《应用数理统计》吴翊李永乐第二章 参数估计课后习题参考
答案
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第二章 参数估计
课后习题参考答案 2.1 设总体X服从二项分布为其子样,求N及p的矩法,,BN,p,1,p,1,X,X,?,X12n估计。
解:
,,,,,,EX,Np,DX,Np1,p
,X,Np,令 ,2,,,S,Np1,p,
解上述关于N、p的方程得:
2,XX ˆN,,,2,ˆpX,S ,2S,ˆp,,1 ,X,
2,,,(),0,xx,,,2fx(;),,,,2.2 对容量为n的子样,对密度函数
,0,0xx,,:,,
其中参数的矩法估计。 ,
,2,,,,aExxxdx()()解: 12,0,
,222,,()xxdx 2,0,,
122123,,,,,,,,,, 2333,,
,1xxxxxxx,,,,??,,,33ax(),,,所以 其中为n个样本的观察值。 121,2nn1n
2.3 使用一测量仪器对同一值进行了12次独立测量,其结果为(单位:mm)
232.50,232.48,232.15,232.52,232.53,232.30 232.48,232.05,232.45,232.60,232.47,232.30 试用矩法估计测量的真值和方差(设仪器无系统差)。
n1EX,X,X,232.4025,,,in,1i解: n212,,,,DX,X,X,S,0.0255,in,1i
12.4 设子样1.3,0.6,1.7,2.2,0.3,1.1是来自具有密度函数的总,,fx,,,,0,,,1,
体,试用矩法估计总体均值、总体方差及参数。 ,解:
n1总体均值:X,X,1.2,in1,i
n212总体方差:S,X,X,0.407,,,in1,i
1,fx,,,,,
,,,x,,,,,EX,xfx,dx,dx,,X,,,200
ˆ参数:,,2X,2.4
22.5 设为的一个字样,求参数,的MLE;又若总体为的X,X,?,X,,N,,1N,,1,,12n
MLE。 ,x,12解:(1) ,fx,e,2i,i,2,,
,,,x,
n,n122,,,,fx,,Lx,e,,iiin,1i,12i,2,,,,,
,,nn21L,x,,,,,x,,ln,ln2ii,1,,,,22i
n,L,xln,i,,,x,,,,0i,,,1,,i
n1,,x,Xˆ,i,1ni
(2)
2,,x1,i,,12,22,,fx,,e,,i2
n2,,1x,i,i,1n,12,,22,2,,,,fx,,Lx,,e,iinn,,i1,2,,2
nn122,,,,,,,lnLx,,,ln2,nln,x,1,ii2,i1,22 2,n,,,lnLx,n12i,,,,,x,1,0,i224,,,1i,,22
2n12ˆ,,,x1,,,ii1,n
,
2.6 设总体X的密度函数为为其样本,求下列情况下的MLE。 fxXXX(;),,,,,?,12n
x,,,,ex,0,1,2,,?,,,0(i) fx(;),,x!,
,0,其它,
,,1,,xx,01,,,,0fx(;), (ii) ,,0,其它,
,,,,,1x,(),0,,xex,, (iii) 已知 ,fx(;),,,0,其它,,
rx,,1,,,,()/(),0xerx,,fx(;),(iv) r已知 ,,0,其它,
x,,1,ex,0,,,,0(v) , fx(;),,,
,0,其它,
nXi,,n,,1i,e()L,解:(i) ,!!!xxx?12n
n
ln()lnln(!!!)LXnxxx,,,,,,?,12in,1i
ndLln()1,,,,Xn0,id,,,1i n1,,Xx,,in,1i
nn,,,,11n(ii) Lxx(),,,,,,,ii,,11ii
n
ln()ln(1)lnLnx,,,,,,,i,1i
ndLnln(),,,,ln0x,id,,,1i nn,1,,11,,,,nxx(ln)(ln),,,iin,,11ii
