I've come accross a term called "statistical efficiency of the median" in a paper and couldn't find any definition in the paper. From my search online, I found that this might mean relative efficiency of the median compared to the mean.

Can anyone shed some light to where I can look for clues?


The statistical efficiency of the median, measured as the ratio of the variance of the mean to the variance of the median, depends on the sample size N=2n+1 as


which tends to the value 2/pi approx 0.637 as N becomes large (see for more information).

Although, the median is less efficient than the mean, it is less sensitive to outliers than the mean.

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  • $\begingroup$ what is the difference between $n$ and $N$ here? $\endgroup$ – Macro Oct 5 '11 at 14:54
  • $\begingroup$ It seems there is a hidden assumption about the underlying distribution, but it's hard to tell exactly what due to the mysterious "n" in the efficiency formula and the fact that the formula diverges with "N". I guess it's a corrupted version of efficiency for Normal distributions only (even though the linked plot is for a Student t distribution). $\endgroup$ – whuber Oct 5 '11 at 15:27
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    $\begingroup$ I think what is being said is that $N = 2n+1$ is an odd integer so that the median is just the $(n+1)$-th ordered sample value. The efficiency is expressed as $4n/(\pi (2n+1))$ in the reference provided and is always smaller than $1$, instead of diverging with $N$. The $N$ is in the denominator, not the numerator, and the efficiency tends to $2/\pi$ as $N$ (or $n$) becomes large, as love-stats (and the link provided) says. $\endgroup$ – Dilip Sarwate Oct 5 '11 at 16:11
  • $\begingroup$ Thanks, @Dilip. That clears some things up. It's still worth noting that this result is specific to one implicitly assumed family of distributions: it's not universal. There is a general formula for distributions of finite variance that have continuous density functions $f$: asymptotically, the efficiency is proportional to the square of $f$ at the median. (I see this (almost) appears in the newly edited link in the reply.) $\endgroup$ – whuber Oct 5 '11 at 17:46
  • $\begingroup$ When the data come from distributions with thick tails, the sample median is more efficient. When the data come from distributions with a thin tail, like the normal, the sample mean is more efficient. $\endgroup$ – love-stats Oct 5 '11 at 18:44

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