?2n?x????当x???1?e ?1?? 1?Fx?????????0 当x??n?的概率密度为 (3) ?f???x??dF???x?dx?2n?x????当x???2ne?? ??0 当x??因为
??E??????xf???x?dx?????2nxe?2n?x???1dx?????
2n?作为?的估计量不具有无偏性. 所以?例4 一商店销售的某类产品来自甲、乙两个厂家,为考察产品性能的差异,现从甲、乙两厂分别抽取了8件和9件产品,测其性能指标X,得到两组数据,经对
2其做相应运算,得到X1?0.190,S12?0.006,X2?0.238,S2?0.008,假设测定结果服
?12从正态分布N??1,??与N??2,??,求2和?1??2的置信度为90%的置信区间,
?22122并对结果加以说明.
?12解 (1)为求2的置信度为90%的置信区间,首先查F分布表.
?2F?/2?n1?1,n2?1??F0.05?7,8?
F1??/2?n1?1,n2?1??F0.95?8,7??则置信区间为
?S12S12?11?2? ??F?n?1,n?1??S2,F?n?1,n?1S?2?221??/2?12??/212即?0.214,2.798?,此区间包含1，故可以认为?12??2,因此,可以利用方差相等的条
11?
F0.05?7,8?3.73件构造?1??2的置信区间.
2 (2)由于?12??2,但其值未知,故关于?1??2的置信度为90%的置信区间为
?1111X?X?tn?n?2S?,X?X?tn?n?2S??????2?/212W12?/212W?1n1n2n1n2???? ?又由于 SW2n1?1?S12??n2?1?S2?7?0.006?8?0.008???0.008
n1?n2?27?8?2t0.05?13??1.77,1111????0.486 n1n28911??0.007 n1n2??t?/2?n1?n2?2?SW故置信区间为
?X1?X2??,X1?X2???,即??0.055,?0.041?
从以上结果可以看出, ?1??2的置信区间不包含0，故可以认为?1??2?0,即两厂家的产品性能有显著差异.
例5 设总体X服从?0,??上的均匀分布, X1,X2,?,Xn是来自X的样本.(1)
?2和T??1;(2)求?的极大似然估计??2;(3)证明??1,T?n?1?求?的矩估计量?12n?1和T有效. ?n?1?minX均是?的无偏估计量;(4)证明T较?1?i?ni12解 (1) EX??令
?0xd?x 2??2 (2)似然函数为
?1?2X. ?X,得?的矩估计量为??1?L?x1,x2,?,xn;?????n??0?1????n??0当0?xi??其他 ?i?1,2,?,n?
当0?x?1??x?2????x?n???其他
又因为
?lnLn???0,所以L?x1,x2,?,xn;??关于?单调减,故当??X?n?时, ???L?x1,x2,?,xn;??取得极大值,因此,?的极大似然估计量是
?2?X?max?X? ?i?n?1?i?n??1?E?2X?2EX?2E?X?2?? (3) E??2?1是?的无偏估计量. 所以?X?n?的密度函数为
?xn?1?nfX?n??x????n?0?当0?x??其他
n?1n?1?xnEX?n??nndx?? 故 ET1??0nn???所以T1是?的无偏估计量.
X?1??min?Xi?的密度函数为
1?i?nfX?1??x??n??1?F?x;????n?1f?x;??
??x?n?11当0?x???n1????? ?????0 其他 ?故 ET2??n?1?EX?1???n?1??所以T2也是?的无偏估计量.
(4) EX??22???0?x?n?1?????n?1x?dx??
?x20?dx??33
DX?EX??EX??2?2?23?4?2?212
?1?D?2X??4DX?? D?n3n?n?1?ET2?12n2?2?0nx2xn?1?n?1??2 dx??nn?n?2?22n?1?2??222DT1?ET1??ET1??????
n?n?2?n?n?2?ET??n?1?222??0?x?nx2?1?????n?1dx2?n?1?2?? ?n?2DT2?2?n?1?2n2???2?? n?2n?2综上,显然有
?2n?n?2???23n?n2? ?n?1? n?2?1和T有效. 所以T1较?2例6 设X1,X2,?,Xn是来自总体X的样本, ?i?0 ?i?1,2,?,n?,??i?1,
i?1n试证:(1)
n??Xii?1ni是EX??的无偏估计量; (2)在?的一切形为??iXi??i?0,
i?1n??i?1i?1)的估计中, X最有效.
nn?n?证 (1)因为 E???iXi????iEX????i ??ii?1?i?1?i?1n所以??iXi是?的无偏估计.特别当?1??2????n?i?11时, X也是?的无偏估n计.
nn?n?222(2) D???iXi????iDX? i???ii?1?i?1?i?1nn要求函数f??1,?2,?,?n????在条件?i?0 ?i?1,2,?n,??,?i?下的12ii?1i?1极小值点,为此令
?n?F??1,?2,?,?n;??????????i?1?
i?1?i?1?2in??F????2?i???0?令 ?in ?i?1,2,?,n?
??F???1?0?i???i?1?21n?解得?i??,??i???1,即???,从而得?i? ?i?1,2,?,n?.因此证明了
nn2i?12n?X最有效.
例7 假设0.50,1.25,0.80,2.00是来自总体X的简单随机样本值,已知
Y?lnX服从正态分布N??,1?,求:(1) X的的数学期望EX(记EX为b);(2)求?的置信度为0.95的置信区间;(3)利用上述结果求b的置信度为0.95的置信区间.
解 (1) Y的密度函数为
1??y???2/2f?y??e ????y????
2?令t?y??,得
b?EX?E?eY???????1y??y???2/21edy?2?2??????et???t/2dt
2?e??1/2?????1??t?1?2/2edt?e??1/2 2?(2) ?的置信度为0.95的置信区间是
11??Y?1.96?,Y?1.96???
44??11??且 P?Y?1.96????Y?1.96???0.95
44??其中 Y?11?ln0.5?ln0.8?ln1.25?ln2??ln1?0 44于是?的置信度为0.95的置信区间为??0.98,0.98?. (3)由ex的严格递增性得
11111??P?Y?1.96??????Y?1.96???
24242???11???1/211?????P?exp?Y?1.96????e?exp?Y?1.96?????0.95
42?42?????故b?e??1/2的置信度为0.95的置信区间为
?11?11?????0.481.48expY?1.96??,expY?1.96???e,e? ???????42?42?????例8 设某种电子器件的寿命(以小时计) T服从双参数的指数分布,其概率
密度函数为
?1??t?c??e?f?t?????0 ?当t?c其他