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I am reading Inadmissibility of the Usual Estimator for the Mean of a Multivariate Normal Distribution seminal paper, which argues that sample mean vector in dimensions three and higher is not always the best choice in terms of minimizing the expected loss, specifically under the squared error loss function. In other words, as it is stated in the paper:

If one observes the real random variables $X_1, X_2, ..., X_n$ independently normally distributed with unknown means $\xi_1, \xi_2, ...\xi_m$ and variance 1, it is customary to estimate $\xi_i$ by $X_i$. If the loss is the sum of squares of the errors, this estimator is admissible for $n < 2$, but inadmissible for $n\geq3$.

Questions

  1. As written in the paper, if one observes $X_i$, it is customary to estimate $\xi_i$ by $X_i$. It is not clear to me whether the author suggests that we observe for $X_i$ random variable a single $x_i$ observations and take $\xi_i=x_i$ or it suggests that for each $X_i$ random variable we can observe a series of observations, $x_1, ... x_k$ and then for mean take the sample mean, i.e., $\xi_i = \frac{x_1+...+x_k}{k}$?
  2. Stien's paper is in the context of Multivariate Normal distribution. Are there any extensions for other distributions, such as Student distribution?
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  • $\begingroup$ Surveys on the Stein phenomenon and books that contain a section on it (such as mine!) do cover much more than the Normal case. There is a paper by L Brown that I cannot trace but which shows that for all integers $p>1$ there exists a distribution for which Stein's improvement only occurs for dimensions larger than $p$. $\endgroup$
    – Xi'an
    Commented Sep 17 at 14:48
  • $\begingroup$ @Xi'an Thank you, I'll have a close look. $\endgroup$
    – Sane
    Commented Sep 17 at 15:49

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I view it as basically a normalization/shorthand when we know the variances. As you correctly point out, in practice we of course typically observe more than one observation, $k$ in your notation, which we then aggregate into the mean.

But then, we may also model the experiment as sampling (scaled) means $$ \sqrt{k}\frac{\bar{X}_i}{\sigma_i}, $$ which will have variance 1 and some mean $\xi_i$.

As to your second question, there is indeed a huge follow-up literature to Stein's seminal paper (if there weren't, we shouldn't be calling it a seminal paper, after all :-)) that is however too vast to review here.

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  • $\begingroup$ So do I understand it correctly, that $\xi_i=X_i$, implies that $X_i$ is the sample mean in general? ($X_i = \frac{x_1+x_2+...x_n}{n}$, where $x_i$x) are observations of $X_i$? $\endgroup$
    – Sane
    Commented Sep 17 at 12:35
  • $\begingroup$ Yes, that would be my understanding too $\endgroup$
    – Christoph Hanck
    Commented Sep 17 at 13:07
  • $\begingroup$ In this en.wikipedia.org/wiki/Stein%27s_example , the last sentence of Formal Statement section it says the following .. "Thus, each parameter is estimated using a single noisy measurement, and each measurement is equally inaccurate." Does this mean that we just have a single observation for $X_i$? $\endgroup$
    – Sane
    Commented Sep 17 at 13:26
  • $\begingroup$ By a sufficiency argument, observing a single $X_i$ or $X_i$ as an average of multiple iid Normal realizations is the same thing, provided the variance is correctly computed. $\endgroup$
    – Xi'an
    Commented Sep 17 at 14:35
  • $\begingroup$ @Xi'an I can see your point ... you mean that given sufficiency argument, it does not really matter if we have $X_i$ as a single observation or a sample mean obtained from a random sample. My confusion comes from the available resources. Here is another example: en.wikipedia.org/wiki/James%E2%80%93Stein_estimator . The second sentence in the section Setting says the following: We are interested in obtaining an estimate, $\hat{\theta}$ of $\theta$ based on a single observation, $y$ of $Y$. Why a single observation? $\endgroup$
    – Sane
    Commented Sep 17 at 15:46

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