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If the sample mean is an efficient estimator of the population mean, what may be an example of an inefficient such estimator?

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    $\begingroup$ It is not that simple. For example sample mean is an efficient estimator for population mean for data coming from Gaussian, but not from Laplace distribution. Similarly sample median will be efficient to estimate population mean for Laplace but will be inefficient to estimate population mean for Gaussian. $\endgroup$ – Cagdas Ozgenc Aug 14 '19 at 12:00
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    $\begingroup$ Pick one element of the sample randomly as your estimator. $\endgroup$ – whuber Aug 14 '19 at 12:36
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    $\begingroup$ OK, let's be more explicit. Your setting is a sample of values $x_1, x_2, \ldots, x_n.$ Roll an n-sided die. Call its outcome $i.$ Let $x_i$ estimate the mean. I hope it's obvious that for $n\gt 1$ this is an inefficient procedure. $\endgroup$ – whuber Aug 14 '19 at 12:56
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    $\begingroup$ What @whuber is telling you is not using your entire data also leads to an inefficient estimator. Estimator is a function of your data such as $f(x_1,x_2,...,x_n)$. This function doesn't have to be in the form of adding elements or doing symmetric/commutative operations. It can be anything. And if this function doesn't use some of its input arguments that will (always, I think) lead to inefficiency. $\endgroup$ – Cagdas Ozgenc Aug 14 '19 at 13:59
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    $\begingroup$ Lars, since my comments aren't making sense to you, we had better back up a bit: could you explain what you understand an "estimator" to be and what your definition of "efficiency" is? $\endgroup$ – whuber Aug 14 '19 at 14:03
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Consider a sample of size $N$ drawn from a normal distribution. The sample median $\tilde{X}$ is an unbiased and consistent estimator for $\mu$. For large $N$ the sample median is approximately normally distributed with mean $\mu$ and variance $\pi/2N$. The efficiency for large $N$ is thus $2/\pi \approx 0.64$. This is the asymptotic efficiency, that is the efficiency in the limit as sample size $N$ tends to infinity.

Edit: The sample mean is a better (more efficient) estimator for the population mean as opposed to the sample median, because the sample median will miss the population mean by more (on average) than the sample mean.

That is, the sample medians (talking asymptotically - as the sample size N reaches infinity) will have a wider distribution compared to the sample means. And since both estimators are unbiased, meaning they will estimate the true population mean - meaning they wont under/overestimate the population mean, all that remains is to look at the variance of these estimators. And as written above, the sample median distribution will have a higher variance compared to the sample mean, which means the sample median is a less efficient estimator for the population mean.

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  • $\begingroup$ That doesn´t answer my question. I was asking for an example of an INEFFICIENT estimator. $\endgroup$ – Lars Ahnland Aug 14 '19 at 12:34
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    $\begingroup$ @LarsAhnland In this case the median is inefficient because the efficiency is roughly 0.64. The efficiency of the mean is 1. $\endgroup$ – user2974951 Aug 14 '19 at 12:38
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    $\begingroup$ @LarsAhnland Yes, in this case the sample median is more variable. $\endgroup$ – user2974951 Aug 14 '19 at 13:16
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    $\begingroup$ @Lars user2974951 was not "using circular logic". When they said "the sample mean is more variable" they were just reframing the statement of your question to a more correct usage of the terminology than what you had in your question. Their response was simply "Yes." and then simply a clarification of what they were saying yes to. If you have looked up the definition of efficiency of estimators, then it should be clear that a comparison of variances (given both estimators are unbiased here) is simply following ... ctd $\endgroup$ – Glen_b Aug 15 '19 at 6:32
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    $\begingroup$ ctd... the definition of efficiency - to be precise, their relative efficiency will be the ratio of the reciprocals of their variances, so the more efficient has a smaller variance.the definition of efficiency - to be precise, their relative efficiency will be the ratio of the reciprocals of their variances, so the more efficient has a smaller variance. $\endgroup$ – Glen_b Aug 15 '19 at 6:38

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