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I am reading Ruppert's Statistics and Data Analysis for Financial Engineering, which contains the following theorem:

Let $Y_1$, $...$ , $Y_n$ be an i.i.d. sample with a CDF $F$. Suppose that $F\;$ has a density $f\;$ that is continuous and positive at $F^{-1}(q)$, 0 < $q$ < 1. Then for large $n\;$, the $q^{th}$ sample quantile is approximately normally distributed with mean equal to the population quantile $F^{-1}(q)$ and variance equal to:

$$ \frac{q(1-q)}{n \left[f\;\left(F^{-1}(q)\right)\right]^2} $$

However, consider a Normal density with mean $\mu$ and variance $\sigma^2$. The distribution of the mean of $n\;$ samples will have mean $\mu$ and variance $\sigma^2/n$. If I am understanding it correctly, for q = 0.5 the above formula gives the variance as $$ \frac{\pi\sigma^2}{2n} $$

What am I missing here? Any insights are appreciated.

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    $\begingroup$ Although the population mean may equal the population median, the sample mean and sample median are not always the same and do not have the same sampling distributions. $\endgroup$ – whuber Aug 27 '11 at 14:53
  • $\begingroup$ Actually it's just the opposite: var(median) > var(mean). $\endgroup$ – Ringold Aug 27 '11 at 19:31
  • $\begingroup$ you're right; I deleted my previous comment $\endgroup$ – Andre Holzner Aug 27 '11 at 20:31
  • $\begingroup$ If n is somewhat smaller like 10 runs, then do these theorems apply ? $\endgroup$ – SriK Jun 25 '18 at 18:03
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As @whuber said in a comment, the formula you cite for the variance of a sample quantile is correct, as is that you give for the sample median in the normal case. The sample mean and sample median have different distributions, and in fact the median has higher variance in the normal case.

In R, you could estimate the SD of the sample median for the normal case as follows:

n <- 100; n.rep <- 1000
x <- matrix(rnorm(n*n.rep), ncol=n)
sd(apply(x, 1, median)) # 0.126
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