Are there non-trivial settings where the MAD statistic has a closed-form density?

Question

The MAD statistic of an iid sample $(x_1,\ldots,x_n)$ is defined as the median of the absolute deviation from the median: $$ \text{mad}(x_1,\ldots,x_n)=\text{med}\left\{|x_i-\text{med}(x_1,\ldots,x_n)|;\ i=1,\ldots,n \right\}\,. $$

I wonder if there exist non-trivial (continuous) distributions on the $X_i$'s such that the distribution of $\text{mad}(X_1,\ldots,X_n)$ can be obtained in closed form (cdf or density).

The next level of unknown is the derivation of the joint density of the median and the MAD statistics for an iid sample of size $n$.

Answer 1

For the uniform distribution, twice the MAD among $2n-1$ samples seems to have the same distribution as the $(n-1)^{th}$-smallest of those samples.

The calculation can be done as $$F(m)=(2n-1)! \int_{A_m} dx_1\ldots dx_{2n-1}$$ where \begin{align} A_m = \{(x_1,\ldots,x_{2n-1}): \ &0< x_{1}< \cdots < x_{2n-1} < 1 \\ &\ \& \, \min(g_1, \ldots, g_n)<m\} \end{align} and $g_i = \max(x_{(i+n-1)}-x_{(n)},x_{(n)}-x_{(i)})$

I conjecture that the result is always $F(m)=I_{2m}(n-1,n+1)$, where $I$ is the incomplete beta function, and I have verified this for $n=2,3,4,5,6$. If so, the intepretation of this in terms of the order statistics is here, and the corresponding pdf is $$f(m)=\frac{2(2n-1)!}{n!(n-2)!}\, (2m)^{n-2}(1-2m)^n.$$

I hope someone will be able to find a simple argument that this conjecture is correct.