Given this weak property is it possible to demonstrate that the difference of expected value is negative?

Question

Let's assume that we have $X,Y$ as random variables and we have as hypothesis that $$X-\mathbb{E}_x \leq Y-\mathbb{E}_y$$ where $\mathbb{E}_x$ is the expected value of x.

Is it possible to demonstrate the following sentence? $$ P(X>l) \leq P(Y>l)$$ A classic result is obtained when $X\leq Y$, but I was wondering if it is possible to obtain using the previous equation.

If no, can you give me a counter example?

What if we restrict $X,Y$ to be only positive random variables?

Answer 1

If $X-\mathbb E_x\leq Y-\mathbb E_y$ then $Z\leq\mathbb E_z$ for $Z:=X-Y$ and consequently $Z=\mathbb E_z$ a.s. or equivalently: $$X=Y+\mathbb E_z\text{ a.s.}$$

Answer 2

Counterexample:

Let $\ell\in\text{supp}\{Y\}\subseteq \mathbb{Z}$ so that $P(Y=\ell)>0.$ Let $X=Y+1$. Then $X-E[X]=Y-E[Y],$ but

$$P(X>\ell)=P(Y>\ell-1)=P(Y=\ell)+P(Y>\ell)>P(Y>\ell).$$