n,n,,xi,1n,,,1i(iii) Lxe()(),,,,,i1i,
nn, ln()ln()(1)lnLnxx,,,,,,,,,,,ii,,11ii
nndLnln,,,,,,,,nxx0,,iid,,,,,,11ii n,1,1,,()x,,in,1i
nn,,xi,1nrn,,1i (iv) Lxer()()/(()),,,,,,i1i,
nnnr ln()ln(1)ln()Lrxxnr,,,,,,,,,,,ii,,11ii
ndLnrln(),,,,x0,id,,,1i
n, nrr1,,,,xX,,innxi,1x,i,1i
n1,xi,1,,1i(v) Le,(),n,
n1 Lnx,,,,,ln()ln(),i,,1i
ndLnln()1,X0,,,,,i2d,,,i1, n,1,,xxX,,,ini1,
,2.7 设总体X的密度函数为,为其子样,求参数X,X,?,X,,,,fx,,,1x,0,x,112n的MLE及矩法估计。今得子样观察值为0.3,0.8,0.27,0.35,0.62,0.55,求参数的,,
估计值。
解:
极大似然估计: n1X,x,0.48,ini,1
nnn,fx,,,Lx,,,,1xiii,ii,,,,,,,1,1,,
n
lnL,x,,,nln,,1,lnx,ii,,,,i,1 n,lnL,x,ni,,,,lnx,0,i,,,,1i,1
,1n1,,ˆ,,,1,lnx,0.234,,,ini,,,1
矩法估计:
11,,1,,ExxfxdxxxdxX,,,1,,,,,,,,,,,,200 1ˆ,,,2,,0.07
X1,
2.8 在处理快艇的6次实验数据中,得到下列的最大速度值(单位:m/s)27 ,38 ,30 ,37 ,35 .31,
求最大艇速的数学期望与方差的无偏估计。
,,XEX,,解:是总体期望的无偏估计
n1,,?EX,,,X,33m/s ,in,1i
22是总体方差,,的无偏估计 SDX,,*
n21222,,S,X,X,18.8m/s ,8in,1,1i
22.9 设总体,为其子样。 X,X,?,XX~N,,,,,12n
2n,1122(1)求k,使为的无偏估计; ˆ,,,,,X,X,i,i1ki,1
n1(2)求k,使为的无偏估计。 ˆ,,,X,X,ik,1i
解:
222111n,n,n,,,,,nn1,112,1,,2,,,,(1) EXXEXX,,,,,,,,,,,,,,,,,,11i,ii,i,,,,kknk,1111i,i,i,,,,,
即k=2(n-1)
(2)
n,11X,X,X,X ii,jnnj,i
,,n,11n,11n,11,,,,,,,,EX,X,EX,EX,EX,EX,,,n,1,,0,,,,iijij,,nnnnnn,,j,ij,i,,
2,,,,n,11n,11,,,,,,,,,,DXXDXDXDXDX,,,,,,,,,iijii22,,nnnn,,j,ij,i,, 2,,n,1n,1n,1222,,,,,,22nnn
n,1,,2XXN所以 ,,~0,,,in,,
n2,1EXX,?,, in,
n,,1n2n,112nn,1: ,,,?E,EX,X,,,i,,kknk,1i
,,2nn,1?k, ,
XNXXX (,1),,,,2.10 设总体为一样本,试证明下述三个估计变量 123
131,,,,XXX11235102
115 ,,,XXX,21233412
111,,,XXX3123,362
都是的无偏估计量,并求出每一估计量的方差,问哪一个最小, ,
131证: ,,,,EEXEXEX()()()()11235102
131 ,,,,,,()5102
115同理:,,,, EEXEXEX()()()()21233412
115,,,,,, ()3412
111,,,, EEXEXEX()()()()3123362
111,,,,,,() 362
是的无偏估计量。 ,,,,,,?123
