l*****0
发帖数: 2 | 1 f(x|theta)是完备分布族,那么0的无偏估计只有0,于是随便哪个统计
量U(X)都与0的无偏估计不相关,是不是就是说U(X)是其期望的UMVUE呢?
望好心人指点~! | w****o
发帖数: 367 | 2 Yes, Here is the statement:
X~f(x|theta), U_0 is unbiased estimator of 0. (i.e., E(U_0) = 0, for every
theta). Delta(X) is UE of g(theta).
Delta(X) is UMVUE of g(theta) iff Cov(U_0, Delta(X) = 0.
Cov(U_0, Delta(X) = 0 is a very strong condition: 如果你在你的Estimator里面
加入白噪音(white noise),也就是U_0, 不会是你的估计变得更好.
换句话说,如果你的Estimator可以被白噪音improved的话,你的estimator就是存在缺陷
的. | l*****0
发帖数: 2 | 3 明白了，Thanks a lot!
【在 w****o 的大作中提到】
: Yes, Here is the statement:
: X~f(x|theta), U_0 is unbiased estimator of 0. (i.e., E(U_0) = 0, for every
: theta). Delta(X) is UE of g(theta).
: Delta(X) is UMVUE of g(theta) iff Cov(U_0, Delta(X) = 0.
: Cov(U_0, Delta(X) = 0 is a very strong condition: 如果你在你的Estimator里面
: 加入白噪音(white noise),也就是U_0, 不会是你的估计变得更好.
: 换句话说,如果你的Estimator可以被白噪音improved的话,你的estimator就是存在缺陷
: 的.
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