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I have read that "Gauss proved that the best estimate for a random variable is the average." Can someone provide Gauss' publication containing that proof? To be clear, I am not looking for you to perform a derivation, instead, I am simply looking for a citation. Thanks.

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  • $\begingroup$ This quotation can be interpreted in different ways: do you mean to estimate a random variable a priori with no data, in which case by "average" you would mean "expectation"? Or do you mean to estimate some property of a random variable based on a sample? If so, which property? Finally, in what sense to you mean "best"? There are some combinations of these interpretations that are (trivially) implied by Gauss's results on least squares, but not all of them are. $\endgroup$ – whuber Oct 20 '18 at 18:47
  • $\begingroup$ Thanks, @whuber, for the helpful clarification. I believe the original (quoted) author's intent was the former option; i.e., in the instance in which there is no sample data, and "average" is meant as "expectation." As for "best," I'm sure that would be outlined in the actual proof, but I think what the original (quoted) author meant was that this result should (at least) not be inferior to any other formulation. Again, thanks. $\endgroup$ – user221772 Oct 20 '18 at 20:43
  • $\begingroup$ I believe that interpretation would be one of the very simplest applications of Gauss's results in Theoria Combinationis Observationum Erroribus Minimum Obnoxiae (1812, 1823, 1826). It consists of fitting the model $Y=\mu+\varepsilon$ under least-squared loss based on a single observation. $\endgroup$ – whuber Oct 20 '18 at 20:56
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    $\begingroup$ I think this is correct, @whuber. If you paste your comment as an answer I can select it. In the meantime, I ordered Stewart's 1987 translation of the Gauss work you mention; I will post page numbers and confirm the answer once I make my way through it. Thanks to all. $\endgroup$ – user221772 Oct 22 '18 at 14:59
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This is a compendium of Gauss's contributions to statistics. If Gauss did indeed accomplish what you assert ... you should be able to find it here.... If not then the assertion may not be true.

I skimmed this rather exhaustive paper and could not precisely find what you are looking for .... but perhaps another scan will pick it up.

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  • $\begingroup$ Thank you, @IrishStat. I will see how this question develops elsewhere before selecting a solution, but regardless this is a very helpful link. Thank you! $\endgroup$ – user221772 Oct 20 '18 at 20:45

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