由于
13119222,,,,,D()()()()1510250
11525222,,,,D()()()(),2341277
1117222,,,,D()()()(),336218?D(),最小。 2
22ˆˆˆ,,,,2.11 设是参数的无偏估计,且有,试证不是的无偏估计。 ,,D,,0
解:
ˆˆ,,是参数的无偏估计,即 ,,E,,,
22ˆˆˆˆ,,,,,,,,,,D,,E,,E,,D,,0又因为
2222ˆˆˆˆ,,,,,,,,,,E,,D,,E,,,,D,,,所以
22ˆ,,综上所述:不是的无偏估计
n1,22.12设总体证明为的,XNXX ?(,),,,,,,已知,为一样本,X,,,ni1n2,1i
1无偏估计,且效率为。 ,,2
2证明:设 则 yx,,,yN (0,),iii
,, Eyyfydy()2(),iiii,0
2yi,,,122, ,yedy2ii,0,,2
2,,
,
n11,, EXEy,,()(),,iinn22,1i
12,,, n ,,n2,n1,即,为的无偏估计 ,X,,,in2,1i
nn11,, DXDX,,, (),,,,ii2nn22,,11ii
n2,2,,,,,,EXEX(),,,,,ii2,,2ni,1
n2,,,22,, ,,,,,22n,,i,1,
(2),2,,,2n
2 由于 XN (,),,
2()x,,,122,fxe,,,(,,)
2,,
则, 2x,11(),,,,,(ln2ln),,2,fxln(,,),,222,,,,,,
21()x,,,,, 3,,
22,,ln(,,)13()fxx,,,,,224,,,, 2,ln(,,)132fx,,2IEEx()()(),,,,,,,,,2242,,,,,
,111 e,,,,()n,,,(2)22,,2DnI()(),, n,2n2,2.13设总体X服从几何分布:
k,1 Pxkppkp()(1),1,2,,01,,,,?,,
n1证明样本均值是的相合,无偏和有效估计量。 EX()XX,,in,1i
证明:(1)相合性
,,,,kk11,, EXkpppkp()(1)(1),,,,,,kk11,,
,k1,Spkp()(1),,,k,1令 ,k1,Spdpkpdp()(1),,,,,k1,
0,,(1)1,pk ,,,,,,,(1)p,1(1),,ppk,0
11, EXp()(),,, pp
22DXEXEX()()(),, k,222221EXpppppkpp()2(1)3(1)(1),,,,,,,,,,,?
22221k, ,,,,,,,,,pppkp(12(1)3(1)(1))??
21kk,,对上式括号中的式子,利用导数,并利用倍差法求和 kpkp(1)((1)),,,,22221k,12(1)3(1)(1),,,,,,,,ppkp??
23k,,,,,,,,,,,((1)2(1)3(1)(1))pppkp?? 12,,pp,,,()23pp
因此,
22,,pp2EXp(),,,32pp 211,,pp222DXEXEX()()()(),,,,,22ppp
?DTXXX((,,))12n相合性: ?,,,,,PTXXXg(,,)(),,12n2,
lim((,,,))0DTXXX?, 当 12n,,n
则是的相合估计。 TXXX(,,,)?g(),12n
,,11,p本题中, lim((,,,))lim0DTXXX?,,,,,12n2,,,,nnnp,,
n1 是的相合估计。 EX()XX,?,in,1i
(2)无偏性
nn111 EXEXEXEX,,,,()()()(),,iinnp,,11ii
(3)有效性
nn111,p DXDXDX()()(),,,,,ii22nnnp,,11ii
x,11ln(,)ln(1)ln(1)ln(1)Pxppppxp,,,,,,11
,,ln(,)1Pxpx111,,,,ppp1
2,,ln(,)1Pxpx111 ,,,222,,ppp(1)
