In article <92geep$2jl$[EMAIL PROTECTED]>, <[EMAIL PROTECTED]> wrote:
>I was wondering if somonone could give me tips about how to go about
>attacking this question, thanks, b1gp1g.
>Show that the estimator:
>d(X) =
>n if X_n > max(X_1,...,X_(n-1))
>_
>X (mean) if X_n <= max(X_1,...,X_(n-1)
>is consistent for 'mu' but not consistent in mean square.
>What is the difference between consistent & mean square
>consistent?
A sequence of extimators d_n of a parameter \theta is consistent
if for any positive number c, P(|d_n - \theta| > c) approaches
0. It is mean square consistent if E(|d_n - \theta|^2)
approaches 0.
As d(X) differs from the mean with probability approaching 0,
each is consistent if and only if the other is. However,
E(|d_n - \theta|^2) >= (n - \theta)^2/n, which is at least
as large as n - 2*\theta. For sufficiently large n, this
is large.
A less contrived situation is that of estimating the
reciprocal of the mean of a normal distribution. The
usual estimate is consistent if the true mean is not
zero, but the expected squared error is infinite for
every value of n.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
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