=+0,当nN3时,42|1hMYYPn,有244|2|2|11hhhMXPMYYPXYYPnn=+0,当nmaxN1,N2,N3时,有hhhYYXPYXXPXYYXPnnnnn=+0,存在M0,使得4|hMXP0,当nN1时,41|hXXPn,因|Xn|=|(XnX)+X|XnX|+|X|,则244|1|1|hhhMXPXXPMXPnn=+0,存在0,当|xy|时,有|g(x)g(y)|0,当nN2时,4|hXXPn0,当nmaxN1,N2时,有|1|)()(|0MXMXXXPXgXgPnnn+UUhhhhMXPMXPXXPnn=+0,有0|lim=+caXPnn,故0|lim=+cacXPnn,即cacXPn5试证:
XXPn的充要条件为:
n+时,有0|1|+XXXXEnn3证:
以连续随机变量为例进行证明,设XnX的密度函数为p(y),必要性:
设XXPn,对任意的0,都有0|lim=+XXPnn,对012+,存在N0,当nN时,+1|2XXPn,则+=+=+|)(|1|)(|1|)(|1|1|yynndyypyydyypyydyypyyXXXXE=+=+11|1)()(12|XXPXXPdyypdyypnnyy,故n+时,有0|1|+XXXXEnn;充分性:
设n+时,有0|1|+XXXXEnn,因+=|)(|1|1)(11)(|yyyndyypyydyypdyypXXP+=+|1|1)(|1|1XXXXEdyypyynn,故0|lim=+XXPnn,即XXPn6设D(x)为退化分布:
x时,有x+n0,D(x+n)=1,即1)(lim=+nxDn,则D(x+n)的极限函数是常量函数f(x)=1,有f()=10,故D(x+n)的极限函数不是分布函数;
(2)若x0,有01+nx,11=+nxD,即11lim=+nxDn,若x时,有01+nx,01=+nxD,即01lim=+nxDn,则=+.0,1;0,01limxxnxDn这是在0点处单点分布的分布函数,满足分布函数的基本性质,4故+nxD1的极限函数是分布函数;(3)若x0,有010,当xn1时,有01nx,11=nxD,即11lim=+nxDn,则=+.0,1;0,01limxxnxDn在x=0处不是右连续,故nxD1的极限函数不是分布函数7设分布函数列Fn(x)弱收敛于连续的分布函数F(x),试证:
Fn(x)在(,+)上一致收敛于分布函数F(x)证:
因F(x)为连续的分布函数,有F()=0,F(+)=1,对任意的0,取正整数2k,则存在分点x1x2xk1,使得1,2,1,)(=kikixFiL,并取x0=,xk=+,可得kkikxFxFii,1,2,1,21)()(1=0,当nN时,1,2,1,2|)()(|=kixFxFiinL,且显然有20|)()(|00=xFxFn,20|)()(|=kknxFxF,对任意实数x,必存在j,1jk,有xj1xxj,因2)()()()
(2)(11+222)()()()(1xFxFxFxFjn,且=+0和任意实数x,总存在N0,当nN时,都有|Fn(x)F(x)|0,存在h0,当|yy0|h时,4|)()(|0+yFyFbaXbaX,又设y是满足|yy0|N1时,4|)()(|xFxFXXn,即4|)()(|N1且|yy0|h时,2|)()(|)()(|)()(|00MFX,4)(=+MFMFXXnn,4)()(limN2时,41)(MFnX,4)(MFnX,可得2)
(1)(|MFMFMXPnnXXn,因数列ana,bnb,存在N3,当nN3时,Mhaan4|,4|hbbnmaxN2,N3时,+=+2|)()(|2|)()(|hbbXaaPhbaXbXaPnnnnnnn2|24|42|=+MXPhhXMhPhbbXaaPnnnnn,则+=+2|)()(|2)(000hbaXbXahybaXPybXaPyFnnnnnnnnbXannnU222|)()(|200+hyFhbaXbXaPhybaXPbaXnnnnnn,且+=+2|)()(|22000hbaXbXaybXaPhybaXPhyFnnnnnnnnbaXnU2)(2|)()(|00+yFhbaXbXaPybXaPnnnbXannnnnnn,即22)(22000+N1且|yy0|h时,2)()
(2)(00+yFyFyFbaXbaXbaXn,在区间+hyhy00,2取FaX+b(y)的任一连续点y1,满足|y1y0|maxN1,N2,N3时,+)
(2)(22)(0100yFyFhyFyFbaXbaXbaXbXannnnn,在区间2,00hyhy取FaX+b(y)的任一连续点y2,满足|y2y0|maxN1,N2,N3时,6+)
