# Variance of the median

For large $$N$$ the sample median is approximately normally distributed with mean $$μ$$ and variance $$π/2N$$. The efficiency for large $$N$$ is thus $$2/π≈0.64$$

• Can somebody explain this for me?
• Where does that variance come from?
• and why then ≈0.64?
• The sample mean has variance $\frac{\sigma^2}N$ (optimal for a sample from a normal distribution) and $\dfrac{\sigma^2/N}{\pi\sigma^2/(2N)} \approx 0.64$ Nov 21 at 15:51
• The distribution of the median is like a beta distributed variable transformed by the quantile function. For increasing $n$ the beta distribution approaches a normal distribution with decreasing standard deviation. Then you can apply the Delta method to describe the distribution of the median. Nov 21 at 16:18
• Here you see that (as Laplace derived) the variance will be $\frac{1}{4nf(m)^2}$. The distribution density pops up because it relates to the derivative of the quantile function. If you fill in the density of a standard normal distribution then you get your result. (So your example counts as the sample median for a sample taken from a normal distributed population). Nov 21 at 16:24
• stats.stackexchange.com/questions/45124
– whuber
Nov 21 at 16:46

The most accessible theoretical demonstrations may be linked in the second Comment of @SextusEmpiricus and in @whuber's link.

Hoping that $$n = 100$$ is large enough to see a suggestive approximation of the ratio $$2/\pi$$ (for normal data), perhaps the following simple simulation in R of $$10^5$$ samples of size $$n=100$$ might give a view of this fact.

set.seed(2021)
n = 100                   # obs per sample (col)
m = 10^5                  # samples (row)
x = rnorm(n*m, 50, 7)
MAT = matrix(x, nrow=m)
a = rowMeans(MAT)         # 10^5 sample means
h = apply(MAT, 1, median) # 10^5 sample medians
var(a)
[1] 0.4879295
7^2/n
[1] 0.49

var(h)
[1] 0.7555167
var(a)/var(h)
[1] 0.6458223  # aprx 2/pi [0.6406 for n=200]
2/pi
[1] 0.6366198


The distribution of sample means is exactly normal; the distribution of sample medians is very nearly normal (ever closer as $$n \rightarrow \infty).$$ R code for the figure is shown below.

par(mfrow=c(2,1))
hist(a, prob=T, br=30, xlim=c(45,55), col="skyblue2",
main="Dist'n of Means")
curve(dnorm(x, 50, 7/10), add=T, col="orange", lwd=2)
hist(h, prob=T, br=30, xlim=c(45,55), col="skyblue2",
main="Dist'n of Medians")
curve(dnorm(x, mean(h), sd(h)), add=T, col="orange", lwd=2)
par(mfrow=c(1,1))