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Consider the following protocol:

Alice has a number $X$, chosen according to a known distribution $\mathcal D$ (e.g., $X\sim U[0,1]$).

She can send a bit to Bob, giving him more information about $X$ (e.g., she can send $Y=\mbox{Bernoulli}(X)$. In turn, Bob estimates $X$ (e.g., as $\widehat X = Y$).

I'm interested in lower bounding the expected (over the choice of $X$) variance of any such protocol.

In the above example, we have $\mbox{Var}[\widehat X | X] = X(1-X)$ and thus $$\mathbb E[\mbox{Var}[\widehat X]] = \mathbb E[\mbox{Var}[\widehat x | X]] = \int_0^1x(1-x)dx = 1/6.$$

For this specific distribution (uniform), this protocol seems to yield the lowest possible expected variance for any unbiased estimation.

Can we use information-theoretic arguments to lower bound the variance for different distributions $\mathcal D$ for any protocol?

I looked into applying conditional differential entropy, but this seems to apply to a specific choice of protocol (how to choose which bit to send).

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    $\begingroup$ Formally, as posed, Bob can just estimate $\widehat X=EX$ without even looking at what Alice sent him (and she can just send him a bit independent of $X$ to avoid tempting him to do anything), thus achieving variance $0$. Apparently you wanted to ask something else, but I'll leave it to you to figure out what exactly it was. $\endgroup$
    – fedja
    Aug 28, 2020 at 21:59
  • $\begingroup$ It seems a better goal to bound from below the conditional variance of $X$ given $\hat{X}$. $\endgroup$ Aug 29, 2020 at 21:04
  • $\begingroup$ Is Alice's alphabet restricted to $\{0,1\}$ or can they send any message with Shannon entropy 1? (Not sure if this matters yet) $\endgroup$ Mar 18, 2021 at 15:30

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This is a specific instance of one-shot lossy source coding which is still open in general. The best work I know of is in this preprint by Elkayam and Feder where they distill it down to the open problem of identifying a convex minimizer $\tilde{D}(z,Q_Y)$.

https://arxiv.org/abs/2001.03983

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