中山大学概率统计课后答案第5章习题解.docx
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中山大学概率统计课后答案第5章习题解.docx
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中山大学概率统计课后答案第5章习题解
中山大学概率统计课后答案第5章习题解
习题五
2是来自正态分布的样本值.2.设xxx,,,N(,),,12n
21)设已知,求的最大似然估计.,,
22)设已知,求的最大似然估计.,,
2解1)设已知.以记参数,似然函数为,,,
n22n,,()/2x,,,,()/2x,,,,,,nnn/2/2/2i,1ii.Lxxee(,,;,)
(2)
(2),,,,,,,,,1ni,1对数似然函数为
nn1n2.,,,,,,,,lnln
(2)ln()LX,i,1i,222
因此,对数似然方程就是,,,ln/0L,
n1n2.,,,,,()0x,i2,1i2,2,
1n222ˆ解得的最大似然估计.,,,,,,x,(),i,1in
22)设已知.似然函数为,
n2222n,,()/2x,,,,()/2x,,,22/2/22/2,,,nnni,1ii.Lxxee(,,;,)
(2)
(2)(),,,,,,,,,1ni,1对数似然函数为
nn1n22.,,,,,,,,lnln
(2)ln()LX,i2,1i222,因此,对数似然方程就是,,,ln/0L,
1n.x,,,()0,i2,1i,
1nˆ解得的最大似然估计.,,,x,i,1in
xx1,X3.设总体服从两点分布:
,其中p,(0,1)是未知参数.PXxpp()
(1),,,x,0,1
如果样本观测值为,求参数的最大似然估计.xxx,,,p12n
解似然函数为
nnnxnx,1xx,,,ii,,11iiiiLxxppppp(,,;)
(1)
(1).,,,,,1n1i,
5-1
对数似然函数为
nn.lnln()ln
(1)Lxpnxp,,,,,,ii,,11ii
对数似然方程就是,,,ln/0Lp
11nn.xnx,,,()0,,ii,,11iipp,1
1nˆ解得的最大似然估计.ppxx,,,i,1in
5.设服从区间上的均匀分布,其中都是未知参数.若是的样本XX[,]abxxx,,,ab,12n值,求的最大似然估计和矩估计.ab,
解先求的最大似然估计.似然函数为ab,
n11nn,,LLxxabIxIx,,,(,,;,)()(),,1[,][,],,nabiabi,,11iibaba,,,,
n,1,,axxbxx,,min{,,),max{,,),11,,nn.,ba,,,,,其他0,
当时,有最大值,故的最大似然估计是Laxxbxx,,min{,,),max{,,)ab,11nn
ˆˆ.axxbxx,,min{,,),max{,,)11nn
现在来求的矩估计.ab,
b2,,,,bxxxab,,(;,)()EXxpxabdxIxdxdx,,,,,[,]ab,,,,,,,a2()2bababa,,,a
b22322,,,,bxxxaabb,,22,(;,)()EXxpxabdxIxdxdx,,,,,[,]ab,,,,,,,a3()3bababa,,,a
222.DXEXEXba,,,,()()/12
解关于的方程组ab,
EXab,,()/2,,,2DXba,,()/12,,
可得,.因而的矩估计是bEXDX,,3ab,aEXDX,,3
ˆˆ,,axm,,3bxm,,322
1n2其中.mxx,,(),2i,1in
5-2
,()x,6.设总体有密度,其中是未知参数,XpxeIx(;)(),,,,,,,,(,)[,),,,
是来自总体的样本.求未知参数的最大似然估计和矩估计.XXX,,,,12n
解先求的最大似然估计.似然函数为,
n,X,n,,i,1in,(),,X,eeXX,min{,,},i1n.LLXXeIX(,,;)(),,,,,,1[,)ni,,,1i,0min{,,}XX,,,1n,当时,有最大值,故的最大似然估计是L,,min{,,)XX,1n
ˆ.,,XXmin{,,)1n
现在来求的矩估计.,
,,,,,,()()xx,,,,,,EXxpxdxxeIxdxxedx,,,(;)(),[,),,,,,,,,,,,
