# Relation between Pi and the Median

For large N the sample median is approximately normally distributed as Then again here I found that for a sample of size N=2n+1 from the normal distribution, the efficiency for large N is and, naturally, as N tends to infinity, 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?
• 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 3, 2020 at 17:10
• My post at math.stackexchange.com/a/3558/1489 describes relationships between $\exp(-z^2),$ polar coordinates, and 2D disks.
– whuber
Apr 3, 2020 at 17:52

### 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.