(2)(22)(0200yFyFhyFyFbaXbaXbaXbXannnnn,即对于FaX+b(y)的任一连续点y0,当nmaxN1,N2,N3时,0,存在h0,当|yy0|h时,4|)()(|0+yFyFaXaX,又设y是满足|yy0|N1时,4|)()(|xFxFXXn,即4|)()(|N1且|yy0|h时,2|)()(|)()(|)()(|00+haYPnn,存在N2,当nN2时,22|haYPn,则+=+2|2)(000haYhyaXPyYXPyFnnnnYXnnU222|200+hyFhaYPhyaXPaXnnn,且+=+2|22000haYyYXPhyaXPhyFnnnnaXnU2)(2|00+yFhaYPyYXPnnYXnnn,即22)(22000+N1且|yy0|h时,2)()
(2)(00+yFyFyFaXaXaXn,在区间+hyhy00,2取FX+a(y)的任一连续点y1,满足|y1y0|maxN1,N2时,+)
(2)(22)(0100yFyFhyFyFaXaXaXYXnnnn,7在区间2,00hyhy取FX+a(y)的任一连续点y2,满足|y2y0|maxN1,N2时,+)
(2)(22)(0200yFyFhyFyFaXaXaXYXnnnn,即对于FX+a(y)的任一连续点y0,当nmaxN1,N2时,0,存在M,使得FX(x)在x=M处连续,且41)(hMFX,4)(hMFX=+,4)()(limhMFMFXXnnN1时,41)(hMFnX,4)(hMFnX,可得2)
(1)(|hMFMFMXPnnXXn,因0PnY,对任意的0,有0|lim=+MYPnn,存在N2,当nN2时,2|hMYPn,则当nmaxN1,N2时,有hMYPMXPMYMXPYXPnnnnnn+|U,故0|lim=+nnnYXP,即0PnnYX11如果XXLn,aYPn,且Yn0,常数a0,试证:
aXYXLnn证:
设y0是FX/a(y)的任一连续点,则对任意的0,存在h0,当|yy0|h时,4|)()(|0/yFyFaXaX,又设y是满足|yy0|N1时,4|)()(|xFxFXXn,即4|)()(|/N1且|yy0|h时,2|)()(|)()(|)()(|0/0/MFX,12)(=+MFMFXXnn,12)()(limN2时,121)(MFnX,12)(MFnX,可得6)
(1)(|MFMFMXPnnXXn,因0aYPn,有02|lim=+haYPnn,存在N30,当nN3时,62|aaYPn,有62|aYPn,且64|2MhaaYPn,可得当nmaxN1,N2,N3时,=2|2)(2hYaaYXPhaYYaXPhaXYXPnnnnnnnnn2|4|2aYMhaaYMXPnnnUU22|4|2+aYPMhaaYPMXPnnn,则+=22)(000/haXYXhyaXPyYXPyFnnnnnnYXnnU22220/0+hyFhaXYXPhyaXPaXnnnnn,且=222000/haXYXyYXPhyaXPhyFnnnnnnaXnU2)(20/0+yFhaXYXPyYXPnnYXnnnnn,即22)(220/0/0/+N1且|yy0|h时,2)()
(2)(0/0/+yFyFyFaXaXaXn,在区间+hyhy00,2取FX/a(y)的任一连续点y1,满足|y1y0|maxN1,N2,N3时,+)
(2)(22)(0/1/0/0/yFyFhyFyFaXaXaXYXnnnn,9在区间2,00hyhy取FX/a(y)的任一连续点y2,满足|y2y0|maxN1,N2,N3时,)
(2)(22)(0/2/0/0/yFyFhyFyFaXaXaXYXnnnn,即对于FX/a(y)的任一连续点y0,当nmaxN1,N2,N3时,|)()(|0/0/yFyFaXYXnn,故)()(/yFyFaXWYXnn,aXYXLnn12设随机变量Xn服从柯西分布,其密度函数为+0,)arctan
(2)arctan
(1)1(|22nnxdxxnnXPn=+=,则122)arctan(lim2|0|lim=+nXPnnn,故0PnX13设随机变量序列Xn独立同分布,其密度函数为0,令Yn=maxX1,X2,Xn,试证:
PnY证:
对任意的0,P|Yn|=PYn=1PmaxX1,X2,Xn=1PX1PX2PXnn=1,则11lim|lim=+nnnnYP,故PnY14设随机变量序列Xn独立同分布,其密度函数为0,P|Yna|=PaYna+=PminX1,X2,Xna+=1PminX1,X2,Xna+=1PX1a+PX2a+PXna+10nnaaxnaaxdx+=e1e1e1)()(,则1)e1(lim|lim=0,221)Var(|)(|nZZEZPnnn=,则01lim|)(|lim02=+nZEZPnnnn,即0|)(|lim=+nnnZEZP,1)(=nPnZEZ,因nZnYe=,且函数ex是直线上的连续函数,根据本节第3题的结论,可得1ee=PZnnY,故cYPn,其中1e=c为常数16设分布函数列Fn(x)弱收敛于分布函数F(x),且Fn(x)和F(x)都是连续、严格单调函数,又设服从(0,1)上的均匀分布,试证:
)()(11FFPn证:
因F(x)为连续的分布函数,有F()=