For what distributions, $X$ is $E(\lfloor c-X \rfloor) = c-1$?

Question

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 $c \in \mathbb R$.

Answer 1

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.

Answer 2

You may modify the uniform distribution. Consider e.g. $U$ with density $f(t)=1/2( \chi_{[-1,0]}(t)+\chi_{[1,2]}(t))$. For $c\in\mathbb{R}$ there exists $x\in[0,1)$ with $c=\lfloor c\rfloor+1$. Tjen we have

\begin{align*} 2E(\lfloor c-U\rfloor)&=\int_{-1}^0 (\lfloor c\rfloor +x-t)dt+\int_0^1(\lfloor c\rfloor +x-t) dt\\ &=\int_{-1}^{x-1}\lfloor c\rfloor +1 dt + \int_{x-1}^0 \lfloor c \rfloor dt + \int_{1}^{1+x}\lfloor c\rfloor -1 dt + \int_{1+x}^2 \lfloor c\rfloor -2 dt\\ &=x(\lfloor c \rfloor +1) + (1-x)\lfloor c\rfloor + x(\lfloor c\rfloor -1)+ (1-x)(\lfloor c\rfloor -2)\\ &=2(\lfloor c\rfloor + x -1)= 2(c-1) \end{align*}