What is the maximum entropy distribution given the median (instead of the mean)? Given that the median seems to be a more robust statistic than the mean/average, I was wondering if there is a solution of the maximum entropy distribution given the median (or the median and some additional constraint).
 A: Lets look at one specific version of this problem.  This is enough to show some problems with this problem formulation. 
Let $X$ be a positive random variable, assumed to have an absolutely continuous distribution with density function $f$ on $(0, \infty)$, and median $m>0$. 
So we are assuming that $\int_0^m f(x) \; dx=\frac12, \quad \int_m^\infty f(x)\; dx=\frac12$.  Note that conditionally on $0<X<m$ $X$ has density $2f(x)$, and likewise conditional on $m<X<\infty$.  The entropy to be maximized is 
$$
   H=-\int_0^\infty f(x) \ln f(x) \; dx = -\int_0^m f(x) \ln f(x) \; dx -\int_m^\infty f(x) \ln f(x) \; dx
$$
Since there is no connection between the two terms, we can maximize separately $H_1 =-\int_0^m 2f(x) \ln (2f(x)) \; dx$ and $H_2=-\int_m^\infty 2f(x) \ln (2f(x)) \; dx$. $H_1$ is maximized by a uniform distribution, the problem is with $H_2$. Without further restrictions there is no maximum, as we can get the entropy as large as we want by diffusing the probability mass more and more.  So to get a solution, you will need more restrictions, one possibility is conditional means on the two subintervals defined by $m$. 
So, to get a more useful answer you need to think about which restriction which can be useful in your problem. In general, median (or more generally quantile) restrictions is not enough to maximize entropy. 