1,1x,111p1IpE()(),,,,,,2222pppp(1)(1),,
1 ,2pp(1),
112,(())4ppe,,,1 n,p11DXnIp()() n22nppp,(1)?XEX()是的有效估计。
22.14 设总体X服从泊松分布,为其子样,试求参数的无偏估计,,,X,X,?,X,,P,12n量的克拉美劳不等式下界。
解:
,X,,,1pXe1,,!X,,1
ln,,,,lnln,,,,pXXX111,,
ln,,ln,,1,,pXpXX111,,,,,,,,22,,2,,2,, 22ln,,ln,,1,,pXpXX111,,,,,,,22234,,,4,2,,
2ln,,,,1,,,pX21,,,,,,,,,,IIe,,23,4,,,,
克拉美劳不等式的下界为:
23'3,,,,g,,1442,,,, 1,,nInn,n,34,
ˆˆˆ,u,a,,b,2.15 设,是的有效估计量,试证明是的有效估计量。 ,,u,a,,ba,0u
解:
:::: ,,,,,,,,E,,,,E,,Ea,,b,aE,,b,a,,b,,
lnpxlnpx1,,,,,, ?,::,a,,,
22:,,,lnp(x)1,lnp(x)1:,,,,,,,,,,?,,,IEEI: ,,:22,,,aa,,,,,,,
,从而的无偏估计量C-R下界
:,1:,122 ,,,,,,,,,,,,nI,(),anI,,aD,,Da,,b,D,
::,是的有效估计量。 ?,,a,,b
,,,,,,X,Y,X,Y,?,X,Y,,X,Y2.16 设有二元总体,而为其样本,证明1122nn
n1ˆ,,CovX,Y,,,,是的无偏估计。 C,X,XY,Y,ii1n,,1i
证明:
nn:11,, ,,,,,,,,,,EC,EX,XY,Y,EX,XY,Y,,,,iiii11n,n,,1,1,,ii
n,11n,11 ,,,,?X,X,X,X,X,EX,X,EXiijij,,nnnnj,ij,i同理
n,11 ,,,,Y,Y,Y,EY,Y,EY,iiknnk,i
,,,,?EX,XY,Yii
2,,,,n,1n,1,,,,,,,,X,EXY,EY,X,EXY,EY,,,iiik22nnk,i,,,E,,,,n,11,,,,,,,,,,,,,X,EXY,EY,X,EXY,EY,,,jijk22,,nn,,j,ij,ik,i,,,,
2,,,,,,n,11,,,,,,,,covX,YEXEXYEY,,,,,,,,jk22,,nnj,ik,i,,,,,,
2,,n,11,,,,,,,covX,Y,EX,XY,Y,jj22nnj,i
2,,n,1n,1,,,,,covX,Y,covX,Y22nn
,n1,,,covX,Yn
nn:11n,1nn,1,,,,,,,,,,,,?EC,EX,XY,Y,covX,Y,covX,Y,covX,Y,,iin,1n,1nn,1n,1,1ii
:?C为,,的无偏统计量。 covX,Y
ˆˆˆˆ,2.17 设和是参数的两个独立的无偏估计量,且的方差为的方差的2倍,试确,,,,1212
ˆˆ,定常数C及C,使得为参数的无偏估计量,并且在所有这样的线性估计中方C,,C,211122
差最小。
ˆˆ,解:和是参数的两个独立的无偏估计量 ,,12
ˆˆˆˆC,C,1则即 ,,,,,,,,,,,,,,,ECCCECE1211221122
ˆˆ又 ,,,,?D,,D,12
2222ˆˆˆˆˆ ,,,,,,,,,,?DC,,C,,CD,,,CD,,C,CD,211221122122
22要使其最小,则要求最小。 2C,C12
212,,2222 2212,,C,C,C,,C,C,,,,1211133,,
21当时,取得最小值,此时 C,C,1233
2.18 设总体X服从指数分布,其密度函数为
,,x,,e,x,0,, ,,,,,0fx,0,x,0,
为其样本,n>2。 X,X,?,X12n
ˆ,(1)求的MLE,;
