Talk:Stein's example

message: In several places this article had things like this:
:[\frac{1}{2}]
where the brackets were too small.  One fixes this by using \left and \right with this result:
:\left[\frac{1}{2}\right].
We also saw
:e^{\frac{-1}{2}}
and the like, i.e., "stacked" fractions in exponents.  These are hideous, and I changed them to
:e^{-(1/2)}\,
etc.  In several cases, TeX's matrix environment was used to align "=" on successive lines.  This has several bad effects, including making fractions look smaller.  In some cases, the matrix environment was used when there was only one line!  I got rid of the matrix environment.  I'd use it for actual matrices, but not for this. Michael Hardy 00:18, 23 October 2005 (UTC)

message: The definition of admissibility requires strict inequality for some \theta. Otherwise no admissible rule would exist if at least two satify "less than or equal" criterion for all theta but none has "less than" for some theta (under the definition of the article before I revised it). Bill Jefferys (talk) 02:19, 21 November 2007 (UTC)
thread_title:  Technical error regarding admissibility/inadmissibility 

message: Does this depend on the variances being known, or unknown but equal, or at least known to be in a given ratio? If so, then that is quite an assumption and needs to be stated explicitly and early in the article. Is it true if the underlying distributions are completely unknown? --Rumping (talk) 14:24, 22 December 2008 (UTC)
replies: text: The answers to many of these questions can be found in James-Stein estimator#Extensions. --Zvika (talk) 18:21, 22 December 2008 (UTC)
replies: text: I had read that.  Hence my question.  It looks to me that the answer to the first question mark is "yes" and the answer to the second might be "no". But I am not sure.--Rumping (talk) 01:45, 23 December 2008 (UTC)
thread_title:  Variances 

message: The implications section provides the example of using a shrinkage estimator to estimate a set of unrelated statistics (US wheat yield, spectators at Wimbledon and the weight of a candy bar). It says that on average, the new estimator is better.
If I understand the shrinkage estimator correctly, this only holds as long as the three statistics are drawn from a distribution whose mean exists. For example, it works if the true US wheat yield, the true number of spectators at Wimbledon, and the true weight of a candy bar are random variables stemming from a normal distribution. However, it would not work if they are random variables stemming from a Cauchy distribution (note that the optimal shrinkage factor alpha goes to zero as the variance of these three statistics goes to infinity). Note that in both cases, they are "unrelated" if unrelated means that their correlation is 0. However, for the shrinkage estimator to work, they must still have "something to do with each other", namely being drawn from a distribution whose mean exists (and probably also must have a finite variance). Unfortunately, that is all based on my own calculations, so these thoughts are not wikipedia-conform. Would be good to have a source for this.  — Preceding unsigned comment added by 2A02:1205:5076:CE60:9495:5865:8674:7A67 (talk) 09:33, 17 June 2015 (UTC)
thread_title:  Implications Section Wrong? 

message: The "intuitive explanation" section doesn't sound right to me. Can anyone note a source for it? A textbook maybe? Wbrinda (talk) 10:09, 10 March 2012 (UTC)
thread_title:  Intuitive Explanation Section 

message: Given n≥3 population means, Stein's method will have a lower mean squared error than just choosing sample values as estimators. But, given n sample values, choosing the sample values is the "optimal" decision. This follows from it being the optimal solution for each mean, and linearity of expectation.
It seems like Stein's method takes advantage of the population means being finite by slightly underestimating their magnitude. Say the sample is 7, then 6 and 8 are not treated as equally likely population means. 6 is "more likely" (and in general the population mean is likely to be slightly lower than the sample).
By multiplying the sample value by a <1 but close enough to 1 factor before using it, it gives us the advantage of reducing large sample values more than it reduces small ones, resulting overall in a lower MSE. 2600:1008:B007:530D:8A6:BA6A:FA37:C28F (talk) 12:11, 4 September 2022 (UTC)
thread_title:  Intuitive Explanation