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Suppose we have a parameter $\theta$ that we want to estimate. We sample an observation (random variable) $X$ from a known distribution $D_{X|\theta}$. Then, we can compute an estimator $\hat\theta(X)$ as a function of $X$.

In normal parameter estimation, $\theta$ is unknown but fixed. This essentially means that we don't make the assumption that we know anything at all about $\theta$. However, it seems reasonable to me that we might have some knowledge about what $\theta$ might look like. To represent this, let $\theta$ be a random variable in some fixed known distribution $D_\theta$, instead of being a fixed unknown value.

I think that the existence of $D_\theta$ and $D_{X|\theta}$ also implies that ($\theta$, $X$) have some fixed known joint distribution $D_{\theta,X}$ (if that's not the case, let's just assume that the distributions are well-behaved enough such that this joint distribution exists). Therefore, there exists a conditional distribution $D_{\theta | X}$ in addition to the usual $D_{X | \theta}$.

With this formulation, $\hat\theta$ is an unbiased estimator (in the usual sense) of $\theta$ iff $E[\hat\theta(X)|\theta] = \theta$. That is, for any fixed true value, the expectation of the estimator is equal to the true value. There have been a lot of things written about how to find unbiased estimators.

However, what I want is an estimator $\hat \theta'(X)$ such that it’s “unbiased in the other way around”. Formally, $E[\theta | X] = \hat\theta'(X)$, that is, for any fixed observation, and therefore fixed estimated value, the expected value of the true value is equal to that fixed estimated value. How would one go about formulating such an estimator? Is there a name for this kind of estimator?

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    $\begingroup$ Is there any motivation for this question? When you talk about $E(\hat X|X) = X$, and $X$ being an r.v. does not conform with the classical definition of a parameter (i.e. a fixed but unknown quantity)... $\endgroup$
    – utobi
    Commented Jan 18, 2023 at 20:55
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    $\begingroup$ Your definition of unbiased is wrong in such a way that the question makes no sense. An unbiased estimator $\hat{\theta}$ is of a parameter $\theta$, not a random variable. An unbiased estimator is one for which $\mathbb{E}\hat{\theta} = \theta$. We don't estimate a random variable unless we are one of several kinds of Bayesians, and even then the random variable is an unobserved parameter. $\endgroup$
    – jbowman
    Commented Jan 18, 2023 at 21:02
  • $\begingroup$ @utobi I started from the issue of estimating a player's skill in a game using some kind of elo system. I want to avoid the issue of regression to the mean, where a person with high elo tends to have skill less than their elo. I then simplified the problem to the stated question. $\endgroup$ Commented Jan 18, 2023 at 21:04
  • $\begingroup$ That's as may be, but you should really think through what you are estimating! A player's performance in any given game may be an r.v., but their underlying skill, which you are trying to estimate, is not. $\endgroup$
    – jbowman
    Commented Jan 18, 2023 at 21:06
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    $\begingroup$ @jbowman " We don't estimate a random variable unless we are one of several kinds of Bayesians" that's not totally right; see e.g. mixed-effects models. $\endgroup$
    – utobi
    Commented Jan 18, 2023 at 21:06

1 Answer 1

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I put my question into ChatGPT, which told me the answer. Turns out that what I've described are called Bayes estimators.

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    $\begingroup$ Unfortunately, it sounds like got an incorrect -- or at least misleading -- answer. When you ask a nonsensical question of that bot, it will gladly still give you a beautiful written response. $\endgroup$
    – whuber
    Commented Jan 19, 2023 at 0:55
  • $\begingroup$ @whuber The linked wikipedia page has the exact same equation as the equation I formulated on my own, and the wikipedia definition uses a distribution of parameters like I did, so I think it's pretty clear that this answer is right. $\endgroup$ Commented Jan 19, 2023 at 0:58
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    $\begingroup$ In which ChatGPT proves that variances are negative: xianblog.wordpress.com/2022/12/21/…. It's one thing to say "this time it's right", in which case it's still up to you to justify that statement; it's another to treat it as in any way an authoritative source. $\endgroup$
    – jbowman
    Commented Jan 19, 2023 at 2:32

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