If $$x + \mathcal{N}(0,\,1) = 6$$ $$y + \mathcal{N}(0,\,1) = 7.5$$ $$\left|{y-x}\right| < 1$$ then what is the median of $y$?
I'm expecting it to be $< 7.5$. If it is not, then why doesn't the $\left|{y-x}\right| < 1$ assertion affect it?
$\mathcal{N}(\mu,\,\sigma^2)$ represents the normal distribution.
We have
$$X = 6 - Z_x,\;\;\; Z_x \sim \mathcal{N}(0,\,1)$$ $$Y = 7.5 - Z_y,\;\;\; Z_y \sim \mathcal{N}(0,\,1)$$ $$|Y-X|<1$$
Using the first two, the 3d relation is written
$$|1.5 - Z_y + Z_x|<1$$
which makes clear that $Z_x$ and $Z_y$ may have identical marginal distributions, but they cannot be the same random variable. Analyzing it further
$$-1< 1.5 - Z_y + Z_x<1 \Rightarrow -2.5-Z_x <-Z_y<-0.5-Z_x$$
$$\Rightarrow 0.5 + Z_x < Z_y < 2.5 + Z_x$$ This inequality restricts $Z_y$ only in relation to $Z_x$. So it provides some information about the dependence between $Z_x$ and $Z_y$ (and so about their joint and conditional distribution) -but it does not affect the marginal distribution of $Z_y$, which is the one that enters in the definition of $Y$ (it would be another matter to investigate for example the distribution of $Y$ conditional on $Z_x$).
$Y$ is an affine function of a standard normal random variable, and so we have that $$Y \sim \mathcal{N}(7.5,\,1)$$ and its median is equal to its expected value, $7.5$.