,,,,,,()()()xxx,,,,,,,,,,,,,,,,,xeedxe,,1,,,,
.,,,EX1
ˆ因而的矩估计是.,,,,X1
7.*设服从区间上的均匀分布,是样本.X[,],,,,ddXX,,1n
1)设已知,是未知参数.求参数的最大似然估计.,dd
2)设已知,是未知参数.求参数的最大似然估计.,,d
3)设和都是未知参数.求参数和的最大似然估计.,,dd
1解1)X有密度,pxdIx,(;)()[,],,,,ddd2
n11nn,,LLXXdIXIX,,,(,,;)()(),,1[,][,],,,,,,,,,,nddiddi,,11iidd22,,
n,1,,,,dXXdXXmin(,,),max(,,),,,,,11,,nn,d2,,,,0其他,
n,1,,dXXXX,,max{min(,,),max(,,)},,,,11,,nn.,d2,,,,0其他,
L当dXXXX,,,max{min(,,),max(,,)},,时,有最大值,故的最大似然估计d11nn
是
ˆ.dXXXX,,,,,max{min(,,),max(,,)}11nn
5-3
12)有密度,XpxIx,,(;)()[,],,,,ddd2
n11nn,,LLXXdIXIX,,,(,,;)()(),,1[,][,],,,,,,,,,,nddiddi,,11iidd22,,
n,1,,XXdXXd,max(,,)min(,,),,,,,11,,nn.,d2,,,,0其他,
当时,有最大值,故任意一个满足Lmax(,,)min(,,)XXdXXd,,,,,11nn
ˆ,,,,XXdXXd[max(,,),min(,,)]11nn
ˆ的估计都是的最大似然然估计.,,
13)有密度,XpxdIx,,(;,)()[,],,,,ddd2
n11nn,,LLXXdIXIX,,,,(,,;,)()(),,1[,][,],,,,,,,,,,nddiddi,,11iidd22,,
n,1,,,,dXXdXXmin{,,),max(,,),,,,,11,,nn,d2,,,,0其他,
n,1,,dXXXX,,max{min(,,),max(,,)},,,,11,,nn.,d2,,,,0其他,
当时,最小,L有最,,,,,,dXXdXXmin{,,),max(,,)2()()ddd,,,,,,11nn
大值.解方程组
,,dXXmin{,,),1n,,,dXXmax(,,),1n,
得,的最大似然估计,d
max(,,)min(,,)XXXX,max(,,)min(,,)XXXX,11nn11nnˆˆ,.,,d,22
X8.设总体服从二项分布,是来自总体的样本.BNp(,)XXX,,,12n
1)求未知参数Np,的矩估计.
2)若已知,求未知参数p的最大似然估计并证明这个估计是无偏估计.N
解1)解关于Np,的方程组
5-4
EXNp,,,DXNpp,,
(1),得
2DX()EX,.p,,1N,EXEXDX,故的矩估计为Np,
2mX2ˆˆ,,N,p,,1Xm,X2
1n2其中.mXX,,(),2i,1in
2)有密度X
xxNx,,CppxN
(1)0,1,,,,,N,pxp(;),,0其他,,似然函数为
nnX,XNXiii.LLXXppXpCpp,,,,(,,;)(;)
(1),,1niN11,,ii对数似然函数为
nnnXilnlnlnln
(1)()LCpXpNX,,,,,.,,,iiN,,,111iii
对数似然方程就是,,,ln/0Lp
11nn.XNX,,,()0,,ii,,11iipp,1
nXXX,ii,1ˆ解得,故的最大似然估计是.pp,,,pNNnN
ˆˆ又因为,所以是的无偏估计.ppEpEXNEXNNPNp,,,,(/)//
10.设总体的样本为,为了估计总体方差,我们利用下面的公式:
XXX,,,12n
n,122,,,kXX().,ii,1i,1
2求的值,使是总体方差的无偏估计.k,
X解设总体为.则
nn,,11222EXXEXXXX()
(2),,,,,,iiiiii,,,111ii,,11
nnn,,,11122,,,EXEXEXEX2,,,iiii,,11iii,,,111
5-5
22222.,,,,,,,,,,,
(1)2
(1)()
(1)2
(1)(())2
(1)nEXnEXnEXnEXEXnDX故
