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A comment in the thread Efficiency of Bayesian Estimator says

Bayesian doesn't assume that there is a true value

(of a parameter).

Question: If it is so, could you give a reference to a branch of Bayesians that do not assume existence of true parameter values? And in absence of true values, what are Bayesian parameter estimators supposed to be targeting?

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  • $\begingroup$ When the question Do some Bayesians assume that true values of parameters do not exist? got closed, the comment said Update the question so it focuses on one problem only. This will help others answer the question. You can edit the question or post a new one. I am posting a new one as suggested. (Given the current situation with moderator activity, the other option might take a long time to take effect. And I do see some current interest from other users.) Judging by the comments under the former question, I hope this one should be fine. $\endgroup$ Dec 6 '19 at 12:40
  • $\begingroup$ de Finetti: "Probability does not exist" $\endgroup$
    – Xi'an
    Dec 6 '19 at 13:27
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    $\begingroup$ Why is this limited to Bayesians? A frequentist can as easily say the same thing. After all, the conditional (on the parameters) model is a construct used by both frequentists and Bayesians, and it is rarely (if ever) the precise reality. By analogy, a map is not the actual terrain, as Bayesians and frequentists would agree. $\endgroup$ Dec 6 '19 at 13:46
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    $\begingroup$ A helpful comments by @Sycorax on the previous thread: stats.stackexchange.com/questions/83731/… $\endgroup$ Dec 6 '19 at 14:11
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    $\begingroup$ A helpful comment by @SextusEmpiricus on the previous thread: To me the questions seems clear now (although the dichtomous name calling of people like either being 'Bayesian' or 'frequentists' is a bit offensive, but so be it). In the question mentioned by Sycorax it is asked whether a 'Bayesian' would "acknowledge that there is one true fixed parameter". The answer is 'typically yes' or (it is irrelevant for Bayesian statistics). But such answers generate new questions "Are there any statisticians to which the answer 'no' applies and that do reject fixed parameters?. $\endgroup$ Dec 6 '19 at 14:12
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That's not quite true, to abuse a term. But where do we begin? Do we dare ask "quod est veritas?" again?

A quantum physicist very much believes that the location of an electron exists, it's merely a probabilistic distribution along the valence of an atom. The random variable is the truth, and its characterizations achieved through observation allow us to calculate electrostatic force.

Particle physics aside, the distinction between a Frequentist and a Bayesian is always, fundamentally the interpretation of probability.

In Frequentist notation, truth is reflected in statements of probability using a subscript. A corollary is that $P_{H_0}(A \ne a) = 0$ reflects the base assumption that $a$ is the true value of $A$. It is a falsifiable statement.

Interestingly, for the Bayesian, the statement that $P(A \ne a)=0$ is a refusal to allow any further evidence to modify belief, because the likelihood for other evidence will always be multiplied by 0 with such a prior.

In summary, these two absolute statements of probability are distinguished insofar as for a frequentist truth is the beginning of science and for a Bayesian truth is the end of science. We are always in the middle.

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  • $\begingroup$ Isn't that equivalent to saying that tomorrow's weather follows a Bernoulli distribution with $p=0.30$? Or is that Bernoulli distribution something like a posterior distribution of $p$? $\endgroup$
    – Dave
    Dec 7 '19 at 0:50
  • $\begingroup$ Looking back at my question, does your answer imply the branch of Bayesians I mention does not exist? $\endgroup$ Dec 7 '19 at 7:49
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    $\begingroup$ @RichardHardy I think it boils down to the powerful suggestiveness of saying something does not exist because it is a random variable. Likely to mislead and confuse. For anyone seeking any depth of understanding on the subject, it is better to say Bayesians think the true value of something is a random variable. (both a rigorous definition and lingo that accords with lay understanding). $\endgroup$
    – AdamO
    Dec 9 '19 at 14:16
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    $\begingroup$ @AdamO, thanks! My understanding seems to align reasonably well with yours. $\endgroup$ Dec 9 '19 at 14:47

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