*ˆˆˆ,(2)计算,求k,使得,,k,为的无偏估计; ,,,E
*ˆ,(3)证明,是的渐进有效估计。
,,n解:(1) xnfxLxe,,,,,,,iin,1,ii,1i,,,,,n
Lxnx,,,,,,ln,lnii,1i,,
nLxn,,ln,ix,,,,,,0 i,1i,,,
n1ˆ,,,nXx,i,1i
::11,,,?E,E,,?(2) XX
,,x,,e,x,0,,~,,,,,,0Xfx ,i0,x,0,
nn,n,1,,x的密度函数 xe,X,i(,1)!,1in
n,,n,1,,x,xe,x,0 ,,fx,,,,,,0,n,2n,,,n,1!Xi,,0,x,0,1i,
,,,,nn,,:1n1nn,1,xn,2,x,,,?E,E,E,nxedx,xedx,,n,,xn,1!n,1,,,,~X00Xi,i,1
n,nn1,n,,,n,2!,,,,,n,1!n,1
9,2,,n,3,,,,n2n2,:2,,1nn,22n1xn3x2,,,,,,,,E,nxedx,xedx,,,,2,,2,n2,,,,,,,,n,1!n,1~n,1n,2x00,,n,4,,,,,,n,1n,2,
222:::22nnn222?D,E,E,,, ,,,,,,,,,,,,,,22,,,,n,1n,2,,,,,,n,1n,1n,2
::::*nn,1*,EkEk?k,,,,,,,,,,,,若,,为的无偏估计,则 ,k,,,nn,111lnfX,,ln,X
(3) ,1,lnLX,11,,X,,,,,2,,,1,lnLX,122,,,,,,,2,1,,,lnLX,1,22I,,E,,,,,,,,,,,, 2,*n,1,,,,ˆˆD,D,,,,,nn,2,,
11n,22e,,,,1*n,,,ˆ,,,,1nDnI,n,2,n,2
*ˆ,,由此可知,是的渐近有效估计。
,X,X,?,X,,P,,,,02.19 设总体X服从泊松分布,为来自X的一个样本。假设有12n
先验分布,其密度为
,,,,e,,0,,,,h, ,0,,,0,
,求在平方损失下的贝叶斯估计。
,解:给定,样本的分布列为:
xnXni,,,,,n,,,0,1,2,,,eei?,,n!,x,1ii,,,|,!,,gxx?xx,,12,ni ,1i,
,0,其它,
nXn,,,1n,,样本的边缘分布列,其中 ,,?,,,gxxx,ed,X,X,12nin,0n,1I!x,i1i,
,,0时的联合密度函数:
nX,,,1n,,,gx,x,,x,gx,x,,x|he,,,,,,?,?, ,,,1212nnn
x!,i,1i,的后验密度为:
nX,,,n1,,,,e,n,!x,i,,,,,,gxx?x,,12ni1,,,,0,1,2,,i?,,,,gxx?,,x,,,12n,|,,, ,,,hxx?x,,,12nnX,,,,,n,,,,,,,eddn,,00,,,!x,,,i,1i,,,,
,0,其它,
,,nX,1ˆ,,,,,,h,|x,x,?,xd,的贝叶斯估计:= 12n,0n,1
2.22 随机的从一批零件中抽取16个,测得长度(单位:cm)为:
2.14 2.10 2.13 2.15 2.13 2.12 2.13 2.10
2.15 2.12 2.14 2.10 2.13 2.11 2.14 2.11
, 以零件长度的分布为正态的,试求总体均值的90%置信区间。
(i)若,,0.01()cm;
,未知 (ii) 若。
解:求得:
x,2.125
S,0.0166
S,0.0171,
,,,,?,,0.11.645,0.95,12
(i)若 ,,0.01()cm
则,的90%置信区间为: ,
,,xnxn,,,,,,,,,,, 11,,,,22
带入数据
,,2.1251.6450.01/16,2.1251.6450.01/16,,,,,,
得:2.121,2.129,,
(ii)若未知 ,
tt(15)(15)1.7531,,,0.951,2
,的的置信区间为:90%
