Talk:James–Stein estimator

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 Field: Probability and statistics

Is the assumption of equal variances fundamental to this? Should say one way or another. —The preceding unsigned comment was added by (talk) .

Thanks for the comment. The assumption of equal variances is not required. I will add some information about this shortly. --Zvika 19:29, 27 September 2006 (UTC)
Looking forward to this addition. Also, what can be done if the variances are not known? After all, if \theta is not known then probably \sigma^2 is not either. (Can you use some version of the sample variances, for instance?) Thanks! Eclecticos (talk) 05:26, 5 October 2008 (UTC)

Thanks for the great article on the James-Stein estimator. I think you may also want to mention the connection to Emprirical Bayes methods (e.g., as discsussed by Effron and Morris in their paper "Stein's Estimation Rule and Its Competitors--An Empirical Bayes Approach"). Personally, I found the Empirical Bayes explanation provided some very useful intuition to the "magic" of this estimator. — Preceding unsigned comment added by (talk) 17:54, 18 April 2007 (UTC)

Thanks for the compliment! Your suggestion sounds like a good idea. User:Billjefferys recently suggested a similar addition to the article Stein's example, but neither of us has gotten around to working on it yet. --Zvika 07:55, 19 April 2007 (UTC)

dimensionality of y[edit]

A confusing point about this article: y is described as "observations" of an m-dimensional vector \theta, suggesting that it should be an m by n matrix, where n is the number of observations. However, this doesn't conform to the use of y in the formula for the James-Stein estimator, where y appears to be a single m-dimensional vector. (Is there some mean involved? Is ||y||^2 computed over all mn scalars?) Furthermore, can we still apply some version of the James-Stein technique in the case where we have more observations of \theta_1 than of \theta_2, i.e., there is not a single n? Thanks for any clarification in the article. Eclecticos (talk) 05:19, 5 October 2008 (UTC)

The setting in the article describes a case where there is one observation per parameter. I have added a clarifying comment to this effect. In the situation you describe, in which several independent observations are given per parameter, the mean of these observations is a sufficient statistic for estimating θ, so that this setting can be reduced to the one in the article. --Zvika (talk) 05:48, 5 October 2008 (UTC)
The wording is still unclear, especially the sentence: "Suppose θ is an unknown parameter vector of length m, and let y be a vector of observations of θ (also of length m)". How can a vector of m-dimensional observations have length m? --StefanVanDerWalt (talk) 11:07, 1 February 2010 (UTC)
Indeed, it does not make sense. I'll give it a shot. (talk) 19:49, 17 February 2010 (UTC)
Is the formula using σ2/ni applicable for different sample sizes in groups?. In Morris, 1983, Parametric Empirical Bayes Inference: Theory and Applications, it is claimed that a more general version (which is also derived there) of Stein's estimator is needed if the variances Vi are unequal, where Vi denotes σ2i/ni so as I understands it, Steins formula is only applicable for equal ni as well.


The estimator is always biased, right? I think this is worth mentioning directly in the article. Lavaka (talk) 02:09, 22 March 2011 (UTC)

Risk functions[edit]

The graph of the MSE functions would need a bit more precisions : we are in the case where ν=0, probably m=10 and σ=1, aren't we ? (I thought that, in this case , for θ = 0, MSE should be equal to 2 ; maybe the red curve represents the positive JS ?) —Preceding unsigned comment added by (talk) 15:40, 10 May 2011 (UTC)