If $X_1,\ldots,X_n$ are continuous and IID with sample median $m$, are $1(X_1\geq m),\ldots,1(X_n\geq m)$ IID?

Question

Suppose $X_1,\ldots,X_n$ are continuous random variables and IID with sample median $m$ for some $n\geq 2$. Is it then true that $1(X_1\geq m),\ldots,1(X_n\geq m)$ are IID?

Here the definition of $1(X_i\geq m)$ is that it is 1 if $X_i\geq m$ and 0 otherwise.

I ask because I often see researchers create these dummy variables (i.e. indicator variables) in regression analyses and I am worried that there is dependence between them by design (i.e. by construction). Mathematically I feel that since the dummy variables are functions of independent random variables, they ought to be independent. On the other hand, by the law of large numbers, for large $n$ we intuitively have that $(1/n)\sum_i1(X_i\geq m)$ is approximately equal to the expectation of $1(X_i\geq m)$ with $m$ being close to the population median, i.e. 0.5, with probability 1, suggesting that there is a linear dependence between the dummy variables for large $n$. (EDIT: Since $m$ is the sample median, we also know that $(1/n)\sum_i1(X_i\geq m)$ is at least 0.5.) I am unsure what I should conclude from this simple line of reasoning and unable to proceed in a more mathematically rigorous manner.

EDIT: I guess the answer to my question is simple. The indicator variables are not independent because all of them cannot be equal to 1 by the definition of the sample median.

Answer 1

A fairly rigorous argument is that $$\mathbb P(X_1\ge m(X_1,\ldots,X_n),\ldots,X_n\ge m(X_1,\ldots,X_n))=0\ne \mathbb P(X_1\ge m(X_1,\ldots,X_n))^n$$

Answer 2

Another argument is that:

1. If $n$ is even, then $\sum_{i=1}^n1(x_i \geq m) = n/2$, therefore the $1(x_i \geq m)$ cannot be independent,

2. If $n$ is odd, then $\sum_{i=1}^n1(x_i \geq m) = (n+1)/2$, therefore the $1(x_i \geq m)$ cannot be independent.