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Let's say we have a game with two players. Both of them know that five samples are drawn from some distribution (not normal). None of them know the parameters of the distribution used to generate the data. The goal of the game is to estimate the mean of the distribution. The player that comes closer to the true mean wins 1\$ (absolute difference between estimated value and actual value is the objective function). If the distribution has a mean that blows up to $\infty$, the player guessing the larger number wins and for $-\infty$, the one guessing the smaller number.

While the first player is given all five samples, the second one is given just the sum of the samples (and they know there were five of them).

What are some examples of distributions where this isn't a fair game and the first player has an advantage? I guess the normal distribution isn't one of them since the sample mean is a sufficient statistic for the true mean.

Note: I asked a similar question here: Mean is not a sufficient statistic for the normal distribution when variance is not known? about the normal distribution and it was suggested I ask a new one for non-normal ones.


EDIT: Two answers with a uniform distribution. I would love to hear about more examples if people know of any.

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    $\begingroup$ The question can be phrased as determining some distributions where the sample mean is Pitman farther than another estimator $\endgroup$
    – user257566
    Jul 21 at 1:41
  • $\begingroup$ Lemma 2.1 here gives a family of dominating estimators for any non negative distributions, simply by transforming the sample mean lib.dr.iastate.edu/cgi/… $\endgroup$
    – user257566
    Jul 21 at 1:48
  • $\begingroup$ This article by Kubokawa gives an example for a particular normal distribution. ism.ac.jp/editsec/aism/pdf/041_3_0477.pdf $\endgroup$
    – user257566
    Jul 21 at 1:56
  • $\begingroup$ @user257566 - can't open the link by pasting. Can you hyper-link? $\endgroup$
    – ryu576
    Jul 21 at 7:15
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For a uniform distribution between $0$ and $2 \mu$, the player who guesses the sample mean would do worse than one which guesses $\frac{3}{5} \max(x_i)$ (the sample maximum is a sufficient statistic for the mean of a uniform distribution lower bounded by 0).

In this particular case, it can be verified numerically. Without loss of generality, we set $\mu = 0.5$ in the simulation. It turns out that about 2/3rds of the time, the 3/5 max estimator does better.

Here is a Python simulation demonstrating this.

import numpy as np
Ntrials = 1000000
xs = np.random.random((5,Ntrials))
sample_mean_error = np.abs(xs.mean(axis=0)-0.5)
better_estimator_error = np.abs(0.6*xs.max(axis=0)-0.5)
print((sample_mean_error > better_estimator_error).sum())
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    $\begingroup$ Could you expand your answer why this happens, mathematically? And what the optimal $c$ is when guessing $c\max(x_i)$ when guessing from $n$ samples rather than just 5? $\endgroup$
    – orlp
    Jul 19 at 10:42
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    $\begingroup$ Do you think $0.5743492$ (half the reciprocal of the median of a Beta$(5,1)$ distribution) might do better than $0.6$? $\endgroup$
    – Henry
    Jul 19 at 14:18
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    $\begingroup$ @soakley only if you insist on an unbiased estimator. But there can be another estimator which is more likely to be closer (and indeed there is in this example) despite being biased $\endgroup$
    – Henry
    Jul 19 at 17:12
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    $\begingroup$ @Henry Based on my calculations, the value of $c$ that minimizes the mean absolute deviation of estimators of the form $W(\mu) = c x_{(n)}$ is $$c = 2^{-n/(n+1)}.$$ So in the case of $n = 5$, this gives $c = 2^{-5/6} \approx 0.561231$. This of course is biased for $\mu$. We get the lower bound $$\operatorname{E}[|W(\mu) - \mu|] \ge \mu (-1 + 2^{1/(n+1)}).$$ By contrast, for $n = 5$, the sample mean $\bar X$ has expected mean absolute deviation $\frac{1199}{5760}\mu$, which is strictly greater. $\endgroup$
    – heropup
    Jul 20 at 9:21
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    $\begingroup$ @heropup My simulations suggest that $0.5743492=2^{-4/5}$ does better than $0.561231=2^{-5/6}$ in about $52.9\%$ of cases $\endgroup$
    – Henry
    Jul 20 at 9:30
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The sum of observations is not sufficient for estimating the mean of a uniform population. The midrange has a smaller expectation of absolute error.

Approximation by simulation in R:

    set.seed(2021)
    a = replicate(10^6, mean(runif(5)))
    mr = replicate(10^6, mean(range(runif(5))))
    mean(a);  mean(mr)
    [1] 0.5000905
    [1] 0.5000926
    mean(abs(a-.5)); mean(abs(mr-.5))
    [1] 0.1040754
    [1] 0.0833201

enter image description here

    par(mfrow=c(2,1))
    hdr1 = "UNIF(0,1): Simulated Dist'n of Mean of 5"
    hist(a, prob=T, xlim=0:1, br=30, col="skyblue2", main=hdr1)
    hdr2 = "UNIF(0,1): Sim. Dist'n of Midrange of 5"
    hist(mr, prob=T, xlim=0:1, br=30, col="skyblue2", main=hdr2)
    par(mfrow=c(1,1))

Note per Comment: Using mean squared error instead of absolute error. Also, with RMSE for comparable units.

    mean((a-.5)^2); mean((mr-.5)^2)
    [1] 0.01665874
    [1] 0.01190478

    sqrt(mean((a-.5)^2)); sqrt(mean((mr-.5)^2))
    [1] 0.1290687
    [1] 0.109109
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  • $\begingroup$ Would your example still apply if we we facing e.g. square loss in place of absolute loss? $\endgroup$ Jul 20 at 17:18
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    $\begingroup$ @RichardHardy because we only care about who was closer, it makes no difference. $\endgroup$
    – user253751
    Jul 20 at 17:33
  • $\begingroup$ @RichardHardy. Using sq err loss, for UNIF, midrange also smaller than mean. See Addendum. $\endgroup$
    – BruceET
    Jul 20 at 18:14
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It might be worth adding that while you can often do better for low-dimensional parametric families, you can't do better if the distribution is completely unknown (or completely unknown apart from knowing it has a finite mean). The mean is the only estimator of the mean that works over all distributions.

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    $\begingroup$ Is there no distribution where the mean blows up and looking at the samples might give you a hint? Even if the $5$ samples became a million? $\endgroup$
    – ryu576
    Jul 19 at 6:36
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    $\begingroup$ Yes, plenty of them. But not if you don't know the distribution (or at least have some sort of prior over it). If you have some function that isn't the sample average, you can find some other distribution where that function doesn't estimate the mean. $\endgroup$ Jul 19 at 7:15
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    $\begingroup$ "The mean is the only estimator of the mean that works over all distributions". I don't know what "works" means in this case, but I assume the Cauchy distribution is a counter example. $\endgroup$
    – Cliff AB
    Jul 19 at 18:59
  • $\begingroup$ You still can't do better than the sample average (ie, infinitely wrong in the worst case) if you allow for distributions with no mean, but I think it's reasonable to leave them out. $\endgroup$ Jul 19 at 19:45
  • $\begingroup$ @ryu576 If you have a million samples, then you can probably make a reasonable guess about the distribution (if it follows some common form). But if you can figure out the distribution, then this answer no longer applies. $\endgroup$
    – NotThatGuy
    Jul 20 at 11:44

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