2
$\begingroup$

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?
$\endgroup$
4
  • 2
    $\begingroup$ The sample mean has variance $\frac{\sigma^2}N$ (optimal for a sample from a normal distribution) and $\dfrac{\sigma^2/N}{\pi\sigma^2/(2N)} \approx 0.64$ $\endgroup$
    – Henry
    Nov 21 at 15:51
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 21 at 16:18
  • 2
    $\begingroup$ 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). $\endgroup$ Nov 21 at 16:24
  • 3
    $\begingroup$ stats.stackexchange.com/questions/45124 $\endgroup$
    – whuber
    Nov 21 at 16:46
0
$\begingroup$

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

enter image description here

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))
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.