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We know that when density (say $f$) exists at the median(say $\theta$) then the median estimator(say $\hat{\theta_n}$) has the following property: $$ \sqrt n(\hat{\theta_n}-\theta) \to^d N(0,1/\{4f(\theta)^2\}). $$ This follows from here (this result is classical and can be found in some reference books as well).

Question: Suppose density at median doesn't exist. Equivalently, suppose we have a point mass at the median. Can we have a similar asymptotic distribution result in this case?

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If there is a central interval where the density is $0,$ then the median of even a large number of observations cannot be anything close to normal.

In the simulation below, the population distribution is a 50:50 mixture of $\mathsf{Unif}(0,1)$ and $\mathsf{Unif}(2,3),$ so that the density is $0$ in $(1,2).$ The simulation shows a histogram of 100,000 medians of samples of size $n = 100$ from this population.

set.seed(2019)
m = 10^5;  n = 100;  h = numeric(m)
for(i in 1:m) {
  b = rbinom(n, 1, .5)
  h[i] = median(b*runif(100) + (1-b)*runif(100, 2,3)) }
hist(h, prob=T, br = 50, col="skyblue2")

If there is a central interval where the density is $0,$ then the median of even a large number of observations cannot be anything close to normal.

In the simulation below, the population distribution is a 50:50 mixture of $\mathsf{Unif}(0,1)$ and $\mathsf{Unif}(2,3),$ so that the density is $0$ in $(1,2).$ The simulation shows a histogram of 100,000 medians of samples of size $n = 100$ from this population.

set.seed(2019)
m = 10^5;  n = 100;  h = numeric(m)
for(i in 1:m) {
  b = rbinom(n, 1, .5)
  h[i] = median(b*runif(100) + (1-b)*runif(100, 2,3)) }
hist(h, prob=T, br = 50, col="wheat")

enter image description here

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