Minimum-variance unbiased estimator to estimate quantiles when the errors are normal distributed

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)?

• 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. – Martijn Weterings Mar 12 at 17:19
• 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 $k.$ – whuber Mar 12 at 18:03
• $k$ 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 $k$). – Martijn Weterings Mar 12 at 18:45
• I see now that I had mistaken $\frac{1}{n}$ as the correction for $\sigma^2$, as well as $\sigma$. – Martijn Weterings Mar 12 at 20:05
• The duplicate I was thinking of was stats.stackexchange.com/questions/382124/… but it is not the same. – Martijn Weterings Mar 12 at 21:15