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What is the minimum-variance unbiased estimator to estimate quantiles when the errors are normal distributed?

  • median

    When we wish to estimate the median, $\mu$, of a normal distributed variable then the sample mean (an efficient estimator of $\mu$) performs better than the sample median. The sample mean has a lower variance than the sample median (and is in fact the minimum-variance unbiased estimator of the median $\mu$).

  • But what about other quantiles?

    My intuiton say that we can view this as estimating the value of $\mu+k\sigma$ for some given value of $k$, and then use the unbiased estimator $\hat \mu + k \hat \sigma$ based on the two (sufficient) statistics below. Is this also the minimum variance unbiased estimator?

    $$\hat \mu = \bar{x} \quad \text{ and } \quad \hat \sigma = c_n s = c_n \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2}$$

    where $c_n$ is a correction factor to make $\hat \sigma$ unbiased.

    Or is there possibly some other statistic, e.g. a combination of two sample quantiles or some version of minimizing the sum of absolute residuals, that could perform better (better as in, unbiased and lower variance)?

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  • $\begingroup$ I vaguely remember it being a duplicate. I believe that a question about the efficiency of the estimator $\bar{x} + k s$ to estimate $\mu + k \sigma$ has occurred before. But I could not find it easily, so I thought that this question, when it is a duplicate after all, may at least be a good pointer for searchers of the same question. $\endgroup$
    – Sextus Empiricus
    Commented Mar 12, 2019 at 17:19
  • $\begingroup$ Could you please state what you think $k$ should be more specifically? After all, the unbiasedness criterion immediately implies $E[m+ks]=\mu+k\sigma\sqrt{2/(n-1)}\Gamma(n/2)/\Gamma((n-1)/2),$ uniquely determining the value of $k.$ $\endgroup$
    – whuber ♦
    Commented Mar 12, 2019 at 18:03
  • $\begingroup$ $k$ is a fixed parameter. It is set according to the quantile of the normal distribution that one wishes to estimate. E.g. if one wishes to estimate the 95th quantile of a population that is normal distributed, then one is indirectly estimating the value of $\mu + 1.645 \sigma$. I wonder whether $\bar{x} + 1.645 s$ is an efficient unbiased estimator of $\mu + 1.645 \sigma$ (and in general the same question for other values of $k$). $\endgroup$
    – Sextus Empiricus
    Commented Mar 12, 2019 at 18:45
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    $\begingroup$ I see now that I had mistaken $\frac{1}{n}$ as the correction for $\sigma^2$, as well as $\sigma$. $\endgroup$
    – Sextus Empiricus
    Commented Mar 12, 2019 at 20:05
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    $\begingroup$ The duplicate I was thinking of was stats.stackexchange.com/questions/382124/… but it is not the same. $\endgroup$
    – Sextus Empiricus
    Commented Mar 12, 2019 at 21:15

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