Let $X_1, X_2,\ldots, X_n$ be i.i.d. random variables with a nice density function (for example the normal(0,1) density). Denote $M_n = \text{median}(X_1,X_2,...X_n)$ and assume that $n$ is odd for simplicity.
Can we write an explicit expression for the density $f_{X_1 - M_n}(x)$ ? Also, if not, how can we rigorously prove that $f_{X_1 - M_n}(x)$ is uniformly bounded in x and n > 2 by a universal constant ?
Intuitively, it is obviously true because the median concentrates around $0$ as $n\to\infty$, but it's not a proof.
Let $n=2k+1$, $Y_1,Y_2,..,Y_n$ be a permutation of $X_1,x_2,..,x_n$ such that $Y_1\le Y_2\le ...\le Y_n$. Then $M_n=Y_k$. $P(Y_k\le x)=P(\text{at least $k$ of the $X_i$ lesser or equal to $x$})$. Let $p=P(X_1\le x)$, then $P(Y_k\le x)=\sum_{j=k}^{n}C_{j}^{n}p^j(1-p)^{n-j}$. You can explicitly find the distribution function of $f_{-Y_k}$ by taking derivative and find $f_{X_1+(-Y_k)}$ by convolution. However, I doubt that the explicit formula will help proving your claim. Another way I can imagine is that when $n$ is very large, I can use central limit theorem to find the approximate distribution of $Y_k$. I think $Y_k$ will tend to the a constant function when $k$ tends to infinity. As a result $f_{X_1-M_n}$ when $n$ is large will be approximately $f_{X_1-\mathbb{E}[X_1]}$.