For what distributions, $X$ is $E(\lfloor c-X \rfloor) = c-1$? Per the question here: Prove that $E([c-U]) = c-1$, it was proven that:
$$E(\lfloor c-U\rfloor) = c-1$$
if $U$ is a uniform random number between $0$ and $1$. I have some reason to suspect that this only holds for this particular distribution of $U$. I tried looking for counter-examples to this conjecture (considering the Beta distribution and some bimodal distributions) and couldn't find any. Is there a way to either prove this conjecture or disprove it via a counter-example?
The random variable, $U$ shouldn't depend on $c$ and the result should hold for all $c \in \mathbb R$.

If $U$ is a r.v.in $(0,1)$ that satisfies the condition of your problem, that is
$$E[\lfloor c-U\rfloor]=c-1$$
for all $c$ then it must be uniform.
Proof:  You have that if $n<c<n+1$,
$$c-1=\mathbb{E}[\textrm{floor}(c-U)]=nP(0<U\leq c-n)+(n-1)P(c-n<U \leq 1])$$
In particular, for any $0<c<1$ ($n=0$)
$$c-1=\mathbb{E}[\textrm{floor}(c-U)]=-P(c<U \leq 1)$$
From this, it follows that
$$c=1-P(c<U\leq 1)=P(0< U\leq c) $$
That shows that $U$ is uniform.