n,122.EkEXXknDX,,,,,()2
(1),ii,1i,1
2为了使得是总体方差的无偏估计,必须使,即要使DX2
(1)knDXDX,,,
1.k,2
(1)n,
11.设服从区间上的均匀分布,其中是未知参数.又设是样本.X[,],,,,ddXX,,,1n
证明和都是的无偏估计.X(min{,,}max{,,})/2XXXX,,11nn
()(),,,dd,,证1,故X是的无偏估计.,EXEX,,,,2
下证是的无偏估计.{max(,,)min(,,)}/2XXXX,,11nn
有密度和分布函数X
1xd,,(),,.pxIx,()()FxIxIx()()(),,Xdd[,),,,,Xddd[,)[,),,,,,,,,d22d进一步有
nxd,,(),,,n,(())()()FxIxIx,,[,)[,),,,,,,,,,,Xddd2d,,
dx,,,,1()()(),,,FxIxIxXddd(,)[,),,,,,,,,2d
ndx,,,,,n(1())()().,,,FxIxIx(,)[,),,,,,,,,,,Xddd2d,,
R设,则有分布函数和密度RXX,min(,,)1n
n,,dr,,,,,n,()1(1())1()()FrFrIrIr,,,,,,,,[,)[,),,,,,,,,,,RXddd2d,,,,,,
n,1ndr,,,,,,()()().prFrIr,,,,,,RRdd[,],,22dd,,
n,1,,,,ndr,,,,,()()ERrprdrrIrdr,,,,,,,Rdd[,],,,,,,,,22dd,,
drt,,,,2,dd,nn11nn,,,,,,,,()(),,rdrdrdttdt,,nn0,d,
(2)
(2)dd
5-6
nn,1,,nddnd
(2)
(2)2,,.,,()dd,,,,,,,,nnnn11,,
(2)d,,,,
设,则有分布函数和密度函数SXX,max(,,)S1n
nsd,,(),,,n,FsFsIsIs()(())()(),,,[,)[,),,,,,,,,,,SXddd2d,,
n,1nsd,,(),,,,.prFsIs()()(),,,,,,SSdd[,],,22dd,,
n,1,,,,nsd,,(),,,ESspsdssIsds,,,()(),,,,Sdd[,],,,,,,,,22dd,,
sdt,,,(),,dd2,nnnn,,11,,,,,,(())(),,ssddstdtdt,,nn,d0,
(2)
(2)dd
nn,1,,nddnd
(2)
(2)2,,.,,()dd,,,,,,,,nnnn11,,
(2)d,,,,
22ndnd,,.,,,,,,,,,,,ERSdd()/2/2,,,,nn11,,
()(),,,dd,,证2,故是的无偏估计.X,EXEX,,,,2
下证是的无偏估计.{max(,,)min(,,)}/2XXXX,,11nn
设,,则独立且都服从区间上的均匀分YdX,,2YY,,[,],,,,ddin,1,2,,ii1n布.因而
EXXEYYEdXdXmin(,,)min(,,)min(2,,2),,,,111nnn
.,,,,EdXXdEXX(2max(,,))2max(,,)11nn
故有
max(,,)min(,,)max(,,)min(,,)XXXXEXXEXX,,1111nnnnE,22
EXXdEXXmax(,,)2max(,,),,11nn.,,d2
XYX12.设总体和总体的分布都依赖于参数,XX,,是来自的样本,YY,,是,1m1n
ˆˆ来自Y的样本.又设统计量和相互独立,都是的无偏估计,分,XX,YY(,,)(,,),11m21n
ˆˆˆ别有方差2和3.求常数k和,使得是的无偏估计且有最小的方差.k,,,,,kk,121122解据题意,求常数k和,满足k12
5-7
ˆˆˆ,,,,,,,,,,,EkEkEkk()112212且使得
2222ˆˆˆˆˆ,,,,,,,,,,,DDkkkDkDkk()231122112212达到最小.
由上面的第一式得,有上面第二式知要使关于的二次三项式kk,,1k211
222fkkkkk()23
(1)563,,,,,,11111
达到最小.为此,只需令,因而也可得.k,3/5k,2/512
13.设,问样本容量至少有多大才能使的置信水平为0.95的置信区间的XN~(,1),,n
长度不大于0.2.