,,SSxx,,,,tntn(1),(1),,,,11,,nn,,1122,,
或
,,SS,,xx,,,,tntn(1),(1),,11,,,,nn22,,
计算得
2.11752.1325,,,
22.23 对方差已知的正态分布总体来说,问需抽取容量n为多大的样本,才能使总体均值,0
2,,,,的置信度为1001,,%置信区间的长度不大于。
证明:
,,X22,,,,,,,T~N0,1,且,0 ,/n
,,,,a,XbA,,, ,,,,,/n/n/n,,
,,估计区间: X,u,/n,X,u,/n,,,0,0,,1122,,
2u/n,2,,区间长度: ,01,2
2,u,,0,1,,,2n?, ,,,,,,,
2.24随机地从A批导线中抽取4根,并从B批导线中抽取5根测得其电阻()为 ,
A批导线 0.143 0.142 0.143 0.137
B批导线 0.140 0.142 0.136 0.138 0.140
22设测试数据分别服从正态分布 NN(,)(,),,,,,和且它们相互独立,12
2。 又未知,试求的置信区间,,,,0.9512
解:此题中,置信度0.95,即 ,,,,0.05,4,5nn12
查得
tnnt(2)(7)2.3646,,,,,120.975,12
2X,,0.1412,S6.1875e-006AA 2X 0.1392,S4.1600e-006,,BB
nnnn(2),,45(452),,,,1212M,,,3.9 nn,,4512
22nSnS,,,,,,46.1875e-00654.1600e-0060.0067 AB12
代入下式得的的置信区间为,,,0.95 12
tnntnn(2)(2),,,,,,,,1212,,112222,,22 XYnSnSXYnSnS,,,,,,,ABAB1212,,MM,,,,
2.36462.3646,,0.14120.13920.0067,0.14120.13920.0067,,,,,,,,3.93.9 ,,
,-0.0021,0.0061 ,,
,,与无明显差异。因0含在此置信区间内,故认为 12
2.26 在一批货物抽100件检查,发现次品16件,求这批货物次品率的0.95置信区间。
u,u,1.96,n,100解:,100件产品中有16件次品,则 X,0.16,0.975,12
X,p使用棣莫弗——拉普拉斯中心极限定理,服从
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
正态分布,于是解下列不等
,,p1,pn
式
2,,X,p2,u ,,1,,1/p,pn2
,,,,2222,,,, n,up,2nX,up,nX,0,,,,,,,,1122,,,,
解上述不等式得:p=0.101,p=0.244,95%的置信区间[0.101 0.244] 2.25随机地取某种炮弹9发作试验,得炮口速度的样本标准差,设炮口速度服Sms,11(/)从正态分布,求这种炮弹炮口速度之标准差的0.95置信区间。 ,
222222,,的点估计量为注意到考虑及Scd,0,01,,,,,S的邻域[cS,dS],使得
222P(A)=PcS,,,,dS1,,,,
变换事件A
111,,A,,,,,222cSdS,,,2,,nnSn =,,,,2cd,,,2nS22由T()= (1),()1,nPA,,,故为使通常取2,,,,nn22,,,,(1) (1)nn,,1,dc,,22解: 于是所求区间为
,,22nSnS,, ,22,,(1)(1)nn,,,,,1,,-22,,,
2本题中,,,,nS9,121,0.05,
22(1)(8)17.535n,,,,,0.975,1,2
22(n-1)=(8) =2.1800.025,,,2
代入得:
,9*1219121,,,, =62.1044499.5413 ,,,,17.5352.180,,
,开根号得这种炮弹炮口速度之标准差的0.95置信区间为[7.8806, 22.3504]