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I was reading about efficiency here and came across something curious.

For large N the sample median is approximately normally distributed as

MedianPi

Then again here I found that for a sample of size N=2n+1 from the normal distribution, the efficiency for large N is

MedianEfficiency and, naturally, as N tends to infinity, MedianEff2

I'm reminded of the YouTuber 3blue1brown in this Ted Talk and his curiosity how sliding blocks and their collisions were related to pi, but he did not see any circles.

My question is twofold:

  1. I'm genuinely curious how pi is related to the Median this much (and probably more).
  2. And since pi is intimately related to circles, can I say that the Median is associated with circles in some way? If so, what sort of association is it?
Sumner18
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    usually with these things i think it's more that $\pi$ is related to the (complex) exponential which truly is ubiquitous rather than circles – jld Apr 03 '20 at 17:10
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    My post at https://math.stackexchange.com/a/3558/1489 describes relationships between $\exp(-z^2),$ polar coordinates, and 2D disks. – whuber Apr 03 '20 at 17:52

1 Answers1

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To answer your specific question about the median, there's a passage in your second link, in the section on sample median:

Sampling distribution

... The distribution of the sample median from a population with a density function $f(x)$ is asymptotically normal with mean $m$ and variance

$\begin{align}\frac{1}{4nf(m)^2}\end{align}$

where $m$ is the median of $f(x)$ and $n$ is the sample size.

The example pulled from your first link assumes a normal population with mean $m$ and unit variance. In this case, $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-(x-m)^2}{2\sigma^2})=\frac{1}{\sqrt{2\pi}}\exp(\frac{-(x-m)^2}{2})$$ The normal distribution's mean is equal to its median, so $f(m)=\frac{1}{\sqrt{2\pi}}$. Putting it together then, we have that the sampling distribution of the median is asymptotically normal with mean $m$, and variance given by

$$\frac{1}{4n(\frac{1}{\sqrt{2\pi}})^2}=\frac{2\pi}{4n}=\frac{\pi}{2n}$$

That's all to say that the reason $\pi$ shows up here is that the formula for the asymptotic variance includes the density function of the population, evaluated at the median. When the population is normally distributed, this brings $\pi$ into the term for the variance, because of the $\pi$ constant in the normal density.

As for why $\pi$ shows up in the normal density, @jld's comment and @whuber's link to his answer offer great insight.

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RyanFrost
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