《应用数理统计》作业题及参考答案(前三章)
第一章 数理统计的基本概念
P26
1.2 设总体X的分布函数为F?x?，密度函数为f?x?，X1，X2，…，Xn为X的子样，求最大顺序统计量X?n?与最小顺序统计量X?1?的分布函数与密度函数。 解：Fn?x??P?Xi?x??P?X1?x，X2?x，?，Xn?x????F?x???.
n?1?fn?x????Fn?x????n??F?x???f?x?.
nF1?x??P?Xi?x??1?P?X1?x，X2?x，?，Xn?x?. ?1?P?X1?x?P?X2?x??P?Xn?x?
?1???1?P?X1?x?????1?P?X2?x??????1?P?Xn?x???
?1???1?F?x???
n??n?1?Fx?n?1fx. f1?x???Fx?????????1??
4?，今抽取容量为5的子样X1，X2，…，X5，试问： 1.3 设总体X服从正态分布N?12，（i）子样的平均值X大于13的概率为多少？
（ii）子样的极小值（最小顺序统计量）小于10的概率为多少？ （iii）子样的极大值（最大顺序统计量）大于15的概率为多少？
??4?，n?5，?X~N?12，?. 解：?X~N?12，5??4（i）
???????13?12?X?1213?12????1???1.12??1?0.8686?0.1314. PX?13?1?PX?13?1?P????1?????444??????5?5??5?????（ii）令Xmin?min?X1，X2，X3，X4，X5?，Xmax?max?X1，X2，X3，X4，X5?.
P?Xmin?10??1?P?Xmin?10??1?P?X1?10，X2?10，?，X5?10?
?1??P?Xi?10??1????1?P?Xi?10????1???1?P?X?10???.
i?1i?1555—1—
《应用数理统计》作业题及参考答案(前三章)
?Y?X?12~N?0，1?， 2?X?1210?12??X?12??P?X?10??P????1??P?Y??1? ??P?2??2?2??1?P?Y?1??1???1??1?0.8413?0.1587.
?P?Xmin?10??1??1?0.1587??1?0.4215?0.5785.
5（iii）
P?Xmax?15??1?P?Xmax?15??1?P?X1?15，X2?15，?，X5?15??1??P?Xi?15??1???P?X?15???.
i?155?P?Xmax?15??1?0.933195?1?0.7077?0.2923.
1.4 试证：
（i）??xi?a?2???xi?x??n?x?a?对任意实数a成立。并由此证明当a?x时，??xi?a?222i?1i?1i?1nnn达到最小。 （ii）??xi?x?i?1nn21n??x?nx，其中x??xi。
ni?1i?12i22n2nn22证明：（i）??xi?a????xi?x?x?a????xi?x??2?xi?x??x?a???x?a?? ???i?1i?1i?1n????xi?x?2x?ai?1nn???2????i?1nxi?x?nx?a???2??xi?x?2x?anx?nx?nx?a
i?1??2??????2??xi?x?nx?a.
i?1?2??2当a?x时，??xi?a????xi?x??n?x?x????xi?x?达到最小。
2222i?1i?1i?1nnn（ii）??xi?x???xi2?2xix?x??xi2?2x?xi?nx??xi2?2x?nx?nx??xi2?nx.
22222i?1i?1i?1i?1i?1i?1nn??nnnn P27
1n1.5 设X1，X2，…，Xn为正态总体X~N??，??的样本，令d??Xi??，试证
ni?122?2??E?d???，D?d???1??。
????n2证明：①X~N??，?2?，则Xi??~N?0，?2?.
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《应用数理统计》作业题及参考答案(前三章)
?1nE?d??E??Xi???ni?1EXi???12??y22?2?1n???EXi??. ?ni?1ye?y22?2??????dy?21?????0ye?y22?2212?2dy?????2???0e?y22?2?y2?d??2? ?2??????2???e?20??.
1n2122?E?d??????n???.
ni?1?n??② E?Xi????D?Xi????E2?Xi?????2.