解
XX,,,,,ZN~(0,1),,21/n/n,
查得,即PZ(||1.96)0.95,,
.PXnXn(1.96/1.96/)0.95,,,,,,故的置信水平为0.95的置信区间为.,[1.96/,1.96/]XnXn,,
要使置信区间的长度不大于0.2,由不等式21.96/,n
21.96/0.2,,n
221.96,,,可得,故样本容量不能少于385.nn,,384.16,,0.2,,
2.设14,样本值为XN~(,),,
3.383.421.612.782.722.975.484.111.903.40
21)已知,求的置信水平为0.95置信区间.,,,1
222)未知,求和方差的置信水平为0.95置信区间.,,,
解1)令
X,,,Z,2/n,
2其中,,则ZN~(0,1).查得n,10,,1
PZ(||1.96)0.95,,,
即
5-8
22.PXnXn([1.96/1.96/])0.95,,,,,,,,
故的置信水平为0.95的置信区间为,
22[1.96/,1.96/]XnXn,,,,
2代入具体的样本值和的值计算得,
x,,,,,,,,,,,(3.383.421.612.782.722.975.484.111.903.40)/103.177
2.1.96/1.961/100.620,n,,
故的置信区间是,即.2.557,3.797,[3.1770.620,3.1770.620],,,,
2)先求的置信区间.令,
X,,,T,2Sn/则其中.查得Tt~n,10n,1
.PT(||2.262)0.95,,
即
22.PXSnXSn([2.776/2.776/]0.95,,,,,,
故的置信区间为,
22.[2.776/,2.776/]XSnXSn,,
代入具体的样本值计算得
22,,,x,3.177s,1.2022.262/0.784sn,
22,.xsn,,2.262/2.393xsn,,2.262/3.961
于是得到的置信水平为0.95的置信区间.,[2.393,3.961]
2现在求的置信区间.令,
1n2,,,,XX(),i2,1i,
2则,其中.查得,.故P(2.700)0.975,,,P(19.023)0.025,,,,,~n,10n,1
1n2,PXX,,,,{2.700()19.023}0.95,i2,1i,
11nn,,222.PXXXX,,,,,,()()0.95,,,,ii,,11ii19.0232.700,,
2的置信水平为0.95的置信区间为,
11nn22XXXX,,.[(),()],,ii,,11ii19.0232.700
5-9
n22以样本值代入,.故的置信水平为0.95的置信区间为()10.820xx,,,,i,1i
10.82010.820,[,]19.0232.700
2即的置信区间为.[0.569,4.007],
15.测得16个零件的长度(毫米)为
12.1512.1212.0112.0812.0912.1612.0312.01
12.0612.1312.0712.1112.0812.0112.0312.06如果零件长度服从正态分布,求
1)零件长度的期望的置信水平为0.95及0.99的置信区间;
2)零件长度的方差和标准差的置信水平为0.95及0.99的置信区间.
2解由题设知零件的长度服从正态分布,设.XXN~(,),,
1)令
X,,,T,2Sn/
则,其中.设置信水平为,满足Tt~n,161,,,n,1
PT(||)1,,,,,
则
22.PXSnXSn([//]1,,,,,,,,,,
而的置信区间为,
22.[/,/]XSnXSn,,,,
代入具体的样本值计算得
2,x,12.075s,0.00244
a)当时,查得,10.95,,,,,2.131
222,,.xsn,,,/12.049xsn,,,/12.101,sn/0.0263,
于是得到,的置信水平为0.95的置信区间[12.049,12.101].
b)当时,查得,10.99,,,,,2.947
222,,.xsn,,,/12.039xsn,,,/12.111,sn/0.0364,
于是得到,的置信水平为0.99的置信区间[12.039,12.111].
22)现在求的置信区间.令,
1n2,,,XX,(),i2,1i,
5-10
2则,其中.设置信水平为,和满足,,~,,n,161,,12n,1
,P()1/2,,,,,,P()/2,,,,,12
则
1n2,PXX,,,,,,,,{()}1,12i2,1i,
,11nn222PXXXX,,.()()1,,,,,,,,,,ii,,11ii,,21,,
2的置信水平为0.95的置信区间为,
,,11nn22.XXXX(),(),,,,ii,,,,11ii,,21,,
n2以样本值代入,()0.0336xx,,.,i,1i
2a)当时,查得,,故的置信水平为0.95的置信,,6.262,,27.48810.95,,,,12
区间为
0.03360.0336,[,]27.4886.262
2即的置信区间为.因而的置信区间为.[0.00133,0.00584][0.0365,0.0765],,
2b)当时,查得,,故的置信水平为0.99的置信区,,4.601,,32.80110.99,,,,12
间为
0.03360.0336,[,]32.8014.601
2即的置信区间为.因而的置信区间为.[0.00112,0.00796][0.0334,0.0892],,
18.随机地从甲批导线中抽取4根,从乙批导线中抽取5根,测得其电阻为(单位:
欧姆):
甲批导线:
0.143,0.142,0.143,0.137
乙批导线:
0.140,0.142,0.136,0.138,0.140
2222设甲、乙两批导线的电阻分别服从,,已知,且两者相互N(,),,N(,),,,,0.002512独立,但,均未知,求,,,的置信水平为0.95的置信区间.,,1221
YX,,,(),,21提示:
把,,,看作参数,枢轴量.ZN,~(0,1)2122/4/5,,,解
YXYX,,,,,,()(),,,,2121,ZN,,~(0,1)222/4/59/20,,,,
5-11
设满足,则PZ(||)0.95,,,,
22.PYXYX(9/209/20)0.95,,,,,,,,,,,,,,21
故的置信水平为0.95的置信区间为,
22.[9/20,9/20]YXYX,,,,,,,,
2查得,代入样本值和的值计算得,,1.9600,
yx,,,0.00205
2.,,9/201.9630.0025/200.003287,,,,
故的置信区间是,即.,0.00534,0.00124,[0.002050.00329,0.002050.00329],,,,,,
5-12
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