This is somewhat ill-defined, but: Why is Wald's decision theory not universally recognized as the foundation of statistics? I gather (or maybe I infer) that it was formulated to put frequentist and Bayesian methods (or any other kind of methods) into a common framework, so that they could be compared in a quantitative way for a wide range of problems.

But I gather that this has not really worked out, in that there are some settings where 'respectable' modern statisticians would still use estimators that are known to be inadmissible in a Waldian sense (like using the usual estimate of the vector mean in three or more dimensions). Why is this? Because loss functions are typically only an approximate stand-in for the true loss function in real problems, which is hard to specify?

I realize this question is rather vague, and I appreciate your patience.

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    $\begingroup$ Not a full answer but here's a few reasons ... 1. Not everything people are interested in when looking at statistics is necessarily best considered as choice under uncertainty. 2. It assumes we "know" more than we do. 3. Results often need to be explainable and interpretable to a wider audience; marginal gains in some guess at a loss function on a model that isn't actually correct surely don't always trump that. $\endgroup$
    – Glen_b
    Jan 5, 2016 at 23:55
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    $\begingroup$ I have knowingly used inadmissible estimators because the cost of computing an admissible one was potentially much greater than the additional expected loss. In other words, when you expand your assessment of loss to include the cost of conducting the statistical analysis in the first place, an inadmissible estimator according to the textbook may become admissible in practice. $\endgroup$
    – whuber
    Jan 6, 2016 at 0:24
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    $\begingroup$ I agree with both comments above, namely that admissibility is not a universal notion in that it depends on the loss function. For instance, the Stein effect boundary $p=3$ only relates to the quadratic loss. Larry Brown proved that there exists a loss function associated with every possible boundary $p$. $\endgroup$
    – Xi'an
    Jan 6, 2016 at 8:34
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    $\begingroup$ See Perlman & Wu (1999), "The Emperor’s new tests", Statist. Sci, 14, 4 (& the comments & rejoinder). $\endgroup$ Jan 6, 2016 at 10:22
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    $\begingroup$ It's possible Fisher's attacks on Wald after he (Wald) died didn't help things. $\endgroup$ May 15, 2017 at 11:14

1 Answer 1


I think whuber gave excellent insight in the comments.

An inadmissible estimator might be dominated by an estimator that takes a long time to compute in practice. Thus, when we expand the utility function to include not just statistical loss (square loss, for example), we might wind up with a dramatically different view of an estimator.

Expanding on this, if we include other (perhaps intangible) facets in our utility function, such as ease of explanation, it might be that the estimator called inadmissible by statistics actually works very well, considering all components: statistical, computational, and interpretability. A modest score on the statistical loss might be made up for when it comes to computational tractability and/or ease of interpretation for explaining results to customers, investors, etc.


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