?DXi???EXi??2?2??E222??Xi????2??2??1???2.
?????1n?D?d??D??Xi???ni?1?1?n??2D??Xi???n?i?11?2?2?2??2?1n?DX????n?1?????1??. i?2?2nn??????ni?1?
1.6 设总体X服从正态N??，?2?，X1，X2，…，Xn为其子样，X与S2分别为子样均值及方差。又设Xn?1与X1，X2，…，Xn独立同分布，试求统计量Y?解：由于Xn?1和X是独立的正态变量，
??2??X~N??，?，Xn?1~N??，?2?，且它们相互独立.
?n?EXn?1?X?E?Xn?1??EX?????0. DXn?1?X?D?Xn?1??DX?2?则Xn?1?X~N??0，??. nXn?1?XSn?1的分布。 n?1????????n?12?. nn?1??Xn?1?X?n~N?01，?. n?12而
nS2?2~??n?1?，且
nS2?2与Xn?1?X相互独立，
n?1~t?n?1?. n?1则T?
Xn?1?X?nn?1nS2Xn?1?X?n?1?2S1.7 设T~t?n?，求证T2~F?1，n?.
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《应用数理统计》作业题及参考答案(前三章)
1?，V~?2?n?，且U与V相互独立，则 证明：又t分布的定义可知，若U~N?0，UU22，其中，U2~?2?1?. T?~t?n?，这时，T?VnVnU2由F分布的定义可知，T?~F?1，n?.
Vn2
1.9 设X1，X2，…，Xn和Y1，Y2，…，Yn分别来自总体N??1，?2?和N??2，?2?，且相互
12独立，?和?是两个已知常数，试求?X??1??Y??22112222????nS?n2S???????n1?n2?2?n1n2?的分布，
其中S12?解：
1n1?Xi?Xn1i?1??2，S22?1n2?Yi?Yn2i?1??2。
??2???2??X~N??1，?，Y~N??2，?，X??1与Y??2相互独立，
n1?n2?????2???2?X??1~N?0，?，Y??2~N?0，?，
?n2??n1???X??1??Y??2?????X??1??Y??2??2?2?2?2?~N?0，?~N?0，1?. ?，n1n2???2?2??????nn2??1~?2?n2?1?，且S12与S22相互独立，
??????n1S12?2~??n1?1?，
22n2S2?2n1S12?2?2n2S2?2~?2?n1?n2?2?.
?X??1??Y??2??2?2?????nn2??1nS????~t?n1?n2?2?，
??2112?nS?2222?n1?n2?2?即?X??1??Y??2????2??2?2?n1S12?n2S2???n1?n2?2?n1n2?~t?n1?n2?2?.
第二章 参数估计（续）
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《应用数理统计》作业题及参考答案(前三章)
P68
2.13 设总体X服从几何分布：P?X?k??p?1?p?k?12，?，，k?1，0?p?1，证明
1n样本均值X??Xi是E?X?的相合、无偏和有效估计量。
ni?1证明：?总体X服从几何分布，
?E?X??11?p，D?X??. 2pp11?1n?1?n?11 ?E?X??E??Xi??E??Xi???n???E?X?.
pp?ni?1?n?i?1?n?1n?样本均值X??Xi是E?X?的无偏估计量。
ni?11?p1?p?1n?1?n?12 D?X??D??Xi??2D??Xi??2?n?2?. 2pnp?ni?1?n?i?1?n?lnf?X1；p??ln?p?1?p??X1?1??lnp??X1?1?ln?1?p?.
??lnf?X1；p?1X1?111?X1. ?????pp1?pp1?p?2lnf?X1；p?1?X11???.
?p2p2?1?p?2?1?1??2lnf?X1；p??1?X1?X1?1?I?p???E??E?2? ???E??2?2?2?2?ppp?1?p???1?p????????????1?1111111?p?EX?1????1??? ??1??p2?1?p?2p2?1?p?2?p?p2?1?p?2p1?p??p?111. ?2??2?2p?1?p?pp?1?p?p?1?p?—5—